From zakseidov at yahoo.com  Wed Nov  1 06:43:12 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Tue, 31 Oct 2006 21:43:12 -0800 (PST)
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
Message-ID: <20061101054313.25333.qmail@web38210.mail.mud.yahoo.com>

Dear seqfans,
Bob, Ivars,

In A065577, 
a(1)=2 because 10=3+7=5+5?
Also a(2)=6 because
100=3+97=11+89=17+83=29+71=41+59=47+53?

The point is that in Ivars Peterson's
"Goldbach's Prime Pairs"
http://www.maa.org/mathland/mathtrek_8_21_00.html
we read:

" 
Note that Goldbach partitions take into account the
order of the primes that are summed. For example, 10
has two Goldbach partitions: 3 + 7 and 7 + 3.

Integer No. of Goldbach partitions 
10 2 
100 6 
1,000 28 
10,000 127 
100,000 810 
1,000,000 5,402 
10,000,000 38,807 
100,000,000 291,400
" 

Me: But in A065577 a(1)=2 and a(2)=6 
(and also other terms as I checked them)
without taking into account 
the order of the primes 
that are summed(?!)

Can anyoone (Bob? Ivars?) clarify this?
Thanks, Zak

BTW Two next terms are calculated by me
(for n=9, 10) as 2274205, 18200487.
Can anyone check these?
Thanks, Zak


%I A065577
%S A065577 2,6,28,127,810,5402,38807,291400
%N A065577 Number of Goldbach partitions of 10^n.
%H A065577 Ivars Peterson's MathTrek, <a
href="http://www.maa.org/mathland/mathtrek_8_21_00.html">Goldbach's
Prime Pairs</a>
%H A065577 Science News Online, week of Aug. 19, 2000;
Vol. 158, No. 8 <a href="http://www.sciencenews.org/
20000819/mathtrek.asp">Goldbach's Prime Pairs</a>
<skip>
%A A065577 Robert G. Wilson v (rgwv(AT)rgwv.com), Dec
01 2001




 
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From ralf at ark.in-berlin.de  Wed Nov  1 10:35:28 2006
From: ralf at ark.in-berlin.de (Ralf Stephan)
Date: Wed, 1 Nov 2006 10:35:28 +0100
Subject: Re^2: Error-finding program
In-Reply-To: <200610312150.k9VLoKfn018613@amer.strw.leidenuniv.nl>
References: <200610312150.k9VLoKfn018613@amer.strw.leidenuniv.nl>
Message-ID: <20061101093528.GA19017@ark.in-berlin.de>

> The definitions of A026468 and A026469 are
>  a(n) = least positive integer > a(n-1) and not a(i)^2 + a(j)^2 for 1<=i<=j<=n.
> and
>  a(n) = least positive integer > a(n-1) and not a(i)^2 + a(j)^2 for 1<=i<=j<=n.  	
> which looks quite identical. So why can A026468/9 be two different
> sequences? Is there a "not" too much or too little, or rather one of the 
> "less-equal" signs to be corrected to "less" or a starting condition on a(2)
> to be added?

My guess is that Clark started the same algorithm with 1, 2 and then
tried it with starting values 1, 3, so the definition of A026469 should 
be amended with ", starting with 1, 3".


ralf






From njas at research.att.com  Wed Nov  1 14:16:47 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 1 Nov 2006 08:16:47 -0500 (EST)
Subject: Re^2: Error-finding program
References: <200610312150.k9VLoKfn018613@amer.strw.leidenuniv.nl>
Message-ID: <200611011316.IAA53531@fry.research.att.com>

Ralf,  I think you've solved it!   Thanks!   Neil


My guess is that Clark started the same algorithm with 1, 2 and then
tried it with starting values 1, 3, so the definition of A026469 should 
be amended with ", starting with 1, 3".





From mathar at strw.leidenuniv.nl  Wed Nov  1 14:17:58 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Wed, 1 Nov 2006 14:17:58 +0100
Subject: A065577 Number of Goldbach partitions of 10^n?
Message-ID: <200611011317.kA1DHwmp018846@amer.strw.leidenuniv.nl>


zs> From seqfan-owner at ext.jussieu.fr  Wed Nov  1 06:44:53 2006
zs> Return-Path: <seqfan-owner at ext.jussieu.fr>
zs> Date: Tue, 31 Oct 2006 21:43:12 -0800 (PST)
zs> From: zak seidov <zakseidov at yahoo.com>
zs> Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
zs> To: seqfan at ext.jussieu.fr, ip at sciserv.org, rgwv at rgwv.com
zs> 
zs> Dear seqfans,
zs> Bob, Ivars,
zs> 
zs> In A065577, 
zs> a(1)=2 because 10=3+7=5+5?
zs> Also a(2)=6 because
zs> 100=3+97=11+89=17+83=29+71=41+59=47+53?
zs> ....

One might call A065577 explicitly the partitions without regard to order. The
conversion between the two counts is simple, see
http://mathworld.wolfram.com/GoldbachPartition.html

The table of n, ordered, and non-ordered partitions is

                                    1, 3, 2
                                   2, 12, 6
                                   3, 56, 28
                                  4, 254, 127
                                 5, 1620, 810
                                6, 10804, 5402
                                7, 77614, 38807

with the ordered partitions in the OEIS as A073610, the non-ordered in A061358 (?)
(Cf from A065577 to these two might also help to show the difference,
A065577(n)=A061358(10^n).)
With the exception of the n=1 leading term, the count of the
ordered partitions is twice that of the non-ordered partitions,
because 10^n/2 is not a prime then.

The Maple program to print the small table above:

A065577 := proc(n,orderd)
	local N,a,i,ip;
	N := 10^n ;
	a := 0 ;
	i := 1 ;
	ip := 2 ;
	while 2*ip <= N do
		if isprime(N-ip) then
			if orderd and ip <> N-ip then
				a := a+ 2;
			else
				a := a+ 1;
			fi ;
		fi ;
		i := i+1 ;
		ip := ithprime(i) ;
	od ;
	RETURN(a) ;
end:

for n from 1 to 20 do
	print(n,A065577(n,true),A065577(n,false)) ;
od ;

In PARI

A065577(n)={
	local(N,a,i,ip);
	N = 10^n ;
	a = 0 ;
	i = 1 ;
	ip = 2 ;
	while(2*ip <= N,
		if(isprime(N-ip),
			a++ ;
		) ;
		i++ ;
		\\ip = prime(i) ; \\ slower and needs -p switch
		ip = nextprime(ip+1) ; \\ pseudoprimes only
	) ;
	return(a) ;
}

{
	for(n=1,20,
		print(n," ",A065577(n)) ;
	)
}

The PARI output confirms the n=9 case quoted by Zak (I've not gone to n=10):

1 2
2 6
3 28
4 127
5 810
6 5402
7 38807
8 291400
9 2274205

-- Richard





From zakseidov at yahoo.com  Wed Nov  1 15:29:37 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 1 Nov 2006 06:29:37 -0800 (PST)
Subject: A065577 Number of Goldbach partitions of 10^n?
In-Reply-To: <200611011317.kA1DHwmp018846@amer.strw.leidenuniv.nl>
Message-ID: <20061101142937.98860.qmail@web38214.mail.mud.yahoo.com>

Richard,
thanks for your reply.
It seems this'd concern mainly Ivars!
Also could you please calculate cases n>=10) -
I guess this'd not take for you hours(days!)
as with my Mathematica.
thanks, Zak


--- Richard Mathar <mathar at strw.leidenuniv.nl> wrote:

> 
> zs> From seqfan-owner at ext.jussieu.fr  Wed Nov  1
> 06:44:53 2006
> zs> Return-Path: <seqfan-owner at ext.jussieu.fr>
> zs> Date: Tue, 31 Oct 2006 21:43:12 -0800 (PST)
> zs> From: zak seidov <zakseidov at yahoo.com>
> zs> Subject: A065577 Number of Goldbach partitions
> of 10^n? More terms?
> zs> To: seqfan at ext.jussieu.fr, ip at sciserv.org,
> rgwv at rgwv.com
> zs> 
> zs> Dear seqfans,
> zs> Bob, Ivars,
> zs> 
> zs> In A065577, 
> zs> a(1)=2 because 10=3+7=5+5?
> zs> Also a(2)=6 because
> zs> 100=3+97=11+89=17+83=29+71=41+59=47+53?
> zs> ....
> 
> One might call A065577 explicitly the partitions
> without regard to order. The
> conversion between the two counts is simple, see
> http://mathworld.wolfram.com/GoldbachPartition.html
> 
> The table of n, ordered, and non-ordered partitions
> is
> 
>                                     1, 3, 2
>                                    2, 12, 6
>                                    3, 56, 28
>                                   4, 254, 127
>                                  5, 1620, 810
>                                 6, 10804, 5402
>                                 7, 77614, 38807
> 
> with the ordered partitions in the OEIS as A073610,
> the non-ordered in A061358 (?)
> (Cf from A065577 to these two might also help to
> show the difference,
> A065577(n)=A061358(10^n).)
> With the exception of the n=1 leading term, the
> count of the
> ordered partitions is twice that of the non-ordered
> partitions,
> because 10^n/2 is not a prime then.
> 
> The Maple program to print the small table above:
> 
> A065577 := proc(n,orderd)
> 	local N,a,i,ip;
> 	N := 10^n ;
> 	a := 0 ;
> 	i := 1 ;
> 	ip := 2 ;
> 	while 2*ip <= N do
> 		if isprime(N-ip) then
> 			if orderd and ip <> N-ip then
> 				a := a+ 2;
> 			else
> 				a := a+ 1;
> 			fi ;
> 		fi ;
> 		i := i+1 ;
> 		ip := ithprime(i) ;
> 	od ;
> 	RETURN(a) ;
> end:
> 
> for n from 1 to 20 do
> 	print(n,A065577(n,true),A065577(n,false)) ;
> od ;
> 
> In PARI
> 
> A065577(n)={
> 	local(N,a,i,ip);
> 	N = 10^n ;
> 	a = 0 ;
> 	i = 1 ;
> 	ip = 2 ;
> 	while(2*ip <= N,
> 		if(isprime(N-ip),
> 			a++ ;
> 		) ;
> 		i++ ;
> 		\\ip = prime(i) ; \\ slower and needs -p switch
> 		ip = nextprime(ip+1) ; \\ pseudoprimes only
> 	) ;
> 	return(a) ;
> }
> 
> {
> 	for(n=1,20,
> 		print(n," ",A065577(n)) ;
> 	)
> }
> 
> The PARI output confirms the n=9 case quoted by Zak
> (I've not gone to n=10):
> 
> 1 2
> 2 6
> 3 28
> 4 127
> 5 810
> 6 5402
> 7 38807
> 8 291400
> 9 2274205
> 
> -- Richard
> 



 
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From tanyakh at TanyaKhovanova.com  Wed Nov  1 16:46:50 2006
From: tanyakh at TanyaKhovanova.com (Tanya Khovanova)
Date: Wed,  1 Nov 2006 07:46:50 -0800
Subject: What is the most famous sequence?
Message-ID: <200611010746.AA12650352@TanyaKhovanova.com>

What is the most famous sequence? The most basic (natural numbers?)? The most simple (0,0,0, ... or 1,1,1, ...?)? The most versatile (Catalan numbers?)?

I would be curious to know your opinion.

I tried to check the number of Google hits for some sequence numbers. Here is what I've got:
A000040 (prime numbers) 68,800
A000045 (Fibonacci)     66,000
A000108 (Catalan)       62,500
A000027 (natural)       47,500

Can someone write a program to find top ten? What is the most referenced sequence inside Encyclopedia? Can we find the most hits outside Encyclopedia? Is it easy to do?

Tanya


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From jvospost3 at gmail.com  Wed Nov  1 18:58:55 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 1 Nov 2006 09:58:55 -0800
Subject: New & old LEGO seqs; production cuts
Message-ID: <5542af940611010958me17f347w3bf2d2f2d6c9dec0@mail.gmail.com>

I like the new seqs about Lego, such as
 A123762 <http://www.research.att.com/%7Enjas/sequences/A123762>
 Number of ways, counted up to symmetry, to build a contiguous building with
n *LEGO* blocks of size 1x2.

Ironically timed, as *"Recent restructuring and production cuts have left
Lego unable to fill
orders<http://www.cbc.ca/money/story/2006/10/31/legoproduction.html>for
the upcoming holiday season. Affected products include Duplo bricks,
Lego City sets, and (horror of horrors!) Star Wars and Lego Technik
sets."*According to the article Lego stands to lose $127 million in
holiday sales.

http://www.cbc.ca/money/story/2006/10/31/legoproduction.html

It might be useful to add crossrefs to earlier seqs such as:

 A007576 <http://www.research.att.com/%7Enjas/sequences/A007576>
 Number of maximally stable towers of 2 X 2 *LEGO* blocks.

 A082679 <http://www.research.att.com/%7Enjas/sequences/A082679>
 Number of *Lego* towers, one piece per floor, where every floor is
perpendicular to the one below it (so we have a kind of 3-dimensional zigzag
pattern).
 -- Jonathan Vos Post
[wondering if my son's earlier intense exposure to Duplo and Lego led him
towards his double B.S. in Math and Computer Science]
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From dean at math.ucdavis.edu  Wed Nov  1 21:12:15 2006
From: dean at math.ucdavis.edu (Dean Hickerson)
Date: Wed, 1 Nov 2006 12:12:15 -0800
Subject: Re^2: Error-finding program
Message-ID: <20061101201215.GA2388@math.ucdavis.edu>

Ralf Stephan wrote:

> My guess is that Clark started the same algorithm with 1, 2 and then
> tried it with starting values 1, 3, so the definition of A026469 should 
> be amended with ", starting with 1, 3".

Actually, A026469 should just say "starting with 1", since the 3 follows
from the rest of the definition.  A026468 should say "starting with 1,2",
since the 2 is artificially added to the sequence.

Dean Hickerson
dean at math.ucdavis.edu





From franktaw at netscape.net  Thu Nov  2 02:04:26 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 01 Nov 2006 20:04:26 -0500
Subject: What is the most famous sequence?
In-Reply-To: <200611010746.AA12650352@TanyaKhovanova.com>
References: <200611010746.AA12650352@TanyaKhovanova.com>
Message-ID: <8C8CC31F9FF2FAD-370-91EC@WEBMAIL-MA05.sysops.aol.com>

I would have to say that the most famous sequence is the primes.

To get the most referenced inside the OEIS, just do a search that 
matches
everything (or almost everything) - I tend to use "-keyword:dead".  Be 
sure
you're sorting by references (for many searches, such as this one, the 
default
sort by relevance works, too).  You'll see the sequences sorted by how 
many
references there are.

The top ten are (with references as of 11/1/06 6 PM CST):

1666 A000040 Primes
1305 A000045 Fibonacci
1177 A000108 Catalan
1009 A000010 Totient function
 883 A000217 Triangular
 870 A000005 Number of Divisors
 808 A000203 Sum of Divisors
 674 A002275 Repunits
 616 A001358 Semiprimes
 594 A000142 Factorials

This tends to under-estimate the importance of sequences like the 
squares
and the powers of 2.  If somebody enters a formula for a sequence that
uses the totient function, they are likely to include a reference to 
A000010.
Adding a formula which squares an integer will not cause people to add 
a
reference to the squares, however.

Incidently, here are some searches, intended to identify how popular 
certain
topics are in the OEIS, with the number of sequences matched:

25267 Prime
17853 keyword:base
10904 Digit -maple:digit
 7087 Square (but this covers a lot more than just the squares)
 6384 Binomial
 5548 Power
 4904 Divisor
 4051 Polynomial
 3901 Keyword:cons
 3722 Transform
 3698 Partition
 3638 Factor
 3526 Keyword:cofr
 3448 Permutation
 3326 Pi
 3325 Fibonacci
 3308 Matrix OR Matrices (search each separately, then subtract number
        where both occur)
 3138 Group
 2601 Phi (includes both golden ratio and totient function)
 2238 Sigma
 2189 Graph
 2116 Composite
 1971 Tree
 1553 Keyword:frac
 1540 Vector
 1549 Catalan
 1511 Reverse
 1197 Factorial (but I can't search for "!" operator)
 1165 Modulo OR Modulus
 1159 Palindrome
 1051 Perfect
 1034 Semiprime

Obviously, many of these searches do not perfectly identify the 
intended
subject.

Franklin T. Adams-Watters


-----Original Message-----
From: tanyakh at TanyaKhovanova.com

What is the most famous sequence? The most basic (natural numbers?)? 
The most
simple (0,0,0, ... or 1,1,1, ...?)? The most versatile (Catalan 
numbers?)?

I would be curious to know your opinion.

I tried to check the number of Google hits for some sequence numbers. 
Here is
what I've got:
A000040 (prime numbers) 68,800
A000045 (Fibonacci)     66,000
A000108 (Catalan)       62,500
A000027 (natural)       47,500

Can someone write a program to find top ten? What is the most 
referenced
sequence inside Encyclopedia? Can we find the most hits outside 
Encyclopedia? Is
it easy to do?

Tanya

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From arndt at jjj.de  Thu Nov  2 02:10:02 2006
From: arndt at jjj.de (Joerg Arndt)
Date: Thu, 2 Nov 2006 12:10:02 +1100
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
Message-ID: <20061102011002.GC6041@cochise.anu.edu.au>

* Jonathan Post <jvospost3 at gmail.com> [Nov 01. 2006 10:14]:
> Eisenstein-Fibonacci sequences.  Cube root of unity
> analogues of square-root of unity A014291  Imaginary
> Rabbits.
> 
> Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
> where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
> = (-1 - i*sqrt(3))/2.
> 
> [...]
> The coefficients of 1 = a(n) = 0, 0, 1, 1, 3, 1, 7, 4,
> 14, 17, 28, 54, 66, 143, 182, 350, ...
> 
> which does not seem to be in OEIS.

Ralf Stephan's ggf() says that the OGF may be:

? g=ggf([1, 1, 3, 1, 7, 4,14, 17, 28, 54, 66, 143, 182, 350])
(x^5 - 3*x^3 + x + 1)/(-x^6 + 3*x^4 - x^3 - 3*x^2 + 1)
? factor(g)

[x - 1 1]

[x^4 + x^3 - 2*x^2 - 2*x - 1 1]

[x^2 + x - 1 -1]

[x^4 - x^3 - x^2 + x + 1 -1]


> [...]







From eclark at math.usf.edu  Thu Nov  2 03:28:27 2006
From: eclark at math.usf.edu (Edwin Clark)
Date: Wed, 1 Nov 2006 21:28:27 -0500 (EST)
Subject: What is the most famous sequence?
In-Reply-To: <200611010746.AA12650352@TanyaKhovanova.com>
References: <200611010746.AA12650352@TanyaKhovanova.com>
Message-ID: <Pine.LNX.4.58.0611012122360.14271@shell.cas.usf.edu>


> What is the most famous sequence?

Since the novel and movie "The Da Vince Code"-- not to mention 
Knuth's "Art of Computer Programming" it would have to be the 
Fibonacci numbers.







From zakseidov at yahoo.com  Thu Nov  2 07:12:28 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 1 Nov 2006 22:12:28 -0800 (PST)
Subject: A124450: no autoreply from OEIS
Message-ID: <20061102061228.7046.qmail@web38206.mail.mud.yahoo.com>

Just submitted:
A124450 (no autoreply from OEIS).
Hope someone may wish to add more terms:
Mathematica isn't the best in such calcs.
Thanks, Zak

%I A124450
%S A124450
5,47,491,4919,49877,499943,4999913,49999757,499999931,4999999937,49999999811,499999999769,4999999998431,49999999999619,499999999999769,4999999999998557,49999999999998887,499999999999999679,4999999999999999661,49999999999999998647
%N A124450 Lesser of pair of closest primes summed to
10^n.
%C A124450 Cf. A065577 Number of Goldbach partitions
of 10^n.
%F A124450 10^n - a(n) is prime.
%e A124450 10^1=5+5; 10^2=47+53; 10^3=491+509;
10^4=4919+5081; 10^5=49877=50123; 10^6=499943+500057;
10^7=4999913+5000087; 10^8=49999757+50000243;
10^9=499999931+500000069;
10^10=4999999937+5000000063}, etc.
%Y A124450 A065577
%O A124450 1
%K A124450 ,nonn,
%A A124450 Zak Seidov  (zakseidov at yahoo.com), Nov 02 2006


 
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From zakseidov at yahoo.com  Thu Nov  2 08:12:42 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 1 Nov 2006 23:12:42 -0800 (PST)
Subject: SEQ FROM Zak Seidov 
In-Reply-To: <200611020706.kA276utj016855@exspider.research.att.com>
Message-ID: <20061102071242.66437.qmail@web38208.mail.mud.yahoo.com>

Neil,
in contrast,
in this case I got immediate autoreply from OEIS.
Does submitting with ascribed or non-ascribed 
sequence number matter?
Zak 
--- The On-Line Encyclopedia of Integer Sequences
<oeis at research.att.com> wrote:

> The following is a copy of the email message that
> was sent to njas
> containing the sequence you submitted.
> 
> All greater than and less than signs have been
> replaced by their html
> equivalents.  They will be changed back when the
> message is processed.
> 
> This copy is just for your records.  No reply is
> expected.
>  Subject: NEW SEQUENCE FROM Zak Seidov 
> 
> 
> %I A000001
> %S A000001
>
5,,3,59,11,17,29,11,71,71,23,11,29,29,11,83,3,11,281,11,101,71,23,257,401,293,107,293,53,11,113,251,47,587,23,179,389,59,173,17,1427,83,431,53,563,593,41,47,239,383,431,1181,701,971,149,593,569,149,191,1973
> %N A000001 Lesser of pair of farsest primes summed
> to 10^n.
> %C A000001 Cf. A065577 Number of Goldbach partitions
> of 10^n, A124450 Lesser of pair of closest primes
> summed to 10^n.
> %F A000001 10^n - a(n) is prime.
> %e A000001 10^1 = 5 + 5, 10^2 = 3 + 97, 10^3 = 3 +
> 997, 10^4 = 59 + 9941, 10^5 = 11 + 99989, 10^6 = 17
> + 999983, 10^7 = 29 + 9999971, 10^8 = 11 + 99999989,
> 10^9 = 
> 71 + 999999929, 10^10 = 23 + 9999999977, etc.
> %Y A000001 A065577, A124450.
> %O A000001 1
> %K A000001 ,nonn,
> %A A000001 Zak Seidov  (zakseidov at yahoo.com), Nov 02
> 2006
> RH 
> RA 192.20.225.32
> RU 
> RI 
> 



 
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From jvospost3 at gmail.com  Thu Nov  2 08:44:39 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 1 Nov 2006 23:44:39 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <20061102011002.GC6041@cochise.anu.edu.au>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <20061102011002.GC6041@cochise.anu.edu.au>
Message-ID: <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>

Thank you, Joerg.  That was interesting.  Does your generating function
hypothesis hold when I extend the sequence by another 5 values, to a(n) = 0,
0, 1, 1, 3, 1, 7, 4,
14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again, assuming
that I got the elementary algebra and arithmetic correct.

If so, I should submit this sequence as by the two us!

Eisenstein-Fibonacci sequences.  Cube root of unity analogues of square-root
of unity A014291  Imaginary Rabbits.

[update of 1 Nov 06]

Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
= (-1 - i*sqrt(3))/2.

There are various integer sequences derived from this
complex sequence (such as the real part, the imaginary
part, the polynomial coefficient is of 1, w, and
w^2)Unless my arithmetic by hand is in error again,
is:

n   b(n)
0   w
1   w^2
2   w(w^2) + w = w^3 + w = 1 + w.
3   w(1 + w) + w^2 = w + w^2 + w^2 = 1 + 2w^2.
4   w(1 + 2w^2) + 1 + w = w + 2w^3 + 1 + w = 3 + 2w.
5   w(3 + 2w) + 1 + 2w^2 = 3w + 2w^2 + 1 + 2w^2.
   = 1 + 3w + 4w^2.
6  w(1 + 3w + 4w^2) + 3 + 2w = w + 3w^2 + 4w^3 + 3 +
2w
   = 7 + 3w + 3w^2.
7  w(7 + 3w + 3w^2) + 1 + 3w + 4w^2 = 7w + 3w^2 + 3w^3

  + 1 + 3w + 4w^2 = 4 + 10w + 7w^2.
8  w(4 + 10w + 7w^2) + 7 + 3w + 3w^2 = 4w + 10w^2 +
  7w^3 + 7 + 3w + 3w^2 = 14 + 7w + 13w^2.
9  w(14 + 7w + 13w^2) + 4 + 10w + 7w^2 = 14w + 7w^2 +
  13w^3 + 4 + 10w + 7w^2 = 17 + 24w + 14w^2.
10 w(17 + 24w + 14w^2) + 14 + 7w + 13w^2 = 17w + 24w^2
  + 14w^3 + 14 + 7w + 13w^2 = 28 + 24w + 37w^2.
11 w(28 + 24w + 37w^2) + 17 + 24w + 14w^2 = 28w +
  24w^2 + 37w^3 + 17 + 24w + 14w^2 = 54 + 52w +
38w^2.
12 w(54 + 52w + 38w^2) + 28 + 24w + 37w^2 = 54w +
  52w^2 + 38w^3 + 28 + 24w + 37w^2 = 66 + 78w +
89w^2.
13 w(66 + 78w + 89w^2) + 54 + 52w + 38w^2 =
  66w + 78w^2 + 89w^3 + 54 + 52w + 38w^2 =
  143 + 118w + 116w^2.
14 w(143 + 118w + 116w^2) + 66 + 78w + 89w^2 =
  143w + 118w^2 + 116w^3 + 66 + 78w + 89w^2 =
  182 + 221w + 207w^2.
15 w(182 + 221w + 207w^2) + 143 + 118w + 116w^2 =
  182w + 221w^2 + 207w^3 + 143 + 118w + 116w^2 =
  350 + 300w + 337w^2.

16 w(350 + 300w + 337w^2) + 350 + 300w + 337w^2 =
    350w + 300w^2 + 337w^3 + 350 + 300w + 337w^2 =
    687 +  650w + 637w^2.

17 w(687 +  650w + 637w^2) + 350 + 300w + 337w^2 =
    687w +  650w^2 + 637w^3 + 350 + 300w + 337w^2 =
    987 + 987w + 987w^2.

18  w(987 + 987w + 987w^2) + 687 +  650w + 637w^2 =
    987w + 987w^2 + 987w^3 + 687 +  650w + 637w^2 =
    1674 + 1637w + 1624w^2.

19  w(1674 + 1637w + 1624w^2) + 987 + 987w + 987w^2 =
     1674w + 1637w^2 + 1624w^3 + 987 + 987w + 987w^2
=
     2611 + 2661w + 2624w^2.

20  w(2611 + 2661w + 2624w^2) + 1674 + 1637w + 1624w^2
=
     2611w + 2661w^2 + 2624w^3 + 1674 + 1637w +
1624w^2 =
     4298 + 4248w + 4285w^2.


The coefficients of 1 = a(n) = 0, 0, 1, 1, 3, 1, 7, 4,
14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611,
4298

which does not seem to be in OEIS.

To excerpt my comment in A107890 (and some other
seqs):

"Eisenstein integers are of the form a + b*omega,
where a and b are ordinary integers, and omega = (-1 +
i*sqrt(3))/2 is a cube root of 1, the other cube roots
of 1 being 1 and omega^2 = (-1 - i*sqrt(3))/2.
Eisenstein integers are complex numbers that are also
members of the imaginary quadratic field Q(sqrt -3) =
Z[omega]. The sums, differences, and products of
Eisenstein integers are other Eisenstein integers."

Other sequences come from:
b(0) = w, b(1) = w^2, b(n) = b(n-1) + w*b(n-2),
and from the O, J notation of D?rrie (1965) in
Weisstein, Eric W. "Eisenstein Integer." From
MathWorld--A Wolfram Web Resource.
http://mathworld.wolfram.com/EisensteinInteger.html


Eisenstein-Fib.doc

=== end 2nd email ===

On 11/1/06, Joerg Arndt <arndt at jjj.de> wrote:
>
> * Jonathan Post <jvospost3 at gmail.com> [Nov 01. 2006 10:14]:
> > Eisenstein-Fibonacci sequences.  Cube root of unity
> > analogues of square-root of unity A014291  Imaginary
> > Rabbits.
> >
> > Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
> > where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
> > = (-1 - i*sqrt(3))/2.
> >
> > [...]
> > The coefficients of 1 = a(n) = 0, 0, 1, 1, 3, 1, 7, 4,
> > 14, 17, 28, 54, 66, 143, 182, 350, ...
> >
> > which does not seem to be in OEIS.
>
> Ralf Stephan's ggf() says that the OGF may be:
>
> ? g=ggf([1, 1, 3, 1, 7, 4,14, 17, 28, 54, 66, 143, 182, 350])
> (x^5 - 3*x^3 + x + 1)/(-x^6 + 3*x^4 - x^3 - 3*x^2 + 1)
> ? factor(g)
>
> [x - 1 1]
>
> [x^4 + x^3 - 2*x^2 - 2*x - 1 1]
>
> [x^2 + x - 1 -1]
>
> [x^4 - x^3 - x^2 + x + 1 -1]
>
>
> > [...]
>
>
>
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From maxale at gmail.com  Thu Nov  2 09:23:51 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 2 Nov 2006 00:23:51 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
Message-ID: <d3dac270611020023t17f583aep351f8e1e187148cc@mail.gmail.com>

On 10/22/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> Eisenstein-Fibonacci sequences.  Cube root of unity
>  analogues of square-root of unity A014291  Imaginary
>  Rabbits.
>
>  Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
>  where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
>  = (-1 - i*sqrt(3))/2.

First off, using the identity w^2 = -w-1, we can uniquely represent
b(n) as u(n)+v(n)*w where u(n) and v(n) are integers.
In particular, u(0)=0, v(0)=1, u(1)=-1, v(1)=-1.

Now the recurrence relation b(n) = w*b(n-1) + b(n-2) implies that
u(n) = u(n-2) - v(n-1)
v(n) = v(n-2) + u(n-1) - v(n-1)
that in turn implies that
[u(n-1),v(n-1),u(n),v(n)] = M*[u(n-2),v(n-2),u(n-1),v(n-1)]
where vectors are column-vectors and the matrix M is defined as
[ 0, 0, 1, 0 ]
[ 0, 0, 0, 1 ]
[ 1, 0, 0, -1 ]
[ 0, 1, 1, -1 ]

The characteristic polynomial of the matrix M is
x^4 + x^3 - x^2 - x + 1
meaning that u(n) and v(n) satisfies the recurrence
u(n) = - u(n-1) + u(n-2) + u(n-3) - u(n-4)
v(n) = - v(n-1) + v(n-2) + v(n-3) - v(n-4)
with u(0)=0, u(1)=-1, u(2)=1, u(3)=-2
and v(0)=1, v(1)=-1, v(2)=1, v(3)=-1.

>  There are various integer sequences derived from this
>  complex sequence (such as the real part, the imaginary
>  part, the polynomial coefficient is of 1, w, and
>  w^2)Unless my arithmetic by hand is in error again,
>  is:
>
>  n   b(n)
>  0   w
>  1   w^2
>  2   w(w^2) + w = w^3 + w = 1 + w.
>  3   w(1 + w) + w^2 = w + w^2 + w^2 = 1 + 2w^2.

It is not clear why you don't reduce the "polynomials" of w to the first degree.
Higher degree "polynomials" are not uniquely defined.
E.g., b(1)=w^2=2w^2+w+1=3w^2+2w+2=...
but if you reduce to the first degree "polynomial", there is an unique
representation:
b(1) = -w-1.

[...]

>  The coefficients of 1 = a(n) = 0, 0, 1, 1, 3, 1, 7, 4,
>  14, 17, 28, 54, 66, 143, 182, 350, ...

This sequence is not well-defined (see above).
Probably, what you are interested in is u(n):
0, -1, 1, -2, 2, -2, 1, 1, -4, 8, -12, 15, -15, 10, 2, -22, 49, -79, 104, -112

Max





From maxale at gmail.com  Thu Nov  2 13:06:51 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 2 Nov 2006 04:06:51 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <d3dac270611020023t17f583aep351f8e1e187148cc@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <d3dac270611020023t17f583aep351f8e1e187148cc@mail.gmail.com>
Message-ID: <d3dac270611020406l10f3deb8jb69c41cb2b267651@mail.gmail.com>

On 11/2/06, Max A. <maxale at gmail.com> wrote:

> >  Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
> >  where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
> >  = (-1 - i*sqrt(3))/2.
>
> First off, using the identity w^2 = -w-1, we can uniquely represent
> b(n) as u(n)+v(n)*w where u(n) and v(n) are integers.

[...]

> u(n) and v(n) satisfies the recurrence
> u(n) = - u(n-1) + u(n-2) + u(n-3) - u(n-4)
> v(n) = - v(n-1) + v(n-2) + v(n-3) - v(n-4)
> with u(0)=0, u(1)=-1, u(2)=1, u(3)=-2
> and v(0)=1, v(1)=-1, v(2)=1, v(3)=-1.

o.g.f. for u(n) is
-x/(1 + x - x^2 - x^3 + x^4)
and
o.g.f. for v(n) is
(1-x^2)/(1 + x - x^2 - x^3 + x^4)

Hence, o.g.f. for b(n) is
(-x + w*(1-x^2)) / (1 + x - x^2 - x^3 + x^4)

Max





From njas at research.att.com  Thu Nov  2 14:24:15 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Thu, 2 Nov 2006 08:24:15 -0500 (EST)
Subject: SEQ FROM Zak Seidov
Message-ID: <200611021324.IAA74292@fry.research.att.com>


Does submitting with ascribed or non-ascribed 
sequence number matter?
Zak 


Me:  I don't know.  But I see you are still using yahoo!
Why not switch to gmail?  Then you won't have these problems.

Best

Neil





From mathar at strw.leidenuniv.nl  Thu Nov  2 14:40:51 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Thu, 2 Nov 2006 14:40:51 +0100
Subject: Re^2:  Eisenstein-Fibonacci sequences
Message-ID: <200611021340.kA2DepPv014946@amer.strw.leidenuniv.nl>


ma> From seqfan-owner at ext.jussieu.fr  Thu Nov  2 13:12:31 2006
ma> Return-Path: <seqfan-owner at ext.jussieu.fr>
ma> Date: Thu, 2 Nov 2006 04:06:51 -0800
ma> From: "Max A." <maxale at gmail.com>
ma> To: "Jonathan Post" <jvospost3 at gmail.com>
ma> Subject: Re: Eisenstein-Fibonacci sequences
ma> Cc: "Sequence Fans Mailing List" <seqfan at ext.jussieu.fr>
ma> 
ma> On 11/2/06, Max A. <maxale at gmail.com> wrote:
ma> 
ma> > >  Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
ma> > >  where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
ma> > >  = (-1 - i*sqrt(3))/2.
ma> >
ma> > First off, using the identity w^2 = -w-1, we can uniquely represent
ma> > b(n) as u(n)+v(n)*w where u(n) and v(n) are integers.
ma> 
ma> [...]
ma> 
ma> > u(n) and v(n) satisfies the recurrence
ma> > u(n) = - u(n-1) + u(n-2) + u(n-3) - u(n-4)
ma> > v(n) = - v(n-1) + v(n-2) + v(n-3) - v(n-4)
ma> > with u(0)=0, u(1)=-1, u(2)=1, u(3)=-2
ma> > and v(0)=1, v(1)=-1, v(2)=1, v(3)=-1.
ma> 
ma> o.g.f. for u(n) is
ma> -x/(1 + x - x^2 - x^3 + x^4)
ma> and
ma> o.g.f. for v(n) is
ma> (1-x^2)/(1 + x - x^2 - x^3 + x^4)
ma> 
ma> Hence, o.g.f. for b(n) is
ma> (-x + w*(1-x^2)) / (1 + x - x^2 - x^3 + x^4)
ma> 
ma> Max

The g.f. above [-x+w(1-x^2)]/[1+x-x^2-x^3+x^4] generates
the reduced  polynomial
                          2             3      4            5               6
w + (-1 - w) x + (1 + w) x  + (-2 - w) x  + 2 x  + (w - 2) x  + (-3 w + 1) x

                  7               8              9      10
     + (1 + 5 w) x  + (-4 - 7 w) x  + (8 + 8 w) x  + O(x  )

If one doesn't reduce the polynomial, one gets the simpler
g(x)=w/[1-w*x-x^2] which generates the non-reduced and more generic
b(0)+b(1) x+ b(2) x^2 + b(3) x^3+ ...
=

     2       3       2     4      2   3     5      3       4
w + w  x + (w  + w) x  + (w  + 2 w ) x  + (w  + 3 w  + w) x  +

      6      4      2   5     7      5      3       6
    (w  + 4 w  + 3 w ) x  + (w  + 5 w  + 6 w  + w) x  +

      8      6       4      2   7     9      7       5       3       8
    (w  + 6 w  + 10 w  + 4 w ) x  + (w  + 7 w  + 15 w  + 10 w  + w) x  +

      10      8       6       4      2   9      10
    (w   + 8 w  + 21 w  + 20 w  + 5 w ) x  + O(x  )

and which becomes (apart from an offset) A000045 for w=1, A052542 for w=2,
A052906 for w=3, A000007 for w=0, and signed variations of these for w=-1,-2 or -3..
The coefficients table of the non-reduced g.f. is in A102426, which is
essentially the same as A098925 and A092865.

R. Mathar





From tanyakh at TanyaKhovanova.com  Thu Nov  2 16:15:08 2006
From: tanyakh at TanyaKhovanova.com (Tanya Khovanova)
Date: Thu,  2 Nov 2006 07:15:08 -0800
Subject: What is the most famous sequence?
Message-ID: <200611020715.AA2085028026@TanyaKhovanova.com>

Would this mean that Fibonacci sequence is the most popularized?

---------- Original Message ----------------------------------
From: Edwin Clark <eclark at math.usf.edu>
Date:  Wed, 1 Nov 2006 21:28:27 -0500 (EST)

>
>> What is the most famous sequence?
>
>Since the novel and movie "The Da Vince Code"-- not to mention 
>Knuth's "Art of Computer Programming" it would have to be the 
>Fibonacci numbers.
>
>
>


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From davidwwilson at comcast.net  Thu Nov  2 16:46:38 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 2 Nov 2006 10:46:38 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
References: <20061101054313.25333.qmail@web38210.mail.mud.yahoo.com>
Message-ID: <008e01c6fe96$19f1bbe0$6401a8c0@yourxhtr8hvc4p>

I wrote a C++ program using sieve method and calculated

a(7) = 38807
a(8) = 291400
a(9) = 2274205
a(10) = 18200488

a(7), a(8) and a(9) agree with previous calculations.

However, Zak Seidov gave

However, my a(10) = 18200487.

So now I have to go root out the discrepancy. I am going to have my computer 
generate a file of all the partitions of 10^10 and do an independent check 
of the results, and hopefully uncover a problem in my program. 






From rsc at swtch.com  Thu Nov  2 17:26:57 2006
From: rsc at swtch.com (Russ Cox)
Date: Thu, 2 Nov 2006 11:26:57 -0500
Subject: SEQ FROM Zak Seidov
In-Reply-To: <20061102071242.66437.qmail@web38208.mail.mud.yahoo.com>
References: <200611020706.kA276utj016855@exspider.research.att.com>
	 <20061102071242.66437.qmail@web38208.mail.mud.yahoo.com>
Message-ID: <ee9e417a0611020826h1daaac8xa85b9f6cc4b0ebdf@mail.gmail.com>

As I said in my earlier mail, the semantics of the
Yahoo "spam protection" are such that if you
submit two sequences within some time period
(maybe 12 hours to a day), the first might get
delayed many hours while the second will go through
immediately.

Out of curiosity, do you have oeis at research.att.com
in your address book?

Russ





From davidwwilson at comcast.net  Thu Nov  2 17:29:52 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 2 Nov 2006 11:29:52 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
Message-ID: <001101c6fe9c$205d0c40$6401a8c0@yourxhtr8hvc4p>

Ok, that last message was incoherent.

For A065577 Zak Seidov got a(10) = 18200488, I got a(10) = 18200487.  I am 
checking my program. 






From pxp at rogers.com  Thu Nov  2 18:36:24 2006
From: pxp at rogers.com (Hans Havermann)
Date: Thu, 2 Nov 2006 12:36:24 -0500
Subject: A124450
In-Reply-To: <20061102061228.7046.qmail@web38206.mail.mud.yahoo.com>
References: <20061102061228.7046.qmail@web38206.mail.mud.yahoo.com>
Message-ID: <D7CB7013-474C-4B78-B336-E5DBA449F167@rogers.com>

zak seidov:

> Hope someone may wish to add more terms:
> Mathematica isn't the best in such calcs.

> %S A124450
> 5,47,491,4919,49877,499943,4999913,49999757,499999931,4999999937,49999 
> 999811,499999999769,4999999998431,49999999999619,499999999999769,49999 
> 99999998557,49999999999998887,499999999999999679,4999999999999999661,4 
> 9999999999999998647
> %N A124450 Lesser of pair of closest primes summed to 10^n.

I think you overstate the Mathematica difficulty. With no effort at  
any sort of efficiency in the programming, it still only takes me 8  
seconds to generate 50 terms.

Table[h =10^n/2; c=0; While[PrimeQ[h-c]==False || PrimeQ[h+c]==False,  
c++]; h-c, {n, 1, 50}]

{5, 47, 491, 4919, 49877, 499943, 4999913, 49999757, 499999931,  
4999999937, 49999999811, 499999999769, 4999999998431, 49999999999619,  
499999999999769, 4999999999998557, 49999999999998887,  
499999999999999679, 4999999999999999661, 49999999999999998647,  
499999999999999999637, 4999999999999999999481,  
49999999999999999998677, 499999999999999999998497,  
4999999999999999999999259, 49999999999999999999998779,  
499999999999999999999999043, 4999999999999999999999998947,  
49999999999999999999999999661, 499999999999999999999999994069,  
4999999999999999999999999997879, 49999999999999999999999999997699,  
499999999999999999999999999997969,  
4999999999999999999999999999995227,  
49999999999999999999999999999995263,  
499999999999999999999999999999989719,  
4999999999999999999999999999999998683,  
49999999999999999999999999999999999871,  
499999999999999999999999999999999996127,  
4999999999999999999999999999999999998557,  
49999999999999999999999999999999999999613,  
499999999999999999999999999999999999988231,  
4999999999999999999999999999999999999991729,  
49999999999999999999999999999999999999994663,  
499999999999999999999999999999999999999997117,  
4999999999999999999999999999999999999999992863,  
49999999999999999999999999999999999999999991807,  
499999999999999999999999999999999999999999991507,  
4999999999999999999999999999999999999999999993199,  
49999999999999999999999999999999999999999999997081}

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From davidwwilson at comcast.net  Thu Nov  2 18:45:17 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 2 Nov 2006 12:45:17 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
References: <20061102171720.72610.qmail@web38203.mail.mud.yahoo.com>
Message-ID: <002b01c6fea6$a96fcae0$6401a8c0@yourxhtr8hvc4p>

Well let's get it right this time

I got a(10) = 18200488, in agreement with Dick Mathar's value.





From jvospost3 at gmail.com  Thu Nov  2 19:01:55 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 2 Nov 2006 10:01:55 -0800
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
In-Reply-To: <002b01c6fea6$a96fcae0$6401a8c0@yourxhtr8hvc4p>
References: <20061102171720.72610.qmail@web38203.mail.mud.yahoo.com>
	 <002b01c6fea6$a96fcae0$6401a8c0@yourxhtr8hvc4p>
Message-ID: <5542af940611021001x27e8951akc9b931928d84333e@mail.gmail.com>

Amusing coincidence that the digits of a(10) = 18200488 consist entirely of
0 and powers of 2.  [base].

-- Jonathan Vos Post

On 11/2/06, David Wilson <davidwwilson at comcast.net> wrote:
>
> Well let's get it right this time
>
> I got a(10) = 18200488, in agreement with Dick Mathar's value.
>
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From jvospost3 at gmail.com  Thu Nov  2 21:54:44 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 2 Nov 2006 12:54:44 -0800
Subject: New & old LEGO seqs; production cuts
In-Reply-To: <C16FD8A8.20890%eilers@math.ku.dk>
References: <5542af940611010958me17f347w3bf2d2f2d6c9dec0@mail.gmail.com>
	 <C16FD8A8.20890%eilers@math.ku.dk>
Message-ID: <5542af940611021254w14b12a38y14021854e772de77@mail.gmail.com>

Dear Soren,

Thank you for a clear and thoughful answer.

My son (in the room with me now) agrees that baby/childhood play with Duplo
and Lego guided him towards his Math/Computer Science expertise today.

He and I agree, however, that it's a shame that the manufacturer doesn't
make 4-D Lego blocks, or, more properly, the ones that they do make are of
extension along the t-axis only in the trivial way, and cannot be freely
rotated in Minkowski space.

So, I wonder, how well does your approach to counting under symmetry work if
extended to an additional spacial dimension?

I suppose that we can consider an AxBxC 4-D Lego to be actually an AxBxCx1
Lego in Z^4. I think that adjacency is well-defined, all rotations are about
planes, and there is a relationship to 4-D analogues of polyominoes, i.e. to
polyhypercubes.

Best,

Jonathan Vos Post

On 11/2/06, S?ren Eilers <eilers at math.ku.dk> wrote:
>
>  Dear Jonathan
>
> Thank you, and yes, I think that LEGO inspires kids to learn about
> fundamental mathematical concepts such as symmetry.
>
> I've also considered adding crossrefs to the sequences you mentioned but
> decided against it.
>
> Regarding A007576: Watson definitely counts a subset of what we count in
> A123818, but as you can see in his paper he only looks at "planar" buildings
> where all blocks could be placed in, say, [0,2]xRxR. This means that one of
> the dimensions of the blocks used is irrelevant, and Watson could just have
> well have counted what we call FLAT structures with 1x2 blocks. Thus, IF one
> was to crossref I would suggest to do it to A123764, but perhaps this is a
> little hard to explain in the limited space of an OEIS entry.
>
> But Watson has even more restrictions; he looks only at MAX HEIGHT
> buildings with one block at each level. This is exactly what LEGO did wrong
> when they claimed that A123828(6)=102981500 [the number they wanted was
> (46^5+2^5)/2, but they got the last digit wrong too].
>
> Of course counting FLAT and MAX HEIGHT structures with 2x2 or 1x2 blocks
> is easy, the formula is 3^(n-1) or (3^(n-1)+1)/2 depending on whether or not
> you want to identify up to symmetry. Watson knows this and tries to count
> how many of these structures would tip over if placed on a flat surface,
> essentially computing the x-coordinate of the center of gravity. So I think
> the kind of LEGO counting done by Watson is quite different from the kind of
> LEGO counting we do.
>
>     {Btw, A007576 is indeed the same as A086821}
>     {If somebody was interested I could probably quite easily compute how
> many of "our" LEGO buildings are stable in the sense of Watson}
>
> Regarding Zeilberger's LEGO counts, they all relate, as far as I recall
> from a relatively hard look some time ago, to buildings with blocks of
> varying size, and hence are quite different in nature from what we do. In
> that case, n is not a number of blocks, but the total volume/weight of the
> structure, so that's a very different ball game.
>
> Best regards,
> Soren
>
>
>
>
>
> On 01/11/06 18:58, "Jonathan Post" <jvospost3 at gmail.com> wrote:
>
> I like the new seqs about Lego, such as
> A123762 <http://www.research.att.com/%7Enjas/sequences/A123762><http://www.research.att.com/%7Enjas/sequences/A123762>
>   Number of ways, counted up to symmetry, to build a contiguous building
> with n *LEGO* blocks of size 1x2.
>
> Ironically timed, as *"Recent restructuring and production cuts have left
> Lego unable to fill orders
> <http://www.cbc.ca/money/story/2006/10/31/legoproduction.html><http://www.cbc.ca/money/story/2006/10/31/legoproduction.html>for the upcoming holiday season. Affected products include Duplo bricks,
> Lego City sets, and (horror of horrors!) Star Wars and Lego Technik sets."
> * According to the article Lego stands to lose $127 million in holiday
> sales.
>
> http://www.cbc.ca/money/story/2006/10/31/legoproduction.html
>
> It might be useful to add crossrefs to earlier seqs such as:
>
> A007576 <http://www.research.att.com/%7Enjas/sequences/A007576><http://www.research.att.com/%7Enjas/sequences/A007576>
>   Number of maximally stable towers of 2 X 2 *LEGO* blocks.
>
>
> A082679 <http://www.research.att.com/%7Enjas/sequences/A082679><http://www.research.att.com/%7Enjas/sequences/A082679>
>   Number of *Lego* towers, one piece per floor, where every floor is
> perpendicular to the one below it (so we have a kind of 3-dimensional zigzag
> pattern).
>  -- Jonathan Vos Post
> [wondering if my son's earlier intense exposure to Duplo and Lego led him
> towards his double B.S. in Math and Computer Science]
>
>
>
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From davidwwilson at comcast.net  Thu Nov  2 22:49:42 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 2 Nov 2006 16:49:42 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
References: <20061102171720.72610.qmail@web38203.mail.mud.yahoo.com> <002b01c6fea6$a96fcae0$6401a8c0@yourxhtr8hvc4p>
Message-ID: <004e01c6fec8$ce34d100$6401a8c0@yourxhtr8hvc4p>

Regarding A065577:

I altered my counting program to generating a file listing, in increasing 
order, all primes p such that there exists prime q with p < q, p+q = 10^10. 
I wrote a verification program to read this file, check that the p are in 
increasing order, count the p, compute q = 10^10-p, check p < q and that p 
and q are both prime (using trial division).

The verification program has finished, and confirms that a(10) >= 18200488. 
This contradicts Zak Seidov's count a(10) = 18200487, but agrees with Dick 
Mathar's count a(10) = 18200488.

If anyone is interested, I am keeping the text file with the p values for a 
little while. It is 195MB, so e-mailing it is probably not a good option.






From zakseidov at yahoo.com  Fri Nov  3 04:09:56 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Thu, 2 Nov 2006 19:09:56 -0800 (PST)
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
In-Reply-To: <004e01c6fec8$ce34d100$6401a8c0@yourxhtr8hvc4p>
Message-ID: <20061103030956.55909.qmail@web38210.mail.mud.yahoo.com>

David, Dick, Bob, Ivars,
seqfans.

After recalculating (at two different PC's)
and rechecking I could finally 
find source of my error
(silly as usual) in one of codes,
and my final value is 
a(10) = 18200488 (NOT 18200487!)
which coincides now with the Dick-David value.

So, Bob, 
you may wish to give in A065577  the correct value 
a(10) = 18200488 with due refer to David Wilson  and
Richard Mathar (not me).
Also Ivars can extend his table.
Thanks, Zak

BTW1 Still, is it possible to find a(11)... with C++?

BIW2 Not all people even in this closed list
may be interested in Mmca specific problems,
so I''l be happy to have a list of you
particularly Mmca-involved.
Then we may circulate relevant messages
not to all seqfan community.
Please anyone interested write me personally
with subject "Me in Mmca", thanks.
Of course it's not intented to be closed list!
Hope this is not against Gerard' policy.

--- David Wilson <davidwwilson at comcast.net> wrote:

> Regarding A065577:
> 
> I altered my counting program to generating a file
> listing, in increasing 
> order, all primes p such that there exists prime q
> with p < q, p+q = 10^10. 
> I wrote a verification program to read this file,
> check that the p are in 
> increasing order, count the p, compute q = 10^10-p,
> check p < q and that p 
> and q are both prime (using trial division).
> 
> The verification program has finished, and confirms
> that a(10) >= 18200488. 
> This contradicts Zak Seidov's count a(10) =
> 18200487, but agrees with Dick 
> Mathar's count a(10) = 18200488.
> 
> If anyone is interested, I am keeping the text file
> with the p values for a 
> little while. It is 195MB, so e-mailing it is
> probably not a good option.
> 
> 



 
____________________________________________________________________________________
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From zakseidov at yahoo.com  Fri Nov  3 05:03:30 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Thu, 2 Nov 2006 20:03:30 -0800 (PST)
Subject: A123998: No primes except 3?
Message-ID: <20061103040330.2777.qmail@web38202.mail.mud.yahoo.com>

Why are there no prime numbers except of 3,
in this sequence?
Zak

A123998
={1,3,9,15,18,39,48,69,78,99,105,114,135,153,165,168,183,189,219,249,273,

288,300,303,309,330,345,363,405,414,438,468,483,498,504,534,585,618,
 639,648,699,714,729,765,804,813,828,879,933}; 
(* %N A123998 Numbers n for which 2n+1 and 4n+1 are
primes. *)



 
__________________________________________________________________________________________
Check out the New Yahoo! Mail - Fire up a more powerful email and get things done faster. 
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From maxale at gmail.com  Fri Nov  3 06:41:20 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 2 Nov 2006 21:41:20 -0800
Subject: A123998: No primes except 3?
In-Reply-To: <20061103040330.2777.qmail@web38202.mail.mud.yahoo.com>
References: <20061103040330.2777.qmail@web38202.mail.mud.yahoo.com>
Message-ID: <d3dac270611022141x4d10f69evdd706982df6e3d28@mail.gmail.com>

Note that if n = 1 mod 3 then 2n+1 is not prime (except n=1);
and if n = 2 mod 3 then 4n+1 is not prime.
Therefore, n must be a multiple of 3, except n=1.

Max

If n>1 and both 2n+1, 4n+1 are primes

On 11/2/06, zak seidov <zakseidov at yahoo.com> wrote:
> Why are there no prime numbers except of 3,
> in this sequence?
> Zak
>
> A123998
> ={1,3,9,15,18,39,48,69,78,99,105,114,135,153,165,168,183,189,219,249,273,
>
> 288,300,303,309,330,345,363,405,414,438,468,483,498,504,534,585,618,
>  639,648,699,714,729,765,804,813,828,879,933};
> (* %N A123998 Numbers n for which 2n+1 and 4n+1 are
> primes. *)
>
>
>
>
> __________________________________________________________________________________________
> Check out the New Yahoo! Mail - Fire up a more powerful email and get things done faster.
> (http://advision.webevents.yahoo.com/mailbeta)
>
>





From zakseidov at yahoo.com  Fri Nov  3 07:01:51 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Thu, 2 Nov 2006 22:01:51 -0800 (PST)
Subject: A123998: No primes except 3?
In-Reply-To: <d3dac270611022141x4d10f69evdd706982df6e3d28@mail.gmail.com>
Message-ID: <20061103060151.3808.qmail@web38207.mail.mud.yahoo.com>

Great!
It'so easy if you are Max, not Zak!
Thanks, 
Zak

--- "Max A." <maxale at gmail.com> wrote:

> Note that if n = 1 mod 3 then 2n+1 is not prime
> (except n=1);
> and if n = 2 mod 3 then 4n+1 is not prime.
> Therefore, n must be a multiple of 3, except n=1.
> 
> Max
> 
> If n>1 and both 2n+1, 4n+1 are primes
> 
> On 11/2/06, zak seidov <zakseidov at yahoo.com> wrote:
> > Why are there no prime numbers except of 3,
> > in this sequence?
> > Zak
> >
> > A123998
> >
>
={1,3,9,15,18,39,48,69,78,99,105,114,135,153,165,168,183,189,219,249,273,
> >
> >
>
288,300,303,309,330,345,363,405,414,438,468,483,498,504,534,585,618,
> >  639,648,699,714,729,765,804,813,828,879,933};
> > (* %N A123998 Numbers n for which 2n+1 and 4n+1
> are
> > primes. *)
> >
> >
> >
> >
> >
>
__________________________________________________________________________________________
> > Check out the New Yahoo! Mail - Fire up a more
> powerful email and get things done faster.
> > (http://advision.webevents.yahoo.com/mailbeta)
> >
> >
> 



 
____________________________________________________________________________________
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From jeremy.gardiner at btinternet.com  Fri Nov  3 09:59:50 2006
From: jeremy.gardiner at btinternet.com (JEREMY GARDINER)
Date: Fri, 3 Nov 2006 08:59:50 +0000 (GMT)
Subject: What is the most famous sequence?
In-Reply-To: <200611020715.AA2085028026@TanyaKhovanova.com>
Message-ID: <20061103085950.5824.qmail@web86610.mail.ird.yahoo.com>

The natural numbers must be the most widely known sequence ...

Tanya Khovanova <tanyakh at TanyaKhovanova.com> wrote:  > What is the most famous sequence?
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From maxale at gmail.com  Fri Nov  3 10:08:28 2006
From: maxale at gmail.com (Max A.)
Date: Fri, 3 Nov 2006 01:08:28 -0800
Subject: querying multidimensional sequences in OEIS
Message-ID: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>

I have a sequence indexed by vectors with non-negative integer components.
Is there a regular way to query such sequence in OEIS ?

As an example, let us consider a sequence of multinomial coefficients.
Any vector with non-negative integer components [k1, k2, ..., kn]
defines an element of this sequence:

a( [k1, k2, ..., kn] ) = (k1 + k2 +  ... + kn)! / (k1! * k2! * ... * kn!)

Assume that I know nothing about these numbers and have only a
black-box computing a(v) for any given vector v with non-negative
integer components. Can I anyhow find this sequence in OEIS and learn
that its elements called "multinomial coefficients" ?

Thanks,
Max





From wouter.meeussen at pandora.be  Fri Nov  3 12:39:40 2006
From: wouter.meeussen at pandora.be (wouter meeussen)
Date: Fri, 3 Nov 2006 12:39:40 +0100
Subject: download seqs resulting from search
Message-ID: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>

dear all,

if I search for all number triangles with row sums equal to the factorials,
I enter
"row sum" "factorial"
and get 9 pages of results in 'short' form.
(1) how can I improve on this query?
(2) how can I download them all in one go?

Wouter





From njas at research.att.com  Fri Nov  3 14:43:47 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Fri, 3 Nov 2006 08:43:47 -0500 (EST)
Subject: download seqs resulting from search
Message-ID: <200611031343.IAA51448@fry.research.att.com>

as i have said before, you can download up to 30 (or is it 50?) sequences
at once by using the old email lookup service

send email to  sequences at research.att.com  saying things like

lookup A000123
lookup A000125
lookup A012345

etc.

Neil





From franktaw at netscape.net  Fri Nov  3 16:18:26 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 10:18:26 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
Message-ID: <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>

I don't know of any standard for functions of an arbitrary finite 
sequence
of positive integers in the OEIS.

In your example, the function is commutative, which makes it 
essentially
a function of the multiset of values - or in other words, of a 
partition.  For
that there is a standard - two of them actually.  One is the 
enumeration
in Abramowitz and Stegun order, as specified in A036036, and the other
is the Mathematica ordering, as specified in A080577.

There are some relevant sequences in the database in both orders, some
only in the A&S order, and a few only in the Mma order.

Since the two orders are the same for partitions of up to 5 objects, 
one
often just search for an early part of the sequence.  If this does not 
suffice,
one should try both before giving up.

For example, the partitions of 5 objects are: [5], [4,1], [3,2], 
[3,1^2],
[2^2,1], [2,1^3], [1^5].  The multinomial coefficients for these are:

1,5,10,20,30,60,120

A search for this turns up A036038, the A&S version, and A078760, the
Mma version.

----
As for arbitrary finite sequences of positive integers, there are two 
main
candidates for an ordering: A066099 and A108244; these are the reverse
of each other.  I would probably go with A066099, as it is closest to 
the
ordering used for the partitions; but you can make an argument for 
either
one.  Possibly both should be used.

Off hand, I can't think of any interesting functions of finite 
sequences of
positive integers that depend on the order of the sequence.

Franklin T. Adams-Watters


-----Original Message-----
From: maxale at gmail.com

I have a sequence indexed by vectors with non-negative integer 
components.?
Is there a regular way to query such sequence in OEIS ??
?
As an example, let us consider a sequence of multinomial coefficients.?
Any vector with non-negative integer components [k1, k2, ..., kn]?
defines an element of this sequence:?
?
a( [k1, k2, ..., kn] ) = (k1 + k2 + ... + kn)! / (k1! * k2! * ... * 
kn!)?
?
Assume that I know nothing about these numbers and have only a?
black-box computing a(v) for any given vector v with non-negative?
integer components. Can I anyhow find this sequence in OEIS and learn?
that its elements called "multinomial coefficients" ??
?
Thanks,?
Max?


________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From rsc at swtch.com  Fri Nov  3 17:06:23 2006
From: rsc at swtch.com (Russ Cox)
Date: Fri, 3 Nov 2006 11:06:23 -0500
Subject: download seqs resulting from search
In-Reply-To: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>
References: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>
Message-ID: <ee9e417a0611030806q1490600ci75384340c6facd67@mail.gmail.com>

> if I search for all number triangles with row sums equal to the factorials,
> I enter
> "row sum" "factorial"
> and get 9 pages of results in 'short' form.
> (1) how can I improve on this query?

Not sure.

> (2) how can I download them all in one go?

Add &n=100 to the URL
and you will get 100 results per page.
Then click on "text" formatting and save the
resulting page.

http://www.research.att.com/~njas/sequences/?q=%22row+sums%22+factorial&n=100&fmt=3

Russ





From franktaw at netscape.net  Fri Nov  3 17:14:55 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 11:14:55 -0500
Subject: download seqs resulting from search
In-Reply-To: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>
References: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>
Message-ID: <8C8CD7A55C060ED-FD0-B6C7@WEBMAIL-MA20.sysops.aol.com>

Try adding "keyword:tabl" to the search.  Then (probably) repeat with 
"keyword:tabf" instead - unless you aren't interested in irregular 
tables.  This won't be all in one go, but two isn't too bad.

Of course, there are undoubtably some tables which sum to factorials 
that don't have the words "row sum" in them.  And maybe some that just 
reference A000142 instead of using the word "factorial".

Franklin T. Adams-Watters


-----Original Message-----
From: wouter.meeussen at pandora.be

dear all,

if I search for all number triangles with row sums equal to the 
factorials,
I enter
"row sum" "factorial"
and get 9 pages of results in 'short' form.
(1) how can I improve on this query?
(2) how can I download them all in one go?

Wouter

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From mlb at well.com  Fri Nov  3 17:26:14 2006
From: mlb at well.com (Marc LeBrun)
Date: Fri, 03 Nov 2006 08:26:14 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <200611031625.kA3GP1mt017249@smtp.well.com>

 >=franktaw at netscape.net

Good advice...maybe it could be made part of the tips-n-tricks section?

 > just search for an early part of the sequence

Or sometimes you can search for a small set of "unusual" later, 
larger, terms that will act as a kind of signature.

 > Off hand, I can't think of any interesting functions of finite sequences of
 > positive integers that depend on the order of the sequence.

Exponents in prime factorizations, say.







From franktaw at netscape.net  Fri Nov  3 19:31:34 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 13:31:34 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <200611031625.kA3GP1mt017249@smtp.well.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com> <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com> <200611031625.kA3GP1mt017249@smtp.well.com>
Message-ID: <8C8CD8D6CD483AD-FD0-C3EF@WEBMAIL-MA20.sysops.aol.com>

Nice suggestion.  That would be A057335.

Franklin T. Adams-Watters


-----Original Message-----
From: mlb at well.com

>=franktaw at netscape.net?
> Off hand, I can't think of any interesting functions of finite 
sequences of?
> positive integers that depend on the order of the sequence.?
?
Exponents in prime factorizations, say.?
?


________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From maxale at gmail.com  Fri Nov  3 21:26:20 2006
From: maxale at gmail.com (Max A.)
Date: Fri, 3 Nov 2006 12:26:20 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <200611031625.kA3GP1mt017249@smtp.well.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <200611031625.kA3GP1mt017249@smtp.well.com>
Message-ID: <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com>

On 11/3/06, Marc LeBrun <mlb at well.com> wrote:

>  > Off hand, I can't think of any interesting functions of finite sequences of
>  > positive integers that depend on the order of the sequence.
>
> Exponents in prime factorizations, say.

Oh, that is a very nice suggestion.
Are there any multidimensional sequences represented this way in OEIS?

I have a number of multidimensional sequences in mind that I'd like to
see in OEIS.
E.g.: the coefficients of incomplete Bell polynomials:

b( [k1, k2, ..., kn] ) = (k1 + 2*k2 + ... + n*kn)! / (k1! * k2! * ...
* kn!) / (1!^k1 * 2!^k2 * ... * n!^kn)

and looking similar the number of permutations with given cycle structure:

c( [k1, k2, ..., kn] ) = (k1 + 2*k2 + ... + n*kn)! / (k1! * k2! * ...
* kn!) / (1^k1 * 2^k2 * ... * n^kn)

Max





From mlb at well.com  Fri Nov  3 22:41:00 2006
From: mlb at well.com (Marc LeBrun)
Date: Fri, 03 Nov 2006 13:41:00 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com
 >
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
 <200611031625.kA3GP1mt017249@smtp.well.com>
 <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com>
Message-ID: <200611032139.kA3LdZSV024106@smtp.well.com>

 > I have a number of multidimensional sequences in mind that I'd like to
 > see in OEIS.  E.g.: the coefficients of incomplete Bell polynomials:

Two possible ways to submit sequences of tuples T(1),T(2),...
where T(i)=[T(i,1),T(i,2),...] are:

A. Define a traversal order by a map (i,j) <--> n and set a(n) = T(i,j).
This gives the tabl/tabf style of entry.

B. Encode each tuple as a single number, say by using it as factor
exponents:  T(i) --> 2^T(i,1) * 3^T(i,2) * 5^T(i,3) *...

For example, for Pascal's triangle

A: 1; 1, 1; 1, 2, 1; 1, 3, 3, 1;...

B: 2, 6,    90,      47250,...

However there are a variety of possible ways to do these things--for example
you might subtract 1 from all the Pascal exponents, B': 1, 1, 3, 225,...

Or traverse in the order (1,1), (2,1), (1,2), (3,1), (1,3), (2,2), (1,4),...
which recursively interleaves sequences instead of the usual diagonalizing.

But this sort of thing can also be abused to breed swarms of silly sequences,
so it should be well-motivated.

It's nicest to use familiar schemes (eg as Frank noted for partitions).







From franktaw at netscape.net  Fri Nov  3 22:41:18 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 16:41:18 -0500
Subject: Possibly more errors in OEIS?
Message-ID: <8C8CDA7EE6FDEF2-12C-B6F@WEBMAIL-MA20.sysops.aol.com>

Looking at sequences with keyword:cons, there is a marked shortage of 
sequences with a given duplicate digit (excluding 0), compared with 
those with a given other pair of digits.  For example, restricting our 
attention to sequences not of the form "Decimal expansion of 1/n", and 
ignoring all pairs involving 0 or 1, there are an average of about 1710 
sequences  with any given digit pair, but an average of only about 1627 
with duplicate pairs.  10 of this difference is explained by the 10 
"selvage" sequences, which have no duplicate digits, but the rest is 
unexplained.

I'm guessing that there may be an input error here; that duplicate 
digits sometimes get combined at some point in the process, whether 
before or after submission.  I'm not sure how to test for this, short 
of examining the "cons" sequences and determining which are correct.

Mostly for Neil,

Sequences A081780 to A081796, "continued cotangent" sequences, all have 
the "cons" keyword, but they should not.

Also, sequence A114721 has the "cons" keyword; it should instead have 
the "frac" keyword.

Franklin T. Adams-Watters

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From jawbrey at att.net  Fri Nov  3 22:51:00 2006
From: jawbrey at att.net (Jon Awbrey)
Date: Fri, 03 Nov 2006 16:51:00 -0500
Subject: querying multidimensional sequences in OEIS
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <200611031625.kA3GP1mt017249@smtp.well.com>
	 <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com> <200611032139.kA3LdZSV024106@smtp.well.com>
Message-ID: <454BB9C4.CCCA8C55@att.net>

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o

you might get some ideas from looking at the
tables of "primal codes" or "primal functions"
that I compiled, for instance, here:

http://stderr.org/pipermail/inquiry/2005-July/002846.html

ja

Marc LeBrun wrote:
> 
>  > I have a number of multidimensional sequences in mind that I'd like to
>  > see in OEIS.  E.g.: the coefficients of incomplete Bell polynomials:
> 
> Two possible ways to submit sequences of tuples T(1),T(2),...
> where T(i)=[T(i,1),T(i,2),...] are:
> 
> A. Define a traversal order by a map (i,j) <--> n and set a(n) = T(i,j).
> This gives the tabl/tabf style of entry.
> 
> B. Encode each tuple as a single number, say by using it as factor
> exponents:  T(i) --> 2^T(i,1) * 3^T(i,2) * 5^T(i,3) *...
> 
> For example, for Pascal's triangle
> 
> A: 1; 1, 1; 1, 2, 1; 1, 3, 3, 1;...
> 
> B: 2, 6,    90,      47250,...
> 
> However there are a variety of possible ways to do these things--for example
> you might subtract 1 from all the Pascal exponents, B': 1, 1, 3, 225,...
> 
> Or traverse in the order (1,1), (2,1), (1,2), (3,1), (1,3), (2,2), (1,4),...
> which recursively interleaves sequences instead of the usual diagonalizing.
> 
> But this sort of thing can also be abused to breed swarms of silly sequences,
> so it should be well-motivated.
> 
> It's nicest to use familiar schemes (eg as Frank noted for partitions).

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o
inquiry e-lab: http://stderr.org/pipermail/inquiry/
citizendium development forums: http://smf.citizendium.org
http://www.textop.org/wiki/index.php?title=User:Jon_Awbrey
wikinfo: http://wikinfo.org/wiki.php?title=User:Jon_Awbrey
wp review: http://wikipediareview.com/index.php?showuser=398
o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o






From maxale at gmail.com  Fri Nov  3 23:02:04 2006
From: maxale at gmail.com (Max A.)
Date: Fri, 3 Nov 2006 14:02:04 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <200611032139.kA3LdZSV024106@smtp.well.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <200611031625.kA3GP1mt017249@smtp.well.com>
	 <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com>
	 <200611032139.kA3LdZSV024106@smtp.well.com>
Message-ID: <d3dac270611031402s62ff2746p3a5c749865678cec@mail.gmail.com>

On 11/3/06, Marc LeBrun <mlb at well.com> wrote:
>  > I have a number of multidimensional sequences in mind that I'd like to
>  > see in OEIS.  E.g.: the coefficients of incomplete Bell polynomials:
>
> Two possible ways to submit sequences of tuples T(1),T(2),...
> where T(i)=[T(i,1),T(i,2),...] are:
>
> A. Define a traversal order by a map (i,j) <--> n and set a(n) = T(i,j).
> This gives the tabl/tabf style of entry.
>
> B. Encode each tuple as a single number, say by using it as factor
> exponents:  T(i) --> 2^T(i,1) * 3^T(i,2) * 5^T(i,3) *...

As I understand, A is applicable only to the cases where the dimension
of tuples is fixed while B is perfectly fine for tuples with variable
dimensions (what I'm currently interested in).

>
> For example, for Pascal's triangle
>
> A: 1; 1, 1; 1, 2, 1; 1, 3, 3, 1;...
>
> B: 2, 6,    90,      47250,...

Does B apply to Pascal triangle?
The 5-th term corresponds to the tuple [0,0,5], what it is in Pascal triangle?

Max





From franktaw at netscape.net  Fri Nov  3 23:06:02 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 17:06:02 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com> <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>

<Me>
>Off hand, I can't think of any interesting functions of finite 
sequences of
>positive integers that depend on the order of the sequence.

Of course, I no sooner wrote that than I started thinking of such 
functions.
I'm not sure any of these are tremendously interesting, but they're at 
least
somewhat so.

In the following, b(i), i = 1..k is the finite sequence.  I'm starting 
all these
sequences with the empty sequence, but I searched for them without
that first term.  All are shown here in A066099 order.

Alternating sum
Sum_i (-1)^{i+1) b(i)
0,1,2,0,3,1,-1,1,4,2,0,2,-2,0,2,0

Binomial sum
Sum_i C(k-1,i-1) b(i)
0,1,2,2,3,3,3,4,4,4,4,5,4,6,5,8

Inverse binomial sum
Sum_i (-1)^{i+1) C(k-1,i-1) b(i)
0,1,2,0,3,1,-1,0,4,2,0,1,-2,-2,1,0

Weighted sum
sum_i i b(i)
0,1,2,3,3,4,5,6,4,5,6,7,7,8,9,10
This is A029931

Zero-based weighted sum
sum_i (i-1) b(i)
0,0,0,1,0,1,2,3,0,1,2,3,3,4,5,6

Sum of products of consecutive elements
sum_{i=1}^{k-1} b(i) b(i+1)
0,0,0,1,0,2,2,2,0,3,4,3,3,4,3,3

Number of rises
sum_{b(i)>b(i-1)} 1
0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,0

Number of falls
sum_{b(i)<b(i-1)} 1
0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0

Number of unchanged
sum_{b(i)=b(i-1)} 1
0,0,0,1,0,0,0,2,0,0,1,1,0,0,1,3

Number of non-rises
sum_{b(i)<=b(i-1)} 1
0,0,0,1,0,0,1,2,0,0,1,1,1,1,2,3

Number of non-falls
sum_{b(i)>=b(i-1)} 1
0,0,0,1,0,1,0,2,0,1,1,2,0,1,1,3

Number of monotonically increasing runs
1 + number of falls, but 0 for empty sequence
0,1,1,1,1,2,1,1,1,2,1,2,1,2,1,1

Number of monotonically decreasing runs
1 + number of rises, but 0 for empty sequence
0,1,1,1,1,1,2,1,1,1,1,1,2,2,2,1

Number of runs of equal terms
1 + number of unchanged, but 0 for empty sequence
0,1,1,2,1,1,1,3,1,1,2,2,1,1,2,4

Number of distinct non-empty subsequences
0,1,1,2,1,3,3,3,1,3,2,5,3,5,5,4
E.g., f(1,1) = 2, for the sequences [1], and [1,1].

Number of distinct subsequences
1 + number of distinct non-empty subsequences
1,2,2,3,2,4,4,4,2,4,3,6,4,6,6,5

Number of set partitions
List parts of set partition by their smallest element, and count the 
part sizes.
E.g., 1|2,4|3 (also known as {1,2,3,2}) would count for the sequence 
[1,2,1].
1,1,1,1,1,2,1,1,1,3,3,3,1,2,1,1

Number of permutations
List permutations in cycle form, sorted by the smallest element in the 
cycle,
and count the cycle lengths.
Number of set partitions * Product_i (b(i)-1)!
1,1,1,1,2,2,1,1,6,6,3,3,2,2,1,1

Number of partially ordered sets (unlabelled) by rank
The rank of an element in a poset is the length of the longest chain of 
which it
is the largest element.
1,1,1,1,1,2,1,1,1,3,4,3,1,2,1,1

Number of partially ordered sets (labelled) by rank
1,1,1,2,1,9,3,6,1,28,54,60,4,36,12,24

Number of partially ordered sets (naturally labelled) by rank
Naturally labelled means labels are consistent with the ordering.
1,1,1,1,1,4,1,1,1,11,13,8,1,4,1,1

Number of forests (unlabelled, unordered) with b(i) nodes at height i.
1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,2,2,1,1,1,1,2,1,1,1,1,1

... and, of course, various other forest options.

If there is agreement that this is the right order for these sequences, 
I'll try
to find time to submit them.
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From tbaruchel at free.fr  Fri Nov  3 23:17:29 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Fri, 3 Nov 2006 23:17:29 +0100
Subject: [math-fun] A fast (written in assembly) library for continued
	fractions
Message-ID: <20061103221729.GB2467@>

I wrote for my own purposes the following X86 assembly code, intended
to be called from GFORTRAN. Such a thing should only work if you have
a similar configuration to mine, but anyway, it may interest you.
I wanted a fast library for computing the length in the continued
fraction expansion of sqrt(n) for any integer n that can fit in
a REAL*8 gfortran type for statistical purposes. I also included
a faster function that tells if this length is odd or even. Though
I am not an assembly expert, there should be a significant speed
improvement if you use this code rather than your own fortran
functions, since the whole process is performed on the FPU stack
with no load/store useless travels.

It may help you, it may not. You know about these kinds of software...

Instructions are at the beginning of the file, as comments.
Read them carefully. Code is below.

Regards.

------------------ FILE quad.s BEGINS BELOW --------------------

/*

  Compile your piece of code with something like
    gfortran -fno-underscoring mycode.f90 quad.s
  and add freely your own flags. Enjoy !
  Of course, if you don't want to use the -fno-underscoring
  option for some reason, you can add an underscore at the end
  of the global functions below, like that:
    .globl initfpu_
    initfpu_:
  it should work.

  Remember this piece of code has the following status:
    Seems to work for me, maybe it will also seem for you,
    maybe your computer will explode, maybe it will make the
    earth explode, maybe God will choose not to have created
    the world, think about it before using it.
  (Anyway, if you see some kind of bug, please let me know ;-)
  Remember that
    a) this is x86 assembly code;
    b) it uses the GNU AS (gas) syntax;
    c) it uses the gfortran calling conventions;
  (now you understand why you can't expect it will works ;-)

  Usage. In your program (or subroutine or whatever), declare
  the function you need like:
    
    logical :: oddperiod
    integer*8 :: period

  (the two subroutines don't need to be declared).

  Then use
    
    integer*4 :: oldfpustatus
    call initfpu(oldfpustatus)

  before any computation; if you don't do it, it won't work.
  When you don't need any more the library,
    
    call restorefpu(oldfpustatus)

  will restore the control register of the FPU.

  During all that time, you shouldn't be disturbed too much, 
  for your own mathematical operations, but
  remember the rounding mode of the FPU has been set to
  "round down" (rather than the usual "nearest").

  How does it work? The whole computation is performed on the
  stack of the FPU; only integer values are needed for these
  tasks, but they are stored as REAL*10, 80 bits reals. They
  are not converted to anything during the whole process, thus
  it should work for great integers stored in your own REAL*8,
  64 bits reals. Of course, it may happen (I think it will never,
  but didn't check) that an intermediate result can not fit in
  such a REAL*10 register; in that case, read above what is
  written about the end of the world...
  
*/



/*
  call initfpu(integer*4)
  =====================
    This subroutine should be called before any other call to the
    library; its purpose is to set the rounding control in the FPU
    control register to the "round down" mode.
    The subroutine puts in the integer variable the initial value
    of the FPU control register.

    If you want to use often this subroutine (which you shouldn't:
    speed being an issue, this subroutine is intended to be used
    once, then the functions billions of times, not the contrary ;-),
    maybe removing the "fnclex" can improve the speed, since I don't
    think it plays a great role here.
*/
.globl initfpu
initfpu:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        fnstcw  (%eax)
	fwait
        movw    (%eax), %dx
	andb	$0xf3, %dh
	orb	$0x4, %dh
        movw    %dx, 2(%eax)
	fnclex
	fldcw   2(%eax)
        fwait
        movw    $0, 2(%eax)
	leave
	ret

/*
  call restorefpu(integer*4)
  ========================
    This subroutine should be used for restoring the initial state of
    the FPU control register; the integer variable must contain the
    right value (which should have been previously set by the
    subroutine initfpu).
*/
.globl restorefpu
restorefpu:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
	fclex
	fldcw   (%eax)
	leave
	ret

/*
  sqrtint( real double precision : REAL*8 )
  =========================================
    This function returns the integer part of the square root
    of the given real. The return value has itself the real type.
*/
.global sqrtint
sqrtint:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
	fldl	(%eax)
	fsqrt
        frndint
	leave
	ret

/*
  oddperiod( real double precision : REAL*8 )
  ===========================================
    Despite the type of the real argument, the variable has to
    be some integer value stored as a double precision.
    The function has the logical type. It returns FALSE if
    the argument is a perfect square or if the continued
    fraction expansion of its square root has an even period
    length; it returns TRUE in the remaining case.
      oddperiod(dble(25)) --> FALSE
      oddperiod(dble(13)) --> TRUE
      oddperiod(dble(12)) --> FALSE
    If it may help, you can declare the function as an integer*4
    type rather than a logical one (output will be 0 or 1).

    Though it seems to work with integer*8, I am not absolutely
    sure it *always* will (unless you add the following instruction
      movl $0, %edx
    at the beginning of the code (for instance between the current
    second and third lines of code), but really why would you
    do it? Maybe something in the calling conventions is such that
    the register EDX already contains 0, but, again, who cares?

    If you read the instructions for the period() function, you
    will learn about some 2^32 limitations, but you don't have to
    care about that here: the length is not kept in any register
    during the computation and the code should work as long as the
    values in the FPU stack are the right one.
    
*/
.global oddperiod
oddperiod:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        fldln2
	fldl	(%eax)
        fld     %st
        fsqrt
        frndint
        fld     %st
        fld     %st
        fmul    %st, %st
        fld     %st(3)
        fsubp   %st, %st(1)
        fcom    %st(4)
        fstsw   %ax
        fwait
        sahf
        ja      oddperiod0
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
oddperiod2:
        movl    $0, %eax
        leave
        ret
oddperiod0:
        fld     %st(1)
        fadd    %st(3), %st
        fdiv    %st(1), %st
        frndint
        fld     %st(2)
        fld     %st(1)
        fmul    %st(3), %st
        fsubp   %st, %st(1)
        fxch    %st(3)
        fsub    %st(3), %st
        fabs
        fcomp   %st(6)
        fstsw   %ax
        fwait
        sahf
        ja      oddperiod1
        fld     %st(3)
        fadd    %st, %st
        fsubp   %st, %st(1)
        fcomp   %st(5)
        fstsw   %ax
        fwait
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        sahf
        ja      oddperiod2
        movl    $1, %eax
        leave
        ret
oddperiod1:
        ffree   %st
        fincstp
        fld     %st(1)
        fmul    %st, %st
        fld     %st(4)
        fsubp   %st, %st(1)
        fdivp   %st, %st(1)
        frndint
        jmp oddperiod0

/*
  period( real double precision : REAL*8 )
  ========================================
    Despite the type of the real argument, the variable has to
    be some integer value stored as a double precision.
    The function has integer*8 type, though the output
    can't be greater than 2^32-1 (this choice has been made
    for avoiding sign problem when the result is greater
    (or equal) then 2^31 and smaller then 2^32.
    The result is the length of period in the continued fraction
    expansion of the square root of the given integer.

    If speed is really for you an obsession, remember that it
    is highly unprobable that you use that library for periods
    between the two fatidic values. Maybe it will increase a little
    the speed to use integer*4 which allows to delete twice the line
        movl    $0,%edx
    (not the first time it appears, but the second and third one).
    If you try such a thing and see some speed improvement, please
    let me know.
*/
.global period
period:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        movl    $0, %edx
        fldln2
	fldl	(%eax)
        fld     %st
        fsqrt
        frndint
        fld     %st
        fld     %st
        fmul    %st, %st
        fld     %st(3)
        fsubp   %st, %st(1)
        fcom    %st(4)
        fstsw   %ax
        fwait
        sahf
        ja      period0
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        movl    %edx, %eax
        movl     $0, %edx
        leave
        ret
period0:
        inc     %edx
        fld     %st(1)
        fadd    %st(3), %st
        fdiv    %st(1), %st
        frndint
        fld     %st(2)
        fld     %st(1)
        fmul    %st(3), %st
        fsubp   %st, %st(1)
        fxch    %st(3)
        fsub    %st(3), %st
        fabs
        fcomp   %st(6)
        fstsw   %ax
        fwait
        sahf
        ja      period1
        fld     %st(3)
        fadd    %st, %st
        fsubp   %st, %st(1)
        fcomp   %st(5)
        fstsw   %ax
        fwait
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        sahf
        jb      period2
        add     %edx, %edx
period2:
        movl    %edx, %eax
        movl    $0,%edx
        leave
        ret
period1:
        ffree   %st
        fincstp
        fld     %st(1)
        fmul    %st, %st
        fld     %st(4)
        fsubp   %st, %st(1)
        fdivp   %st, %st(1)
        frndint
        jmp period0

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From tbaruchel at free.fr  Fri Nov  3 23:17:29 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Fri, 3 Nov 2006 23:17:29 +0100
Subject: A fast (written in assembly) library for continued fractions
Message-ID: <20061103221729.GB2467@>

I wrote for my own purposes the following X86 assembly code, intended
to be called from GFORTRAN. Such a thing should only work if you have
a similar configuration to mine, but anyway, it may interest you.
I wanted a fast library for computing the length in the continued
fraction expansion of sqrt(n) for any integer n that can fit in
a REAL*8 gfortran type for statistical purposes. I also included
a faster function that tells if this length is odd or even. Though
I am not an assembly expert, there should be a significant speed
improvement if you use this code rather than your own fortran
functions, since the whole process is performed on the FPU stack
with no load/store useless travels.

It may help you, it may not. You know about these kinds of software...

Instructions are at the beginning of the file, as comments.
Read them carefully. Code is below.

Regards.

------------------ FILE quad.s BEGINS BELOW --------------------

/*

  Compile your piece of code with something like
    gfortran -fno-underscoring mycode.f90 quad.s
  and add freely your own flags. Enjoy !
  Of course, if you don't want to use the -fno-underscoring
  option for some reason, you can add an underscore at the end
  of the global functions below, like that:
    .globl initfpu_
    initfpu_:
  it should work.

  Remember this piece of code has the following status:
    Seems to work for me, maybe it will also seem for you,
    maybe your computer will explode, maybe it will make the
    earth explode, maybe God will choose not to have created
    the world, think about it before using it.
  (Anyway, if you see some kind of bug, please let me know ;-)
  Remember that
    a) this is x86 assembly code;
    b) it uses the GNU AS (gas) syntax;
    c) it uses the gfortran calling conventions;
  (now you understand why you can't expect it will works ;-)

  Usage. In your program (or subroutine or whatever), declare
  the function you need like:
    
    logical :: oddperiod
    integer*8 :: period

  (the two subroutines don't need to be declared).

  Then use
    
    integer*4 :: oldfpustatus
    call initfpu(oldfpustatus)

  before any computation; if you don't do it, it won't work.
  When you don't need any more the library,
    
    call restorefpu(oldfpustatus)

  will restore the control register of the FPU.

  During all that time, you shouldn't be disturbed too much, 
  for your own mathematical operations, but
  remember the rounding mode of the FPU has been set to
  "round down" (rather than the usual "nearest").

  How does it work? The whole computation is performed on the
  stack of the FPU; only integer values are needed for these
  tasks, but they are stored as REAL*10, 80 bits reals. They
  are not converted to anything during the whole process, thus
  it should work for great integers stored in your own REAL*8,
  64 bits reals. Of course, it may happen (I think it will never,
  but didn't check) that an intermediate result can not fit in
  such a REAL*10 register; in that case, read above what is
  written about the end of the world...
  
*/



/*
  call initfpu(integer*4)
  =====================
    This subroutine should be called before any other call to the
    library; its purpose is to set the rounding control in the FPU
    control register to the "round down" mode.
    The subroutine puts in the integer variable the initial value
    of the FPU control register.

    If you want to use often this subroutine (which you shouldn't:
    speed being an issue, this subroutine is intended to be used
    once, then the functions billions of times, not the contrary ;-),
    maybe removing the "fnclex" can improve the speed, since I don't
    think it plays a great role here.
*/
.globl initfpu
initfpu:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        fnstcw  (%eax)
	fwait
        movw    (%eax), %dx
	andb	$0xf3, %dh
	orb	$0x4, %dh
        movw    %dx, 2(%eax)
	fnclex
	fldcw   2(%eax)
        fwait
        movw    $0, 2(%eax)
	leave
	ret

/*
  call restorefpu(integer*4)
  ========================
    This subroutine should be used for restoring the initial state of
    the FPU control register; the integer variable must contain the
    right value (which should have been previously set by the
    subroutine initfpu).
*/
.globl restorefpu
restorefpu:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
	fclex
	fldcw   (%eax)
	leave
	ret

/*
  sqrtint( real double precision : REAL*8 )
  =========================================
    This function returns the integer part of the square root
    of the given real. The return value has itself the real type.
*/
.global sqrtint
sqrtint:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
	fldl	(%eax)
	fsqrt
        frndint
	leave
	ret

/*
  oddperiod( real double precision : REAL*8 )
  ===========================================
    Despite the type of the real argument, the variable has to
    be some integer value stored as a double precision.
    The function has the logical type. It returns FALSE if
    the argument is a perfect square or if the continued
    fraction expansion of its square root has an even period
    length; it returns TRUE in the remaining case.
      oddperiod(dble(25)) --> FALSE
      oddperiod(dble(13)) --> TRUE
      oddperiod(dble(12)) --> FALSE
    If it may help, you can declare the function as an integer*4
    type rather than a logical one (output will be 0 or 1).

    Though it seems to work with integer*8, I am not absolutely
    sure it *always* will (unless you add the following instruction
      movl $0, %edx
    at the beginning of the code (for instance between the current
    second and third lines of code), but really why would you
    do it? Maybe something in the calling conventions is such that
    the register EDX already contains 0, but, again, who cares?

    If you read the instructions for the period() function, you
    will learn about some 2^32 limitations, but you don't have to
    care about that here: the length is not kept in any register
    during the computation and the code should work as long as the
    values in the FPU stack are the right one.
    
*/
.global oddperiod
oddperiod:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        fldln2
	fldl	(%eax)
        fld     %st
        fsqrt
        frndint
        fld     %st
        fld     %st
        fmul    %st, %st
        fld     %st(3)
        fsubp   %st, %st(1)
        fcom    %st(4)
        fstsw   %ax
        fwait
        sahf
        ja      oddperiod0
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
oddperiod2:
        movl    $0, %eax
        leave
        ret
oddperiod0:
        fld     %st(1)
        fadd    %st(3), %st
        fdiv    %st(1), %st
        frndint
        fld     %st(2)
        fld     %st(1)
        fmul    %st(3), %st
        fsubp   %st, %st(1)
        fxch    %st(3)
        fsub    %st(3), %st
        fabs
        fcomp   %st(6)
        fstsw   %ax
        fwait
        sahf
        ja      oddperiod1
        fld     %st(3)
        fadd    %st, %st
        fsubp   %st, %st(1)
        fcomp   %st(5)
        fstsw   %ax
        fwait
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        sahf
        ja      oddperiod2
        movl    $1, %eax
        leave
        ret
oddperiod1:
        ffree   %st
        fincstp
        fld     %st(1)
        fmul    %st, %st
        fld     %st(4)
        fsubp   %st, %st(1)
        fdivp   %st, %st(1)
        frndint
        jmp oddperiod0

/*
  period( real double precision : REAL*8 )
  ========================================
    Despite the type of the real argument, the variable has to
    be some integer value stored as a double precision.
    The function has integer*8 type, though the output
    can't be greater than 2^32-1 (this choice has been made
    for avoiding sign problem when the result is greater
    (or equal) then 2^31 and smaller then 2^32.
    The result is the length of period in the continued fraction
    expansion of the square root of the given integer.

    If speed is really for you an obsession, remember that it
    is highly unprobable that you use that library for periods
    between the two fatidic values. Maybe it will increase a little
    the speed to use integer*4 which allows to delete twice the line
        movl    $0,%edx
    (not the first time it appears, but the second and third one).
    If you try such a thing and see some speed improvement, please
    let me know.
*/
.global period
period:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        movl    $0, %edx
        fldln2
	fldl	(%eax)
        fld     %st
        fsqrt
        frndint
        fld     %st
        fld     %st
        fmul    %st, %st
        fld     %st(3)
        fsubp   %st, %st(1)
        fcom    %st(4)
        fstsw   %ax
        fwait
        sahf
        ja      period0
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        movl    %edx, %eax
        movl     $0, %edx
        leave
        ret
period0:
        inc     %edx
        fld     %st(1)
        fadd    %st(3), %st
        fdiv    %st(1), %st
        frndint
        fld     %st(2)
        fld     %st(1)
        fmul    %st(3), %st
        fsubp   %st, %st(1)
        fxch    %st(3)
        fsub    %st(3), %st
        fabs
        fcomp   %st(6)
        fstsw   %ax
        fwait
        sahf
        ja      period1
        fld     %st(3)
        fadd    %st, %st
        fsubp   %st, %st(1)
        fcomp   %st(5)
        fstsw   %ax
        fwait
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        sahf
        jb      period2
        add     %edx, %edx
period2:
        movl    %edx, %eax
        movl    $0,%edx
        leave
        ret
period1:
        ffree   %st
        fincstp
        fld     %st(1)
        fmul    %st, %st
        fld     %st(4)
        fsubp   %st, %st(1)
        fdivp   %st, %st(1)
        frndint
        jmp period0





From jvospost3 at gmail.com  Fri Nov  3 23:36:07 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 3 Nov 2006 14:36:07 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com>

I think the example givem without A number, is:

 A007188 <http://www.research.att.com/%7Enjas/sequences/A007188>
 Multiplicative encoding of Pascal triangle: Product p(i+1)^C(n,i).
(Formerly M1722)
 +0
2
2, 6, 90, 47250, 66852843750, 2806877704512541816406250,
1216935896582703898519354781702537118597533386230468750
There might be other Godel-coded sequences, and the like, besides Conway's
brilliant "Fractran", above and beyond those with "decimal Godelization" --
but usually these quickly lead to numbers too large for OEIS.

-- Jonathan Vos Post

On 11/3/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
> <Me>
> >Off hand, I can't think of any interesting functions of finite
> sequences of
> >positive integers that depend on the order of the sequence.
>
> Of course, I no sooner wrote that than I started thinking of such
> functions.
> I'm not sure any of these are tremendously interesting, but they're at
> least
> somewhat so.
>
> In the following, b(i), i = 1..k is the finite sequence.  I'm starting
> all these
> sequences with the empty sequence, but I searched for them without
> that first term.  All are shown here in A066099 order.
>
> Alternating sum
> Sum_i (-1)^{i+1) b(i)
> 0,1,2,0,3,1,-1,1,4,2,0,2,-2,0,2,0
>
> Binomial sum
> Sum_i C(k-1,i-1) b(i)
> 0,1,2,2,3,3,3,4,4,4,4,5,4,6,5,8
>
> Inverse binomial sum
> Sum_i (-1)^{i+1) C(k-1,i-1) b(i)
> 0,1,2,0,3,1,-1,0,4,2,0,1,-2,-2,1,0
>
> Weighted sum
> sum_i i b(i)
> 0,1,2,3,3,4,5,6,4,5,6,7,7,8,9,10
> This is A029931
>
> Zero-based weighted sum
> sum_i (i-1) b(i)
> 0,0,0,1,0,1,2,3,0,1,2,3,3,4,5,6
>
> Sum of products of consecutive elements
> sum_{i=1}^{k-1} b(i) b(i+1)
> 0,0,0,1,0,2,2,2,0,3,4,3,3,4,3,3
>
> Number of rises
> sum_{b(i)>b(i-1)} 1
> 0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,0
>
> Number of falls
> sum_{b(i)<b(i-1)} 1
> 0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0
>
> Number of unchanged
> sum_{b(i)=b(i-1)} 1
> 0,0,0,1,0,0,0,2,0,0,1,1,0,0,1,3
>
> Number of non-rises
> sum_{b(i)<=b(i-1)} 1
> 0,0,0,1,0,0,1,2,0,0,1,1,1,1,2,3
>
> Number of non-falls
> sum_{b(i)>=b(i-1)} 1
> 0,0,0,1,0,1,0,2,0,1,1,2,0,1,1,3
>
> Number of monotonically increasing runs
> 1 + number of falls, but 0 for empty sequence
> 0,1,1,1,1,2,1,1,1,2,1,2,1,2,1,1
>
> Number of monotonically decreasing runs
> 1 + number of rises, but 0 for empty sequence
> 0,1,1,1,1,1,2,1,1,1,1,1,2,2,2,1
>
> Number of runs of equal terms
> 1 + number of unchanged, but 0 for empty sequence
> 0,1,1,2,1,1,1,3,1,1,2,2,1,1,2,4
>
> Number of distinct non-empty subsequences
> 0,1,1,2,1,3,3,3,1,3,2,5,3,5,5,4
> E.g., f(1,1) = 2, for the sequences [1], and [1,1].
>
> Number of distinct subsequences
> 1 + number of distinct non-empty subsequences
> 1,2,2,3,2,4,4,4,2,4,3,6,4,6,6,5
>
> Number of set partitions
> List parts of set partition by their smallest element, and count the
> part sizes.
> E.g., 1|2,4|3 (also known as {1,2,3,2}) would count for the sequence
> [1,2,1].
> 1,1,1,1,1,2,1,1,1,3,3,3,1,2,1,1
>
> Number of permutations
> List permutations in cycle form, sorted by the smallest element in the
> cycle,
> and count the cycle lengths.
> Number of set partitions * Product_i (b(i)-1)!
> 1,1,1,1,2,2,1,1,6,6,3,3,2,2,1,1
>
> Number of partially ordered sets (unlabelled) by rank
> The rank of an element in a poset is the length of the longest chain of
> which it
> is the largest element.
> 1,1,1,1,1,2,1,1,1,3,4,3,1,2,1,1
>
> Number of partially ordered sets (labelled) by rank
> 1,1,1,2,1,9,3,6,1,28,54,60,4,36,12,24
>
> Number of partially ordered sets (naturally labelled) by rank
> Naturally labelled means labels are consistent with the ordering.
> 1,1,1,1,1,4,1,1,1,11,13,8,1,4,1,1
>
> Number of forests (unlabelled, unordered) with b(i) nodes at height i.
> 1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,2,2,1,1,1,1,2,1,1,1,1,1
>
> ... and, of course, various other forest options.
>
> If there is agreement that this is the right order for these sequences,
> I'll try
> to find time to submit them.
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and
> industry-leading spam and email virus protection.
>
>
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From jvospost3 at gmail.com  Sat Nov  4 00:59:34 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 3 Nov 2006 15:59:34 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
	 <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com>
Message-ID: <5542af940611031559x7a45e57euad1d783fb83d82b@mail.gmail.com>

To give a specific example:

NEW SEQUENCE FROM Jonathan Vos Post

%I A000001
%S A000001 2, 6, 450, 2836181250, 81492043057751910481759423160156250,
4561157026363824997482074305569280581505536351717093893927260661169357729871499327113563125890139588096951624677718591308593750
%N A000001 Multiplicative encoding of Catalan's triangle: Product
p(i+1)^T(n,i).
%C A000001 This is to A009766 "Catalan's triangle T(n,k) (read by
rows)"  as A007188 "Multiplicative encoding of Pascal triangle: Product
p(i+1)^C(n,i)" is to A007318 "Pascal's triangle read by rows."
%F A000001 a(n) = PRODUCT[i=i..n] p(i+1)^T(n,i), where T(n,i), are
Cataln's triangle as in a009766.
%e A000001 a(1) = p(1)^T(1,1) = 2^1 = 2.
a(2) = p(1)^T(2,1) * p(2)^T(2,2) = 2^1 * 3^1 = 6.
a(3) = p(1)^T(3,1) * p(2)^T(3,2) * p(3)^T(3,3) = 2^1 * 3^2 * 5^2 = 450.
a(4) = 2^1 * 3^3 * 5^5 * 7^5 = 2836181250.
a(5) = 2^1 * 3^4 * 5^9 * 7^14 * 11^14 =
81492043057751910481759423160156250.
a(6) = 2^1 * 3^5 * 5^14 * 7^28 * 11^42 * 13^42.
%Y A000001 Cf. A007188, A007318, A009766.
%O A000001 1
%K A000001 ,easy,nonn,
%A A000001 Jonathan Vos Post (jvospost2 at yahoo.com
<http://us.f551.mail.yahoo.com/ym/Compose?To=jvospost2 at yahoo.com&YY=14244&y5beta=yes&y5beta=yes&order=down&sort=date&pos=0>),
Nov 03 2006
RH
RA 192.20.225.32



On 11/3/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> I think the example givem without A number, is:
>
>  A007188 <http://www.research.att.com/%7Enjas/sequences/A007188>
>  Multiplicative encoding of Pascal triangle: Product p(i+1)^C(n,i).
> (Formerly M1722)
>  +0
> 2
> 2, 6, 90, 47250, 66852843750, 2806877704512541816406250,
> 1216935896582703898519354781702537118597533386230468750
> There might be other Godel-coded sequences, and the like, besides Conway's
> brilliant "Fractran", above and beyond those with "decimal Godelization" --
> but usually these quickly lead to numbers too large for OEIS.
>
> -- Jonathan Vos Post
>
> On 11/3/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
> >
> > <Me>
> > >Off hand, I can't think of any interesting functions of finite
> > sequences of
> > >positive integers that depend on the order of the sequence.
> >
> > Of course, I no sooner wrote that than I started thinking of such
> > functions.
> > I'm not sure any of these are tremendously interesting, but they're at
> > least
> > somewhat so.
> >
> > In the following, b(i), i = 1..k is the finite sequence.  I'm starting
> > all these
> > sequences with the empty sequence, but I searched for them without
> > that first term.  All are shown here in A066099 order.
> >
> > Alternating sum
> > Sum_i (-1)^{i+1) b(i)
> > 0,1,2,0,3,1,-1,1,4,2,0,2,-2,0,2,0
> >
> > Binomial sum
> > Sum_i C(k-1,i-1) b(i)
> > 0,1,2,2,3,3,3,4,4,4,4,5,4,6,5,8
> >
> > Inverse binomial sum
> > Sum_i (-1)^{i+1) C(k-1,i-1) b(i)
> > 0,1,2,0,3,1,-1,0,4,2,0,1,-2,-2,1,0
> >
> > Weighted sum
> > sum_i i b(i)
> > 0,1,2,3,3,4,5,6,4,5,6,7,7,8,9,10
> > This is A029931
> >
> > Zero-based weighted sum
> > sum_i (i-1) b(i)
> > 0,0,0,1,0,1,2,3,0,1,2,3,3,4,5,6
> >
> > Sum of products of consecutive elements
> > sum_{i=1}^{k-1} b(i) b(i+1)
> > 0,0,0,1,0,2,2,2,0,3,4,3,3,4,3,3
> >
> > Number of rises
> > sum_{b(i)>b(i-1)} 1
> > 0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,0
> >
> > Number of falls
> > sum_{b(i)<b(i-1)} 1
> > 0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0
> >
> > Number of unchanged
> > sum_{b(i)=b(i-1)} 1
> > 0,0,0,1,0,0,0,2,0,0,1,1,0,0,1,3
> >
> > Number of non-rises
> > sum_{b(i)<=b(i-1)} 1
> > 0,0,0,1,0,0,1,2,0,0,1,1,1,1,2,3
> >
> > Number of non-falls
> > sum_{b(i)>=b(i-1)} 1
> > 0,0,0,1,0,1,0,2,0,1,1,2,0,1,1,3
> >
> > Number of monotonically increasing runs
> > 1 + number of falls, but 0 for empty sequence
> > 0,1,1,1,1,2,1,1,1,2,1,2,1,2,1,1
> >
> > Number of monotonically decreasing runs
> > 1 + number of rises, but 0 for empty sequence
> > 0,1,1,1,1,1,2,1,1,1,1,1,2,2,2,1
> >
> > Number of runs of equal terms
> > 1 + number of unchanged, but 0 for empty sequence
> > 0,1,1,2,1,1,1,3,1,1,2,2,1,1,2,4
> >
> > Number of distinct non-empty subsequences
> > 0,1,1,2,1,3,3,3,1,3,2,5,3,5,5,4
> > E.g., f(1,1) = 2, for the sequences [1], and [1,1].
> >
> > Number of distinct subsequences
> > 1 + number of distinct non-empty subsequences
> > 1,2,2,3,2,4,4,4,2,4,3,6,4,6,6,5
> >
> > Number of set partitions
> > List parts of set partition by their smallest element, and count the
> > part sizes.
> > E.g., 1|2,4|3 (also known as {1,2,3,2}) would count for the sequence
> > [1,2,1].
> > 1,1,1,1,1,2,1,1,1,3,3,3,1,2,1,1
> >
> > Number of permutations
> > List permutations in cycle form, sorted by the smallest element in the
> > cycle,
> > and count the cycle lengths.
> > Number of set partitions * Product_i (b(i)-1)!
> > 1,1,1,1,2,2,1,1,6,6,3,3,2,2,1,1
> >
> > Number of partially ordered sets (unlabelled) by rank
> > The rank of an element in a poset is the length of the longest chain of
> > which it
> > is the largest element.
> > 1,1,1,1,1,2,1,1,1,3,4,3,1,2,1,1
> >
> > Number of partially ordered sets (labelled) by rank
> > 1,1,1,2,1,9,3,6,1,28,54,60,4,36,12,24
> >
> > Number of partially ordered sets (naturally labelled) by rank
> > Naturally labelled means labels are consistent with the ordering.
> > 1,1,1,1,1,4,1,1,1,11,13,8,1,4,1,1
> >
> > Number of forests (unlabelled, unordered) with b(i) nodes at height i.
> > 1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,2,2,1,1,1,1,2,1,1,1,1,1
> >
> > ... and, of course, various other forest options.
> >
> > If there is agreement that this is the right order for these sequences,
> > I'll try
> > to find time to submit them.
> > ________________________________________________________________________
> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > industry-leading spam and email virus protection.
> >
> >
>
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From davidwwilson at comcast.net  Sat Nov  4 01:40:21 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Fri, 3 Nov 2006 19:40:21 -0500
Subject: A124015 thoughts
Message-ID: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p>

The description of A124015 is

%N A124015 Number of words with n letters in the National Scrabble Association Dictionary.

Some notes:

- It should be noted that the NSAD includes a restricted set of English words (words of 2 to 14 letters, no proper nouns or derivatives, no words with non-alphabetic characters (e.g, contractions, hyphenated words), and it is this restricted set that is being counted.

- The edition of the NSAD used to create A124015 should be specified, since the NSAD is continually edited.  While I do not think that A124015 should change with each new edition of the NSAD, because it may be referenced in other literature. On the other hand, I don't think that a new sequence should be created for each new edition of the NSAD unless the new edition exhibits some interesting statistical departure from the current NSAD (e.g, in some future NSAD, the number of 9-letter words exceeds the number of 8-letter words). This is because I believe that the value of A124015 is mainly to indicate a distribution of English word lengths.

- On the other hand, I have always understood that 4-letter words are most common in literature (where small words appear more often). It might be interesting to create a sequence counting words of length n in the King James Bible, War and Peace, or some other suitably large and stable piece of English literature (we don't have to go overboard on this, just one or two examples indicating word length distributions in literature).

- Another interesting idea: In various languages, what is the distribution of the number of letters in the names of numbers from one to (say) one million?

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From franktaw at netscape.net  Sat Nov  4 02:49:48 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 20:49:48 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <5542af940611031559x7a45e57euad1d783fb83d82b@mail.gmail.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>	 <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com> <5542af940611031559x7a45e57euad1d783fb83d82b@mail.gmail.com>
Message-ID: <8C8CDCAA57554FD-ED8-5165@FWM-M11.sysops.aol.com>

This is OK, but it is the opposite of what we started out talking 
about.  This is a function from integers to sequences; but we are 
talking about functions from sequences to integers.  And not every 
integer encodes a sequence in this way; only those in A055932.

Franklin T. Adams-Watters


-----Original Message-----
From: jvospost3 at gmail.com

  To give a specific example:

 NEW SEQUENCE FROM Jonathan Vos Post

%I A000001
%S A000001 2, 6, 450, 2836181250, 81492043057751910481759423160156250,
4561157026363824997482074305569280581505536351717093893927260661169357729
871499327113563125890139588096951624677718591308593750

%N A000001 Multiplicative encoding of Catalan's triangle: Product
p(i+1)^T(n,i).
%C A000001 This is to A009766 "Catalan's triangle T(n,k) (read by
rows)"  as A007188 "Multiplicative encoding of Pascal triangle: Product

p(i+1)^C(n,i)" is to A007318 "Pascal's triangle read by rows."
%F A000001 a(n) = PRODUCT[i=i..n] p(i+1)^T(n,i), where T(n,i), are
Cataln's triangle as in a009766.
%e A000001 a(1) = p(1)^T(1,1) = 2^1 = 2.

a(2) = p(1)^T(2,1) * p(2)^T(2,2) = 2^1 * 3^1 = 6.
a(3) = p(1)^T(3,1) * p(2)^T(3,2) * p(3)^T(3,3) = 2^1 * 3^2 * 5^2 = 450.
a(4) = 2^1 * 3^3 * 5^5 * 7^5 = 2836181250.
a(5) = 2^1 * 3^4 * 5^9 * 7^14 * 11^14 =

81492043057751910481759423160156250.
a(6) = 2^1 * 3^5 * 5^14 * 7^28 * 11^42 * 13^42.
%Y A000001 Cf. A007188, A007318, A009766.
%O A000001 1
%K A000001 ,easy,nonn,
%A A000001 Jonathan Vos Post (
jvospost2 at yahoo.com), Nov 03 2006
RH
RA 192.20.225.32

On 11/3/06, Jonathan Post <jvospost3 at gmail.com> wrote: I think the 
example givem without A number, is:

   A007188
   Multiplicative encoding of Pascal triangle: Product p(i+1)^C(n,i).
(Formerly M1722)
   +0
2

  2, 6, 90, 47250, 66852843750, 2806877704512541816406250, 
1216935896582703898519354781702537118597533386230468750

  There might be other Godel-coded sequences, and the like, besides 
Conway's brilliant "Fractran", above and beyond those with "decimal 
Godelization" -- but usually these quickly lead to numbers too large 
for OEIS.

 -- Jonathan Vos Post

On 11/3/06, franktaw at netscape.net < franktaw at netscape.net> wrote: <Me>
>Off hand, I can't think of any interesting functions of finite
 sequences of
>positive integers that depend on the order of the sequence.

Of course, I no sooner wrote that than I started thinking of such
functions.
I'm not sure any of these are tremendously interesting, but they're at
least
somewhat so.

In the following, b(i), i = 1..k is the finite sequence. I'm starting
all these
sequences with the empty sequence, but I searched for them without
that first term. All are shown here in A066099 order.

Alternating sum
Sum_i (-1)^{i+1) b(i)
0,1,2,0,3,1,-1,1,4,2,0,2,-2,0,2,0

Binomial sum
Sum_i C(k-1,i-1) b(i)
0,1,2,2,3,3,3,4,4,4,4,5,4,6,5,8

Inverse binomial sum
Sum_i (-1)^{i+1) C(k-1,i-1) b(i)
0,1,2,0,3,1,-1,0,4,2,0,1,-2,-2,1,0

Weighted sum
sum_i i b(i)
0,1,2,3,3,4,5,6,4,5,6,7,7,8,9,10
This is A029931

Zero-based weighted sum
sum_i (i-1) b(i)
0,0,0,1,0,1,2,3,0,1,2,3,3,4,5,6

 Sum of products of consecutive elements
sum_{i=1}^{k-1} b(i) b(i+1)
0,0,0,1,0,2,2,2,0,3,4,3,3,4,3,3

Number of rises
sum_{b(i)>b(i-1)} 1
0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,0

Number of falls
sum_{b(i)<b(i-1)} 1
0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0

Number of unchanged
sum_{b(i)=b(i-1)} 1
0,0,0,1,0,0,0,2,0,0,1,1,0,0,1,3

Number of non-rises
sum_{b(i)<=b(i-1)} 1
0,0,0,1,0,0,1,2,0,0,1,1,1,1,2,3

Number of non-falls
sum_{b(i)>=b(i-1)} 1
0,0,0,1,0,1,0,2,0,1,1,2,0,1,1,3

Number of monotonically increasing runs
1 + number of falls, but 0 for empty sequence
0,1,1,1,1,2,1,1,1,2,1,2,1,2,1,1

Number of monotonically decreasing runs
1 + number of rises, but 0 for empty sequence
0,1,1,1,1,1,2,1,1,1,1,1,2,2,2,1

Number of runs of equal terms
1 + number of unchanged, but 0 for empty sequence
0,1,1,2,1,1,1,3,1,1,2,2,1,1,2,4

Number of distinct non-empty subsequences
0,1,1,2,1,3,3,3,1,3,2,5,3,5,5,4
E.g., f(1,1) = 2, for the sequences [1], and [1,1].

Number of distinct subsequences
1 + number of distinct non-empty subsequences
1,2,2,3,2,4,4,4,2,4,3,6,4,6,6,5

 Number of set partitions
List parts of set partition by their smallest element, and count the
part sizes.
E.g., 1|2,4|3 (also known as {1,2,3,2}) would count for the sequence
[1,2,1].
1,1,1,1,1,2,1,1,1,3,3,3,1,2,1,1

Number of permutations
List permutations in cycle form, sorted by the smallest element in the
cycle,
and count the cycle lengths.
Number of set partitions * Product_i (b(i)-1)!
1,1,1,1,2,2,1,1,6,6,3,3,2,2,1,1

Number of partially ordered sets (unlabelled) by rank
The rank of an element in a poset is the length of the longest chain of
which it
is the largest element.
1,1,1,1,1,2,1,1,1,3,4,3,1,2,1,1

Number of partially ordered sets (labelled) by rank
1,1,1,2,1,9,3,6,1,28,54,60,4,36,12,24

Number of partially ordered sets (naturally labelled) by rank
Naturally labelled means labels are consistent with the ordering.
1,1,1,1,1,4,1,1,1,11,13,8,1,4,1,1

 Number of forests (unlabelled, unordered) with b(i) nodes at height i.
1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,2,2,1,1,1,1,2,1,1,1,1,1

... and, of course, various other forest options.

If there is agreement that this is the right order for these sequences,
I'll try
to find time to submit them.
________________________________________________________________________
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From maxale at gmail.com  Sat Nov  4 03:09:57 2006
From: maxale at gmail.com (Max A.)
Date: Fri, 3 Nov 2006 18:09:57 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <200611032139.kA3LdZSV024106@smtp.well.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <200611031625.kA3GP1mt017249@smtp.well.com>
	 <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com>
	 <200611032139.kA3LdZSV024106@smtp.well.com>
Message-ID: <d3dac270611031809j6a1f850dj5a1846b40153dc0@mail.gmail.com>

On 11/3/06, Marc LeBrun <mlb at well.com> wrote:

> B. Encode each tuple as a single number, say by using it as factor
> exponents:  T(i) --> 2^T(i,1) * 3^T(i,2) * 5^T(i,3) *...

The sequences I've mentioned encoded this way:

I) Multinomial coefficients:

a( [k1, k2, ..., kn] ) = (k1 + k2 +  ... + kn)! / (k1! * k2! * ... * kn!)

1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 2, 1, 1, 3, 1, 3, 2, 2, 1,
4, 1, 2, 1, 3, 1, 6, 1, 1, 2, 2, 2, 6, 1, 2, 2, 4, 1, 6, 1, 3, 3, 2,
1, 5, 1, 3, 2, 3, 1, 4, 2, 4, 2, 2, 1, 12, 1, 2, 3, 1, 2, 6, 1, 3, 2,
6, 1, 10, 1, 2, 3, 3, 2, 6, 1, 5, 1, 2, 1, 12, 2, 2, 2, 4, 1, 12, 2,
3, 2, 2, 2, 6, 1, 3, 3, 6

II) Coefficients of incomplete Bell polynomials:

b( [k1, k2, ..., kn] ) = (k1 + 2*k2 + ... + n*kn)! / (k1! * k2! * ...
* kn!) / (1!^k1 * 2!^k2 * ... * n!^kn)

1, 1, 1, 1, 1, 3, 1, 1, 3, 4, 1, 6, 1, 5, 10, 1, 1, 15, 1, 10, 15, 6,
1, 10, 10, 7, 15, 15, 1, 60, 1, 1, 21, 8, 35, 45, 1, 9, 28, 20, 1,
105, 1, 21, 105, 10, 1, 15, 35, 70, 36, 28, 1, 105, 56, 35, 45, 11, 1,
210, 1, 12, 210, 1, 84, 168, 1, 36, 55, 280, 1, 105, 1, 13, 280, 45,
126, 252, 1, 35, 105, 14, 1, 420, 120, 15, 66, 56, 1, 840, 210, 55,
78, 16, 165, 21, 1, 315, 378, 280

III) The number of permutations with given cycle structure:

c( [k1, k2, ..., kn] ) = (k1 + 2*k2 + ... + n*kn)! / (k1! * k2! * ...
* kn!) / (1^k1 * 2^k2 * ... * n^kn)

1, 1, 1, 1, 2, 3, 6, 1, 3, 8, 24, 6, 120, 30, 20, 1, 720, 15, 5040,
20, 90, 144, 40320, 10, 40, 840, 15, 90, 362880, 120, 3628800, 1, 504,
5760, 420, 45, 39916800, 45360, 3360, 40, 479001600, 630, 6227020800,
504, 210, 403200, 87178291200, 15, 1260, 280, 25920, 3360,
1307674368000, 105, 2688, 210, 226800, 3991680, 20922789888000, 420,
355687428096000, 43545600, 1260, 1, 20160, 4032, 6402373705728000,
25920, 2217600, 3360, 121645100408832000, 105, 2432902008176640000,
518918400, 1120, 226800, 18144, 30240, 51090942171709440000, 70, 105,
6706022400, 1124000727777607680000, 2520, 172800, 93405312000,
23950080, 1344, 25852016738884976640000, 1680, 151200, 2217600,
283046400, 1394852659200, 1663200, 21, 620448401733239439360000,
11340, 9072, 1120

Max





From jawbrey at att.net  Sat Nov  4 03:26:36 2006
From: jawbrey at att.net (Jon Awbrey)
Date: Fri, 03 Nov 2006 21:26:36 -0500
Subject: querying multidimensional sequences in OEIS
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>	 <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com> <5542af940611031559x7a45e57euad1d783fb83d82b@mail.gmail.com> <8C8CDCAA57554FD-ED8-5165@FWM-M11.sysops.aol.com>
Message-ID: <454BFA5C.ACD4F79B@att.net>

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o

but every positive integer in N = {1, 2, 3, ...}
uniquely codes a finite partial function f : N -> N

http://stderr.org/pipermail/inquiry/2005-July/002846.html

ja

franktaw at netscape.net wrote:
> 
> This is OK, but it is the opposite of what we started out talking
> about.  This is a function from integers to sequences; but we are
> talking about functions from sequences to integers.  And not every
> integer encodes a sequence in this way; only those in A055932.
> 
> Franklin T. Adams-Watters

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o
inquiry e-lab: http://stderr.org/pipermail/inquiry/
citizendium development forums: http://smf.citizendium.org
http://www.textop.org/wiki/index.php?title=User:Jon_Awbrey
wikinfo: http://wikinfo.org/wiki.php?title=User:Jon_Awbrey
wp review: http://wikipediareview.com/index.php?showuser=398
o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o






From jvospost3 at gmail.com  Sat Nov  4 03:49:46 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 3 Nov 2006 18:49:46 -0800
Subject: A124015 thoughts
In-Reply-To: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p>
Message-ID: <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>

I mostly agree.  A large sample of text as actually used in a language, such
as a set of books, newspaper articles, transcripts of phone conversations,
or the like, is often called a "corpus" by those who undertake statistical
analyses.  Various government agencies have done so since well before World
War II, tapping the minds of as many of half the Mathematicians in the
world, with computational resources in advance of corporate and personal
users, and with the help of Mathematicians such as Turing.

There are statistical analyses of various editions of the Bible, of the
works of Shakespeare, and the like, with phalanxes of statistical analysts
on Literature.

Many of these are covered in specialized journals, of questionable value to
seqfans, and unnoticed here except when, for instance, a minor poem is
claimed by such methods to be by Shakespeare.

Hence I second the motion by David Wilson to start with an extremely
well-defined if artificial corpus, such as "In various languages, what is
the distribution of the number of letters in the names of numbers from one
to (say) one million?"

I have commented more extensively on Scrabble in other venues.  Do not ever
agree to use the OED (Oxford English Dictionary) as the official dictionary
when playing Scrabble.  Many misspellings by Tournament rumes are there.
Even though the OED is the definitive dictionary of the largest language on
Earth, and fascinating reading for various purposes, it is just too big and
too weird for game use.  Also, it makes it hard to distinguish a bluff from
a variant.

Now that I think of it, logarithmic trends can be interesting.  How about "what
is the distribution of the number of letters in the names of numbers from
one to one billion?"

-- Jonathan Vos Post

On 11/3/06, David Wilson <davidwwilson at comcast.net> wrote:
>
>  The description of A124015 is
>
> %N A124015 Number of words with n letters in the National Scrabble
> Association Dictionary.
>
> Some notes:
>
> - It should be noted that the NSAD includes a restricted set of English
> words (words of 2 to 14 letters, no proper nouns or derivatives, no words
> with non-alphabetic characters (e.g, contractions, hyphenated words), and
> it is this restricted set that is being counted.
>
> - The edition of the NSAD used to create A124015 should be specified,
> since the NSAD is continually edited.  While I do not think that A124015
> should change with each new edition of the NSAD, because it may be
> referenced in other literature. On the other hand, I don't think that a new
> sequence should be created for each new edition of the NSAD unless the new
> edition exhibits some interesting statistical departure from the current
> NSAD (e.g, in some future NSAD, the number of 9-letter words exceeds the
> number of 8-letter words). This is because I believe that the value of
> A124015 is mainly to indicate a distribution of English word lengths.
>
> - On the other hand, I have always understood that 4-letter words are most
> common in literature (where small words appear more often). It might be
> interesting to create a sequence counting words of length n in the King
> James Bible, War and Peace, or some other suitably large and stable piece of
> English literature (we don't have to go overboard on this, just one or two
> examples indicating word length distributions in literature).
>
> - Another interesting idea: In various languages, what is the distribution
> of the number of letters in the names of numbers from one to (say) one
> million?
>
>
>
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From davidwwilson at comcast.net  Sat Nov  4 17:58:32 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sat, 4 Nov 2006 11:58:32 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
References: <20061103030956.55909.qmail@web38210.mail.mud.yahoo.com>
Message-ID: <001701c70032$75ba7ff0$6401a8c0@yourxhtr8hvc4p>

----- Original Message ----- 
From: "zak seidov" <zakseidov at yahoo.com>
To: <seqfan at ext.jussieu.fr>
Sent: Thursday, November 02, 2006 10:09 PM
Subject: Re: A065577 Number of Goldbach partitions of 10^n? More terms?


> David, Dick, Bob, Ivars,
> seqfans.
>
> After recalculating (at two different PC's)
> and rechecking I could finally
> find source of my error
> (silly as usual) in one of codes,
> and my final value is
> a(10) = 18200488 (NOT 18200487!)
> which coincides now with the Dick-David value.
>
> So, Bob,
> you may wish to give in A065577  the correct value
> a(10) = 18200488 with due refer to David Wilson  and
> Richard Mathar (not me).

I don't remember if Richard Mathar or myself was first to find a(10) = 
18200488. It's not really important to me, since any value would be suspect 
until independently confirmed. The fact that your suspect value led you to a 
bug in your program simply serves as additional confirmation. I have no 
problem with all three of us getting attribution. But then, I have no 
problem with none of us getting attribution, as this computation strikes me 
as on the level of an undergraduate homework problem.

> Also Ivars can extend his table.
> Thanks, Zak
>
> BTW1 Still, is it possible to find a(11)... with C++?

Well, it is possible to find a(11) using a C++ sieving program on an aging 
PC in a few hours, and if you were to do so, you would find that a(11) = 
149091160. Methinks that finding a(12) using the same program on the same PC 
would take about a week, except that said PC is a shared resource.

> BIW2 Not all people even in this closed list
> may be interested in Mmca specific problems,
> so I''l be happy to have a list of you
> particularly Mmca-involved.
> Then we may circulate relevant messages
> not to all seqfan community.
> Please anyone interested write me personally
> with subject "Me in Mmca", thanks.
> Of course it's not intented to be closed list!
> Hope this is not against Gerard' policy.

I truly wish I could afford Mma (or better yet, a math-enabled PDF editor). 
If I'm on anybody's Christmas list...






From mathar at strw.leidenuniv.nl  Sat Nov  4 18:20:07 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Sat, 4 Nov 2006 18:20:07 +0100
Subject: primes in arithmetic progression
Message-ID: <200611041720.kA4HK70T027399@amer.strw.leidenuniv.nl>


Is there an OEIS table that shows the smallest prime p
of k (not necessarily consecutive) primes in arithmetic progression
with common difference d? The table would look similar to this one
below, and contain rather large primes where I am leaving blanks:

    k=1 k=2  k=3  k=4  k=5  k=6
d
0     2   2    2    2    2    2
2     2   3    3
4     2   3    3
6     2   5    5    5    5
8     2   3    3
10    2   3    3
12    2   5    5    5    5
14    2   3    3 
16    2   3
18    2   5    5    5
20    2   3
22    2   7
24    2   5    5   59
26    2   3
28    2   3    3
30    2   7    7    7     7    7
32    2   5
34    2   3    3
36    2   5    7   31

The row d=0 and the column k=1 are degenerate and filled with the
prime 2. All strides d are even. Example for row d=24 and column k=4:
The 4 numbers 59,59+24,59+2*24 and 59+3*24 are all primes.

http://www.research.att.com/~njas/sequences/Sindx_Pri.html#primes_AP

--Richard





From rsc at swtch.com  Sat Nov  4 18:24:53 2006
From: rsc at swtch.com (Russ Cox)
Date: Sat, 4 Nov 2006 12:24:53 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
In-Reply-To: <001701c70032$75ba7ff0$6401a8c0@yourxhtr8hvc4p>
References: <20061103030956.55909.qmail@web38210.mail.mud.yahoo.com>
	 <001701c70032$75ba7ff0$6401a8c0@yourxhtr8hvc4p>
Message-ID: <ee9e417a0611040924h7c62e2a6hc2ba0ec199c7a175@mail.gmail.com>

I have confirmed the values so far
and also computed that a(12) = 1243722370.
Submitted (with a(11)) to OEIS via form.

Russ





From jvospost3 at gmail.com  Sat Nov  4 19:33:14 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 4 Nov 2006 10:33:14 -0800
Subject: primes in arithmetic progression
In-Reply-To: <200611041720.kA4HK70T027399@amer.strw.leidenuniv.nl>
References: <200611041720.kA4HK70T027399@amer.strw.leidenuniv.nl>
Message-ID: <5542af940611041033s7a722e9fjf6ad6643f988477d@mail.gmail.com>

The semiprime analogue of this is a table that shows the smallest semiprime
S
of k (not necessarily consecutive) primes in arithmetic progression
with common difference d (once a 0 appears, the row has ended nonzero
values):

d   k+1  k=2  k=3  k=4  k=5  k=6  k=7 k=8
0   4      4       4      4       4       4       4 ...
1   4      9       33    0
2   4      4       91    213   1383 8129 3091 0
3   4      6       115
4   4      6       6      111
5   4      4       77
6   4      4
7   4      14    51
8   4      6      6       69

-- Fairly easy to extend, fairly easy to prove the first 0 in a row.

The same can be done for 3-almost primes, 4-almost primes, and so forth,
this making a 3-dimensional array.



On 11/4/06, Richard Mathar <mathar at strw.leidenuniv.nl> wrote:
>
>
> Is there an OEIS table that shows the smallest prime p
> of k (not necessarily consecutive) primes in arithmetic progression
> with common difference d? The table would look similar to this one
> below, and contain rather large primes where I am leaving blanks:
>
>     k=1 k=2  k=3  k=4  k=5  k=6
> d
> 0     2   2    2    2    2    2
> 2     2   3    3
> 4     2   3    3
> 6     2   5    5    5    5
> 8     2   3    3
> 10    2   3    3
> 12    2   5    5    5    5
> 14    2   3    3
> 16    2   3
> 18    2   5    5    5
> 20    2   3
> 22    2   7
> 24    2   5    5   59
> 26    2   3
> 28    2   3    3
> 30    2   7    7    7     7    7
> 32    2   5
> 34    2   3    3
> 36    2   5    7   31
>
> The row d=0 and the column k=1 are degenerate and filled with the
> prime 2. All strides d are even. Example for row d=24 and column k=4:
> The 4 numbers 59,59+24,59+2*24 and 59+3*24 are all primes.
>
> http://www.research.att.com/~njas/sequences/Sindx_Pri.html#primes_AP
>
> --Richard
>
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From jvospost3 at gmail.com  Sat Nov  4 19:37:19 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 4 Nov 2006 10:37:19 -0800
Subject: primes in arithmetic progression
In-Reply-To: <5542af940611041033s7a722e9fjf6ad6643f988477d@mail.gmail.com>
References: <200611041720.kA4HK70T027399@amer.strw.leidenuniv.nl>
	 <5542af940611041033s7a722e9fjf6ad6643f988477d@mail.gmail.com>
Message-ID: <5542af940611041037s636536abnf98a7eee7feb349e@mail.gmail.com>

Sorry, I mean: "The semiprime analogue of this is a table that shows the
smallest semiprime S of k (not necessarily consecutive) SEMIprimes in
arithmetic progression
 with common difference d."

On 11/4/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> The semiprime analogue of this is a table that shows the smallest
> semiprime S
> of k (not necessarily consecutive) primes in arithmetic progression
> with common difference d (once a 0 appears, the row has ended nonzero
> values):
>
> d   k+1  k=2  k=3  k=4  k=5  k=6  k=7 k=8
> 0   4      4       4      4       4       4       4 ...
> 1   4      9       33    0
> 2   4      4       91    213   1383 8129 3091 0
> 3   4      6       115
> 4   4      6       6      111
> 5   4      4       77
> 6   4      4
> 7   4      14    51
> 8   4      6      6       69
>
> -- Fairly easy to extend, fairly easy to prove the first 0 in a row.
>
> The same can be done for 3-almost primes, 4-almost primes, and so forth,
> this making a 3-dimensional array.
>
>
>
> On 11/4/06, Richard Mathar <mathar at strw.leidenuniv.nl> wrote:
> >
> >
> > Is there an OEIS table that shows the smallest prime p
> > of k (not necessarily consecutive) primes in arithmetic progression
> > with common difference d? The table would look similar to this one
> > below, and contain rather large primes where I am leaving blanks:
> >
> >     k=1 k=2  k=3  k=4  k=5  k=6
> > d
> > 0     2   2    2    2    2    2
> > 2     2   3    3
> > 4     2   3    3
> > 6     2   5    5    5    5
> > 8     2   3    3
> > 10    2   3    3
> > 12    2   5    5    5    5
> > 14    2   3    3
> > 16    2   3
> > 18    2   5    5    5
> > 20    2   3
> > 22    2   7
> > 24    2   5    5   59
> > 26    2   3
> > 28    2   3    3
> > 30    2   7    7    7     7    7
> > 32    2   5
> > 34    2   3    3
> > 36    2   5    7   31
> >
> > The row d=0 and the column k=1 are degenerate and filled with the
> > prime 2. All strides d are even. Example for row d=24 and column k=4:
> > The 4 numbers 59,59+24,59+2*24 and 59+3*24 are all primes.
> >
> > http://www.research.att.com/~njas/sequences/Sindx_Pri.html#primes_AP<http://www.research.att.com/%7Enjas/sequences/Sindx_Pri.html#primes_AP>
> >
> > --Richard
> >
>
>
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From njas at research.att.com  Sat Nov  4 20:10:10 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sat, 4 Nov 2006 14:10:10 -0500 (EST)
Subject: primes in arithmetic progression
Message-ID: <200611041910.OAA62978@fry.research.att.com>

Just to get the ball rolling, I am adding this entry:

%I A124064
%S A124064 2,2,2,2,3,2,2,3,3,2,2,5,3
%N A124064 Array read by antidiagonals: T(d,k) (k >= 1, d = 2,4,6,8,...) = smallest prime p of k (not necessarily consecutive) primes in arithmetic progression with common difference d.
%C A124064 The row d=0 and the column k=1 are degenerate and are filled with the prime 2.
%O A124064 1,1
%K A124064 nonn,tabl,more
%e A124064 Array begins:
%e A124064 d.\...k=1.k=2..k=3..k=4..k=5..k=6
%e A124064 0..|..2...2....2....2....2....2
%e A124064 2..|..2...3....3
%e A124064 4..|..2...3....3
%e A124064 6..|..2...5....5....5....5
%e A124064 8..|..2...3....3
%e A124064 10.|..2...3....3
%e A124064 12.|..2...5....5....5....5
%e A124064 14.|..2...3....3.
%e A124064 16.|..2...3
%e A124064 18.|..2...5....5....5
%e A124064 20.|..2...3
%e A124064 22.|..2...7
%e A124064 24.|..2...5....5...59
%e A124064 26.|..2...3
%e A124064 28.|..2...3....3
%e A124064 30.|..2...7....7....7.....7....7
%e A124064 32.|..2...5
%e A124064 34.|..2...3....3
%e A124064 36.|..2...5....7...31
%e A124064 Example for row d=24 and column k=4: the 4 numbers 59,59+24,59+2*24 and 59+3*24 are all primes.
%A A124064 Richard Mathar (mathar(AT)strw.leidenuniv.nl), Nov 04 2006




NJAS





From davidwwilson at comcast.net  Sat Nov  4 21:10:45 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sat, 4 Nov 2006 15:10:45 -0500
Subject: Observation on A053669
Message-ID: <000601c7004d$507dc510$6401a8c0@yourxhtr8hvc4p>

It seems to me that

%N A053669 Smallest prime co-prime to n.

could just as easily be described

%N A053669 Smallest number > 1 coprime to n.

and the fact that a(n) is prime falls out.  This definition would also simplify the program.
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From mathar at strw.leidenuniv.nl  Sat Nov  4 21:45:17 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Sat, 4 Nov 2006 21:45:17 +0100
Subject: Observation on A053669
Message-ID: <200611042045.kA4KjHeV011396@amer.strw.leidenuniv.nl>


dww> From seqfan-owner at ext.jussieu.fr  Sat Nov  4 21:12:32 2006
dww> Return-Path: <seqfan-owner at ext.jussieu.fr>
dww> From: "David Wilson" <davidwwilson at comcast.net>
dww> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
dww> Subject: Observation on A053669
dww> ...
dww> It seems to me that
dww> 
dww> %N A053669 Smallest prime co-prime to n.
dww> 
dww> could just as easily be described
dww> 
dww> %N A053669 Smallest number > 1 coprime to n.
dww> 
dww> and the fact that a(n) is prime falls out.  This definition would also =
dww> simplify the program.
dww>...

Yes. If the smallest prime co-prime to n would not be a prime,
it would be built by at least two prime factors, the smaller of which
would be the correct entry (proof by reductio ad absurdum).

It would still be useful to keep in a comment that the numbers are all prime
(this might speed up some searches for the numbers...)

I'd propose to list in a comment the first occurrences of the first primes:
first 5 at a(6)
first 7 is a(30)
first 11 is a(210)
first 13 is a(2310)
first 17 is a(30030)
first 19 is a(510510)
where the indices are 6=2*3, 30=2*3*5, 210=2*3*5*7, 2*3*5*7*11 etc,
(the primorials, I guess, as in A002110)

-- RJM





From jvospost3 at gmail.com  Sat Nov  4 22:55:31 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 4 Nov 2006 13:55:31 -0800
Subject: primes in arithmetic progression
In-Reply-To: <200611041910.OAA62978@fry.research.att.com>
References: <200611041910.OAA62978@fry.research.att.com>
Message-ID: <5542af940611041355vd65aa5cr3a4a9df3cc348c15@mail.gmail.com>

The analogue for 3-almost primes (or, as njas prefers, numbers which are the
product of exactly 3 primes) is left as an exercise for the reader...  I've
just submitted the following.

Subject: PRE-NUMBERED NEW SEQUENCE A124570 FROM Jonathan Vos Post


%I A124570
%S A124570 4, 4, 4, 4, 9, 4, 4, 4, 33, 4, 4, 6, 91, 0, 4, 4, 6, 115,
213, 0, 4, 4, 4, 6, 0, 1383, 0, 4, 4, 4, 77, 111, 0, 8129, 0, 4, 4, 14
%N A124570 Array read by antidiagonals: T(d,k) (k >= 1, d =
1,2,3,4,5,6,...) = smallest semiprime s of k (not necessarily consecutive)
semiprimes in arithmetic progression with common difference d.
%C A124570 Semiprime analogue of A124064 Array read by antidiagonals:
T(d,k) (k >= 1, d = 2,4,6,8,...) = smallest prime p of k (not
necessarily consecutive) primes in arithmetic progression with common
difference d. The row d=0 and the column k=1 are degenerate and are filled with
the semiprime 4. Once a 0 occurs in a row, there are no more nonzero
vales in the row. For example, There cannot be 4 semiprimes in arithmetic
progression with common difference 3, starting with k, because modulo 4
we have {k, k+3, k+6, k+9} == {k+0, k+3, k+2, k+1} and one of these
must be divisible by 4, hence a nonsemiprime (eliminating k = 4 by
inspection).
%e A124570 Array begins:
d.\...k=1.k=2..k=3..k=4..k=5..k=6..k=7..k=8
0..|..4...4....4....4....4....4....4....4
1..|..4...9....33...0....0....0....0....0
2..|..4...4....91...213..1383.8129.3091.0
3..|..4...6....115..0....0....0....0....0
4..|..4...6....6....6....111...
5..|..4...4....77...
6..|..4...4....
7..|..4...14...51...
8..|..4...6....6....69.
Example for row 3: 115 = 5 * 23 is semiprime, 115+3 = 118 = 2 * 59 is
semiprime, and 115+3+3 = 121 = 11^2 is semiprime, so T(3,3) = 115.
%Y A124570 Cf. A000040, A001358, A056809, A070552, A092125, A092126,
A092127, A092128, A092129, A124064.
%O A124570 1,1
%K A124570 ,easy,more,nonn,tabl,
%A A124570 Jonathan Vos Post (jvospost2 at yahoo.com
<http://us.f551.mail.yahoo.com/ym/Compose?To=jvospost2 at yahoo.com&YY=99063&y5beta=yes&y5beta=yes&order=down&sort=date&pos=0&view=a&head=b>),
Nov 04 2006
RH
RA 192.20.225.32
RU
RI



On 11/4/06, N. J. A. Sloane <njas at research.att.com> wrote:
>
> Just to get the ball rolling, I am adding this entry:
>
> %I A124064
> %S A124064 2,2,2,2,3,2,2,3,3,2,2,5,3
> %N A124064 Array read by antidiagonals: T(d,k) (k >= 1, d = 2,4,6,8,...) =
> smallest prime p of k (not necessarily consecutive) primes in arithmetic
> progression with common difference d.
> %C A124064 The row d=0 and the column k=1 are degenerate and are filled
> with the prime 2.
> %O A124064 1,1
> %K A124064 nonn,tabl,more
> %e A124064 Array begins:
> %e A124064 d.\...k=1.k=2..k=3..k=4..k=5..k=6
> %e A124064 0..|..2...2....2....2....2....2
> %e A124064 2..|..2...3....3
> %e A124064 4..|..2...3....3
> %e A124064 6..|..2...5....5....5....5
> %e A124064 8..|..2...3....3
> %e A124064 10.|..2...3....3
> %e A124064 12.|..2...5....5....5....5
> %e A124064 14.|..2...3....3.
> %e A124064 16.|..2...3
> %e A124064 18.|..2...5....5....5
> %e A124064 20.|..2...3
> %e A124064 22.|..2...7
> %e A124064 24.|..2...5....5...59
> %e A124064 26.|..2...3
> %e A124064 28.|..2...3....3
> %e A124064 30.|..2...7....7....7.....7....7
> %e A124064 32.|..2...5
> %e A124064 34.|..2...3....3
> %e A124064 36.|..2...5....7...31
> %e A124064 Example for row d=24 and column k=4: the 4 numbers
> 59,59+24,59+2*24 and 59+3*24 are all primes.
> %A A124064 Richard Mathar (mathar(AT)strw.leidenuniv.nl), Nov 04 2006
>
>
>
>
> NJAS
>
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From jrbibers at indiana.edu  Sat Nov  4 23:17:45 2006
From: jrbibers at indiana.edu (Joseph Biberstine)
Date: Sat, 04 Nov 2006 17:17:45 -0500
Subject: A124015 thoughts
In-Reply-To: <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
Message-ID: <454D1189.8090102@indiana.edu>

	For anyone interested, this Mathematica function, eng[], will (as
usual, barring bugs) give the American English name for integers on 0 to
10^21-1.  Extension of the upper bound should be plain to implement
(just one line of code -- observe the trend).

	All printing is lowercase and without hyphens.  In holding that "and"
should only ever be read for radix points, it is never printed here.
Note this implementation uses a text postprocessor and so is inherently
ugly.  Also note that the helper memoizes for better or worse.  Please,
of course, report all bugs.

	Attached are some diagrams of the predictable distribution of
StringLength[eng[n]] (note this includes spaces).  The third ranges up
to 10^9, sampling every 10^6.

	Here is sample output from random naturals on 0,999999:

 40573, forty thousand five hundred seventy three
108097, one hundred eight thousand ninety seven
539799, five hundred thirty nine thousand seven hundred ninety nine
136868, one hundred thirty six thousand eight hundred sixty eight
597841, five hundred ninety seven thousand eight hundred forty one
272460, two hundred seventy two thousand four hundred sixty
135550, one hundred thirty five thousand five hundred fifty
402424, four hundred two thousand four hundred twenty four
543617, five hundred forty three thousand six hundred seventeen
868945, eight hundred sixty eight thousand nine hundred forty five
951779, nine hundred fifty one thousand seven hundred seventy nine
742845, seven hundred forty two thousand eight hundred forty five
 84140, eighty four thousand one hundred forty
680145, six hundred eighty thousand one hundred forty five
 82734, eighty two thousand seven hundred thirty four
154758, one hundred fifty four thousand seven hundred fifty eight
961084, nine hundred sixty one thousand eighty four
360944, three hundred sixty thousand nine hundred forty four
574416, five hundred seventy four thousand four hundred sixteen
164099, one hundred sixty four thousand ninety nine

	Here is the code:

Clear[eng, engRaw, n];
eng[n_] := (
  pre = engRaw[n];
  If[pre == "", pre = "zero"]; (* fixes "zero" *)
  pre = FixedPoint[StringReplace[#, "  " -> " "] &, pre] ; (* no
consecutive spaces *)
  pre = FixedPoint[StringReplace[#, RegularExpression[" $"] :> ""] &,
pre] ; (* no trailing spaces *)
  pre);
(* This raw function relies heavily on the postprocessor, which fixes
"zero" and deals with whitespace *)
engRaw[n_] := engRaw[n] = (* memo *)
  Which[
    n == 0, "",(* fixed in postprocessing *)
    n == 1, "one",
    n == 2, "two",
    n == 3, "three",
    n == 4, "four",
    n == 5, "five",
    n == 6, "six",
    n == 7, "seven",
    n == 8, "eight",
    n == 9, "nine",
    n == 10, "ten",
    n == 11, "eleven",
    n == 12, "twelve",
    n == 13, "thirteen",
    n == 14, "fourteen",
    n == 15, "fifteen",
    n == 16, "sixteen",
    n == 17, "seventeen",
    n == 18, "eighteen",
    n == 19, "nineteen",
    20 <= n <= 29, "twenty " <> engRaw[Mod[n, 10]],
    30 <= n <= 39, "thirty " <> engRaw[Mod[n, 10]],
    40 <= n <= 49, "forty " <> engRaw[Mod[n, 10]],
    50 <= n <= 59, "fifty " <> engRaw[Mod[n, 10]],
    60 <= n <= 69, "sixty " <> engRaw[Mod[n, 10]],
    70 <= n <= 79, "seventy " <> engRaw[Mod[n, 10]],
    80 <= n <= 89, "eighty " <> engRaw[Mod[n, 10]],
    90 <= n <= 99, "ninety " <> engRaw[Mod[n, 10]],
    10^2 <= n <= 10^3 - 1, engRaw[Quotient[n, 10^2]] <> " hundred " <>
engRaw[Mod[n, 10^2]],
    10^3 <= n <= 10^6 - 1, engRaw[Quotient[n, 10^3]] <> " thousand " <>
engRaw[Mod[n, 10^3]],
    10^6 <= n <= 10^9 - 1, engRaw[Quotient[n, 10^6]] <> " million " <>
engRaw[Mod[n, 10^6]],
    10^9 <= n <= 10^12 - 1, engRaw[Quotient[n, 10^9]] <> " billion " <>
engRaw[Mod[n, 10^9]],
    10^12 <= n <= 10^15 - 1, engRaw[Quotient[n, 10^12]] <> " trillion "
<> engRaw[Mod[n, 10^12]],
    10^15 <= n <= 10^18 - 1, engRaw[Quotient[n, 10^15]] <> " quadrillion
" <> engRaw[Mod[n, 10^15]],
    10^18 <= n <= 10^21 - 1, engRaw[Quotient[n, 10^18]] <> " quintillion
" <> engRaw[Mod[n, 10^18]],
    True, Return["Naturals beyond 10^21-1 are not supported"];];

-JRB

Jonathan Post wrote:
> Now that I think of it, logarithmic trends can be interesting.  How
> about "what is the distribution of the number of letters in the names of
> numbers from one to one billion?"
> 
> -- Jonathan Vos Post
> 
> On 11/3/06, *David Wilson* <davidwwilson at comcast.net
> <mailto:davidwwilson at comcast.net>> wrote:
> 
>     The description of A124015 is
>      
>     %N A124015 Number of words with n letters in the National Scrabble
>     Association Dictionary.
>      
>     Some notes:
>      
>     - It should be noted that the NSAD includes a restricted set of
>     English words (words of 2 to 14 letters, no proper nouns or
>     derivatives, no words with non-alphabetic characters (e.g,
>     contractions, hyphenated words), and it is this restricted set that
>     is being counted.
>      
>     - The edition of the NSAD used to create A124015 should be
>     specified, since the NSAD is continually edited.  While I do not
>     think that A124015 should change with each new edition of the NSAD,
>     because it may be referenced in other literature. On the other hand,
>     I don't think that a new sequence should be created for each new
>     edition of the NSAD unless the new edition exhibits some interesting
>     statistical departure from the current NSAD (e.g, in some future
>     NSAD, the number of 9-letter words exceeds the number of 8-letter
>     words). This is because I believe that the value of A124015 is
>     mainly to indicate a distribution of English word lengths.
>      
>     - On the other hand, I have always understood that 4-letter words
>     are most common in literature (where small words appear more often).
>     It might be interesting to create a sequence counting words of
>     length n in the King James Bible, War and Peace, or some other
>     suitably large and stable piece of English literature (we don't have
>     to go overboard on this, just one or two examples indicating word
>     length distributions in literature).
>      
>     - Another interesting idea: In various languages, what is the
>     distribution of the number of letters in the names of numbers from
>     one to (say) one million?
>      
>      
> 
> 
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From jvospost3 at gmail.com  Sun Nov  5 02:05:28 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 4 Nov 2006 17:05:28 -0800
Subject: A124015 thoughts
In-Reply-To: <454D1189.8090102@indiana.edu>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p>
	 <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
	 <454D1189.8090102@indiana.edu>
Message-ID: <5542af940611041705u66643b27u66c836a95a2b9a1@mail.gmail.com>

Nice code.  And what fascinating graphs! They have both periodicities and a
fractal appearance.  Somehwere in there is the equivalent of Benford's Law,
and Zipf's Law, on lengths names of numbers (as opposed to digit
distribution and power law).

On 11/4/06, Joseph Biberstine <jrbibers at indiana.edu> wrote:
>
>         For anyone interested, this Mathematica function, eng[], will (as
> usual, barring bugs) give the American English name for integers on 0 to
> 10^21-1.  Extension of the upper bound should be plain to implement
> (just one line of code -- observe the trend).
>
>         All printing is lowercase and without hyphens.  In holding that
> "and"
> should only ever be read for radix points, it is never printed here.
> Note this implementation uses a text postprocessor and so is inherently
> ugly.  Also note that the helper memoizes for better or worse.  Please,
> of course, report all bugs.
>
>         Attached are some diagrams of the predictable distribution of
> StringLength[eng[n]] (note this includes spaces).  The third ranges up
> to 10^9, sampling every 10^6.
>
>         Here is sample output from random naturals on 0,999999:
>
> 40573, forty thousand five hundred seventy three
> 108097, one hundred eight thousand ninety seven
> 539799, five hundred thirty nine thousand seven hundred ninety nine
> 136868, one hundred thirty six thousand eight hundred sixty eight
> 597841, five hundred ninety seven thousand eight hundred forty one
> 272460, two hundred seventy two thousand four hundred sixty
> 135550, one hundred thirty five thousand five hundred fifty
> 402424, four hundred two thousand four hundred twenty four
> 543617, five hundred forty three thousand six hundred seventeen
> 868945, eight hundred sixty eight thousand nine hundred forty five
> 951779, nine hundred fifty one thousand seven hundred seventy nine
> 742845, seven hundred forty two thousand eight hundred forty five
> 84140, eighty four thousand one hundred forty
> 680145, six hundred eighty thousand one hundred forty five
> 82734, eighty two thousand seven hundred thirty four
> 154758, one hundred fifty four thousand seven hundred fifty eight
> 961084, nine hundred sixty one thousand eighty four
> 360944, three hundred sixty thousand nine hundred forty four
> 574416, five hundred seventy four thousand four hundred sixteen
> 164099, one hundred sixty four thousand ninety nine
>
>         Here is the code:
>
> Clear[eng, engRaw, n];
> eng[n_] := (
>   pre = engRaw[n];
>   If[pre == "", pre = "zero"]; (* fixes "zero" *)
>   pre = FixedPoint[StringReplace[#, "  " -> " "] &, pre] ; (* no
> consecutive spaces *)
>   pre = FixedPoint[StringReplace[#, RegularExpression[" $"] :> ""] &,
> pre] ; (* no trailing spaces *)
>   pre);
> (* This raw function relies heavily on the postprocessor, which fixes
> "zero" and deals with whitespace *)
> engRaw[n_] := engRaw[n] = (* memo *)
>   Which[
>     n == 0, "",(* fixed in postprocessing *)
>     n == 1, "one",
>     n == 2, "two",
>     n == 3, "three",
>     n == 4, "four",
>     n == 5, "five",
>     n == 6, "six",
>     n == 7, "seven",
>     n == 8, "eight",
>     n == 9, "nine",
>     n == 10, "ten",
>     n == 11, "eleven",
>     n == 12, "twelve",
>     n == 13, "thirteen",
>     n == 14, "fourteen",
>     n == 15, "fifteen",
>     n == 16, "sixteen",
>     n == 17, "seventeen",
>     n == 18, "eighteen",
>     n == 19, "nineteen",
>     20 <= n <= 29, "twenty " <> engRaw[Mod[n, 10]],
>     30 <= n <= 39, "thirty " <> engRaw[Mod[n, 10]],
>     40 <= n <= 49, "forty " <> engRaw[Mod[n, 10]],
>     50 <= n <= 59, "fifty " <> engRaw[Mod[n, 10]],
>     60 <= n <= 69, "sixty " <> engRaw[Mod[n, 10]],
>     70 <= n <= 79, "seventy " <> engRaw[Mod[n, 10]],
>     80 <= n <= 89, "eighty " <> engRaw[Mod[n, 10]],
>     90 <= n <= 99, "ninety " <> engRaw[Mod[n, 10]],
>     10^2 <= n <= 10^3 - 1, engRaw[Quotient[n, 10^2]] <> " hundred " <>
> engRaw[Mod[n, 10^2]],
>     10^3 <= n <= 10^6 - 1, engRaw[Quotient[n, 10^3]] <> " thousand " <>
> engRaw[Mod[n, 10^3]],
>     10^6 <= n <= 10^9 - 1, engRaw[Quotient[n, 10^6]] <> " million " <>
> engRaw[Mod[n, 10^6]],
>     10^9 <= n <= 10^12 - 1, engRaw[Quotient[n, 10^9]] <> " billion " <>
> engRaw[Mod[n, 10^9]],
>     10^12 <= n <= 10^15 - 1, engRaw[Quotient[n, 10^12]] <> " trillion "
> <> engRaw[Mod[n, 10^12]],
>     10^15 <= n <= 10^18 - 1, engRaw[Quotient[n, 10^15]] <> " quadrillion
> " <> engRaw[Mod[n, 10^15]],
>     10^18 <= n <= 10^21 - 1, engRaw[Quotient[n, 10^18]] <> " quintillion
> " <> engRaw[Mod[n, 10^18]],
>     True, Return["Naturals beyond 10^21-1 are not supported"];];
>
> -JRB
>
> Jonathan Post wrote:
> > Now that I think of it, logarithmic trends can be interesting.  How
> > about "what is the distribution of the number of letters in the names of
> > numbers from one to one billion?"
> >
> > -- Jonathan Vos Post
> >
> > On 11/3/06, *David Wilson* <davidwwilson at comcast.net
> > <mailto:davidwwilson at comcast.net>> wrote:
> >
> >     The description of A124015 is
> >
> >     %N A124015 Number of words with n letters in the National Scrabble
> >     Association Dictionary.
> >
> >     Some notes:
> >
> >     - It should be noted that the NSAD includes a restricted set of
> >     English words (words of 2 to 14 letters, no proper nouns or
> >     derivatives, no words with non-alphabetic characters (e.g,
> >     contractions, hyphenated words), and it is this restricted set that
> >     is being counted.
> >
> >     - The edition of the NSAD used to create A124015 should be
> >     specified, since the NSAD is continually edited.  While I do not
> >     think that A124015 should change with each new edition of the NSAD,
> >     because it may be referenced in other literature. On the other hand,
> >     I don't think that a new sequence should be created for each new
> >     edition of the NSAD unless the new edition exhibits some interesting
> >     statistical departure from the current NSAD (e.g, in some future
> >     NSAD, the number of 9-letter words exceeds the number of 8-letter
> >     words). This is because I believe that the value of A124015 is
> >     mainly to indicate a distribution of English word lengths.
> >
> >     - On the other hand, I have always understood that 4-letter words
> >     are most common in literature (where small words appear more often).
> >     It might be interesting to create a sequence counting words of
> >     length n in the King James Bible, War and Peace, or some other
> >     suitably large and stable piece of English literature (we don't have
> >     to go overboard on this, just one or two examples indicating word
> >     length distributions in literature).
> >
> >     - Another interesting idea: In various languages, what is the
> >     distribution of the number of letters in the names of numbers from
> >     one to (say) one million?
> >
> >
> >
> >
>
>
>
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From pxp at rogers.com  Sun Nov  5 02:29:38 2006
From: pxp at rogers.com (Hans Havermann)
Date: Sat, 4 Nov 2006 20:29:38 -0500
Subject: A124015 thoughts
In-Reply-To: <454D1189.8090102@indiana.edu>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com> <454D1189.8090102@indiana.edu>
Message-ID: <3E87B5F0-736C-466D-B4EC-C2D438AA1BA3@rogers.com>

Joseph Biberstine:

> 	For anyone interested, this Mathematica function, eng[], will (as  
> usual, barring bugs) give the American English name for integers on  
> 0 to 10^21-1.  Extension of the upper bound should be plain to  
> implement (just one line of code -- observe the trend).

I forget where I found the following bit of elegant code, but it  
essentially does the same thing as yours (except use a hyphen instead  
of a space between the tens and units for tens > 20):


Ones[n_, pad_:0] := If[Mod[pad, 10] > 0, "-", ""] <> {"", "one",  
"two", "three", "four", "five", "six", "seven", "eight", "nine",  
"ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen",  
"sixteen", "seventeen", "eighteen", "nineteen"}[[n + 1]]

Tens[n_] := {"", "twenty", "thirty", "forty", "fifty", "sixty",  
"seventy", "eighty", "ninety"}[[n]]

LastThree[n_] := LastThree[n] = Module[{h = Quotient[n, 100], m = Mod 
[n, 100]}, If[n > 99, Ones[h, 0] <> " hundred ", ""] <> If[m < 20,  
Ones[m], Tens[Quotient[m, 10]] <> Ones[Mod[m, 10], m]]]

Powers[n_] := Powers[n] = {"", " thousand", " million", " billion", "  
trillion", " quadrillion", " quintillion", " sextillion", "  
septillion", " octillion", "nonillion", " decillion", " undecillion",  
" duodecillion", " tredecillion", " quattuordecillion", "  
quindecillion", " sexdecillion", " septendecillion", "  
octodecillion", " novemdecillion", " vigintillion"}[[n]]

(*NumberName[0] := "zero"*)

NumberName[n_Integer] := Module[{db = Reverse[MapIndexed[If[#1 == 0,  
"", LastThree[#1] <> Powers[#2[[1]]] <> " "] &, FromDigits[#, 10] & / 
@ Reverse /@ Partition[Reverse[IntegerDigits[n, 10]], 3, 3, {1, 1},  
{}]]]}, StringDrop[StringJoin @@ db, -1]]

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From zbi74583 at boat.zero.ad.jp  Sun Nov  5 04:58:03 2006
From: zbi74583 at boat.zero.ad.jp (koh)
Date: Sun, 05 Nov 2006 12:58:03 +0900
Subject: terms lack
Message-ID: <20061105035802.zbi74583@boat.zero.ad.jp>

    Recently I updated more terms of A054572, which are calculated by Richard Mathar.
    And Neil claimed or got angry.
    He said, "You are red card. Don't submit a wrong sequences"
    I am exaggerating a little.
    These words are not exactly what he wrote.



    To Neil
    They are correct, not "wrong". 
    It is only not complete.

    In early time, I didn't do a computer search, so there is a possibility that my old submitted sequences are not complete.
    Now, I calculate both theoretically and with a computer.



    To Seqfans
    I defined many "divisor" functions and calculated many Amicable Numbers and Perfect Numbers using those functions.
    And I submitted them as sequences to  OEIS.

        http://www.research.att.com/~njas/sequences/?q=kohmoto+amicable&sort=0&fmt=0&language=english&go=Search        
        
        http://www.research.att.com/~njas/sequences/?q=kohmoto+perfect&sort=0&fmt=0&language=english&go=Search
   
    I suppose that they are almost complete but some of them might not be complete.

    I wish some one do a computer search about them.
    I think that A054572 is the worst case, because I gave only two terms up to 10^4.
    But sometimes my theory gives a complete set of examples.
    See A099105  , I was sure that these three terms are all, and a computer search also said it is right.

    Yasutoshi
    





From deutsch at duke.poly.edu  Sun Nov  5 05:30:30 2006
From: deutsch at duke.poly.edu (Emeric Deutsch)
Date: Sat, 4 Nov 2006 23:30:30 -0500 (EST)
Subject: No subject
Message-ID: <Pine.GSO.4.62.0611041740160.21855@duke.poly.edu>

Dear Seqfans,
We (Lou Shapiro and myself) would like to submit a few
sequences similar to A124497 (shown at the bottom of this
message). They will refer to 1-2-3-4 trees, 1-2-3-4-5
trees, etc. rather than to 1-2-3 trees. We can describe
the corresponding g.f.'s but we need help in finding the
terms themselves.

Here is the description of all the g.f.'s. Let M[k](z) be
the g.f. of the 1-2-...-k trees with thinning limbs. I will
describe these functions; consequently, we may disregard
the objects counted by them.

Denote by C[k](z) the g.f. of the k-ary trees. C[k]
satisfies C[k]=1+zC[k]^k and the coefficient of z^n
in its series expansion is
(*)      binom(kn,n)/[(k-1)n+1].

I will write Mk for M[k] and Ck for C[k].

Obviously, M1(z)=1/(1-z) (path-trees). The next one is 
M2(z)=M1(z)C2(z^2*M1), i.e. M1 times C2 evaluated at
z^2*M1. This gives no problem since
C2(z)=[1-sqrt(1-4z)]/(2z). We obtain A090344.

Next M3(z)=M2(z)C3(z^3*M2^2); since there is an
explicit expression for C3(z), namely
C3(z)=2/sqrt(3z)*sin[(1/3)*arcsin(sqrt(27z/4))],
this yields easily A124497.

 From here on we need help.
M4(z)=M3(z)C4(z^4*M3^3); 
Most probably sequence starts with
1,1,2,4,10,24,62,160,425,1140,3105 (I used some terms
of the series of C4(z); see (*)).

M5(z)=M4(z)C5(z^5*M4^4);
Most probably sequence starts with
1,1,2,4,10,25,67,180,495,1375,3871 (I used some terms
of the series of C5(z); see (*)).

And so on.

These sequences approach A124344:
1,1,2,4,10,25,68,187,530,1523,4447,13121,39107,...

Many thanks.
Emeric

--------------------------------
A124497  Number of 1-2-3 trees with n edges and with thinning limbs. A 
1-2-3 tree is an ordered tree with vertices of outdegree at most 3. A 
rooted tree with thinning limbs is such that if a node has k children, all 
its children have at most k children.

1, 1, 2, 4, 9, 20, 48, 116, 288, 724, 1849, 4768, 12423, 32628, 86342, 
229952, 616042, 1659012, 4489101, 12199521, 33284546, 91140797, 250396629, 
690043032, 1907022197, 5284167884, 14677681554, 40862469713, 114001697975 
(list; graph; listen)
OFFSET  0,3
FORMULA  G.f.=H*T(H^2*z^3), where 
T=2/sqrt(3*x)*sin((1/3)*arcsin(sqrt(27*x/4))) (solution of T=1+zT^3, 
T(0)=1), H=C(z^2/(1-z))/(1-z), and C(x)=[1-sqrt(1-4x)]/(2x) is the Catalan 
function. More generally, if M[k](z) is the g.f. of the 1-2-...-k trees 
with thinning limbs and C[k](z)=1+z*{C[k](z)}^k is the g.f. of the k-ary 
trees, then M[k](z)=M[k-1](z)C[k](M[k-1]^(k-1)*z^k).
MAPLE  C:=x->(1-sqrt(1-4*x))/2/x: 
T:=x->2/sqrt(3*x)*sin((1/3)*arcsin(sqrt(27*x/4))): M2:=C(z^2/(1-z))/(1-z): 
G:=M2*T(M2^2*z^3): Gser:=series(G, z=0, 40): seq(coeff(Gser, z, n), 
n=0..33);
CROSSREFS  Cf. A090344, A124344.
KEYWORD  nonn,new
AUTHOR  Emeric Deutsch and Louis Shapiro (deutsch(AT)duke.poly.edu, 
lshapiro(AT)Howard.edu), Nov 04 2006





From zakseidov at yahoo.com  Sun Nov  5 07:28:49 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sat, 4 Nov 2006 22:28:49 -0800 (PST)
Subject: Observation on A053669
In-Reply-To: <200611042045.kA4KjHeV011396@amer.strw.leidenuniv.nl>
Message-ID: <20061105062849.51615.qmail@web38208.mail.mud.yahoo.com>

Yes, 
and because a(n_odd)=2
keep only even n's.
Zak

--- Richard Mathar <mathar at strw.leidenuniv.nl> wrote:

> 
> dww> From seqfan-owner at ext.jussieu.fr  Sat Nov  4
> 21:12:32 2006
> dww> Return-Path: <seqfan-owner at ext.jussieu.fr>
> dww> From: "David Wilson" <davidwwilson at comcast.net>
> dww> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
> dww> Subject: Observation on A053669
> dww> ...
> dww> It seems to me that
> dww> 
> dww> %N A053669 Smallest prime co-prime to n.
> dww> 
> dww> could just as easily be described
> dww> 
> dww> %N A053669 Smallest number > 1 coprime to n.
> dww> 
> dww> and the fact that a(n) is prime falls out. 
> This definition would also =
> dww> simplify the program.
> dww>...
> 
> Yes. If the smallest prime co-prime to n would not
> be a prime,
> it would be built by at least two prime factors, the
> smaller of which
> would be the correct entry (proof by reductio ad
> absurdum).
> 
> It would still be useful to keep in a comment that
> the numbers are all prime
> (this might speed up some searches for the
> numbers...)
> 
> I'd propose to list in a comment the first
> occurrences of the first primes:
> first 5 at a(6)
> first 7 is a(30)
> first 11 is a(210)
> first 13 is a(2310)
> first 17 is a(30030)
> first 19 is a(510510)
> where the indices are 6=2*3, 30=2*3*5, 210=2*3*5*7,
> 2*3*5*7*11 etc,
> (the primorials, I guess, as in A002110)
> 
> -- RJM
> 



 
__________________________________________________________________________________________
Check out the New Yahoo! Mail - Fire up a more powerful email and get things done faster. 
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From davidwwilson at comcast.net  Sun Nov  5 07:38:34 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sun, 5 Nov 2006 01:38:34 -0500
Subject: Notes on A124064 (was Re: primes in arithmetic progression)
References: <200611041910.OAA62978@fry.research.att.com>
Message-ID: <003101c700a5$05184930$6401a8c0@yourxhtr8hvc4p>

This is one of those cases where I think we have jumped the gun, creating a 
sequence without first determining the best structure for that sequence.

Richard Mather proposed a table of T(d,k) = the smallest possible initial 
element of a nondecreasing AP of k primes with difference d. Clearly, T(d,k) 
is defined only for d >= 0 and k >= 1.

For d = 0, it is pretty easy to show T(0, k) = 2 for any k >= 1. For d >= 1, 
however, T(d,k) is defined only for a finite number of k.

Proof: Suppose there exists AP P of primes with difference d >= 1 and 2d+2 
elements. Let c = d+1. Because P is an arithmetic progression of 2c elements 
with difference d, and c is coprime to d, P will have two elements divisible 
by c, call them cx and cy. Whereas P is an increasing sequence of positive 
integers, we can assume that 0 < cx < cy. Hence 0 < x < y, and y >= 2. But 
this makes cy composite, contrary to fact that P is a sequence of primes. 
Hence P does not exist, that is to say, any AP of primes with difference d 
 >= 1 has at most 2d+1 elements, and T(d,k) is undefined for k > 2d+1. QED.

In fact, assuming the k-tuple conjecture is true, we can give the precise 
domain of T for d >= 1: If p is the smallest prime not dividing d, then 
T(d,k) is defined precisely for 1 <= k <= p if the AP of p elements with 
initial element p and difference d is a prime AP, and for 1 <= k <= p-1 
otherwise.

This means that the table of T(d,k) looks as follows:

  \k|  1  2  3  4  5  6  7  8  9 10
  d\|
----+------------------------------
  0 |  2  2  2  2  2  2  2  2  2  2 ...
  1 |  2  2
  2 |  2  3  3
  3 |  2  2
  4 |  2  3  3
  5 |  2  2
  6 |  2  5  5  5  5
  7 |  2
  8 |  2  3  3
  9 |  2  2
 10 |  2  3  3
 11 |  2  2
 12 |  2  5  5  5  5
 13 |  2
 14 |  2  3  3
 15 |  2  2
 16 |  2  3
 17 |  2  2
 18 |  2  5  5  5
 19 |  2
 20 |  2  3  3
 ...

Two important points:

- Sometimes in the OEIS, we are tempted to pare down sequences to remove 
uninteresting values in order to emphasize the interesting values. In such 
cases, I think that the OEIS should include a main sequence that is 
complete, and additional pared-down sequences if they seem to augment our 
understanding of the main sequence. In this case, Mather's sequence included 
T(d,k) only for even d, presumably because T(d,k) is uninteresting for odd 
d. My table above includes T(d,k) for both even and odd d. In fact, it turns 
out that for odd d, T(d,1) = 2 and T(d,2) = 2 if 2+d is prime, not 
scintillating behavior, but not trivial either. My feeling is that the main 
sequence for T(d,k) (presumably A124064) should include all the values of 
T(d,k), and if restricting T(d,k) to even d seems to augment our 
understanding, a second pared-down sequence should be made referring back to 
the complete main sequence.

- For d >= 1 and sufficient k, T(d,k) is undefined. The standard 
diagonalization technique won't work for this irregular table. That is why 
NJAS's proposed A124064 ends where it does, because the next value, T(2,4), 
does not exist (it is not a matter of doing more computation, T(2,4) does 
not exist because there is no AP of 4 primes with difference 2). I can see 
two possible ways to address this problem, the (icky) option of making 
T(d,k) = 0 if undefined, or the (less icky) option of requiring the APs in 
question to be strictly increasing. This would dispense with the infinite d 
= 0 line and leave us with a list of finite sequences which we could 
concatenate to build A124064, vis:

2,2,2,3,3,2,2,2,3,3,2,2,2,5,5,5,5,2,2,3,3,2,2,2,3,3,2,2,2,5,5,5,5,2,2,
3,3,2,2,2,3,2,2,2,5,5,5,2,2,3,3,2,2,2,7,2,2,5,5,59,2,2,3,2,2,2,3,3,2,2,
2,7,7,7,7,7,2,2,5,2,2,3,3,2,2,2,5,7,31,2,2,3,3,2,2,2,3,3,2,2,2,5,5,5,5

We could put the line lengths in another sequence.

2,3,2,3,2,5,1,3,2,3,2,5,1,3,2,2,2,4,1,3,2,2,1,4,1,2,2,3,2,6,1,2,1,3,2,4,
1,3,2,3,2,5,1,2,2,2,1,5,1,3,2,2,1,4,1,2,2,2,2,6,1,2,1,3,2,4,1,3,2,2,2,4,
1,2,1,2,2,4,1,3,2,2,1,4,1,2,2,2,1,6,1,2,1,3,2,5,1,3,2,2,2,4,1,3,2,2,2,4

----- Original Message ----- 
From: "N. J. A. Sloane" <njas at research.att.com>
To: <seqfan at ext.jussieu.fr>; "Richard Mathar" <mathar at strw.leidenuniv.nl>
Sent: Saturday, November 04, 2006 2:10 PM
Subject: Re: primes in arithmetic progression


> Just to get the ball rolling, I am adding this entry:
>
> %I A124064
> %S A124064 2,2,2,2,3,2,2,3,3,2,2,5,3
> %N A124064 Array read by antidiagonals: T(d,k) (k >= 1, d = 2,4,6,8,...) = 
> smallest prime p of k (not necessarily consecutive) primes in arithmetic 
> progression with common difference d.
> %C A124064 The row d=0 and the column k=1 are degenerate and are filled 
> with the prime 2.
> %O A124064 1,1
> %K A124064 nonn,tabl,more
> %e A124064 Array begins:
> %e A124064 d.\...k=1.k=2..k=3..k=4..k=5..k=6
> %e A124064 0..|..2...2....2....2....2....2
> %e A124064 2..|..2...3....3
> %e A124064 4..|..2...3....3
> %e A124064 6..|..2...5....5....5....5
> %e A124064 8..|..2...3....3
> %e A124064 10.|..2...3....3
> %e A124064 12.|..2...5....5....5....5
> %e A124064 14.|..2...3....3.
> %e A124064 16.|..2...3
> %e A124064 18.|..2...5....5....5
> %e A124064 20.|..2...3
> %e A124064 22.|..2...7
> %e A124064 24.|..2...5....5...59
> %e A124064 26.|..2...3
> %e A124064 28.|..2...3....3
> %e A124064 30.|..2...7....7....7.....7....7
> %e A124064 32.|..2...5
> %e A124064 34.|..2...3....3
> %e A124064 36.|..2...5....7...31
> %e A124064 Example for row d=24 and column k=4: the 4 numbers 
> 59,59+24,59+2*24 and 59+3*24 are all primes.
> %A A124064 Richard Mathar (mathar(AT)strw.leidenuniv.nl), Nov 04 2006
>
>
>
>
> NJAS
>
>
> -- 
> No virus found in this incoming message.
> Checked by AVG Free Edition.
> Version: 7.5.430 / Virus Database: 268.13.27/517 - Release Date: 11/3/2006 
> 4:30 PM
> 






From davidwwilson at comcast.net  Sun Nov  5 07:40:23 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sun, 5 Nov 2006 01:40:23 -0500
Subject: Observation on A053669
References: <200611042045.kA4KjHeV011396@amer.strw.leidenuniv.nl>
Message-ID: <003b01c700a5$45a9d3b0$6401a8c0@yourxhtr8hvc4p>

Another possible description of A053669:

Smallest prime non-divisor of n.

----- Original Message ----- 
From: "Richard Mathar" <mathar at strw.leidenuniv.nl>
To: <seqfan at ext.jussieu.fr>
Sent: Saturday, November 04, 2006 3:45 PM
Subject: Re: Observation on A053669


>
> dww> From seqfan-owner at ext.jussieu.fr  Sat Nov  4 21:12:32 2006
> dww> Return-Path: <seqfan-owner at ext.jussieu.fr>
> dww> From: "David Wilson" <davidwwilson at comcast.net>
> dww> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
> dww> Subject: Observation on A053669
> dww> ...
> dww> It seems to me that
> dww>
> dww> %N A053669 Smallest prime co-prime to n.
> dww>
> dww> could just as easily be described
> dww>
> dww> %N A053669 Smallest number > 1 coprime to n.
> dww>
> dww> and the fact that a(n) is prime falls out.  This definition would 
> also =
> dww> simplify the program.
> dww>...
>
> Yes. If the smallest prime co-prime to n would not be a prime,
> it would be built by at least two prime factors, the smaller of which
> would be the correct entry (proof by reductio ad absurdum).
>
> It would still be useful to keep in a comment that the numbers are all 
> prime
> (this might speed up some searches for the numbers...)
>
> I'd propose to list in a comment the first occurrences of the first 
> primes:
> first 5 at a(6)
> first 7 is a(30)
> first 11 is a(210)
> first 13 is a(2310)
> first 17 is a(30030)
> first 19 is a(510510)
> where the indices are 6=2*3, 30=2*3*5, 210=2*3*5*7, 2*3*5*7*11 etc,
> (the primorials, I guess, as in A002110)
>
> -- RJM
>
>
> -- 
> No virus found in this incoming message.
> Checked by AVG Free Edition.
> Version: 7.5.430 / Virus Database: 268.13.27/517 - Release Date: 11/3/2006 
> 4:30 PM
>
> 






From pauldhanna at juno.com  Sun Nov  5 07:49:40 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Sun, 5 Nov 2006 01:49:40 -0500
Subject: A124497: 1-2-3-...-k Trees
Message-ID: <20061105.014940.1648.0.pauldhanna@juno.com>

Emeric (and Seqfans),
     What a fascinating problem.
Although I will not be using this formula now, 
let's extend your formula for k-ary trees: 
(*)  [z^n] C[k](z)^m  =  binomial(kn+m-1,n)*m/[(k-1)n+m] 
to give the coefficient of z^n in the m-th power of C[k](z). 
Note that the logarithm is obtained similarly: 
(**) [z^n] log( C[k](z) ) =  binomial(kn-1,n)/[(k-1)n], for n>0.
No doubt you are already aware of this lovely facts; 
and they may simplify some series representations involved here. 
 
I will not attempt to give a simpler g.f. for your M[n], where:
   M[n+1](z) = M[n](z)*C[n](z^n*M[n-1]^(n-1)); 
 
but using PARI I will list the first 31 terms for n=2..10 (below). 
 
You have some impressive results so far. 
     Paul 
 
(PARI)
{M=1/(1-x+O(x^31));
 for(n=2,10,
   M=M*sum(k=0,30,binomial(n*k,k)/((n-1)*k+1)*(x^n*M^(n-1))^k);
   print("n="n":");print(Vec(M))
 )
}

n=2: 
[1, 1, 2, 3, 6, 11, 23, 47, 102, 221, 493, 1105, 2516, 5763, 13328, 
30995, 72556, 170655, 403351, 957135, 2279948, 5449013, 13063596, 
31406517, 75701508, 182902337, 442885683, 1074604289, 2612341856, 
6361782007, 15518343597]
n=3: 
[1, 1, 2, 4, 9, 20, 48, 116, 288, 724, 1849, 4768, 12423, 32628, 
86342, 229952, 616042, 1659012, 4489101, 12199521, 33284546, 
91140797, 250396629, 690043032, 1907022197, 5284167884, 14677681554, 
40862469713, 114001697975, 318681491431, 892495078849]
n=4: 
[1, 1, 2, 4, 10, 24, 62, 160, 425, 1140, 3105, 8528, 23643, 66008, 
185526, 524384, 1489810, 4251852, 12184745, 35048405, 101156752, 
292865417, 850314803, 2475327088, 7223400899, 21126670372, 
61920289652, 181838859665, 534980035009, 1576653291583, 
4654122552827]
n=5: 
[1, 1, 2, 4, 10, 25, 67, 180, 495, 1375, 3871, 10993, 31493, 90843, 
263686, 769466, 2256135, 6643082, 19634705, 58232350, 173242381, 
516860717, 1546035258, 4635543843, 13929569399, 41943013047, 
126532961332, 382396277940, 1157548532794, 3509416800398, 
10655157352027]
n=6: 
[1, 1, 2, 4, 10, 25, 68, 186, 522, 1479, 4246, 12289, 35872, 105411, 
311662, 926270, 2765778, 8292296, 24953437, 75338686, 228140842, 
692733127, 2108652750, 6433255041, 19668210742, 60247367313, 
184879648441, 568281131800, 1749490180723, 5393738767856, 
16651705792875]
n=7: 
[1, 1, 2, 4, 10, 25, 68, 187, 529, 1514, 4393, 12856, 37944, 112740, 
337017, 1012671, 3056978, 9265765, 28187304, 86028764, 263339173, 
808256040, 2486804090, 7668374150, 23694936466, 73354869098, 
227489892754, 706643494680, 2198340096395, 6848600059320, 
21363953916171]
n=8: 
[1, 1, 2, 4, 10, 25, 68, 187, 530, 1522, 4437, 13056, 38766, 115908, 
348733, 1054775, 3205251, 9780165, 29951916, 92029644, 283606809, 
876333336, 2714448570, 8426790446, 26213887250, 81699432762, 
255072076566, 797641572824, 2498068134817, 7834443329344, 
24602515926899]
n=9: 
[1, 1, 2, 4, 10, 25, 68, 187, 530, 1523, 4446, 13110, 39030, 117060, 
353404, 1072811, 3272526, 10024884, 30825490, 95103360, 294300321, 
913203240, 2840650881, 8856206258, 27667828196, 86601978030, 
271545341505, 852829532176, 2682483667081, 8449321174753, 
26648690288909]
n=10: 
[1, 1, 2, 4, 10, 25, 68, 187, 530, 1523, 4447, 13120, 39095, 117400, 
354974, 1079493, 3299426, 10128934, 31216245, 96538200, 299477771, 
931628670, 2905496841, 9082358938, 28450647731, 89294760932, 
280759352295, 884216374006, 2788990466531, 8809540396813, 
27863492984436]

END. 
----------------------------------------------------------
On Sat, 4 Nov 2006 23:30:30 -0500 (EST) Emeric Deutsch
<deutsch at duke.poly.edu> writes:
> Dear Seqfans,
> We (Lou Shapiro and myself) would like to submit a few
> sequences similar to A124497 (shown at the bottom of this
> message). They will refer to 1-2-3-4 trees, 1-2-3-4-5
> trees, etc. rather than to 1-2-3 trees. We can describe
> the corresponding g.f.'s but we need help in finding the
> terms themselves.
> 
> Here is the description of all the g.f.'s. Let M[k](z) be
> the g.f. of the 1-2-...-k trees with thinning limbs. I will
> describe these functions; consequently, we may disregard
> the objects counted by them.
> 
> Denote by C[k](z) the g.f. of the k-ary trees. C[k]
> satisfies C[k]=1+zC[k]^k and the coefficient of z^n
> in its series expansion is
> (*)      binom(kn,n)/[(k-1)n+1].
> 
> I will write Mk for M[k] and Ck for C[k].
> 
> Obviously, M1(z)=1/(1-z) (path-trees). The next one is 
> M2(z)=M1(z)C2(z^2*M1), i.e. M1 times C2 evaluated at
> z^2*M1. This gives no problem since
> C2(z)=[1-sqrt(1-4z)]/(2z). We obtain A090344.
> 
> Next M3(z)=M2(z)C3(z^3*M2^2); since there is an
> explicit expression for C3(z), namely
> C3(z)=2/sqrt(3z)*sin[(1/3)*arcsin(sqrt(27z/4))],
> this yields easily A124497.
> 
> From here on we need help.
> M4(z)=M3(z)C4(z^4*M3^3); 
> Most probably sequence starts with
> 1,1,2,4,10,24,62,160,425,1140,3105 (I used some terms
> of the series of C4(z); see (*)).
> 
> M5(z)=M4(z)C5(z^5*M4^4);
> Most probably sequence starts with
> 1,1,2,4,10,25,67,180,495,1375,3871 (I used some terms
> of the series of C5(z); see (*)).
> 
> And so on.
> 
> These sequences approach A124344:
> 1,1,2,4,10,25,68,187,530,1523,4447,13121,39107,...
> 
> Many thanks.
> Emeric





From pauldhanna at juno.com  Sun Nov  5 08:04:15 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Sun, 5 Nov 2006 02:04:15 -0500
Subject: A124497: 1-2-3-...-k Trees
Message-ID: <20061105.020415.1648.1.pauldhanna@juno.com>

Emeric (and Seqfans),
     Correction to my typo in prior e-mail:
> I will not attempt to give a simpler g.f. for your M[n], where:
>    M[n+1](z) = M[n](z)*C[n](z^n*M[n-1]^(n-1)); 
 
that should have been:
 
M[1](z) = 1/(1-z);
 
M[n+1](z) = M[n](z)*C[n+1](z^(n+1)*M[n]^n); 
 
This was more clearly given by Emeric, but I correct myself for the sake
of accuracy. 
    Paul





From zakseidov at yahoo.com  Sun Nov  5 09:49:35 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 00:49:35 -0800 (PST)
Subject: A124057 is subset of A045940
Message-ID: <20061105084935.7574.qmail@web38212.mail.mud.yahoo.com>

A124057 is subset of A045940
from which more terms for A124057 
can be retrieved:

602,603,1083,2012,2091,2522,2523,2524,2634,2763,3243,3355,4202,4203,4921,4922,4923,5034,5035,5132,5203,5282,5283,5785,5882,5954,5972,6092,6212,6476,6962,6985,7730,7731,7945,8393,8825,8956,8972


%N A045940 Numbers n such that factorizations of n
through n+3 have same number of primes (including 
               multiplicities).

%N A124057 Numbers n such that n, n+1, n+2 and n+3 are
products of exactly 3 primes.



 
____________________________________________________________________________________
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From jvospost3 at gmail.com  Sun Nov  5 17:53:45 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 5 Nov 2006 08:53:45 -0800
Subject: A124057 is subset of A045940
In-Reply-To: <20061105084935.7574.qmail@web38212.mail.mud.yahoo.com>
References: <20061105084935.7574.qmail@web38212.mail.mud.yahoo.com>
Message-ID: <5542af940611050853h50af8ae5ue5004d32f982bef0@mail.gmail.com>

Dear Zak.

You're right.  I knew that; it's in my handwritten notes.  I should have you
submit that as a comment, and also have A045940 added as a cf to A124057,
and also show your extension. Thank you!

Shalom,

Jonathan

On 11/5/06, zak seidov <zakseidov at yahoo.com> wrote:
>
> A124057 is subset of A045940
> from which more terms for A124057
> can be retrieved:
>
>
> 602,603,1083,2012,2091,2522,2523,2524,2634,2763,3243,3355,4202,4203,4921,4922,4923,5034,5035,5132,5203,5282,5283,5785,5882,5954,5972,6092,6212,6476,6962,6985,7730,7731,7945,8393,8825,8956,8972
>
>
> %N A045940 Numbers n such that factorizations of n
> through n+3 have same number of primes (including
>                multiplicities).
>
> %N A124057 Numbers n such that n, n+1, n+2 and n+3 are
> products of exactly 3 primes.
>
>
>
>
>
> ____________________________________________________________________________________
> Access over 1 million songs - Yahoo! Music Unlimited
> (http://music.yahoo.com/unlimited)
>
>
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From franktaw at netscape.net  Sun Nov  5 18:35:13 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Sun, 05 Nov 2006 12:35:13 -0500
Subject: A124015 thoughts
In-Reply-To: <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
Message-ID: <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com>

Some years ago, researchers at Brown University put together a 
collection of text from a large variety of source to serve as a 
cross-section of English usage, for study of this kind of question.  If 
you're going to add a sequence with statistics of this sort, that would 
be the place to start.

See, e.g., Wikipedia: http://en.wikipedia.org/wiki/Brown_Corpus.

Franklin T. Adams-Watters

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From bdm at cs.anu.edu.au  Sun Nov  5 23:40:24 2006
From: bdm at cs.anu.edu.au (Brendan McKay)
Date: Mon, 6 Nov 2006 09:40:24 +1100
Subject: A124015 thoughts
In-Reply-To: <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com> <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com>
Message-ID: <20061105224024.GC28253@cs.anu.edu.au>

* franktaw at netscape.net <franktaw at netscape.net> [061106 04:35]:
> Some years ago, researchers at Brown University put together a 
> collection of text from a large variety of source to serve as a 
> cross-section of English usage, for study of this kind of question.  If 
> you're going to add a sequence with statistics of this sort, that would 
> be the place to start.
> 
> See, e.g., Wikipedia: http://en.wikipedia.org/wiki/Brown_Corpus.
> 
> Franklin T. Adams-Watters

Yes, but I don't know if statistics based on sampling belong in OEIS.
They are not precise well-defined quantities.  Counts derived from
single texts are more concrete, but even then there are multiple
editions, even changes between printings of the same edition.

Brendan.





From zakseidov at yahoo.com  Sun Nov  5 23:53:40 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 14:53:40 -0800 (PST)
Subject: Don't restrict OEIS sequence domains!
In-Reply-To: <001501c70116$e2fabd20$6401a8c0@yourxhtr8hvc4p>
Message-ID: <20061105225340.19195.qmail@web38215.mail.mud.yahoo.com>

David, 

I simply remember Neil's recommendation
(somewhere in his tip's) to omit trivial cases
e.g. if each second term is zero.

And you may wish to 
put in %C A053699 
some of your observations, 
e.g., a(n_odd)=2.
Zak

--- David Wilson <davidwwilson at comcast.net> wrote:

> ----- Original Message ----- 
> From: "zak seidov" <zakseidov at yahoo.com>
> To: <seqfan at ext.jussieu.fr>
> Sent: Sunday, November 05, 2006 1:28 AM
> Subject: Re: Observation on A053669
> 
> 
> > Yes,
> > and because a(n_odd)=2
> > keep only even n's.
> > Zak
> 
> > [Earlier included message elided]
> ---------------------------
> 
> Let's take a look at Zak's recommendation to
> restrict the domain of A053699 
> to even n:
> 
> A053699(n) currently looks like:
> 
>
2,3,2,3,2,5,2,3,2,3,2,5,2,3,2,3,2,5,2,3,2,3,2,5,2,3,2,3,2,5,2,3,2,3,2,7,...
> 
> But a(k) = 2 for odd k, these values are
> uninteresting, so list only a(2n):
> 
>
3,3,5,3,3,5,3,3,5,3,3,5,3,3,7,3,3,5,3,3,5,3,3,5,3,3,5,3,3,7,3,3,5,3,3,5,...
> 
> But now a(k) = 3 unless 6 divides k, so we should
> list only a(6n):
> 
>
5,5,5,5,7,5,5,5,5,7,5,5,5,5,7,5,5,5,5,7,5,5,5,5,7,5,5,5,5,7,5,5,5,5,11,..
> 
> But again, a(k) = 5 except when 30 divides k, so
> let's eject these boring 
> values and keep only a(30n):
> 
>
7,7,7,7,7,7,11,7,7,7,7,7,7,11,7,7,7,7,7,7,11,7,7,7,7,7,7,11,7,7,7,7,7,...
> 
> But look at all the uninteresting 7's. Maybe we
> should list only a(210n)...
> 
> I think you get the picture. In the case of this
> sequence, we repeatedly 
> eliminate uninteresting domain values, but we never
> flush out the elusive 
> "interesting domain" of the sequence. Indeed, we can
> eliminate the entire 
> domain as uninteresting, whereas I think the
> original sequence is indeed 
> quite interesting.
> 
> Clearly, there are times when restricted domains are
> useful. For example, 
> since the areas of Pythagorean triangles are always
> divisible by 6, it would 
> seem reasonable, in the name of compactness, to
> prefer "Number of 
> Pythagorean triangles with area 6n" over "Number of
> Pythagorean triangles of 
> area n", since the latter would have a(n) = 0 for n
> != 6k. Nevertheless, I 
> would advocate having both sequences in the OEIS,
> because we never know if 
> the user will be searching for the compact or
> inflated version of the 
> sequence.
> 
> Similarly, if we adopt Zak's recommendation to
> restrict A065577 to a domain 
> of even values, and then someone looks up
> 
> 2,3,2,3,2,5,2,3,2,3,2,5
> 
> they will probably miss the most straightforward
> explanation of their 
> sequence.
> 
> In short: Don't restrict domains of OEIS sequences
> unless it makes clear 
> sense to do so, and if you do include a sequence on
> some restricted 
> "interesting" domain, have pity on the OEIS users
> and also include the same 
> sequence on its fullest possible domain.
> 
> ----- Original Message ----- 
> From: "zak seidov" <zakseidov at yahoo.com>
> To: <seqfan at ext.jussieu.fr>
> Sent: Sunday, November 05, 2006 1:28 AM
> Subject: Re: Observation on A053669
> 
> 
> > Yes,
> > and because a(n_odd)=2
> > keep only even n's.
> > Zak 
> 
> 



 
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From franktaw at netscape.net  Mon Nov  6 00:26:46 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Sun, 05 Nov 2006 18:26:46 -0500
Subject: A124015 thoughts
In-Reply-To: <20061105224024.GC28253@cs.anu.edu.au>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com> <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com> <20061105224024.GC28253@cs.anu.edu.au>
Message-ID: <8C8CF48FECF25AB-958-F877@FWM-M05.sysops.aol.com>

I'm not saying one way or the other whether this sort of thing belongs
in the OEIS; just that, if it is going to be there, the Brown Corpus
would be the best source for it.

The Brown Corpus is about as concrete as you can get in this area.  It
is very well defined and a standard research reference.

Franklin T. Adams-Watters


-----Original Message-----
From: bdm at cs.anu.edu.au

  * franktaw at netscape.net <franktaw at netscape.net> [061106 04:35]:
> Some years ago, researchers at Brown University put together a
> collection of text from a large variety of source to serve as a
> cross-section of English usage, for study of this kind of question.  
If
> you're going to add a sequence with statistics of this sort, that 
would
> be the place to start.
>
> See, e.g., Wikipedia: http://en.wikipedia.org/wiki/Brown_Corpus.
>
> Franklin T. Adams-Watters

Yes, but I don't know if statistics based on sampling belong in OEIS.
They are not precise well-defined quantities.  Counts derived from
single texts are more concrete, but even then there are multiple
editions, even changes between printings of the same edition.

Brendan.


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From zakseidov at yahoo.com  Mon Nov  6 00:51:18 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 15:51:18 -0800 (PST)
Subject: A124057 is subset of A045940
Message-ID: <20061105235118.90653.qmail@web38203.mail.mud.yahoo.com>

Just submitted:

My Q to seqfans:
Really, there are no numbers n such that n, n+1, n+2
and n+3 are products of 6 primes?
No for n<7258290.
Thanks, Zak


%I A124728
%S A124728
4023,7314,9162,12122,12123,16674,19434,19940,23874,24723,29094,33234,35124,35125,39234,42182,42183,44163,45175,46988,49147,51793,52854,52855,54584,54585,54663,58375,63594,64074,64075,64323,64491,64712,64713,65943,67507,67962,70262,71124,75123,75446,77283,80792,80793,81025,81963,81975,86823
%N A124728 Numbers n such that n, n+1, n+2 and n+3 are
products of 4 primes.
%C A124728 Subset of A045940  Numbers n such that
factorizations of n through n+3 have same number of
primes (including multiplicities).
Cf. A124057, A124729  Numbers n such that n, n+1, n+2
and n+3 are products of exactly 3,5 primes. 
There are no numbers n such that n, n+1, n+2 and n+3
are products of exactly 6 primes(?)
%e A124728 4023=3^3*149, 4024=2^3*503, 4025=5^2*7*23,
4026=2*3*11*61 (all products of 4 primes).
%Y A124728 A045940, A124057, A124729.
%O A124728 1
%K A124728 ,nonn,
%A A124728 Zak Seidov  (zakseidov at yahoo.com), Nov 05
2006


%I A124729
%S A124729
57967,491875,543303,584647,632148,632149,715374,824523,878875,914823,930123,931623,955448,964143,995874
%N A124729 Numbers n such that n, n+1, n+2 and n+3 are
products of 5 primes.
%C A124729 Subset of A045940  Numbers n such that
factorizations of n through n+3 have same number of
primes (including multiplicities).
Cf. A124057, A124728  Numbers n such that n, n+1, n+2
and n+3 are products of exactly 3,4 primes. 
There are no numbers n such that n, n+1, n+2 and n+3
are products of exactly 6 primes(?)
%e A124729 57967=7^3*13^2, 57968=2^4*3623,
57969=3^3*19*113, 57970=2*5*11*17*31 (all product of 5
primes (including multiplicities)};
632148 is the first number such that n through n+4 are
5-almost primes.
%Y A124729 A045940, A124057, A124728.
%O A124729 1
%K A124729 ,nonn,
%A A124729 Zak Seidov  (zakseidov at yahoo.com), Nov 05
2006



 
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From zakseidov at yahoo.com  Mon Nov  6 02:02:27 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 17:02:27 -0800 (PST)
Subject: n..n+6 are all N-almost primes
Message-ID: <20061106010227.58041.qmail@web38206.mail.mud.yahoo.com>

Just dubmitted.

My Q to gurus:
Are there longer set(s) of successive numbers
n..n+m (m>6)
which are products of the same number (N) of primes
(including multiplicities)?

It seems that for  larger N's, 
the larger m's may be(?).

Thanks, Zak
  
%I A000001
%S A000001
211673,298433,355923,381353,460801,506521,540292,568729,690593,705953,737633,741305,921529
%N A000001 Numbers n such that factorizations of
n..n+6 have same number of primes (including
multiplicities).
%C A000001 Subset of A045940  Numbers n such that
factorizations of n..n+3 have same number of primes
(including multiplicities).
%e A000001 211673=7*11*2749, 211674=2*3*35279,
211675=5^2*8467, 211676=2^2*52919, 211677=3*37*1907,
211677=2*109*971, 211679=13*19*857 all 3-almost
primes;
355923=3^2*71*557, 355924=2^2*101*881,
355925=5^22*23*619, 355926=2*3*137*433,
355927=11*13*19*131, 355928=2^3*44491,
355929=3*7*17*997 all 4-almost primes.
%Y A000001 A045940.
%O A000001 1
%K A000001 ,nonn,
%A A000001 Zak Seidov  (zakseidov at yahoo.com), Nov 05
2006



 
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From zakseidov at yahoo.com  Mon Nov  6 03:49:26 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 18:49:26 -0800 (PST)
Subject: n..n+6 are all N-almost primes
In-Reply-To: <20061106010227.58041.qmail@web38206.mail.mud.yahoo.com>
Message-ID: <20061106024926.64870.qmail@web38203.mail.mud.yahoo.com>

--- zak seidov <zakseidov at yahoo.com> wrote:
> Are there longer set(s) of successive numbers
> n..n+m (m>6)
> which are products of the same number (N) of primes
> (including multiplicities)?

Me: Just submitted:
%I A000001
%S A000001
3405122,3405123,6612470,8360103,8520321,9306710
%N A000001 Numbers n such that n..n+7 are products of
the same number of primes.
%C A000001 Note that because 3405130=2*5*167*2039 is
also 4-almost prime,
3405122 is the first n such that numbers n..n+8 are
products of the same number N of primes (N=4).
%e A000001 3405122=2*7*29*8387, 3405123=3^2*19*19913,
3405124=2^2*127*6703, 3405125=5^3*27241,
3405126=2*3*59*9619, 3405127=11*23*43*313,
3405128=2^3*425641, 3405129=3*7*13*12473 all 4-almost
primes.
%O A000001 1
%K A000001 ,nonn,
%A A000001 Zak Seidov  (zakseidov at yahoo.com), Nov 05
2006

Now we need to find numbers n such that n..n+m (m>8)
are N-almost primes (n'd be >10^7).

Thanks, Zak

BTW The "annoying" ad at the bottom of message is 
from Yahoo!Mail - not me, sorry.

<original message>
> Just dubmitted.
> 
> My Q to gurus:
> Are there longer set(s) of successive numbers
> n..n+m (m>6)
> which are products of the same number (N) of primes
> (including multiplicities)?
> 
> It seems that for  larger N's, 
> the larger m's may be(?).
> 
> Thanks, Zak
>   
> %I A000001
> %S A000001
>
211673,298433,355923,381353,460801,506521,540292,568729,690593,705953,737633,741305,921529
> %N A000001 Numbers n such that factorizations of
> n..n+6 have same number of primes (including
> multiplicities).
> %C A000001 Subset of A045940  Numbers n such that
> factorizations of n..n+3 have same number of primes
> (including multiplicities).
> %e A000001 211673=7*11*2749, 211674=2*3*35279,
> 211675=5^2*8467, 211676=2^2*52919, 211677=3*37*1907,
> 211677=2*109*971, 211679=13*19*857 all 3-almost
> primes;
> 355923=3^2*71*557, 355924=2^2*101*881,
> 355925=5^22*23*619, 355926=2*3*137*433,
> 355927=11*13*19*131, 355928=2^3*44491,
> 355929=3*7*17*997 all 4-almost primes.
> %Y A000001 A045940.
> %O A000001 1
> %K A000001 ,nonn,
> %A A000001 Zak Seidov  (zakseidov at yahoo.com), Nov 05
> 2006
> 




 
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From davidwwilson at comcast.net  Mon Nov  6 04:10:17 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sun, 5 Nov 2006 22:10:17 -0500
Subject: A124015 thoughts
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com> <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com> <20061105224024.GC28253@cs.anu.edu.au> <8C8CF48FECF25AB-958-F877@FWM-M05.sysops.aol.com>
Message-ID: <001901c70151$16106360$6401a8c0@yourxhtr8hvc4p>

Sounds great. Where do I get the Brown Corpus in computer-readable form?

----- Original Message ----- 
From: <franktaw at netscape.net>
To: <seqfan at ext.jussieu.fr>
Sent: Sunday, November 05, 2006 6:26 PM
Subject: Re: A124015 thoughts


> I'm not saying one way or the other whether this sort of thing belongs
> in the OEIS; just that, if it is going to be there, the Brown Corpus
> would be the best source for it.
>
> The Brown Corpus is about as concrete as you can get in this area.  It
> is very well defined and a standard research reference.
>
> Franklin T. Adams-Watters
>
>
> -----Original Message-----
> From: bdm at cs.anu.edu.au
>
>  * franktaw at netscape.net <franktaw at netscape.net> [061106 04:35]:
>> Some years ago, researchers at Brown University put together a
>> collection of text from a large variety of source to serve as a
>> cross-section of English usage, for study of this kind of question.
> If
>> you're going to add a sequence with statistics of this sort, that
> would
>> be the place to start.
>>
>> See, e.g., Wikipedia: http://en.wikipedia.org/wiki/Brown_Corpus.
>>
>> Franklin T. Adams-Watters
>
> Yes, but I don't know if statistics based on sampling belong in OEIS.
> They are not precise well-defined quantities.  Counts derived from
> single texts are more concrete, but even then there are multiple
> editions, even changes between printings of the same edition.
>
> Brendan.
>
>
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading 
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>
> -- 
> No virus found in this incoming message.
> Checked by AVG Free Edition.
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> 5:30 PM
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> 






From dean at math.ucdavis.edu  Mon Nov  6 10:39:09 2006
From: dean at math.ucdavis.edu (Dean Hickerson)
Date: Mon, 6 Nov 2006 01:39:09 -0800
Subject: terms lack
Message-ID: <20061106093909.GA16454@math.ucdavis.edu>

Mostly to Yasutoshi Kohmoto <zbi74583 at boat.zero.ad.jp>:

> Recently I updated more terms of A054572, which are calculated by
> Richard Mathar.
> And Neil claimed or got angry.
> He said, "You are red card. Don't submit a wrong sequences"
> I am exaggerating a little.
> These words are not exactly what he wrote.
>
> To Neil
> They are correct, not "wrong".
> It is only not complete.

The terms that are listed in a sequence should be the start of the sequence.
If there are missing terms in between the listed ones, then the sequence
entry is not just incomplete, it's wrong.

If you've computed some terms of a sequence, but you're not sure if there
are smaller ones that you've missed, then you should only include the ones
that you're sure are at the start of the sequence.  You can add a comment
about the larger terms, stating that it's not known if there are smaller ones.
For example, the "EXTENSIONS" lines for the sequence A000043 (Mersenne
exponents) mention some recently discovered terms.  But there might be
some smaller ones, so these are not included in the sequence.

> I defined many "divisor" functions and calculated many Amicable Numbers and
> Perfect Numbers using those functions.
> And I submitted them as sequences to  OEIS.
>
> http://www.research.att.com/~njas/sequences/?q=kohmoto+amicable&sort=0&fmt=0&language=english&go=Search
>
> http://www.research.att.com/~njas/sequences/?q=kohmoto+perfect&sort=0&fmt=0&language=english&go=Search
>
> I suppose that they are almost complete but some of them might not be
> complete.

I've found missing terms for several of your sequences:


Sequences A045613 and A045614 ("Super Unitary Amicable Number") consist
of pairs of distinct numbers a and b such that

    usigma(usigma(a)) = usigma(usigma(b)) = a+b.

(The definitions don't make this clear, but A045613 contains the smaller
members of the pairs and A045614 contains the larger members.  Also they're
sorted by their smallest members, so 155 comes before 142 in A045614.)  The
first 3 pairs that you've listed are (105, 155), (110, 142) and (2145, 3055).
But you've missed several smaller ones:

    (33, 35), (208, 224), (268, 272), (455, 601), (695, 1033), (812, 956),
    (1609, 1847), (1808, 2512), (1892, 3004), (1913, 2407), (2096, 2224)


A051594 and A051595 are "(-1)sigma amicable numbers".  (Again, the definition
doesn't make it clear that A051595 has the smaller terms and A051594 has the
larger terms.)  The first pair that you've listed is (1969706592, 2236072608).
But there are many smaller ones, starting with:

    (429552, 466200), (497808, 604656), (800496, 1103760).


A036471-A036474 are defined as "Amicable quadruple: 4 different numbers
which satisfy sigma(a)=sigma(b)=sigma(c)=sigma(d)=a+b+c+d".  The first
quadruple that you've listed has smallest element
342151462276356306033089201934180, but here's a much smaller one:

    (3270960, 3361680, 3461040, 3834000)

(I think that's the smallest one, but I could be wrong.)  If someone can
determine the first several terms of this sequence, then it should be
edited; otherwise it should be deleted.


In some of your sequences, I wasn't able to find any smaller terms, but the
ones that you've listed are so large that I'm not at all confident that they
are the smallest ones.  For example, A038362 and A038363 are defined as
"A Rational Amicable Number: two different numbers a, b which satisfies the
following equation: sigma(a)=sigma(b)=(a+b)^3/(a^2+b^2)", and the smallest
pair that you've included is:

    (26403469440047700, 30193441130006700)

Do you have any reason to think that that's the smallest such pair?  If not,
and if you can't determine what the first few terms really are, then the
sequence should be deleted.


I've only looked at a few of your sequences, but that's enough to convince
me that they should all be checked.

Neil, I'll submit corrections for A045613, A045614, A051594, and A051595,
but I'm not going to work on the others.  Maybe they should be deleted or
marked as "probation" until the author or someone else fixes them.

Dean Hickerson
dean at math.ucdavis.edu





From bdm at cs.anu.edu.au  Mon Nov  6 12:45:30 2006
From: bdm at cs.anu.edu.au (Brendan McKay)
Date: Mon, 6 Nov 2006 22:45:30 +1100
Subject: terms lack
In-Reply-To: <20061106093909.GA16454@math.ucdavis.edu>
References: <20061106093909.GA16454@math.ucdavis.edu>
Message-ID: <20061106114530.GB4232@cs.anu.edu.au>

* Dean Hickerson <dean at math.ucdavis.edu> [061106 20:40]:
> The terms that are listed in a sequence should be the start of the sequence.
> If there are missing terms in between the listed ones, then the sequence
> entry is not just incomplete, it's wrong.
> 
> If you've computed some terms of a sequence, but you're not sure if there
> are smaller ones that you've missed, then you should only include the ones
> that you're sure are at the start of the sequence.  You can add a comment
> about the larger terms, stating that it's not known if there are smaller ones.
> For example, the "EXTENSIONS" lines for the sequence A000043 (Mersenne
> exponents) mention some recently discovered terms.  But there might be
> some smaller ones, so these are not included in the sequence.
 
Is there an official rule about this?  I'm not sure it is correct.
Of course we should not submit sequences with possible gaps, but
suppose we can compute a(10) thru a(20) but have trouble with
a(1) thru a(9).  I don't see why we can't submit a(10..20) with
an OFFSET of 10 and a comment explaining why a(1..9) are absent.

Brendan.





From Annette.Warlich at t-online.de  Mon Nov  6 13:05:20 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Mon, 06 Nov 2006 13:05:20 +0100
Subject: Fibbinary sequence
Message-ID: <454F2500.2020107@t-online.de>

A courious result by chance:

A003714	Fibbinary numbers:
       if n = F_i1+F_i2+...+F_ik is the Zeckendorf representation of n
       (i.e. write n in Fibonacci number system)
       then
         a(n) = 2^{i1-2}+2^{i2-2}+...+2^{ik-2}.

	0, 1, 2, 4, 5, 8, 9, 10, 16, 17, 18, 20, 21, 32, 33, 34, 36, 37,
      40, 41, 42, 64, 65, 66, 68, 69, 72, 73, 74, 80, 81, 82, 84, 85,
     128, 129, 130, 132, 133, 136, 137, 138, 144, 145, 146, 148, 149,
     160, 161, 162, 164, 165, 168, 169, 170, 256, 257, 258, 260

I was checking some variants of the Pascal-matrix and got the
following
Let Px =
      1       .    .     .     .    .  .  .
      1      -1    .     .     .    .  .  .
    3/2      -3    1     .     .    .  .  .
      4      -9    6    -1     .    .  .  .
     15     -35   30   -10     1    .  .  .
  137/2  -345/2  165   -75    15   -1  .  .
 1491/4 -2009/2 1050  -560 315/2  -21  1  .
   2374   -6720 7588 -4515  1540 -294 28 -1

then the index r of the rows, r=1..inf(??) , which contain only integers, is A003714
      1, 2, 4, 5, 8, 9, 10, 16, 17, 18, 20, 21, 32, 33, 34, 36, 37,
     40, 41, 42, 64,

(although I checked this only up to N=64)


Explanation: -----------------------------------------------------------

The matrix Px is constructed as a variant of Pj
Pj:
 1  .  .   .  .   . .  .
 1 -1  .   .  .   . .  .
 1 -2  1   .  .   . .  .
 1 -3  3  -1  .   . .  .
 1 -4  6  -4  1   . .  .
 1 -5 10 -10  5  -1 .  .
 1 -6 15 -20 15  -6 1  .
 1 -7 21 -35 35 -21 7 -1

which is the column-signed version of the Pascalmatrix P
   Pj = P*J
where J is diagonal unit with alternating signs (diag(1,-1,1,-1,...))

The matrix S2 of the stirling-numbers of 2'nd kind
 1   .   .    .    .   .  . .
 1   1   .    .    .   .  . .
 1   3   1    .    .   .  . .
 1   7   6    1    .   .  . .
 1  15  25   10    1   .  . .
 1  31  90   65   15   1  . .
 1  63 301  350  140  21  1 .
 1 127 966 1701 1050 266 28 1

and S1 of the stirling numbers of first kind (where also S1 = S2^-1)
     1     .      .    .     .   .   . .
    -1     1      .    .     .   .   . .
     2    -3      1    .     .   .   . .
    -6    11     -6    1     .   .   . .
    24   -50     35  -10     1   .   . .
  -120   274   -225   85   -15   1   . .
   720 -1764   1624 -735   175 -21   1 .
 -5040 13068 -13132 6769 -1960 322 -28 1

with the eigenvalues Z(1) = diag(1,1/2,1/3,1/4) such that

   G = S2 * Z(1) * S2^-1

G:
     1     .    .     .   .    .   .   .
   1/2   1/2    .     .   .    .   .   .
   1/6   1/2  1/3     .   .    .   .   .
     0   1/4  1/2   1/4   .    .   .   .
 -1/30     0  1/3   1/2 1/5    .   .   .
     0 -1/12    0  5/12 1/2  1/6   .   .
  1/42     0 -1/6     0 1/2  1/2 1/7   .
     0  1/12    0 -7/24   0 7/12 1/2 1/8

form an eigensystem of G.
G ist interesting as it contains the bernoulli-
numbers, and coefficients of the integrals of the
bernoulli-functions.



Also G is Eigenmatrix of the signed Pascal-matrix Pj,
with the eigenvalues J = diag(1,-1,1,-1,...)

  Pj = G * J * G^-1

and Pj is used in divergent Euler-summation, Hasse's
zeta-summation and elsewhere.




Now I tried a variant of P by varying G, where the eigenvalues
are not Z(1), but are V(1/2) = diag(1,1/2,1/4,1/8,....) for some
analyses about summing zeta series, since Z(1) and V(1/2) are
closely related in this subject and complementary/mutually supplementary
in certain manners.

Now with the G-variant

  Gx  = S2 * V(1/2) * S2^-1
Gx:
     1      .       .        .      .      .    .     .
   1/2    1/2       .        .      .      .    .     .
     0    3/4     1/4        .      .      .    .     .
  -1/4    3/8     3/4      1/8      .      .    .     .
     0   -5/8   15/16      5/8   1/16      .    .     .
   1/2 -15/16  -15/32    45/32  15/32   1/32    .     .
     0    7/4 -105/32    35/64 105/64  21/64 1/64     .
 -17/8 147/32    7/32 -735/128  35/16 105/64 7/32 1/128

I come to

  Px =         Gx               * J *       Gx^-1
     =  ( S2 * V(1/2) * S2^-1 ) * J * (S2 * V(2) * S2^-1)

(while Pj is
  Pj =         G                * J *       G ^-1
     =  ( S2 * Z(-1) * S2^-1 )  * J * (S2 * Z(1) * S2^-1)
  )

and as shown above:

Px:
      1       .    .     .     .    .  .  .
      1      -1    .     .     .    .  .  .
    3/2      -3    1     .     .    .  .  .
      4      -9    6    -1     .    .  .  .
     15     -35   30   -10     1    .  .  .
  137/2  -345/2  165   -75    15   -1  .  .
 1491/4 -2009/2 1050  -560 315/2  -21  1  .
   2374   -6720 7588 -4515  1540 -294 28 -1

and the rows containing only integers, are uniquely
indicated by A003714, up to r=64, where the rownumber r
starts with 1.

Don't have an idea, how this coincidence could be interesting -
I thought I'd just sentd the observation..

Regards -

Gottfried Helms







From Labos at ana.sote.hu  Mon Nov  6 13:38:39 2006
From: Labos at ana.sote.hu (=?ISO-8859-2?Q?L=DFbos_Elem=DAr?=)
Date: Mon, 6 Nov 2006 13:38:39 MET-1MEST
Subject: terms lack
In-Reply-To: <20061106114530.GB4232@cs.anu.edu.au>
References: <20061106093909.GA16454@math.ucdavis.edu>
Message-ID: <6578DDB2A59@ana.sote.hu>

On 6 Nov 06, at 22:45, Brendan McKay wrote:

> * Dean Hickerson <dean at math.ucdavis.edu> [061106 20:40]:
> > The terms that are listed in a sequence should be the start of the sequence.
> > If there are missing terms in between the listed ones, then the sequence
> > entry is not just incomplete, it's wrong.
> > 
> > If you've computed some terms of a sequence, but you're not sure if there
> > are smaller ones that you've missed, then you should only include the ones
> > that you're sure are at the start of the sequence.  You can add a comment
> > about the larger terms, stating that it's not known if there are smaller ones.
> > For example, the "EXTENSIONS" lines for the sequence A000043 (Mersenne
> > exponents) mention some recently discovered terms.  But there might be
> > some smaller ones, so these are not included in the sequence.
>  
> Is there an official rule about this?  I'm not sure it is correct.
> Of course we should not submit sequences with possible gaps, but
> suppose we can compute a(10) thru a(20) but have trouble with
> a(1) thru a(9).  I don't see why we can't submit a(10..20) with
> an OFFSET of 10 and a comment explaining why a(1..9) are absent.
> 
> Brendan.

There are 2 cases :
 1  -- You cannot even know how many terms are missing.
          This is essentally the case of eg Mersenne primes..
-------------------
 2. --  You can specify the missing terms by list-
          position. In this second case you can introduce 
          a marker signal to point to missing position.

Handling of the cases can follow different standard..
The general case can be worthwhile to publish if 
so many hard computation is behind. 
Ordering terms somehow it can be mentioned that
the sequence is lacunar.
With correct description the situation, the case is
neither wrong, nor good. Perhaps unnecessary...

The empty sequence represents the full lacunarity.
Quite correctly. However the empty sequence is
not unique. 

Labos 






From dean at math.ucdavis.edu  Mon Nov  6 14:22:58 2006
From: dean at math.ucdavis.edu (Dean Hickerson)
Date: Mon, 6 Nov 2006 05:22:58 -0800
Subject: terms lack
Message-ID: <20061106132258.GA19536@math.ucdavis.edu>

Mostly to Brendan McKay

> Is there an official rule about this?  I'm not sure it is correct.
> Of course we should not submit sequences with possible gaps, but
> suppose we can compute a(10) thru a(20) but have trouble with
> a(1) thru a(9).  I don't see why we can't submit a(10..20) with
> an OFFSET of 10 and a comment explaining why a(1..9) are absent.

Sure, that's fine.  In that case the sequence that's in the OEIS starts
with a(10), and it doesn't have any gaps.  It's part of a longer, partially
unknown, one that starts with a(1), but the sequence that it's part of
isn't in the OEIS.

But in cases like Kohmoto's sequences that wouldn't work.  For example,
suppose we know from hand calculation (as Kohmoto did for A045613), that
105, 110, and 2145 are in the sequence.  Presumably we can test that
there are no terms between 105 and 110, but we can't tell if there are
terms below 105 or between 110 and 2145.  So (ignoring the 4-term rule for
submissions) we could submit a sequence with just 105 and 110, and add
comments saying that there might be earlier terms and that 2145 is also
in the sequence, but we shouldn't submit the sequence as starting with
105, 110, 2145.

Dean Hickerson
dean at math.ucdavis.edu





From tbaruchel at free.fr  Mon Nov  6 17:31:32 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Mon, 6 Nov 2006 17:31:32 +0100
Subject: A fast (written in assembly) library for continued fractions
In-Reply-To: <20061103221729.GB2467@>
References: <20061103221729.GB2467@>
Message-ID: <20061106163132.GA30587@>

On Ven, nov 03, 2006 at 11:17:29 +0100, Thomas Baruchel wrote:
> I wrote for my own purposes the following X86 assembly code, intended

The final version of my code can be found at
  http://baruchel.free.fr/~thomas/wp/?p=10
I give also (in the comments of the file) advices for using the functions
from gcc, gfortran, g95. The fastest continued fractions in the west...

Next step will be functions for computing the Pell equation, but
pure FPU stack will not be possible this time (only eight registers), thus
some memory access will be required :-(

Regards,

-- 
Thomas Baruchel
  http://baruchel.free.fr/~thomas/wp/





From Eric.Angelini at kntv.be  Mon Nov  6 18:57:00 2006
From: Eric.Angelini at kntv.be (Eric Angelini)
Date: Mon, 6 Nov 2006 18:57:00 +0100
Subject: Langford/Skolem-like sequence : odd + even terms
Message-ID: <F05775148D000C4B9321A5113D53CB7D191001@KNTV-SERVER01.kntv.local>


Hello SeqFans,
Another Langford/Skolem-like sequence, with a mixture of odd and
even terms -- plus two questions:
http://www.cetteadressecomportecinquantesignes.com/OddEvenSkol.htm
Best,
?.






From noe at sspectra.com  Mon Nov  6 19:04:49 2006
From: noe at sspectra.com (T. D. Noe)
Date: Mon, 6 Nov 2006 10:04:49 -0800
Subject: Cyclotomic polynomial for n=0
Message-ID: <p06230985c175281c5eee@[192.168.1.43]>

For n=1,2,3,... the cyclotomic polynomials are

x-1, x+1, x^2+x+1,...

It seems that there are two conventions for n=0:

x or 1

Mathematica says it is 1.  I think Maple uses x.  Can someone explain this?

Thanks,

Tony





From franktaw at netscape.net  Mon Nov  6 22:27:14 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 06 Nov 2006 16:27:14 -0500
Subject: Cyclotomic polynomial for n=0
In-Reply-To: <p06230985c175281c5eee@[192.168.1.43]>
References: <p06230985c175281c5eee@[192.168.1.43]>
Message-ID: <8C8D001766F0D57-BD0-2F58@WEBMAIL-MA14.sysops.aol.com>

Strictly speaking, the cyclotomic polynomial for n=0 is undefned.  The 
two packages have chosen different ways to extend the concept to n=0; 
it would have been perfectly OK for them to have returned some kind of 
error.

There are reasonable arguments for either choice.  A lot comes down to 
the basic property that the degree is phi(n); the question then is, 
what is phi(0)?  Again, this is strictly speaking undefined, but there 
are two reasonable choices: 0 and 1.  phi(0) = 0 gives 1 as the value 
of the 0th cyclotomic polynomial.  If we take phi(0) = 1, then we need 
to choose a root for the polynomial.  Choosing 0 as the root gives the 
nice property that the cyclotomic polynomials for n >= 0 constitute all 
the primitive monic polynomials all of whose roots lie in the unit 
disk; hence the selection of x as the polynomial.

If you take the formula Phi_n(x) = Product_{0<=k<n, gcd(k,n) = 1} e^{2 
k Pi i / n}, you get Phi_0(x) = 1.  On the other hand, if we regard 
this as a product over distinct residues modulo n that are relatively 
prime to n, we get x - e^{2 Pi i / 0}, which is undefined.

You can also make a case for 0 as the "correct" value.  Phi_n(x) is 
x^n-1, divided by all x - u for primitive dth roots of unity u for d a 
proper divisor of n.  x^0-1 is 0; you can divide it by anything you 
want, and the result is still 0.  (Of course, this becomes an infinite 
product in the denominator - but this is only a plausibility argument.)

Franklin T. Adams-Watters


-----Original Message-----
From: noe at sspectra.com

For n=1,2,3,... the cyclotomic polynomials are

x-1, x+1, x^2+x+1,...

It seems that there are two conventions for n=0:

x or 1

Mathematica says it is 1.  I think Maple uses x.  Can someone explain 
this?

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From zbi74583 at boat.zero.ad.jp  Tue Nov  7 02:34:32 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Tue, 7 Nov 2006 10:34:32 +0900
Subject: terms lack
References: <20061106093909.GA16454@math.ucdavis.edu>
Message-ID: <018901c7020c$e2322a30$ef73d7dc@FMC57937B6BE00>

    To Dean Hicherson

    Thanks for editing my sequences.
    And thanks for a good explanation about "completeness".
    Now, I understand that  the  purpose of OEIS is "Identification of 
sequences", it is necessary.

    Should A038362 be deleted?
    Still, I think that it should exist , because a sequence is a mapping 
from N to N, so a subset of the images is also a good information for 
understanding the mapping.

    Yasutoshi


----- Original Message ----- 
From: "Dean Hickerson" <dean at math.ucdavis.edu>
To: <seqfan at ext.jussieu.fr>
Sent: Monday, November 06, 2006 6:39 PM
Subject: Re: terms lack


> Mostly to Yasutoshi Kohmoto <zbi74583 at boat.zero.ad.jp>:
>
>> Recently I updated more terms of A054572, which are calculated by
>> Richard Mathar.
>> And Neil claimed or got angry.
>> He said, "You are red card. Don't submit a wrong sequences"
>> I am exaggerating a little.
>> These words are not exactly what he wrote.
>>
>> To Neil
>> They are correct, not "wrong".
>> It is only not complete.
>
> The terms that are listed in a sequence should be the start of the 
> sequence.
> If there are missing terms in between the listed ones, then the sequence
> entry is not just incomplete, it's wrong.
>
> If you've computed some terms of a sequence, but you're not sure if there
> are smaller ones that you've missed, then you should only include the ones
> that you're sure are at the start of the sequence.  You can add a comment
> about the larger terms, stating that it's not known if there are smaller 
> ones.
> For example, the "EXTENSIONS" lines for the sequence A000043 (Mersenne
> exponents) mention some recently discovered terms.  But there might be
> some smaller ones, so these are not included in the sequence.
>
>> I defined many "divisor" functions and calculated many Amicable Numbers 
>> and
>> Perfect Numbers using those functions.
>> And I submitted them as sequences to  OEIS.
>>
>> http://www.research.att.com/~njas/sequences/?q=kohmoto+amicable&sort=0&fmt=0&language=english&go=Search
>>
>> http://www.research.att.com/~njas/sequences/?q=kohmoto+perfect&sort=0&fmt=0&language=english&go=Search
>>
>> I suppose that they are almost complete but some of them might not be
>> complete.
>
> I've found missing terms for several of your sequences:
>
>
> Sequences A045613 and A045614 ("Super Unitary Amicable Number") consist
> of pairs of distinct numbers a and b such that
>
>    usigma(usigma(a)) = usigma(usigma(b)) = a+b.
>
> (The definitions don't make this clear, but A045613 contains the smaller
> members of the pairs and A045614 contains the larger members.  Also 
> they're
> sorted by their smallest members, so 155 comes before 142 in A045614.) 
> The
> first 3 pairs that you've listed are (105, 155), (110, 142) and (2145, 
> 3055).
> But you've missed several smaller ones:
>
>    (33, 35), (208, 224), (268, 272), (455, 601), (695, 1033), (812, 956),
>    (1609, 1847), (1808, 2512), (1892, 3004), (1913, 2407), (2096, 2224)
>
>
> A051594 and A051595 are "(-1)sigma amicable numbers".  (Again, the 
> definition
> doesn't make it clear that A051595 has the smaller terms and A051594 has 
> the
> larger terms.)  The first pair that you've listed is (1969706592, 
> 2236072608).
> But there are many smaller ones, starting with:
>
>    (429552, 466200), (497808, 604656), (800496, 1103760).
>
>
> A036471-A036474 are defined as "Amicable quadruple: 4 different numbers
> which satisfy sigma(a)=sigma(b)=sigma(c)=sigma(d)=a+b+c+d".  The first
> quadruple that you've listed has smallest element
> 342151462276356306033089201934180, but here's a much smaller one:
>
>    (3270960, 3361680, 3461040, 3834000)
>
> (I think that's the smallest one, but I could be wrong.)  If someone can
> determine the first several terms of this sequence, then it should be
> edited; otherwise it should be deleted.
>
>
> In some of your sequences, I wasn't able to find any smaller terms, but 
> the
> ones that you've listed are so large that I'm not at all confident that 
> they
> are the smallest ones.  For example, A038362 and A038363 are defined as
> "A Rational Amicable Number: two different numbers a, b which satisfies 
> the
> following equation: sigma(a)=sigma(b)=(a+b)^3/(a^2+b^2)", and the smallest
> pair that you've included is:
>
>    (26403469440047700, 30193441130006700)
>
> Do you have any reason to think that that's the smallest such pair?  If 
> not,
> and if you can't determine what the first few terms really are, then the
> sequence should be deleted.
>
>
> I've only looked at a few of your sequences, but that's enough to convince
> me that they should all be checked.
>
> Neil, I'll submit corrections for A045613, A045614, A051594, and A051595,
> but I'm not going to work on the others.  Maybe they should be deleted or
> marked as "probation" until the author or someone else fixes them.
>
> Dean Hickerson
> dean at math.ucdavis.edu
> 






From maxale at gmail.com  Tue Nov  7 22:14:38 2006
From: maxale at gmail.com (Max A.)
Date: Tue, 7 Nov 2006 13:14:38 -0800
Subject: square arrangements with bit-wise constraints
Message-ID: <d3dac270611071314i6023ba68h4e7e15d63da8c8b3@mail.gmail.com>

There is a problem at MathPages about arrangements of numbers
0,1,...,15 into 4x4 array with certain bit-wise constraints:
http://www.mathpages.com/home/kmath352.htm

This problem (and its variants) can be easily generalized to
(2^n)x(2^n) arrays, e.g.:

In how many distinct ways can the integers 0 through 2^(2n)-1 be arranged
in a (2^n)x(2^n) array such that the bit-wise OR over each row, each column,
and each diagonal is 2^(2n)-1, and the bit-wise AND over each row, column,
and diagonal is 0?

Does anybody wish to compute corresponding sequences and submit them to OEIS?

Regards,
Max





From maxale at gmail.com  Tue Nov  7 23:49:04 2006
From: maxale at gmail.com (Max A.)
Date: Tue, 7 Nov 2006 14:49:04 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>

On 11/3/06, franktaw at netscape.net <franktaw at netscape.net> wrote:

> In the following, b(i), i = 1..k is the finite sequence.  I'm starting
> all these
> sequences with the empty sequence, but I searched for them without
> that first term.  All are shown here in A066099 order.

What is good about this order comparing, say, to the prime
factorization order suggested by Marc?

I would vote for having a standard way to represent such sequences in
OEIS (well documented there, and maybe even having a dedicated
keyword). That is important when one come up with a multidimensional
sequence and want to query it in OEIS. Then it will be enough for
him/her to encode the sequence in the standard way and make a single
query.

Of course, there are many representations possible but one of them
should be called "standard", and every multidimensional sequence
should have this standard representation (among maybe others) in OEIS.

Neil, does that sound reasonable?

Regards,
Max





From mlb at well.com  Wed Nov  8 00:02:52 2006
From: mlb at well.com (Marc LeBrun)
Date: Tue, 07 Nov 2006 15:02:52 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com
 >
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
 <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>
Message-ID: <200611072304.kA7N3xSr010827@smtp.well.com>

 > does that sound reasonable?

It does, but in practice there are already several orderings in wide 
use (as Frank indicated).

Of course it's always possible to submit more than one ordering.

In fact sometimes different ones would be preferred in different contexts.

Or fancier (perhaps special-purpose) lookups could be used.

"That's the great thing about standards--there's so many to choose from!"







From franktaw at netscape.net  Wed Nov  8 00:56:58 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Tue, 07 Nov 2006 18:56:58 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com> <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>
Message-ID: <8C8D0DF8BAA6EAF-124-5E95@FWM-M16.sysops.aol.com>

Exactly what factorization order?  For most numbers, the sequence of
prime power exponents is not a sequence of positive integers.  If 2
represents the sequence (1), what sequence does 3 represent?

If doesn't work any better to use prime power exponents to represent
sequences of non-negative integers.  If 2 is (1) and 3 is (0,1), what 
do
you use to represent (1,0)?

There are ways around these difficulties, but they are pretty 
artificial.

A066099 sorts, first of all, by the number being composed; I think any
reasonable standard candidate should do that.  Second, it is simple,
with a definition tied in to the binary numbers.  Third, it is similar 
to the
Mathematica ordering for partitions (subsort in reverse lexicographic
order), which is one of the two standards for partition ordering in the
OEIS.

(The equivalent of the other, the Abramowitz and Stegun ordering,
was not yet in the OEIS; I did submit it the other day with A-number
A124734.  It is, in any event, not as simple to define as A066099.)

In any event, based on Neil saying to go ahead and submit these
sequences, I  have submitted them with comments stating that
A066099 is the standard ordering.

Franklin T. Adams-Watters


-----Original Message-----
From: maxale at gmail.com

  On 11/3/06, franktaw at netscape.net <franktaw at netscape.net> wrote:

  > In the following, b(i), i = 1..k is the finite sequence. I'm 
starting
 > all these
 > sequences with the empty sequence, but I searched for them without
 > that first term. All are shown here in A066099 order.

 What is good about this order comparing, say, to the prime
 factorization order suggested by Marc?

 I would vote for having a standard way to represent such sequences in
 OEIS (well documented there, and maybe even having a dedicated
 keyword). ...
________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From maxale at gmail.com  Wed Nov  8 01:42:29 2006
From: maxale at gmail.com (Max A.)
Date: Tue, 7 Nov 2006 16:42:29 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8D0DF8BAA6EAF-124-5E95@FWM-M16.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
	 <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>
	 <8C8D0DF8BAA6EAF-124-5E95@FWM-M16.sysops.aol.com>
Message-ID: <d3dac270611071642x66119c57o13f7bf6df6d2436a@mail.gmail.com>

On 11/7/06, franktaw at netscape.net <franktaw at netscape.net> wrote:

> Exactly what factorization order?  For most numbers, the sequence of
> prime power exponents is not a sequence of positive integers.  If 2
> represents the sequence (1), what sequence does 3 represent?
>
> If doesn't work any better to use prime power exponents to represent
> sequences of non-negative integers.  If 2 is (1) and 3 is (0,1), what
> do you use to represent (1,0)?

Actually, the sequences that I mentioned recently does not depend on
trailing zeros, so index (1,0) for them is the same as (1). That's why
the factorization order works well for them.

> A066099 sorts, first of all, by the number being composed; I think any
> reasonable standard candidate should do that.  Second, it is simple,
> with a definition tied in to the binary numbers.  Third, it is similar
> to the
> Mathematica ordering for partitions (subsort in reverse lexicographic
> order), which is one of the two standards for partition ordering in the
> OEIS.

I see now the advantages of the A066099 ordering. And I agree that it
is a good candidate for the "standard".

Max





From arndt at jjj.de  Wed Nov  8 03:54:45 2006
From: arndt at jjj.de (Joerg Arndt)
Date: Wed, 8 Nov 2006 13:54:45 +1100
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com> <20061102011002.GC6041@cochise.anu.edu.au> <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
Message-ID: <20061108025445.GC6252@cochise.anu.edu.au>

* Jonathan Post <jvospost3 at gmail.com> [Nov 08. 2006 13:29]:
> Thank you, Joerg.  That was interesting.  Does your generating function
> hypothesis hold when I extend the sequence by another 5 values, to a(n) = 0, 0,
> 1, 1, 3, 1, 7, 4,
> 14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again, assuming that
> I got the elementary algebra and arithmetic correct.

Alas, the function I gave was spurious, with the
additional terms no OGF is found.

The gp/script can be found here
http://www.jjj.de/pari/ggf.inc.gp
(cannot find Ralf original at the moment).


> If so, I should submit this sequence as by the two us!

Definitely your invention!
But thanks for the kind offer.


> [...]

best regards,    jj





From zbi74583 at boat.zero.ad.jp  Wed Nov  8 06:33:16 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Wed, 8 Nov 2006 14:33:16 +0900
Subject: N(2)=1+1/2^2+2/3^2+5/4^2+4/5^2+2/6^2+6/7^2+....
Message-ID: <02ba01c702f7$69237ea0$ef73d7dc@FMC57937B6BE00>

    Hi, Seqfans

    I asked Mathar to calculate the following
 number.
    N(s)=   Sum_{m=1 to infinity} (-1)Sigma(m)/m^s

     See the end of this mail

     I tried to find an identity like the following

         Zeta(s)Zeta(s-1)=Sum_{m=1 to infinity} Sigma(m)/m^s
         Zeta(s)/Zeta(s+1)=Sum_{m=1 to infinity} Phi(m)/m^s

    But it is too difficult.
    I transformed it a little as follows.

         N(s)=   Sum_{m=1 to infinity} (-1)Sigma(m)/m^s
         =   Product_{p_i} ( Sum_{k=0 to infinity}
(-1)Sigma(p_i^k)/p_i^(k*s) )
         =   Product_{p_i} ( Sum_{k=0 to infinity}
(Sigma(p_i^k)-2)/p_i^(k*s) )
         =   Product_{p_i} ( Sum_{k=0 to infinity}
 Sigma(p_i^k)/p_i^(k*s) -2/(p_i^s-1) )

        Some members of Seqfan might solve it easily.

    Yasutsoshi

----- Original Message ----- 
From: "Richard Mathar" <mathar at strw.leidenuniv.nl>
To: <zbi74583 at boat.zero.ad.jp>
Cc: <mathar at mail.strw.leidenuniv.nl>
Sent: Sunday, November 05, 2006 10:35 PM
Subject: Re: N(2)=1+1/2^2+2/3^2+5/4^2+4/5^2+2/6^2+6/7^2+....


>
> These are 5 tables of partial sums of N(s) up to some maximum
> m of sum_{m=1...max(m)) (-1)Sigma(m)/m^s. The first number
> is s, the second the maximum m up to which the sum has been accumulated,
> and finally the floating point number the partial sum.
> The case s=1 seems not to be a converging series; the partial sums are
> almost
> linear to the upper limit of m. (If your formula with the zeta-function is
> correct, this is expected since zeta(1)=infinty.) I am including some
> values
> for small maximum m so you can check the results for the basic cases.
> N(2) seems to be a converging series. For N(3) you could already get some
> pretty accurate estimate of the series limit.
>
> zeta(2)=1.644934067
> zeta(3)=1.202056903
> zeta(4)=1.082323234
> zeta(3)*zeta(2)=1.977304350
> zeta(4)*zeta(3)=1.301014115
> zeta(5)*zeta(4)=1.122291001
>
>
> PARI:
>
> \\ computes (-1)sigma(n)
> A049060(n)={
>        local(i,resul,rmax,p) ;
>        if(n==1,
>                return(1)
>        ) ;
>        i=factor(n) ;
>        rmax=matsize(i)[1] ;
>        resul=1 ;
>        for(r=1,rmax,
>                p=0 ;
>                for(j=1,i[r,2],
>                        p += i[r,1]^j ;
>                ) ;
>                resul *= p-1 ;
>        ) ;
>        return(resul) ;
> }
>
>
> N(s,mmax)={
> local(resul) ;
> resul=0.0 ;
> for(m=1,mmax,
> resul += A049060(m)/m^s ;
> if(m<10 || (m % 10000 == 0),
> print(s," ",m," ",resul) ;
> ) ;
> ) ;
> }
>
> {
> mmax=1000000 ;
> for(s=1,5,
> N(s,mmax) ;
> ) ;
> }
>
> Tables:
>
> 1 1 1.000000000000000000000000000
> 1 2 1.500000000000000000000000000
> 1 3 2.166666666666666666666666667
> 1 4 3.416666666666666666666666667
> 1 5 4.216666666666666666666666667
> 1 6 4.550000000000000000000000000
> 1 7 5.407142857142857142857142857
> 1 8 7.032142857142857142857142857
> 1 9 8.254365079365079365079365079
> 1 10000 8956.254813409229652481598757
> 1 20000 17913.54053798274101959639246
> 1 30000 26870.58900461199889420226700
> 1 40000 35827.55225952432056619824099
> 1 50000 44784.92216728336777991954546
> 1 60000 53741.30998083683291788901077
> 1 70000 62698.82877172292127401143322
> 1 80000 71656.42458494024747176547617
> 1 90000 80613.31691563743683765060924
> 1 100000 89570.20047616274019702521482
> 1 110000 98527.44018335441754406085900
> 1 120000 107484.2058258699552949703883
> 1 130000 116441.3450587298928159003627
> 1 140000 125399.1600101618061663172293
> 1 150000 134355.0716257765731578804073
> 1 160000 143312.6869188725594888173896
> 1 170000 152270.1691945778107893068833
> 1 180000 161227.3673088893702787242209
> 1 190000 170184.2283671443801174189235
> 1 200000 179141.9177161676803505348327
> 1 210000 188098.1905736334290746346704
> 1 220000 197055.5778178213299226211954
> 1 230000 206012.3243951074382665848232
> 1 240000 214969.0180191522602672455303
> 1 250000 223927.0626654355275622417046
> 1 260000 232884.0479228396547382227555
> 1 270000 241841.6897312605543987962985
> 1 280000 250798.1260711861562596104630
> 1 290000 259755.4115589713808307871029
> 1 300000 268712.2473108133199765189322
> 1 310000 277669.3610844430796369732363
> 1 320000 286626.7360084187087594659096
> 1 330000 295583.0702328420388811375076
> 1 340000 304540.9324662583651406115180
> 1 350000 313497.9488508493337302810705
> 1 360000 322455.2848872162239393519675
> 1 370000 331411.9952274305516780284586
> 1 380000 340369.3878688847801453424847
> 1 390000 349326.5442385613025662494469
> 1 400000 358283.7198151948112652311070
> 1 410000 367240.8596635115166082534173
> 1 420000 376197.2486760902447184900822
> 1 430000 385154.6524560274677028589761
> 1 440000 394112.0480841471143112935114
> 1 450000 403069.2697966051366843904756
> 1 460000 412026.1487890321657369929892
> 1 470000 420983.4928713199267860102478
> 1 480000 429940.0597652682271439943634
> 1 490000 438897.5981659322519223492273
> 1 500000 447855.4516433798944040122164
> 1 510000 456811.4127981064830642426529
> 1 520000 465769.1531758862533175249594
> 1 530000 474726.4930671377953578153300
> 1 540000 483683.7925672078781534825684
> 1 550000 492640.2315981062235193108053
> 1 560000 501597.6949987133986923353009
> 1 570000 510554.3958474732897001847922
> 1 580000 519511.5431567859764988621906
> 1 590000 528468.5920184442509092919859
> 1 600000 537425.2057958757909274862160
> 1 610000 546383.3300255139505983850944
> 1 620000 555340.2697592667964924572269
> 1 630000 564297.7073054137624068690112
> 1 640000 573254.2920689474814450061694
> 1 650000 582211.4873730487914773223047
> 1 660000 591168.5211525597522567544916
> 1 670000 600125.8589826114248021714650
> 1 680000 609083.0330009002048275663287
> 1 690000 618039.4919349372313479054900
> 1 700000 626997.2222632460664089943429
> 1 710000 635954.0917690439691078847726
> 1 720000 644911.6290093784488180074886
> 1 730000 653868.6304875870104467000556
> 1 740000 662825.3447013751015479514900
> 1 750000 671782.5558801465303249818702
> 1 760000 680739.9011662731435592584669
> 1 770000 689696.9475041453715157412648
> 1 780000 698653.1437839886472918561286
> 1 790000 707611.0985532863373721850039
> 1 800000 716568.7987693866776749447598
> 1 810000 725525.6575122445531974385646
> 1 820000 734482.4569919820693082893669
> 1 830000 743439.4779916042359124217226
> 1 840000 752396.3611940500092646796766
> 1 850000 761353.5394438445571430835057
> 1 860000 770310.9434931792736520368474
> 1 870000 779267.1927223886953325647496
> 1 880000 788225.1436674517253585125689
>
>
> 2 1 1.000000000000000000000000000
> 2 2 1.250000000000000000000000000
> 2 3 1.472222222222222222222222222
> 2 4 1.784722222222222222222222222
> 2 5 1.944722222222222222222222222
> 2 6 2.000277777777777777777777778
> 2 7 2.122726757369614512471655329
> 2 8 2.325851757369614512471655329
> 2 9 2.461654226505416981607457798
> 2 10000 8.619813177975507784214344698
> 2 20000 9.240635095871089404115753095
> 2 30000 9.603791897573761358292380061
> 2 40000 9.861458278261452185262848858
> 2 50000 10.06133043757656995196303496
> 2 60000 10.22462154552329778519365923
> 2 70000 10.36270038005636494621123847
> 2 80000 10.48231016697477680498962023
> 2 90000 10.58780522013752853231672862
> 2 100000 10.68217402235127296288785820
> 2 110000 10.76754463607533705360724580
> 2 120000 10.84547789729425765394094918
> 2 130000 10.91717268365027978460835188
> 2 140000 10.98355652885066079792761960
> 2 150000 11.04534545539722125239381859
> 2 160000 11.10315629233238669062836862
> 2 170000 11.15746025373895516747888760
> 2 180000 11.20865775307751148407150182
> 2 190000 11.25708463818456648077093060
> 2 200000 11.30303121795283056763425981
> 2 210000 11.34672871948779384291766276
> 2 220000 11.38839827063844308711114745
> 2 230000 11.42821240662906506074836109
> 2 240000 11.46633165190016188934526342
> 2 250000 11.50290004647923375585461222
> 2 260000 11.53802980091084626273394586
> 2 270000 11.57183602344116902681335304
> 2 280000 11.60440829561909536077171934
> 2 290000 11.63584045378869260629085740
> 2 300000 11.66620539683007648663354504
> 2 310000 11.69557551635064846235471407
> 2 320000 11.72401391041322342493145640
> 2 330000 11.75157395832622563415629569
> 2 340000 11.77831576251751436463182945
> 2 350000 11.80427986574616427809478607
> 2 360000 11.82951337960016320803463309
> 2 370000 11.85405377158818097821428014
> 2 380000 11.87794149361683265940241605
> 2 390000 11.90120806565142413868457075
> 2 400000 11.92388559358277288937420440
> 2 410000 11.94600303970164868968234173
> 2 420000 11.96758569454234998141373716
> 2 430000 11.98866287480559328733313634
> 2 440000 12.00925544388189350449936925
> 2 450000 12.02938482345842327394291528
> 2 460000 12.04907101416111155199518539
> 2 470000 12.06833481184766752219695603
> 2 480000 12.08719139777266186606623586
> 2 490000 12.10566116650605692754670219
> 2 500000 12.12375840131307130932177405
> 2 510000 12.14149351071909408890611230
> 2 520000 12.15888769016783248485026059
> 2 530000 12.17594975401341942285000014
> 2 540000 12.19269280126406301090981599
> 2 550000 12.20912705098196862278557687
> 2 560000 12.22526702472305560991043337
> 2 570000 12.24111996896626519121471994
> 2 580000 12.25669798166764128539680162
> 2 590000 12.27200952384435815319104573
> 2 600000 12.28706299338740013449483108
> 2 610000 12.30187012252190226181814868
> 2 620000 12.31643453756242668558457904
> 2 630000 12.33076670711890682146429120
> 2 640000 12.34487182648867321005497632
> 2 650000 12.35875920337989470739102627
> 2 660000 12.37243430644503889140869087
> 2 670000 12.38590421689421153982279985
> 2 680000 12.39917431914306915108546756
> 2 690000 12.41224965443445935379058168
> 2 700000 12.42513867918653815467126754
> 2 710000 12.43784365088579118428209464
> 2 720000 12.45037185509338005878377820
> 2 730000 12.46272650898502725121177200
> 2 740000 12.47491268376206272216237202
> 2 750000 12.48693594951832440956683550
> 2 760000 12.49880013638169511883229569
> 2 770000 12.51050884124405914256853750
> 2 780000 12.52206537108093533871333277
> 2 790000 12.53347692259745252540675575
> 2 800000 12.54474459716977128685280547
> 2 810000 12.55587125001645305770022096
> 2 820000 12.56686130746829341595469366
> 2 830000 12.57771842011223665163037795
> 2 840000 12.58844534265072464040923243
> 2 850000 12.59904566706465055360427300
> 2 860000 12.60952226922408394518254911
> 2 870000 12.61987642201166975414741555
> 2 880000 12.63011418303496833362065684
> 2 890000 12.64023571583693921823484395
> 2 900000 12.65024408926657285403820450
> 2 910000 12.66014049405210257196085418
> 2 920000 12.66993013960070545251999449
> 2 930000 12.67961329658082585267888159
> 2 940000 12.68919341076487592672709130
> 2 950000 12.69867217785123447166660908
> 2 960000 12.70805040456023104531710849
> 2 970000 12.71733330696500792660758365
> 2 980000 12.72652037909829933006161382
> 2 990000 12.73561385879130711822105048
> 2 1000000 12.74461639944508312563225494
>
>
> 3 1 1.000000000000000000000000000
> 3 2 1.125000000000000000000000000
> 3 3 1.199074074074074074074074074
> 3 4 1.277199074074074074074074074
> 3 5 1.309199074074074074074074074
> 3 6 1.318458333333333333333333333
> 3 7 1.335951044703595724003887269
> 3 8 1.361341669703595724003887269
> 3 9 1.376430832940907109463420877
> 3 10000 1.463913982110741753040591678
> 3 20000 1.463958761182127253891639226
> 3 30000 1.463973687899719174955107484
> 3 40000 1.463981151485566123839035784
> 3 50000 1.463985629911768534082474807
> 3 60000 1.463988615257603046255509048
> 3 70000 1.463990747941060935932260523
> 3 80000 1.463992347452072574407883095
> 3 90000 1.463993591418911296987275373
> 3 100000 1.463994586598910692322299337
> 3 110000 1.463995400874372718599240224
> 3 120000 1.463996079403452056486658683
> 3 130000 1.463996653569731083112186861
> 3 140000 1.463997145746936259101012298
> 3 150000 1.463997572212239449568643565
> 3 160000 1.463997945442508018944894819
> 3 170000 1.463998274756427172013405468
> 3 180000 1.463998567470471281896562070
> 3 190000 1.463998829362945706567178177
> 3 200000 1.463999065087923587578410681
> 3 210000 1.463999278329572278322382185
> 3 220000 1.463999472210290336272529678
> 3 230000 1.463999649219477985324795694
> 3 240000 1.463999811477588227637144335
> 3 250000 1.463999960777271571969049099
> 3 260000 1.464000098575628251036076097
> 3 270000 1.464000226175768714632044135
> 3 280000 1.464000344645861174695942201
> 3 290000 1.464000454956284414358238477
> 3 300000 1.464000557907605374141443674
> 3 310000 1.464000654220015163939180059
> 3 320000 1.464000744515464470055773720
> 3 330000 1.464000829328753859822237621
> 3 340000 1.464000909166687500211857215
> 3 350000 1.464000984435262908416401907
> 3 360000 1.464001055524739843653406744
> 3 370000 1.464001122766895437258623701
> 3 380000 1.464001186474841361410807209
> 3 390000 1.464001246914091200041356493
> 3 400000 1.464001304331492624240843891
> 3 410000 1.464001358947841911631088786
> 3 420000 1.464001410959152183436079651
> 3 430000 1.464001460557008749462202957
> 3 440000 1.464001507900298795165105625
> 3 450000 1.464001553138542176224706874
> 3 460000 1.464001596408268761316169011
> 3 470000 1.464001637838876141306278665
> 3 480000 1.464001677539798346550656570
> 3 490000 1.464001715624418180233792011
> 3 500000 1.464001752186865804286414230
> 3 510000 1.464001787308096983826068464
> 3 520000 1.464001821085252219783666269
> 3 530000 1.464001853586288170141300728
> 3 540000 1.464001884883421221311137792
> 3 550000 1.464001915039618326560450277
> 3 560000 1.464001944122159842425557477
> 3 570000 1.464001972181866862157721242
> 3 580000 1.464001999275380923272745020
> 3 590000 1.464002025450188768071703730
> 3 600000 1.464002050751299694154935270
> 3 610000 1.464002075226966770079630464
> 3 620000 1.464002098909928297454328876
> 3 630000 1.464002121842320025653578842
> 3 640000 1.464002144055977272720855659
> 3 650000 1.464002165587616796774276693
> 3 660000 1.464002186466412269783936185
> 3 670000 1.464002206722643544544226821
> 3 680000 1.464002226382732127695424983
> 3 690000 1.464002245471462412103956612
> 3 700000 1.464002264017435494080639785
> 3 710000 1.464002282039246813434225341
> 3 720000 1.464002299561748289079351576
> 3 730000 1.464002316603153605988980695
> 3 740000 1.464002333183468992396940641
> 3 750000 1.464002349322542550155865090
> 3 760000 1.464002365037130225625255935
> 3 770000 1.464002380343037616331619950
> 3 780000 1.464002395255086253184010138
> 3 790000 1.464002409792474848650673824
> 3 800000 1.464002423965997236326475471
> 3 810000 1.464002437788253358795504488
> 3 820000 1.464002451273302166423360831
> 3 830000 1.464002464433742258170070393
> 3 840000 1.464002477280648906020788669
> 3 850000 1.464002489825692224210435982
> 3 860000 1.464002502079291985535094261
> 3 870000 1.464002514049668308408128899
> 3 880000 1.464002525750211478039722139
> 3 890000 1.464002537187197706443946991
> 3 900000 1.464002548369947911638829284
> 3 910000 1.464002559305408714470469907
> 3 920000 1.464002570004676399632165939
> 3 930000 1.464002580473145840547517414
> 3 940000 1.464002590719441673840433620
> 3 950000 1.464002600750056430793286939
> 3 960000 1.464002610570359301053681776
> 3 970000 1.464002620190111512591642127
> 3 980000 1.464002629612902400536905838
> 3 990000 1.464002638845007329954067381
> 3 1000000 1.464002647892926094713436024
>
>
> 4 1 1.000000000000000000000000000
> 4 2 1.062500000000000000000000000
> 4 3 1.087191358024691358024691358
> 4 4 1.106722608024691358024691358
> 4 5 1.113122608024691358024691358
> 4 6 1.114665817901234567901234568
> 4 7 1.117164776668414909425599416
> 4 8 1.120338604793414909425599416
> 4 9 1.122015178486449507809992039
> 4 10000 1.126342859836933119635341727
> 4 20000 1.126342863195068461987333491
> 4 30000 1.126342863816977724856173635
> 4 40000 1.126342864034657367890755384
> 4 50000 1.126342864135419045299551969
> 4 60000 1.126342864190149270391012376
> 4 70000 1.126342864223154637804140480
> 4 80000 1.126342864244576264194082203
> 4 90000 1.126342864259261709833237965
> 4 100000 1.126342864269766231843710036
> 4 110000 1.126342864277538770311620766
> 4 120000 1.126342864283450137689488953
> 4 130000 1.126342864288050791700191891
> 4 140000 1.126342864291701518257416350
> 4 150000 1.126342864294646135946293287
> 4 160000 1.126342864297056569571605552
> 4 170000 1.126342864299054229967212718
> 4 180000 1.126342864300728234359584700
> 4 190000 1.126342864302144893188134103
> 4 200000 1.126342864303354525085913547
> 4 210000 1.126342864304395340180474300
> 4 220000 1.126342864305297594007409955
> 4 230000 1.126342864306084686590084820
> 4 240000 1.126342864306775457553279160
> 4 250000 1.126342864307385095784161975
> 4 260000 1.126342864307925686556878209
> 4 270000 1.126342864308407364997509003
> 4 280000 1.126342864308838305443671084
> 4 290000 1.126342864309225477016399704
> 4 300000 1.126342864309574563445429650
> 4 310000 1.126342864309890425673710622
> 4 320000 1.126342864310177149073405791
> 4 330000 1.126342864310438174074903581
> 4 340000 1.126342864310676548757732228
> 4 350000 1.126342864310894763694630232
> 4 360000 1.126342864311095054783294446
> 4 370000 1.126342864311279313873956099
> 4 380000 1.126342864311449231412772564
> 4 390000 1.126342864311606242436985681
> 4 400000 1.126342864311751625747966668
> 4 410000 1.126342864311886501047539562
> 4 420000 1.126342864312011847410167188
> 4 430000 1.126342864312128564203286214
> 4 440000 1.126342864312237413459871362
> 4 450000 1.126342864312339084988654298
> 4 460000 1.126342864312434194531858928
> 4 470000 1.126342864312523302682800842
> 4 480000 1.126342864312606892652019146
> 4 490000 1.126342864312685425831403550
> 4 500000 1.126342864312759296677644308
> 4 510000 1.126342864312828850304390296
> 4 520000 1.126342864312894443054768726
> 4 530000 1.126342864312956355217879326
> 4 540000 1.126342864313014859446213372
> 4 550000 1.126342864313070196425470463
> 4 560000 1.126342864313122601535952824
> 4 570000 1.126342864313172268506265819
> 4 580000 1.126342864313219391141266622
> 4 590000 1.126342864313264137597311253
> 4 600000 1.126342864313306663426890322
> 4 610000 1.126342864313347121770685419
> 4 620000 1.126342864313385633102777798
> 4 630000 1.126342864313422327185152818
> 4 640000 1.126342864313457311416964776
> 4 650000 1.126342864313490695748042037
> 4 660000 1.126342864313522573574899694
> 4 670000 1.126342864313553035740126664
> 4 680000 1.126342864313582163334291352
> 4 690000 1.126342864313610031536993753
> 4 700000 1.126342864313636717733889822
> 4 710000 1.126342864313662281830870319
> 4 720000 1.126342864313686789977325667
> 4 730000 1.126342864313710296430980716
> 4 740000 1.126342864313732855692110296
> 4 750000 1.126342864313754519823275656
> 4 760000 1.126342864313775334724375007
> 4 770000 1.126342864313795343268423855
> 4 780000 1.126342864313814585400999315
> 4 790000 1.126342864313833105102330919
> 4 800000 1.126342864313850934103761246
> 4 810000 1.126342864313868105235437633
> 4 820000 1.126342864313884651902605718
> 4 830000 1.126342864313900604514476466
> 4 840000 1.126342864313915990566003172
> 4 850000 1.126342864313930837276585851
> 4 860000 1.126342864313945169444268968
> 4 870000 1.126342864313959008481394737
> 4 880000 1.126342864313972380956143569
> 4 890000 1.126342864313985304497313295
> 4 900000 1.126342864313997799554982470
> 4 910000 1.126342864314009883288627525
> 4 920000 1.126342864314021576813991216
> 4 930000 1.126342864314032894395637362
> 4 940000 1.126342864314043853302500574
> 4 950000 1.126342864314054467992200910
> 4 960000 1.126342864314064751304010854
> 4 970000 1.126342864314074720218563997
> 4 980000 1.126342864314084384860927016
> 4 990000 1.126342864314093757784864329
> 4 1000000 1.126342864314102851387025112
>
>
> 5 1 1.000000000000000000000000000
> 5 2 1.031250000000000000000000000
> 5 3 1.039480452674897119341563786
> 5 4 1.044363265174897119341563786
> 5 5 1.045643265174897119341563786
> 5 6 1.045900466820987654320987654
> 5 7 1.046257460930584845967325490
> 5 8 1.046654189446209845967325490
> 5 9 1.046840475412102579121146892
> 5 10000 1.047127646962981639236350714
> 5 20000 1.047127646963242803683634380
> 5 30000 1.047127646963269060444667980
> 5 40000 1.047127646963275452376929668
> 5 50000 1.047127646963277728778610486
> 5 60000 1.047127646963278734909173763
> 5 70000 1.047127646963279246707166974
> 5 80000 1.047127646963279534023699805
> 5 90000 1.047127646963279707590251386
> 5 100000 1.047127646963279818572046343
> 5 110000 1.047127646963279892819669335
> 5 120000 1.047127646963279944352143863
> 5 130000 1.047127646963279981235720216
> 5 140000 1.047127646963280008327383488
> 5 150000 1.047127646963280028667159600
> 5 160000 1.047127646963280044239866946
> 5 170000 1.047127646963280056361640181
> 5 180000 1.047127646963280065937720739
> 5 190000 1.047127646963280073602739981
> 5 200000 1.047127646963280079811373591
> 5 210000 1.047127646963280084892514359
> 5 220000 1.047127646963280089092048700
> 5 230000 1.047127646963280092592526331
> 5 240000 1.047127646963280095533744603
> 5 250000 1.047127646963280098023437624
> 5 260000 1.047127646963280100144477685
> 5 270000 1.047127646963280101962983953
> 5 280000 1.047127646963280103530722491
> 5 290000 1.047127646963280104889770905
> 5 300000 1.047127646963280106073563423
> 5 310000 1.047127646963280107109545028
> 5 320000 1.047127646963280108020080532
> 5 330000 1.047127646963280108823485592
> 5 340000 1.047127646963280109535261396
> 5 350000 1.047127646963280110167943658
> 5 360000 1.047127646963280110732291221
> 5 370000 1.047127646963280111237235413
> 5 380000 1.047127646963280111690454876
> 5 390000 1.047127646963280112098366088
> 5 400000 1.047127646963280112466502499
> 5 410000 1.047127646963280112799594529
> 5 420000 1.047127646963280113101691720
> 5 430000 1.047127646963280113376369686
> 5 440000 1.047127646963280113626641318
> 5 450000 1.047127646963280113855154497
> 5 460000 1.047127646963280114064219587
> 5 470000 1.047127646963280114255879066
> 5 480000 1.047127646963280114431883567
> 5 490000 1.047127646963280114593830274
> 5 500000 1.047127646963280114743084164
> 5 510000 1.047127646963280114880831757
> 5 520000 1.047127646963280115008212055
> 5 530000 1.047127646963280115126153891
> 5 540000 1.047127646963280115235519959
> 5 550000 1.047127646963280115337066821
> 5 560000 1.047127646963280115431500449
> 5 570000 1.047127646963280115519415563
> 5 580000 1.047127646963280115601376090
> 5 590000 1.047127646963280115677873082
> 5 600000 1.047127646963280115749351712
> 5 610000 1.047127646963280115816230987
> 5 620000 1.047127646963280115878856405
> 5 630000 1.047127646963280115937571799
> 5 640000 1.047127646963280115992669514
> 5 650000 1.047127646963280116044432219
> 5 660000 1.047127646963280116093104355
> 5 670000 1.047127646963280116138915493
> 5 680000 1.047127646963280116182070550
> 5 690000 1.047127646963280116222756887
> 5 700000 1.047127646963280116261156889
> 5 710000 1.047127646963280116297420392
> 5 720000 1.047127646963280116331699689
> 5 730000 1.047127646963280116364124370
> 5 740000 1.047127646963280116394819088
> 5 750000 1.047127646963280116423900164
> 5 760000 1.047127646963280116451471135
> 5 770000 1.047127646963280116477627542
> 5 780000 1.047127646963280116502457450
> 5 790000 1.047127646963280116526050682
> 5 800000 1.047127646963280116548478239
> 5 810000 1.047127646963280116569809891
> 5 820000 1.047127646963280116590113535
> 5 830000 1.047127646963280116609450953
> 5 840000 1.047127646963280116627878223
> 5 850000 1.047127646963280116645449098
> 5 860000 1.047127646963280116662212610
> 5 870000 1.047127646963280116678212198
> 5 880000 1.047127646963280116693495678
> 5 890000 1.047127646963280116708099149
> 5 900000 1.047127646963280116722060660
> 5 910000 1.047127646963280116735413376
> 5 920000 1.047127646963280116748193681
> 5 930000 1.047127646963280116760429366
> 5 940000 1.047127646963280116772150557
> 5 950000 1.047127646963280116783383436
> 5 960000 1.047127646963280116794151686
> 5 970000 1.047127646963280116804482528
> 5 980000 1.047127646963280116814395317
> 5 990000 1.047127646963280116823911289
> 5 1000000 1.047127646963280116833050882
>
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From ralf at ark.in-berlin.de  Wed Nov  8 09:26:05 2006
From: ralf at ark.in-berlin.de (Ralf Stephan)
Date: Wed, 8 Nov 2006 09:26:05 +0100
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <20061108025445.GC6252@cochise.anu.edu.au>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com> <20061102011002.GC6041@cochise.anu.edu.au> <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com> <20061108025445.GC6252@cochise.anu.edu.au>
Message-ID: <20061108082605.GA21119@ark.in-berlin.de>

You wrote 
> * Jonathan Post <jvospost3 at gmail.com> [Nov 08. 2006 13:29]:
> > Thank you, Joerg.  That was interesting.  Does your generating function
> > hypothesis hold when I extend the sequence by another 5 values, to a(n) = 0, 0,
> > 1, 1, 3, 1, 7, 4,
> > 14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again, assuming that
> > I got the elementary algebra and arithmetic correct.
> 
> Alas, the function I gave was spurious, with the
> additional terms no OGF is found.

Quite possible. Always treat them as hypotheses, as Jonathan said.


ralf






From maxale at gmail.com  Wed Nov  8 11:27:40 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 8 Nov 2006 02:27:40 -0800
Subject: A038379 is all wrong
Message-ID: <d3dac270611080227lc6e0ad2qde71bd834c86d0f2@mail.gmail.com>

A038379 is defined as the "number of positive semi-definite real {0,1}
n X n matrices".
But the sequence and comments have nothing to do with the description.

In particular, all given values do not match the definition.
It lists 1, 3, 27, 729, 52649 while the correct values (that match the
definition) are 2, 7, 59, 1468, 102751 (I will be grateful if somebody
check these values).
For the simplest case n=1, it is clear that there two positive
semi-definite real {0,1}-matrices:
(0) and (1).

There is also a comment:
%C A038379 For n <= 4 a(n) = the upper bound 3^C(n,2).
Why the upper bound is 3^C(n,2) ? We have only two options for each
element of the matrix.
There would be an upper bound of this type if:
1) The main diagonal of a matrix is fixed, and the matrix is
symmetric. Then there are C(n,2) distinct non-diagonal elements whose
values should be assigned.
2) There are three (not two!) possible values for each non-diagonal
element of the matrix.

It seems that the definition of A038379 does not match the rest. Can
anybody find a definition of A038379 that is consistent with its
current content? If so, I will submit the real "number of positive
semi-definite real {0,1} n X n matrices" as a separate sequence.

Regards,
Max





From njas at research.att.com  Wed Nov  8 15:54:53 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 8 Nov 2006 09:54:53 -0500 (EST)
Subject: A038379 is all wrong
Message-ID: <200611081454.JAA83283@fry.research.att.com>

Max,   maybe I meant to say "Number of positive semi-definite real
matrices with entries {-1, 0, +1}." - can you check?

Neil





From maxale at gmail.com  Wed Nov  8 22:59:13 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 8 Nov 2006 13:59:13 -0800
Subject: A038379 is all wrong
In-Reply-To: <200611081454.JAA83283@fry.research.att.com>
References: <200611081454.JAA83283@fry.research.att.com>
Message-ID: <d3dac270611081359t117aab4ag1545e9c3130fa53a@mail.gmail.com>

On 11/8/06, N. J. A. Sloane <njas at research.att.com> wrote:
> Max,   maybe I meant to say "Number of positive semi-definite real
> matrices with entries {-1, 0, +1}." - can you check?

For the comment to make sense, the main diagonal should be fixed as well.
I will check most obvious variants.

Max





From jvospost3 at gmail.com  Thu Nov  9 02:51:54 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 8 Nov 2006 17:51:54 -0800
Subject: primes in arithmetic progression
In-Reply-To: <335cf8920611042302v7db1ca58xb8b67f872981b204@mail.gmail.com>
References: <200611041910.OAA62978@fry.research.att.com>
	 <5542af940611041355vd65aa5cr3a4a9df3cc348c15@mail.gmail.com>
	 <335cf8920611042302v7db1ca58xb8b67f872981b204@mail.gmail.com>
Message-ID: <5542af940611081751y69ee576as6473da25e91a2175@mail.gmail.com>

Sorry, Olivier.  I hope not to repeat the mistake.

Thank you again, Richard J. Mathar, for your corrections and extensions of 6
Nov 2006, which appear online today for A124570.  I'm surprised that nobody
looked in this direction, before you started it.  NJAS was right to see the
value, and 'kick things off" -- and I hope that it does not stop with my
clumsy shot at the semiprime case, which you so gallantly repaired.

I'm mostly satisfied with the results of myself and tens of millions of
others voting in the U.S. elections Tuesday 7 Nove 2006.  I feel as if the
whole country is being corrected and extended, to benefits both domestic and
global. But this is not the venue to discuss that.



On 11/4/06, Olivier Gerard <olivier.gerard at gmail.com> wrote:
>
> Hello Jonathan,
>
> Thanks for this message.
>
> Please remember NOT to send your posts to seqfan as HTML,
> even with a pure text version.
>
> regards,
>
> Olivier
>
>
> On 11/4/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> >
> > The analogue for 3-almost primes (or, as njas prefers, numbers which are
> > the product of exactly 3 primes) is left as an exercise for the reader...
> > I've just submitted the following.
> >
> > Subject: PRE-NUMBERED NEW SEQUENCE A124570 FROM Jonathan Vos Post
> >
> >
> >
>
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From jvospost3 at gmail.com  Thu Nov  9 04:11:11 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 8 Nov 2006 19:11:11 -0800
Subject: Scrabblomics; COMMENT FROM Jonathan Vos Post RE A113172
In-Reply-To: <20061108213820.28640.qmail@web55109.mail.re4.yahoo.com>
References: <20061108213820.28640.qmail@web55109.mail.re4.yahoo.com>
Message-ID: <5542af940611081911h2ff3a639v3a4b53b5c2927de2@mail.gmail.com>

Here's more fun, with a silly if plausibly motivated name.

Subject: NEW SEQUENCE FROM Jonathan Vos Post


%I A000001
%S A000001 1,
3, 3,
8, 10, 8,
9, 12, 12, 9,
4, 15, 19, 15, 4,
7, 15, 19, 19, 15, 7,
8, 18, 19, 21, 19, 18, 8,
9, 22, 20, 11, 11, 20, 22, 9,
4, 15, 17, 15, 18, 15, 17, 15, 4,
7, 15, 18, 18, 20, 20, 18, 18, 15, 7,
8, 18, 20, 22, 21, 11, 21, 22, 20, 18, 8,
9, 22, 21, 17, 19, 18, 18, 19, 17, 21, 22, 9
%N A000001 Pascrabble Triangle, by rows.
%C A000001 The apex of the triangle is 1.  Any other value is the
Scrabble value of English name for the number which is the sum of the
numbers above. This is generated the same way as A007318 Pascal's
triangle read by rows, except apply A113172 to each sum. The first
column of this triangle is 1, 3, 8, 9, 4, 7, 8, 9, 4, 7, 8, 9, 4, 7...
= iterations 1, A113172(1), A113172(A113172(1)),
A113172(A113172(A113172(1))).
The central pascrabble numbers T(2n+1,n) = 1, 10, 19, 21, 18, 11, 22, ...
%F A000001 T(1,1) = 1; for i > 1, T(i,j) = A113172(T(i-1, j-1)+T(i-1, j).
%e A000001 Triangle begins:
row.|.values in row
.1..|01
.2..|03.03
.3..|08.10.08
.4..|09.12.12.09
.5..|04.15.19.15.04
.6..|07.15.19.19.15.07
.7..|08.18.19.21.19.18.08
.8..|09.22.20.11.11.20.22.09
.9..|04.15.17.15.18.15.17.15.04
10..|07.15.18.18.20.20.18.18.15.07
11..|08.18.20.22.21.11.21.22.20.18.8
12..|09.22.21.17.19.18.18.19.17.21.22.09
%Y A000001 cf. A007318 Pascal's triangle read by rows, A113172.
%O A000001 1
%K A000001 ,easy,nonn,tabl,word,
%A A000001 Jonathan Vos Post (jvospost2 at yahoo.com), Nov 08 2006

I think that this cut & paste is pure text, and not HTML (as Olivier asked)
but I am not sure.

On 11/8/06, jonathan post <jvospost2 at yahoo.com> wrote:
>
> My backup copy (as no autoreply yet).
>
> Subject: COMMENT FROM Jonathan Vos Post RE A113172
>
>
> %I A113172
> %S A113172
> 13,3,6,8,7,10,10,8,9,4,3,9,12,11,11,13,14,12,12,8,12,15,18,20,19,22,22
> %N A113172 Scrabble value of English word for the
> number n.
> %C A113172 Is 12 the unique fixed point of Scrabble
> value of English word for the number n, where
> A113172(n) = n? 12 --> "TWELVE" --> 1+4+1+1+4+1 -->
> 12. For what n are A113172(n) prime? For what n are
> there cycles of what length for iterations n,
> A113172(n), A113172(A113172(n)),
> A113172(A113172(A113172(n))), ...? For what n does
> A113172(n) | n? These are key questions in general
> Scrabblomics.
> %F A113172 Scrabble value of English word for the
> number n.
> %O A113172 1
> %K A113172 ,easy,nonn,word,
> %A A113172 Jonathan Vos Post (jvospost2 at yahoo.com),
> Nov 08 2006
>
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From rkg at cpsc.ucalgary.ca  Thu Nov  9 20:26:26 2006
From: rkg at cpsc.ucalgary.ca (Richard Guy)
Date: Thu, 9 Nov 2006 12:26:26 -0700 (MST)
Subject: Small notation clean-up
In-Reply-To: <5542af940611081751y69ee576as6473da25e91a2175@mail.gmail.com>
References: <200611041910.OAA62978@fry.research.att.com> 
 <5542af940611041355vd65aa5cr3a4a9df3cc348c15@mail.gmail.com> 
 <335cf8920611042302v7db1ca58xb8b67f872981b204@mail.gmail.com>
 <5542af940611081751y69ee576as6473da25e91a2175@mail.gmail.com>
Message-ID: <Pine.LNX.4.63.0611091221220.23740@csl>

Would someone more competent & energetic than I
clean up the (usually unexplained)   S_n
notation that appears sporadically in OEIS ?

Evidently it means a ``star'' with  n  vertices,
and so is in fact the complete bipartite graph
K_(1,n-1), a notation which is more widely
known.   Best to all,   R.





From maxale at gmail.com  Thu Nov  9 20:35:39 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 9 Nov 2006 11:35:39 -0800
Subject: Small notation clean-up
In-Reply-To: <Pine.LNX.4.63.0611091221220.23740@csl>
References: <200611041910.OAA62978@fry.research.att.com>
	 <5542af940611041355vd65aa5cr3a4a9df3cc348c15@mail.gmail.com>
	 <335cf8920611042302v7db1ca58xb8b67f872981b204@mail.gmail.com>
	 <5542af940611081751y69ee576as6473da25e91a2175@mail.gmail.com>
	 <Pine.LNX.4.63.0611091221220.23740@csl>
Message-ID: <d3dac270611091135u58ec268au1f85498bf1ea978b@mail.gmail.com>

On 11/9/06, Richard Guy <rkg at cpsc.ucalgary.ca> wrote:
> Would someone more competent & energetic than I
> clean up the (usually unexplained)   S_n
> notation that appears sporadically in OEIS ?
>
> Evidently it means a ``star'' with  n  vertices,

It is also often used for the symmetric group of order n.

Max





From maxale at gmail.com  Thu Nov  9 23:11:00 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 9 Nov 2006 14:11:00 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <20061108025445.GC6252@cochise.anu.edu.au>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <20061102011002.GC6041@cochise.anu.edu.au>
	 <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
	 <20061108025445.GC6252@cochise.anu.edu.au>
Message-ID: <d3dac270611091411t71b55c5fl1087e070509b8fee@mail.gmail.com>

On 11/7/06, Joerg Arndt <arndt at jjj.de> wrote:
> * Jonathan Post <jvospost3 at gmail.com> [Nov 08. 2006 13:29]:
> > Thank you, Joerg.  That was interesting.  Does your generating function
> > hypothesis hold when I extend the sequence by another 5 values, to a(n) = 0, 0,
> > 1, 1, 3, 1, 7, 4,
> > 14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again, assuming that
> > I got the elementary algebra and arithmetic correct.
>
> Alas, the function I gave was spurious, with the
> additional terms no OGF is found.

Not surprising at all!
This sequence is not well-defined (see above in this discussion), so
one may view its terms as random numbers. How can they have an OGF? ;)

Max





From jvospost3 at gmail.com  Thu Nov  9 23:15:12 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 9 Nov 2006 14:15:12 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <d3dac270611091411t71b55c5fl1087e070509b8fee@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <20061102011002.GC6041@cochise.anu.edu.au>
	 <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
	 <20061108025445.GC6252@cochise.anu.edu.au>
	 <d3dac270611091411t71b55c5fl1087e070509b8fee@mail.gmail.com>
Message-ID: <5542af940611091415m4db107dei8659adb11bdcbffa@mail.gmail.com>

Max's position is well taken.  I think that "random" overstates it
somewhat.  I think that even careful definition leaves a residuum of chaos
in the proper sense.  I have not yet posted any seq based on this, out of
caution.  When I do, I'll probably post just one and see what happens.

On 11/9/06, Max A. <maxale at gmail.com> wrote:
>
> On 11/7/06, Joerg Arndt <arndt at jjj.de> wrote:
> > * Jonathan Post <jvospost3 at gmail.com> [Nov 08. 2006 13:29]:
> > > Thank you, Joerg.  That was interesting.  Does your generating
> function
> > > hypothesis hold when I extend the sequence by another 5 values, to
> a(n) = 0, 0,
> > > 1, 1, 3, 1, 7, 4,
> > > 14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again,
> assuming that
> > > I got the elementary algebra and arithmetic correct.
> >
> > Alas, the function I gave was spurious, with the
> > additional terms no OGF is found.
>
> Not surprising at all!
> This sequence is not well-defined (see above in this discussion), so
> one may view its terms as random numbers. How can they have an OGF? ;)
>
> Max
>
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From maxale at gmail.com  Thu Nov  9 23:41:15 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 9 Nov 2006 14:41:15 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <5542af940611091415m4db107dei8659adb11bdcbffa@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <20061102011002.GC6041@cochise.anu.edu.au>
	 <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
	 <20061108025445.GC6252@cochise.anu.edu.au>
	 <d3dac270611091411t71b55c5fl1087e070509b8fee@mail.gmail.com>
	 <5542af940611091415m4db107dei8659adb11bdcbffa@mail.gmail.com>
Message-ID: <d3dac270611091441la822b41v6a324c49ec2a1a76@mail.gmail.com>

On 11/9/06, Jonathan Post <jvospost3 at gmail.com> wrote:

> Max's position is well taken.  I think that "random" overstates it somewhat.

OK, it may not be random from your point of view since you followed
some rules (I believe) defining the terms of this sequence. The
problem is that you did not let us know these rules!
Remember the example I gave above? It was about different
representations of b(1) as 2-nd degree polynomials of w:
b(1) = w^2 = 2w^2+w+1 = 3w^2+2w+2 = ...
They all are equal (since w=(-1 + i*sqrt(3))/2) but at the same time
they all have different coefficients as polynomials of w.
For some (unknown to us) reason, you chose the first representation
that gave the 1st term of your sequence equal 0 (as the coefficient of
w^0 in b(1)=w^2). But you chose b(1)=2w^2+w+1, the first term would be
1; if you chose b(1)=3w^2+2w+2, it would be 2, etc.

There is yet another way to define your sequence precisely. This time
let's forget the exact value of w and keep in mind only that w^3 = 1.
Then every element of the sequence b(0) = w, b(1) = w^2, b(n) =
w*b(n-1) + b(n-2) have unique representation as a 2-nd degree
polynomial of w:
b(2) = -x + 1
b(3) = -2*x^2 + x
b(4) = x^2 + x - 3
b(5) = 3*x^2 - 4*x + 1
b(6) = -5*x^2 + 6
b(7) = -3*x^2 + 10*x - 6
b(8) = 15*x^2 - 6*x - 9
...
This gives us the following sequence of coefficients of w^0:
0, 0, 1, 0, -3, 1, 6, -6, -9
But it is different from your original sequence (again).

Max





From franktaw at netscape.net  Fri Nov 10 00:40:10 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 09 Nov 2006 18:40:10 -0500
Subject: Incorrect conjecture
Message-ID: <8C8D26F876C6EFF-76C-A121@WEBMAIL-MA17.sysops.aol.com>

The conjecture in http://www.research.att.com/~njas/sequences/A092903 
is wrong.  This conjecture is that A092903 is not the same as A005153.  
Translated, the opposite of this conjecture is that if every number up 
to m can be expressed as the sum of distinct divisors of m, then every 
number up to sigma(m) can be.  (The latter condition is known as m is 
practical.)

Following the link from A005153 to 
http://planetmath.org/encyclopedia/PracticalNumber.html, we find the 
lemma (paraphrased for non-graphical presentation): an integer m >= 2 
with factorization Product_{i=1}^k p_i^e_i with the p_i in ascending 
order is practical if and only if p_1 = 2 and, for 1 < i <= k, p_i <= 
sigma(Product_{j < i} p_j^e_j) + 1.

Proof: Let m =  Product_{i=1}^k p_i^e_i  be a number not of this form.  
If p_1 != 2, 2 cannot be represented as a sum.  If p_i > 
sigma(Product_{j<i} p_j^e_j) + 1, then sigma(Product_{j<i} p_j^e_j) + 1 
cannot be represented as the sum, and since this number is less than 
p_i, it is less than m.  Hence not every number up to m can be 
expressed as the sum of distinct divisors of m.   Thus if every number 
up to m is expressible as the sum of distinct divisors, so is every 
number up to sigma(m).  The converse is trivial, so the conditions are 
equivalent.

Hence A092903 is a duplicate of A005153.

Franklin T. Adams-Watters

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From davidwwilson at comcast.net  Fri Nov 10 03:06:32 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 9 Nov 2006 21:06:32 -0500
Subject: Explanation of A123937
Message-ID: <002f01c7046c$d7f7f880$6401a8c0@yourxhtr8hvc4p>

This is a hurried attempt to provide background on the genesis of A123937.  I am sure that a most of what I write here is a poor rehash of basic theory, and this whole post probably equates to a few lines of generating function theory.

For d and n in Z+, let A_d(n) be the set of integer points on the d-sphere of radius sqrt(n) centered at the origin, that is:

    [1] a_d(n) = #({P in Z^d : |P|^2 = n}).

A table of a_d(n) for a few small values of d and n:

Table of a_d(n)

 \d|  0    1    2    3    4    5     6     7      8      9      10
 n\|
---+--------------------------------------------------------------
 0 |  1    1    1    1    1    1     1     1      1      1       1
 1 |  0    2    4    6    8   10    12    14     16     18      20
 2 |  0    0    4   12   24   40    60    84    112    144     180
 3 |  0    0    0    8   32   80   160   280    448    672     960
 4 |  0    2    4    6   24   90   252   574   1136   2034    3380
 5 |  0    0    8   24   48  112   312   840   2016   4320    8424
 6 |  0    0    0   24   96  240   544  1288   3136   7392   16320
 7 |  0    0    0    0   64  320   960  2368   5504  12672   28800
 8 |  0    0    4   12   24  200  1020  3444   9328  22608   52020
 9 |  0    2    4   30  104  250   876  3542  12112  34802   88660
10 |  0    0    8   24  144  560  1560  4424  14112  44640  129064

Each sequence a_d is a column of this table. All of these column sequences already exist in the OEIS.

It turns out that a_d can be gotten by convolving the sequence j = (1,2,0,0,2,...) = A000122 d times with the convolutional identity sequence e = (1,0,0,0,0,...) = A000007. We could say that a_d = j^d as a "convolutional power".

The generating function of a_d is t(x)^d where t is the generating function of j. t turns out to be the Jacobi theta function theta_3, as noted on A000122. If we apply the finite difference transform to the sequence of generating functions {t(x)^d}, we obtain the sequence {(t(x)-1)^d}. Expanding these functions back to sequences gives the sequence of sequences {b_d} where b_d = (j-e)^d as a convolutional power. This allows us to constuct a table of b_d:

Table of b_d(n)

 \d|  0    1    2    3    4    5    6    7    8    9    10
 n\|
---+------------------------------------------------------
 0 |  1    0    0    0    0    0    0    0    0    0     0
 1 |  0    2    0    0    0    0    0    0    0    0     0
 2 |  0    0    4    0    0    0    0    0    0    0     0
 3 |  0    0    0    8    0    0    0    0    0    0     0
 4 |  0    2    0    0   16    0    0    0    0    0     0
 5 |  0    0    8    0    0   32    0    0    0    0     0
 6 |  0    0    0   24    0    0   64    0    0    0     0
 7 |  0    0    0    0   64    0    0  128    0    0     0
 8 |  0    0    4    0    0  160    0    0  256    0     0
 9 |  0    2    0   24    0    0  384    0    0  512     0
10 |  0    0    8    0   96    0    0  896    0    0  1024

It turns out that the process we used to get from the a_d table to the b_d table, namely, converting columns of a_d to generating functions, applying the finite difference transform to the gf's, and expanding the resulting functions back to columns of b_d, is the same as applying the finite difference transform to the rows of the a_d table to obtain rows of the b_d table.

An inductive proof starting with b_d = (j-e)^d where j-e = (0,2,0,0,2,...) shows that b_d(d) = 2^d and b_d(n) = 0 for n < d. This means that the nth element of the nth row of the b_d table is the last nonzero element in that row, which in turn implies that the nth row of the a_d table, that is, the number of integer points P with |P|^2 = n, is a polynomial of degree n in d.

If, instead of [1], we had started with

    [1] a_d'(n) = #({P in Z^d : |P|^2 <= n})

which counts points inside as well as on the d-sphere, the nth row of a_d' table would be gotten by adding termwise the rows 0 through n of the a_d table. The sequence a_d' has generating function t(x)^d/(1-x). Similarly, the nth row of the corresponding b_d' table is the sum of rows 0 through n of the b_d table, and sequence b_d has generating function (t(x)-1)^d)/(1-x). The table of b_d', omitting the superdiagonal zeroes, is sequence A123937.

The recurrence given in the title of A123937 follows from the relation

    b_d = (1,1,1,1,1,...) = A000012 if d = 0; (j-e) * b_{d-1}' if d >= 1.

I know this is a spotty exposition, but I'm sure the loose ends could be tied together by a real number theorist.

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From zakseidov at yahoo.com  Sat Nov 11 14:53:18 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sat, 11 Nov 2006 05:53:18 -0800 (PST)
Subject: A052530: to be edited
Message-ID: <20061111135318.71618.qmail@web38206.mail.mud.yahoo.com>

Both %F's and %t don't accord with entries %S 
(mainly with first term of %S),
and note this misprint(?):
"a(n)-4*a(n+1)+a(n+2)})"

%S A052530
1,2,8,30,112,418,1560,5822,21728,81090,302632,

%F A052530 a(0) = c, a(1) = p*c^3; a(n+2) =
p*c^2*a(n+1) - a(n), for p = 1, c = 2.

%F A052530 Recurrence: {a(0)=1, a(1)=2, a(2)=8, 
a(n)-4*a(n+1)+a(n+2)}

%t A052530 a[0] = c; a[1] = p*c^3; a[n_] := a[n] =
p*c^2*a[n - 1] - a[n - 2]; p = 1; c = 2; 
Table[ a[n], {n, 0, 20} ]

Anyone may wish to fix it? 
Zak


 
____________________________________________________________________________________
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From davidwwilson at comcast.net  Sat Nov 11 16:34:08 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sat, 11 Nov 2006 10:34:08 -0500
Subject: A124057 is subset of A045940
References: <20061105235118.90653.qmail@web38203.mail.mud.yahoo.com>
Message-ID: <004501c705a6$d4c0bea0$6501a8c0@yourxhtr8hvc4p>

Nah, there's lots of them:

Yah, lots:

8706123 24463374 32442848 32942943 36782289 48580623 55486248 57476573
59600365 59757774 62481222 62664810 62884590 63262374 63728124 64724373
65159575 65450824 69362487 70302087 70370223 70785924 71494773 72060272
72503682 73256910 73638422 74066874 74361858 75356070 75414702 78676623
83349123 84405123 88094895 92349423 93803982 96192782 97757955 97822374
98474661 99104550 100708374 100997575 101302623 101905622 101905623
102956103 105596334 110476143 111673374 112529493 112801974 113067422
114811332 114940375 115338894 115466874 116079774 118269150 119076174
120919623 123052110 123821970 123952542 124045207 124232373 124465624
124581534 126115623 126134874 126619623 127151127 128095749 129079248
129764789 131404623 131422575 132568225 132617814 133891623 133918134
134350623 134421174 135691143 138468069 142680174 142904430 142904431
145863150 148286886 151757655 152779624 153101583 153129248 153930510
154322523 155740023 155826423 158252994


----- Original Message ----- 
From: "zak seidov" <zakseidov at yahoo.com>
To: <seqfan at ext.jussieu.fr>
Sent: Sunday, November 05, 2006 6:51 PM
Subject: Re: A124057 is subset of A045940


> Just submitted:
> 
> My Q to seqfans:
> Really, there are no numbers n such that n, n+1, n+2
> and n+3 are products of 6 primes?
> No for n<7258290.
> Thanks, Zak






From njas at research.att.com  Sat Nov 11 22:52:30 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sat, 11 Nov 2006 16:52:30 -0500 (EST)
Subject: querying multidimensional sequences in OEIS
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
    <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
    <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <200611112152.QAA93259@fry.research.att.com>

Max, the other day you said:


Of course, there are many representations possible but one of them
should be called "standard", and every multidimensional sequence
should have this standard representation (among maybe others) in OEIS.

Neil, does that sound reasonable?


Me:  it is very hard to force a standard representation - there are
too many people who use the OEIS and don't spend much time
reading rules

even such a simple thing as the "offset" and "keywords" gives
a lot of people trouble.

in short, it would not work

Neil





From njas at research.att.com  Sun Nov 12 04:40:14 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sat, 11 Nov 2006 22:40:14 -0500 (EST)
Subject: Thanks to Tony Noe.
Message-ID: <200611120340.WAA18989@fry.research.att.com>


Seqfans:

in the past few weeks Tony Noe has corrected several hundred entries 
in the OEIS.  the kinds of errors he has found include numbers
broken by commas, numbers run together, dropped digits, duplicated 
numbers, and just plain wrong entries.

It is astonishing to me that there were so many errors.

And it suggests that there must be thousands of other errors
waiting to be found.

I would like to thank him for all his work.   Incidentally he
has also added hundreds of b-files.

Right now I am in the middle of a large batch of updates - there
will be a new version of the OEIS by dawn.

Tony:  thanks!


Neil





From neoneye at gmail.com  Sun Nov 12 08:27:28 2006
From: neoneye at gmail.com (Simon Strandgaard)
Date: Sun, 12 Nov 2006 08:27:28 +0100
Subject: growing binary trees
Message-ID: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>

I made up an interesting sequence:

1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 26, 33, 41,
50, 62, 77, 94, 115, 142, 174, 212, 260, 319,
389, 475, 582, 711, 867, 1060, 1296


Its a binary tree where leafs is getting
added when a nodes internal counter reaches 3.





below is my slow ruby program:

class Node
  def initialize
    @n = 1
    @childs = []
  end
  def count
    @childs.inject(@n){|n,child| n + child.count}
  end
  def grow
    return @n += 1 if @n < 3
    @childs.each{|child| child.grow }
    @childs << Node.new if @childs.size < 2
  end
end

result = []
node = Node.new
30.times do |i|
  result << node.count
  node.grow
end
p result







# btw: the grow method below outputs A000071: Fibonacci numbers - 1.
def grow
  return @n += 1 if @n < 1
  @childs.each{|child| child.grow }
  @childs << Node.new if @childs.size < 2
end


# btw: the grow method below outputs A054405: Row sums of array T as in A055215.
def grow
  return @n += 1 if @n < 2
  @childs.each{|child| child.grow }
  @childs << Node.new if @childs.size < 2
end




--
Simon Strandgaard





From ralf at ark.in-berlin.de  Sun Nov 12 10:09:02 2006
From: ralf at ark.in-berlin.de (Ralf Stephan)
Date: Sun, 12 Nov 2006 10:09:02 +0100
Subject: growing binary trees
In-Reply-To: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
Message-ID: <20061112090902.GA1205@ark.in-berlin.de>

> I made up an interesting sequence:
> 
> 1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 26, 33, 41,
> 50, 62, 77, 94, 115, 142, 174, 212, 260, 319,
> 389, 475, 582, 711, 867, 1060, 1296

According to superseeker:

                                     2      3      4
                        6 + 2 x + 2 x  - 3 x  - 4 x
                     [- ----------------------------, ogf]
                                5            3
                              -x  - 1 + x + x

Second cumulative sums of:

%I A017817
%S A017817 1,0,0,1,1,0,1,2,1,1,3,3,2,4,6,5,6,10,11,11,16,21,22,27,37,43,49,64,80,
%T A017817 92,113,144,172,205,257,316,377,462,573,693,839,1035,1266,1532,1874,
%U A017817 2301,2798,3406,4175,5099,6204,7581,9274,11303,13785,16855,20577,25088
%N A017817 a(0)=1, a(1)=a(2)=0, a(3)=1; a(n)=a(n-3)+a(n-4).
%F A017817 G.f.: 1/(1-x^3-x^4).
%F A017817 a(n)/a(n-1) tends to r = 1.2207440846..., a real root of x^4 - x - 1 = 0. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Oct 22 2006

First sums of:

Transformation T018 gave a match with:
%I A122511
%S A122511 1,1,1,2,2,2,3,4,4,5,7,8,9,12,15,17,21,27,32,38,48,59,70,86,107,129,156,
%T A122511 193,236,285,349,429,521,634,778,950,1155,1412,1728,2105,2567,3140,3833,
%U A122511 4672,5707,6973,8505,10379,12680,15478
%N A122511 Alternative method for A079398 using vector matrirx Markov.
%K A122511 nonn,uned
%A A122511 Roger Bagula (rlbagula(AT)sbcglobal.net), Sep 16 2006

which is IDENTICAL with

%I A079398
%S A079398 0,1,1,1,1,2,2,2,3,4,4,5,7,8,9,12,15,17,21,27,32,38,48,59,70,86,107,129,
%T A079398 156,193,236,285,349,429,521,634,778,950,1155,1412,1728,2105,2567,3140,
%U A079398 3833,4672,5707,6973,8505,10379,12680,15478,18884,23059,28158,34362
%N A079398 Conjectured values of d(n), the dimension of a Z-module in MZV(conv).
%C A079398 From the conjecture of Zagier, Drinfeld, Kontsevich and Goncharov (see link).
%C A079398 P(0)=P(1)=P(2)=P(3)=1, for m>3: P(m) = P(m-3) + P(m-4) is the 3rd sequence in the series: Fibonacci sequence, Padovan sequence, ... The Padovan sequence (whose ratio of successive terms approaches the plastic constant) is similar to the Perrin sequence. - Jonathan Vos Post (jvospost2(AT)yahoo.com), Jan 23 2005
%D A079398 Michel Waldschmidt, "Multiple Zeta values and Euler-Zagier numbers", in Number theory and discrete mathematics, International conference in honour of Srinivasa Ramanujan, Center for Advanced Study in Mathematics, Panjab University, Chandigarh, (Oct 02, 2000).
%H A079398 Michel Waldschmidt, <a href="http://www.institut.math.jussieu.fr/~miw/articles/pdf/Transparents.pdf">Multiple Zeta values and Euler-Zagier numbers</a>
%H A079398 Eric W. Weisstein, <a href="http://mathworld.wolfram.com/PadovanSequence.html">Padovan Sequence</a>.
%F A079398 a(1)=0 a(2)=a(3)=a(4)=1 for n>=4 a(n)=a(n-2)+a(n-3)
%F A079398 a(n)=sum{k=0..floor((n-1)/2), binomial(floor((n-k-1)/3), k)} (offset 0); a(n)=sum{k=0..floor(n/2), binomial(floor((n-k-1)/3), k)}-0^n. (offset 0). - Paul Barry (pbarry(AT)wit.ie), Jul 06 2004
%F A079398 For n>1, a(n) = P(n-2) where P(n) is defined by: P(0)=P(1)=P(2)=P(3)=1, for m>3: P(m) = P(m-3) + P(m-4). - Jonathan Vos Post (jvospost2(AT)yahoo.com), Jan 23 2005
%Y A079398 Cf. A000931.
%K A079398 nonn
%O A079398 1,6
%A A079398 Benoit Cloitre (abmt(AT)wanadoo.fr), Feb 16 2003


ralf





From neoneye at gmail.com  Sun Nov 12 16:54:56 2006
From: neoneye at gmail.com (Simon Strandgaard)
Date: Sun, 12 Nov 2006 16:54:56 +0100
Subject: growing binary trees
In-Reply-To: <20061112090902.GA1205@ark.in-berlin.de>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
	 <20061112090902.GA1205@ark.in-berlin.de>
Message-ID: <df1390cc0611120754n68a94a44k76bc07615c3bdb98@mail.gmail.com>

On 11/12/06, Ralf Stephan <ralf at ark.in-berlin.de> wrote:
> > I made up an interesting sequence:
> >
> > 1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 26, 33, 41,
> > 50, 62, 77, 94, 115, 142, 174, 212, 260, 319,
> > 389, 475, 582, 711, 867, 1060, 1296
>
> According to superseeker:
[snip]

I don't see how its already in oeis?

How is A017817, A122511, A079398 related to it?
There is no match between my numbers and their numbers.
There doesn't seem to be any constant I can add to
make them match. Im an amateur.


--
Simon Strandgaard





From ralf at ark.in-berlin.de  Sun Nov 12 17:18:14 2006
From: ralf at ark.in-berlin.de (Ralf Stephan)
Date: Sun, 12 Nov 2006 17:18:14 +0100
Subject: growing binary trees
In-Reply-To: <20061112090902.GA1205@ark.in-berlin.de>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com> <20061112090902.GA1205@ark.in-berlin.de>
Message-ID: <20061112161814.GA11423@ark.in-berlin.de>

Simon, sorry for the cryptic reply.

In other words:

Let n>=0, and your sequence a(n) = 1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 
26, 33, 41, 50, 62, 77, 94, 115, 142, 174, 212, 260, 319, ...

then 

A079398(n) = a(n-2) - a(n-1) 
A122511(n) = a(n+1) - a(n)      
A017817(n) = A122511(n+3) - A122511(n+2)

All of these sequences satisfy the recurrence
a(n) = a(n-2) + a(n-3) + c

and are thus related to the Padovan sequence A000931.
With your sequence, c appears to be 3 but this has
to be proved, and all of the above is only a conjecture,
based on solid evidence, though.


ralf







From neoneye at gmail.com  Sun Nov 12 17:30:14 2006
From: neoneye at gmail.com (Simon Strandgaard)
Date: Sun, 12 Nov 2006 17:30:14 +0100
Subject: growing binary trees
In-Reply-To: <20061112161814.GA11423@ark.in-berlin.de>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
	 <20061112090902.GA1205@ark.in-berlin.de>
	 <20061112161814.GA11423@ark.in-berlin.de>
Message-ID: <df1390cc0611120830u79249ff5ue1f9c588cde122d2@mail.gmail.com>

On 11/12/06, Ralf Stephan <ralf at ark.in-berlin.de> wrote:
[snip]
> All of these sequences satisfy the recurrence
> a(n) = a(n-2) + a(n-3) + c
>
> and are thus related to the Padovan sequence A000931.
> With your sequence, c appears to be 3 but this has
> to be proved, and all of the above is only a conjecture,
> based on solid evidence, though.
[snip]

I get it. Thanks for the explaination :-)

--
Simon Strandgaard





From njas at research.att.com  Sun Nov 12 17:38:06 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sun, 12 Nov 2006 11:38:06 -0500 (EST)
Subject: growing binary trees
Message-ID: <200611121638.LAA08650@fry.research.att.com>

I will add Simon S.'s sequence to the OEIS - it will be A123015 
in a few minutes

Can Ruby lines be combined onto one line?

NJAS





From njas at research.att.com  Sun Nov 12 18:22:45 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sun, 12 Nov 2006 12:22:45 -0500 (EST)
Subject: growing binary trees
Message-ID: <200611121722.MAA17909@fry.research.att.com>

About this sequence (it will be A123015):


1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 26, 33, 41,
50, 62, 77, 94, 115, 142, 174, 212, 260, 319,
389, 475, 582, 711, 867, 1060, 1296


Its a binary tree where leafs is getting
added when a nodes internal counter reaches 3.

- I really don't understand the definition.  What exactly is the
n-th term?

Neil Sloane





From neoneye at gmail.com  Sun Nov 12 19:10:23 2006
From: neoneye at gmail.com (Simon Strandgaard)
Date: Sun, 12 Nov 2006 19:10:23 +0100
Subject: growing binary trees
In-Reply-To: <200611121638.LAA08650@fry.research.att.com>
References: <200611121638.LAA08650@fry.research.att.com>
Message-ID: <df1390cc0611121010r2cf66135gfa52aff18c944540@mail.gmail.com>

On 11/12/06, N. J. A. Sloane <njas at research.att.com> wrote:
> I will add Simon S.'s sequence to the OEIS - it will be A123015
> in a few minutes
>
> Can Ruby lines be combined onto one line?


I cannot express this thing as math, and its difficult for me to
compact the code without loss of readability.


below is a 5 line version:

class Node; def initialize; @n = 1; @c = [] end
def count; @c.inject(@n){|n,c| n + c.count} end
def grow; return @n += 1 if @n < 3; @c.each{|c| c.grow }
@c << Node.new if @c.size < 2; end; end; r = []; node = Node.new
30.times { r << node.count; node.grow }; p r


here is the oneliner (less readable):

class C;def initialize;@n=1;@c=[] end;def n;@c.inject(@n){|n,c|n+c.n}
end;def g;return @n+=1 if @n<3; @c.each{|c|c.g};@c<<C.new if
@c.size<2;end;end;r=[];c=C.new;30.times{r<<c.n;c.g};p r



the n-th term is:
a(0)=1, a(1)=2, a(2)=3, a(3)=4, a(4)=6, a(5)=8, a(6)=10, a(7)=13, ...


code for n'th term is below:

class Node
 def initialize
   @n = 1
   @childs = []
 end
 def count
   @childs.inject(@n){|n,child| n + child.count}
 end
 def grow
   return @n += 1 if @n < 3
   @childs.each{|child| child.grow }
   @childs << Node.new if @childs.size < 2
 end
end

def a(n)
  node = Node.new
  n.times{ node.grow }
  node.count
end

0.upto(30) {|n| p a(n) }



visually it works like this:

step#0
  1


step#1
  2


step#2
  3


step#3
  3
 /
1


step#4
  3
 / \
2   1


step#5
  3
 / \
3   2


step#6
    3
   / \
  3   3
 /
1


step#7
      3
    /   \
  3       3
 / \     /
2   1   1


step#8
      3
    /   \
  3       3
 / \     / \
3   2   2   1



--
Simon Strandgaard





From maxale at gmail.com  Mon Nov 13 07:53:16 2006
From: maxale at gmail.com (Max A.)
Date: Sun, 12 Nov 2006 22:53:16 -0800
Subject: growing binary trees
In-Reply-To: <20061112090902.GA1205@ark.in-berlin.de>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
	 <20061112090902.GA1205@ark.in-berlin.de>
Message-ID: <d3dac270611122253m4930fb58qdf5d06ce4fbf8026@mail.gmail.com>

On 11/12/06, Ralf Stephan <ralf at ark.in-berlin.de> wrote:

> %I A122511
> %S A122511 1,1,1,2,2,2,3,4,4,5,7,8,9,12,15,17,21,27,32,38,48,59,70,86,107,129,156,
> %T A122511 193,236,285,349,429,521,634,778,950,1155,1412,1728,2105,2567,3140,3833,
> %U A122511 4672,5707,6973,8505,10379,12680,15478
> %N A122511 Alternative method for A079398 using vector matrirx Markov.
> %K A122511 nonn,uned
> %A A122511 Roger Bagula (rlbagula(AT)sbcglobal.net), Sep 16 2006
>
> which is IDENTICAL with
>
> %I A079398
> %S A079398 0,1,1,1,1,2,2,2,3,4,4,5,7,8,9,12,15,17,21,27,32,38,48,59,70,86,107,129,
> %T A079398 156,193,236,285,349,429,521,634,778,950,1155,1412,1728,2105,2567,3140,
> %U A079398 3833,4672,5707,6973,8505,10379,12680,15478,18884,23059,28158,34362
> %N A079398 Conjectured values of d(n), the dimension of a Z-module in MZV(conv).
> %C A079398 From the conjecture of Zagier, Drinfeld, Kontsevich and Goncharov (see link).
> %C A079398 P(0)=P(1)=P(2)=P(3)=1, for m>3: P(m) = P(m-3) + P(m-4) is the 3rd sequence in the series: Fibonacci sequence, Padovan sequence, ... The Padovan sequence (whose ratio of successive terms approaches the plastic constant) is similar to the Perrin sequence. - Jonathan Vos Post (jvospost2(AT)yahoo.com), Jan 23 2005
> %D A079398 Michel Waldschmidt, "Multiple Zeta values and Euler-Zagier numbers", in Number theory and discrete mathematics, International conference in honour of Srinivasa Ramanujan, Center for Advanced Study in Mathematics, Panjab University, Chandigarh, (Oct 02, 2000).
> %H A079398 Michel Waldschmidt, <a href="http://www.institut.math.jussieu.fr/~miw/articles/pdf/Transparents.pdf">Multiple Zeta values and Euler-Zagier numbers</a>
> %H A079398 Eric W. Weisstein, <a href="http://mathworld.wolfram.com/PadovanSequence.html">Padovan Sequence</a>.
> %F A079398 a(1)=0 a(2)=a(3)=a(4)=1 for n>=4 a(n)=a(n-2)+a(n-3)
> %F A079398 a(n)=sum{k=0..floor((n-1)/2), binomial(floor((n-k-1)/3), k)} (offset 0); a(n)=sum{k=0..floor(n/2), binomial(floor((n-k-1)/3), k)}-0^n. (offset 0). - Paul Barry (pbarry(AT)wit.ie), Jul 06 2004
> %F A079398 For n>1, a(n) = P(n-2) where P(n) is defined by: P(0)=P(1)=P(2)=P(3)=1, for m>3: P(m) = P(m-3) + P(m-4). - Jonathan Vos Post (jvospost2(AT)yahoo.com), Jan 23 2005
> %Y A079398 Cf. A000931.
> %K A079398 nonn
> %O A079398 1,6
> %A A079398 Benoit Cloitre (abmt(AT)wanadoo.fr), Feb 16 2003

Should not then A122511 be marked as a duplicate of A079398 with an
appropriate comment?
I do not see much sense in keeping them separate.

Max





From davidwwilson at comcast.net  Mon Nov 13 18:50:32 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Mon, 13 Nov 2006 12:50:32 -0500
Subject: A102567 = A106498
Message-ID: <000601c7074c$375f4d50$6501a8c0@yourxhtr8hvc4p>

A102567 and A106498 are duplicates and should be merged. The latter sequence has an associated b-file.
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From maxale at gmail.com  Mon Nov 13 19:24:29 2006
From: maxale at gmail.com (Max A.)
Date: Mon, 13 Nov 2006 10:24:29 -0800
Subject: A050259 and a new solution to 2^n = 3 (mod n)
Message-ID: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>

SeqFans,

In course of exploring all solutions of the congruence 2^n = 3 (mod n)
below 10^16, I have found a new solution:

n = 3468371109448915

This came as a big surprise to me since my original goal was to prove
that 8365386194032363 is the second term of A050259. But now it
appears to be at least the third term.

So far I've submitted the new solution as a comment to A050259, but
hopefully it will take its true place in this sequence soon.

Regards,
Max

P.S. See also http://mathworld.wolfram.com/2.html





From rkg at cpsc.ucalgary.ca  Mon Nov 13 20:15:20 2006
From: rkg at cpsc.ucalgary.ca (Richard Guy)
Date: Mon, 13 Nov 2006 12:15:20 -0700 (MST)
Subject: A050259 and a new solution to 2^n = 3 (mod n)
In-Reply-To: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>
References: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>
Message-ID: <Pine.LNX.4.63.0611131211270.24385@csl>

Congratulations!  May I know Max's full name,
so that he can earn undying fame in UPINT as
well as in OEIS ?   Thanks,   R.

On Mon, 13 Nov 2006, Max A. wrote:

> SeqFans,
>
> In course of exploring all solutions of the congruence 2^n = 3 
> (mod n)
> below 10^16, I have found a new solution:
>
> n = 3468371109448915
>
> This came as a big surprise to me since my original goal was 
> to prove
> that 8365386194032363 is the second term of A050259. But now 
> it
> appears to be at least the third term.
>
> So far I've submitted the new solution as a comment to 
> A050259, but
> hopefully it will take its true place in this sequence soon.
>
> Regards,
> Max
>
> P.S. See also http://mathworld.wolfram.com/2.html






From jvospost3 at gmail.com  Mon Nov 13 21:34:20 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Mon, 13 Nov 2006 12:34:20 -0800
Subject: A050259 and a new solution to 2^n = 3 (mod n)
In-Reply-To: <Pine.LNX.4.63.0611131211270.24385@csl>
References: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>
	 <Pine.LNX.4.63.0611131211270.24385@csl>
Message-ID: <5542af940611131234l54af8bcar76fcba8e87d0fb1b@mail.gmail.com>

Is this a hierarchy of fame, or a graph, or directed graph, or a lattice, or
what? What percentage of "nice" sequences are referenced in each successive
edition of UPINT (Unsolved problems in Number Theory)? How many OEIS seqs
per edition of UPINT, now that the two are more tightly interlinked? What
estimates does one make to extrapolate for future editions? Should there be
a "very nice" keyword for seqs that pose problems in UPINT, when they get
solved in later editions?

Only hypothetical for me, as I've had over 1400 seqs in OEIS and not one is
yet "nice", although some are comments on "nice" seqs. So if there is a
hierarchy of fame, I'm just above the lowest  level (or highest, depending
on indexing) in quality, which is most assuredly not correlated well with
quantity.  In Science Fiction, for instance, there are indeed a hierarchies
of fame, based on awards (Hugo, Nebula), coauthorships, pay, anthologies,
and the like.  I'm also just somewhat above the bottom, in part for my
coauthorships with Asimov, Bradbury, Clarke, and Heinlein.  I've done
research on Asimov Number as the Science Fiction equivalent of Erdos Number,
but that's another story.

By the way, I have the penultimate edition (so far) of UPINT as a gift from
the wonderful Tony Noe, when he bought the ultimate edition (so far). I echo
the applause by njas for all the hundreds of errors in OEIS found and
corrected by Tony, and the many many b-files.

Best,

Jonathan Vos Post

On 11/13/06, Richard Guy <rkg at cpsc.ucalgary.ca> wrote:
>
> Congratulations!  May I know Max's full name,
> so that he can earn undying fame in UPINT as
> well as in OEIS ?   Thanks,   R.
>
> On Mon, 13 Nov 2006, Max A. wrote:
>
> > SeqFans,
> >
> > In course of exploring all solutions of the congruence 2^n = 3
> > (mod n)
> > below 10^16, I have found a new solution:
> >
> > n = 3468371109448915
> >
> > This came as a big surprise to me since my original goal was
> > to prove
> > that 8365386194032363 is the second term of A050259. But now
> > it
> > appears to be at least the third term.
> >
> > So far I've submitted the new solution as a comment to
> > A050259, but
> > hopefully it will take its true place in this sequence soon.
> >
> > Regards,
> > Max
> >
> > P.S. See also http://mathworld.wolfram.com/2.html
>
>
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From r.rosenthal at web.de  Mon Nov 13 23:12:57 2006
From: r.rosenthal at web.de (Rainer Rosenthal)
Date: Mon, 13 Nov 2006 23:12:57 +0100
Subject: A050259 and a new solution to 2^n = 3 (mod n)
In-Reply-To: <Pine.LNX.4.63.0611131211270.24385@csl>
References: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com> <Pine.LNX.4.63.0611131211270.24385@csl>
Message-ID: <4558EDE9.2080404@web.de>

Richard Guy wrote:
> Congratulations!  May I know Max's full name,
> so that he can earn undying fame in UPINT as
> well as in OEIS ?   Thanks,   R.
> 
> On Mon, 13 Nov 2006, Max A. wrote:
>>In course of exploring all solutions of the congruence 2^n = 3 
>>(mod n) below 10^16, I have found a new solution:
>>
>>n = 3468371109448915
>>P.S. See also http://mathworld.wolfram.com/2.html

This is a lovely finding, which I will have to tell
immediately in the German Newsgroup de.sci.mathematik.

Congratulations from a friend of OEIS and UPINT and
of things that are not too difficult to comprehend :-)
Not only congratulations to Max for finding, but to all
of us for being witnesses.

Best regards,
Rainer Rosenthal
r.rosenthal at web.de






From franktaw at netscape.net  Tue Nov 14 00:01:52 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 13 Nov 2006 18:01:52 -0500
Subject: A102567 = A106498
In-Reply-To: <000601c7074c$375f4d50$6501a8c0@yourxhtr8hvc4p>
References: <000601c7074c$375f4d50$6501a8c0@yourxhtr8hvc4p>
Message-ID: <8C8D58ED7DCDA34-128-9CD7@FWM-R16.sysops.aol.com>

A116148 (n concatenated with n-1 gives the product of two numbers which 
differ by 2) is also a duplicate of these two.  The only possible 
exceptions would be numbers of the form 10^n-1; but then the 
concatenation is 10^{2n}+10^n-1, and 10^{2n}+10^n is never a square.

Franklin T. Adams-Watters


-----Original Message-----
From: davidwwilson at comcast.net

A102567 and A106498 are duplicates and should be merged.?The latter 
sequence has an associated b-file.




________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From maxale at gmail.com  Tue Nov 14 00:52:58 2006
From: maxale at gmail.com (Max A.)
Date: Mon, 13 Nov 2006 15:52:58 -0800
Subject: A050259 and a new solution to 2^n = 3 (mod n)
In-Reply-To: <Pine.LNX.4.63.0611131211270.24385@csl>
References: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>
	 <Pine.LNX.4.63.0611131211270.24385@csl>
Message-ID: <d3dac270611131552v64b33034n8ed43b34faf905a4@mail.gmail.com>

I answered Richard privately.
But if other people also want to refer to my full name, it is
Max Alekseyev

Regards,
Max

P.S. It is not a conspiracy but rather my way to beat spamming. So
far, spam emails never addressed me by my full name, and I can easily
recognize important emails that do contain my full name in the To:
field.

On 11/13/06, Richard Guy <rkg at cpsc.ucalgary.ca> wrote:
> Congratulations!  May I know Max's full name,
> so that he can earn undying fame in UPINT as
> well as in OEIS ?   Thanks,   R.
>
> On Mon, 13 Nov 2006, Max A. wrote:
>
> > SeqFans,
> >
> > In course of exploring all solutions of the congruence 2^n = 3
> > (mod n)
> > below 10^16, I have found a new solution:
> >
> > n = 3468371109448915
> >
> > This came as a big surprise to me since my original goal was
> > to prove
> > that 8365386194032363 is the second term of A050259. But now
> > it
> > appears to be at least the third term.
> >
> > So far I've submitted the new solution as a comment to
> > A050259, but
> > hopefully it will take its true place in this sequence soon.
> >
> > Regards,
> > Max
> >
> > P.S. See also http://mathworld.wolfram.com/2.html
>
>





From bowerc at usa.net  Tue Nov 14 02:10:04 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Mon, 13 Nov 2006 17:10:04 -0800
Subject: growing binary trees
Message-ID: <397kkNBJe5074S28.1163466604@cmsweb28.cms.usa.net>

 From Simon's diagrams, I can see this is analogous to Fibonacci's
rabbits. Basically, the behavior of the node is given by its age.
A node at age 0 or 1 grows and one at age 2 or 3 produces a new node.

 From this its easy to get a g.f. of

(1+x+x^2)/(1-x-x^3+x^5)

1 2 3 4 6 8 10 13 17 21 26 33 41 50 62 77 94 115 142 174 212 260 319 389
475 582 711 867 1060 1296 1581 1930 2359 2880 3514 4292 5242 6397 7809
9537

Also, the number of nodes has g.f.

1/(1-x-x^3+x^5)
1 1 1 2 3 3 4 6 7 8 11 14 16 20 26 31 37 47 58 69 85 106 128 155 192 235
284 348 428 520 633 777 949 1154 1411 1727 2104 2566 3139 3832

which isn't in the EIS either.

------ Original Message ------
From: "Simon Strandgaard" <neoneye at gmail.com>
To: njas at research.att.comCc: seqfan at ext.jussieu.fr
Subject: Re: growing binary trees

...

> visually it works like this:
> 
> step#0
>   1
> 
> 
> step#1
>   2
> 
> 
> step#2
>   3
> 
> 
> step#3
>   3
>  /
> 1
> 
> 
> step#4
>   3
>  / \
> 2   1
> 
> 
> step#5
>   3
>  / \
> 3   2
> 
> 
> step#6
>     3
>    / \
>   3   3
>  /
> 1
> 
> 
> step#7
>       3
>     /   \
>   3       3
>  / \     /
> 2   1   1
> 
> 
> step#8
>       3
>     /   \
>   3       3
>  / \     / \
> 3   2   2   1
> 
> 
> 
> --
> Simon Strandgaard
> 









From zbi74583 at boat.zero.ad.jp  Tue Nov 14 06:50:02 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Tue, 14 Nov 2006 14:50:02 +0900
Subject: A036471
Message-ID: <013d01c707b0$bedd9660$d967d7dc@FMC57937B6BE00>

    Hi, Seqfans

    An exhaustive search of A036471  with a computer up to 10^7 is almost 
impossible.

    So, using my algorithm, I searched small examples which are ?regular 
type?.



    See this site, there is an explanation of "type".



              http://amicable.homepage.dk/apstat.htm#typesys







    I found these examples.



    2^9*3^2*13*31*(5*11,71)*(7*23,191) 
11 digits

    2^15*3*5^2*11*31*257*(1439,19*71,23*59,29*47) 
16 digits

    2^3*3^4*5^2*(7*23,191)*(17*19,359) 
9 digits

    (2^3*19*41,2^5*199)*(3*5*7*13,3^2*5*7*139) 
8 digits

    (2^3*19*41,2^5*199)*(3^3*5*7*71,3^3*5*17*31) 
9 digits

    (2^3*19*41,2^5*199)*(3^2*7*13*5*17,3^2*7*13*107) 
9 digits

    2^7*3^2*13*(5*11,71)*(7*59,479) 
9 digits

    2^5*3^4*5^2*(17*19,359)*(23*29,719) 
11 digits







    Where, x*(y,z) means (x*z,x*y). (x,y)*(z,u)=(x*z,x*u,y*z,y*z)





    I think that if a smaller one than 3270960  exists, then it must be 
?irregular".



    Regurer type Amicable Quadruple are generated from ?seeds? which are 
two pairs or one quadruple of primes or almost primes, {x,y} and {z,u} or 
{x,y,z,u}.

    The numbers x,y,z,u of small Amicable Quadruple must satisfy the 
following conditions.

             .

    o Sigma(x)=Sigma(y) and Sigma(z)=Sigma(u), or 
Sigma(x)=Sigma(y)=Sigma(z)=Sigma(u)

    o all prime factors of x,y,z,u are small

    o all prime factors of x+y and z+u are small, or all  prime factors of 
x+y+z+u are small



    For example, if {x,y}={7*23,191} and {z,u}={17*19,359} then

    o Sigma(7*23)=Sigma(191), Sigma(17*19)=Sigma(359)

    o all prime factors of 7*23,191,17*19,359  are small

    o x+y=2^5*11, z+u=2*11*31

    So, (7*23,191)*(17*19,359) generates 2^3*3^4*5^2*(7*23,191)*(17*19,359)





    If we obtain all small seeds then we will calculate all small regular 
terms of A036471.



    Yasutoshi


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From g.resta at iit.cnr.it  Tue Nov 14 14:40:49 2006
From: g.resta at iit.cnr.it (Giovanni Resta)
Date: Tue, 14 Nov 2006 14:40:49 +0100
Subject: A036471
In-Reply-To: <013d01c707b0$bedd9660$d967d7dc@FMC57937B6BE00>
References: <013d01c707b0$bedd9660$d967d7dc@FMC57937B6BE00>
Message-ID: <4559C761.70305@iit.cnr.it>

kohmoto wrote:
>
> Hi, Seqfans
>
> An exhaustive search of A036471 with a computer up to 10^7 is almost
> impossible.
>
Here it is the result of an exhaustive search
with a computer up to 10^7, in the format
sigma -> {a,b,c,d} (with a,b,c,d<10^7)

13927680 -> {3270960, 3361680, 3461040, 3834000}
16248960 -> {3767400, 4090320, 4150440, 4240800}
19498752 -> {4651920, 4839120, 4918320, 5089392}
21228480 -> {4969440, 5116320, 5475960, 5666760}
24373440 -> {5682600, 5831280, 5920200, 6939360}
24998400 -> {5405400, 6029100, 6421800, 7142100}
27855360 -> {6514200, 6640200, 6768720, 7932240}
28304640 -> {6126120, 6541920, 7559640, 8076960}
28304640 -> {6126120, 6922080, 7280280, 7976160}
29030400 -> {6320160, 6902280, 7685496, 8122464}
29030400 -> {6977880, 7087080, 7227360, 7738080}
29877120 -> {7013160, 7436520, 7688520, 7738920}
29998080 -> {6819120, 7327320, 7715400, 8136240}
29998080 -> {6966960, 7054320, 7840560, 8136240}
32497920 -> {7706160, 7722000, 7948080, 9121680}
33022080 -> {7731360, 7852320, 8125920, 9312480}
33868800 -> {7469280, 8157240, 8873760, 9368520}
33868800 -> {7469280, 8157240, 9098460, 9143820}

Note that there are values which admit more than one quadruple,
and there are also quadruples which share 1 or 2 common values.

Here are some larger quadruples, but now there can be some
missing among these, because here I used an heuristic.

{8353800, 8920800, 10063200, 10159800}
{8288280, 9360540, 9830520, 10260180}
{8316000, 8759520, 10260180, 10403820}
{9258480, 9621360, 9661680, 10455984}
{9009000, 10174500, 10521000, 11179980}
{10048500, 10174500, 10188360, 10473120}
{9840600, 10098000, 10914120, 10930320}
{9923760, 10316880, 10747440, 10794960}
{9563400, 10085040, 11241720, 11964240}
{9646560, 10667160, 11502540, 11729340}
{9767520, 11202840, 11217960, 11357280}
{9828000, 10796940, 11362680, 12041820}
{10417680, 10538640, 10979280, 13061520}
{10450440, 11614680, 11802420, 13306860}
{10450440, 12179160, 12204360, 12340440}

etc.etc. (program is running).

Giovanni Resta






From davidwwilson at comcast.net  Tue Nov 14 18:00:27 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Tue, 14 Nov 2006 12:00:27 -0500
Subject: A023153 through A023161
References: <p06230918c17f1b479bcf@[192.168.1.43]>
Message-ID: <006d01c7080e$632ff480$6501a8c0@yourxhtr8hvc4p>

Tony Noe has found that A023153 through A023161 are not multiplicative as I 
had marked them.

At the time, I was convinced that in sequence {s_k(n) = k^(2^n)}, a cycle of 
period x (mod a) and a cycle of period y (mod b) implied a unique cycle of 
period xy (mod ab). However, if gcd(x, y) > 1, the cycles (mod a) and (mod 
b) can align in more than one way, producing more than one cycle (mod ab). 
For example, the cycle (2,4) (mod 7) and the cycle (4,7) (mod 9) produce two 
cycles (4,16) and (25,58) (mod 63). This means that sometimes f(ab) > 
f(a)f(b).

This problem affects A023153 through A023161, and Tony Noe has shown that 
indeed none of these are multiplicative.

NJAS: Please remove the "mult" keyword and any comments related to 
multiplicativity from A023153-A023161.

Tony: Since these sequences are almost multiplicative, but are not, and you 
seem to be investigating these sequences, could you please add lines such as

%C A023153 Not multiplicative: a(63) = 10, but a(7)*a(9) = 3*3 = 9.

Thanks for spotting my error, Tony. Seems I have been bitten by the dragon I 
released.

----- Original Message ----- 
From: "T. D. Noe" <noe at sspectra.com>
To: <davidwwilson at comcast.net>
Sent: Tuesday, November 14, 2006 2:06 AM
Subject: Question about A023161


> David, according to my calculations, this sequence fails to be
> multiplicative at a(667).
>
> Best regards,
>
> Tony
>
>
> -- 
> No virus found in this incoming message.
> Checked by AVG Free Edition.
> Version: 7.5.430 / Virus Database: 268.14.5/533 - Release Date: 11/13/2006 
> 8:56 PM
>
> 






From jvospost3 at gmail.com  Tue Nov 14 20:46:00 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Tue, 14 Nov 2006 11:46:00 -0800
Subject: A023153 through A023161
In-Reply-To: <006d01c7080e$632ff480$6501a8c0@yourxhtr8hvc4p>
References: <p06230918c17f1b479bcf@192.168.1.43>
	 <006d01c7080e$632ff480$6501a8c0@yourxhtr8hvc4p>
Message-ID: <5542af940611141146j1a3212b0u2d31815c6bee8387@mail.gmail.com>

"almost multiplicative" -- interesting notion.  What other "almost
multiplicative" are there in OEIS masquerading as multiplicative (i'm
guessing that the amazing Tony Noe is searching), and what are but with no
keyword on that? Can "almost multiplicative" be axiomatizeed, or is it just
a subjective interpretation of "almost"? What pattens are there in "almost
multiplicative" -- new sequences on where multiplicativity breaks down?

-- Jonathan Vos Post

On 11/14/06, David Wilson <davidwwilson at comcast.net> wrote:
>
> Tony Noe has found that A023153 through A023161 are not multiplicative as
> I
> had marked them.
>
> At the time, I was convinced that in sequence {s_k(n) = k^(2^n)}, a cycle
> of
> period x (mod a) and a cycle of period y (mod b) implied a unique cycle of
> period xy (mod ab). However, if gcd(x, y) > 1, the cycles (mod a) and (mod
> b) can align in more than one way, producing more than one cycle (mod ab).
> For example, the cycle (2,4) (mod 7) and the cycle (4,7) (mod 9) produce
> two
> cycles (4,16) and (25,58) (mod 63). This means that sometimes f(ab) >
> f(a)f(b).
>
> This problem affects A023153 through A023161, and Tony Noe has shown that
> indeed none of these are multiplicative.
>
> NJAS: Please remove the "mult" keyword and any comments related to
> multiplicativity from A023153-A023161.
>
> Tony: Since these sequences are almost multiplicative, but are not, and
> you
> seem to be investigating these sequences, could you please add lines such
> as
>
> %C A023153 Not multiplicative: a(63) = 10, but a(7)*a(9) = 3*3 = 9.
>
> Thanks for spotting my error, Tony. Seems I have been bitten by the dragon
> I
> released.
>
> ----- Original Message -----
> From: "T. D. Noe" <noe at sspectra.com>
> To: <davidwwilson at comcast.net>
> Sent: Tuesday, November 14, 2006 2:06 AM
> Subject: Question about A023161
>
>
> > David, according to my calculations, this sequence fails to be
> > multiplicative at a(667).
> >
> > Best regards,
> >
> > Tony
> >
> >
> > --
> > No virus found in this incoming message.
> > Checked by AVG Free Edition.
> > Version: 7.5.430 / Virus Database: 268.14.5/533 - Release Date:
> 11/13/2006
> > 8:56 PM
> >
> >
>
>
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From pauldhanna at juno.com  Wed Nov 15 05:04:51 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Tue, 14 Nov 2006 23:04:51 -0500
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
Message-ID: <20061114.230452.536.0.pauldhanna@juno.com>

Seqfans, 
    Could someone compute the initial terms of the following. 
I do not believe that it is in the OEIS (?). 
  
Number of labelled nodes in generation n of a rooted tree 
where a node with label k has k child nodes such that 
the label of each child node is the least unused label in 
the path from the root to that child node, 
and where root is labeled '1'. 
  
Thus every distinct path in the tree eventually forms 
a permutation of the positive integers. 
 
The wording needs to be made more precise and concise. 
 
Here are the initial nodes of the tree for generations 0..4: 
Gen 0: [1];
Gen 1: (1)->[2];
Gen 2: (2)->[3,4];
Gen 3: (3)->[4,5,6], (4)->[3,5,6,7];
Gen 4: (4)->[5,6,7,8], (5)->[4,6,7,8,9], (6)->[4,5,7,8,9,10], 
(3)->[5,6,7], (5)->[3,6,7,8,9], (6)->[3,5,7,8,9,10], 
(7)->[3,5,6,8,9,10,11]; ...
 
Thus, the number of nodes in generation n begins:
[1, 1, 2, 7, 36, 248, ...]
 
The maximum node in generation n begins:
[1, 2, 4, 7, 11, ...]
 
Would appreciate it if someone could compute more terms.
Thanks,
      Paul





From franktaw at netscape.net  Wed Nov 15 05:13:17 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Tue, 14 Nov 2006 23:13:17 -0500
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
In-Reply-To: <20061114.230452.536.0.pauldhanna@juno.com>
References: <20061114.230452.536.0.pauldhanna@juno.com>
Message-ID: <8C8D68382FC1E11-FEC-53E1@FWM-M20.sysops.aol.com>

You need to reword that a bit.  Your example makes clear that the
children of a given node must all have distinct labels, but you don't
say so.

Without that condition, the problem is pretty easy.  :-)

Franklin T. Adams-Watters


-----Original Message-----
From: pauldhanna at juno.com

...

Number of labelled nodes in generation n of a rooted tree
where a node with label k has k child nodes such that
the label of each child node is the least unused label in
the path from the root to that child node,
and where root is labeled '1'.

...
________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From njas at research.att.com  Wed Nov 15 06:13:02 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 15 Nov 2006 00:13:02 -0500 (EST)
Subject: king chickens
Message-ID: <200611150513.AAA96507@fry.research.att.com>

can anyone extend this?

%I A123553
%S A123553 1,2,12,104
%N A123553 A "king chicken" in a tournament graph (a directed labeled graph on n nodes) is a player A who for any other player B either beats B directly or beats someone who beats B.  Sequence gives total number of king chickens in all 2^(n(n-1)/2) tourna
%C A123553 H. G. Landau showed in 1951 that a player is a king chicken if and only if he has the highest score. There may be several king chickens in a tournament.
%D A123553 S. B. Maurer, The king chicken theorems, Math. Mag., 53 (1980), 67-80.
%e A123553 For n = 3 there are 8 tournaments: six of the form A beats B and C, and B beats C, with one king chicken (A), and two of the form A beats B beats C beats A, with three king chickens each (A or B or C), for a total of 6*1 + 2*3 = 12.
%K A123553 nonn,more,new
%O A123553 1,2
%A A123553 njas, Nov 14 2006

Neil





From bdm at cs.anu.edu.au  Wed Nov 15 06:37:18 2006
From: bdm at cs.anu.edu.au (Brendan McKay)
Date: Wed, 15 Nov 2006 16:37:18 +1100
Subject: king chickens
In-Reply-To: <200611150513.AAA96507@fry.research.att.com>
References: <200611150513.AAA96507@fry.research.att.com>
Message-ID: <20061115053717.GA15829@cs.anu.edu.au>


I don't have time right now, so if someone who knows how to use nauty
wants to jump into this, here is how to do it:
1. Fetch the unlabelled tournaments from
      http://cs.anu.edu.au/~bdm/data/digraphs.html
2. For each unlabelled tournament compute the automorphism group size A
   and the number of king chickens K.
3. The answer is the sum of K*(n!/A) over the unlabelled tournaments.

That will get the answer up to n=10 fairly easily.  The next case n=11
is plausible but more effort as there are nearly a billion tournaments.

Incidentally, the definition of "tournament" in the entry is incomplete.
It has to add that exactly one of the directed edges u->v, v->u is
present for each pair u,v of distinct vertices.

Brendan.


* N. J. A. Sloane <njas at research.att.com> [061115 16:13]:
> can anyone extend this?
> 
> %I A123553
> %S A123553 1,2,12,104
> %N A123553 A "king chicken" in a tournament graph (a directed labeled graph on n nodes) is a player A who for any other player B either beats B directly or beats someone who beats B.  Sequence gives total number of king chickens in all 2^(n(n-1)/2) tourna
> %C A123553 H. G. Landau showed in 1951 that a player is a king chicken if and only if he has the highest score. There may be several king chickens in a tournament.
> %D A123553 S. B. Maurer, The king chicken theorems, Math. Mag., 53 (1980), 67-80.
> %e A123553 For n = 3 there are 8 tournaments: six of the form A beats B and C, and B beats C, with one king chicken (A), and two of the form A beats B beats C beats A, with three king chickens each (A or B or C), for a total of 6*1 + 2*3 = 12.
> %K A123553 nonn,more,new
> %O A123553 1,2
> %A A123553 njas, Nov 14 2006
> 
> Neil





From g.resta at iit.cnr.it  Wed Nov 15 11:32:25 2006
From: g.resta at iit.cnr.it (Giovanni Resta)
Date: Wed, 15 Nov 2006 11:32:25 +0100
Subject: A question on A068480 
Message-ID: <455AECB9.5050308@iit.cnr.it>

A068480 is the sequence on numbers n such that gcd(n!-1,2^n+1)>1.

I noticed that for all the numbers listed gcd(n!-1,2^n+1) = 2n+1.

Is this a rule (proof ?) or a coincidence (counterexample ?) .

thanks,
giovanni.







From mathar at strw.leidenuniv.nl  Wed Nov 15 13:40:44 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Wed, 15 Nov 2006 13:40:44 +0100
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
Message-ID: <200611151240.kAFCeiKc009825@amer.strw.leidenuniv.nl>


pdh> From seqfan-owner at ext.jussieu.fr  Wed Nov 15 05:08:51 2006
pdh> To: seqfan at ext.jussieu.fr
pdh> Date: Tue, 14 Nov 2006 23:04:51 -0500
pdh> Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
pdh> From: "Paul D. Hanna" <pauldhanna at juno.com>
pdh> ...
pdh> Number of labelled nodes in generation n of a rooted tree 
pdh> where a node with label k has k child nodes such that 
pdh> the label of each child node is the least unused label in 
pdh> the path from the root to that child node, 
pdh> and where root is labeled '1'. 
pdh>   
pdh> Thus every distinct path in the tree eventually forms 
pdh> a permutation of the positive integers. 
pdh>  
pdh> The wording needs to be made more precise and concise. 
pdh>  
pdh> Here are the initial nodes of the tree for generations 0..4: 
pdh> Gen 0: [1];
pdh> Gen 1: (1)->[2];
pdh> Gen 2: (2)->[3,4];
pdh> Gen 3: (3)->[4,5,6], (4)->[3,5,6,7];
pdh> Gen 4: (4)->[5,6,7,8], (5)->[4,6,7,8,9], (6)->[4,5,7,8,9,10], 
pdh> (3)->[5,6,7], (5)->[3,6,7,8,9], (6)->[3,5,7,8,9,10], 
pdh> (7)->[3,5,6,8,9,10,11]; ...
pdh>  
pdh> Thus, the number of nodes in generation n begins:
pdh> [1, 1, 2, 7, 36, 248, ...]
pdh>  
pdh> The maximum node in generation n begins:
pdh> [1, 2, 4, 7, 11, ...]

My count of node numbers and maximum nodes is
Gen 0 1 1
Gen 1 1 2
Gen 2 2 4
Gen 3 7 7
Gen 4 36 11
Gen 5 248 16
Gen 6 2157 22
Gen 7 22761 29
Gen 8 283220 37
Gen 9 4068411 46

This list is limited by the simplicity of my program which keeps all the
trees in memory at the same time and runs into allocation problems if the
tenth generations are attempted.





From pauldhanna at juno.com  Wed Nov 15 14:34:15 2006
From: pauldhanna at juno.com (Paul D Hanna)
Date: Wed, 15 Nov 2006 13:34:15 GMT
Subject: Trees with Labeled Nodes that form Permutation of Positive Integer
	s
Message-ID: <20061115.053511.1712.764552@webmail42.nyc.untd.com>

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From franktaw at netscape.net  Wed Nov 15 18:34:02 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 15 Nov 2006 12:34:02 -0500
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
In-Reply-To: <20061114.230452.536.0.pauldhanna@juno.com>
References: <20061114.230452.536.0.pauldhanna@juno.com>
Message-ID: <8C8D6F3606C258E-3A4-7467@FWM-R09.sysops.aol.com>

From: pauldhanna at juno.com

>    Could someone compute the initial terms of the following.
>I do not believe that it is in the OEIS (?).
>
>Number of labelled nodes in generation n of a rooted tree
>where a node with label k has k child nodes such that
>the label of each child node is the least unused label in
>the path from the root to that child node,
excluding those used by earlier children of the same parent,
>and where root is labeled '1'.
>
>Thus every distinct path in the tree eventually forms
>a permutation of the positive integers.
>
>...

This isn't really true.  There are many paths in this tree
that do not form a permutation of the positive integers.
You get such a permutation iff, infinitely often, you choose
the first child from the current node.

It is true, of course, that any finite initial path can be
extended to a path that is a permuatation of the positive
integers.

It is an interesting problem to construct an interesting tree
where every path from the root is a permutation of the
positive integers.  I think the following construction works:
start with node 1, which arbitrarily has node 2 as a child
(by the construction that follows, 1 would have no children).
For each node with label m, add BigOmega(m) children, where
BigOmega(m) is the number of prime factors of m with
repetition; label these chilidren with the first positive integers
not occurring in the path from the root to the current node.

The tree starts:

......1
......|
......2
......|
......3
......|
......4
...../.\
....5...6
.../....|\
..6.....5.7
./.\....|.|
7...8...7.5
|../|\..|.|
8.7.9.A.8.8

Any time you hit a prime, there is only one child in the next
generation.  If you keep choosing non-prime children, the
primes start accumulating in the list of positive integers not
occurring in the path, until eventually all children of a node will
be prime.

One could of course use Omega(m), the number of distinct
prime divisors, instead of BigOmega(m).  I think tau(m)-1 would
also work.  A001511(m) (the power of 2 dividing 2m) works, too
- in this case, the single children are under the odd nodes.
________________________________________________________________________
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From qq-quet at mindspring.com  Wed Nov 15 19:23:30 2006
From: qq-quet at mindspring.com (Leroy Quet)
Date: Wed, 15 Nov 06 11:23:30 -0700
Subject: Max Score For A Game
Message-ID: <E1GkPUS-0005H1-00@pop04.mail.atl.earthlink.net>

Below is a game I invented (as posted to sci.math and rec.puzzles -- it 
is just one of many stupid silly simple games I have posted there).

I post it here to seq.fans because I wonder what the sequence is where 
the nth term is the maximum possible score for an n-by-n grid. (We could 
also have a table where the element in the mth column and nth row is the 
maximum score for an m-by-n grid.)
Of course, if there is a way to always get the score n^2 -1 for each 
n-by-n case, or m*n -1 for each m-by-n case (n or m equal 1 excluded), 
then there is be no need to calculate the sequence.

Thanks,
Leroy Quet

=====

Here is a solitaire game played on a square grid.
(I hope this game is more fun than my last game.)

Start by drawing an r-by-r grid on paper.
(I suggest an r of 5 or so for beginners.)
 
Put a "1" in each square of the left-most column
and in each square of the top-most row of the grid.

On each move the player chooses any one of the grid's
empty squares. This square is (m,n), which is the
square in the mth column from the left side of the
grid and in the nth row from the top.

The player then sums up all the integers already
written in the mth column and nth row.
So, if the grid looks like this, where the * is
the square the player has chosen to fill in next
with an integer, 
1 1 1 1
1 3
1 * 6
1 6

then the sum of the column 2's terms and row 3's
terms is 1+3+6 +1+6 = 17. Let this sum be s.

Next the player counts the number of integers in the
squares to the left of (m,n) and above (m,n)
(ie, the squares with coordinates (j,k), 1 <=j <=m,
1 <=k <=n) which are coprime to s.
If the count is c integers in the given rectangle
which are coprime to s, then the player writes c
in square (m,n).

Play continues until there is an integer in every
square of the grid.
The player's score is the number in the last square
he/she fills in.

Example game:
(r=4)
(View with fixed-width font.)

1 1 1 1   1 1 1 1   1 1 1 1   1 1 1 1 
1         1 3       1 3       1 3 5  
1         1         1   6     1   6
1         1         1         1

1 1 1 1   1 1 1 1   1 1 1 1
1 3 5 6   1 3 5 6   1 3 5 6
1   6     1   6     1 5 6
1         1 6       1 6

1 1 1 1   1 1 1 1   1 1 1 1
1 3 5 6   1 3 5 6   1 3 5 6
1 5 6     1 5 6     1 5 6 11
1 6   10  1 6 1110  1 6 1110

So the player gets 11 points.

(It seems that a good strategy is to
{unlike in my example} finish in the
lower right square, and to {as in my
example, 1+6+10+1+5+6=29} have a row
and column sum on the last move which
is a prime.)

Question: Is it always possible to get,
for an r-by-r grid (r >= 2), r^2 -1 points?

Thanks,
Leroy Quet





From davidwwilson at comcast.net  Wed Nov 15 19:48:18 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Wed, 15 Nov 2006 13:48:18 -0500
Subject: Domino questions
Message-ID: <001501c708e6$9e07a570$6501a8c0@yourxhtr8hvc4p>

1. On an n x n (m x n) checkerboard, what is the smallest number of dominoes that can be placed so that no further dominoes can be placed?

2. Suppose two players, A and B, alternately place dominoes on an n x n (m x n) checkerboard until no further play is possible. A plays to maximize the final number of dominoes on the board, B to minimize. What is the number of dominoes on the board at the end of the game if A moves first? If B moves first?
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From franktaw at netscape.net  Wed Nov 15 21:01:19 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 15 Nov 2006 15:01:19 -0500
Subject: Domino questions
In-Reply-To: <001501c708e6$9e07a570$6501a8c0@yourxhtr8hvc4p>
References: <001501c708e6$9e07a570$6501a8c0@yourxhtr8hvc4p>
Message-ID: <8C8D707F39EB04F-934-6F47@FWM-R01.sysops.aol.com>

From: davidwwilson at comcast.net
>1. On an n x n (m x n) checkerboard, what is the smallest number
>of dominoes that can be placed so that no further dominoes can
>be placed?

This might be ceiling(n^2/3), A008810.  Can someone please
check the reference there: "J. H. Conway and R. K. Guy, The Book
of Numbers, Copernicus Press, NY, 1996, number of red blocks in
Fig 2.5."
?
>2. Suppose two players, A and B, alternately place dominoes on an
>n x n (m x n) checkerboard until no further play is possible. A plays
>to maximize the final number of dominoes on the board, B to
>minimize. What is the number of dominoes on the board at the end
>of the game if A moves first? If B moves first?

You don't want much, do you?

(Note for anyone who may be confused - that comment means that
I think this is likely to be a hard problem.)

Franklin T. Adams-Watters
________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From njas at research.att.com  Wed Nov 15 23:40:20 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 15 Nov 2006 17:40:20 -0500 (EST)
Subject: request for a b-file for a core file
Message-ID: <200611152240.RAA60703@fry.research.att.com>


Could someone help produce a b-file for A000014?
(and A001678 along the way)

Here is the sequence:

%I A000014 M0320 N0118
%S A000014 0,1,1,0,1,1,2,2,4,5,10,14,26,42,78,132,249,445,842,1561,2988,5671,10981,
%T A000014 21209,41472,81181,160176,316749,629933,1256070,2515169,5049816,10172638,
%U A000014 20543579,41602425,84440886,171794492,350238175,715497037,1464407113
%N A000014 Number of series-reduced trees with n nodes.


I could use 200, or 500 or so terms

Notes on computing A14 and A1678:

looking at the H'book of Graph Theory, p. 525:

they define f(x) (essentially A1678 but with a different offset):
f(x) = 1*x+1*x^3+1*x^4+2*x^5+3*x^6+6*x^7+10*x^8+19*x^9+35*x^10+...

by

f = (x/(x+1))* exp( sum( f(x^i) / i, i=1..infinity ) )

then A14 is given by

(1+x)f(x) - ((1+x)/2) f(x)^2 + ((1-x)/2) f(x^2)

= x + x^2 + x^4 + x^5 + 2 x^6 + ...

= A14

I actually haven't checked these formulae - there are
slightly different versions in the entry for A14 itself.

Thanks!

Neil






From bowerc at usa.net  Thu Nov 16 01:35:58 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 15 Nov 2006 16:35:58 -0800
Subject: request for a b-file for a core file
Message-ID: <964kkPaI79512S23.1163637358@cmsweb23.cms.usa.net>

I just sent a b-file of terms 0-500 to Neil for A001678.
I'm still working on a version for A000014

------ Original Message ------
From: "N. J. A. Sloane" <njas at research.att.com>
To: seqfans at seqfan.net, seqfan at ext.jussieu.frCc: njas at research.att.com
Subject: request for a b-file for a core file

> 
> Could someone help produce a b-file for A000014?
> (and A001678 along the way)
> 
> Here is the sequence:
> 
> %I A000014 M0320 N0118
> %S A000014
0,1,1,0,1,1,2,2,4,5,10,14,26,42,78,132,249,445,842,1561,2988,5671,10981,
> %T A000014
21209,41472,81181,160176,316749,629933,1256070,2515169,5049816,10172638,
> %U A000014
20543579,41602425,84440886,171794492,350238175,715497037,1464407113
> %N A000014 Number of series-reduced trees with n nodes.
> 
> 
> I could use 200, or 500 or so terms
> 
> Notes on computing A14 and A1678:
> 
> looking at the H'book of Graph Theory, p. 525:
> 
> they define f(x) (essentially A1678 but with a different offset):
> f(x) = 1*x+1*x^3+1*x^4+2*x^5+3*x^6+6*x^7+10*x^8+19*x^9+35*x^10+...
> 
> by
> 
> f = (x/(x+1))* exp( sum( f(x^i) / i, i=1..infinity ) )
> 
> then A14 is given by
> 
> (1+x)f(x) - ((1+x)/2) f(x)^2 + ((1-x)/2) f(x^2)
> 
> = x + x^2 + x^4 + x^5 + 2 x^6 + ...
> 
> = A14
> 
> I actually haven't checked these formulae - there are
> slightly different versions in the entry for A14 itself.
> 
> Thanks!
> 
> Neil
> 
> 









From rkg at cpsc.ucalgary.ca  Thu Nov 16 01:50:35 2006
From: rkg at cpsc.ucalgary.ca (Richard Guy)
Date: Wed, 15 Nov 2006 17:50:35 -0700 (MST)
Subject: Domino questions
In-Reply-To: <8C8D707F39EB04F-934-6F47@FWM-R01.sysops.aol.com>
References: <001501c708e6$9e07a570$6501a8c0@yourxhtr8hvc4p>
 <8C8D707F39EB04F-934-6F47@FWM-R01.sysops.aol.com>
Message-ID: <Pine.LNX.4.63.0611151749420.25970@csl>

1.  Checked!  Unfortunately, in the printing,
red and orange are barely distinguishable.   R.

On Wed, 15 Nov 2006, franktaw at netscape.net wrote:

> From: davidwwilson at comcast.net
>> 1. On an n x n (m x n) checkerboard, what is the smallest number
>> of dominoes that can be placed so that no further dominoes can
>> be placed?
>
> This might be ceiling(n^2/3), A008810.  Can someone please
> check the reference there: "J. H. Conway and R. K. Guy, The Book
> of Numbers, Copernicus Press, NY, 1996, number of red blocks in
> Fig 2.5."
> ?
>> 2. Suppose two players, A and B, alternately place dominoes on an
>> n x n (m x n) checkerboard until no further play is possible. A plays
>> to maximize the final number of dominoes on the board, B to
>> minimize. What is the number of dominoes on the board at the end
>> of the game if A moves first? If B moves first?
>
> You don't want much, do you?
>
> (Note for anyone who may be confused - that comment means that
> I think this is likely to be a hard problem.)
>
> Franklin T. Adams-Watters
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading 
> spam and email virus protection.
>

From bowerc at usa.net  Thu Nov 16 02:01:10 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 15 Nov 2006 17:01:10 -0800
Subject: request for a b-file for a core file
Message-ID: <888kkPBak9480S24.1163638870@cmsweb24.cms.usa.net>

I just sent the b-file for A000014

Christian

------ Original Message ------
From: "Christian G. Bower" <bowerc at usa.net>
To: "seqfan" <seqfan at ext.jussieu.fr>
Subject: Re: request for a b-file for a core file

> I just sent a b-file of terms 0-500 to Neil for A001678.
> I'm still working on a version for A000014
> 








From zbi74583 at boat.zero.ad.jp  Thu Nov 16 04:45:46 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Thu, 16 Nov 2006 12:45:46 +0900
Subject: A036471
Message-ID: <011301c70931$b92d7500$ca73d7dc@FMC57937B6BE00>

    Thanks, Giovanni Resta.



    The smallest term!!

    It is great!

       .

    Now human being became to know much about A036471.       

    The first term is 3270960 and known last term is       

   

    



    

    

    n=32583657,    19.6million digits.



    Yasutoshi

    


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From maxale at gmail.com  Thu Nov 16 10:48:46 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 16 Nov 2006 01:48:46 -0800
Subject: A question on A068480
In-Reply-To: <455AECB9.5050308@iit.cnr.it>
References: <455AECB9.5050308@iit.cnr.it>
Message-ID: <d3dac270611160148q5f52517gbc511cc7b9fd042c@mail.gmail.com>

It is unlikely for n!-1 and 2^n+1 to have a common prime factor
(unless in a special case described below), especially since
1) all prime factors of n!-1 are greater than n;
2) all prime factors of 2^n+1 are of the form p=2kq+1 where q is a
divisor of n, and the multiplicative of 2 modulo p divides 2q.

The special case is when n is even and p=2n+1 is prime. In this case,
p divides both n!-1 and 2^n+1, and thus gcd(n!-1,2^n+1). And it is
very likely that gcd(n!-1,2^n+1) = p.

Overall, the statement "if gcd(n!-1,2^n+1)>1 then
gcd(n!-1,2^n+1)=2n+1" sounds plausible and I verified it for n up to
45,000. But it may be very hard to give a proof.

Max

On 11/15/06, Giovanni Resta <g.resta at iit.cnr.it> wrote:
> A068480 is the sequence on numbers n such that gcd(n!-1,2^n+1)>1.
>
> I noticed that for all the numbers listed gcd(n!-1,2^n+1) = 2n+1.
>
> Is this a rule (proof ?) or a coincidence (counterexample ?) .
>
> thanks,
> giovanni.
>
>
>





From maxale at gmail.com  Thu Nov 16 11:49:25 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 16 Nov 2006 02:49:25 -0800
Subject: A question on A068480
In-Reply-To: <d3dac270611160148q5f52517gbc511cc7b9fd042c@mail.gmail.com>
References: <455AECB9.5050308@iit.cnr.it>
	 <d3dac270611160148q5f52517gbc511cc7b9fd042c@mail.gmail.com>
Message-ID: <d3dac270611160249k774410b5p3b07705ed642b8ae@mail.gmail.com>

On 11/16/06, Max A. <maxale at gmail.com> wrote:

> The special case is when n is even and p=2n+1 is prime. In this case,
> p divides both n!-1 and 2^n+1, and thus gcd(n!-1,2^n+1). And it is
> very likely that gcd(n!-1,2^n+1) = p.

Oh, my bad! Correct statement:

The special case is when n is of the form 4k+1 and p=2n+1 is prime
(implying that 2 and -1 are non-squares modulo p), and p also belongs
to A058302.
In this case, p divides both n!-1 and 2^n+1, and thus gcd(n!-1,2^n+1).
And it is very likely that gcd(n!-1,2^n+1) = p.

The hypothesis about gcd(n!-1,2^n+1) implies that:
8k+3 belong to A058302 if and only if 4k+1 belong to A068480.

Max





From davidwwilson at comcast.net  Thu Nov 16 17:47:04 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 16 Nov 2006 11:47:04 -0500
Subject: b-files for power series expansions
Message-ID: <000601c7099e$d8c5e880$6501a8c0@yourxhtr8hvc4p>

There is a seqload of OEIS sequences that are expansions of real functions, which I guess means power series coefficients. An example is A000810.  Would Mma or some other math power tool be able to quickly generate b-files for these monsters?
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From mathar at strw.leidenuniv.nl  Thu Nov 16 18:44:15 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Thu, 16 Nov 2006 18:44:15 +0100
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
Message-ID: <200611161744.kAGHiFGX005987@amer.strw.leidenuniv.nl>


pdh> From seqfan-owner at ext.jussieu.fr  Wed Nov 15 05:08:51 2006
pdh> Return-Path: <seqfan-owner at ext.jussieu.fr>
pdh> To: seqfan at ext.jussieu.fr
pdh> Date: Tue, 14 Nov 2006 23:04:51 -0500
pdh> Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
pdh> From: "Paul D. Hanna" <pauldhanna at juno.com>
pdh> ...
pdh> Number of labelled nodes in generation n of a rooted tree 
pdh> where a node with label k has k child nodes such that 
pdh> the label of each child node is the least unused label in 
pdh> the path from the root to that child node, 
pdh> and where root is labeled '1'. 
pdh>   ...
pdh> Here are the initial nodes of the tree for generations 0..4: 
pdh> Gen 0: [1];
pdh> Gen 1: (1)->[2];
pdh> Gen 2: (2)->[3,4];
pdh> Gen 3: (3)->[4,5,6], (4)->[3,5,6,7];
pdh> Gen 4: (4)->[5,6,7,8], (5)->[4,6,7,8,9], (6)->[4,5,7,8,9,10], 
pdh> (3)->[5,6,7], (5)->[3,6,7,8,9], (6)->[3,5,7,8,9,10], 
pdh> (7)->[3,5,6,8,9,10,11]; ...
pdh>  
pdh> Thus, the number of nodes in generation n begins:
pdh> [1, 1, 2, 7, 36, 248, ...]
pdh>  
pdh> The maximum node in generation n begins:
pdh> [1, 2, 4, 7, 11, ...]
pdh>  
pdh> Would appreciate it if someone could compute more terms.
pdh> Thanks,
pdh>       Paul

rjm> My count of node numbers and maximum nodes is
rjm> Gen 0 1 1
rjm> Gen 1 1 2
rjm> Gen 2 2 4
rjm> Gen 3 7 7
rjm> Gen 4 36 11
rjm> Gen 5 248 16
rjm> Gen 6 2157 22
rjm> Gen 7 22761 29
rjm> Gen 8 283220 37
rjm> Gen 9 4068411 46

The maximum node label is trivially A000124 because the maximum node will be
generated from the child with itself the highest label, and this adds 'n' to
the maximum label for each new generation. This produces the arithmetic sum
sequence that is shown above as the last number in each row.

For those who have some spare computer cycles and Maple: below is a program
that runs recursively ("memory friendly") through the entire tree up to 
some maximum generation 'maxgen' (to be set/edited in the fourth but last line
below, maxgen=10 equivalent to Gen up to 9 above). It prints a film of
progresses made to count the nodes found so far at all levels, and should show
all numbers in the list monotonically increasing slowly towards
[1, 1, 2, 7, 36, 248, 2157, 22761, 283220, 4068411, 66367834, ...]
                                                    ^^^ 66367834 is Gen 10
I can also provide my older (but non-recursive, memory-limited) C++ version.

Maple V:

gen := proc(parents,maxgen,ocounts,lvl)
	local thislbl,lbl,childlbl,counts,npar ;
	counts := ocounts ;
	counts[lvl] := counts[lvl]+1 ;
	if nops(parents) < maxgen then
		thislbl := op(-1,parents) ;
		childlbl := 1 ;
		for lbl from 1 to thislbl do
			while ( childlbl in parents ) or ( childlbl = thislbl ) do
				childlbl := childlbl+1 ;
			od ;
			npar := [op(parents),childlbl] ;
			# this would generate a full list of all paths:
			# print("route",npar) ;
			if nops(counts) < lvl+1 then
				counts := [op(counts),0] ;
			fi ;
			counts := gen(npar,maxgen,counts,lvl+1) ;
			childlbl := childlbl+1 ;
		od ;
	fi ;
	# this produces a snapshot of the counts accumulated so far:
	# if the current node creation level is not too high, print it...
	if lvl <= maxgen -3 then
		print(counts) ;
	fi ;
	RETURN(counts) ;
end:

# Edit the maximum number below...total run time probably grows exponentially
# with the number chosen here.
maxgen := 8 ;
parents := [1,2] ;
n := [1,0] ;
gen(parents,maxgen,n,2) ;
print(%) ;





From franktaw at netscape.net  Thu Nov 16 18:59:28 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 16 Nov 2006 12:59:28 -0500
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
In-Reply-To: <200611161744.kAGHiFGX005987@amer.strw.leidenuniv.nl>
References: <200611161744.kAGHiFGX005987@amer.strw.leidenuniv.nl>
Message-ID: <8C8D7C018493BAF-87C-1422@FWM-M21.sysops.aol.com>

I don't know Maple well enough - or have enough time - to tell: does 
this keep each path separately, or does it keep a count of the number 
of paths with a given set of integers on the path?

The latter approach should be much faster and less memory intensive, 
once you get past the first few generations.  (Probably still 
exponential, though.)

Franklin T. Adams-Watters

-----Original Message-----
From: mathar at strw.leidenuniv.nl

...

Maple V:

gen := proc(parents,maxgen,ocounts,lvl)
    local thislbl,lbl,childlbl,counts,npar ;
    counts := ocounts ;
    counts[lvl] := counts[lvl]+1 ;
    if nops(parents) < maxgen then
        thislbl := op(-1,parents) ;
        childlbl := 1 ;
        for lbl from 1 to thislbl do
            while ( childlbl in parents ) or ( childlbl = thislbl ) do
                childlbl := childlbl+1 ;
            od ;
            npar := [op(parents),childlbl] ;
            # this would generate a full list of all paths:
            # print("route",npar) ;
            if nops(counts) < lvl+1 then
                counts := [op(counts),0] ;
            fi ;
            counts := gen(npar,maxgen,counts,lvl+1) ;
            childlbl := childlbl+1 ;
        od ;
    fi ;
    # this produces a snapshot of the counts accumulated so far:
    # if the current node creation level is not too high, print it...
    if lvl <= maxgen -3 then
        print(counts) ;
    fi ;
    RETURN(counts) ;
end:

# Edit the maximum number below...total run time probably grows 
exponentially
# with the number chosen here.
maxgen := 8 ;
parents := [1,2] ;
n := [1,0] ;
gen(parents,maxgen,n,2) ;
print(%) ;

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From zakseidov at yahoo.com  Thu Nov 16 19:33:20 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Thu, 16 Nov 2006 10:33:20 -0800 (PST)
Subject: b-files for power series expansions
In-Reply-To: <000601c7099e$d8c5e880$6501a8c0@yourxhtr8hvc4p>
Message-ID: <20061116183320.50052.qmail@web38201.mail.mud.yahoo.com>

Coefficients of the series
1, 1, 4, 13/3, 44/3, 242/15, 2408/45, 18581/315
differ from A000810 entries (?!),
Zak

A000810
1,1,8,26,352,1936,38528,297296,7869952,78098176
%N A000810 Expansion of (sin x + cos x)/cos 3x.


--- David Wilson <davidwwilson at comcast.net> wrote:

> There is a seqload of OEIS sequences that are
> expansions of real functions, which I guess means
> power series coefficients. An example is A000810. 
> Would Mma or some other math power tool be able to
> quickly generate b-files for these monsters?



 
____________________________________________________________________________________
Sponsored Link

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From franktaw at netscape.net  Thu Nov 16 20:25:13 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 16 Nov 2006 14:25:13 -0500
Subject: b-files for power series expansions
In-Reply-To: <20061116183320.50052.qmail@web38201.mail.mud.yahoo.com>
Message-ID: <8C8D7CC12CFC338-87C-19DE@FWM-M21.sysops.aol.com>

A000810 is the exponential generating function:

1/0!, 1/1!, 8/2!, 26/3!, 352/4!, ...

Franklin T. Adams-Watters


-----Original Message-----
From: zakseidov at yahoo.com

Coefficients of the series
1, 1, 4, 13/3, 44/3, 242/15, 2408/45, 18581/315
differ from A000810 entries (?!),
Zak

A000810
1,1,8,26,352,1936,38528,297296,7869952,78098176
%N A000810 Expansion of (sin x + cos x)/cos 3x.


--- David Wilson <davidwwilson at comcast.net> wrote:

> There is a seqload of OEIS sequences that are
> expansions of real functions, which I guess means
> power series coefficients. An example is A000810.
> Would Mma or some other math power tool be able to
> quickly generate b-files for these monsters?




_________________________________________________________________________
___________
Sponsored Link

Mortgage rates near 39yr lows.
$420k for $1,399/mo. Calculate new payment!
www.LowerMyBills.com/lre

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From mathar at strw.leidenuniv.nl  Thu Nov 16 20:31:50 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Thu, 16 Nov 2006 20:31:50 +0100
Subject: b-files for power series expansions
Message-ID: <200611161931.kAGJVoHb015119@amer.strw.leidenuniv.nl>


dww> From seqfan-owner at ext.jussieu.fr  Thu Nov 16 17:49:20 2006
dww> From: "David Wilson" <davidwwilson at comcast.net>
dww> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
dww> Subject: b-files for power series expansions
dww> Date: Thu, 16 Nov 2006 11:47:04 -0500
dww> ...
dww> There is a seqload of OEIS sequences that are expansions of real =
dww> functions, which I guess means power series coefficients. An example is =
dww> A000810.  Would Mma or some other math power tool be able to quickly =
dww> generate b-files for these monsters?
dww>...

These are usually taylor series coefficients multiplied by n!
(the E.g.f. given in the definition). Where the functions are even,
the odd terms (coefficients equal to 0) are omitted.

I put some of them into
http://www.strw.leidenuniv.nl/~mathar/progs/b000810.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000813.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000816.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000819.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000822.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000825.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000828.txt

Unrelated are
http://www.strw.leidenuniv.nl/~mathar/progs/b007504.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b049060.txt





From mathar at strw.leidenuniv.nl  Thu Nov 16 22:05:41 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Thu, 16 Nov 2006 22:05:41 +0100
Subject: Re^2: Trees with Labeled Nodes that form Permutation of Positive Integers
Message-ID: <200611162105.kAGL5fkI022493@amer.strw.leidenuniv.nl>


ftaw> From seqfan-owner at ext.jussieu.fr  Thu Nov 16 19:01:33 2006
ftaw> Return-Path: <seqfan-owner at ext.jussieu.fr>
ftaw> To: seqfan at ext.jussieu.fr
ftaw> Subject: Re: Trees with Labeled Nodes that form Permutation of Positive Integers
ftaw> 
ftaw> I don't know Maple well enough - or have enough time - to tell: does 
ftaw> this keep each path separately, or does it keep a count of the number 
ftaw> of paths with a given set of integers on the path?
ftaw> 
ftaw> The latter approach should be much faster and less memory intensive, 
ftaw> once you get past the first few generations.  (Probably still 
ftaw> exponential, though.)
ftaw> 
ftaw> Franklin T. Adams-Watters
ftaw> 

There is no such optimization in the program shown below. For each
node, it descends into all branches/leafs independent of whether the
set of labels in the current node and higher up in the trunk is the same
or not the same as any set seen earlier. The state vector is
only the vector of counts (one number per generation) which is updated
at each level of recursion, and the single list of labels of the
ancestors of the current node.
So yes, there is room for optimization by introducing some memory
in here, certainly at the cost of increasing the machine memory requirement
to remember all the sub-counts of partial tree in some hash area.

The first action to become faster is to return to a compiled language.
Since this here does not involve any type of functionality which needs a number
theoretic package, one would try to exploit the fact that C would typically be a
factor of 5 to 10 faster than Maple. I don't want to appear selfish and
will therefore leave all this as an exercise to others :-)

RJM

rjm> -----Original Message-----
rjm> From: mathar at strw.leidenuniv.nl
rjm> 
rjm> ...
rjm> 
rjm> Maple V:
rjm> 
rjm> gen := proc(parents,maxgen,ocounts,lvl)
rjm>     local thislbl,lbl,childlbl,counts,npar ;
rjm>     counts := ocounts ;
rjm>     counts[lvl] := counts[lvl]+1 ;
rjm>     if nops(parents) < maxgen then
rjm>         thislbl := op(-1,parents) ;
rjm>         childlbl := 1 ;
rjm>         for lbl from 1 to thislbl do
rjm>             while ( childlbl in parents ) or ( childlbl = thislbl ) do
rjm>                 childlbl := childlbl+1 ;
rjm>             od ;
rjm>             npar := [op(parents),childlbl] ;
rjm>             # this would generate a full list of all paths:
rjm>             # print("route",npar) ;
rjm>             if nops(counts) < lvl+1 then
rjm>                 counts := [op(counts),0] ;
rjm>             fi ;
rjm>             counts := gen(npar,maxgen,counts,lvl+1) ;
rjm>             childlbl := childlbl+1 ;
rjm>         od ;
rjm>     fi ;
rjm>     # this produces a snapshot of the counts accumulated so far:
rjm>     # if the current node creation level is not too high, print it...
rjm>     if lvl <= maxgen -3 then
rjm>         print(counts) ;
rjm>     fi ;
rjm>     RETURN(counts) ;
rjm> end:
rjm> 
rjm> # Edit the maximum number below...total run time probably grows 
rjm> exponentially
rjm> # with the number chosen here.
rjm> maxgen := 8 ;
rjm> parents := [1,2] ;
rjm> n := [1,0] ;
rjm> gen(parents,maxgen,n,2) ;
rjm> print(%) ;





From franktaw at netscape.net  Fri Nov 17 00:46:06 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 16 Nov 2006 18:46:06 -0500
Subject: Re^2: Trees with Labeled Nodes that form Permutation of Positive
 Integers
In-Reply-To: <200611162105.kAGL5fkI022493@amer.strw.leidenuniv.nl>
References: <200611162105.kAGL5fkI022493@amer.strw.leidenuniv.nl>
Message-ID: <8C8D7F084CDAF74-554-265F@FWM-M18.sysops.aol.com>

No hashing is necessary.  There are (n-1)! possible sets of labels.  
Each
sorted sequence of labels has an increase of at most k between the kth
and k+1st label.  So you can take the first differences, subtract 1,
reverse, and treat as a base factorial number.  E.g., for 1,2,4,5,9, 
the
differences are 1,2,1,4; subtract one and reverse to get 3010,
3*3!+0*2!+1*1! = 19, so this will be at (zero-based) index 19.  (This 
is
easily reversible to get the label set from the index and level.)

For each label set, you'll need to keep a count of how many nodes
there are with each possible final label, to tell what to generate in
the next level.  Since 1 and 2 can't be labels after level 2, you only
need to keep n-2 such counters for each set.

We can tell that this algorithm is on the order of n! to compute level 
n.
The naive approach is proportional to the size of the nth level, which 
is
certainly greater; since the largest term at level n is C(n,2)+1 ~ 
n^2/2,
it is certainly no more than (n^2/2)!.  I'm guessing that is it more 
like
c^n n!, for some c between 1.6 and 2 (probably closer to 2, from the
data, though convergence is slow; k 2^n n!/n^2 gives a fairly good
fit to the available data).

Based on this, the eventual savings should be much larger than merely
switching to a compiled language.  I'll also leave this as an exercise,
though. :-)

Franklin T. Adams-Watters


-----Original Message-----
From: mathar at strw.leidenuniv.nl

ftaw> From seqfan-owner at ext.jussieu.fr  Thu Nov 16 19:01:33 2006
ftaw>
ftaw> I don't know Maple well enough - or have enough time - to tell: 
does
ftaw> this keep each path separately, or does it keep a count of the 
number
ftaw> of paths with a given set of integers on the path?
ftaw>
ftaw> The latter approach should be much faster and less memory 
intensive,
ftaw> once you get past the first few generations.  (Probably still
ftaw> exponential, though.)
ftaw>
ftaw> Franklin T. Adams-Watters
ftaw>

There is no such optimization in the program shown below. For each
node, it descends into all branches/leafs independent of whether the
set of labels in the current node and higher up in the trunk is the same
or not the same as any set seen earlier. The state vector is
only the vector of counts (one number per generation) which is updated
at each level of recursion, and the single list of labels of the
ancestors of the current node.
So yes, there is room for optimization by introducing some memory
in here, certainly at the cost of increasing the machine memory 
requirement
to remember all the sub-counts of partial tree in some hash area.

The first action to become faster is to return to a compiled language.
Since this here does not involve any type of functionality which needs 
a number
theoretic package, one would try to exploit the fact that C would 
typically be a
factor of 5 to 10 faster than Maple. I don't want to appear selfish and
will therefore leave all this as an exercise to others :-)

RJM

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From pauldhanna at juno.com  Fri Nov 17 07:13:46 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Fri, 17 Nov 2006 01:13:46 -0500
Subject: Sub-Fibonacci Tree
Message-ID: <20061117.011346.692.1.pauldhanna@juno.com>

Seqfans, 
     Consider the following "sub-Fibonacci tree" and in particular 
its associated sequence defined by: 
 
Sum of labels of nodes in generation n of a rooted tree in which a node 
with label k has child nodes assigned labels of successive integers 
beginning with k+1 such that the number of child nodes for each node 
is given by the label of the parent node, starting at generation n=0 
with a root node with label '1' followed by a child node with label '2'. 
 
(Perhaps someone could make this description more concise?!)
 
EXAMPLE.
The initial nodes of the tree for generations 0..5 are: 
gen 0: [1];
gen 1: [2];
gen 2: [3];
gen 3: [4,5];
gen 4: (4)->[5,6,7],(5)->[6,7,8];
gen 5: (5)->[6,7,8,9],(6)->[7,8,9,10],(7)->[8,9,10,11], 
(6)->[7,8,9,10,11],(7)->[8,9,10,11,12],(8)->[9,10,11,12,13];
  
By definition, there are 2 child nodes for node [3] since 
the parent of node [3] is [2]; likewise, 
there are 3 child nodes for nodes [4] and [5] (in gen.3) 
since the parent of both nodes has label [3] (in gen.2). 
 
Notice the minimum label in generation n is n+1, and 
the maximum label in generation n is Fibonacci(n+2).  
  
 From this tree I am interested in gleaning 2 sequences. 
  
SEQUENCE 1. 
I compute the number of nodes in generation n to begin: 
[1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368, ...]
 
which is equal (with offset) to A005270:
"Sub-Fibonacci sequences of length n."
1, 1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368 
 
SEQUENCE 2. 
But the following related sequence is NOT found in the OEIS: 
 
Sum of labels of nodes in generation n:
[1, 2, 3, 9, 39, 252, 2361, 32077, 631058, ...]
 
EXTENSION?
Could someone compute more terms of sequences 1 and 2? 
 
Thanks,
    Paul





From pauldhanna at juno.com  Fri Nov 17 08:55:47 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Fri, 17 Nov 2006 02:55:47 -0500
Subject: Sub-Fibonacci Tree
Message-ID: <20061117.025548.692.2.pauldhanna@juno.com>

Seqfans,
    Here I extend A005270 by one more term and also extend the sequence 
a(n) = sum of labels of nodes in generation n in the sub-Fibonacci tree.
I demonstrate a slightly faster method to compute these sequences. 
 
First, I am changing the description to be (hopefully) more clear: 
 
"Sum of labels of nodes in generation n of the sub-Fibonacci tree in
which 
a node with label k and parent node with label g, has g child nodes that 
are given labels beginning with k+1 through k+g; the tree starts at 
generation n=0 with a root node labeled '1' and a child node labeled
'2'."
 
Would appreciate any suggestions to change in wording.
  
 
EXTENDING A005270.
There is a faster way to compute b(n)=A005270(n+1): 
b(n) = "number of nodes in generation n of the sub-Fibonacci tree"
 
by employing the following statistic:
 
b(n+1) = sum of (parent label)*(label) over all nodes in generation n. 
 
For example, 
b(4) = 27 = 3*(4+5);
b(5) = 177 = 4*(5+6+7) + 5*(6+7+8);
b(6) = 1680 = 5*(6+7+8+9) + 6*(7+8+9+10) + 7*(8+9+10+11)+ 
 6*(7+8+9+10+11) + 7*(8+9+10+11+12) + 8*(9+10+11+12+13); ...

Using this method, I extend sequence A005270 by one more term:
1, 1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368, 13067353,
 

SUM OF LABELS SEQUENCE. 
In like manner, the computation of:
a(n) = "sum of labels of nodes in generation n of the sub-Fibonacci tree"

can be accomplished by the statistic:
 
a(n+1) = sum of (parent label)*(label) over all nodes in generation n 
+ sum of (parent label)*[label*(label+1)/2] over all nodes in generation
n-1. 
 
For example, 
a(2) = 3 = 1*2  + 1*( 1*2/2 ) ;
a(3) = 9 = 2*3  + 1*( 2*3/2 ) ;
a(4) = 39 = 3*(4+5)  + 2*( 3*4/2 ) ;
a(5) = 252 = 4*(5+6+7) + 5*(6+7+8)  + 3*( 4*5/2 + 5*6/2 ) ;
a(6) = 2361 = 5*(6+7+8+9) + 6*(7+8+9+10) + 7*(8+9+10+11) + 
 6*(7+8+9+10+11) + 7*(8+9+10+11+12) + 8*(9+10+11+12+13)  +
 4*( 5*6/2 + 6*7/2 + 7*8/2 ) + 5*( 6*7/2 + 7*8/2 + 8*9/2) ;
...
 
With this method, I extend the sum of labels in generation n to: 
1, 2, 3, 9, 39, 252, 2361, 32077, 631058, 18035534, 
 

Would someone please confirm my calculation of the terms: 
b(10) = A005270(11) = 13067353  and 
a(10) = 18035534
 
and maybe compute a few more terms? 
Thanks,
    Paul
----------------------------------------------------
> EXAMPLE.
> The initial nodes of the tree for generations 0..5 are: 
> gen 0: [1];
> gen 1: [2];
> gen 2: [3];
> gen 3: [4,5];
> gen 4: (4)->[5,6,7],(5)->[6,7,8];
> gen 5: (5)->[6,7,8,9],(6)->[7,8,9,10],(7)->[8,9,10,11], 
> (6)->[7,8,9,10,11],(7)->[8,9,10,11,12],(8)->[9,10,11,12,13];
>   
> By definition, there are 2 child nodes for node [3] since 
> the parent of node [3] is [2]; likewise, 
> there are 3 child nodes for nodes [4] and [5] (in gen.3) 
> since the parent of both nodes has label [3] (in gen.2). 
>  
> Notice the minimum label in generation n is n+1, and 
> the maximum label in generation n is Fibonacci(n+2).  
>   
> From this tree I am interested in gleaning 2 sequences. 
>   
> SEQUENCE 1. 
> I compute the number of nodes in generation n to begin: 
> [1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368, ...]
>  
> which is equal (with offset) to A005270:
> "Sub-Fibonacci sequences of length n."
> 1, 1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368 
>  
> SEQUENCE 2. 
> But the following related sequence is NOT found in the OEIS: 
>  
> Sum of labels of nodes in generation n:
> [1, 2, 3, 9, 39, 252, 2361, 32077, 631058, ...]
>  
> EXTENSION?
> Could someone compute more terms of sequences 1 and 2? 
>  
> Thanks,
>     Paul





From zbi74583 at boat.zero.ad.jp  Sat Nov 18 02:27:22 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Sat, 18 Nov 2006 10:27:22 +0900
Subject: A085058
Message-ID: <019201c70ab0$b44b3ce0$ca73d7dc@FMC57937B6BE00>

    Hi, Seqfans

    If 2^k(n) is the largest power of two dividing tangent number T(n) then 
numbers such that 2*n-k(n) are as follows.



    S : 2,3,2,4,2,3,2,5,2,....



    It looks like A085058.

    Are they the same seqences ?



    Yasutoshi


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From alec at mihailovs.com  Sun Nov 19 11:38:43 2006
From: alec at mihailovs.com (Alec Mihailovs)
Date: Sun, 19 Nov 2006 04:38:43 -0600
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
References: <200611161744.kAGHiFGX005987@amer.strw.leidenuniv.nl>
Message-ID: <000c01c70bc6$e36ead60$03fea8c0@Media>

From: "Richard Mathar" <mathar at strw.leidenuniv.nl>
Sent: Thursday, November 16, 2006 11:44 AM
>
> pdh> Date: Tue, 14 Nov 2006 23:04:51 -0500
> pdh> From: "Paul D. Hanna" <pauldhanna at juno.com>
> pdh> ...
> pdh> Number of labelled nodes in generation n of a rooted tree
> pdh> where a node with label k has k child nodes such that
> pdh> the label of each child node is the least unused label in
> pdh> the path from the root to that child node,
> pdh> and where root is labeled '1'.
> pdh>   ...
> pdh> Thus, the number of nodes in generation n begins:
> pdh> [1, 1, 2, 7, 36, 248, ...]
>
> [1, 1, 2, 7, 36, 248, 2157, 22761, 283220, 4068411, 66367834, ...]
>                                                    ^^^ 66367834 is Gen 10

Maple 10 can compile (some) procedures in C. I used the following 
back-tracking procedure,

f:=proc(n::integer[4],A::Array(datatype=integer[4]),
B::Array(datatype=integer[4]))::integer[4]; local c::integer[4],
i::integer[4],len::integer[4],m::integer[4]; c,len,m:=0,3,3;
while len>1 do if len=n then c:=c+1;m:=A[len];B[m]:=0;len:=len-1;
B[A[len]]:=B[A[len]]+1 elif B[A[len]]<=A[len] then for i from m+1
do if B[i]=0 then break fi od; len:=len+1;A[len]:=i;B[i]:=1;m:=2
else m:=A[len];B[m]:=0;len:=len-1;B[A[len]]:=B[A[len]]+1 fi od; c end:

cf:=Compiler:-Compile(f):

F:=proc(n::posint) local A,B;
if n<3 then 1 elif n=3 then 2 else
A:=Array([$1..3,0$(n-3)],datatype=integer[4]);
B:=Array([1$3,0$((n-2)*(n+1)/2)],datatype=integer[4]);
cf(n,A,B) fi end:

seq(F(n),n=1..12);

  1, 1, 2, 7, 36, 248, 2157, 22761, 283220, 4068411, 66367834,

        1213504295

That confirms Richard Mathar's and Paul D. Hanna's calculations and gives 
one more element of the sequence. Further elements could not be calculated 
on my 32-bit computer using this procedure, because they are greater than 
2^32. Changing the types of c and f from integer[4] to float[8] allows 
further calculations. I did that and after about an hour got the next 
element,

F(13);

                      24606397802


Alec Mihailovs
http://mihailovs.com/Alec/








From deutsch at duke.poly.edu  Mon Nov 20 17:33:59 2006
From: deutsch at duke.poly.edu (Emeric Deutsch)
Date: Mon, 20 Nov 2006 11:33:59 -0500 (EST)
Subject: help with a new sequence 
Message-ID: <Pine.GSO.4.62.0611201046020.12396@duke.poly.edu>

Seqfans,

I intend to submit to OEIS the following new sequence (triangle):

1,1,1,1,3,1,1,4,2,6,1,1,5,5,10,10,10,1,

Triangle read by rows: T(n,k) is the number of Dyck paths of
semilength n whose ascent lengths form the k-th partition of
the integer n; the partitions of n are ordered in the way
exemplified by [5], [4,1], [3,2], [3,1,1], [2,2,1], [2,1,1,1],
[1,1,1,1,1] (the "Mathematica" ordering).

Equivalently, T(n,k) is the number of ordered trees with n
edges whose node degrees form the k-th partition of the
integer n.

Example: T(5,3)=5 because the 3rd partition of 5 is [3,2] and
we have (UU)DD(UUU)DDD, (UUU)DDD(UU)DD, (UU)D(UUU)DDDD,
(UUU)D(UU)DDDD, and (UUU)DD(UU)DDD; here U=(1,1), D=(1,-1) and
the ascents are shown between parentheses.

Triangle starts:
1;
1,1;
1,3,1;
1,4,2,6,1;
1,5,5,10,10,10,1;

Row n has A000041(n) terms (=number of partitions of n).

------------

So far this is all I have; terms have been found by
straightforward counting. I am sure you can find more
terms, more facts, etc. I'd appreciate any collaboration.

Thanks, Emeric





From pauldhanna at juno.com  Mon Nov 20 19:39:48 2006
From: pauldhanna at juno.com (Paul D Hanna)
Date: Mon, 20 Nov 2006 18:39:48 GMT
Subject: help with a new sequence 
Message-ID: <20061120.104022.757.551642@webmail37.nyc.untd.com>

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From franktaw at netscape.net  Mon Nov 20 19:49:27 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 20 Nov 2006 13:49:27 -0500
Subject: help with a new sequence 
In-Reply-To: <20061120.104022.757.551642@webmail37.nyc.untd.com>
References: <20061120.104022.757.551642@webmail37.nyc.untd.com>
Message-ID: <8C8DAEBBD6AFC69-91C-B5F3@FWM-M20.sysops.aol.com>

This is A008284, a famous function.

Franklin T. Adams-Watters


-----Original Message-----
From: pauldhanna at juno.com

Seqfans,
??????Row n of a?related triangle?could?count?the number of terms in 
row n
of?Emerics triangle that?sum to?form the respective?terms of the 
Narayana triangle.
?
This triangle P?would begin:
1;
1, 1;
1, 1, 1;
1, 2, 1, 1;
1, 2, 2, 1, 1; ...

where row sums equal the partition numbers.
?
For example, row 3 of P?is [1, 2, 1, 1]
since Narayana row 3 is [1, 4+2, 6, 1]
formed from row 3 of?Emerics triangle: [1, 4, 2, 6, 1].
?
Is the resulting partition-related triangle P?already in the OEIS?

It?might have an interesting matrix inverse.
??? Paul
?
--?Emeric?Deutsch?<deutsch at duke.poly.edu>?wrote:
...
Triangle?starts:
1;
1,1;
1,3,1;
1,4,2,6,1;
1,5,5,10,10,10,1;

Row?n?has?A000041(n)?terms?(=number?of?partitions?of?n).
...
Thanks,?Emeric




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From franktaw at netscape.net  Mon Nov 20 20:55:23 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 20 Nov 2006 14:55:23 -0500
Subject: help with a new sequence 
In-Reply-To: <Pine.GSO.4.62.0611201046020.12396@duke.poly.edu>
References: <Pine.GSO.4.62.0611201046020.12396@duke.poly.edu>
Message-ID: <8C8DAF4F355FB9E-B7C-B33B@FWM-M19.sysops.aol.com>

How about if you look at the sequence (composition) of ascents, instead 
of the partition?

I think this starts (using the A066099 ordering):

1
1,1
1,2,1,1
1,3,2,3,1,2,1,1
1,4,3,6,2,5,3,1,1,3,2,2,1,3,4,1

Amazingly, this corresponds to Stern's diatomic array (version 3, 
A070879) for the first 18 terms.
Makes me wonder if A070879 can be interpreted as a function on 
compositions.

Franklin T. Adams-Watters


-----Original Message-----
From: deutsch at duke.poly.edu

Seqfans,?
?
I intend to submit to OEIS the following new sequence (triangle):?
?
1,1,1,1,3,1,1,4,2,6,1,1,5,5,10,10,10,1,?
?
Triangle read by rows: T(n,k) is the number of Dyck paths of?
semilength n whose ascent lengths form the k-th partition of?
the integer n; the partitions of n are ordered in the way?
exemplified by [5], [4,1], [3,2], [3,1,1], [2,2,1], [2,1,1,1],?
[1,1,1,1,1] (the "Mathematica" ordering).?
?
Equivalently, T(n,k) is the number of ordered trees with n?
edges whose node degrees form the k-th partition of the?
integer n.?
?
Example: T(5,3)=5 because the 3rd partition of 5 is [3,2] and?
we have (UU)DD(UUU)DDD, (UUU)DDD(UU)DD, (UU)D(UUU)DDDD,?
(UUU)D(UU)DDDD, and (UUU)DD(UU)DDD; here U=(1,1), D=(1,-1) and?
the ascents are shown between parentheses.?
?
Triangle starts:?
1;?
1,1;?
1,3,1;?
1,4,2,6,1;?
1,5,5,10,10,10,1;?
?
Row n has A000041(n) terms (=number of partitions of n).?
?
------------?
?
So far this is all I have; terms have been found by?
straightforward counting. I am sure you can find more?
terms, more facts, etc. I'd appreciate any collaboration.?
?
Thanks, Emeric?


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From franktaw at netscape.net  Mon Nov 20 22:54:07 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 20 Nov 2006 16:54:07 -0500
Subject: help with a new sequence 
In-Reply-To: <8C8DAF4F355FB9E-B7C-B33B@FWM-M19.sysops.aol.com>
References: <Pine.GSO.4.62.0611201046020.12396@duke.poly.edu> <8C8DAF4F355FB9E-B7C-B33B@FWM-M19.sysops.aol.com>
Message-ID: <8C8DB058980377A-B7C-BB3C@FWM-M19.sysops.aol.com>

Let f(a_0, a_1, ..., a_m) be the number of Dyck paths associated with 
the composition
[a_0, a_1, ..., a_m].   We have f(a_0) = 1; and for m > 0,

f(a_0, a_1, ..., a_m) = Sum_{i=0}^{a_0-1} f(a_1+i, a_2, ..., a_m).

Using this, I get the following table (including row 0, the empty 
composition):

1,
1,
1,1,
1,2,1,1,
1,3,2,3,1,2,1,1,
1,4,3,6,2,5,3,4,1,3,2,3,1,2,1,1,
1,5,4,10,3,9,6,10,2,7,5,9,3,7,4,5,1,4,3,6,2,5,3,4,1,3,2,3,1,2,1,1,
1,6,5,15,4,14,10,20,3,12,9,19,6,16,10,15,2,9,7,16,5,14,9,14,3,10,7,12,4,9
,5,6,1,5,4,
   10,3,9,6,10,2,7,5,9,3,7,4,5,1,4,3,6,2,5,3,4,1,3,2,3,1,2,1,1

And accumulating by partition, for Emeric's original table (but also 
including row 0):

1,
1,
1,1,
1,3,1,
1,4,2,6,1,
1,5,5,10,10,10,1,
1,6,6,15,3,30,20,5,30,15,1,
1,7,7,21,7,42,35,21,21,105,35,35,70,21,1,
1,8,8,28,8,56,56,4,56,28,168,70,28,84,168,280,56,14,140,140,28,1,
1,9,9,36,9,72,84,9,72,36,252,126,36,72,252,252,504,126,12,252,252,84,756,
630,84,
   126,420,252,36,1

Given a partition of n into k parts, [a_1^e_1,...,a_k^e_k], the number 
of
associated Dyck paths appears to be n!/(n-k+1)!/(Product (e_i)!).

Now, maybe somebody can prove that.

Franklin T. Adams-Watters


-----Original Message-----
From: franktaw at netscape.net

How about if you look at the sequence (composition) of ascents, instead 
of the partition??
?
I think this starts (using the A066099 ordering):?
?
1?
1,1?
1,2,1,1?
1,3,2,3,1,2,1,1?
1,4,3,6,2,5,3,1,1,3,2,2,1,3,4,1?
?
Amazingly, this corresponds to Stern's diatomic array (version 3, 
A070879) for the first 18 terms.?
Makes me wonder if A070879 can be interpreted as a function on 
compositions.?
?
Franklin T. Adams-Watters?
?
-----Original Message-----?
From: deutsch at duke.poly.edu?
?
Seqfans,??
??
I intend to submit to OEIS the following new sequence (triangle):??
??
1,1,1,1,3,1,1,4,2,6,1,1,5,5,10,10,10,1,??
??
Triangle read by rows: T(n,k) is the number of Dyck paths of??
semilength n whose ascent lengths form the k-th partition of??
the integer n; the partitions of n are ordered in the way??
exemplified by [5], [4,1], [3,2], [3,1,1], [2,2,1], [2,1,1,1],??
[1,1,1,1,1] (the "Mathematica" ordering).??
??
Equivalently, T(n,k) is the number of ordered trees with n??
edges whose node degrees form the k-th partition of the??
integer n.??
??
Example: T(5,3)=5 because the 3rd partition of 5 is [3,2] and??
we have (UU)DD(UUU)DDD, (UUU)DDD(UU)DD, (UU)D(UUU)DDDD,??
(UUU)D(UU)DDDD, and (UUU)DD(UU)DDD; here U=(1,1), D=(1,-1) and??
the ascents are shown between parentheses.??
??
Triangle starts:??
1;??
1,1;??
1,3,1;??
1,4,2,6,1;??
1,5,5,10,10,10,1;??
??
Row n has A000041(n) terms (=number of partitions of n).??
??
------------??
??
So far this is all I have; terms have been found by??
straightforward counting. I am sure you can find more??
terms, more facts, etc. I'd appreciate any collaboration.??
??
Thanks, Emeric??
?
________________________________________________________________________?

Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From njas at research.att.com  Mon Nov 20 23:25:04 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Mon, 20 Nov 2006 17:25:04 -0500 (EST)
Subject: two "find the next term" puzzles from Knuth Vol 4
Message-ID: <200611202225.RAA93655@fry.research.att.com>


the two puzzles (and their solutions) are

%I A123896
%S A123896 0,1,1,1,12,12,12,12,12,12,100,121,122,123,123
%N A123896 A123895(n^2).
...

%I A123902
%S A123902 0,1,12,100,112,121,122,123
%N A123902 A123896 sorted and uniqued.
...

which are based on this:

%I A123895
%S A123895 0,1,1,1,1,1,1,1,1,1,10,11,12,12,12,12,12,12,12,12,10,12,11,12,12,
%T A123895 12,12,12,12,12,10,12,12,11,12,12,12,12,12,12,10,12,12,12,11,12,12,
%U A123895 12,12,12,10,12,12,12,12,11,12,12,12,12,10,12,12,12,12,12,11,12,12
%N A123895 Restricted growth string for the (decimal expansion of the) number n.
%C A123895 Write n in base 10 prefixed with a 0. Read this string from left to r
ight. Write a 0 each time you see the first distinct digit (which is 0), write a
1 each time you see the second distinct digit, write a 2 each time you see the t
hird distinct digit, and so on. Finally, delete the leading zeros.
%D A123895 D. E. Knuth, TAOCP, Vol. 4, Section 7.2.1.5, Problems 4 and 5.
%e A123895 To find a(66041171): 066041171 -> 011023343 -> 11023343.
%Y A123895 Cf. A123896, A123902.
%K A123895 nonn,base,new
%O A123895 0,11
%A A123895 njas, Nov 20 2006

It would be nice if someone could extend the first two, and provide
maple or mma or other code for the third one

Neil






From franktaw at netscape.net  Mon Nov 20 23:59:41 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 20 Nov 2006 17:59:41 -0500
Subject: two "find the next term" puzzles from Knuth Vol 4
In-Reply-To: <200611202225.RAA93655@fry.research.att.com>
References: <200611202225.RAA93655@fry.research.att.com>
Message-ID: <8C8DB0EB222DED3-B7C-BF60@FWM-M19.sysops.aol.com>

A123896:

0 1 1 1 12 12 12 12 12 12 100 121 122 123 123 112 123 123 123 123 100 
112 121 123 123 123 121 123 123 123 100 123 1023 1023 1123 1223 1234 
1234 1222 1231 1200 1231 1234 1234 1234 1012 1223 1102 1203 1203 1200 
1203 1203 1203 1234 1023 1213 1234 1123 1234 1200 1234 1233 1232 1023 
1223 1234 1123 1231 1234 1200 1023 1234 1234 1234 1231 1223 1232 1023 
1234 1200 1213 1234 1223 1023 1223 1234 1234 1122 1234 1200

A123902:

0 1 12 100 112 121 122 123 1012 1023 1102 1122 1123 1200 1201 1203 1213 
1222 1223 1231 1232 1233 1234 10000 10012 10023 10102 10123 10201 10202 
10203 10213 10223 10231 10232 10234 11023 11102 11200 11203 11211 11213 
11221 11223 11232 11234 12003 12013 12023 12032 12033 12034 12100 12113 
12121 12131 12133 12134 12200 12203 12212 12213 12231 12232 12233 12234 
12300 12301 12304 12312 12313 12314 12321 12323 12324 12331 12332 12334 
12341 12342

Sorry, I didn't use Mathematica, Maple, or even PARI; I used Excel and 
VBA.

Here's a VBA program:

Public Function RestrictedGrowthString(ByVal x As String) As String
    Dim i As Long
    Dim dig As Integer
    Dim pos As Long

    For i = 1 To Len(x)
        If Mid(x, i, 1) = "0" Then
            RestrictedGrowthString = RestrictedGrowthString & "0"
        Else
            pos = InStr(x, Mid(x, i, 1))
            If pos = i Then
                dig = dig + 1
                 RestrictedGrowthString = RestrictedGrowthString & 
Format(dig)
            Else
                 RestrictedGrowthString = RestrictedGrowthString & 
Mid(RestrictedGrowthString, pos, 1)
            End If
        End If
    Next i
End Function

Franklin T. Adams-Watters


-----Original Message-----
From: njas at research.att.com

the two puzzles (and their solutions) are

%I A123896
%S A123896 0,1,1,1,12,12,12,12,12,12,100,121,122,123,123
%N A123896 A123895(n^2).
...

%I A123902
%S A123902 0,1,12,100,112,121,122,123
%N A123902 A123896 sorted and uniqued.
...

which are based on this:

%I A123895
%S A123895 
0,1,1,1,1,1,1,1,1,1,10,11,12,12,12,12,12,12,12,12,10,12,11,12,12,
%T A123895 
12,12,12,12,12,10,12,12,11,12,12,12,12,12,12,10,12,12,12,11,12,12,
%U A123895 
12,12,12,10,12,12,12,12,11,12,12,12,12,10,12,12,12,12,12,11,12,12
%N A123895 Restricted growth string for the (decimal expansion of the) 
number n.
%C A123895 Write n in base 10 prefixed with a 0. Read this string from 
left to r
ight. Write a 0 each time you see the first distinct digit (which is 
0), write a
1 each time you see the second distinct digit, write a 2 each time you 
see the t
hird distinct digit, and so on. Finally, delete the leading zeros.
%D A123895 D. E. Knuth, TAOCP, Vol. 4, Section 7.2.1.5, Problems 4 and 
5.
%e A123895 To find a(66041171): 066041171 -> 011023343 -> 11023343.
%Y A123895 Cf. A123896, A123902.
%K A123895 nonn,base,new
%O A123895 0,11
%A A123895 njas, Nov 20 2006

It would be nice if someone could extend the first two, and provide
maple or mma or other code for the third one

Neil


________________________________________________________________________
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From njas at research.att.com  Tue Nov 21 00:43:03 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Mon, 20 Nov 2006 18:43:03 -0500 (EST)
Subject: King chickens - I was wrong!
Message-ID: <200611202343.SAA88240@fry.research.att.com>


Martin Fuller pointed out that I was wrong.
A highest scorer in a tournament is a king chicken,
but not conversely.

So there are really two sequences, A123553 (K.C.'s),
now corrected by Martin, and A125031, the total number
of highest scorers, computed by Martin Fuller to n=11
(this is what Gordon Royle computed out to n=10).

Maurer also defines "Emperors", chickens
who peck everybody else, which is n*2^((n-1)(n-2)/2)
(it will be A123903).

Neil





From zakseidov at yahoo.com  Tue Nov 21 06:19:11 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Mon, 20 Nov 2006 21:19:11 -0800 (PST)
Subject: A125097.gif: (Very) nice patterns
Message-ID: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>

Neil, 
seqfans,

1. Have a look at this full graph -
aint these patterns (very) nice?!
It may be known, but still it's nice...

2. Is it possible to 
put "the extended graph" into OEIS?
In the standard graph no patterns are seen at all...

3. Neil removed  keyword "nice" from A125097
(as he did regularly with ny sequences)-
as if I insist that it's me who is "nice"
not it is the seq which is nice.

Thanks, Zak


 
____________________________________________________________________________________
Sponsored Link

$200,000 mortgage for $660/ mo - 
30/15 yr fixed, reduce debt - 
http://yahoo.ratemarketplace.com
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From maxale at gmail.com  Tue Nov 21 07:03:55 2006
From: maxale at gmail.com (Max A.)
Date: Mon, 20 Nov 2006 22:03:55 -0800
Subject: A125097.gif: (Very) nice patterns
In-Reply-To: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>
References: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>
Message-ID: <d3dac270611202203p2b9fff6s2bc522bc6662e735@mail.gmail.com>

Zak,

I do not see anything special about this graph.
Yes, it includes some points from the line y=2x+1, and, yes, it is
symmetric the region [0,500]x[0,1000] (i.e., a(n)=k iff
a((k-1)/2)=2n+1). But that does not make it any nicer.

Regards,
Max

On 11/20/06, zak seidov <zakseidov at yahoo.com> wrote:
> Neil,
> seqfans,
>
> 1. Have a look at this full graph -
> aint these patterns (very) nice?!
> It may be known, but still it's nice...
>
> 2. Is it possible to
> put "the extended graph" into OEIS?
> In the standard graph no patterns are seen at all...
>
> 3. Neil removed  keyword "nice" from A125097
> (as he did regularly with ny sequences)-
> as if I insist that it's me who is "nice"
> not it is the seq which is nice.
>
> Thanks, Zak
>
>
>
> ____________________________________________________________________________________
> Sponsored Link
>
> $200,000 mortgage for $660/ mo -
> 30/15 yr fixed, reduce debt -
> http://yahoo.ratemarketplace.com
>
>





From pauldhanna at juno.com  Tue Nov 21 11:07:54 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Tue, 21 Nov 2006 05:07:54 -0500
Subject: Formula for New Sequence A125054 ?
Message-ID: <20061121.050754.1448.0.pauldhanna@juno.com>

Seqfans,
     Could someone find a formula for this new sequence? 
Sequence: A125054  begins: 
 
1,3,21,327,9129,396363,24615741,2068052367,225742096209,
31048132997523,5252064083753061,1071525520294178007,
259439870666594250489,73542221109962636293083,
24125551094579137082039181,9068240688454120376775401247,
3871645204706420218816959159969,
(I can supply many more terms if needed).
 
The sequence forms the Central terms of a new triangle A125053 
(a variant of triangle A008301 - enumeration of binary trees). 
 
Triangle A125053 is nice since the first column (and row sums) 
form the Secant numbers A000364 (an unexpected result!). 
  
Below I define the triangle.
 
I would be very interested in a formula, perhaps an E.g.f.? 
(I know I am asking a lot, but Seqfans surprise me all the time 
with amazing formulas). 
  
Thanks, 
      Paul 
---------------------------------------------------------
  
If we write triangle A125053 like this: 
.......................... ...1; 
.................... ...1, ...3, ...1; 
.............. ...5, ..15, ..21, ..15, ...5; 
........ ..61, .183, .285, .327, .285, .183, ..61; 
.. 1385, 4155, 6681, 8475, 9129, 8475, 6681, 4155, 1385; 

then the first nonzero term is the sum of the previous row: 
1385 = 61 + 183 + 285 + 327 + 285 + 183 + 61,

the next term is 3 times the first:
4155 = 3*1385,

and the remaining terms in each row are obtained by the rule 
illustrated by: 
6681 = 2*4155 - 1385 - 4*61 ;
8475 = 2*6681 - 4155 - 4*183 ;
9129 = 2*8475 - 6681 - 4*285 ;
8475 = 2*9129 - 8475 - 4*327 ;
6681 = 2*8475 - 9129 - 4*285 ;
4155 = 2*6681 - 8475 - 4*183 ;
1385 = 2*4155 - 6681 - 4*61 .
 
An alternate recurrence is illustrated by:
4155 = 1385 + 2*(61 + 183 + 285 + 327 + 285 + 183 + 61);
6681 = 4155 + 2*(183 + 285 + 327 + 285 + 183);
8475 = 6681 + 2*(285 + 327 + 285);
9129 = 8475 + 2*(327); 
 
and then for k>n, k<=2n, T(n,k) = T(n,2*n-k).
 
END.





From pauldhanna at juno.com  Tue Nov 21 14:58:26 2006
From: pauldhanna at juno.com (Paul D Hanna)
Date: Tue, 21 Nov 2006 13:58:26 GMT
Subject: Formula for New Sequence A125054 ?
Message-ID: <20061121.055901.20413.609605@webmail43.nyc.untd.com>

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From antti.karttunen at gmail.com  Tue Nov 21 15:27:22 2006
From: antti.karttunen at gmail.com (Antti Karttunen)
Date: Tue, 21 Nov 2006 16:27:22 +0200
Subject: Binary tree terminology, again. Re: Formula for New Sequence A125054
 ?
In-Reply-To: <20061121.055901.20413.609605@webmail43.nyc.untd.com>
References: <20061121.055901.20413.609605@webmail43.nyc.untd.com>
Message-ID: <45630CCA.8000103@gmail.com>

Paul D Hanna wrote:

> Seqfans,
>
>      Could someone find a formula for this new sequence?
> Sequence: A125054  begins:
>  
> 1,3,21,327,9129,396363,24615741,2068052367,225742096209,
> 31048132997523,5252064083753061,1071525520294178007,
> 259439870666594250489,73542221109962636293083,
> 24125551094579137082039181,9068240688454120376775401247,
> 3871645204706420218816959159969,
> (I can supply many more terms if needed).
>  
> The sequence forms the Central terms of a new triangle A125053
> (a variant of triangle A008301 - enumeration of binary trees).
>
Is there any more helpful description for A008301 and A125053 ?
E.g. do they relate to non-planar binary trees
(as http://www.research.att.com/~njas/sequences/A001190 )
or to (rooted, unlabeled) planar binary trees, as Catalan numbers, 
A000108, do?
Labeled or unlabeled? Rooted or non-rooted?


Cumprimentos,

Antti Karttunen
writing from Porto, Portugal.

>  
> Triangle A125053 is nice since the first column (and row sums)
> form the Euler numbers A000364 (an unexpected result!).
>  
>
> Seqfans,
>     One should always try Superseeker first!
> Superseeker says:
>
> A125054 = Binomial transform of A000182 (e.g.f. tan(x))
>
> and it holds true.
>
> I did not expect such a simple answer! 
>     Paul
>

>  
>






From tbaruchel at free.fr  Tue Nov 21 15:47:28 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Tue, 21 Nov 2006 15:47:28 +0100 (CET)
Subject: Almost the number of divisors of n+1
Message-ID: <20061121154018.X1354@localhost.my.domain>

Hi, I have some number-theory related empirically collected datas to study.
As long as I can see, I have a link between each integer and the number
of divisors of its successor. But though the rule seems to work very
well in some cases: predecessors of prime numbers seem to appear 0 or 1 times,
predecessors of numbers like 12 or 24 seem to appear much more, etc. I am not
sure this is the very exact rule. Of course, the trouble may come from
the fact that my datas is quite hard to "extract", but maybe you know some
variants of
   a(n) is the number of divisors of (n+1)
Have you ever seen closely-related sequences like that one?

Regards,

-- 
Thomas Baruchel





From davidwwilson at comcast.net  Tue Nov 21 18:18:36 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Tue, 21 Nov 2006 12:18:36 -0500
Subject: Confirmation
Message-ID: <004601c70d91$14d8f3d0$6501a8c0@yourxhtr8hvc4p>

Define

    f(1) = 1
    f(n+1) = f(n) + (f(n) mod n+3364).

I wrote a computer program to find the smallest x with

    x+3365 divides f(x)

The value of x found is just under 8,350,000,000 and the corresponding f(x) just under 2^64. Although my program included an overflow check, I am still not comfortable because this value of f(x) is close to my 64-bit unsigned integer size.

Could someone with a fast computer, preferably a UNIX box, please confirm or refute my values?
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From maxale at gmail.com  Tue Nov 21 20:47:20 2006
From: maxale at gmail.com (Max A.)
Date: Tue, 21 Nov 2006 11:47:20 -0800
Subject: Confirmation
In-Reply-To: <004601c70d91$14d8f3d0$6501a8c0@yourxhtr8hvc4p>
References: <004601c70d91$14d8f3d0$6501a8c0@yourxhtr8hvc4p>
Message-ID: <d3dac270611211147i7b7ef94ev729014f094d24dc4@mail.gmail.com>

I've got

x = 8334201245
f(x) = 17364786429187398690

I've also checked that the value f(n) was non-decreasing during
computations. Since f(n+1) < f(n)+n+3364 and f(x)+x+3364 < 2^64, that
means no overflow in 64-bit unsigned integer happened.

Max

On 11/21/06, David Wilson <davidwwilson at comcast.net> wrote:
>
>
> Define
>
>     f(1) = 1
>     f(n+1) = f(n) + (f(n) mod n+3364).
>
> I wrote a computer program to find the smallest x with
>
>     x+3365 divides f(x)
>
> The value of x found is just under 8,350,000,000 and the corresponding f(x)
> just under 2^64. Although my program included an overflow check, I am still
> not comfortable because this value of f(x) is close to my 64-bit unsigned
> integer size.
>
> Could someone with a fast computer, preferably a UNIX box, please confirm or
> refute my values?





From all at abouthugo.de  Tue Nov 21 22:22:29 2006
From: all at abouthugo.de (Hugo Pfoertner)
Date: Tue, 21 Nov 2006 22:22:29 +0100
Subject: Wrong new terms in A085000
Message-ID: <45636E15.FD6E8D3F@abouthugo.de>

Seqfans, Neil,

the sequence A085000 "Maximal determinant of an n X n matrix using the
integers 1 to n^2."

that had previously the terms a(1)...a(6)
1,10,412,40800,6839492,1865999570,

has recently been extendend as follows:

762074188050,440477621828418, 345964972957674150

The comment says:
a(7) - a(9) were found by Richard Mueller
(richard.mueller(AT)informatik.hu-berlin.de) and Johannes Dewender
(johannes.dewender(AT)informatik.hu-berlin.de) in cooperation, Oct 22
2006 
Entry updated by Richard Mueller  Oct 22 2006

Unfortunately bigger counterexamples exist for all 3 terms:

d7:=det mat
((28, 9,34,41, 1,40,22),
 (47,25, 7,36,35,16, 8),
 (3, 27, 6,39,31,26,43),
 (38,21,32,14,18, 4,48),
 (19, 5,37,13,49,33,20),
 (29,42,15, 2,17,45,24),
 (11,46,44,30,23,12,10));
 
 Reduce: d7 := 762139309293

d8:= det mat
((13, 53, 40, 25,  2, 45, 58, 23),
 (21, 22,  8, 54, 24, 56, 18, 57),
 (60, 11, 36, 44, 12,  7, 49, 41),
 (15, 19, 48,  4, 52, 26, 38, 59),
 (20, 62, 34, 51, 42,  1, 17, 33),
 (37, 29,  3, 31, 61, 35, 55, 10),
 (30, 14, 63, 46, 39, 47, 16,  5),
 (64, 50, 28,  6, 27, 43,  9, 32));

Reduce: d8 := 440857916120379

d9:= det mat
((40, 48, 72, 26,  6, 50, 39, 78, 10),
 (23, 77, 42,  7, 37, 41, 74, 12, 56),
 (52, 47,  8, 54,  5, 64, 19, 44, 75),
 (59, 25, 13, 43, 58, 76, 63, 30,  3),
 (79, 36, 71, 60, 31, 18, 34,  2, 38),
 (32, 81, 22, 61, 67, 15, 17, 53, 21),
 (20, 24, 66, 27, 73, 68,  4, 33, 55),
 ( 1, 14, 45, 80, 35, 29, 69, 46, 49),
 (65, 16, 28, 11, 57,  9, 51, 70, 62));

Reduce: d9 := 346202123011922653

(Thanks to Rainer Rosenthal for X-checking the determinants with Maple).

None of the counterexamples is known to be the biggest possible
solution. The best value I've found in 2 CPU years for
a(7)=762140212575, but currently I don't have access to the computer
where I've found this solution and the corresponding arrangement of
matrix elements.

My suggestion is to remove the 3 new terms and the new Mathematica
program and to restore the previous state of the sequence. The
determination of a(8) and a(9) is currently beyond our available
computational capabilities. I will try to confirm a(7) and to find
improved lower bounds for a(8) and a(9) by running the slightly improved
program given at the link for some months on a cluster of Intel Itanium
2 CPUs.

In the meantime I suggest to add a comment:

Known lower bounds for a(7)...a(9) are
762140212575, 440857916120379, 346202123011922653,

Hugo Pfoertner





From mlb at well.com  Wed Nov 22 00:25:34 2006
From: mlb at well.com (Marc LeBrun)
Date: Tue, 21 Nov 2006 15:25:34 -0800
Subject: Almost the number of divisors of n+1
In-Reply-To: <20061121154018.X1354@localhost.my.domain>
References: <20061121154018.X1354@localhost.my.domain>
Message-ID: <200611212325.kALNPQh3019729@smtp.well.com>

 >  a(n) is the number of divisors of (n+1)
 > Have you ever seen closely-related sequences like that one?

Some random and rather vague ideas:

1. I faintly recall seeing something about algorithms for
factoring n involving properties of n-1 (I mean besides
the obvious connection to "Fermat's little theorem").
Perhaps in Knuth's TAOCP?

2. It sounds like you might have a case where the relationship
you're seeing, a(n) ~ d(n), is actually a first approximation
to some identity like

   a(n) = d(n) + F1(n) + F2(n) + ...

where the Fk depend on the factor structure of n or something.
So perhaps you might try to "take out" the d(n) and then look
further into the "error" and see what you can guess about how
the Fk's specifically vary with primes, highly composites, etc.
For example you might try "fitting" a(n) with Moebius-style sums.

3. There's some rather surprising known relationships involving
divisors--my favorite is one of Legendre's that is basically an
identity found by expanding elliptic functions in power series.
This suggests that you might want to express your relationships
as generating functions and see what emerges.

4. Sometimes there is simple underlying "combinatorial machinery".
For example I was once puzzled by some coefficients that turned
out to be formed by taking each divisor of n, *adding 1*, and then
multiplying them all together!  It took me a while to see that's
algebraically identical to summing the products of every possible
subset of the divisors.  But the "add 1" way of expressing this
was totally opaque, whereas the "every subset" explained them.

Happy hunting!







From zbi74583 at boat.zero.ad.jp  Wed Nov 22 02:15:26 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Wed, 22 Nov 2006 10:15:26 +0900
Subject: N(2)
Message-ID: <00a201c70dd3$b538af40$977fd7dc@FMC57937B6BE00>

    I wonder if N(2) goes infinity.

    Where N(s)=   Sum_{m=1 to infinity} (-1)Sigma(m)/m^s

    Mathar computed up to m=1000000



  s m N(s)

> 2 1 1.000000000000000000000000000
> 2 2 1.250000000000000000000000000
> 2 3 1.472222222222222222222222222
> 2 4 1.784722222222222222222222222
> 2 5 1.944722222222222222222222222
> 2 6 2.000277777777777777777777778
> 2 7 2.122726757369614512471655329
> 2 8 2.325851757369614512471655329
> 2 9 2.461654226505416981607457798
> 2 10000 8.619813177975507784214344698
> 2 20000 9.240635095871089404115753095
> 2 30000 9.603791897573761358292380061
> 2 40000 9.861458278261452185262848858
> 2 50000 10.06133043757656995196303496
> 2 60000 10.22462154552329778519365923
> 2 70000 10.36270038005636494621123847
> 2 80000 10.48231016697477680498962023
> 2 90000 10.58780522013752853231672862
> 2 100000 10.68217402235127296288785820
> 2 110000 10.76754463607533705360724580
> 2 120000 10.84547789729425765394094918
> 2 130000 10.91717268365027978460835188
> 2 140000 10.98355652885066079792761960
> 2 150000 11.04534545539722125239381859
> 2 160000 11.10315629233238669062836862
> 2 170000 11.15746025373895516747888760
> 2 180000 11.20865775307751148407150182
> 2 190000 11.25708463818456648077093060
> 2 200000 11.30303121795283056763425981
> 2 210000 11.34672871948779384291766276
> 2 220000 11.38839827063844308711114745
> 2 230000 11.42821240662906506074836109
> 2 240000 11.46633165190016188934526342
> 2 250000 11.50290004647923375585461222
> 2 260000 11.53802980091084626273394586
> 2 270000 11.57183602344116902681335304
> 2 280000 11.60440829561909536077171934
> 2 290000 11.63584045378869260629085740
> 2 300000 11.66620539683007648663354504
> 2 310000 11.69557551635064846235471407
> 2 320000 11.72401391041322342493145640
> 2 330000 11.75157395832622563415629569
> 2 340000 11.77831576251751436463182945
> 2 350000 11.80427986574616427809478607
> 2 360000 11.82951337960016320803463309
> 2 370000 11.85405377158818097821428014
> 2 380000 11.87794149361683265940241605
> 2 390000 11.90120806565142413868457075
> 2 400000 11.92388559358277288937420440
> 2 410000 11.94600303970164868968234173
> 2 420000 11.96758569454234998141373716
> 2 430000 11.98866287480559328733313634
> 2 440000 12.00925544388189350449936925
> 2 450000 12.02938482345842327394291528
> 2 460000 12.04907101416111155199518539
> 2 470000 12.06833481184766752219695603
> 2 480000 12.08719139777266186606623586
> 2 490000 12.10566116650605692754670219
> 2 500000 12.12375840131307130932177405
> 2 510000 12.14149351071909408890611230
> 2 520000 12.15888769016783248485026059
> 2 530000 12.17594975401341942285000014
> 2 540000 12.19269280126406301090981599
> 2 550000 12.20912705098196862278557687
> 2 560000 12.22526702472305560991043337
> 2 570000 12.24111996896626519121471994
> 2 580000 12.25669798166764128539680162
> 2 590000 12.27200952384435815319104573
> 2 600000 12.28706299338740013449483108
> 2 610000 12.30187012252190226181814868
> 2 620000 12.31643453756242668558457904
> 2 630000 12.33076670711890682146429120
> 2 640000 12.34487182648867321005497632
> 2 650000 12.35875920337989470739102627
> 2 660000 12.37243430644503889140869087
> 2 670000 12.38590421689421153982279985
> 2 680000 12.39917431914306915108546756
> 2 690000 12.41224965443445935379058168
> 2 700000 12.42513867918653815467126754
> 2 710000 12.43784365088579118428209464
> 2 720000 12.45037185509338005878377820
> 2 730000 12.46272650898502725121177200
> 2 740000 12.47491268376206272216237202
> 2 750000 12.48693594951832440956683550
> 2 760000 12.49880013638169511883229569
> 2 770000 12.51050884124405914256853750
> 2 780000 12.52206537108093533871333277
> 2 790000 12.53347692259745252540675575
> 2 800000 12.54474459716977128685280547
> 2 810000 12.55587125001645305770022096
> 2 820000 12.56686130746829341595469366
> 2 830000 12.57771842011223665163037795
> 2 840000 12.58844534265072464040923243
> 2 850000 12.59904566706465055360427300
> 2 860000 12.60952226922408394518254911
> 2 870000 12.61987642201166975414741555
> 2 880000 12.63011418303496833362065684
> 2 890000 12.64023571583693921823484395
> 2 900000 12.65024408926657285403820450
> 2 910000 12.66014049405210257196085418
> 2 920000 12.66993013960070545251999449
> 2 930000 12.67961329658082585267888159
> 2 940000 12.68919341076487592672709130
> 2 950000 12.69867217785123447166660908
> 2 960000 12.70805040456023104531710849
> 2 970000 12.71733330696500792660758365
> 2 980000 12.72652037909829933006161382
> 2 990000 12.73561385879130711822105048
> 2 1000000 12.74461639944508312563225494



         UnitaryPhi(m) < (-1)Sigma(m) < Sigma(m)

         So, U(2) < N(2) < S(2)=Zeta(1):*Zeta(2)

         Where,U(s) = Sum_{m=1 to infinity} UnitaryPhi(m)/m^s , S(s) = 
Sum_{m=1 to infinity} Sigma(m)/m^s



    I suppose that U(2) is already known.

    Does anyone know it?



    Yasutoshi
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From zbi74583 at boat.zero.ad.jp  Wed Nov 22 02:15:24 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Wed, 22 Nov 2006 10:15:24 +0900
Subject: A085058
References: <019201c70ab0$b44b3ce0$ca73d7dc@FMC57937B6BE00> <007501c70ac2$a6b28900$6303f0d5@speedy>
Message-ID: <00a101c70dd3$b38cd400$977fd7dc@FMC57937B6BE00>

   Is it an easy well known fact?

   I would like to know the reason why they are the same.



   Yasutoshi



  ----- Original Message ----- 
  From: Vladeta Jovovic
  To: kohmoto
  Sent: Saturday, November 18, 2006 12:35 PM
  Subject: Re: A085058



  Yes, they are the same sequence: see A101921.

  Best wishes,
  Vladeta Jovovic

    ----- Original Message ----- 
    From: kohmoto
    To: seqfan at ext.jussieu.fr
    Sent: Saturday, November 18, 2006 2:27 AM
    Subject: A085058


        Hi, Seqfans

        If 2^k(n) is the largest power of two dividing tangent number T(n) 
then numbers such that 2*n-k(n) are as follows.



        S : 2,3,2,4,2,3,2,5,2,....



        It looks like A085058.

        Are they the same seqences ?



        Yasutoshi




    __________ NOD32 1.1454 (20060321) Information __________

    This message was checked by NOD32 antivirus system.
    http://www.nod32.com
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From zbi74583 at boat.zero.ad.jp  Wed Nov 22 02:15:22 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Wed, 22 Nov 2006 10:15:22 +0900
Subject: A020955
Message-ID: <009801c70dd3$b0eb0000$977fd7dc@FMC57937B6BE00>

    Hi, Seafans.

    I suppose that no one understand what A020955 means.



    So, I explain it a little.

    First of all, see my home page and click ?Free Class?, recently I 
rewrote it.



         http://boat.zero.ad.jp/~zbi74583/another02.htm



    The idea ?ma??.an abbreviation of ?make??.is represented by a 
mapping from subsets of a set to members of the set..

    Because the idea ?ma? has the same property as a mapping.



    Number of subsets of a set which has N members is 2^N, and the 
possibility of images of each subsets is N, so number of the mapping 
subsets ->members is N^(2^N).



    But if the set is a model of Free Class then the mapping must satisfy 
some conditions.         .

    They are as follows.



    If the set is {0,1,2,3,?.n} then { } ma 0, {0} ma1, {0,1} ma 2, {0,1,2} 
ma 3,?.,{0,1,2,3,?.n} ma n .



    So, images of n+1 subsets are already decided.

    Hence the number of the mapping or the number of models which has N 
members is N^(2^N-n-1) .



    Yasutoshi


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From aplewe at sbcglobal.net  Wed Nov 22 02:59:20 2006
From: aplewe at sbcglobal.net (Andrew Plewe)
Date: Tue, 21 Nov 2006 17:59:20 -0800
Subject: Almost the number of divisors of n+1
In-Reply-To: <200611212325.kALNPQh3019729@smtp.well.com>
Message-ID: <200611220159.kAM1xMZG084485@shiva.jussieu.fr>

Marc said:

Some random and rather vague ideas:

1. I faintly recall seeing something about algorithms for
factoring n involving properties of n-1 (I mean besides
the obvious connection to "Fermat's little theorem").
Perhaps in Knuth's TAOCP? 



I have that book in front of me right now; he demonstrates how to use the
properties of n-1 to verify primality (pg. 375, in vol. 2, 2nd edition)
after applying Fermat's theorm. 

One a similar note, one could also try to solve:

n - 1 = a mod (a + 1) (or, more generally, n - x = a mod (a + x))

in which case a + 1 (or x) will be a factor of n, but I don't think that's a
particularly practical method -- finding a "mod difference" for n - x seems
to be just as hard as factoring n, even if you know the factors of n - x.

	-Andrew Plewe-

-----Original Message-----
From: Marc LeBrun [mailto:mlb at well.com] 
Sent: Tuesday, November 21, 2006 3:26 PM
To: seqfan at ext.jussieu.fr
Subject: Re: Almost the number of divisors of n+1

 >  a(n) is the number of divisors of (n+1)
 > Have you ever seen closely-related sequences like that one?

Some random and rather vague ideas:

1. I faintly recall seeing something about algorithms for
factoring n involving properties of n-1 (I mean besides
the obvious connection to "Fermat's little theorem").
Perhaps in Knuth's TAOCP?

2. It sounds like you might have a case where the relationship
you're seeing, a(n) ~ d(n), is actually a first approximation
to some identity like

   a(n) = d(n) + F1(n) + F2(n) + ...

where the Fk depend on the factor structure of n or something.
So perhaps you might try to "take out" the d(n) and then look
further into the "error" and see what you can guess about how
the Fk's specifically vary with primes, highly composites, etc.
For example you might try "fitting" a(n) with Moebius-style sums.

3. There's some rather surprising known relationships involving
divisors--my favorite is one of Legendre's that is basically an
identity found by expanding elliptic functions in power series.
This suggests that you might want to express your relationships
as generating functions and see what emerges.

4. Sometimes there is simple underlying "combinatorial machinery".
For example I was once puzzled by some coefficients that turned
out to be formed by taking each divisor of n, *adding 1*, and then
multiplying them all together!  It took me a while to see that's
algebraically identical to summing the products of every possible
subset of the divisors.  But the "add 1" way of expressing this
was totally opaque, whereas the "every subset" explained them.

Happy hunting!







From bowerc at usa.net  Wed Nov 22 03:31:46 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Tue, 21 Nov 2006 18:31:46 -0800
Subject: Assembly of nothing
Message-ID: <129kkVcEU2922S01.1164162706@cmsweb01.cms.usa.net>

This is about a problem with sequence transforms I've been thinking
about for a couple of years now. I call it the "assembly of nothing"
problem.

Assembly refers to creating a set of combinatorial structures
classified by number of nodes of the form A(B) where B is a
preexisting set of structures and A is a rule for how to place
the B-structures into the new structure.

One type of assembly is combinatorial substitution (or composition).
In this instance the "A" rule is itself a collection of structures.
The assembly is to take copies of "B" structures and use them as
the points of the "A" structure.

A variant of this assembly is one where certain pairs of points in
the "A" structure are set aside to be filled with the same "B"
structure (i.e. 2 copies of the same structure) or different
"B" structures.

What's particularly notable about these assemblies is that they
can be described by index series which are formal polynomial
series of the indexed variables x_1, x_2, x_3, ....

The "nothing" part of the title refers to "B" structures that have
a size (or number of points) of zero. The formulas detailing the
counting of these structures frequently forbid the "B" structure
from containing any zero sized members. In many cases this seems
like a necessary restriction, but in others, it can get in the
way of some useful combinatorial questions.

Let's consider 2 types of assemblies.

Sets of structures, allowing the same structure multiple times
(as implemented by the Euler transform for unlabeled structures
and the Exponential transform for labeled structures.)

Sets of structures, requiring all structures to be different
(as implemented by the Weigh transform for unlabeled structures.)

http://www.research.att.com/~njas/sequences/A000041
The partition sequence A000041 counts unlabeled sets of nonempty sets
hence it has g.f. EULER(x/(1-x)) if we consider EULER as a g.f.
transformation.

http://www.research.att.com/~njas/sequences/A000009
A000009 counts sets of nonempty sets all different from one another
with g.f. WEIGH(x/(1-x)).

in the first case, it makes sense that the "B" structure (the
nonempty sets) are not allowed to have 0 sized elements (i.e.
be empty). If we tried to count partitions of n and allowed n
to be partitioned into 0s, we would have infinitely many
possibilities. 1, 1+0, 1+0+0, ....

In the second case, we can allow empty "B" structures, since
each structure can only be used once giving us 4 partitions
of 3 into distinct parts: 3, 3+0, 1+2, 1+2+0, or in general
2*A000009(n) partitions of n into distinct parts allowing 0
as a part. Hence we would like to say that
WEIGH(1/(1-x)) = 2*WEIGH(x/(1-x)).

The commonly used formulas appear to support these observations.

The Euler transform as applied to a sequence is often rendered:

(1-x)^(-a1)*(1-x^2)^(-a2)*(1-x^3)^(-a3)*(1-x^4)^(-a4)*...

with the Weigh transform as

(1+x)^a1*(1+x^2)^a2*(1+x^3)^a3*(1+x^4)^a4*...

both of these formulas ignore the a0 term, but if we were to extend
these one term backward we would have Euler start

(1-1)^(-a0)*...

and Weigh

(1+1)^a0*...

and we can see that in Euler's case given any positive a0 we begin
with 0 raised to a negative power giving an infinite value and for
Weigh's case we start the product with 2^a0.

A similar effect is seen from the index series.

An index series is applied to a g.f. as follows:

If s(x_1, x_2, x_3, ...) is the index series, then it maps the
g.f. A(x) to s(A(x), A(x^2), A(x^3), ...) thus x_2^2 + x_3*x_4
maps as A(x^2)^2+A(x^3)*A(x^4).

Euler's i.s. is exp(x_1/1 + x_2/2 + x_3/3 + ...)

Weigh's i.s. is exp(x_1/1 - x_2/2 + x_3/3 - ...)

We can see that if A(x) has a positive constant term, then
that term will be multiplied by the divergent harmonic series
1+1/2+1/3+... and blow up. With Weigh we have the alternating
harmonic series 1-1/2+1/3-...=ln 2 giving it the 2^a0 character.

So in these 2 examples, the problem seems manageable, but if one
is looking for a general theory to cover all such assemblies or
all transforms defined by i.s., things are not so easy.

Consider 2 other assembly rules:

Ordered lists of structures, allowing the same structure multiple times
(as implemented by the Invert transform for unlabeled structures.)

Ordered lists of structures, requiring a structure not have a copy
of itself as a neighbor
(I call the unlabeled version of this the Carlitz transform. It's
not in the EIS canon yet.)

Invert's i.s. is 1/(1-x_1)
Carlitz's i.s. is 1/(1-x_1+x_2-x_3+...)

Invert creates compositions (or ordered partitions) the way Euler
creates partitions.

Hence we have INVERT(x/(1-x)) = (1-x)/(1-2x) = 1,1,2,4,8,16,...

Just as in the unordered partitions, we would not allow one to
compose n into 0s, there would be infinitely many choices.
Note that unlike the Euler transform which blows up for a0>0,
The Invert transform can be calculated for any a0 != 1, although
it's difficult to give a combinatorial interpretation to any
nonzero value.

http://www.research.att.com/~njas/sequences/A003242
Carlitz(x/(1-x)) gives us A003242 Carlitz compositions, those
where no two adjacent parts are equal. We can allow Carlitz
compositions to contain 0s since the separation of the 0s by other
numbers keeps it finite: Carlitz(1/(1-x)) = A114900.
http://www.research.att.com/~njas/sequences/A114900
However, the i.s. does not tell us how to do it. If A(x)=1 what
does one make of Carlitz(A(x)) = 1/(1-1+1-1+...)

I recently had an idea about how to deal with issues like this:
use an analytic interpretation of the g.f. of a sequence.

If a sequence represents structures contains only finitely many
structures, then the number of structures is A(1) where A(x) is
the g.f. My idea is to take the case where A(x) has a radius of
convergence (ROC) >0 and <=1 so that it cannot be evaluated by
adding the terms, and associating its size with A(1) of the
analytic extension of the g.f. Here in the Carlitz case we have
1-x+x^2-x^3+... = 1/(1+x) hence A(1) = 1/2 and 1/(1/2)=2, the
expected value for Carlitz compositions of 0 allowing 0 as a part.
(The 2 are the empty composition and zero.)

In this way I can evaluate a transform of a sequence whose only
nonzero term is a_0 as T(a_0*x)(1) and for more general sequences
I can apply the transform to a 2 dimensional sequence as follows.

If sequence has g.f. A(x) then let C(x,y) = y*A(x).

Apply T to C(x,y) getting D(x,y) as output:

Let d_n = [x^n] D(x,1)

I suspect this will be an adequate solution to many problems of
this sort, but I'm not sure how far this method can be trusted.
Part of the problem is I don't know my analytic function theory
as well as I should. (If anyone out there does, please chime in.)
More importantly, I've run across some examples that challenge
the method.

My first example involves the Euler transform. It's basically only
a problem because I would expect that if f(1)=g(1) then
(Tf)(1) = (Tg)(1) for any transform with both f and g in its
domain. It seems reasonable as 1^n=1 for all n so that i.s.'s
behave like g.f.'s when evaluated at 1. More importantly, the
choice to use the linear term in "C(x,y) = y*A(x)" as opposed
to "C(x,y) = y^2*A(x)" was arnitrary and it seemingly should
not matter which term I chose, so I would like some assurance
that I'll get the same answer following any such path.

Consider

EULER(0) = 1
EULER(x-x^2) = 1+x
EULER(x-x^3) = 1+x+x^2

For all of these, the inputs have f(1)=0, but the outputs have
g(x)=1,2 or 3

I think the reason for this anamoly may be in the instability
Euler has at 0. Just as Weigh behaves like 2^x, Euler behaves
like 0^(-x) giving 0 for negative values of x, infinity for
positive values and doing what we see here at x=0. 0^0 is
generally defined as 1, but analytically it can be anything and
that may be the limitation I'm running into.

My other example concerns me, because I'm interested in formulas
that can be applied to arbitrary sequences, not just integer
sequences. Most of our familiar transform formulas work fine for
arbitrary sequences, and it would be nice if all of them did,
however...


Ordered lists of structures, requiring all structures to be different
(as implemented by the AGK transform for unlabeled structures
found on
http://www.research.att.com/~njas/sequences/transforms2.html )

Obviously, this assembly makes sense for empty structures since
each structure can appear at most once. If I allow there to be
n types of empty structures in the input then I have
A003149(n) empty structures in the output.
http://www.research.att.com/~njas/sequences/A003149

While it seems quite plausible that this sequence can be
interpolated into a nice analytic function, the method I illustrated
above offers no help as for n anything other that a
nonnegative integer gives me a sequence with ROC of zero.

So the idea in posting this is to present some new ideas to the
sequence community, get some opinions from people who know some
things that I don't know and get some ideas about how to handle
difficult cases.

Christian








From davidwwilson at comcast.net  Wed Nov 22 04:22:55 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Tue, 21 Nov 2006 22:22:55 -0500
Subject: Confirmation
References: <200611211857.kALIvmai029436@amer.strw.leidenuniv.nl>
Message-ID: <001001c70de5$812f2190$6501a8c0@yourxhtr8hvc4p>

Thanks for your help, your results agree with my values.





From jawbrey at att.net  Wed Nov 22 04:28:31 2006
From: jawbrey at att.net (Jon Awbrey)
Date: Tue, 21 Nov 2006 22:28:31 -0500
Subject: Assembly of nothing
References: <129kkVcEU2922S01.1164162706@cmsweb01.cms.usa.net>
Message-ID: <4563C3DE.FB5134DC@att.net>

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o

This is one of the impulses that led to the construction of riffs and rotes.
It was probably in my freshman chemistry class, or later when I worked as a
hospital orderly and used to run errands for the nuclear medicine unit that
I was thinking about the techniques of molecular tagging, which is like the
business of labelling graphs, only you have to label molecules with bits of
physical structure, an extra neutron or the like, that alters the resulting
structure.

So the question arises whether you can have a class of structures that is
closed under self-labeling.  Somewhat roughly trying to use your notation,
if A(B) is the set of all structures where those of type A are labeled by
those of type B, can you have the doamin equation A = A(A) for any type A?

Or something like that ...

Jon Awbrey

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o
inquiry e-lab: http://stderr.org/pipermail/inquiry/
wikinfo: http://wikinfo.org/wiki.php?title=User:Jon_Awbrey
wp review: http://wikipediareview.com/index.php?showuser=398
o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o






From Annette.Warlich at t-online.de  Wed Nov 22 13:37:30 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Wed, 22 Nov 2006 13:37:30 +0100
Subject: Eigensequences
Message-ID: <4564448A.5010200@t-online.de>

In  a recent treatise of the pascal matrix and its
connections to the bernoulli-numbers I restated
their connection as an eigenvector/eigensequence-
relation.

Say P is the lower triangular matrix of binomial-
coefficients, Pj the matrix, where the column-signs
alternate and B the vector of bernoulli-numbers
(b_1 = +1/2)
then
   Pj * B = B

Now in connection with the subject of divergent
summmation I investigated powerseries, say
  V(x) = [1,x,x^2,x^3,... ]
in connection with Pj, like

  ??? = Pj * V(x)      // binomial-transform
  ??? = V(x)~ * Pj

(using ~ for the transpose)

Now since

   Pj * B = B

also

   V(x) * Pj * B = V(x) * B

and the effect of the coefficients of Pj
disappear, if I consider powerseries with
the bernoulli-numbers as coefficients:

 f(x) = b0 + b1*x + b2*x^2 + b3*x^3 + ...
      =    b0
         + (1*b0 - 1*b1)*x
         + (1*b0 - 2*b1 + 1*b2)*x^2
         + ...

---------
Now incidently I also found, that, using J as a diagonal-
unit-matrix with units with alternating signs
 J = diag(1,-1,1,-1,...)

such that also

 Pj  = P * J

then a matrix G, containing the bernoulli-numbers
in the first column (and shifted and scaled
in a certain manner in the follwing columns)

 Pj = G * J * G^-1

and thus, for instance, the above eigensequence-
property of the bernoulli-numbers can be extended
to an infinite set of coefficients arising from

   V(x) * Pj * G = V(x) * G * J

in the infinite number of columns of the result.

The second column, for instance adds to the above
f(x) the function g(x)

 g(x) =    1*b0*x + 2*b1*x^2 + 3*b2*x^3 + 4*b3*x^4 + ...
      = -  (    1* 0
             + (1* 0 - 1*1*b0)*x
             + (1* 0 - 2*1*b0 + 1*2*b1)*x^2
             + (1* 0 - 3*1*b0 + 3*2*b1 -  1*3*b2)*x^3
             + ...
      = -  (
               ( - 1*b0)*x
             + ( - 2*b0 + 2*b1)*x^2
             + ( - 3*b0 + 6*b1 -  3*b2)*x^3
             + ...

or, using x=1, the said eigensequence-operation, here
with the associated eigenvalue -1.

-------------------------------------
Even more interesting seems to me, that the said matrix
G can be decomposed into an eigensystem, where

  G = St2 * Z(1) * St2^-1

with St2 = matrix of stirling-numbers of 2'nd kind
     Z(s)= diag(1,  1/2^s,  1/3^s, 1/4^s,... )
the diagonal-matrix of the zeta-summands of power s.

That means, that even the bernoulli-numbers have
eigensequences related to them (even infinite many),
namely the columns of St2 , associated with the
eigenvalues (1,1/2,1/3,...) such that

  V(x) * G * St2*Z(1) = V(x)*St2

and for instance,

  f(x) = 1 + 1*x + 1*x^2 + 1*x^3 + ....  // taken from V(x)*St2

       =       1*b0                     // taken from V(x) * G * St2 * Z(1)
          + (1/2*b0 + b1)*m
          + (1/3*b0 + b1 +     b2 )*m^2
          + (1/4*b0 + b1 + 3/2*b2 +      b3 )*m^3
          + (1/5*b0 + b1 +   2*b2 +    2*b3 +     b4)*m^4
          + (1/6*b0 + b1 + 5/2*b2 + 10/3*b3 + 5/2*b4 +      b5)*m^5
           ...

which may look kind trivial, since the involved Stirling-
coefficients are all 1
Other coefficients occur in the next columns 1,2,3,..., for instance
in column 1:

   g(x)  =.  x + S2_1*x^2 + S3_1*x^3 + S4_1*x^4 + S5_1*x^5 + ....
         =              b0 * x
             ( 2/3*S2_1*b0 +      2*b1 ) * x^2
             ( 1/2*S3_1*b0 + 2*S2_1*b1 + 3     *b2 )*m^3
             ( 2/5*S4_1*b0 + 2*S3_1*b1 + 4*S2_1*b2 +    4     *b3 ) *m^4
             ( 1/3*S5_1*b0 + 2*S4_1*b1 + 5*S3_1*b2 + 20/3*S2_1*b3  +  5*b4) *m^5

and so on for the next columns.
With the involved Stirling-numbers evaluated we have then for this 2'nd
column:

   g(x)  =   .    2*x + 6*x^2 + 14*x^3 + 30*x^4  +  62*x^5
         =          2*b0 * x
            +   (   4*b0 +  4*b1)*x^2
            +   (   7*b0 + 12*b1 +  6*b2) *x^3
            +   (  12*b0 + 28*b1 + 24*b2 +  8*b3) *m^4
            +   (62/3*b0 + 60*b1 + 70*b2 + 40*b3 + 10*b4))))*m^5
              ...
which, if the bernoulli-numbers are inserted is obvious.
Using x=1 and written as sequences, we have another
eigensequence-expression and implicitely infinitely many.

This description was animated by Neil's article in the
integer-sequences-journal, which I came across recently -
though I do not know, if the above is actually of interest
related to the article (my current context is originally
the study of divergent summation, where these identities
popped up)

Regards-

Gottfried Helms







From Annette.Warlich at t-online.de  Wed Nov 22 13:46:11 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Wed, 22 Nov 2006 13:46:11 +0100
Subject: Eigensequences /editing corrections
Message-ID: <45644693.1000503@t-online.de>

// upps. In the previous article I've overlooked some
// needed editings, replacing the m,m^2,... by x,x^2 ...
// sorry. Repost here:


In  a recent treatise of the pascal matrix and its
connections to the bernoulli-numbers I restated
their connection as an eigenvector/eigensequence-
relation.

Say P is the lower triangular matrix of binomial-
coefficients, Pj the matrix, where the column-signs
alternate and B the vector of bernoulli-numbers
(b_1 = +1/2)
then
   Pj * B = B

Now in connection with the subject of divergent
summmation I investigated powerseries, say
  V(x) = [1,x,x^2,x^3,... ]
in connection with Pj, like

  ??? = Pj * V(x)      // binomial-transform
  ??? = V(x)~ * Pj

(using ~ for the transpose)

Now since

   Pj * B = B

also

   V(x) * Pj * B = V(x) * B

and the effect of the coefficients of Pj
disappear, if I consider powerseries with
the bernoulli-numbers as coefficients:

 f(x) = b0 + b1*x + b2*x^2 + b3*x^3 + ...
      =    b0
         + (1*b0 - 1*b1)*x
         + (1*b0 - 2*b1 + 1*b2)*x^2
         + ...

---------
Now incidently I also found, that, using J as a diagonal-
unit-matrix with units with alternating signs
 J = diag(1,-1,1,-1,...)

such that also

 Pj  = P * J

then a matrix G, containing the bernoulli-numbers
in the first column (and shifted and scaled
in a certain manner in the follwing columns)

 Pj = G * J * G^-1

and thus, for instance, the above eigensequence-
property of the bernoulli-numbers can be extended
to an infinite set of coefficients arising from

   V(x) * Pj * G = V(x) * G * J

in the infinite number of columns of the result.

The second column, for instance adds to the above
f(x) the function g(x)

 g(x) =    1*b0*x + 2*b1*x^2 + 3*b2*x^3 + 4*b3*x^4 + ...
      = -  (    1* 0
             + (1* 0 - 1*1*b0)*x
             + (1* 0 - 2*1*b0 + 1*2*b1)*x^2
             + (1* 0 - 3*1*b0 + 3*2*b1 -  1*3*b2)*x^3
             + ...
      = -  (
               ( - 1*b0)*x
             + ( - 2*b0 + 2*b1)*x^2
             + ( - 3*b0 + 6*b1 -  3*b2)*x^3
             + ...

or, using x=1, the said eigensequence-operation, here
with the associated eigenvalue -1.

-------------------------------------
Even more interesting seems to me, that the said matrix
G can be decomposed into an eigensystem, where

  G = St2 * Z(1) * St2^-1

with St2 = matrix of stirling-numbers of 2'nd kind
     Z(s)= diag(1,  1/2^s,  1/3^s, 1/4^s,... )
the diagonal-matrix of the zeta-summands of power s.

That means, that even the bernoulli-numbers have
eigensequences related to them (even infinite many),
namely the columns of St2 , associated with the
eigenvalues (1,1/2,1/3,...) such that

  V(x) * G * St2*Z(1) = V(x)*St2

and for instance,

  f(x) = 1 + 1*x + 1*x^2 + 1*x^3 + ....  // taken from V(x)*St2

       =       1*b0                     // taken from V(x) * G * St2 * Z(1)
          + (1/2*b0 + b1)*x
          + (1/3*b0 + b1 +     b2 )*x^2
          + (1/4*b0 + b1 + 3/2*b2 +      b3 )*x^3
          + (1/5*b0 + b1 +   2*b2 +    2*b3 +     b4)*x^4
          + (1/6*b0 + b1 + 5/2*b2 + 10/3*b3 + 5/2*b4 +      b5)*x^5
           ...

which may look kind trivial, since the involved Stirling-
coefficients are all 1
Other coefficients occur in the next columns 1,2,3,..., for instance
in column 1:

   g(x)  =.  x + S2_1*x^2 + S3_1*x^3 + S4_1*x^4 + S5_1*x^5 + ....
         =              b0 * x
             ( 2/3*S2_1*b0 +      2*b1 ) * x^2
             ( 1/2*S3_1*b0 + 2*S2_1*b1 + 3     *b2 )*x^3
             ( 2/5*S4_1*b0 + 2*S3_1*b1 + 4*S2_1*b2 +    4     *b3 ) *x^4
             ( 1/3*S5_1*b0 + 2*S4_1*b1 + 5*S3_1*b2 + 20/3*S2_1*b3  +  5*b4) *x^5

and so on for the next columns.
With the involved Stirling-numbers evaluated we have then for this 2'nd
column:

   g(x)  =   .    2*x + 6*x^2 + 14*x^3 + 30*x^4  +  62*x^5
         =          2*b0 * x
            +   (   4*b0 +  4*b1)*x^2
            +   (   7*b0 + 12*b1 +  6*b2) *x^3
            +   (  12*b0 + 28*b1 + 24*b2 +  8*b3) *x^4
            +   (62/3*b0 + 60*b1 + 70*b2 + 40*b3 + 10*b4) *x^5
              ...
which, if the bernoulli-numbers are inserted is obvious.
Using x=1 and written as sequences, we have another
eigensequence-expression and implicitely infinitely many.

This description was animated by Neil's article in the
integer-sequences-journal, which I came across recently -
though I do not know, if the above is actually of interest
related to the article (my current context is originally
the study of divergent summation, where these identities
popped up)

Regards-

Gottfried Helms








From njas at research.att.com  Wed Nov 22 17:58:32 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 22 Nov 2006 11:58:32 -0500 (EST)
Subject: A125097.gif: (Very) nice patterns
References: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>
Message-ID: <200611221658.LAA41364@fry.research.att.com>

The other day Zak said:
> Neil,
> seqfans,
>
> 1. Have a look at this full graph -
> aint these patterns (very) nice?!
> It may be known, but still it's nice...
>
> 2. Is it possible to
> put "the extended graph" into OEIS?
> In the standard graph no patterns are seen at all...
>
> 3. Neil removed  keyword "nice" from A125097
> (as he did regularly with ny sequences)-
> as if I insist that it's me who is "nice"
> not it is the seq which is nice.
>

Here is my reply:
1. i will use the b-file
2. we don't need the gif since the graph button will use the b-file
3. the definition of the sequence is
     a(n) is k such that k*(2n+1) mod 1000 = 123, or 2n+1 if no such k exists.
Now there is rule that a sequence should not depend on an arbitrary 
parameter. Here we have two of them. Why not:
a(n) is k such that k*(2n+1) mod 44445555 = 321345, or 2n+1 if no such k exists. ?
4. the sequence is not finite (a(1000) = 123)
5. although i know you are a nice guy, i do not think this sequence is nice.
Best regards
Neil





From grafix at csl.pl  Wed Nov 22 18:58:30 2006
From: grafix at csl.pl (Artur)
Date: Wed, 22 Nov 2006 18:58:30 +0100
Subject: All primes
In-Reply-To: <200611221658.LAA41364@fry.research.att.com>
References: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com> <200611221658.LAA41364@fry.research.att.com>
Message-ID: <op.tjfivsouat030q@tata>

Dear Members
I'm looking for members which have Mathematica Programme and can help in
finding the first number such that
(PrimeQ[2(n) + 1] == True) && (PrimeQ[4(
       n) + 1] == True) && (PrimeQ[6(n) + 1] == True) && (PrimeQ[8(
       n) + 1] == True) && (PrimeQ[10(n) + 1] == True) && (PrimeQ[12(
       n) + 1] == True) && (PrimeQ[14(n) + 1] == True) && (PrimeQ[16(
       n) + 1] == True) && (PrimeQ[18(n) + 1] == True) && (PrimeQ[20(
       n) + 1] == True) && (PrimeQ[22(n) + 1] == True)
these n is bigger than 4850000000000
I have Mathematica procedure which is 4000 much more quickest as above but  
for as big numbers PrimeQ working slow and will be good divided range of  
searching on few subranges and run each other on different computer.
If somebody can help let me know. I need information how long time  
computer can run without reset.
BEST REGARDS
ADRTUR JASINSKI






From rgwv at rgwv.com  Wed Nov 22 19:22:17 2006
From: rgwv at rgwv.com (Robert G. Wilson v)
Date: Wed, 22 Nov 2006 13:22:17 -0500
Subject: Whereabouts of Joe K. Crump & www.spacefire.com
Message-ID: <45649559.9030304@rgwv.com>

,

Does anyone on this list know the whereabouts of Joe K. Crump or the
status of the website www.spacefire.com ? Thanks in advance for any
information.

Bob.





From noe at sspectra.com  Wed Nov 22 19:27:50 2006
From: noe at sspectra.com (T. D. Noe)
Date: Wed, 22 Nov 2006 10:27:50 -0800
Subject: All primes
In-Reply-To: <op.tjfivsouat030q@tata>
References: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>
 <200611221658.LAA41364@fry.research.att.com> <op.tjfivsouat030q@tata>
Message-ID: <p06230905c18a46fef974@[192.168.1.43]>

At 7:04 PM +0100 11/22/06, Artur wrote:
>Dear Members
>I'm looking for members which have Mathematica Programme and can help in
>finding the first number such that
>(PrimeQ[2(n) + 1] == True) && (PrimeQ[4(
>       n) + 1] == True) && (PrimeQ[6(n) + 1] == True) && (PrimeQ[8(
>       n) + 1] == True) && (PrimeQ[10(n) + 1] == True) && (PrimeQ[12(
>       n) + 1] == True) && (PrimeQ[14(n) + 1] == True) && (PrimeQ[16(
>       n) + 1] == True) && (PrimeQ[18(n) + 1] == True) && (PrimeQ[20(
>       n) + 1] == True) && (PrimeQ[22(n) + 1] == True)
>these n is bigger than 4850000000000

See A071576.

Tony






From mathar at strw.leidenuniv.nl  Wed Nov 22 19:39:07 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Wed, 22 Nov 2006 19:39:07 +0100
Subject: Whereabouts of Joe K. Crump & www.spacefire.com
Message-ID: <200611221839.kAMId7db031776@amer.strw.leidenuniv.nl>


> From seqfan-owner at ext.jussieu.fr  Wed Nov 22 19:09:21 2006
> Date: Wed, 22 Nov 2006 13:22:17 -0500
> From: "Robert G. Wilson v" <rgwv at rgwv.com>
> To: "seqfan at ext.jussieu.fr" <SeqFan at ext.jussieu.fr>
> Subject: Whereabouts of Joe K. Crump & www.spacefire.com
> 
> Does anyone on this list know the whereabouts of Joe K. Crump or the
> status of the website www.spacefire.com ? Thanks in advance for any
> information.
..
According to Max (whom we now know as Alekseyev), there is a 
archive.org backup of his web page:

http://groups-beta.google.com/group/sci.math/browse_thread/thread/8d061b676bd7e32b/071ae01f53f9a1f5?hl=en#071ae01f53f9a1f5
(sci.math of Nov 14)
http://web.archive.org/web/20040609202646/www.spacefire.com/numbertheory/

An e-mail address of 2002 is shown in
http://www.primepuzzles.net/thepuzzlers/Crump.htm
and becomes
ssaywer1 at rr.carolina.com
in the archived web page of 2000.





From franktaw at netscape.net  Wed Nov 22 22:20:32 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 22 Nov 2006 16:20:32 -0500
Subject: Categories
Message-ID: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>

How many categories are there?

First, how many categories are there with n morphisms and k objects?  
This table starts:

 1
 2  1
 7  3 1
35 16 3 1

The first column is A058129, the number of monoids; the main diagonal 
is all 1's.  I am not
100% certain of the 16 in the final row.

Taking the row sums, we get:

1,3,11,55

the number of categories with n morphisms.  This is probably not in the 
OEIS (only
A001776 is possible - other matches become less than A058129).  The 
inverse Euler
transform,

1,2,8,41

is the number of connected categories with n morphisms; this is 
likewise probably not
in the OEIS (only A052447 is possible).

Can somebody generate more data?

Franklin T. Adams-Watters

A category is a collection of objects and morphisms; each morphism is 
from one object
to another (not necessarily different) object.  Where the destination 
of one morphism
is the source of a second, their composition is defined; composition is 
associative where
it is defined.  Each object has an identity morphism, which connects it 
to itself; this
is an identity when composed with morphisms coming in and with 
morphisms going
out.
________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From jvospost3 at gmail.com  Wed Nov 22 23:55:44 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 22 Nov 2006 14:55:44 -0800
Subject: Categories
In-Reply-To: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
Message-ID: <5542af940611221455r6e693b45kae8a9af38c6ee9eb@mail.gmail.com>

This actually relates to the enumeration results I've made on "prime"
endofunctions, namely endofunctions which are not unions of other prime
endofunctions nor categorical products of endofunctions. My partial results
here are in A125024, A124023, and a comment on A048888 that I submitted a
day or two ago about the enumeration of prime endofunctions in terms of
numbrals.

I have been working on enumerating not just endofunctions, but
(categorically) endomorphisms of endofunctions, (2-categorically)
endomorphisms of endomorphisms of endofunctions.  I've been posting partial
results on John Baez's "n-category cafe" and apparently just annoyed John
Baez, who didn't yet see that I was finding a new combinatorial result at
the base level (endofunctions) and working on the n-category hierarchy above
that, so those comments may already have been deleted from his blog.

That is, in counting all categories on n objects, most are trivially the
union of smaller disjoint categories (that is, the sagittal digraph is not
connected), or categorical products of smaller categories.  They key to
enumeration involves thus both a tree-enumeration, and the numbral part
which, per a recent paper in the Journal on Integer Sequences, relate to
balanced ordered trees.

Hard to summarize less cryptically, here, but I stand by in support of what
you're doing, and think my approach is genuinely related.

-- Jonathan Vos Post

On 11/22/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
> How many categories are there?
>
> First, how many categories are there with n morphisms and k objects?
> This table starts:
>
> 1
> 2  1
> 7  3 1
> 35 16 3 1
>
> The first column is A058129, the number of monoids; the main diagonal
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
>
> Taking the row sums, we get:
>
> 1,3,11,55
>
> the number of categories with n morphisms.  This is probably not in the
> OEIS (only
> A001776 is possible - other matches become less than A058129).  The
> inverse Euler
> transform,
>
> 1,2,8,41
>
> is the number of connected categories with n morphisms; this is
> likewise probably not
> in the OEIS (only A052447 is possible).
>
> Can somebody generate more data?
>
> Franklin T. Adams-Watters
>
> A category is a collection of objects and morphisms; each morphism is
> from one object
> to another (not necessarily different) object.  Where the destination
> of one morphism
> is the source of a second, their composition is defined; composition is
> associative where
> it is defined.  Each object has an identity morphism, which connects it
> to itself; this
> is an identity when composed with morphisms coming in and with
> morphisms going
> out.
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and
> industry-leading spam and email virus protection.
>
>
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From jvospost3 at gmail.com  Thu Nov 23 01:39:16 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 22 Nov 2006 16:39:16 -0800
Subject: Categories
In-Reply-To: <5542af940611221455r6e693b45kae8a9af38c6ee9eb@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <5542af940611221455r6e693b45kae8a9af38c6ee9eb@mail.gmail.com>
Message-ID: <5542af940611221639n2fde8e43j93dc69594ae9392b@mail.gmail.com>

Next question is thus: "How many 2-categories (bicategories) are there whose
objects are the categories on n objects?"

See:
Basic Bicategories
Tom Leinster (1998-10-04)  arXiv.org:math/9810017 A concise guide to very
basic bicategory theory, from the definition of a bicategory to the
coherence theorem. Comment: 11 pages; LaTeX 2e with Paul Taylor's diagram
macros

This is even harder to answer, because there seem to be at least 7 different
definitions of n-categories, of which categories = 1-categories, and I've
just raised the enumeration probem for n=2.

www.ima.umn.edu/categories/abstracts.html
Abstracts for the IMA 2004 Summer Program: n-Categories.
Ross Street has proposed two related definitions of n-category, one in 1987
and one in 2002.

See also:
Tom Leinster (2003-05-02) arXiv.org:math/0305049 Higher-dimensional category
theory is the study of n-categories, operads, braided monoidal categories,
and other such exotic structures. It draws its inspiration from areas as
diverse as topology, quantum algebra, mathematical physics, logic, and
theoretical computer science. This is the first book on ... Comment: Book,
410 pages

http://www.lepp.cornell.edu/spr/2002-04/msg0041086.html

In the beginning, there was nothing.
But with nothing came nothingness, the vacuity, which was something.
So now there was nothingness and somethingness,
vacuity and triviality, falsehood and truth,
which were 2 things.
So now there was a set of nothingness and somethingness,
and the whole realm of sets of elements
sprang up out of the set of nothingness and somethingness,
and sprouted functions between them
to relate them back to the set of nothingness and somethingness.
So now there was a category of sets and functions,
and the whole realm of categories of objects and morphisms
sprang up out of the category of sets and functions,
and sprouted functors between them,
to relate them back to the category of sets and functions,
which sprouted natural transformations between *them*,
to relate the relationships.
So now there was ...


-- Toby
   toby at math.ucr.edu

But your question is fundamental, and I am eager to see an answer.  I hope
it is not rude for me to be anticipating follow-ups.

Happy Thanksgiving,

Jonathan Vos Post

On 11/22/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> This actually relates to the enumeration results I've made on "prime"
> endofunctions, namely endofunctions which are not unions of other prime
> endofunctions nor categorical products of endofunctions. My partial results
> here are in A125024, A124023, and a comment on A048888 that I submitted a
> day or two ago about the enumeration of prime endofunctions in terms of
> numbrals.
>
> I have been working on enumerating not just endofunctions, but
> (categorically) endomorphisms of endofunctions, (2-categorically)
> endomorphisms of endomorphisms of endofunctions.  I've been posting partial
> results on John Baez's "n-category cafe" and apparently just annoyed John
> Baez, who didn't yet see that I was finding a new combinatorial result at
> the base level (endofunctions) and working on the n-category hierarchy above
> that, so those comments may already have been deleted from his blog.
>
> That is, in counting all categories on n objects, most are trivially the
> union of smaller disjoint categories (that is, the sagittal digraph is not
> connected), or categorical products of smaller categories.  They key to
> enumeration involves thus both a tree-enumeration, and the numbral part
> which, per a recent paper in the Journal on Integer Sequences, relate to
> balanced ordered trees.
>
> Hard to summarize less cryptically, here, but I stand by in support of
> what you're doing, and think my approach is genuinely related.
>
> -- Jonathan Vos Post
>
> On 11/22/06, franktaw at netscape.net <franktaw at netscape.net > wrote:
> >
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> > 1
> > 2  1
> > 7  3 1
> > 35 16 3 1
> >
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
> > Taking the row sums, we get:
> >
> > 1,3,11,55
> >
> > the number of categories with n morphisms.  This is probably not in the
> > OEIS (only
> > A001776 is possible - other matches become less than A058129).  The
> > inverse Euler
> > transform,
> >
> > 1,2,8,41
> >
> > is the number of connected categories with n morphisms; this is
> > likewise probably not
> > in the OEIS (only A052447 is possible).
> >
> > Can somebody generate more data?
> >
> > Franklin T. Adams-Watters
> >
> > A category is a collection of objects and morphisms; each morphism is
> > from one object
> > to another (not necessarily different) object.  Where the destination
> > of one morphism
> > is the source of a second, their composition is defined; composition is
> > associative where
> > it is defined.  Each object has an identity morphism, which connects it
> > to itself; this
> > is an identity when composed with morphisms coming in and with
> > morphisms going
> > out.
> > ________________________________________________________________________
> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > industry-leading spam and email virus protection.
> >
> >
>
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From maxale at gmail.com  Thu Nov 23 04:52:55 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 22 Nov 2006 19:52:55 -0800
Subject: Categories
In-Reply-To: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
Message-ID: <d3dac270611221952h2853749ev8fcb5ced9e2f80a7@mail.gmail.com>

On 11/22/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
> How many categories are there?
>
> First, how many categories are there with n morphisms and k objects?

Does that correspond to transitively closed digraphs with k (labeled?)
vertices and n edges?

Max





From maxale at gmail.com  Thu Nov 23 05:10:51 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 22 Nov 2006 20:10:51 -0800
Subject: Whereabouts of Joe K. Crump & www.spacefire.com
In-Reply-To: <200611221839.kAMId7db031776@amer.strw.leidenuniv.nl>
References: <200611221839.kAMId7db031776@amer.strw.leidenuniv.nl>
Message-ID: <d3dac270611222010m17a14a53x75b9de0cf2ffbe3a@mail.gmail.com>

On 11/22/06, Richard Mathar <mathar at strw.leidenuniv.nl> wrote:

> http://web.archive.org/web/20040609202646/www.spacefire.com/numbertheory/

I've put an updated version of Joe's Database of 2^x-3 factorizations at
http://www-cse.ucsd.edu/users/maxal/2nm3_db.txt

Please contribute if you like.

Max





From pauldhanna at juno.com  Thu Nov 23 07:19:41 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Thu, 23 Nov 2006 01:19:41 -0500
Subject: Formula for A112319? - Diagonals in Self-Compositions of (x+x^2)
Message-ID: <20061123.011941.272.1.pauldhanna@juno.com>

Seqfans, 
     Would someone like to try to find a formula for A112319:  
"Coefficients of x^n in the (n-1)-th self-composition of (x + x^2) for
n>=1."
1, 1, 2, 9, 64, 630, 7916, 121023, 2179556, 45179508, 1059312264, 
 
The related triangles A122888 and A122890 seem to indicate that 
A112319 may have a formula involving the Catalan numbers: 
http://www.research.att.com/~njas/sequences/A122890
 
Below I copy A112319 and triangle A122888 for your convenience. 
Other diagonals in self-compositions of (x+x^2) are A112317, A112320.
 
Be it a recurrence, series reversion, etc., any formula for these
diagonals 
could lead to other discoveries involving self-compositions of functions.

     Paul 
----------------------------------------------------------------

A122888
Triangle, read by rows, where row n lists the coefficients of x^k, 
k=1..2^n, in the n-th self-composition of (x + x^2) for n>=0. 

EXAMPLE  
Triangle begins: 
1; 
1, 1; 
1, 2, 2, 1; 
1, 3, 6, 9, 10, 8, 4, 1; 
1, 4, 12, 30, 64, 118, 188, 258, 302, 298, 244, 162, 84, 32, 8, 1; 

Multiplying the g.f. of column k by (1-x)^k, k>=1, with leading zeros, 
yields the g.f. of row k in the triangle A122890: 
1; 
0, 1; 
0, 0, 2; 
0, 0, 1, 5; 
0, 0, 0, 10, 14; 
0, 0, 0, 8, 70, 42; 
0, 0, 0, 4, 160, 424, 132; 
0, 0, 0, 1, 250, 1978, 2382, 429; 
0, 0, 0, 0, 302, 6276, 19508, 12804, 1430; ... 

in which the main diagonal is the Catalan numbers 
and the row sums form the factorials. 

----------------------------------------------------------------

A112319  
Coefficients of x^n in the (n-1)-th self-composition of (x + x^2) for
n>=1. 

1, 1, 2, 9, 64, 630, 7916, 121023, 2179556, 45179508, 1059312264, 
27715541568, 800423573676, 25289923553700, 867723362137464, 
32128443862364255, 1276818947065793736, 54208515369076658640, 
2448636361058495090816
 
FORMULA  
a(n) = [x^n] F_{n-1}(x) where F_n(x) = F_{n-1}(x+x^2) 
with F_1(x) = x+x^2 and F_0(x)=x for n>=1. 
 
EXAMPLE  
Initial terms in self-compositions of (x+x^2) are: 
F(x) = x + (1)*x^2 
F(F(x)) = x + 2*x^2 + (2)*x^3 + x^4 
F(F(F(x))) = x + 3*x^2 + 6*x^3+ (9)*x^4 +... 
F(F(F(F(x)))) = x + 4*x^2 + 12*x^3 + 30*x^4 + (64)*x^5 +... 
F(F(F(F(F(x))))) = x + 5*x^2 + 20*x^3 + 70*x^4 + 220*x^5 + (630)*x^6 +...
 
----------------------------------------------------------------
END.
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From jvospost3 at gmail.com  Thu Nov 23 07:34:45 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 22 Nov 2006 22:34:45 -0800
Subject: Categories
In-Reply-To: <d3dac270611221952h2853749ev8fcb5ced9e2f80a7@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <d3dac270611221952h2853749ev8fcb5ced9e2f80a7@mail.gmail.com>
Message-ID: <5542af940611222234r3ae28d48h1c3e6541143896a7@mail.gmail.com>

First, John Baez agrees that an earlier post of his partially answered my
follow-up question about 2-categories (some assembly required):

http://golem.ph.utexas.edu/category/2006/10/klein_2geometry_vi.html#c005157

 Re: Klein 2-Geometry VI

Tim writes:

So I'll wrap up and say what I think I've discovered. Just as
finite-dimensional vector spaces and their subspaces are intimately related
to finite sets and their subsets, so finite-dimensional 2-vector spaces and
their 2-subspaces are intimately related to finite graphs and their
subgraphs. These are obviously much more complicated and I'm not at all
confident my analyses here are correct, particularly the identification of
the various 2-Grassmannians associated to various graph maps. The basic idea
seems sound though.

Yeah, that's a great idea!

The way I see it, your fundamental idea is this. There's a 2-functor from
the 2-category of directed graphs to the 2-category of vector 2-spaces
(?2-term chain complexes). And, it goes like this:

   - Any directed graph gives a 2-term chain complex.
   - Any map between directed graphs gives a chain map between 2-term
   chain complexes.
   - And, there's also a kind of "homotopy between maps between directed
   graphs", which gives a chain homotopy between chain maps between 2-term
   chain complexes.

(Do graph theorists ever think about those homotopies? They should!)

All this generalizes further, to an (n+1)-functor from n-dimensional cell
complexes to (n+1)-term chain complexes.

I'm not sure how to get some amazing new insights into projective n-geometry
from this way of thinking - but that's not surprising, since I just read
your post 2 minutes ago. We should mull on it.

Thanks!
Posted by: John Baez <http://math.ucr.edu/home/baez/> on October 8, 2006
7:25 PM |

Second, I took Category theory in grad school 1976 or 1977, and forgot what
little I knew, and have relearned only some. So anyone in seqfans who
learned more and remebered more should jump in here, and don't worry about
making me look foolish, as I take the risk of exposing my ignorance every
time I post here.

If one asks: "how many categories are there with n morphisms and k objects?"
one is asking up to isomorphism. I'm making a first cut here off the top of
my head, which may be no more than an ill-informed guess.

The sagittal graph of a general morphism a.k.a. homomorphism (not restricted
to a monomorphism, epimorphism, isomorphism, endomorphism, or automorphism)
leaves us wondering what the structure of the objects are.  I think, by the
way that the question is asked, that they have no structure, i.e. that they
are just points. If they have structure, that affects everything.
unstructrured objects (points), then we still get different counts if the
vertices and arcs are labeled or unlabeled, and whether one allows loops or
not. if the edges are unlabeled, and loops are allowed, then any of the k
objects can be mapped to any of the k objects including itself, giving k^k,
if n = k.  If n>k then we can 't use more than k of the n because the
mapping must be functional, that is, only one arrow out of each point (but
multiple arrows to a point is allowed).

 If n<k then the first point can be mapped to any of the k points, and we
run out of arrows when we reach k^n.  Now we see why labeling matters, as if
they are unlabeled, we count only once in equivalence classes, but if
labeled each such mapping is distinct.  Each loop and arc is de facto
labeled by its source point and target point, as an ordered pair.  If you
want to allow other labels, you have to multiply by the proper combinational
multiplier.

So we're taking a summation of n=0 (just the k points and none map to any),
n=1 (one of k points maps to one of k), n=2 and so forth.

That starts the answer.  For k=n and any point can map to any, we have the
count of endomorphisms being the same as the count of endofunctions already
in OEIS.

Category theory has differing approaches depending on whether we have
digraphs, multgraphs,  reversible digraphs, reflexive digraphs, irreflexive
digraphs, and various tricks to represent undirected graphs with directed
graphs.

So exact definitions are necessary.

Sorry, just a first cut.  Not enough sleep.  Some time tomorrow before gests
and Thanksgiving dinner kicks in, but no promises.

Thanks for the dialog.

Best,

Jonathan Vos Post
<http://golem.ph.utexas.edu/cgi-bin/MT-3.0/sxp-comments.pl?entry_id=955;parent_id=5157>


On 11/22/06, Max A. <maxale at gmail.com> wrote:
>
> On 11/22/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
>
> Does that correspond to transitively closed digraphs with k (labeled?)
> vertices and n edges?
>
> Max
>
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From Roland.Bacher at ujf-grenoble.fr  Thu Nov 23 07:54:39 2006
From: Roland.Bacher at ujf-grenoble.fr (Roland Bacher)
Date: Thu, 23 Nov 2006 07:54:39 +0100
Subject: Categories
In-Reply-To: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
Message-ID: <20061123065439.GA28473@fourier.ujf-grenoble.fr>


This numbers can also be obtained as the numbers
of the following quotients of quivers:

Associate to a category with n morphisms and k objects 
the quiver with k vertices corresponding to the objects 
and a(X,Y) arrows directed from the object X to the object
Y if there are a(X,Y) morphisms from X into Y.

Given two morphism g:X-->Y, f:Y-->Z with composition
h=f o g:X-->Z, put the relation gf=h on the quiver algebra

The resulting quotient algebra has dimension n and a basis
given by the (simple) arrows.

This leads to the following "algorithm" for enumerating
all categories with n morphisms and k objects:

(a) enumerate all quivers (directed graphs) with k vertices and
n oriented edges.

(b) Associate the following "weight" to such a quiver as follows :

(b1)given a triplet of vertices X,Y,Z,
set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
(this counts the number of ways which associate a morphism
X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the 
number of oriented arrows starting at U and ending at W.

(b2) associate to a given quiver the weight 
\prod_{X,Y,Z} w(X,Y,Z)
where the product is over all triplets of vertices.

The total sum of such weighted quivers yields then the solution.

Roland Bacher



   


On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.net wrote:
> How many categories are there?
> 
> First, how many categories are there with n morphisms and k objects?  
> This table starts:
> 
> 1
> 2  1
> 7  3 1
> 35 16 3 1
> 
> The first column is A058129, the number of monoids; the main diagonal 
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
> 
> Taking the row sums, we get:
> 
> 1,3,11,55
> 
> the number of categories with n morphisms.  This is probably not in the 
> OEIS (only
> A001776 is possible - other matches become less than A058129).  The 
> inverse Euler
> transform,
> 
> 1,2,8,41
> 
> is the number of connected categories with n morphisms; this is 
> likewise probably not
> in the OEIS (only A052447 is possible).
> 
> Can somebody generate more data?
> 
> Franklin T. Adams-Watters
> 
> A category is a collection of objects and morphisms; each morphism is 
> from one object
> to another (not necessarily different) object.  Where the destination 
> of one morphism
> is the source of a second, their composition is defined; composition is 
> associative where
> it is defined.  Each object has an identity morphism, which connects it 
> to itself; this
> is an identity when composed with morphisms coming in and with 
> morphisms going
> out.
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and 
> industry-leading spam and email virus protection.
> 





From jvospost3 at gmail.com  Thu Nov 23 18:14:47 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 23 Nov 2006 09:14:47 -0800
Subject: Categories
In-Reply-To: <20061123065439.GA28473@fourier.ujf-grenoble.fr>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
Message-ID: <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>

Roland Bacher correctly refers to quivers as summarized in wikipedia at

http://en.wikipedia.org/wiki/Quiver_%28mathematics%29

On 11/22/06, Roland Bacher <Roland.Bacher at ujf-grenoble.fr> wrote:
>
>
> This numbers can also be obtained as the numbers
> of the following quotients of quivers:
>
> Associate to a category with n morphisms and k objects
> the quiver with k vertices corresponding to the objects
> and a(X,Y) arrows directed from the object X to the object
> Y if there are a(X,Y) morphisms from X into Y.
>
> Given two morphism g:X-->Y, f:Y-->Z with composition
> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
>
> The resulting quotient algebra has dimension n and a basis
> given by the (simple) arrows.
>
> This leads to the following "algorithm" for enumerating
> all categories with n morphisms and k objects:
>
> (a) enumerate all quivers (directed graphs) with k vertices and
> n oriented edges.
>
> (b) Associate the following "weight" to such a quiver as follows :
>
> (b1)given a triplet of vertices X,Y,Z,
> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> (this counts the number of ways which associate a morphism
> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> number of oriented arrows starting at U and ending at W.
>
> (b2) associate to a given quiver the weight
> \prod_{X,Y,Z} w(X,Y,Z)
> where the product is over all triplets of vertices.
>
> The total sum of such weighted quivers yields then the solution.
>
> Roland Bacher
>
>
>
>
>
>
> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.net wrote:
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> > 1
> > 2  1
> > 7  3 1
> > 35 16 3 1
> >
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
> > Taking the row sums, we get:
> >
> > 1,3,11,55
> >
> > the number of categories with n morphisms.  This is probably not in the
> > OEIS (only
> > A001776 is possible - other matches become less than A058129).  The
> > inverse Euler
> > transform,
> >
> > 1,2,8,41
> >
> > is the number of connected categories with n morphisms; this is
> > likewise probably not
> > in the OEIS (only A052447 is possible).
> >
> > Can somebody generate more data?
> >
> > Franklin T. Adams-Watters
> >
> > A category is a collection of objects and morphisms; each morphism is
> > from one object
> > to another (not necessarily different) object.  Where the destination
> > of one morphism
> > is the source of a second, their composition is defined; composition is
> > associative where
> > it is defined.  Each object has an identity morphism, which connects it
> > to itself; this
> > is an identity when composed with morphisms coming in and with
> > morphisms going
> > out.
> > ________________________________________________________________________
> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > industry-leading spam and email virus protection.
> >
>
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From jvospost3 at gmail.com  Thu Nov 23 21:10:51 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 23 Nov 2006 12:10:51 -0800
Subject: Categories
In-Reply-To: <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
Message-ID: <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>

*"Formalized Proof, Computation, and the Construction Problem in Algebraic
geometry", by Carlos Simpson.

[PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
2004<http://arxiv.org/pdf/math/0410224> File
Format: PDF/Adobe Acrobat - View as
HTML<http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5>
finite integer N, *how many categories* are there with N morphisms? What *
...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
(1965), 450-464.



On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> Roland Bacher correctly refers to quivers as summarized in wikipedia at
>
> http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
>
> On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> >
> >
> > This numbers can also be obtained as the numbers
> > of the following quotients of quivers:
> >
> > Associate to a category with n morphisms and k objects
> > the quiver with k vertices corresponding to the objects
> > and a(X,Y) arrows directed from the object X to the object
> > Y if there are a(X,Y) morphisms from X into Y.
> >
> > Given two morphism g:X-->Y, f:Y-->Z with composition
> > h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> >
> > The resulting quotient algebra has dimension n and a basis
> > given by the (simple) arrows.
> >
> > This leads to the following "algorithm" for enumerating
> > all categories with n morphisms and k objects:
> >
> > (a) enumerate all quivers (directed graphs) with k vertices and
> > n oriented edges.
> >
> > (b) Associate the following "weight" to such a quiver as follows :
> >
> > (b1)given a triplet of vertices X,Y,Z,
> > set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> > (this counts the number of ways which associate a morphism
> > X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> > number of oriented arrows starting at U and ending at W.
> >
> > (b2) associate to a given quiver the weight
> > \prod_{X,Y,Z} w(X,Y,Z)
> > where the product is over all triplets of vertices.
> >
> > The total sum of such weighted quivers yields then the solution.
> >
> > Roland Bacher
> >
> >
> >
> >
> >
> >
> > On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.net wrote:
> > > How many categories are there?
> > >
> > > First, how many categories are there with n morphisms and k objects?
> > > This table starts:
> > >
> > > 1
> > > 2  1
> > > 7  3 1
> > > 35 16 3 1
> > >
> > > The first column is A058129, the number of monoids; the main diagonal
> > > is all 1's.  I am not
> > > 100% certain of the 16 in the final row.
> > >
> > > Taking the row sums, we get:
> > >
> > > 1,3,11,55
> > >
> > > the number of categories with n morphisms.  This is probably not in
> > the
> > > OEIS (only
> > > A001776 is possible - other matches become less than A058129).  The
> > > inverse Euler
> > > transform,
> > >
> > > 1,2,8,41
> > >
> > > is the number of connected categories with n morphisms; this is
> > > likewise probably not
> > > in the OEIS (only A052447 is possible).
> > >
> > > Can somebody generate more data?
> > >
> > > Franklin T. Adams-Watters
> > >
> > > A category is a collection of objects and morphisms; each morphism is
> > > from one object
> > > to another (not necessarily different) object.  Where the destination
> > > of one morphism
> > > is the source of a second, their composition is defined; composition
> > is
> > > associative where
> > > it is defined.  Each object has an identity morphism, which connects
> > it
> > > to itself; this
> > > is an identity when composed with morphisms coming in and with
> > > morphisms going
> > > out.
> > >
> > ________________________________________________________________________
> > > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > > industry-leading spam and email virus protection.
> > >
> >
>
>
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From jvospost3 at gmail.com  Thu Nov 23 21:20:58 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 23 Nov 2006 12:20:58 -0800
Subject: Categories
In-Reply-To: <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
	 <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
Message-ID: <5542af940611231220oe45dce2rbee869c1dea94b9e@mail.gmail.com>

FOM: Set theory vs category theory: Representability of categories *Vaughan
Pratt* pratt at cs.Stanford.EDU
<fom%40cs.nyu.edu?Subject=FOM%3A%20Set%20theory%20vs%20category%20theory%3A%20Representability%20of%20categories&In-Reply-To=199801271058.LAA14876%40brics.dk>
*Tue Jan 27 13:37:42 EST 1998

http://cs.nyu.edu/pipermail/fom/1998-January/001021.html
*



A number of representation theorems were found by Czech algebraists and
category theorists during the latter half of the 1960's, representing
arbitrary small categories as categories of familiar concrete structures
and their homomorphisms.  Two notable representations are in terms
of semigroups, and of directed graphs.  In each case the theorem is
that every small category is representable as a category of semigroups
(directed graphs, etc.) and their homomorphisms (of the standard kind
in each case).

The criteria used here for representability of objects  a  and morphisms
f:a-&gt;b of a category C by respectively structures F(a) and homomorphisms
F(f): F(a)->F(b) of the concrete representing category D are as follows.

(i) F:C->D is a *functor*, that is, a graph homomorphism (viewing the
morphisms of C and D as edges of their respective underlying graphs)
that preserves composition and identities (a condition exactly analogous
to those of monoid homomorphism and group homomorphism).

(ii) F is *faithful*, i.e. for any pair a,b of objects of C, distinct
morphisms f:a-&gt;b of C are represented by distinct homomorphisms
F(f):F(a)->F(b) of D.

(iii) F is *full*, i.e. every homomorphism from F(a) to F(b) in D is
the representation under F of some morphism from a to b in C.

The intuitive meaning of full and faithful is that the representing
object is just as stiff as the object it represents when you wiggle both.
Fullness means it is at least as stiff (D offers no new transformations),
while faithfulness means it is at most as stiff (no transformations are
lost by identification).

Unfortunately the project of embedding arbitrary categories fully and
faithfully in familiar concrete categories has turned out not to have
the impact of its counterparts in group theory (permutations), Boolean
algebras (fields of sets), distributive lattices (rings of sets), etc.
These results have in consequence not been widely publicized in the
introductory categorical literature.

The most objectionable feature of these embeddings in my view (and I
am aware of no other serious objection) is their lack of respect for
concreteness itself.  If one takes for C a small *concrete* category,
such as all quotient groups of the additive group of natural numbers
and their group homomorphisms, the above-mentioned embedding represents
the finite group Z_2 as an infinite semigroup!  And if &quot;natural
numbers&quot;

is replaced by "reals", the corresponding embedding represents Z_2 as an
uncountable semigroup.  Such a lack of respect for concreteness would
seem to undermine the very point of representing abstract objects as
concrete structures.

In contrast a finite group of order n is representable by permutations of
n things, a finite Boolean algebra as a field of subsets of a finite set,
and so on.  In these better-known representation theorems, cardinality
is respected, typically to within at most two exponentials (which in
the infinite case means to within two beth numbers) and often much better.

This unsatisfactory situation with the concrete representation of
objects of arbitrary categories prompts the speculation that the notion
of category is too broad to admit of any improvement in the situation.
After all, any random graph G is made a category by adding to its
edges all paths in G, with composition defined as path concatenation
(the free category generated by G).  How could there possibly be a
cardinality-respecting representation of the vertices of a random graph
in terms of the objects of a familiar concrete category?  When one can
do any violence one wishes to the structure by adding an edge at random
between any two vertices, it would seem that any concrete representation
of each object would inevitably have a cardinality on the order of the
whole category.

But in fact there *is* a representation that does the job.  Let us make
two easily understood modifications to the usual notion of topological
space.

First, drop the requirement that the open sets be closed under arbitrary
union and finite intersection.  (The role of this traditional restriction
is in effect to limit the &quot;signature&quot; of a topological space
to just that
needed to express the notion of limit, or even less with the more discrete
spaces; dropping it greatly broadens the range of possible signatures.)

Second, in place of the customary two degrees of membership in the open
sets (ordinary sets), permit a set K of possible degrees of membership,
call such open sets fuzzy (Zadeh's fuzzy sets are the case K the reals).

Define continuity as usual for functions between topological spaces,
wording the definition in such a way however that its extension to
fuzzy sets is made clear.  Specifically, f:X-&gt;Y is continuous when for
every open set g:Y-&gt;K (here expressed as a characteristic function),
the composite

		   f      g
		X ---> Y ---> K

is an open set of X.  When K = {0,1} it is easily verified that this is
the standard notion of continuity.

Call such generalized topological spaces Chu spaces over K.  An even more
general notion was first studied in detail by Peter Chu in the 1970's,
a master's student of Michael Barr.  The more elementary version above
was first studied by Lafont and Streicher in 1991 (LICS) under the rubric
of games but the term &quot;Chu construction&quot; was already in the computer
science air in the late 1980's making it too late to change the name.

Definitions.

1.  A concrete category is a pair (C,U) where C is a category and U:C-&gt;Set
is a faithful functor, the "forgetful" functor giving the *underlying
set* of each object along with the realization of each morphism of C as
a function (a morphism of Set).

2.  A concrete functor F:(C,U)->(D,V) between two concrete categories
is one satisfying VF = U.  That is, the underlying set VF(a) of the
representation F(a) of a is the same as the underlying set U(a) of
a itself.

Remark.  The elements of all the representing sets of a small concrete
category (C,U) themselves form a single set, namely the disjoint or
marked union of all the sets in the category.  Call this set Elt(C,U),
the set of elements of (C,U).

A very minor technicality requires the notion of an *honest* concrete
category, namely one having, for all objects a and b, a morphism from
a to b if U(a) is empty (necessarily at most one, by concreteness).
Dishonest concrete categories cannot be represented as below because
the empty topological space has nowhere to store the information as to
which other spaces it does or does not have functions to.  (Thanks to
Peter Freyd for pointing out the lacuna in my proof that necessitated
this condition.)

Theorem.  Every small honest concrete category (C,U) embeds fully,
faithfully, and concretely in the category of Chu spaces over Elt(C,U).

(This theorem enjoys all the advantages of the previous embeddings, but
in addition satisfies the strongest possible concreteness requirement:
the representing object has the *same* underlying set as the object it
represents, and the representing morphisms viewed concretely are exactly
the same functions.)

Proof.  Let b be an object of C with underlying set X = U(b).  Represent
b by a generalized topological space X, having one open set Y_h for
each morphism h:b-&gt;c in C for some c.  Form Y_h by taking the degree of
membership of each point x in Y_h to be U(h)(x), i.e. the element of U(c)
which is the image of x under the underlying function U(h) of h.

It is then straightforward to show that this representation is faithful,
full, and concrete.

Obviously the burden of the obstacle that we have overcome has merely
been shifted to the open sets, the number of which must now be on the
order of the cardinality of the category.  But since when has the number
of open sets been an obstacle to topology?  The points of a space are
clearly visible, but our embedding has left these untouched.  Who has
ever seen an open set or cares about how many there are?

For more information on Chu spaces consult

	<A HREF="http://boole.stanford.edu/chuguide.html">http://boole.stanford.edu/chuguide.html</A>

My original interest in Chu spaces was as a model of concurrent
behavior.  More recently I have become interested in them as a primarily
set-theoretic foundation for mathematics that more directly emulates
what category theory has to offer than the extant litany of single-sorted
and and multisorted relational structures and algebras with and without
(ordinary) topology and their standard homomorphisms, all of which are
fully, faithfully, and concretely representable by Chu spaces.

---
Name: Vaughan Pratt
Position: Professor of Computer Science
Institution: Stanford University
Research interests: Foundations of computation and mathematics
For more information: <A
HREF="http://boole.stanford.edu">http://boole.stanford.edu</A>
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From maxale at gmail.com  Thu Nov 23 23:13:45 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 23 Nov 2006 14:13:45 -0800
Subject: connected components in subgraphs of the complete graph: A125205 - A125207
Message-ID: <d3dac270611231413v4ea7e950ifc9405a7ed04e057@mail.gmail.com>

SeqFans,

I've just sent to Neil the following three sequences that you may find
interesting.

It would be also interesting to compute unlabeled variants of these
sequences. If you like to contribute, please do.

Regards,
Max


%I A125205
%S A125205 1,2,1,3,6,3,1,4,18,30,24,15,6,1,5,40,135,250,295,282,215,120,45,10,1,6,
%T A125205 75,420,1385,3015,4800,6365,7170,6705,5065,3009,1365,455,105,15,1,7,126,
%U A125205 1050,5355,18690,47880,96796,166890,251370,329945,373947,362292,297115
%N A125205 Triangular array T(n,k) (n>=1, 0<=k<=n(n-1)/2) giving the
total number
           of connected components in all subgraphs (V,E') with |E'|=k of
           the complete labeled graph K_n=(V,E).
%F A125205 g.f.: Sum_{n,k}
T(n,k)*x^n/n!*y^k=(F(x,y)-1)*exp(F(x,y)-1)=G(x,y)*ln(G(x,y))
           where G(x,y)=Sum_{n=0..oo} (1+y)^(n(n-1)/2)*x^n/n!, and
F(x,y)=1+ln(G(x,y))
           is g.f. of A062734.
%e A125205 The array starts with
           1
           2, 1
           3, 6, 3, 1
           4, 18, 30, 24, 15, 6, 1
           5, 40, 135, 250, 295, 282, 215, 120, 45, 10, 1
           ...
%e A125205 T(3,1)=6 since there are three different subgraphs of K_3
with one edge,
           and each subgraph has two connected components.
%o A125205 (PARI) { reverse(v)=vector(length(v),i,v[length(v)+1-i]) }
           G=sum(n=0,6,(1+y)^(n*(n-1)/2)*x^n/n!); K=G*log(G);
           for(n=1,6,print(reverse(Vec(n!*polcoeff(K,n,x)))))
%Y A125205 Cf. A062734
%Y A125205 Cf. A125206 (row-reversed version), A125207 (row sums)
%K A125205 nonn,tabf
%O A125205 1,2
%A A125205 Max Alekseyev (maxal(AT)cs.ucsd.edu), Nov 23 2006



%I A125206
%S A125206 1,1,2,1,3,6,3,1,6,15,24,30,18,4,1,10,45,120,215,282,295,250,135,40,5,1,
%T A125206 15,105,455,1365,3009,5065,6705,7170,6365,4800,3015,1385,420,75,6,1,21,210,
%U A125206 1330,5985,20349,54271,116385,204225,297115,362292,373947,329945,251370
%N A125206 Triangular array T(n,k) (n>=1, 0<=k<=n(n-1)/2) giving the
total number
           of connected components in all subgraphs obtained from the complete
           labeled graph K_n by removing k edges.
%C A125206 Row-reversed version of A125205, see A125205 for further details
%e A125206 The array starts with
           1
           1, 2
           1, 3, 6, 3
           1, 6, 15, 24, 30, 18, 4
           1, 10, 45, 120, 215, 282, 295, 250, 135, 40, 5
           ...
%Y A125206 Cf. A125205 (row-reversed version), A125207 (row sums)
%K A125206 nonn,tabf
%O A125206 1,3
%A A125206 Max Alekseyev (maxal(AT)cs.ucsd.edu), Nov 23 2006




%I A125207
%S A125207 1,3,13,98,1398,39956,2354240,286394544,71225744048,35884971729760,36419817759267072,
%T A125207 74221711070826087424,303193538300703211111936,2480118087478081928075065344,
%U A125207 40601989279034990139321984265216,1329877330680067685563700135615633408
%N A125207 The total number of connected components in all subgraphs
obtained from
           the complete labeled graph K_n by removing zero or more edges.
%F A125207 E.g.f.: (F(x)-1)*exp(F(x)-1) = G(x)*ln(G(x)) where
           G(x)=Sum_{n=0..oo} 2^(n(n-1)/2)*x^n/n! and F(x)=1+ln(G(x))
is e.g.f. of A001187
%e A125207 For n=2, we have two graph on two vertices: complete and
empty, the former has
           one connected component while the latter has two connected
components. The total
           number of connected components is 3, which is a(2).
%o A125207 (PARI) G=sum(n=0,30,2^(n*(n-1)/2)*x^n/n!) + O(x31); v=Vec(G*log(G));
           for(i=1,length(v),v[i]*=i!); print(v)
%Y A125207 Cf. A001187, A125205, A125206
%K A125207 nonn
%O A125207 1,2
%A A125207 Max Alekseyev (maxal(AT)cs.ucsd.edu), Nov 23 2006





From davidwwilson at comcast.net  Fri Nov 24 01:39:43 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 23 Nov 2006 19:39:43 -0500
Subject: b-files for power series expansions
Message-ID: <01ad01c70f61$09624cf0$6501a8c0@yourxhtr8hvc4p>

I guess my point was, if someone has access to Mma or some other symbolic math package, it would be nice if they could write a program to generate b-files for sequences that are e.g.f. expansions of real functions. We could then generate a whole slew of b-files fairly easily.
 
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From maxale at gmail.com  Fri Nov 24 02:04:09 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 23 Nov 2006 17:04:09 -0800
Subject: Chestnut
In-Reply-To: <000601c6f538$1ca30fe0$6401a8c0@yourxhtr8hvc4p>
References: <000601c6f538$1ca30fe0$6401a8c0@yourxhtr8hvc4p>
Message-ID: <d3dac270611231704ua5487a7s547879c8e77174b7@mail.gmail.com>

On 10/21/06, David Wilson <davidwwilson at comcast.net> wrote:
>
>
>
>
> This is a chestnut that I am convinced it is true, but I cannot prove it. I
> have posed it on math-fun occasionally, and never got an answer. I would
> like to see it posed to some better minds that some of you might know, e.g,
> JHC or other number theorist.
>
> Let S and T be sets of real numbers. Call T a divider of S if some element
> of T lies strictly between any two elements of S.
>
> For integer n >= 1, let Fence(n) be the set of all rationals with
> denominator n, that is, { k/n : k in Z }.
>
> For real set S, let f(S) be the least n such that Fence(n) is a divider of
> S, if such an n exists.
>
> Let Recip(n) be the set of all integer reciprocals on [0,1] with denominator
> <= n, that is, { 1/b : 1 <= b <= n }
>
> Let Farey(n) be the set of all rationals on [0,1] with denominator <= n,
> that is, { a/b : 0 <= a <= b, 1 <= b <= n }
>
> Is f(Farey(n)) = f(Recip(n)) for every n?
>
> The apparently common sequences a(n) = f(Recip(n)) =? f(Farey(n)) is in the
> OEIS, I cannot find it in the short time I have.





From Roland.Bacher at ujf-grenoble.fr  Fri Nov 24 08:53:08 2006
From: Roland.Bacher at ujf-grenoble.fr (Roland Bacher)
Date: Fri, 24 Nov 2006 08:53:08 +0100
Subject: Categories
In-Reply-To: <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com> <20061123065439.GA28473@fourier.ujf-grenoble.fr> <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com> <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
Message-ID: <20061124075307.GA17864@fourier.ujf-grenoble.fr>


I realised that the formula below for enumerating categories
using quivers is completely wrong. It gives only an upper bound
since one has to require moreover associativity for the composition of 
morphisms. The quiver approach works however in principle but
is much more complex (and I guess this approach has been 
used in some of the previous approaches).   Roland Bacher

On Thu, Nov 23, 2006 at 12:10:51PM -0800, Jonathan Post wrote:
> *"Formalized Proof, Computation, and the Construction Problem in Algebraic
> geometry", by Carlos Simpson.
> 
> [PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
> 2004<http://arxiv.org/pdf/math/0410224> File
> Format: PDF/Adobe Acrobat - View as
> HTML<http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5>
> finite integer N, *how many categories* are there with N morphisms? What *
> ...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
> (1965), 450-464.
> 
> 
> 
> On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> >
> >Roland Bacher correctly refers to quivers as summarized in wikipedia at
> >
> >http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
> >
> >On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> >>
> >>
> >> This numbers can also be obtained as the numbers
> >> of the following quotients of quivers:
> >>
> >> Associate to a category with n morphisms and k objects
> >> the quiver with k vertices corresponding to the objects
> >> and a(X,Y) arrows directed from the object X to the object
> >> Y if there are a(X,Y) morphisms from X into Y.
> >>
> >> Given two morphism g:X-->Y, f:Y-->Z with composition
> >> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> >>
> >> The resulting quotient algebra has dimension n and a basis
> >> given by the (simple) arrows.
> >>
> >> This leads to the following "algorithm" for enumerating
> >> all categories with n morphisms and k objects:
> >>
> >> (a) enumerate all quivers (directed graphs) with k vertices and
> >> n oriented edges.
> >>
> >> (b) Associate the following "weight" to such a quiver as follows :
> >>
> >> (b1)given a triplet of vertices X,Y,Z,
> >> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> >> (this counts the number of ways which associate a morphism
> >> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> >> number of oriented arrows starting at U and ending at W.
> >>
> >> (b2) associate to a given quiver the weight
> >> \prod_{X,Y,Z} w(X,Y,Z)
> >> where the product is over all triplets of vertices.
> >>
> >> The total sum of such weighted quivers yields then the solution.
> >>
> >> Roland Bacher
> >>
> >>
> >>
> >>
> >>
> >>
> >> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.net wrote:
> >> > How many categories are there?
> >> >
> >> > First, how many categories are there with n morphisms and k objects?
> >> > This table starts:
> >> >
> >> > 1
> >> > 2  1
> >> > 7  3 1
> >> > 35 16 3 1
> >> >
> >> > The first column is A058129, the number of monoids; the main diagonal
> >> > is all 1's.  I am not
> >> > 100% certain of the 16 in the final row.
> >> >
> >> > Taking the row sums, we get:
> >> >
> >> > 1,3,11,55
> >> >
> >> > the number of categories with n morphisms.  This is probably not in
> >> the
> >> > OEIS (only
> >> > A001776 is possible - other matches become less than A058129).  The
> >> > inverse Euler
> >> > transform,
> >> >
> >> > 1,2,8,41
> >> >
> >> > is the number of connected categories with n morphisms; this is
> >> > likewise probably not
> >> > in the OEIS (only A052447 is possible).
> >> >
> >> > Can somebody generate more data?
> >> >
> >> > Franklin T. Adams-Watters
> >> >
> >> > A category is a collection of objects and morphisms; each morphism is
> >> > from one object
> >> > to another (not necessarily different) object.  Where the destination
> >> > of one morphism
> >> > is the source of a second, their composition is defined; composition
> >> is
> >> > associative where
> >> > it is defined.  Each object has an identity morphism, which connects
> >> it
> >> > to itself; this
> >> > is an identity when composed with morphisms coming in and with
> >> > morphisms going
> >> > out.
> >> >
> >> ________________________________________________________________________
> >> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> >> > industry-leading spam and email virus protection.
> >> >
> >>
> >
> >





From tbaruchel at free.fr  Fri Nov 24 15:12:13 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Fri, 24 Nov 2006 15:12:13 +0100 (CET)
Subject: Quadratic residues
Message-ID: <20061124150520.F1510@localhost.my.domain>


I asked recently about
>  a(n) is the number of divisors of (n+1)
> Have you ever seen closely-related sequences like that one?

After some hints and a
> Happy hunting!
I finally noticed this :
   a) prime numbers 4n+3 lead to the set of their quadratic residues
   b) prime numbers 4n+1 lead to the set of their quadratic non-residues
   c) odd composite numbers :
      15 --> quadratic residues
      21 --> difficult to understand at first sight
Since my datas are empirical and need much work it is difficult to get
much information. Even numbers are difficult to study also.
Have you ever heard about such a behaviour? In that case, what would
you predict for 21?

WARNING. When I speak of "quadratic residues", I mean the residue has
a GCD = 1 with the number (convention may vary).

Thank you for everything,

-- 
Thomas Baruchel





From tbaruchel at free.fr  Fri Nov 24 15:59:29 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Fri, 24 Nov 2006 15:59:29 +0100 (CET)
Subject: Quadratic residues
In-Reply-To: <20061124150520.F1510@localhost.my.domain>
References: <20061124150520.F1510@localhost.my.domain>
Message-ID: <20061124155641.B1631@localhost.my.domain>

On Fri, 24 Nov 2006, Thomas Baruchel wrote:
> Have you ever heard about such a behaviour? In that case, what would
> you predict for 21?

Forget it, I just found : Jacobi symbol + quadratic reciprocity theorem

Thanks.





From eclark at math.usf.edu  Fri Nov 24 16:28:29 2006
From: eclark at math.usf.edu (Edwin Clark)
Date: Fri, 24 Nov 2006 10:28:29 -0500 (EST)
Subject: Categories
In-Reply-To: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
Message-ID: <Pine.LNX.4.58.0611240958550.32273@shell.cas.usf.edu>

On Wed, 22 Nov 2006 franktaw at netscape.net wrote:

> How many categories are there?
> 
> First, how many categories are there with n morphisms and k objects?  
> This table starts:
> 
>  1
>  2  1
>  7  3 1
> 35 16 3 1
> 
> The first column is A058129, the number of monoids; the main diagonal 
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
> 

As you say, the number of one object categories on n objects is M(n)=the 
number of monoids with n elements. But this is known only up to n = 7
according to the OEIS. This seems to make the number of TWO object
categories with n morphisms a very difficult problem...Let alone THREE
object categories...

MathSciNet gives only one hit on "two object categor*":
------------------------------------------------------------------
Rosick\'y, Ji\v r?
Codensity and binding categories.
Comment. Math. Univ. Carolinae 16 (1975), no. 3, 515--529.

If there is no measurable cardinal, the author constructs a two-object 
category all of whose well-powered complete extensions are binding. 
-------------------------------------------------------------------
which is clearly not related to the enumeration problem. 

A search on "categor* with two objects" give a few more hits
but as far as I can see still no papers related to enumeration..







From jvospost3 at gmail.com  Fri Nov 24 17:51:12 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 24 Nov 2006 08:51:12 -0800
Subject: Categories
In-Reply-To: <Pine.LNX.4.58.0611240958550.32273@shell.cas.usf.edu>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <Pine.LNX.4.58.0611240958550.32273@shell.cas.usf.edu>
Message-ID: <5542af940611240851l23b0d23di56bf7bf28fe71f82@mail.gmail.com>

One should also count loops, i.e. an arrow from an object to itself.

Then there are 2 categories on one object, with and without the loop:
Notating V(C) = the vertices in category C (which we label in this
illustrartion but enumerate up to isomorphism, i.e. on forgetting labels;
and L(C) is the set of loops of C, and A(C) is the set of (proper) arcs.
That is, I've separated the loops out from their being a mere subset of
arcs, to aid in enumeration. Then each category is, for this enumeration, a
triple: C= V(C) U L(C) U A(C).

C_1,1 = (1,null,null);
C_1,2 = (1, 1->1,null).

There are 6 categories on 2 objects:
C_2,1 = ({1,2},null,null);
C_2,2 = ({1,2},null,(1->1))
note that the above is isomorphic to C_2,2 = ({1,2},null,(2->2))
C_2,3 = ({1,2},{(1->1),(2->2)},null);
C_2,4 = ({1,2},null,(1->2)) which is isomorphic to C_2,4 =
({1,2},null,(2->1));
C_2,5 = ({1,2},(1>1),(1->2)) which is isomorphic to ({1,2},(2>2),(2->1));
C_2,6 = ({1,2},null,(1>1),(2->1)).

There do seem to be 35 categories on 3 objects, which are essentially the
same as the 7 endomorphisms on 3 objects, with each loop either left in, or
thrown away.

The combinatorics are now straightforward.  We have the endofunction
enumeration, and binary choices on each loop.

I have many, many pages of drawings of these, but not wanting to either scan
them (not having a scanner) nor write out the explicit sets as above.

Is this clear enough, though, to proceed?

-- Jonathan
C_2,2 = ({1,2},(1->1),null)


On 11/24/06, Edwin Clark <eclark at math.usf.edu> wrote:
>
> On Wed, 22 Nov 2006 franktaw at netscape.net wrote:
>
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> >  1
> >  2  1
> >  7  3 1
> > 35 16 3 1
> >
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
>
> As you say, the number of one object categories on n objects is M(n)=the
> number of monoids with n elements. But this is known only up to n = 7
> according to the OEIS. This seems to make the number of TWO object
> categories with n morphisms a very difficult problem...Let alone THREE
> object categories...
>
> MathSciNet gives only one hit on "two object categor*":
> ------------------------------------------------------------------
> Rosick\'y, Ji\v r?
> Codensity and binding categories.
> Comment. Math. Univ. Carolinae 16 (1975), no. 3, 515--529.
>
> If there is no measurable cardinal, the author constructs a two-object
> category all of whose well-powered complete extensions are binding.
> -------------------------------------------------------------------
> which is clearly not related to the enumeration problem.
>
> A search on "categor* with two objects" give a few more hits
> but as far as I can see still no papers related to enumeration..
>
>
>
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From jvospost3 at gmail.com  Fri Nov 24 20:39:20 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 24 Nov 2006 11:39:20 -0800
Subject: Categories
In-Reply-To: <20061124075307.GA17864@fourier.ujf-grenoble.fr>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
	 <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
	 <20061124075307.GA17864@fourier.ujf-grenoble.fr>
Message-ID: <5542af940611241139sa6472eoc31f5980db42a918@mail.gmail.com>

Offline, I'm coordinating with Edwin Clark on definitions.

But I'm counting as lower bounds  (many pages of paper with drawings, too
lengthy for me to turn into ascii, don't have scanner):

2 categories on 1 point;
6 categories on 2 points;
16 categories on 3 points;
46 categories on 4 points;
116 categories on 2 points;

in each case where the underlying digraph is of an endofunction. The full
count on categories on n points is a weighted partial sum on these, related
to my enumeration of "prime" endofunctions being the sum of Fobonacci(n),
tribonacci(n), teranacci(n), ...

-- I can give examples on request.  Of course, I can always be off by 1 or
more on each count, through sheer blunder.  But I am being self-consistent
in methodology, and it can be done by software using the same algorithm,
carefully translated.

-- Jonathan


On 11/23/06, Roland Bacher <Roland.Bacher at ujf-grenoble.fr> wrote:
>
>
> I realised that the formula below for enumerating categories
> using quivers is completely wrong. It gives only an upper bound
> since one has to require moreover associativity for the composition of
> morphisms. The quiver approach works however in principle but
> is much more complex (and I guess this approach has been
> used in some of the previous approaches).   Roland Bacher
>
> On Thu, Nov 23, 2006 at 12:10:51PM -0800, Jonathan Post wrote:
> > *"Formalized Proof, Computation, and the Construction Problem in
> Algebraic
> > geometry", by Carlos Simpson.
> >
> > [PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
> > 2004<http://arxiv.org/pdf/math/0410224> File
> > Format: PDF/Adobe Acrobat - View as
> > HTML<
> http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5
> >
> > finite integer N, *how many categories* are there with N morphisms? What
> *
> > ...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
> > (1965), 450-464.
> >
> >
> >
> > On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> > >
> > >Roland Bacher correctly refers to quivers as summarized in wikipedia at
> > >
> > >http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
> > >
> > >On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> > >>
> > >>
> > >> This numbers can also be obtained as the numbers
> > >> of the following quotients of quivers:
> > >>
> > >> Associate to a category with n morphisms and k objects
> > >> the quiver with k vertices corresponding to the objects
> > >> and a(X,Y) arrows directed from the object X to the object
> > >> Y if there are a(X,Y) morphisms from X into Y.
> > >>
> > >> Given two morphism g:X-->Y, f:Y-->Z with composition
> > >> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> > >>
> > >> The resulting quotient algebra has dimension n and a basis
> > >> given by the (simple) arrows.
> > >>
> > >> This leads to the following "algorithm" for enumerating
> > >> all categories with n morphisms and k objects:
> > >>
> > >> (a) enumerate all quivers (directed graphs) with k vertices and
> > >> n oriented edges.
> > >>
> > >> (b) Associate the following "weight" to such a quiver as follows :
> > >>
> > >> (b1)given a triplet of vertices X,Y,Z,
> > >> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> > >> (this counts the number of ways which associate a morphism
> > >> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> > >> number of oriented arrows starting at U and ending at W.
> > >>
> > >> (b2) associate to a given quiver the weight
> > >> \prod_{X,Y,Z} w(X,Y,Z)
> > >> where the product is over all triplets of vertices.
> > >>
> > >> The total sum of such weighted quivers yields then the solution.
> > >>
> > >> Roland Bacher
> > >>
> > >>
> > >>
> > >>
> > >>
> > >>
> > >> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.netwrote:
> > >> > How many categories are there?
> > >> >
> > >> > First, how many categories are there with n morphisms and k
> objects?
> > >> > This table starts:
> > >> >
> > >> > 1
> > >> > 2  1
> > >> > 7  3 1
> > >> > 35 16 3 1
> > >> >
> > >> > The first column is A058129, the number of monoids; the main
> diagonal
> > >> > is all 1's.  I am not
> > >> > 100% certain of the 16 in the final row.
> > >> >
> > >> > Taking the row sums, we get:
> > >> >
> > >> > 1,3,11,55
> > >> >
> > >> > the number of categories with n morphisms.  This is probably not in
> > >> the
> > >> > OEIS (only
> > >> > A001776 is possible - other matches become less than A058129).  The
> > >> > inverse Euler
> > >> > transform,
> > >> >
> > >> > 1,2,8,41
> > >> >
> > >> > is the number of connected categories with n morphisms; this is
> > >> > likewise probably not
> > >> > in the OEIS (only A052447 is possible).
> > >> >
> > >> > Can somebody generate more data?
> > >> >
> > >> > Franklin T. Adams-Watters
> > >> >
> > >> > A category is a collection of objects and morphisms; each morphism
> is
> > >> > from one object
> > >> > to another (not necessarily different) object.  Where the
> destination
> > >> > of one morphism
> > >> > is the source of a second, their composition is defined;
> composition
> > >> is
> > >> > associative where
> > >> > it is defined.  Each object has an identity morphism, which
> connects
> > >> it
> > >> > to itself; this
> > >> > is an identity when composed with morphisms coming in and with
> > >> > morphisms going
> > >> > out.
> > >> >
> > >>
> ________________________________________________________________________
> > >> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > >> > industry-leading spam and email virus protection.
> > >> >
> > >>
> > >
> > >
>
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From jvospost3 at gmail.com  Fri Nov 24 23:41:12 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 24 Nov 2006 14:41:12 -0800
Subject: Categories
In-Reply-To: <5542af940611241139sa6472eoc31f5980db42a918@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
	 <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
	 <20061124075307.GA17864@fourier.ujf-grenoble.fr>
	 <5542af940611241139sa6472eoc31f5980db42a918@mail.gmail.com>
Message-ID: <5542af940611241441w68d51ab5pc6ba0832f740d4e8@mail.gmail.com>

Somehow the reply to Frank went astray.

The wikipedia entry on directed graph makes some of the definitional
distinctions I've made, and then some.  Further, it makes the useful
statement:

In Category theory,
a category can be considered a directed multigraph
with the objects as vertices and the morphisms as directed edges.
The functors between categories induce then some,
but not necessarily all, of the digraph morphisms.


-- Jonathan


On 11/24/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> Offline, I'm coordinating with Edwin Clark on definitions.
>
> But I'm counting as lower bounds  (many pages of paper with drawings, too
> lengthy for me to turn into ascii, don't have scanner):
>
> 2 categories on 1 point;
> 6 categories on 2 points;
> 16 categories on 3 points;
> 46 categories on 4 points;
> 116 categories on 2 points;
>
> in each case where the underlying digraph is of an endofunction. The full
> count on categories on n points is a weighted partial sum on these, related
> to my enumeration of "prime" endofunctions being the sum of Fobonacci(n),
> tribonacci(n), teranacci(n), ...
>
> -- I can give examples on request.  Of course, I can always be off by 1 or
> more on each count, through sheer blunder.  But I am being self-consistent
> in methodology, and it can be done by software using the same algorithm,
> carefully translated.
>
> -- Jonathan
>
>
> On 11/23/06, Roland Bacher <Roland.Bacher at ujf-grenoble.fr> wrote:
> >
> >
> > I realised that the formula below for enumerating categories
> > using quivers is completely wrong. It gives only an upper bound
> > since one has to require moreover associativity for the composition of
> > morphisms. The quiver approach works however in principle but
> > is much more complex (and I guess this approach has been
> > used in some of the previous approaches).   Roland Bacher
> >
> > On Thu, Nov 23, 2006 at 12:10:51PM -0800, Jonathan Post wrote:
> > > *"Formalized Proof, Computation, and the Construction Problem in
> > Algebraic
> > > geometry", by Carlos Simpson.
> > >
> > > [PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
> > > 2004<http://arxiv.org/pdf/math/0410224> File
> > > Format: PDF/Adobe Acrobat - View as
> > > HTML<http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5
> > >
> > > finite integer N, *how many categories* are there with N morphisms?
> > What *
> > > ...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
> > > (1965), 450-464.
> > >
> > >
> > >
> > > On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> > > >
> > > >Roland Bacher correctly refers to quivers as summarized in wikipedia
> > at
> > > >
> > > >http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
> > > >
> > > >On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> > > >>
> > > >>
> > > >> This numbers can also be obtained as the numbers
> > > >> of the following quotients of quivers:
> > > >>
> > > >> Associate to a category with n morphisms and k objects
> > > >> the quiver with k vertices corresponding to the objects
> > > >> and a(X,Y) arrows directed from the object X to the object
> > > >> Y if there are a(X,Y) morphisms from X into Y.
> > > >>
> > > >> Given two morphism g:X-->Y, f:Y-->Z with composition
> > > >> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> > > >>
> > > >> The resulting quotient algebra has dimension n and a basis
> > > >> given by the (simple) arrows.
> > > >>
> > > >> This leads to the following "algorithm" for enumerating
> > > >> all categories with n morphisms and k objects:
> > > >>
> > > >> (a) enumerate all quivers (directed graphs) with k vertices and
> > > >> n oriented edges.
> > > >>
> > > >> (b) Associate the following "weight" to such a quiver as follows :
> > > >>
> > > >> (b1)given a triplet of vertices X,Y,Z,
> > > >> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> > > >> (this counts the number of ways which associate a morphism
> > > >> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> > > >> number of oriented arrows starting at U and ending at W.
> > > >>
> > > >> (b2) associate to a given quiver the weight
> > > >> \prod_{X,Y,Z} w(X,Y,Z)
> > > >> where the product is over all triplets of vertices.
> > > >>
> > > >> The total sum of such weighted quivers yields then the solution.
> > > >>
> > > >> Roland Bacher
> > > >>
> > > >>
> > > >>
> > > >>
> > > >>
> > > >>
> > > >> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.netwrote:
> > > >> > How many categories are there?
> > > >> >
> > > >> > First, how many categories are there with n morphisms and k
> > objects?
> > > >> > This table starts:
> > > >> >
> > > >> > 1
> > > >> > 2  1
> > > >> > 7  3 1
> > > >> > 35 16 3 1
> > > >> >
> > > >> > The first column is A058129, the number of monoids; the main
> > diagonal
> > > >> > is all 1's.  I am not
> > > >> > 100% certain of the 16 in the final row.
> > > >> >
> > > >> > Taking the row sums, we get:
> > > >> >
> > > >> > 1,3,11,55
> > > >> >
> > > >> > the number of categories with n morphisms.  This is probably not
> > in
> > > >> the
> > > >> > OEIS (only
> > > >> > A001776 is possible - other matches become less than
> > A058129).  The
> > > >> > inverse Euler
> > > >> > transform,
> > > >> >
> > > >> > 1,2,8,41
> > > >> >
> > > >> > is the number of connected categories with n morphisms; this is
> > > >> > likewise probably not
> > > >> > in the OEIS (only A052447 is possible).
> > > >> >
> > > >> > Can somebody generate more data?
> > > >> >
> > > >> > Franklin T. Adams-Watters
> > > >> >
> > > >> > A category is a collection of objects and morphisms; each
> > morphism is
> > > >> > from one object
> > > >> > to another (not necessarily different) object.  Where the
> > destination
> > > >> > of one morphism
> > > >> > is the source of a second, their composition is defined;
> > composition
> > > >> is
> > > >> > associative where
> > > >> > it is defined.  Each object has an identity morphism, which
> > connects
> > > >> it
> > > >> > to itself; this
> > > >> > is an identity when composed with morphisms coming in and with
> > > >> > morphisms going
> > > >> > out.
> > > >> >
> > > >>
> > ________________________________________________________________________
> > > >> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > > >> > industry-leading spam and email virus protection.
> > > >> >
> > > >>
> > > >
> > > >
> >
>
>
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From njas at research.att.com  Sat Nov 25 11:51:53 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sat, 25 Nov 2006 05:51:53 -0500 (EST)
Subject: request for new member
Message-ID: <200611251051.FAA87094@fry.research.att.com>


Olivier:

Richard Guy sent me this:

>From rkg at cpsc.ucalgary.ca  Fri Nov 24 17:07:00 2006
Delivered-To: njas at research.att.com
Date: Fri, 24 Nov 2006 15:02:50 -0700 (MST)
From: Richard Guy <rkg at cpsc.ucalgary.ca>
X-X-Sender: rkg at csl
To: "Neil J. A. Sloane" <njas at research.att.com>
cc: "Saff, Kevin" <kevin.saff at gmail.com>
Subject: New Subscriber
X-Virus-Scanned: by amavisd-new at cpsc.ucalgary.ca
X-Spam-Checker-Version: SpamAssassin 3.0.6 (2005-12-07) on 
        mail-purple.research.att.com
X-Spam-Level: 
X-Spam-Status: No, score=-2.9 required=4.0 tests=AWL,BAYES_00 autolearn=ham 
        version=3.0.6

Neil,
      May Kevin Saff join the seq-fan list?

      His email address is as above.

      Thanks & best wishes,    R.

Neil







From Annette.Warlich at t-online.de  Sat Nov 25 20:26:49 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Sat, 25 Nov 2006 20:26:49 +0100
Subject: Curious binomial-identity /A002720
Message-ID: <456898F9.8020608@t-online.de>


By chance I came across this curious identity
involving the Pascal-triangle.

Assume a row n, say n=4 and the column n, combined
each weighted with the running factorial as in the example:


           1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
 ratio =   -----------------------------------------  -------------------
           1/0! + 4/1! + 6/2! + 4/3! + 1/4!           weighted row-sum


then

  ratio = e  (=exp(1))

The actual sums are the entries of A002720
 http://www.research.att.com/~njas/sequences/A002720

Regards -

Gottfried Helms






From Annette.Warlich at t-online.de  Sat Nov 25 20:41:53 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Sat, 25 Nov 2006 20:41:53 +0100
Subject: Curious binomial-identity /A002720 (small correction of prev.
 post)
In-Reply-To: <456898F9.8020608@t-online.de>
References: <456898F9.8020608@t-online.de>
Message-ID: <45689C81.2020302@t-online.de>

Am 25.11.2006 20:26 schrieb Gottfried Helms:
> By chance I came across this curious identity
> involving the Pascal-triangle.
> 
> Assume a row n, say n=4 and the column n, combined
> each weighted with the running factorial as in the example:
> 
> 
>            1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
>  ratio =   -----------------------------------------  -------------------
>            1/0! + 4/1! + 6/2! + 4/3! + 1/4!           weighted row-sum
> 
> 
> then
> 
>   ratio = e  (=exp(1))
> 

------------------------------------

> The actual sums are the entries of A002720
>  http://www.research.att.com/~njas/sequences/A002720
> 
that should be corrected; the entries in A002720 are

 A(n)  =  weighted-rowsum(n) * n!
       =  weighted-colsum(n) * n! / exp(1)

I forgot to mention the additional n!, since in numerator and
denominator of the above fraction they cancel out, sorry.

Gottfried Helms






From jvospost3 at gmail.com  Sat Nov 25 21:10:27 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 25 Nov 2006 12:10:27 -0800
Subject: Curious binomial-identity /A002720 (small correction of prev. post)
In-Reply-To: <45689C81.2020302@t-online.de>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de>
Message-ID: <5542af940611251210g5e4916bcvee22981abd609ca1@mail.gmail.com>

There are several formulae for A002720, including a(n) = Sum k!C(n, k)^2,
k=0..n.

What you write might be true in the asymptotic limit, but not for any term,
as each term is rational and dividing by e would make each term
transcendental.  I'm sure that you meant the limit, right?

On 11/25/06, Gottfried Helms <Annette.Warlich at t-online.de> wrote:
>
> Am 25.11.2006 20:26 schrieb Gottfried Helms:
> > By chance I came across this curious identity
> > involving the Pascal-triangle.
> >
> > Assume a row n, say n=4 and the column n, combined
> > each weighted with the running factorial as in the example:
> >
> >
> >            1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
> >  ratio =
> -----------------------------------------  -------------------
> >            1/0! + 4/1! + 6/2! + 4/3! + 1/4!           weighted row-sum
> >
> >
> > then
> >
> >   ratio = e  (=exp(1))
> >
>
> ------------------------------------
>
> > The actual sums are the entries of A002720
> >  http://www.research.att.com/~njas/sequences/A002720
> >
> that should be corrected; the entries in A002720 are
>
> A(n)  =  weighted-rowsum(n) * n!
>        =  weighted-colsum(n) * n! / exp(1)
>
> I forgot to mention the additional n!, since in numerator and
> denominator of the above fraction they cancel out, sorry.
>
> Gottfried Helms
>
>
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From Annette.Warlich at t-online.de  Sat Nov 25 22:41:35 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Sat, 25 Nov 2006 22:41:35 +0100
Subject: Curious binomial-identity /A002720 (small correction of prev.
 post)
In-Reply-To: <5542af940611251210g5e4916bcvee22981abd609ca1@mail.gmail.com>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de> <5542af940611251210g5e4916bcvee22981abd609ca1@mail.gmail.com>
Message-ID: <4568B88F.1020800@t-online.de>

Am 25.11.2006 21:10 schrieb Jonathan Post:
> There are several formulae for A002720, including  a(n) = Sum k!C(n,
> k)^2, k=0..n.
> 
> What you write might be true in the asymptotic limit, but not for any
> term, as each term is rational and dividing by e would make each term
> transcendental.  I'm sure that you meant the limit, right?

Yes; as the +... in the numerator should indicate. The sum of columns
have infinitely many terms.

I've seen the generation-formula involving the exp()-function; maybe
this translates in an obvious way (... only with gf's I don't have
*any* experience... sigh...) and the asymptotic formula, which looks
a bit more complicated, but additionally relates it to a pi-expression.

But that we have

   1/3! +  3/2! + 3/1! + 1/0!  = A(3) /3!
                       + 3/1!
                       + 6/2!
                       +10/3!
                       + ...
                        -----
    limit k->oo         A(3) /3! * exp(1)

 or

   1/4! +  4/3! + 6/2! + 4/1! + 1/0!  = A(4)/4!
                              + 4/1!
                              +10/2!
                              +20/3!
                              + ...
                              -----
    limit k->oo               A(4)/4! *exp(1)

has something of a certain beauty... :-)


Gottfried Helms







From jrbibers at indiana.edu  Sat Nov 25 22:53:40 2006
From: jrbibers at indiana.edu (Joseph Biberstine)
Date: Sat, 25 Nov 2006 16:53:40 -0500
Subject: Curious binomial-identity /A002720 (small correction of prev.
 post)
In-Reply-To: <45689C81.2020302@t-online.de>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de>
Message-ID: <4568BB64.9050703@indiana.edu>

The example you gave has an off-by-one row index as follows:

Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n+1)

Note that the row index is incremented.  Mathematica simplifies this
fraction to LaguerreL[-(n+1), 1]/LaguerreL[n+1, -1].  This is not equal
to Exp[1], but interestingly seems to approach Exp[1] monotonically from
below as n approaches infinity.

I'm sure you meant:

Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n)

Mathematica simplifies this to LaguerreL[-(n+1), 1]/LaguerreL[n, -1].
Although I can't get it to explicitly resolve that this is exactly
Exp[1], the values do seem to bear this truth out.

So, I believe your example should read: e.g. for n=3,

    1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
e = -----------------------------------------  -------------------
           1/0! + 3/1! + 3/2! + 1/3!           weighted row-sum

This does seem to hold for every col/row pair in the triangle.

Nice find :)
-JRB

Gottfried Helms wrote:
> Am 25.11.2006 20:26 schrieb Gottfried Helms:
>> By chance I came across this curious identity
>> involving the Pascal-triangle.
>>
>> Assume a row n, say n=4 and the column n, combined
>> each weighted with the running factorial as in the example:
>>
>>
>>            1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
>>  ratio =   -----------------------------------------  -------------------
>>            1/0! + 4/1! + 6/2! + 4/3! + 1/4!           weighted row-sum
>>
>>
>> then
>>
>>   ratio = e  (=exp(1))
>>
> 
> ------------------------------------
> 
>> The actual sums are the entries of A002720
>>  http://www.research.att.com/~njas/sequences/A002720
>>
> that should be corrected; the entries in A002720 are
> 
>  A(n)  =  weighted-rowsum(n) * n!
>        =  weighted-colsum(n) * n! / exp(1)
> 
> I forgot to mention the additional n!, since in numerator and
> denominator of the above fraction they cancel out, sorry.
> 
> Gottfried Helms
> 





From Annette.Warlich at t-online.de  Sun Nov 26 00:43:25 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Sun, 26 Nov 2006 00:43:25 +0100
Subject: Curious binomial-identity /A002720 (small correction of prev.
 post)
In-Reply-To: <4568BB64.9050703@indiana.edu>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de> <4568BB64.9050703@indiana.edu>
Message-ID: <4568D51D.8000305@uni-kassel.de>

Am 25.11.2006 22:53 schrieb Joseph Biberstine:
> The example you gave has an off-by-one row index as follows:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n+1)
> 
> Note that the row index is incremented.  Mathematica simplifies this
> fraction to LaguerreL[-(n+1), 1]/LaguerreL[n+1, -1].  This is not equal
> to Exp[1], but interestingly seems to approach Exp[1] monotonically from
> below as n approaches infinity.
> 
> I'm sure you meant:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n)
> 
> Mathematica simplifies this to LaguerreL[-(n+1), 1]/LaguerreL[n, -1].
> Although I can't get it to explicitly resolve that this is exactly
> Exp[1], the values do seem to bear this truth out.
> 
> So, I believe your example should read: e.g. for n=3,
> 
>     1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
> e = -----------------------------------------  -------------------
>            1/0! + 3/1! + 3/2! + 1/3!           weighted row-sum
> 
> This does seem to hold for every col/row pair in the triangle.

Yes, your correction is right - I didn't look right at the
matrix.


The connection to the Laguerre-polynomials is not too
much surprising; I'm dealing with a ceratin classes
of lower-triangular matrices, which are derived from the
logic of the computing of the Pascal-matrix by matrix-
exponentiation.

While

  Pascalmatrix     = Pk(1)=  matrixexp( subdiagonal(1; [1,2,3,4,5,...]))

the Laguerre-matrix occurs from

  Laguerrematrix(*) = Pk(2) = matrixexp( subdiagonal(1; [1,2^2,3^2,4^2,5^2,...]))
           ((*) means: in a signed and simply scaled form)


I also have
                      Pk(0) = matrixexp( subdiagonal(1; [1,2^0,3^0,4^0,5^0,...]))

(where the vectors are filled in the first principal subdiagonal of an
infinite square-matrix)




Now the current matrix was created by a somehow routinely
"completion of my toolbox" of pascal-like-matrices, as a
hadamard-product of

      Pf =    Pk(1) (x) Pk(0)

         =    1/0!
              1/1!  1/0!
              1/2!  2/1!  1/0!
              1/3!  3/2!  3/1!  1/0!
              1/4!  4/3!  6/2!  4/1!   1/0!
                ....


Then with a powerseries-column-vector

      V(x) = [1,x,x^2,x^3,....] ~

and a diagonal factorial-matrix (used for scaling)

     dF(m) = diag(0! , 1!^m, 2!^m .... )

I got

   dF(1)  *   Pf  * V(1) = A002720       // A002720 read as column-vector
                                         // Due to the entry of Paul Berry

and
   dF(1) *    Pf~ * V(1) = A002720  *exp(1)  // added today



As a table it looks like:

                                                 Row-Sum
         =    1/0!                              =  1/0!    =  A(0)/0!
              1/1!  1/0!                        =  2/1!    =  A(1)/1!
              1/2!  2/1!  1/0!                  =  7/2!    =  A(2)/2!
              1/3!  3/2!  3/1!  1/0!            = 34/3!    =  A(3)/3!
              1/4!  4/3!  6/2!  4/1!   1/0!     =209/4!    =  A(4)/4!
                ....
             --------------------------------------------------------
Col-sum   e*[  1    2/1!  7/2! 34/3! 209/4! ]

Note, that using

   dF(1)  *   Pf  * V(2) =          ???
   dF(1) *    Pf~ * V(2) = exp(2) * ???

  in the second an exponentiation of 2 occurs, and generally for an s

   dF(1) *    Pf~ * V(s) = exp(s) * ???

 with unknown, but -heuristically- integer sequences of coefficients in ???
 for integer s>=0 .


------------------------------------------------------------------------


The Pascal- and (unsigned,scaled) Laguerre-matrix are also
related by

  Pk(0)                              // "base"-matrix
  Pk(1) =   dF(1)* Pk(0) * dF(-1)    // Pascal-matrix
  Pk(2) =   dF(1)* Pk(1) * dF(-1)    // unsigned,scaled Laguerrematrix
 ...
(which is due to the construction via matrix-exponentiation
of the related subdiagonal-matrices)
so finally the occurence of the relation to the Laguerre-matrix
using Mathematica is not too surprising...

Regards -

Gottfried Helms







From zakseidov at yahoo.com  Sun Nov 26 20:58:31 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 26 Nov 2006 11:58:31 -0800 (PST)
Subject: A124080 10 times triangular numbers
Message-ID: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>

Neil, Zerinvary, seqfans,
Why this sequence?!
Thanks, Zak 

%I A124080
%S A124080
0,10,30,60,100,150,210,280,360,450,550,660,780,910,1050,1200,1360,1530,
%T A124080
1710,1900,2100,2310,2530,2760,3000,3250,3510,3780,4060,4350,4650,4960,
%U A124080
5280,5610,5950,6300,6660,7030,7410,7800,8200,8610,9030,9460,9900,10350
%N A124080 10 times triangular numbers.
%F A124080 a(n)=10*C(n,2), n>=1
%p A124080 [seq(10*binomial(n,2) ,n=1..51)];
%Y A124080 Cf. A028895, A046092, A045943, A002378,
A028896, A024966, A033996, A027468.
%K A124080 easy,nonn,new
%O A124080 0,2
%A A124080 Zerinvary Lajos
(zerinvarylajos(AT)yahoo.com), Nov 24 2006



 
____________________________________________________________________________________
Do you Yahoo!?
Everyone is raving about the all-new Yahoo! Mail beta.
http://new.mail.yahoo.com





From jvospost3 at gmail.com  Sun Nov 26 22:23:25 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 26 Nov 2006 13:23:25 -0800
Subject: Categories
In-Reply-To: <20061124075307.GA17864@fourier.ujf-grenoble.fr>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
	 <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
	 <20061124075307.GA17864@fourier.ujf-grenoble.fr>
Message-ID: <5542af940611261323y13ada5f4p9e8f6e5d9051f93d@mail.gmail.com>

I can, with Edwin Clark's permission, forward to Frank my complete
correspondance with Edwin Clark about category enumeration via careful
definition.  I am asked not to cut & paste out of context.  But my emails to
Franklin T. Adams-Watters alone went astray, for whatever reason. so I am
not sure what to do.  I've also asked some Haskell programming experts about
Frank's enumeration question (properly citing Frank), as Haskell has monads
and functors and other categorical stuff built in.  I'll share any useful
answers that I get.

On 11/23/06, Roland Bacher <Roland.Bacher at ujf-grenoble.fr> wrote:
>
>
> I realised that the formula below for enumerating categories
> using quivers is completely wrong. It gives only an upper bound
> since one has to require moreover associativity for the composition of
> morphisms. The quiver approach works however in principle but
> is much more complex (and I guess this approach has been
> used in some of the previous approaches).   Roland Bacher
>
> On Thu, Nov 23, 2006 at 12:10:51PM -0800, Jonathan Post wrote:
> > *"Formalized Proof, Computation, and the Construction Problem in
> Algebraic
> > geometry", by Carlos Simpson.
> >
> > [PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
> > 2004<http://arxiv.org/pdf/math/0410224> File
> > Format: PDF/Adobe Acrobat - View as
> > HTML<
> http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5
> >
> > finite integer N, *how many categories* are there with N morphisms? What
> *
> > ...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
> > (1965), 450-464.
> >
> >
> >
> > On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> > >
> > >Roland Bacher correctly refers to quivers as summarized in wikipedia at
> > >
> > >http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
> > >
> > >On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> > >>
> > >>
> > >> This numbers can also be obtained as the numbers
> > >> of the following quotients of quivers:
> > >>
> > >> Associate to a category with n morphisms and k objects
> > >> the quiver with k vertices corresponding to the objects
> > >> and a(X,Y) arrows directed from the object X to the object
> > >> Y if there are a(X,Y) morphisms from X into Y.
> > >>
> > >> Given two morphism g:X-->Y, f:Y-->Z with composition
> > >> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> > >>
> > >> The resulting quotient algebra has dimension n and a basis
> > >> given by the (simple) arrows.
> > >>
> > >> This leads to the following "algorithm" for enumerating
> > >> all categories with n morphisms and k objects:
> > >>
> > >> (a) enumerate all quivers (directed graphs) with k vertices and
> > >> n oriented edges.
> > >>
> > >> (b) Associate the following "weight" to such a quiver as follows :
> > >>
> > >> (b1)given a triplet of vertices X,Y,Z,
> > >> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> > >> (this counts the number of ways which associate a morphism
> > >> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> > >> number of oriented arrows starting at U and ending at W.
> > >>
> > >> (b2) associate to a given quiver the weight
> > >> \prod_{X,Y,Z} w(X,Y,Z)
> > >> where the product is over all triplets of vertices.
> > >>
> > >> The total sum of such weighted quivers yields then the solution.
> > >>
> > >> Roland Bacher
> > >>
> > >>
> > >>
> > >>
> > >>
> > >>
> > >> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.netwrote:
> > >> > How many categories are there?
> > >> >
> > >> > First, how many categories are there with n morphisms and k
> objects?
> > >> > This table starts:
> > >> >
> > >> > 1
> > >> > 2  1
> > >> > 7  3 1
> > >> > 35 16 3 1
> > >> >
> > >> > The first column is A058129, the number of monoids; the main
> diagonal
> > >> > is all 1's.  I am not
> > >> > 100% certain of the 16 in the final row.
> > >> >
> > >> > Taking the row sums, we get:
> > >> >
> > >> > 1,3,11,55
> > >> >
> > >> > the number of categories with n morphisms.  This is probably not in
> > >> the
> > >> > OEIS (only
> > >> > A001776 is possible - other matches become less than A058129).  The
> > >> > inverse Euler
> > >> > transform,
> > >> >
> > >> > 1,2,8,41
> > >> >
> > >> > is the number of connected categories with n morphisms; this is
> > >> > likewise probably not
> > >> > in the OEIS (only A052447 is possible).
> > >> >
> > >> > Can somebody generate more data?
> > >> >
> > >> > Franklin T. Adams-Watters
> > >> >
> > >> > A category is a collection of objects and morphisms; each morphism
> is
> > >> > from one object
> > >> > to another (not necessarily different) object.  Where the
> destination
> > >> > of one morphism
> > >> > is the source of a second, their composition is defined;
> composition
> > >> is
> > >> > associative where
> > >> > it is defined.  Each object has an identity morphism, which
> connects
> > >> it
> > >> > to itself; this
> > >> > is an identity when composed with morphisms coming in and with
> > >> > morphisms going
> > >> > out.
> > >> >
> > >>
> ________________________________________________________________________
> > >> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > >> > industry-leading spam and email virus protection.
> > >> >
> > >>
> > >
> > >
>
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From jvospost3 at gmail.com  Sun Nov 26 22:27:11 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 26 Nov 2006 13:27:11 -0800
Subject: A124080 10 times triangular numbers
In-Reply-To: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
References: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
Message-ID: <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>

I wondered too.  It intersects the much more interesting
 A000537 <http://www.research.att.com/%7Enjas/sequences/A000537>
 Sum of first n cubes; or n-th triangular number squared.  at 100.

Not that all of my roughly 1 kilophi sequences and comments are interesting,
either.

-- Jonathan

On 11/26/06, zak seidov <zakseidov at yahoo.com> wrote:
>
> Neil, Zerinvary, seqfans,
> Why this sequence?!
> Thanks, Zak
>
> %I A124080
> %S A124080
> 0,10,30,60,100,150,210,280,360,450,550,660,780,910,1050,1200,1360,1530,
> %T A124080
> 1710,1900,2100,2310,2530,2760,3000,3250,3510,3780,4060,4350,4650,4960,
> %U A124080
> 5280,5610,5950,6300,6660,7030,7410,7800,8200,8610,9030,9460,9900,10350
> %N A124080 10 times triangular numbers.
> %F A124080 a(n)=10*C(n,2), n>=1
> %p A124080 [seq(10*binomial(n,2) ,n=1..51)];
> %Y A124080 Cf. A028895, A046092, A045943, A002378,
> A028896, A024966, A033996, A027468.
> %K A124080 easy,nonn,new
> %O A124080 0,2
> %A A124080 Zerinvary Lajos
> (zerinvarylajos(AT)yahoo.com), Nov 24 2006
>
>
>
>
>
> ____________________________________________________________________________________
> Do you Yahoo!?
> Everyone is raving about the all-new Yahoo! Mail beta.
> http://new.mail.yahoo.com
>
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From jvospost3 at gmail.com  Sun Nov 26 23:43:48 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 26 Nov 2006 14:43:48 -0800
Subject: A124080 10 times triangular numbers
In-Reply-To: <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>
References: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
	 <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>
Message-ID: <5542af940611261443g32cd3bf3lf2174cfc2d42e716@mail.gmail.com>

On the other hand, every sequence is related to some interesting sequence...
I just submitted by form the following:

 NEW SEQUENCE FROM Jonathan Vos Post


%I A000001
%S A000001 11, 29, 31, 59, 61, 101, 149, 151, 211, 281, 359, 449, 659,
661, 911, 1049, 1051, 1201, 1361, 1531, 1709, 1901, 2099, 2309, 2311,
2531, 2999, 3001
%N A000001 Primes of the form A124080 (10 times triangular numbers), +1
or -1.
%C A000001 The j, such that A124080(j)-1 is prime or A124080(j)+1 is
prime, where repetition means a twin prime, are 1, 2, 2, 3, 3, 4, 5, 5,
6, 7, 8, 9, 11, 11, 13, 14, 14, 15, 16, 17, 18, 19, 20, 21, 22, 24, 24.
%F A000001 {A124080(j)-1 when prime} U {A124080(j)+1 when prime} =
{i = 10*T(j)-1 such that i is prime} U {i = 10*T(j)+1 such that i is
prime} where T(j) = A000217(j) = j*(j+1)/2.
%e A000001 a(1) = A124080(1)+1 = (10*T(1)) - 1 = 10*(1*(1+1)/2) + 1 =
10+1 = 11 is prime.
a(2) = A124080(2)-1 = (10*T(2))-1 = 10*(2*(2+1)/2) - 1 = 30-1 = 29 is
prime.
a(3) = A124080(2)+1 = (10*T(2))+1 = 10*(2*(2+1)/2) + 1 = 30+1 = 31 is
prime.
%Y A000001 Cf. A000040, A000217, A028895, A046092, A045943, A002378,
A028896, A024966, A033996, A027468.
%O A000001 1,1
%K A000001 ,easy,nonn,
%A A000001 Jonathan Vos Post (jvospost2 at yahoo.com
<http://us.f551.mail.yahoo.com/ym/Compose?To=jvospost2 at yahoo.com&YY=86457&y5beta=yes&y5beta=yes&order=down&sort=date&pos=0&view=a&head=b>),
Nov 26 2006
RH
RA 192.20.225.32
RU
RI



On 11/26/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> I wondered too.  It intersects the much more interesting
>  A000537 <http://www.research.att.com/%7Enjas/sequences/A000537>
>  Sum of first n cubes; or n-th triangular number squared.  at 100.
>
> Not that all of my roughly 1 kilophi sequences and comments are
> interesting, either.
>
> -- Jonathan
>
> On 11/26/06, zak seidov <zakseidov at yahoo.com> wrote:
> >
> > Neil, Zerinvary, seqfans,
> > Why this sequence?!
> > Thanks, Zak
> >
> > %I A124080
> > %S A124080
> > 0,10,30,60,100,150,210,280,360,450,550,660,780,910,1050,1200,1360,1530,
> > %T A124080
> > 1710,1900,2100,2310,2530,2760,3000,3250,3510,3780,4060,4350,4650,4960,
> > %U A124080
> > 5280,5610,5950,6300,6660,7030,7410,7800,8200,8610,9030,9460,9900,10350
> > %N A124080 10 times triangular numbers.
> > %F A124080 a(n)=10*C(n,2), n>=1
> > %p A124080 [seq(10*binomial(n,2) ,n=1..51)];
> > %Y A124080 Cf. A028895, A046092, A045943, A002378,
> > A028896, A024966, A033996, A027468.
> > %K A124080 easy,nonn,new
> > %O A124080 0,2
> > %A A124080 Zerinvary Lajos
> > (zerinvarylajos(AT)yahoo.com), Nov 24 2006
> >
> >
> >
> >
> >
> > ____________________________________________________________________________________
> > Do you Yahoo!?
> > Everyone is raving about the all-new Yahoo! Mail beta.
> > http://new.mail.yahoo.com
> >
>
>
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From njas at research.att.com  Mon Nov 27 03:23:49 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sun, 26 Nov 2006 21:23:49 -0500 (EST)
Subject: A124080 10 times triangular numbers
References: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
    <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>
Message-ID: <200611270223.VAA41276@fry.research.att.com>

JVP, you submitted this sequence:
%N A000001 Primes of the form A124080 (10 times triangular numbers), +1
or -1.
I will use it, but I do not find it interesting.  Why?  Because it 
is derivative.  It is based on a sequence of little interest - acceptable
only because it was sent in by someone who perhaps does not know
what is worth being included in the database - and is a step further 
down the ladder towards the basement.

I seem to have to play the role of the singing teacher
at times:  so be it - this song was off-key!

Neil





From jvospost3 at gmail.com  Mon Nov 27 04:10:47 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 26 Nov 2006 19:10:47 -0800
Subject: A124080 10 times triangular numbers
In-Reply-To: <200611270223.VAA41276@fry.research.att.com>
References: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
	 <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>
	 <200611270223.VAA41276@fry.research.att.com>
Message-ID: <5542af940611261910x10195634teeac87bdf4bfd1e7@mail.gmail.com>

As usual, you're right, Neil.

I was trying to be nice to the newcomer, by showing him the collaborative
nature of the endeavor. I'm sure that his enthusiasm can be made fruitful.
I've certainly erred in judgment, many times, and you and your associate
editors have been splendid in guiding me towards better work.

Sometimes, as they sing: "you've got to be cruel to be kind."

Hence the "probation" keyword and, one of my useful suggestions to you and
your board, the "less" keyword.

Sorry for where that road leads, which I help pave with good intentions.

-- Jonathan

On 11/26/06, N. J. A. Sloane <njas at research.att.com> wrote:
>
> JVP, you submitted this sequence:
> %N A000001 Primes of the form A124080 (10 times triangular numbers), +1
> or -1.
> I will use it, but I do not find it interesting.  Why?  Because it
> is derivative.  It is based on a sequence of little interest - acceptable
> only because it was sent in by someone who perhaps does not know
> what is worth being included in the database - and is a step further
> down the ladder towards the basement.
>
> I seem to have to play the role of the singing teacher
> at times:  so be it - this song was off-key!
>
> Neil
>
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From pauldhanna at juno.com  Mon Nov 27 04:58:25 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Sun, 26 Nov 2006 22:58:25 -0500
Subject: Missing Table?
Message-ID: <20061126.225826.332.1.pauldhanna@juno.com>

Seqfans,
     Where is the rectangular table referred to by these sequences?
 
A040027  Main diagonal of an array of binomial recurrence coefficients.  
 0, 1, 1, 3, 9, 31, 121, 523, 2469, 12611, 69161, 404663,  
A045501  Second diagonal in table of binomial recurrence coefficients.  
 1, 1, 4, 14, 54, 233, 1101, 5625, 30846, 
A045499  Third diagonal in table of binomial recurrence coefficients. 
 1, 1, 5, 20, 85, 400, 2046, 11226, 65676, 407787, 2675410, 
A045500  Fourth diagonal in table of binomial recurrence coefficients. 
 1, 1, 6, 27, 125, 635, 3488, 20425, 126817, 831915, 5744784, 41618459 
(AUTHOR of these sequences:  H. W. Gould) 
 
The closest thing I can find is Emeric Deutsch's recent triangle A124496,

in which these sequences appear as columns. 
 
The lower half of the table (I guess) must look like this:
1, ...
1, 1, ...
1, 1, 3, ...
1, 1, 4, 9, ...
1, 1, 5, 14, 31, ...
1, 1, 6, 20, 54, 121, ...
1, 1, 7, 27, 85, 233, 523, ...
1, 1, 8, 35, 125, 400, 1101, 2469, ...
1, 1, 9, 44, 175, 635, 2046, 5625, 12611, ...
 
Is the complete table in the OEIS?  
If not, it certainly should be. 
     Paul
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From mathar at strw.leidenuniv.nl  Mon Nov 27 10:47:35 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Mon, 27 Nov 2006 10:47:35 +0100
Subject: Missing Table?
Message-ID: <200611270947.kAR9lZBJ001619@amer.strw.leidenuniv.nl>


The master table of A040027 could also be A046936..but this does not fit the
2nd, 3rd and 4th diagonal as well as A124496 does.

RJM

> From seqfan-owner at ext.jussieu.fr  Mon Nov 27 05:03:23 2006
> Return-Path: <seqfan-owner at ext.jussieu.fr>
> To: Seqfan at ext.jussieu.fr
> Cc: gould at math.wvu.edu
> Date: Sun, 26 Nov 2006 22:58:25 -0500
> Subject: Missing Table?
> From: "Paul D. Hanna" <pauldhanna at juno.com>
> 
> Seqfans,
>      Where is the rectangular table referred to by these sequences?
>  
> A040027  Main diagonal of an array of binomial recurrence coefficients.  
>  0, 1, 1, 3, 9, 31, 121, 523, 2469, 12611, 69161, 404663,  
> A045501  Second diagonal in table of binomial recurrence coefficients.  
>  1, 1, 4, 14, 54, 233, 1101, 5625, 30846, 
> A045499  Third diagonal in table of binomial recurrence coefficients. 
>  1, 1, 5, 20, 85, 400, 2046, 11226, 65676, 407787, 2675410, 
> A045500  Fourth diagonal in table of binomial recurrence coefficients. 
>  1, 1, 6, 27, 125, 635, 3488, 20425, 126817, 831915, 5744784, 41618459 
> (AUTHOR of these sequences:  H. W. Gould) 
>  
> The closest thing I can find is Emeric Deutsch's recent triangle A124496,
> 
> in which these sequences appear as columns. 
>  
> The lower half of the table (I guess) must look like this:
> 1, ...
> 1, 1, ...
> 1, 1, 3, ...
> 1, 1, 4, 9, ...
> 1, 1, 5, 14, 31, ...
> 1, 1, 6, 20, 54, 121, ...
> 1, 1, 7, 27, 85, 233, 523, ...
> 1, 1, 8, 35, 125, 400, 1101, 2469, ...
> 1, 1, 9, 44, 175, 635, 2046, 5625, 12611, ...
>  
> Is the complete table in the OEIS?  
> If not, it certainly should be. 
>      Paul





From tanyakh at TanyaKhovanova.com  Mon Nov 27 17:05:13 2006
From: tanyakh at TanyaKhovanova.com (Tanya Khovanova)
Date: Mon, 27 Nov 2006 08:05:13 -0800
Subject: practical numbers
Message-ID: <200611270805.AA74449328@TanyaKhovanova.com>


Hello seqfans,

There is some confusion with practical numbers. Wikipedia article on practical numbers: http://en.wikipedia.org/wiki/Practical_number
gives the following definition of a practical number:
A practical number or panarithmic number is a positive integer n such that all preceding positive integers are a sum of distinct divisors of n.

At the same time it refers to the sequence A005153 which gives a slightly different definition:
Practical numbers (first definition): all k <= sigma(n) are sums of distinct divisors of n. Also called panarithmic numbers.

Browsing other websites (including my own :-) ) I've found that the wiki definition is used more often. Moreover, my calculation with the my definition exactly matches the sequence A005153, though they are supposed to be different.

Please, clarify my confusion.

Tanya Khovanova






From mathar at strw.leidenuniv.nl  Mon Nov 27 17:54:41 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Mon, 27 Nov 2006 17:54:41 +0100
Subject: practical numbers
Message-ID: <200611271654.kARGsfOX022242@amer.strw.leidenuniv.nl>

> From seqfan-owner at ext.jussieu.fr  Mon Nov 27 17:07:28 2006
> Date: Mon, 27 Nov 2006 08:05:13 -0800
> From: "Tanya Khovanova" <tanyakh at tanyakhovanova.com>
> Reply-To: <tanyakh at tanyakhovanova.com>
> To: <seqfan at ext.jussieu.fr>
> Subject: practical numbers
> ...
> There is some confusion with practical numbers. Wikipedia article on practical numbers: http://en.wikipedia.org/wiki/Practical_numbe
> r
> gives the following definition of a practical number:
> A practical number or panarithmic number is a positive integer n such that all preceding positive integers are a sum of distinct div
> isors of n.
> 
> At the same time it refers to the sequence A005153 which gives a slightly different definition:
> Practical numbers (first definition): all k <= sigma(n) are sums of distinct divisors of n. Also called panarithmic numbers.
> 
> Browsing other websites (including my own :-) ) I've found that the wiki definition is used more often. Moreover, my calculation wit
> h the my definition exactly matches the sequence A005153, though they are supposed to be different.
> 
> Please, clarify my confusion.
> ...

The A005153 definition [with reference to the sigma(n)] is also used
in
http://www.dm.unipi.it/gauss-pages/melfi/public_html/articoli/jnt.ps
but the same author uses in his web page
http://members.unine.ch/giuseppe.melfi/pratica.html
just the other definition.
see also http://citeseer.ist.psu.edu/285.html
http://arxiv.org/abs/math.NT/0404555

B. M Stewart in Am J. Math 76 (4) (1954) p 779, available from
http://www.jstor.org, quotes Srinivasan [Current Science (1948), pp 179] with
the definition without the sigma.

E J Scourfield in J. Number Theory 62 (1) (1997) p 163 uses the definition
including the sigma:
http://www.emis.de/projects/EULER/detail?ide=1997saiaentidividens&matchno=7&matchtotal=18&q=cr%3AE*+Saias+

I have to stop here. The recent IAU resolution on what a useful definition of a
"planet" might be, in particular the status of Pluto, be has already caused
enough confusion, and we're still recovering from this.

Richard





From franktaw at netscape.net  Mon Nov 27 18:04:03 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 27 Nov 2006 12:04:03 -0500
Subject: practical numbers
In-Reply-To: <200611270805.AA74449328@TanyaKhovanova.com>
References: <200611270805.AA74449328@TanyaKhovanova.com>
Message-ID: <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com>

The definitions are equivalent.  The condition cited in the Wikipedia 
article, to make everything up to n a sum of the divisors of n, is 
sufficient to make everything up to sigma(n) be such a sum.

Franklin T. Adams-Watters


-----Original Message-----
From: tanyakh at TanyaKhovanova.com

Hello seqfans,

There is some confusion with practical numbers. Wikipedia article on 
practical
numbers: http://en.wikipedia.org/wiki/Practical_number
gives the following definition of a practical number:
A practical number or panarithmic number is a positive integer n such 
that all
preceding positive integers are a sum of distinct divisors of n.

At the same time it refers to the sequence A005153 which gives a 
slightly
different definition:
Practical numbers (first definition): all k <= sigma(n) are sums of 
distinct
divisors of n. Also called panarithmic numbers.

Browsing other websites (including my own :-) ) I've found that the 
wiki
definition is used more often. Moreover, my calculation with the my 
definition
exactly matches the sequence A005153, though they are supposed to be 
different.

Please, clarify my confusion.

Tanya Khovanova


________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From davidwwilson at comcast.net  Mon Nov 27 18:30:31 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Mon, 27 Nov 2006 12:30:31 -0500
Subject: practical numbers
References: <200611270805.AA74449328@TanyaKhovanova.com> <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com>
Message-ID: <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p>

Assuming Franklin's observation to be correct, why not

%C A005153 Also, n such that all k <= n are sums of distinct divisors of n.

Some poorly-thought-out questions before I head off to work:

Are there not-quite-practical numbers, that is, numbers such that n-1 is the 
only number < n which is not the sum of distinct divisors of n?

If not, is there some interesting bound f(n) < n such that if all k <= f(n) 
are sums of distinct divisors of n, then n is also a sum of distinct 
divisors of n?

----- Original Message ----- 
From: <franktaw at netscape.net>
To: <tanyakh at TanyaKhovanova.com>; <seqfan at ext.jussieu.fr>
Sent: Monday, November 27, 2006 12:04 PM
Subject: Re: practical numbers


> The definitions are equivalent.  The condition cited in the Wikipedia 
> article, to make everything up to n a sum of the divisors of n, is 
> sufficient to make everything up to sigma(n) be such a sum.
>
> Franklin T. Adams-Watters






From davidwwilson at comcast.net  Mon Nov 27 18:42:56 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Mon, 27 Nov 2006 12:42:56 -0500
Subject: practical numbers
References: <200611270805.AA74449328@TanyaKhovanova.com> <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com> <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p>
Message-ID: <000601c7124b$7c8b0040$6501a8c0@yourxhtr8hvc4p>

Simce any number n is a repeated sum of any of its divisors, "sum of 
distinct divisors" seems like overkill.

Anyway, another question: What is the smallest number a(n) that is a sum of 
its distinct divisors in n ways? 






From franktaw at netscape.net  Mon Nov 27 20:12:37 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 27 Nov 2006 14:12:37 -0500
Subject: practical numbers
In-Reply-To: <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p>
References: <200611270805.AA74449328@TanyaKhovanova.com> <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com> <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p>
Message-ID: <8C8E06F230775CD-7BC-A6B4@FWM-M18.sysops.aol.com>

To answer the second question first, it suffices to show that every 
number up to and including ceiling(n/2) is a sum of the divisors of n.  
Except for n=3, that can be floor(n/2) instead.  (If n is odd, 2 is not 
a sum; if n is even, every m with n/2 <= m < n is a sum n/2 plus a 
number less than n/2.)  This answers the first question, essentially in 
the negative: 3 is the only such number.

More interesting, perhaps: what about numbers such that every number up 
to n, with only 1 exception, are the sum of divisors of n?  (Note that 
this has to be n in this case, not sigma(n), since the possible sums 
are symmetric with respect to sigma(n) - if a set S of divisors sums to 
m, then the complement of S, with respect to the set of divisors of n, 
sums to sigma(n) - m.)

70 is one such number, skipping only 4; I think 70p for any prime 5 <= 
p < 70 does the same.  945 is another, skipping 2.  (945 is the 
smallest odd abundant number.)  I think that these values (70, 350, 
490, 770, 910, 945) are the only ones up to 1000.

The number skipped is always even; in fact, it always one less than a 
prime divisor p of n (and 1 more than the sum of the divisors of a 
practical divisor of n - which implies that that practical divisor is a 
square or twice a square).  Except for n=3, there must be another prime 
divisor q of n with p < q < 2p.

Franklin T. Adams-Watters


-----Original Message-----
From: davidwwilson at comcast.net

Assuming Franklin's observation to be correct, why not?
?
%C A005153 Also, n such that all k <= n are sums of distinct divisors 
of n.?
?
Some poorly-thought-out questions before I head off to work:?
?
Are there not-quite-practical numbers, that is, numbers such that n-1 
is the only number < n which is not the sum of distinct divisors of n??
?
If not, is there some interesting bound f(n) < n such that if all k <= 
f(n) are sums of distinct divisors of n, then n is also a sum of 
distinct divisors of n??
?
----- Original Message ----- From: <franktaw at netscape.net>?
To: <tanyakh at TanyaKhovanova.com>; <seqfan at ext.jussieu.fr>?
Sent: Monday, November 27, 2006 12:04 PM?
Subject: Re: practical numbers?
?
> The definitions are equivalent. The condition cited in the Wikipedia 
 > article, to make everything up to n a sum of the divisors of n, is > 
sufficient to make everything up to sigma(n) be such a sum.?
>?
> Franklin T. Adams-Watters?
?


________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From franktaw at netscape.net  Mon Nov 27 20:25:04 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 27 Nov 2006 14:25:04 -0500
Subject: practical numbers
In-Reply-To: <000601c7124b$7c8b0040$6501a8c0@yourxhtr8hvc4p>
References: <200611270805.AA74449328@TanyaKhovanova.com> <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com> <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p> <000601c7124b$7c8b0040$6501a8c0@yourxhtr8hvc4p>
Message-ID: <8C8E070DFC523C7-7BC-A7AB@FWM-M18.sysops.aol.com>

I don't think "sum of distinct divisors" is really overkill.  If you 
leave out the "distinct", you'll get people complaining "but every 
number is a sum of its divisors" - or worse, misunderstanding.

A096356.  (Which should really be edited to say "proper divisors" 
instead of just "divisors".  Adding an initial 1 at offset 0 would also 
make sense to me.)

Franklin T. Adams-Watters


-----Original Message-----
From: davidwwilson at comcast.net

Simce any number n is a repeated sum of any of its divisors, "sum of 
distinct divisors" seems like overkill.?
?
Anyway, another question: What is the smallest number a(n) that is a 
sum of its distinct divisors in n ways? ?


________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From mathar at strw.leidenuniv.nl  Mon Nov 27 21:46:49 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Mon, 27 Nov 2006 21:46:49 +0100
Subject: Re^2: practical numbers
Message-ID: <200611272046.kARKknOV015522@amer.strw.leidenuniv.nl>


> From seqfan-owner at ext.jussieu.fr  Mon Nov 27 18:44:49 2006
> From: "David Wilson" <davidwwilson at comcast.net>
> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
> Subject: Re: practical numbers
> ...
> Anyway, another question: What is the smallest number a(n) that is a sum of 
> its distinct divisors in n ways? 

For each n we get the "trivial" solution where the sum consists of n itself
as the only term. Further solutions, where the sum contains more than one term
(all of which are distinct divisors), are listed below in the format
n [list of terms that sum to n]:

6 [1, 2, 3]
12 [1, 2, 3, 6]
12 [2, 4, 6]
18 [1, 2, 6, 9]
18 [3, 6, 9]
20 [1, 4, 5, 10]
24 [1, 2, 3, 4, 6, 8]
24 [1, 2, 3, 6, 12]
24 [2, 4, 6, 12]
24 [1, 3, 8, 12]
24 [4, 8, 12]
28 [1, 2, 4, 7, 14]
30 [1, 3, 5, 6, 15]
30 [2, 3, 10, 15]
30 [5, 10, 15]
36 [2, 3, 4, 6, 9, 12]
36 [2, 3, 4, 9, 18]
36 [1, 2, 6, 9, 18]
36 [3, 6, 9, 18]
36 [1, 2, 3, 12, 18]
36 [2, 4, 12, 18]
36 [6, 12, 18]
40 [1, 2, 4, 5, 8, 20]
40 [1, 4, 5, 10, 20]
40 [2, 8, 10, 20]
42 [1, 6, 14, 21]
42 [7, 14, 21]


n=6 has 1 extra solution, n=12 or 18 have 2 extra solutions, n=20 has one,
n=24 has five etc.

Maple program to create that one (not checked!):

a := proc(n)
	local findx,d,f,found,i,count ;
	count :=0 :
	# decompose n into all sums that are candidates for the decomposition
	d := combinat[partition](n) ;
	# take each sum indivdually
	for findx from 1 to nops(d) do
		f := op(findx,d) ;
		# start with the assumption that it works
		found := true ;
		for i from 2 to nops(f) do
			# if two terms are equal in the Maple list, 
			# we flag this case as "not admitted"
			if op(i,f) = op(i-1,f) then
				found := false ;
				break ;
			fi ;
		od ;
		for i from 1 to nops(f) do
			# if any term is not a divisor,
			# we flag this case as "not admitted"
			if n mod op(i,f) <> 0 then
				found := false ;
				break ;
			fi ;
		od ;
		# If this sum seems to work and has more than one term, we list it
		if found and nops(f) > 1 then
			printf("%a %a\n",n,f) ;
			count := count+1 ;
		fi ;
	od ;
	RETURN(count) ;
end:

# loop over the most fundamental cases..
for n from 1 to 47 do
	a(n) ;
od ;

# Richard Mathar





From zbi74583 at boat.zero.ad.jp  Tue Nov 28 07:18:47 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Tue, 28 Nov 2006 15:18:47 +0900
Subject: A038260
Message-ID: <025d01c712b5$16134290$977fd7dc@FMC57937B6BE00>

    Neil wrote on comment line of A038260.

    >Surely there are smaller members! njas, Nov 07 2006



    Why do you think that?

    I searched the ?seeds? for ?regular? Rational Amicable Number again.

       .

    And {11*19, 239 } is the smallest one.           .

    So, I think that at least it is the first terms of regular Rational 
Amicable Pair.

            .

    Such a Diophantine equation which is high degree and is involved with 
Sigma function has many conditions  , so the solution must have many prime 
factors for having good property, hence it becomes to have many digits.



    I guess that 2^2*7*3*5^2*13*31*139*277*3877*11*19 is the smallest and if 
an irregular example exists then it has around 16 digits.



    Yasutoshi


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From Annette.Warlich at t-online.de  Wed Nov 29 08:38:17 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Wed, 29 Nov 2006 08:38:17 +0100
Subject: Curious binomial-identity /A002720
In-Reply-To: <4568BB64.9050703@indiana.edu>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de> <4568BB64.9050703@indiana.edu>
Message-ID: <456D38E9.8030900@t-online.de>

Am 25.11.2006 22:53 schrieb Joseph Biberstine:
> The example you gave has an off-by-one row index as follows:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n+1)
> 
> Note that the row index is incremented.  Mathematica simplifies this
> fraction to LaguerreL[-(n+1), 1]/LaguerreL[n+1, -1].  This is not equal
> to Exp[1], but interestingly seems to approach Exp[1] monotonically from
> below as n approaches infinity.
> 
> I'm sure you meant:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n)
> 
> Mathematica simplifies this to LaguerreL[-(n+1), 1]/LaguerreL[n, -1].
> Although I can't get it to explicitly resolve that this is exactly
> Exp[1], the values do seem to bear this truth out.


I think I got it. As usual I like to write that problem in
matrix-mode.
So with

     V(x)~ * Pf = B

defined as the (infinite) matrix product as set of formal powerseries
in x:

                     1/0!   .     .     .
                     1/1!  1/0!
                     1/2!  2/1!  1/0!
                     1/3!  3/2!  3/1!  1/0!
                 .....
                   -------------------------------
[1 x x^2 x^3...]  = [ b0   b1    b2     b3 .... ]

we have, using the apostroph ' for derivative wrt x:

   b0  =            exp(x)      /0!
   b1  =  x*  [ x  *exp(x) ]'   /1!
   b2 =   x^2*[ x^2*exp(x) ]''  /2!
   b3 =   x^3*[ x^3*exp(x) ]''' /3!
 ...
   bk =   x^k*[ x^k*exp(x) ]'(k)' /k!

Expanding the derivatives gives

(1):

   b0 *0! / exp(x) =   1
   b1 *1! / exp(x) =   1 + x

   b2 *2! / exp(x) =   (2 + 4x + x^2)
                   =   2!/0! + 2*2!/1!  + 1*2!/2! x^2

   b3 *3! / exp(x) =   (6 + 18x + 9x^2 + x^3)
                        3! + 3*3!/1! x + 3*3!/2! x^2 + 1 *3!/3! x^3

   b4 *4! / exp(x) =   (24   + 96x  + 72x^2   + 16x^3 + 1x^4)
                       (1*4! + 4*4!/1! x + 6*4!/2!x^2 + 4*4!/3!x^3 + 1*4!/4!*x^4)



and the rhs written as matrix, where also the k! of the k'th row is cancelled , gives

(2):
   b0 / exp(x)      =   1                              *  [1,  x,  x^2,  x^3, ....]~
   b1 / exp(x)      =   1/0!  1/1!
   b2 / exp(x)      =   1/0!  2/1!  1/2!
   b3 / exp(x)      =   1/0!  3/1!  3/2!   1/3!
     ...            =     ....

If we now consider the result when letting x->1 we can further use
the horizontal symmetry of the binomial-coefficients to reorder
the rows and just get the matrix Pf again as coefficients

(3):

 [b0 b1 b2 b3...]~ /exp(1) =    1/0!                            * [1, 1, 1, 1, 1, ...]~
                                1/1!   1/0!
                                1/2!   2/1!   1/0!
                                1/3!   3/2!   3/1!   1/0!


 so

   V(1)~ * Pf = B~
 or
           Pf~ * V(1) = B
 and
           Pf  * V(1) = B / exp(x)

The systematic of the derivation of matrix Pf as resultant
of the derivative-proceeding in (1) and (2) needs now a better
systematic proof, but I think, that could be done...


--------------------------------------------------

One could proceed one further step and state

 for a current row and column m of Pf we have

(4):
    V(x)~ * Pf[0..inf,m] = exp(x) * Pf[m,0..m] * x^(2m) * V(1/x)

or (Pari:)

   suminf( k=0, x^k*Pf[k+1,m+1]) = exp(x) * x^(2*m) *  sum(k=0,m,Pf[m+1,k+1]/x^k)

where for a row r and col c the entry Pf[r+1,c+1] may be replaced by

      Pf[r+1,c+1] = ch(r,c) / (r-c)!          // for r>=c
                  = 0                         // for r<c

where ch(a,b) is the binomial-coefficient.

----------------

The common notation for the above observation may be given as:

    oo            x^k               m    m              x^(m-k)
   sum   ch(k,m)*------   = exp(x) x  * sum   ch(m,k)* ---------
    k=m          (k-m)!                 k=0              (m-k)!

                where ch(a,b) is the binomial-coefficient.


Gottfried Helms






From bowerc at usa.net  Wed Nov 29 10:49:58 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 29 Nov 2006 01:49:58 -0800
Subject: Categories
Message-ID: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>


I have one more row to add

------ Original Message ------
From: franktaw at netscape.net
To: seqfan at ext.jussieu.fr
Subject: Categories

> How many categories are there?
> 
> First, how many categories are there with n morphisms and k objects?  
> This table starts:
> 
>  1
>  2  1
>  7  3 1
> 35 16 3 1
> 
228 77 20 3 1

> The first column is A058129, the number of monoids; the main diagonal 
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
> 
> Taking the row sums, we get:
> 
> 1,3,11,55
329
> 
> the number of categories with n morphisms.  This is probably not in the 
> OEIS (only
> A001776 is possible - other matches become less than A058129).

While the 329 is eerily close to A001776's 330, EULER(A058129) is a
lower limit for this sequence and we have 2982 vs 2345 for #6 there.
>  The 
> inverse Euler
> transform,
> 
> 1,2,8,41
258
> 
> is the number of connected categories with n morphisms; this is 
> likewise probably not
> in the OEIS (only A052447 is possible).
No longer possible
> 
> Can somebody generate more data?
> 
> Franklin T. Adams-Watters
> 

Christian

PS

I don't know if it will be feasible to collect enough data to do
this, but the columns of the triangle converge and that convergence
would make an interesting sequence in and of itself.








From zakseidov at yahoo.com  Wed Nov 29 13:05:28 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 29 Nov 2006 04:05:28 -0800 (PST)
Subject: Difference between neighbor square and cube
Message-ID: <566064.14049.qm@web38215.mail.mud.yahoo.com>

Just submitted (with no autoreply).
My Q is:
Is the Conjectured list (see %C).. 
old hat or what?
Thanks, Zak

%I A125643
%S A125643
0,0,1,1,4,8,9,16,25,27,36,49,64,64,81,100,121,125,144,169,196,216,225,256,289,324,343,361,400,441,484,512,529,576,625,676,729,729,784,841,900,961,1000,1024,1089,1156,1225,1296,1331,1369,1444,1521,1600,1681,1728,1764,1849,1936,2025,2116,2197
%N A125643 Squares and cubes (with repetitions)
%C A125643 Cf. A002760 Squares and cubes (without
repetitions).
Conjectured list of numbers not appeared as difference
between neighbor terms:
5,6,10,14,16,22,23,29,32,34,42,44,46,50,52,54,58,62,64,66,70,72,78,82,84,86,88,90,92,93,94,96,98,102,105,110,112,114,117,118,120,122,124,126,130,132,134,136,140,141,142,144,153,156,158,160,162,164,165,166,172,176,178,179,182,188,190,192,194
%Y A125643 A002760
%O A125643 0
%K A125643 ,nonn,
%A A125643 Zak Seidov  (zakseidov at gmail.com), Nov 29 2006


 
____________________________________________________________________________________
Do you Yahoo!?
Everyone is raving about the all-new Yahoo! Mail beta.
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From maxale at gmail.com  Wed Nov 29 13:37:32 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 29 Nov 2006 04:37:32 -0800
Subject: Difference between neighbor square and cube
In-Reply-To: <566064.14049.qm@web38215.mail.mud.yahoo.com>
References: <566064.14049.qm@web38215.mail.mud.yahoo.com>
Message-ID: <d3dac270611290437x23c81f2fq6fb37e5d662b801d@mail.gmail.com>

Zak,

Your conjecture is related to Mordell curves of the form y^2 = x^3 + n, see
http://mathworld.wolfram.com/MordellCurve.html
http://tnt.math.metro-u.ac.jp/simath/MORDELL/

The latter link can help to resolve the most difficult case of whether
a given number is the difference of a square and a cube. In
particular, for n=5 and n=-5 the corresponding equation y^2 = x^3 + n
has no positive solutions, meaning that 5 is not a difference of
positive square and cube (not necessary neighbors, so this is even
stronger than needed by conjecture).
It is also easy to see that 5 is not the difference of two cubes. And
5 can be represented in an unique way as the difference of two
squares: 5 = 3^2 - 2^2 but 2^2 and 3^2 are not neighbors in A125643
since they are separated by 2^3.
Therefore, 5 indeed never appears as the difference between neighbor
terms of A125643.

Max


On 11/29/06, zak seidov <zakseidov at yahoo.com> wrote:
> Just submitted (with no autoreply).
> My Q is:
> Is the Conjectured list (see %C)..
> old hat or what?
> Thanks, Zak
>
> %I A125643
> %S A125643
> 0,0,1,1,4,8,9,16,25,27,36,49,64,64,81,100,121,125,144,169,196,216,225,256,289,324,343,361,400,441,484,512,529,576,625,676,729,729,784,841,900,961,1000,1024,1089,1156,1225,1296,1331,1369,1444,1521,1600,1681,1728,1764,1849,1936,2025,2116,2197
> %N A125643 Squares and cubes (with repetitions)
> %C A125643 Cf. A002760 Squares and cubes (without
> repetitions).
> Conjectured list of numbers not appeared as difference
> between neighbor terms:
> 5,6,10,14,16,22,23,29,32,34,42,44,46,50,52,54,58,62,64,66,70,72,78,82,84,86,88,90,92,93,94,96,98,102,105,110,112,114,117,118,120,122,124,126,130,132,134,136,140,141,142,144,153,156,158,160,162,164,165,166,172,176,178,179,182,188,190,192,194
> %Y A125643 A002760
> %O A125643 0
> %K A125643 ,nonn,
> %A A125643 Zak Seidov  (zakseidov at gmail.com), Nov 29 2006
>
>
>
> ____________________________________________________________________________________
> Do you Yahoo!?
> Everyone is raving about the all-new Yahoo! Mail beta.
> http://new.mail.yahoo.com
>





From zakseidov at yahoo.com  Wed Nov 29 14:44:52 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 29 Nov 2006 05:44:52 -0800 (PST)
Subject: Difference between neighbor square and cube
In-Reply-To: <d3dac270611290437x23c81f2fq6fb37e5d662b801d@mail.gmail.com>
Message-ID: <948080.43313.qm@web38204.mail.mud.yahoo.com>

Max,
thanks a lot for your (excellent as usual) reply!!

Using it I found:
A054504(*
Numbers n such that Mordell's equation y^2 = x^3 + n
has no integral solutions. *)=
{6, 7, 11, 13, 14, 20, 21, 23, 29, 32, 34, 39, 42, 45,
46, 47, 51, 53, 58, 59, 60, 61, 62, 66, 67, 69, 70,
74, 75, 77, 78, 83, 84, 85, 86, 87, 88, 90, 93, 95,
96, 102, 103, 104, 109, 110, 111, 114, 115, 116, 118,
123, 124, 130, 133, 135, 137, 139, 140, 146, 147, 149,
153, 155};

A081121=(*
Numbers n such that Mordell's equation y^2 = x^3 - n
has no integral solutions. *){
3, 5, 6, 9, 10, 12, 14, 16, 17, 21, 22, 24, 29, 30,
31, 32, 33, 34, 36, 37, 38, 41, 42, 43, 46, 50, 51,
52, 57, 58, 59, 62, 65, 66, 68, 69, 70, 73, 75, 77,
78, 80, 82, 84, 85, 86, 88, 90, 91, 92, 93, 94, 96,
97, 98, 99};

Intersection[A081121,A054504]
ND={6,14,21,29,32,34,42,46,51,58,59,62,66,69,70,75,77,78,84,85,86,88,90,93,96}

And ND list gives 
(Absolute values of) Non-difference between square and
cube (not only neighbor as in %C A125643).

Hence any number in ID should be in %C A125643.
But 21 is in ND and absent in %C A125643!

Indeed in A125643,
we have:
..,64,81,100,121,125,...,
and 121-100=11^2-10^2=21!!

21 is not difference between neighbor square and cube,
but it is difference between neighbor terms in
A125643.

As a result: my subject 'Difference between neighbor
square and cube' is not correct,
and in %C A125643 we have 
'Difference between neighbor square/cube and
square/cube'.

Sorry and thanks to Max & all seqfans,
Zak  


--- "Max A." <maxale at gmail.com> wrote:

> Zak,
> 
> Your conjecture is related to Mordell curves of the
> form y^2 = x^3 + n, see
> http://mathworld.wolfram.com/MordellCurve.html
> http://tnt.math.metro-u.ac.jp/simath/MORDELL/
> 
> The latter link can help to resolve the most
> difficult case of whether
> a given number is the difference of a square and a
> cube. In
> particular, for n=5 and n=-5 the corresponding
> equation y^2 = x^3 + n
> has no positive solutions, meaning that 5 is not a
> difference of
> positive square and cube (not necessary neighbors,
> so this is even
> stronger than needed by conjecture).
> It is also easy to see that 5 is not the difference
> of two cubes. And
> 5 can be represented in an unique way as the
> difference of two
> squares: 5 = 3^2 - 2^2 but 2^2 and 3^2 are not
> neighbors in A125643
> since they are separated by 2^3.
> Therefore, 5 indeed never appears as the difference
> between neighbor
> terms of A125643.
> 
> Max
> 
> 
> On 11/29/06, zak seidov <zakseidov at yahoo.com> wrote:
> > Just submitted (with no autoreply).
> > My Q is:
> > Is the Conjectured list (see %C)..
> > old hat or what?
> > Thanks, Zak
> >
> > %I A125643
> > %S A125643
> >
>
0,0,1,1,4,8,9,16,25,27,36,49,64,64,81,100,121,125,144,169,196,216,225,256,289,324,343,361,400,441,484,512,529,576,625,676,729,729,784,841,900,961,1000,1024,1089,1156,1225,1296,1331,1369,1444,1521,1600,1681,1728,1764,1849,1936,2025,2116,2197
> > %N A125643 Squares and cubes (with repetitions)
> > %C A125643 Cf. A002760 Squares and cubes (without
> > repetitions).
> > Conjectured list of numbers not appeared as
> difference
> > between neighbor terms:
> >
>
5,6,10,14,16,22,23,29,32,34,42,44,46,50,52,54,58,62,64,66,70,72,78,82,84,86,88,90,92,93,94,96,98,102,105,110,112,114,117,118,120,122,124,126,130,132,134,136,140,141,142,144,153,156,158,160,162,164,165,166,172,176,178,179,182,188,190,192,194
> > %Y A125643 A002760
> > %O A125643 0
> > %K A125643 ,nonn,
> > %A A125643 Zak Seidov  (zakseidov at gmail.com), Nov
> 29 2006
> >
> >
> >
> >
>
____________________________________________________________________________________
> > Do you Yahoo!?
> > Everyone is raving about the all-new Yahoo! Mail
> beta.
> > http://new.mail.yahoo.com
> >
> 



 
____________________________________________________________________________________
Do you Yahoo!?
Everyone is raving about the all-new Yahoo! Mail beta.
http://new.mail.yahoo.com





From jvospost3 at gmail.com  Wed Nov 29 19:52:44 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 29 Nov 2006 10:52:44 -0800
Subject: Categories
In-Reply-To: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
References: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
Message-ID: <5542af940611291052l1a3b2654k7467715f3073dfde@mail.gmail.com>

Thank you, Christian.  That looks right.

Note that the partial sums of the first column are A118601 Number of monoids
(semigroups with identity) of order <=n.  This first column also equals the
row sums of A058137, which partitions the monoids of order n by the number k
of idempotents.






On 11/29/06, Christian G. Bower <bowerc at usa.net> wrote:
>
>
> I have one more row to add
>
> ------ Original Message ------
> Received: Wed, 22 Nov 2006 01:21:01 PM PST
> From: franktaw at netscape.net
> To: seqfan at ext.jussieu.fr
> Subject: Categories
>
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> >  1
> >  2  1
> >  7  3 1
> > 35 16 3 1
> >
> 228 77 20 3 1
>
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
> > Taking the row sums, we get:
> >
> > 1,3,11,55
> 329
> >
> > the number of categories with n morphisms.  This is probably not in the
> > OEIS (only
> > A001776 is possible - other matches become less than A058129).
>
> While the 329 is eerily close to A001776's 330, EULER(A058129) is a
> lower limit for this sequence and we have 2982 vs 2345 for #6 there.
> >  The
> > inverse Euler
> > transform,
> >
> > 1,2,8,41
> 258
> >
> > is the number of connected categories with n morphisms; this is
> > likewise probably not
> > in the OEIS (only A052447 is possible).
> No longer possible
> >
> > Can somebody generate more data?
> >
> > Franklin T. Adams-Watters
> >
>
> Christian
>
> PS
>
> I don't know if it will be feasible to collect enough data to do
> this, but the columns of the triangle converge and that convergence
> would make an interesting sequence in and of itself.
>
>
>
>
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From franktaw at netscape.net  Wed Nov 29 20:09:00 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 29 Nov 2006 14:09:00 -0500
Subject: Categories
In-Reply-To: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
References: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
Message-ID: <8C8E200F6031B3A-794-8706@FWM-M14.sysops.aol.com>

Thanks, Christian.

I think we need to look at one more table:  the number of connected
categories with n morphisms and k objects.  This starts:

  1
  2
  7 1
 35 6
228 28 2

(Row n has length ceiling(n/2).)

The table of the number of categories that I started with is a two-
dimensional Euler transform of this table.

The downward-sloping diagonal sums of this table are 1,3,15,???.
Shifting this left and taking the Euler transform gives the limiting
values Christian is referring to; starting with b(0):

1,3,21,???

For the table above, a(2n-1,n) has a purely algebraic or graph-
theoretic interpretation.  It is the number of connected
anti-transitive relations on n objects (meaning that if a R b and
b R c, then NOT (a R c)); equivalently, the number of bipartite
oriented trees, where each edge origin is in the same part.

If I haven't made any mistakes, this sequence starts:

1,1,2,3,6,10

This is not enough data to determine whether it is in the OEIS.

Franklin T. Adams-Watters


-----Original Message-----
From: bowerc at usa.net

I have one more row to add

------ Original Message ------

> How many categories are there?
>
> First, how many categories are there with n morphisms and k objects?
> This table starts:
>
>  1
>  2  1
>  7  3 1
> 35 16 3 1
>
228 77 20 3 1

> The first column is A058129, the number of monoids; the main diagonal
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
>
> Taking the row sums, we get:
>
> 1,3,11,55
329
>
> the number of categories with n morphisms.  This is probably not in 
the
> OEIS (only
> A001776 is possible - other matches become less than A058129).

While the 329 is eerily close to A001776's 330, EULER(A058129) is a
lower limit for this sequence and we have 2982 vs 2345 for #6 there.
>  The
> inverse Euler
> transform,
>
> 1,2,8,41
258
>
> is the number of connected categories with n morphisms; this is
> likewise probably not
> in the OEIS (only A052447 is possible).
No longer possible
>
> Can somebody generate more data?
>
> Franklin T. Adams-Watters
>

Christian

PS

I don't know if it will be feasible to collect enough data to do
this, but the columns of the triangle converge and that convergence
would make an interesting sequence in and of itself.

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From jvospost3 at gmail.com  Wed Nov 29 20:23:47 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 29 Nov 2006 11:23:47 -0800
Subject: Categories
In-Reply-To: <8C8E200F6031B3A-794-8706@FWM-M14.sysops.aol.com>
References: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
	 <8C8E200F6031B3A-794-8706@FWM-M14.sysops.aol.com>
Message-ID: <5542af940611291123l4f9c3385l283f3b2243dab1c2@mail.gmail.com>

The issue of "connected" in the underlying functional graph, as a subset of
the set of all digraphs, is further revealed by the distinction (definitions
at MathWorld) of Weakly Conncted versus Strongly Conncted Digraphs.

Every functional digraph is the disjoint union of connected digraphs.
That's why there is a recursion in the enumeration of F(n) = functional
graphs on n points. F(n) is equal or greater to the number of functional
graphs on n-1 points, unioned with an isolated point looping to itself, plus
the number of functional graphs on n-2 points unioned with any of the 3
functional graphs on 3 points, ..., throwing out any double or multiple
counts by equivalence class on isomorphisms of the unlabelled digraphs.

Note that MathWorld's correct definition of functional graph, and its
Mathematica, is identical to what I'd called endofunctions in earlier
posting of this thread.  Hence the enumeration problem that I solved for
"prime" endofunctions is identical to a subset of your category enumeration
problem, relating to the number of categories that are not unions nor
categorical products of smaller categories.

However, when I referred to 2-categories before, I should more correctly
have said Bicategories, the difference being based on whether associativity
is strict or up to 2-isomorphism.  Your enumeration of categories is more
formally explained in terms of Cat, the category of categories and functors,
but this is not the right venue for me to go into excruciating details.

On 11/29/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
> Thanks, Christian.
>
> I think we need to look at one more table:  the number of connected
> categories with n morphisms and k objects.  This starts:
>
>   1
>   2
>   7 1
> 35 6
> 228 28 2
>
> (Row n has length ceiling(n/2).)
>
> The table of the number of categories that I started with is a two-
> dimensional Euler transform of this table.
>
> The downward-sloping diagonal sums of this table are 1,3,15,???.
> Shifting this left and taking the Euler transform gives the limiting
> values Christian is referring to; starting with b(0):
>
> 1,3,21,???
>
> For the table above, a(2n-1,n) has a purely algebraic or graph-
> theoretic interpretation.  It is the number of connected
> anti-transitive relations on n objects (meaning that if a R b and
> b R c, then NOT (a R c)); equivalently, the number of bipartite
> oriented trees, where each edge origin is in the same part.
>
> If I haven't made any mistakes, this sequence starts:
>
> 1,1,2,3,6,10
>
> This is not enough data to determine whether it is in the OEIS.
>
> Franklin T. Adams-Watters
>
>
> -----Original Message-----
> From: bowerc at usa.net
>
> I have one more row to add
>
> ------ Original Message ------
> Received: Wed, 22 Nov 2006 01:21:01 PM PST
>
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> >  1
> >  2  1
> >  7  3 1
> > 35 16 3 1
> >
> 228 77 20 3 1
>
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
> > Taking the row sums, we get:
> >
> > 1,3,11,55
> 329
> >
> > the number of categories with n morphisms.  This is probably not in
> the
> > OEIS (only
> > A001776 is possible - other matches become less than A058129).
>
> While the 329 is eerily close to A001776's 330, EULER(A058129) is a
> lower limit for this sequence and we have 2982 vs 2345 for #6 there.
> >  The
> > inverse Euler
> > transform,
> >
> > 1,2,8,41
> 258
> >
> > is the number of connected categories with n morphisms; this is
> > likewise probably not
> > in the OEIS (only A052447 is possible).
> No longer possible
> >
> > Can somebody generate more data?
> >
> > Franklin T. Adams-Watters
> >
>
> Christian
>
> PS
>
> I don't know if it will be feasible to collect enough data to do
> this, but the columns of the triangle converge and that convergence
> would make an interesting sequence in and of itself.
>
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and
> industry-leading spam and email virus protection.
>
>
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From njas at research.att.com  Wed Nov 29 22:34:56 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 29 Nov 2006 16:34:56 -0500 (EST)
Subject: OEIS on vacation in December
Message-ID: <200611292134.QAA53261@fry.research.att.com>


This means:
- do send in sequences in published works or 
   that are important for your work
- do send important corrections, extensions, updates
- do send EDITED versions

- don't send in sequences that you made up (unless
   they are really beautiful - and if they
   are base dependent or involve primes 
   they are probably not)
- don't send very minor corrections

I will be doing occasional updates, but I won't be
working on the OEIS 24/7 !

Thanks!

Neil





From maxale at gmail.com  Wed Nov 29 22:45:38 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 29 Nov 2006 13:45:38 -0800
Subject: Difference between neighbor square and cube
In-Reply-To: <948080.43313.qm@web38204.mail.mud.yahoo.com>
References: <d3dac270611290437x23c81f2fq6fb37e5d662b801d@mail.gmail.com>
	 <948080.43313.qm@web38204.mail.mud.yahoo.com>
Message-ID: <d3dac270611291345n2ad5dae4id319f3bce8fca71c@mail.gmail.com>

On 11/29/06, zak seidov <zakseidov at yahoo.com> wrote:

> Intersection[A081121,A054504]
> ND={6,14,21,29,32,34,42,46,51,58,59,62,66,69,70,75,77,78,84,85,86,88,90,93,96}

That's A110223. But Robert G. Wilson posed as "conjectured".
I believe this sequence can be confirmed/verified using the
information about Mordell curves from
http://tnt.math.metro-u.ac.jp/simath/MORDELL/

Max





From bowerc at usa.net  Thu Nov 30 00:42:31 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 29 Nov 2006 15:42:31 -0800
Subject: Categories
Message-ID: <623kkCXpf3086S27.1164843751@cmsweb27.cms.usa.net>

ftaw
> I think we need to look at one more table:  the number of connected
> categories with n morphisms and k objects.  This starts:

>   1
>   2
>   7 1
>  35 6
> 228 28 2

> (Row n has length ceiling(n/2).)
...
> For the table above, a(2n-1,n) has a purely algebraic or graph-
> theoretic interpretation.  It is the number of connected
> anti-transitive relations on n objects (meaning that if a R b and
> b R c, then NOT (a R c)); equivalently, the number of bipartite
> oriented trees, where each edge origin is in the same part.
> 
> If I haven't made any mistakes, this sequence starts:
> 
> 1,1,2,3,6,10
> 
> This is not enough data to determine whether it is in the OEIS.

For n>1 it's A122086
http://www.research.att.com/~njas/sequences/A122086
A sequence which could use a nicer description and formula such as
a(n) = A000055(n) - A000081(n/2) where a sequence evaluated at a
noninteger is 0.
http://www.research.att.com/~njas/sequences/A000055
http://www.research.att.com/~njas/sequences/A000081

Christian








From bowerc at usa.net  Thu Nov 30 01:36:23 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 29 Nov 2006 16:36:23 -0800
Subject: Categories
Message-ID: <209kkdajX8954S10.1164846983@cmsweb10.cms.usa.net>

I have row 6 now:

> How many categories are there?
>
> First, how many categories are there with n morphisms and k objects?
> This table starts:
>
>  1
>  2  1
>  7  3 1
> 35 16 3 1
>
228 77 20 3 1
2237 485 111 21 3 1

> Taking the row sums, we get:
>
> 1,3,11,55
329
2858

>  The
> inverse Euler
> transform,
>
> 1,2,8,41
258
2407

> I think we need to look at one more table:  the number of connected
> categories with n morphisms and k objects.  This starts:

>   1
>   2
>   7 1
>  35 6
> 228 28 2
2237 159 11
31559 ?   ? 3

> (Row n has length ceiling(n/2).)
> 
> The table of the number of categories that I started with is a two-
> dimensional Euler transform of this table.
> 
> The downward-sloping diagonal sums of this table are 1,3,15,???.
77
> Shifting this left and taking the Euler transform gives the limiting
> values Christian is referring to; starting with b(0):
> 
> 1,3,21,???
132








From franktaw at netscape.net  Thu Nov 30 14:53:38 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 30 Nov 2006 08:53:38 -0500
Subject: Categories
In-Reply-To: <623kkCXpf3086S27.1164843751@cmsweb27.cms.usa.net>
References: <623kkCXpf3086S27.1164843751@cmsweb27.cms.usa.net>
Message-ID: <8C8E29E12708888-5D8-4EB@WEBMAIL-RA17.sysops.aol.com>

A typo there; that should be a(n) = 2*A000055(n) - A000081(n/2).

Franklin T. Adams-Watters


-----Original Message-----
From: bowerc at usa.net

...
For n>1 it's A122086
http://www.research.att.com/~njas/sequences/A122086
A sequence which could use a nicer description and formula such as
a(n) = A000055(n) - A000081(n/2) where a sequence evaluated at a
noninteger is 0.
http://www.research.att.com/~njas/sequences/A000055
http://www.research.att.com/~njas/sequences/A000081

Christian





________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From jvospost3 at gmail.com  Thu Nov 30 19:34:51 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 30 Nov 2006 10:34:51 -0800
Subject: Categories
In-Reply-To: <8C8E29E12708888-5D8-4EB@WEBMAIL-RA17.sysops.aol.com>
References: <623kkCXpf3086S27.1164843751@cmsweb27.cms.usa.net>
	 <8C8E29E12708888-5D8-4EB@WEBMAIL-RA17.sysops.aol.com>
Message-ID: <5542af940611301034t3bba13fdnadc86411facc535d@mail.gmail.com>

a(n) = 2*A000055(n) - A000081(n/2) = (2* Number of trees with n unlabeled
nodes) - Number of rooted trees with n/2 nodes (or connected functions with
a fixed point)

where a(n) is downwards sloping diagonal sums of main category table, or
what?

It isn't until we look at bicategories that we are forced to look at the
symmetries and structure of the individual categories being enumerated here,
as objects in the higher categories, and have to enumerate the morphisms of
the functions. A (finite, concrete) category whose underlying directed graph
has no symmetry (i.e. automorphism group of order 1) can only map to itself
or an identical copy of itself, point to point, loop to loop, arc to arc.
But any cycle of length >1 in that directed graph allows morphisms to other
directed graphs.  The directed graph C_n, a cycle of length n, can thus map
to the same graph but mapping each point to the next, and each arc to the
next, and so forth. That requires then analysis of cycles in directed
graphs, as by de Bruijn in 1952 and more deeply understood since then. This
is another enumeration problem, unnecessary to completing the answer to
Frank's original question, and more tied to my endofunctions analysis. Sorry
if my interruptions distracted you from the straightforward question.  Good
work, guys!

Funny how Category Theory folks tend to be addicted to theory for its own
sake, and less willing to "get their hands dirty" with anything as old
fashioned as enumeration.  OEIS helps us focus here, by demanding the
integer sequences!

On 11/30/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
> A typo there; that should be a(n) = 2*A000055(n) - A000081(n/2).
>
> Franklin T. Adams-Watters
>
>
> -----Original Message-----
> From: bowerc at usa.net
>
> ...
> For n>1 it's A122086
> http://www.research.att.com/~njas/sequences/A122086
> A sequence which could use a nicer description and formula such as
> a(n) = A000055(n) - A000081(n/2) where a sequence evaluated at a
> noninteger is 0.
> http://www.research.att.com/~njas/sequences/A000055
> http://www.research.att.com/~njas/sequences/A000081
>
> Christian
>
>
>
>
>
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and
> industry-leading spam and email virus protection.
>
>
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From all at abouthugo.de  Thu Nov 30 21:28:06 2006
From: all at abouthugo.de (Hugo Pfoertner)
Date: Thu, 30 Nov 2006 21:28:06 +0100
Subject: Difference between neighbor square and cube
References: <948080.43313.qm@web38204.mail.mud.yahoo.com>
Message-ID: <456F3ED6.D0619322@abouthugo.de>

zak seidov wrote:
> 
> Max,
> thanks a lot for your (excellent as usual) reply!!
> 
> Using it I found:
> A054504(*
> Numbers n such that Mordell's equation y^2 = x^3 + n
> has no integral solutions. *)=
> {6, 7, 11, 13, 14, 20, 21, 23, 29, 32, 34, 39, 42, 45,
> 46, 47, 51, 53, 58, 59, 60, 61, 62, 66, 67, 69, 70,
> 74, 75, 77, 78, 83, 84, 85, 86, 87, 88, 90, 93, 95,
> 96, 102, 103, 104, 109, 110, 111, 114, 115, 116, 118,
> 123, 124, 130, 133, 135, 137, 139, 140, 146, 147, 149,
> 153, 155};
> 
> A081121=(*
> Numbers n such that Mordell's equation y^2 = x^3 - n
> has no integral solutions. *){
> 3, 5, 6, 9, 10, 12, 14, 16, 17, 21, 22, 24, 29, 30,
> 31, 32, 33, 34, 36, 37, 38, 41, 42, 43, 46, 50, 51,
> 52, 57, 58, 59, 62, 65, 66, 68, 69, 70, 73, 75, 77,
> 78, 80, 82, 84, 85, 86, 88, 90, 91, 92, 93, 94, 96,
> 97, 98, 99};
> 
> Intersection[A081121,A054504]
> ND={6,14,21,29,32,34,42,46,51,58,59,62,66,69,70,75,77,78,84,85,86,88,90,93,96}
> 
> And ND list gives
> (Absolute values of) Non-difference between square and
> cube (not only neighbor as in %C A125643).
> 
> Hence any number in ID should be in %C A125643.
> But 21 is in ND and absent in %C A125643!
> 
> Indeed in A125643,
> we have:
> ..,64,81,100,121,125,...,
> and 121-100=11^2-10^2=21!!
> 
> 21 is not difference between neighbor square and cube,
> but it is difference between neighbor terms in
> A125643.
> 
> As a result: my subject 'Difference between neighbor
> square and cube' is not correct,
> and in %C A125643 we have
> 'Difference between neighbor square/cube and
> square/cube'.
> 
> Sorry and thanks to Max & all seqfans,
> Zak

Also closely related:

http://www.research.att.com/~njas/sequences/A087285
http://www.research.att.com/~njas/sequences/A087286
http://www.research.att.com/~njas/sequences/A088017

Hugo





From bowerc at usa.net  Thu Nov 30 22:29:20 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Thu, 30 Nov 2006 13:29:20 -0800
Subject: Categories
Message-ID: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>



------ Original Message ------
From: "Jonathan Post" <jvospost3 at gmail.com>
To: "franktaw at netscape.net" <franktaw at netscape.net>Cc: bowerc at usa.net,
seqfan at ext.jussieu.fr, jvospost2 at yahoo.com
Subject: Re: Categories

> a(n) = 2*A000055(n) - A000081(n/2) = (2* Number of trees with n unlabeled
> nodes) - Number of rooted trees with n/2 nodes (or connected functions with
> a fixed point)
> 
> where a(n) is downwards sloping diagonal sums of main category table, or
> what?

Actually it's the "corner" values from this table

>   *1*
>   2
>   7 *1*
>  35 6
> 228 28 *2*
> 2237 159 11
> 31559 ?   ? *3*

The downward sloping sums were an aide in calculating the convergent
sequence of the columns

The interpretation of these corners comes from the fact we have a
saturated category. It has 2n-1 elements, n of which are object
identities. It's connected, so it has a minimum of n-1 "crossover"
morphisms, it also has a maximum of n-1 cm's since there are only
n-1 elements left. Any connected graph with n points and n-1 edges is
a tree. The associativity of categories states that if we have an
a-b morphism (an edge on the graph) and a b-c we must have an a-c.
But if we have all 3, it's no longer a tree, hence we never have the
a-b and the b-c, hence each point has all edges either pointing in or
pointing out, hence the biparite graph analogy, hence a little
combinatorial magic.

...
> Funny how Category Theory folks tend to be addicted to theory for its > own
> sake, and less willing to "get their hands dirty" with anything as old
> fashioned as enumeration.  OEIS helps us focus here, by demanding the
> integer sequences!
> 

I try to stay out of philosophical stuff here, but sometimes it's too
hard to resist. I figure I'm guilty of spending too much time counting
stuff and not taking the theory in. Most of the world would not give a
hoot about either, so I'm glad there are folk spending time on what I
would not give any time to.

Still, I find it jarring any time I read a math paper and they write
about several things that can be represented as sequences, yet all they
give is a mathematical description, rarely do they cite the OEIS
reference or even give me the first n terms leaving me to calculate them
myself and hope I didn't make a mistake. If I wrote a paper, I'd
include that stuff, not just because I'm an OEIS nerd, but also because
I would want people to be able to find my paper who didn't know they
were looking for it.

Just a little more self indulgence here with my list of semigroup pipe
dreams. After doing sequences on semigroups
http://www.research.att.com/~njas/sequences/A027851
and monoids
http://www.research.att.com/~njas/sequences/A058129
I knew categories was one of the next logical steps, but I held off
because I knew it would be difficult. Perhaps I can spur someone else's
interest in related problems that may prove even more difficult.

There's transformation semigroups, the semigroup version of permutation
groups
http://www.research.att.com/~njas/sequences/A000638
I've seen TSs written about, but no attempts to enumerate them.
(Note, a TS is any set of endofunctions closed under composition)
There TS's close cousins which could be called:
transformation monoids
relation semigroups
relation monoids
(Composition of relations is such that (a,c) is in R1(R2) iff there
is b such that (a,b) in R1 and (b,c) in R2)
Then there are the generalizations of categories to semigroups and
groupoids. Note I mean the groupoid that is a set and closed operation
http://www.research.att.com/~njas/sequences/A001329
Not the group like category groupoid which might also be interesting
to count and not too difficult; oh terminology.
Anyway, with the semigroup and groupoid categories, there comes the
issue of treating objects and morphims. The ordinary (monoid like)
categories avoid this problem by having object identities among the
morphisms that can be associatied with the objects themselves. Without
that restriction, it's possible to have objects untouched by any
morphism or two structures with identical (or isomorphic) multiplication
tables, but not isomorphic when you take the objects into account.
Basically many of the same counting problems one has with multigraphs
(and probably the same solutions too.)








From jvospost3 at gmail.com  Thu Nov 30 22:53:42 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 30 Nov 2006 13:53:42 -0800
Subject: Categories
In-Reply-To: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>
References: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>
Message-ID: <5542af940611301353m486a4788wff2e4eb539b76943@mail.gmail.com>

I strongly agree with Christian.  My doctoral dissertation involved taking
the Krohn-Rhodes decomposition of the semigroup of differential operators of
a system of nonlinear differential equations that had been in the
mathematical Biology literature for decades, and routinely dismissed as not
capable of closed-form solution. Marvin Minsky (who had been the PhD advisor
of the eponymous John Rhodes) thought my work was the best use yet of his
student's theorem.  The great Stan Ulam (coinventor of cellular automata,
the H-bomb, and the nuclear pulse rocket) also thought it was cool. But only
half a dozen biologists in theb world knew that much about semigroups, and
some of them wrote to me from Edinburgh, Scotland, and places in the USSR.
Ulam died before we could coauthor; Semigroup Forum accepted my paper with
delight for a special issue, but wanted it reformatted, and I was laid off
from the company where I worked, and they erased on only machine-readable
version of the paper.  Equations were harder to typeset those decades
ago...  But yes, lots of so-called professionals go out of their way to
avoid citing the OEIS, or give any useful examples.  Also yes: semigroup
enumerations are worthwhile.  Also, you nailed the differences between those
and categories.

See for instance:

  A118581 <http://www.research.att.com/~njas/sequences/A118581> Number of
nonisomorphic *semigroups* of order <= n.

  A118099 <http://www.research.att.com/~njas/sequences/A118099> Number of
inverse *semigroups* of order <= n.

  A118100 <http://www.research.att.com/~njas/sequences/A118100>

Number of commutative *semigroups* of order <= n.

  A118601 <http://www.research.att.com/~njas/sequences/A118601> Number of
monoids (*semigroups* with identity) of order <= n.
  A113534 <http://www.research.att.com/~njas/sequences/A113534> Ascending
descending base exponent transform of the flipped tribonacci substitution (
A092782 <http://www.research.att.com/~njas/sequences/A092782>).
with its semigroup reference, and some others.

-- Jonathan


On 11/30/06, Christian G. Bower <bowerc at usa.net> wrote:
>
>
>
> ------ Original Message ------
> Received: Thu, 30 Nov 2006 10:34:55 AM PST
> From: "Jonathan Post" <jvospost3 at gmail.com>
> To: "franktaw at netscape.net" <franktaw at netscape.net>Cc: bowerc at usa.net,
> seqfan at ext.jussieu.fr, jvospost2 at yahoo.com
> Subject: Re: Categories
>
> > a(n) = 2*A000055(n) - A000081(n/2) = (2* Number of trees with n
> unlabeled
> > nodes) - Number of rooted trees with n/2 nodes (or connected functions
> with
> > a fixed point)
> >
> > where a(n) is downwards sloping diagonal sums of main category table, or
> > what?
>
> Actually it's the "corner" values from this table
>
> >   *1*
> >   2
> >   7 *1*
> >  35 6
> > 228 28 *2*
> > 2237 159 11
> > 31559 ?   ? *3*
>
> The downward sloping sums were an aide in calculating the convergent
> sequence of the columns
>
> The interpretation of these corners comes from the fact we have a
> saturated category. It has 2n-1 elements, n of which are object
> identities. It's connected, so it has a minimum of n-1 "crossover"
> morphisms, it also has a maximum of n-1 cm's since there are only
> n-1 elements left. Any connected graph with n points and n-1 edges is
> a tree. The associativity of categories states that if we have an
> a-b morphism (an edge on the graph) and a b-c we must have an a-c.
> But if we have all 3, it's no longer a tree, hence we never have the
> a-b and the b-c, hence each point has all edges either pointing in or
> pointing out, hence the biparite graph analogy, hence a little
> combinatorial magic.
>
> ...
> > Funny how Category Theory folks tend to be addicted to theory for its >
> own
> > sake, and less willing to "get their hands dirty" with anything as old
> > fashioned as enumeration.  OEIS helps us focus here, by demanding the
> > integer sequences!
> >
>
> I try to stay out of philosophical stuff here, but sometimes it's too
> hard to resist. I figure I'm guilty of spending too much time counting
> stuff and not taking the theory in. Most of the world would not give a
> hoot about either, so I'm glad there are folk spending time on what I
> would not give any time to.
>
> Still, I find it jarring any time I read a math paper and they write
> about several things that can be represented as sequences, yet all they
> give is a mathematical description, rarely do they cite the OEIS
> reference or even give me the first n terms leaving me to calculate them
> myself and hope I didn't make a mistake. If I wrote a paper, I'd
> include that stuff, not just because I'm an OEIS nerd, but also because
> I would want people to be able to find my paper who didn't know they
> were looking for it.
>
> Just a little more self indulgence here with my list of semigroup pipe
> dreams. After doing sequences on semigroups
> http://www.research.att.com/~njas/sequences/A027851
> and monoids
> http://www.research.att.com/~njas/sequences/A058129
> I knew categories was one of the next logical steps, but I held off
> because I knew it would be difficult. Perhaps I can spur someone else's
> interest in related problems that may prove even more difficult.
>
> There's transformation semigroups, the semigroup version of permutation
> groups
> http://www.research.att.com/~njas/sequences/A000638
> I've seen TSs written about, but no attempts to enumerate them.
> (Note, a TS is any set of endofunctions closed under composition)
> There TS's close cousins which could be called:
> transformation monoids
> relation semigroups
> relation monoids
> (Composition of relations is such that (a,c) is in R1(R2) iff there
> is b such that (a,b) in R1 and (b,c) in R2)
> Then there are the generalizations of categories to semigroups and
> groupoids. Note I mean the groupoid that is a set and closed operation
> http://www.research.att.com/~njas/sequences/A001329
> Not the group like category groupoid which might also be interesting
> to count and not too difficult; oh terminology.
> Anyway, with the semigroup and groupoid categories, there comes the
> issue of treating objects and morphims. The ordinary (monoid like)
> categories avoid this problem by having object identities among the
> morphisms that can be associatied with the objects themselves. Without
> that restriction, it's possible to have objects untouched by any
> morphism or two structures with identical (or isomorphic) multiplication
> tables, but not isomorphic when you take the objects into account.
> Basically many of the same counting problems one has with multigraphs
> (and probably the same solutions too.)
>
>
>
>
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From franktaw at netscape.net  Thu Nov 30 22:54:39 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 30 Nov 2006 16:54:39 -0500
Subject: Categories
In-Reply-To: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>
References: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>
Message-ID: <8C8E2E14509A6B2-934-A97D@FWM-M10.sysops.aol.com>

I'd like to put my plug in for more application in math.

I heard of a case some years ago where a paper was published about some 
kind of mathematical object (I think it was a kind of topological 
space, but it really doesn't matter).  Three or four more papers were 
published, establishing more and more properties for this type of 
object - until finally it was proved that they don't exist!  This 
wouldn't have happened if somebody had asked for an example at an early 
stage.

There is a strong tendency in mathematics to start at the end.  The 
researcher pursues a line of thought, which eventually leads to a 
spiffy proof.  The proof is then published, with no hint of the process 
by which it was reached.  This is a disservice to anybody who might use 
a similar approach to solve some other problem.  It is especially a 
disservice when presented to students.

On a more personal level, I find when looking a math paper, I want to 
know how this relates to problems that I am already interested in or at 
least familiar with.  If I can't get an answer to that, I have a hard 
time maintaining any interest.

Franklin T. Adams-Watters

________________________________________________________________________
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From zakseidov at yahoo.com  Wed Nov  1 06:43:12 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Tue, 31 Oct 2006 21:43:12 -0800 (PST)
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
Message-ID: <20061101054313.25333.qmail@web38210.mail.mud.yahoo.com>

Dear seqfans,
Bob, Ivars,

In A065577, 
a(1)=2 because 10=3+7=5+5?
Also a(2)=6 because
100=3+97=11+89=17+83=29+71=41+59=47+53?

The point is that in Ivars Peterson's
"Goldbach's Prime Pairs"
http://www.maa.org/mathland/mathtrek_8_21_00.html
we read:

" 
Note that Goldbach partitions take into account the
order of the primes that are summed. For example, 10
has two Goldbach partitions: 3 + 7 and 7 + 3.

Integer No. of Goldbach partitions 
10 2 
100 6 
1,000 28 
10,000 127 
100,000 810 
1,000,000 5,402 
10,000,000 38,807 
100,000,000 291,400
" 

Me: But in A065577 a(1)=2 and a(2)=6 
(and also other terms as I checked them)
without taking into account 
the order of the primes 
that are summed(?!)

Can anyoone (Bob? Ivars?) clarify this?
Thanks, Zak

BTW Two next terms are calculated by me
(for n=9, 10) as 2274205, 18200487.
Can anyone check these?
Thanks, Zak


%I A065577
%S A065577 2,6,28,127,810,5402,38807,291400
%N A065577 Number of Goldbach partitions of 10^n.
%H A065577 Ivars Peterson's MathTrek, <a
href="http://www.maa.org/mathland/mathtrek_8_21_00.html">Goldbach's
Prime Pairs</a>
%H A065577 Science News Online, week of Aug. 19, 2000;
Vol. 158, No. 8 <a href="http://www.sciencenews.org/
20000819/mathtrek.asp">Goldbach's Prime Pairs</a>
<skip>
%A A065577 Robert G. Wilson v (rgwv(AT)rgwv.com), Dec
01 2001




 
____________________________________________________________________________________
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From ralf at ark.in-berlin.de  Wed Nov  1 10:35:28 2006
From: ralf at ark.in-berlin.de (Ralf Stephan)
Date: Wed, 1 Nov 2006 10:35:28 +0100
Subject: Re^2: Error-finding program
In-Reply-To: <200610312150.k9VLoKfn018613@amer.strw.leidenuniv.nl>
References: <200610312150.k9VLoKfn018613@amer.strw.leidenuniv.nl>
Message-ID: <20061101093528.GA19017@ark.in-berlin.de>

> The definitions of A026468 and A026469 are
>  a(n) = least positive integer > a(n-1) and not a(i)^2 + a(j)^2 for 1<=i<=j<=n.
> and
>  a(n) = least positive integer > a(n-1) and not a(i)^2 + a(j)^2 for 1<=i<=j<=n.  	
> which looks quite identical. So why can A026468/9 be two different
> sequences? Is there a "not" too much or too little, or rather one of the 
> "less-equal" signs to be corrected to "less" or a starting condition on a(2)
> to be added?

My guess is that Clark started the same algorithm with 1, 2 and then
tried it with starting values 1, 3, so the definition of A026469 should 
be amended with ", starting with 1, 3".


ralf






From njas at research.att.com  Wed Nov  1 14:16:47 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 1 Nov 2006 08:16:47 -0500 (EST)
Subject: Re^2: Error-finding program
References: <200610312150.k9VLoKfn018613@amer.strw.leidenuniv.nl>
Message-ID: <200611011316.IAA53531@fry.research.att.com>

Ralf,  I think you've solved it!   Thanks!   Neil


My guess is that Clark started the same algorithm with 1, 2 and then
tried it with starting values 1, 3, so the definition of A026469 should 
be amended with ", starting with 1, 3".





From mathar at strw.leidenuniv.nl  Wed Nov  1 14:17:58 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Wed, 1 Nov 2006 14:17:58 +0100
Subject: A065577 Number of Goldbach partitions of 10^n?
Message-ID: <200611011317.kA1DHwmp018846@amer.strw.leidenuniv.nl>


zs> From seqfan-owner at ext.jussieu.fr  Wed Nov  1 06:44:53 2006
zs> Return-Path: <seqfan-owner at ext.jussieu.fr>
zs> Date: Tue, 31 Oct 2006 21:43:12 -0800 (PST)
zs> From: zak seidov <zakseidov at yahoo.com>
zs> Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
zs> To: seqfan at ext.jussieu.fr, ip at sciserv.org, rgwv at rgwv.com
zs> 
zs> Dear seqfans,
zs> Bob, Ivars,
zs> 
zs> In A065577, 
zs> a(1)=2 because 10=3+7=5+5?
zs> Also a(2)=6 because
zs> 100=3+97=11+89=17+83=29+71=41+59=47+53?
zs> ....

One might call A065577 explicitly the partitions without regard to order. The
conversion between the two counts is simple, see
http://mathworld.wolfram.com/GoldbachPartition.html

The table of n, ordered, and non-ordered partitions is

                                    1, 3, 2
                                   2, 12, 6
                                   3, 56, 28
                                  4, 254, 127
                                 5, 1620, 810
                                6, 10804, 5402
                                7, 77614, 38807

with the ordered partitions in the OEIS as A073610, the non-ordered in A061358 (?)
(Cf from A065577 to these two might also help to show the difference,
A065577(n)=A061358(10^n).)
With the exception of the n=1 leading term, the count of the
ordered partitions is twice that of the non-ordered partitions,
because 10^n/2 is not a prime then.

The Maple program to print the small table above:

A065577 := proc(n,orderd)
	local N,a,i,ip;
	N := 10^n ;
	a := 0 ;
	i := 1 ;
	ip := 2 ;
	while 2*ip <= N do
		if isprime(N-ip) then
			if orderd and ip <> N-ip then
				a := a+ 2;
			else
				a := a+ 1;
			fi ;
		fi ;
		i := i+1 ;
		ip := ithprime(i) ;
	od ;
	RETURN(a) ;
end:

for n from 1 to 20 do
	print(n,A065577(n,true),A065577(n,false)) ;
od ;

In PARI

A065577(n)={
	local(N,a,i,ip);
	N = 10^n ;
	a = 0 ;
	i = 1 ;
	ip = 2 ;
	while(2*ip <= N,
		if(isprime(N-ip),
			a++ ;
		) ;
		i++ ;
		\\ip = prime(i) ; \\ slower and needs -p switch
		ip = nextprime(ip+1) ; \\ pseudoprimes only
	) ;
	return(a) ;
}

{
	for(n=1,20,
		print(n," ",A065577(n)) ;
	)
}

The PARI output confirms the n=9 case quoted by Zak (I've not gone to n=10):

1 2
2 6
3 28
4 127
5 810
6 5402
7 38807
8 291400
9 2274205

-- Richard





From zakseidov at yahoo.com  Wed Nov  1 15:29:37 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 1 Nov 2006 06:29:37 -0800 (PST)
Subject: A065577 Number of Goldbach partitions of 10^n?
In-Reply-To: <200611011317.kA1DHwmp018846@amer.strw.leidenuniv.nl>
Message-ID: <20061101142937.98860.qmail@web38214.mail.mud.yahoo.com>

Richard,
thanks for your reply.
It seems this'd concern mainly Ivars!
Also could you please calculate cases n>=10) -
I guess this'd not take for you hours(days!)
as with my Mathematica.
thanks, Zak


--- Richard Mathar <mathar at strw.leidenuniv.nl> wrote:

> 
> zs> From seqfan-owner at ext.jussieu.fr  Wed Nov  1
> 06:44:53 2006
> zs> Return-Path: <seqfan-owner at ext.jussieu.fr>
> zs> Date: Tue, 31 Oct 2006 21:43:12 -0800 (PST)
> zs> From: zak seidov <zakseidov at yahoo.com>
> zs> Subject: A065577 Number of Goldbach partitions
> of 10^n? More terms?
> zs> To: seqfan at ext.jussieu.fr, ip at sciserv.org,
> rgwv at rgwv.com
> zs> 
> zs> Dear seqfans,
> zs> Bob, Ivars,
> zs> 
> zs> In A065577, 
> zs> a(1)=2 because 10=3+7=5+5?
> zs> Also a(2)=6 because
> zs> 100=3+97=11+89=17+83=29+71=41+59=47+53?
> zs> ....
> 
> One might call A065577 explicitly the partitions
> without regard to order. The
> conversion between the two counts is simple, see
> http://mathworld.wolfram.com/GoldbachPartition.html
> 
> The table of n, ordered, and non-ordered partitions
> is
> 
>                                     1, 3, 2
>                                    2, 12, 6
>                                    3, 56, 28
>                                   4, 254, 127
>                                  5, 1620, 810
>                                 6, 10804, 5402
>                                 7, 77614, 38807
> 
> with the ordered partitions in the OEIS as A073610,
> the non-ordered in A061358 (?)
> (Cf from A065577 to these two might also help to
> show the difference,
> A065577(n)=A061358(10^n).)
> With the exception of the n=1 leading term, the
> count of the
> ordered partitions is twice that of the non-ordered
> partitions,
> because 10^n/2 is not a prime then.
> 
> The Maple program to print the small table above:
> 
> A065577 := proc(n,orderd)
> 	local N,a,i,ip;
> 	N := 10^n ;
> 	a := 0 ;
> 	i := 1 ;
> 	ip := 2 ;
> 	while 2*ip <= N do
> 		if isprime(N-ip) then
> 			if orderd and ip <> N-ip then
> 				a := a+ 2;
> 			else
> 				a := a+ 1;
> 			fi ;
> 		fi ;
> 		i := i+1 ;
> 		ip := ithprime(i) ;
> 	od ;
> 	RETURN(a) ;
> end:
> 
> for n from 1 to 20 do
> 	print(n,A065577(n,true),A065577(n,false)) ;
> od ;
> 
> In PARI
> 
> A065577(n)={
> 	local(N,a,i,ip);
> 	N = 10^n ;
> 	a = 0 ;
> 	i = 1 ;
> 	ip = 2 ;
> 	while(2*ip <= N,
> 		if(isprime(N-ip),
> 			a++ ;
> 		) ;
> 		i++ ;
> 		\\ip = prime(i) ; \\ slower and needs -p switch
> 		ip = nextprime(ip+1) ; \\ pseudoprimes only
> 	) ;
> 	return(a) ;
> }
> 
> {
> 	for(n=1,20,
> 		print(n," ",A065577(n)) ;
> 	)
> }
> 
> The PARI output confirms the n=9 case quoted by Zak
> (I've not gone to n=10):
> 
> 1 2
> 2 6
> 3 28
> 4 127
> 5 810
> 6 5402
> 7 38807
> 8 291400
> 9 2274205
> 
> -- Richard
> 



 
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From tanyakh at TanyaKhovanova.com  Wed Nov  1 16:46:50 2006
From: tanyakh at TanyaKhovanova.com (Tanya Khovanova)
Date: Wed,  1 Nov 2006 07:46:50 -0800
Subject: What is the most famous sequence?
Message-ID: <200611010746.AA12650352@TanyaKhovanova.com>

What is the most famous sequence? The most basic (natural numbers?)? The most simple (0,0,0, ... or 1,1,1, ...?)? The most versatile (Catalan numbers?)?

I would be curious to know your opinion.

I tried to check the number of Google hits for some sequence numbers. Here is what I've got:
A000040 (prime numbers) 68,800
A000045 (Fibonacci)     66,000
A000108 (Catalan)       62,500
A000027 (natural)       47,500

Can someone write a program to find top ten? What is the most referenced sequence inside Encyclopedia? Can we find the most hits outside Encyclopedia? Is it easy to do?

Tanya


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From jvospost3 at gmail.com  Wed Nov  1 18:58:55 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 1 Nov 2006 09:58:55 -0800
Subject: New & old LEGO seqs; production cuts
Message-ID: <5542af940611010958me17f347w3bf2d2f2d6c9dec0@mail.gmail.com>

I like the new seqs about Lego, such as
 A123762 <http://www.research.att.com/%7Enjas/sequences/A123762>
 Number of ways, counted up to symmetry, to build a contiguous building with
n *LEGO* blocks of size 1x2.

Ironically timed, as *"Recent restructuring and production cuts have left
Lego unable to fill
orders<http://www.cbc.ca/money/story/2006/10/31/legoproduction.html>for
the upcoming holiday season. Affected products include Duplo bricks,
Lego City sets, and (horror of horrors!) Star Wars and Lego Technik
sets."*According to the article Lego stands to lose $127 million in
holiday sales.

http://www.cbc.ca/money/story/2006/10/31/legoproduction.html

It might be useful to add crossrefs to earlier seqs such as:

 A007576 <http://www.research.att.com/%7Enjas/sequences/A007576>
 Number of maximally stable towers of 2 X 2 *LEGO* blocks.

 A082679 <http://www.research.att.com/%7Enjas/sequences/A082679>
 Number of *Lego* towers, one piece per floor, where every floor is
perpendicular to the one below it (so we have a kind of 3-dimensional zigzag
pattern).
 -- Jonathan Vos Post
[wondering if my son's earlier intense exposure to Duplo and Lego led him
towards his double B.S. in Math and Computer Science]
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From dean at math.ucdavis.edu  Wed Nov  1 21:12:15 2006
From: dean at math.ucdavis.edu (Dean Hickerson)
Date: Wed, 1 Nov 2006 12:12:15 -0800
Subject: Re^2: Error-finding program
Message-ID: <20061101201215.GA2388@math.ucdavis.edu>

Ralf Stephan wrote:

> My guess is that Clark started the same algorithm with 1, 2 and then
> tried it with starting values 1, 3, so the definition of A026469 should 
> be amended with ", starting with 1, 3".

Actually, A026469 should just say "starting with 1", since the 3 follows
from the rest of the definition.  A026468 should say "starting with 1,2",
since the 2 is artificially added to the sequence.

Dean Hickerson
dean at math.ucdavis.edu





From franktaw at netscape.net  Thu Nov  2 02:04:26 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 01 Nov 2006 20:04:26 -0500
Subject: What is the most famous sequence?
In-Reply-To: <200611010746.AA12650352@TanyaKhovanova.com>
References: <200611010746.AA12650352@TanyaKhovanova.com>
Message-ID: <8C8CC31F9FF2FAD-370-91EC@WEBMAIL-MA05.sysops.aol.com>

I would have to say that the most famous sequence is the primes.

To get the most referenced inside the OEIS, just do a search that 
matches
everything (or almost everything) - I tend to use "-keyword:dead".  Be 
sure
you're sorting by references (for many searches, such as this one, the 
default
sort by relevance works, too).  You'll see the sequences sorted by how 
many
references there are.

The top ten are (with references as of 11/1/06 6 PM CST):

1666 A000040 Primes
1305 A000045 Fibonacci
1177 A000108 Catalan
1009 A000010 Totient function
 883 A000217 Triangular
 870 A000005 Number of Divisors
 808 A000203 Sum of Divisors
 674 A002275 Repunits
 616 A001358 Semiprimes
 594 A000142 Factorials

This tends to under-estimate the importance of sequences like the 
squares
and the powers of 2.  If somebody enters a formula for a sequence that
uses the totient function, they are likely to include a reference to 
A000010.
Adding a formula which squares an integer will not cause people to add 
a
reference to the squares, however.

Incidently, here are some searches, intended to identify how popular 
certain
topics are in the OEIS, with the number of sequences matched:

25267 Prime
17853 keyword:base
10904 Digit -maple:digit
 7087 Square (but this covers a lot more than just the squares)
 6384 Binomial
 5548 Power
 4904 Divisor
 4051 Polynomial
 3901 Keyword:cons
 3722 Transform
 3698 Partition
 3638 Factor
 3526 Keyword:cofr
 3448 Permutation
 3326 Pi
 3325 Fibonacci
 3308 Matrix OR Matrices (search each separately, then subtract number
        where both occur)
 3138 Group
 2601 Phi (includes both golden ratio and totient function)
 2238 Sigma
 2189 Graph
 2116 Composite
 1971 Tree
 1553 Keyword:frac
 1540 Vector
 1549 Catalan
 1511 Reverse
 1197 Factorial (but I can't search for "!" operator)
 1165 Modulo OR Modulus
 1159 Palindrome
 1051 Perfect
 1034 Semiprime

Obviously, many of these searches do not perfectly identify the 
intended
subject.

Franklin T. Adams-Watters


-----Original Message-----
From: tanyakh at TanyaKhovanova.com

What is the most famous sequence? The most basic (natural numbers?)? 
The most
simple (0,0,0, ... or 1,1,1, ...?)? The most versatile (Catalan 
numbers?)?

I would be curious to know your opinion.

I tried to check the number of Google hits for some sequence numbers. 
Here is
what I've got:
A000040 (prime numbers) 68,800
A000045 (Fibonacci)     66,000
A000108 (Catalan)       62,500
A000027 (natural)       47,500

Can someone write a program to find top ten? What is the most 
referenced
sequence inside Encyclopedia? Can we find the most hits outside 
Encyclopedia? Is
it easy to do?

Tanya

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From arndt at jjj.de  Thu Nov  2 02:10:02 2006
From: arndt at jjj.de (Joerg Arndt)
Date: Thu, 2 Nov 2006 12:10:02 +1100
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
Message-ID: <20061102011002.GC6041@cochise.anu.edu.au>

* Jonathan Post <jvospost3 at gmail.com> [Nov 01. 2006 10:14]:
> Eisenstein-Fibonacci sequences.  Cube root of unity
> analogues of square-root of unity A014291  Imaginary
> Rabbits.
> 
> Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
> where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
> = (-1 - i*sqrt(3))/2.
> 
> [...]
> The coefficients of 1 = a(n) = 0, 0, 1, 1, 3, 1, 7, 4,
> 14, 17, 28, 54, 66, 143, 182, 350, ...
> 
> which does not seem to be in OEIS.

Ralf Stephan's ggf() says that the OGF may be:

? g=ggf([1, 1, 3, 1, 7, 4,14, 17, 28, 54, 66, 143, 182, 350])
(x^5 - 3*x^3 + x + 1)/(-x^6 + 3*x^4 - x^3 - 3*x^2 + 1)
? factor(g)

[x - 1 1]

[x^4 + x^3 - 2*x^2 - 2*x - 1 1]

[x^2 + x - 1 -1]

[x^4 - x^3 - x^2 + x + 1 -1]


> [...]







From eclark at math.usf.edu  Thu Nov  2 03:28:27 2006
From: eclark at math.usf.edu (Edwin Clark)
Date: Wed, 1 Nov 2006 21:28:27 -0500 (EST)
Subject: What is the most famous sequence?
In-Reply-To: <200611010746.AA12650352@TanyaKhovanova.com>
References: <200611010746.AA12650352@TanyaKhovanova.com>
Message-ID: <Pine.LNX.4.58.0611012122360.14271@shell.cas.usf.edu>


> What is the most famous sequence?

Since the novel and movie "The Da Vince Code"-- not to mention 
Knuth's "Art of Computer Programming" it would have to be the 
Fibonacci numbers.







From zakseidov at yahoo.com  Thu Nov  2 07:12:28 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 1 Nov 2006 22:12:28 -0800 (PST)
Subject: A124450: no autoreply from OEIS
Message-ID: <20061102061228.7046.qmail@web38206.mail.mud.yahoo.com>

Just submitted:
A124450 (no autoreply from OEIS).
Hope someone may wish to add more terms:
Mathematica isn't the best in such calcs.
Thanks, Zak

%I A124450
%S A124450
5,47,491,4919,49877,499943,4999913,49999757,499999931,4999999937,49999999811,499999999769,4999999998431,49999999999619,499999999999769,4999999999998557,49999999999998887,499999999999999679,4999999999999999661,49999999999999998647
%N A124450 Lesser of pair of closest primes summed to
10^n.
%C A124450 Cf. A065577 Number of Goldbach partitions
of 10^n.
%F A124450 10^n - a(n) is prime.
%e A124450 10^1=5+5; 10^2=47+53; 10^3=491+509;
10^4=4919+5081; 10^5=49877=50123; 10^6=499943+500057;
10^7=4999913+5000087; 10^8=49999757+50000243;
10^9=499999931+500000069;
10^10=4999999937+5000000063}, etc.
%Y A124450 A065577
%O A124450 1
%K A124450 ,nonn,
%A A124450 Zak Seidov  (zakseidov at yahoo.com), Nov 02 2006


 
____________________________________________________________________________________
Everyone is raving about the all-new Yahoo! Mail 
(http://advision.webevents.yahoo.com/mailbeta/)






From zakseidov at yahoo.com  Thu Nov  2 08:12:42 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 1 Nov 2006 23:12:42 -0800 (PST)
Subject: SEQ FROM Zak Seidov 
In-Reply-To: <200611020706.kA276utj016855@exspider.research.att.com>
Message-ID: <20061102071242.66437.qmail@web38208.mail.mud.yahoo.com>

Neil,
in contrast,
in this case I got immediate autoreply from OEIS.
Does submitting with ascribed or non-ascribed 
sequence number matter?
Zak 
--- The On-Line Encyclopedia of Integer Sequences
<oeis at research.att.com> wrote:

> The following is a copy of the email message that
> was sent to njas
> containing the sequence you submitted.
> 
> All greater than and less than signs have been
> replaced by their html
> equivalents.  They will be changed back when the
> message is processed.
> 
> This copy is just for your records.  No reply is
> expected.
>  Subject: NEW SEQUENCE FROM Zak Seidov 
> 
> 
> %I A000001
> %S A000001
>
5,,3,59,11,17,29,11,71,71,23,11,29,29,11,83,3,11,281,11,101,71,23,257,401,293,107,293,53,11,113,251,47,587,23,179,389,59,173,17,1427,83,431,53,563,593,41,47,239,383,431,1181,701,971,149,593,569,149,191,1973
> %N A000001 Lesser of pair of farsest primes summed
> to 10^n.
> %C A000001 Cf. A065577 Number of Goldbach partitions
> of 10^n, A124450 Lesser of pair of closest primes
> summed to 10^n.
> %F A000001 10^n - a(n) is prime.
> %e A000001 10^1 = 5 + 5, 10^2 = 3 + 97, 10^3 = 3 +
> 997, 10^4 = 59 + 9941, 10^5 = 11 + 99989, 10^6 = 17
> + 999983, 10^7 = 29 + 9999971, 10^8 = 11 + 99999989,
> 10^9 = 
> 71 + 999999929, 10^10 = 23 + 9999999977, etc.
> %Y A000001 A065577, A124450.
> %O A000001 1
> %K A000001 ,nonn,
> %A A000001 Zak Seidov  (zakseidov at yahoo.com), Nov 02
> 2006
> RH 
> RA 192.20.225.32
> RU 
> RI 
> 



 
____________________________________________________________________________________
Access over 1 million songs - Yahoo! Music Unlimited 
(http://music.yahoo.com/unlimited)






From jvospost3 at gmail.com  Thu Nov  2 08:44:39 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 1 Nov 2006 23:44:39 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <20061102011002.GC6041@cochise.anu.edu.au>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <20061102011002.GC6041@cochise.anu.edu.au>
Message-ID: <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>

Thank you, Joerg.  That was interesting.  Does your generating function
hypothesis hold when I extend the sequence by another 5 values, to a(n) = 0,
0, 1, 1, 3, 1, 7, 4,
14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again, assuming
that I got the elementary algebra and arithmetic correct.

If so, I should submit this sequence as by the two us!

Eisenstein-Fibonacci sequences.  Cube root of unity analogues of square-root
of unity A014291  Imaginary Rabbits.

[update of 1 Nov 06]

Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
= (-1 - i*sqrt(3))/2.

There are various integer sequences derived from this
complex sequence (such as the real part, the imaginary
part, the polynomial coefficient is of 1, w, and
w^2)Unless my arithmetic by hand is in error again,
is:

n   b(n)
0   w
1   w^2
2   w(w^2) + w = w^3 + w = 1 + w.
3   w(1 + w) + w^2 = w + w^2 + w^2 = 1 + 2w^2.
4   w(1 + 2w^2) + 1 + w = w + 2w^3 + 1 + w = 3 + 2w.
5   w(3 + 2w) + 1 + 2w^2 = 3w + 2w^2 + 1 + 2w^2.
   = 1 + 3w + 4w^2.
6  w(1 + 3w + 4w^2) + 3 + 2w = w + 3w^2 + 4w^3 + 3 +
2w
   = 7 + 3w + 3w^2.
7  w(7 + 3w + 3w^2) + 1 + 3w + 4w^2 = 7w + 3w^2 + 3w^3

  + 1 + 3w + 4w^2 = 4 + 10w + 7w^2.
8  w(4 + 10w + 7w^2) + 7 + 3w + 3w^2 = 4w + 10w^2 +
  7w^3 + 7 + 3w + 3w^2 = 14 + 7w + 13w^2.
9  w(14 + 7w + 13w^2) + 4 + 10w + 7w^2 = 14w + 7w^2 +
  13w^3 + 4 + 10w + 7w^2 = 17 + 24w + 14w^2.
10 w(17 + 24w + 14w^2) + 14 + 7w + 13w^2 = 17w + 24w^2
  + 14w^3 + 14 + 7w + 13w^2 = 28 + 24w + 37w^2.
11 w(28 + 24w + 37w^2) + 17 + 24w + 14w^2 = 28w +
  24w^2 + 37w^3 + 17 + 24w + 14w^2 = 54 + 52w +
38w^2.
12 w(54 + 52w + 38w^2) + 28 + 24w + 37w^2 = 54w +
  52w^2 + 38w^3 + 28 + 24w + 37w^2 = 66 + 78w +
89w^2.
13 w(66 + 78w + 89w^2) + 54 + 52w + 38w^2 =
  66w + 78w^2 + 89w^3 + 54 + 52w + 38w^2 =
  143 + 118w + 116w^2.
14 w(143 + 118w + 116w^2) + 66 + 78w + 89w^2 =
  143w + 118w^2 + 116w^3 + 66 + 78w + 89w^2 =
  182 + 221w + 207w^2.
15 w(182 + 221w + 207w^2) + 143 + 118w + 116w^2 =
  182w + 221w^2 + 207w^3 + 143 + 118w + 116w^2 =
  350 + 300w + 337w^2.

16 w(350 + 300w + 337w^2) + 350 + 300w + 337w^2 =
    350w + 300w^2 + 337w^3 + 350 + 300w + 337w^2 =
    687 +  650w + 637w^2.

17 w(687 +  650w + 637w^2) + 350 + 300w + 337w^2 =
    687w +  650w^2 + 637w^3 + 350 + 300w + 337w^2 =
    987 + 987w + 987w^2.

18  w(987 + 987w + 987w^2) + 687 +  650w + 637w^2 =
    987w + 987w^2 + 987w^3 + 687 +  650w + 637w^2 =
    1674 + 1637w + 1624w^2.

19  w(1674 + 1637w + 1624w^2) + 987 + 987w + 987w^2 =
     1674w + 1637w^2 + 1624w^3 + 987 + 987w + 987w^2
=
     2611 + 2661w + 2624w^2.

20  w(2611 + 2661w + 2624w^2) + 1674 + 1637w + 1624w^2
=
     2611w + 2661w^2 + 2624w^3 + 1674 + 1637w +
1624w^2 =
     4298 + 4248w + 4285w^2.


The coefficients of 1 = a(n) = 0, 0, 1, 1, 3, 1, 7, 4,
14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611,
4298

which does not seem to be in OEIS.

To excerpt my comment in A107890 (and some other
seqs):

"Eisenstein integers are of the form a + b*omega,
where a and b are ordinary integers, and omega = (-1 +
i*sqrt(3))/2 is a cube root of 1, the other cube roots
of 1 being 1 and omega^2 = (-1 - i*sqrt(3))/2.
Eisenstein integers are complex numbers that are also
members of the imaginary quadratic field Q(sqrt -3) =
Z[omega]. The sums, differences, and products of
Eisenstein integers are other Eisenstein integers."

Other sequences come from:
b(0) = w, b(1) = w^2, b(n) = b(n-1) + w*b(n-2),
and from the O, J notation of D?rrie (1965) in
Weisstein, Eric W. "Eisenstein Integer." From
MathWorld--A Wolfram Web Resource.
http://mathworld.wolfram.com/EisensteinInteger.html


Eisenstein-Fib.doc

=== end 2nd email ===

On 11/1/06, Joerg Arndt <arndt at jjj.de> wrote:
>
> * Jonathan Post <jvospost3 at gmail.com> [Nov 01. 2006 10:14]:
> > Eisenstein-Fibonacci sequences.  Cube root of unity
> > analogues of square-root of unity A014291  Imaginary
> > Rabbits.
> >
> > Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
> > where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
> > = (-1 - i*sqrt(3))/2.
> >
> > [...]
> > The coefficients of 1 = a(n) = 0, 0, 1, 1, 3, 1, 7, 4,
> > 14, 17, 28, 54, 66, 143, 182, 350, ...
> >
> > which does not seem to be in OEIS.
>
> Ralf Stephan's ggf() says that the OGF may be:
>
> ? g=ggf([1, 1, 3, 1, 7, 4,14, 17, 28, 54, 66, 143, 182, 350])
> (x^5 - 3*x^3 + x + 1)/(-x^6 + 3*x^4 - x^3 - 3*x^2 + 1)
> ? factor(g)
>
> [x - 1 1]
>
> [x^4 + x^3 - 2*x^2 - 2*x - 1 1]
>
> [x^2 + x - 1 -1]
>
> [x^4 - x^3 - x^2 + x + 1 -1]
>
>
> > [...]
>
>
>
-------------- next part --------------
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From maxale at gmail.com  Thu Nov  2 09:23:51 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 2 Nov 2006 00:23:51 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
Message-ID: <d3dac270611020023t17f583aep351f8e1e187148cc@mail.gmail.com>

On 10/22/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> Eisenstein-Fibonacci sequences.  Cube root of unity
>  analogues of square-root of unity A014291  Imaginary
>  Rabbits.
>
>  Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
>  where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
>  = (-1 - i*sqrt(3))/2.

First off, using the identity w^2 = -w-1, we can uniquely represent
b(n) as u(n)+v(n)*w where u(n) and v(n) are integers.
In particular, u(0)=0, v(0)=1, u(1)=-1, v(1)=-1.

Now the recurrence relation b(n) = w*b(n-1) + b(n-2) implies that
u(n) = u(n-2) - v(n-1)
v(n) = v(n-2) + u(n-1) - v(n-1)
that in turn implies that
[u(n-1),v(n-1),u(n),v(n)] = M*[u(n-2),v(n-2),u(n-1),v(n-1)]
where vectors are column-vectors and the matrix M is defined as
[ 0, 0, 1, 0 ]
[ 0, 0, 0, 1 ]
[ 1, 0, 0, -1 ]
[ 0, 1, 1, -1 ]

The characteristic polynomial of the matrix M is
x^4 + x^3 - x^2 - x + 1
meaning that u(n) and v(n) satisfies the recurrence
u(n) = - u(n-1) + u(n-2) + u(n-3) - u(n-4)
v(n) = - v(n-1) + v(n-2) + v(n-3) - v(n-4)
with u(0)=0, u(1)=-1, u(2)=1, u(3)=-2
and v(0)=1, v(1)=-1, v(2)=1, v(3)=-1.

>  There are various integer sequences derived from this
>  complex sequence (such as the real part, the imaginary
>  part, the polynomial coefficient is of 1, w, and
>  w^2)Unless my arithmetic by hand is in error again,
>  is:
>
>  n   b(n)
>  0   w
>  1   w^2
>  2   w(w^2) + w = w^3 + w = 1 + w.
>  3   w(1 + w) + w^2 = w + w^2 + w^2 = 1 + 2w^2.

It is not clear why you don't reduce the "polynomials" of w to the first degree.
Higher degree "polynomials" are not uniquely defined.
E.g., b(1)=w^2=2w^2+w+1=3w^2+2w+2=...
but if you reduce to the first degree "polynomial", there is an unique
representation:
b(1) = -w-1.

[...]

>  The coefficients of 1 = a(n) = 0, 0, 1, 1, 3, 1, 7, 4,
>  14, 17, 28, 54, 66, 143, 182, 350, ...

This sequence is not well-defined (see above).
Probably, what you are interested in is u(n):
0, -1, 1, -2, 2, -2, 1, 1, -4, 8, -12, 15, -15, 10, 2, -22, 49, -79, 104, -112

Max





From maxale at gmail.com  Thu Nov  2 13:06:51 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 2 Nov 2006 04:06:51 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <d3dac270611020023t17f583aep351f8e1e187148cc@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <d3dac270611020023t17f583aep351f8e1e187148cc@mail.gmail.com>
Message-ID: <d3dac270611020406l10f3deb8jb69c41cb2b267651@mail.gmail.com>

On 11/2/06, Max A. <maxale at gmail.com> wrote:

> >  Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
> >  where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
> >  = (-1 - i*sqrt(3))/2.
>
> First off, using the identity w^2 = -w-1, we can uniquely represent
> b(n) as u(n)+v(n)*w where u(n) and v(n) are integers.

[...]

> u(n) and v(n) satisfies the recurrence
> u(n) = - u(n-1) + u(n-2) + u(n-3) - u(n-4)
> v(n) = - v(n-1) + v(n-2) + v(n-3) - v(n-4)
> with u(0)=0, u(1)=-1, u(2)=1, u(3)=-2
> and v(0)=1, v(1)=-1, v(2)=1, v(3)=-1.

o.g.f. for u(n) is
-x/(1 + x - x^2 - x^3 + x^4)
and
o.g.f. for v(n) is
(1-x^2)/(1 + x - x^2 - x^3 + x^4)

Hence, o.g.f. for b(n) is
(-x + w*(1-x^2)) / (1 + x - x^2 - x^3 + x^4)

Max





From njas at research.att.com  Thu Nov  2 14:24:15 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Thu, 2 Nov 2006 08:24:15 -0500 (EST)
Subject: SEQ FROM Zak Seidov
Message-ID: <200611021324.IAA74292@fry.research.att.com>


Does submitting with ascribed or non-ascribed 
sequence number matter?
Zak 


Me:  I don't know.  But I see you are still using yahoo!
Why not switch to gmail?  Then you won't have these problems.

Best

Neil





From mathar at strw.leidenuniv.nl  Thu Nov  2 14:40:51 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Thu, 2 Nov 2006 14:40:51 +0100
Subject: Re^2:  Eisenstein-Fibonacci sequences
Message-ID: <200611021340.kA2DepPv014946@amer.strw.leidenuniv.nl>


ma> From seqfan-owner at ext.jussieu.fr  Thu Nov  2 13:12:31 2006
ma> Return-Path: <seqfan-owner at ext.jussieu.fr>
ma> Date: Thu, 2 Nov 2006 04:06:51 -0800
ma> From: "Max A." <maxale at gmail.com>
ma> To: "Jonathan Post" <jvospost3 at gmail.com>
ma> Subject: Re: Eisenstein-Fibonacci sequences
ma> Cc: "Sequence Fans Mailing List" <seqfan at ext.jussieu.fr>
ma> 
ma> On 11/2/06, Max A. <maxale at gmail.com> wrote:
ma> 
ma> > >  Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
ma> > >  where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
ma> > >  = (-1 - i*sqrt(3))/2.
ma> >
ma> > First off, using the identity w^2 = -w-1, we can uniquely represent
ma> > b(n) as u(n)+v(n)*w where u(n) and v(n) are integers.
ma> 
ma> [...]
ma> 
ma> > u(n) and v(n) satisfies the recurrence
ma> > u(n) = - u(n-1) + u(n-2) + u(n-3) - u(n-4)
ma> > v(n) = - v(n-1) + v(n-2) + v(n-3) - v(n-4)
ma> > with u(0)=0, u(1)=-1, u(2)=1, u(3)=-2
ma> > and v(0)=1, v(1)=-1, v(2)=1, v(3)=-1.
ma> 
ma> o.g.f. for u(n) is
ma> -x/(1 + x - x^2 - x^3 + x^4)
ma> and
ma> o.g.f. for v(n) is
ma> (1-x^2)/(1 + x - x^2 - x^3 + x^4)
ma> 
ma> Hence, o.g.f. for b(n) is
ma> (-x + w*(1-x^2)) / (1 + x - x^2 - x^3 + x^4)
ma> 
ma> Max

The g.f. above [-x+w(1-x^2)]/[1+x-x^2-x^3+x^4] generates
the reduced  polynomial
                          2             3      4            5               6
w + (-1 - w) x + (1 + w) x  + (-2 - w) x  + 2 x  + (w - 2) x  + (-3 w + 1) x

                  7               8              9      10
     + (1 + 5 w) x  + (-4 - 7 w) x  + (8 + 8 w) x  + O(x  )

If one doesn't reduce the polynomial, one gets the simpler
g(x)=w/[1-w*x-x^2] which generates the non-reduced and more generic
b(0)+b(1) x+ b(2) x^2 + b(3) x^3+ ...
=

     2       3       2     4      2   3     5      3       4
w + w  x + (w  + w) x  + (w  + 2 w ) x  + (w  + 3 w  + w) x  +

      6      4      2   5     7      5      3       6
    (w  + 4 w  + 3 w ) x  + (w  + 5 w  + 6 w  + w) x  +

      8      6       4      2   7     9      7       5       3       8
    (w  + 6 w  + 10 w  + 4 w ) x  + (w  + 7 w  + 15 w  + 10 w  + w) x  +

      10      8       6       4      2   9      10
    (w   + 8 w  + 21 w  + 20 w  + 5 w ) x  + O(x  )

and which becomes (apart from an offset) A000045 for w=1, A052542 for w=2,
A052906 for w=3, A000007 for w=0, and signed variations of these for w=-1,-2 or -3..
The coefficients table of the non-reduced g.f. is in A102426, which is
essentially the same as A098925 and A092865.

R. Mathar





From tanyakh at TanyaKhovanova.com  Thu Nov  2 16:15:08 2006
From: tanyakh at TanyaKhovanova.com (Tanya Khovanova)
Date: Thu,  2 Nov 2006 07:15:08 -0800
Subject: What is the most famous sequence?
Message-ID: <200611020715.AA2085028026@TanyaKhovanova.com>

Would this mean that Fibonacci sequence is the most popularized?

---------- Original Message ----------------------------------
From: Edwin Clark <eclark at math.usf.edu>
Date:  Wed, 1 Nov 2006 21:28:27 -0500 (EST)

>
>> What is the most famous sequence?
>
>Since the novel and movie "The Da Vince Code"-- not to mention 
>Knuth's "Art of Computer Programming" it would have to be the 
>Fibonacci numbers.
>
>
>


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From davidwwilson at comcast.net  Thu Nov  2 16:46:38 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 2 Nov 2006 10:46:38 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
References: <20061101054313.25333.qmail@web38210.mail.mud.yahoo.com>
Message-ID: <008e01c6fe96$19f1bbe0$6401a8c0@yourxhtr8hvc4p>

I wrote a C++ program using sieve method and calculated

a(7) = 38807
a(8) = 291400
a(9) = 2274205
a(10) = 18200488

a(7), a(8) and a(9) agree with previous calculations.

However, Zak Seidov gave

However, my a(10) = 18200487.

So now I have to go root out the discrepancy. I am going to have my computer 
generate a file of all the partitions of 10^10 and do an independent check 
of the results, and hopefully uncover a problem in my program. 






From rsc at swtch.com  Thu Nov  2 17:26:57 2006
From: rsc at swtch.com (Russ Cox)
Date: Thu, 2 Nov 2006 11:26:57 -0500
Subject: SEQ FROM Zak Seidov
In-Reply-To: <20061102071242.66437.qmail@web38208.mail.mud.yahoo.com>
References: <200611020706.kA276utj016855@exspider.research.att.com>
	 <20061102071242.66437.qmail@web38208.mail.mud.yahoo.com>
Message-ID: <ee9e417a0611020826h1daaac8xa85b9f6cc4b0ebdf@mail.gmail.com>

As I said in my earlier mail, the semantics of the
Yahoo "spam protection" are such that if you
submit two sequences within some time period
(maybe 12 hours to a day), the first might get
delayed many hours while the second will go through
immediately.

Out of curiosity, do you have oeis at research.att.com
in your address book?

Russ





From davidwwilson at comcast.net  Thu Nov  2 17:29:52 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 2 Nov 2006 11:29:52 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
Message-ID: <001101c6fe9c$205d0c40$6401a8c0@yourxhtr8hvc4p>

Ok, that last message was incoherent.

For A065577 Zak Seidov got a(10) = 18200488, I got a(10) = 18200487.  I am 
checking my program. 






From pxp at rogers.com  Thu Nov  2 18:36:24 2006
From: pxp at rogers.com (Hans Havermann)
Date: Thu, 2 Nov 2006 12:36:24 -0500
Subject: A124450
In-Reply-To: <20061102061228.7046.qmail@web38206.mail.mud.yahoo.com>
References: <20061102061228.7046.qmail@web38206.mail.mud.yahoo.com>
Message-ID: <D7CB7013-474C-4B78-B336-E5DBA449F167@rogers.com>

zak seidov:

> Hope someone may wish to add more terms:
> Mathematica isn't the best in such calcs.

> %S A124450
> 5,47,491,4919,49877,499943,4999913,49999757,499999931,4999999937,49999 
> 999811,499999999769,4999999998431,49999999999619,499999999999769,49999 
> 99999998557,49999999999998887,499999999999999679,4999999999999999661,4 
> 9999999999999998647
> %N A124450 Lesser of pair of closest primes summed to 10^n.

I think you overstate the Mathematica difficulty. With no effort at  
any sort of efficiency in the programming, it still only takes me 8  
seconds to generate 50 terms.

Table[h =10^n/2; c=0; While[PrimeQ[h-c]==False || PrimeQ[h+c]==False,  
c++]; h-c, {n, 1, 50}]

{5, 47, 491, 4919, 49877, 499943, 4999913, 49999757, 499999931,  
4999999937, 49999999811, 499999999769, 4999999998431, 49999999999619,  
499999999999769, 4999999999998557, 49999999999998887,  
499999999999999679, 4999999999999999661, 49999999999999998647,  
499999999999999999637, 4999999999999999999481,  
49999999999999999998677, 499999999999999999998497,  
4999999999999999999999259, 49999999999999999999998779,  
499999999999999999999999043, 4999999999999999999999998947,  
49999999999999999999999999661, 499999999999999999999999994069,  
4999999999999999999999999997879, 49999999999999999999999999997699,  
499999999999999999999999999997969,  
4999999999999999999999999999995227,  
49999999999999999999999999999995263,  
499999999999999999999999999999989719,  
4999999999999999999999999999999998683,  
49999999999999999999999999999999999871,  
499999999999999999999999999999999996127,  
4999999999999999999999999999999999998557,  
49999999999999999999999999999999999999613,  
499999999999999999999999999999999999988231,  
4999999999999999999999999999999999999991729,  
49999999999999999999999999999999999999994663,  
499999999999999999999999999999999999999997117,  
4999999999999999999999999999999999999999992863,  
49999999999999999999999999999999999999999991807,  
499999999999999999999999999999999999999999991507,  
4999999999999999999999999999999999999999999993199,  
49999999999999999999999999999999999999999999997081}

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From davidwwilson at comcast.net  Thu Nov  2 18:45:17 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 2 Nov 2006 12:45:17 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
References: <20061102171720.72610.qmail@web38203.mail.mud.yahoo.com>
Message-ID: <002b01c6fea6$a96fcae0$6401a8c0@yourxhtr8hvc4p>

Well let's get it right this time

I got a(10) = 18200488, in agreement with Dick Mathar's value.





From jvospost3 at gmail.com  Thu Nov  2 19:01:55 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 2 Nov 2006 10:01:55 -0800
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
In-Reply-To: <002b01c6fea6$a96fcae0$6401a8c0@yourxhtr8hvc4p>
References: <20061102171720.72610.qmail@web38203.mail.mud.yahoo.com>
	 <002b01c6fea6$a96fcae0$6401a8c0@yourxhtr8hvc4p>
Message-ID: <5542af940611021001x27e8951akc9b931928d84333e@mail.gmail.com>

Amusing coincidence that the digits of a(10) = 18200488 consist entirely of
0 and powers of 2.  [base].

-- Jonathan Vos Post

On 11/2/06, David Wilson <davidwwilson at comcast.net> wrote:
>
> Well let's get it right this time
>
> I got a(10) = 18200488, in agreement with Dick Mathar's value.
>
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From jvospost3 at gmail.com  Thu Nov  2 21:54:44 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 2 Nov 2006 12:54:44 -0800
Subject: New & old LEGO seqs; production cuts
In-Reply-To: <C16FD8A8.20890%eilers@math.ku.dk>
References: <5542af940611010958me17f347w3bf2d2f2d6c9dec0@mail.gmail.com>
	 <C16FD8A8.20890%eilers@math.ku.dk>
Message-ID: <5542af940611021254w14b12a38y14021854e772de77@mail.gmail.com>

Dear Soren,

Thank you for a clear and thoughful answer.

My son (in the room with me now) agrees that baby/childhood play with Duplo
and Lego guided him towards his Math/Computer Science expertise today.

He and I agree, however, that it's a shame that the manufacturer doesn't
make 4-D Lego blocks, or, more properly, the ones that they do make are of
extension along the t-axis only in the trivial way, and cannot be freely
rotated in Minkowski space.

So, I wonder, how well does your approach to counting under symmetry work if
extended to an additional spacial dimension?

I suppose that we can consider an AxBxC 4-D Lego to be actually an AxBxCx1
Lego in Z^4. I think that adjacency is well-defined, all rotations are about
planes, and there is a relationship to 4-D analogues of polyominoes, i.e. to
polyhypercubes.

Best,

Jonathan Vos Post

On 11/2/06, S?ren Eilers <eilers at math.ku.dk> wrote:
>
>  Dear Jonathan
>
> Thank you, and yes, I think that LEGO inspires kids to learn about
> fundamental mathematical concepts such as symmetry.
>
> I've also considered adding crossrefs to the sequences you mentioned but
> decided against it.
>
> Regarding A007576: Watson definitely counts a subset of what we count in
> A123818, but as you can see in his paper he only looks at "planar" buildings
> where all blocks could be placed in, say, [0,2]xRxR. This means that one of
> the dimensions of the blocks used is irrelevant, and Watson could just have
> well have counted what we call FLAT structures with 1x2 blocks. Thus, IF one
> was to crossref I would suggest to do it to A123764, but perhaps this is a
> little hard to explain in the limited space of an OEIS entry.
>
> But Watson has even more restrictions; he looks only at MAX HEIGHT
> buildings with one block at each level. This is exactly what LEGO did wrong
> when they claimed that A123828(6)=102981500 [the number they wanted was
> (46^5+2^5)/2, but they got the last digit wrong too].
>
> Of course counting FLAT and MAX HEIGHT structures with 2x2 or 1x2 blocks
> is easy, the formula is 3^(n-1) or (3^(n-1)+1)/2 depending on whether or not
> you want to identify up to symmetry. Watson knows this and tries to count
> how many of these structures would tip over if placed on a flat surface,
> essentially computing the x-coordinate of the center of gravity. So I think
> the kind of LEGO counting done by Watson is quite different from the kind of
> LEGO counting we do.
>
>     {Btw, A007576 is indeed the same as A086821}
>     {If somebody was interested I could probably quite easily compute how
> many of "our" LEGO buildings are stable in the sense of Watson}
>
> Regarding Zeilberger's LEGO counts, they all relate, as far as I recall
> from a relatively hard look some time ago, to buildings with blocks of
> varying size, and hence are quite different in nature from what we do. In
> that case, n is not a number of blocks, but the total volume/weight of the
> structure, so that's a very different ball game.
>
> Best regards,
> Soren
>
>
>
>
>
> On 01/11/06 18:58, "Jonathan Post" <jvospost3 at gmail.com> wrote:
>
> I like the new seqs about Lego, such as
> A123762 <http://www.research.att.com/%7Enjas/sequences/A123762><http://www.research.att.com/%7Enjas/sequences/A123762>
>   Number of ways, counted up to symmetry, to build a contiguous building
> with n *LEGO* blocks of size 1x2.
>
> Ironically timed, as *"Recent restructuring and production cuts have left
> Lego unable to fill orders
> <http://www.cbc.ca/money/story/2006/10/31/legoproduction.html><http://www.cbc.ca/money/story/2006/10/31/legoproduction.html>for the upcoming holiday season. Affected products include Duplo bricks,
> Lego City sets, and (horror of horrors!) Star Wars and Lego Technik sets."
> * According to the article Lego stands to lose $127 million in holiday
> sales.
>
> http://www.cbc.ca/money/story/2006/10/31/legoproduction.html
>
> It might be useful to add crossrefs to earlier seqs such as:
>
> A007576 <http://www.research.att.com/%7Enjas/sequences/A007576><http://www.research.att.com/%7Enjas/sequences/A007576>
>   Number of maximally stable towers of 2 X 2 *LEGO* blocks.
>
>
> A082679 <http://www.research.att.com/%7Enjas/sequences/A082679><http://www.research.att.com/%7Enjas/sequences/A082679>
>   Number of *Lego* towers, one piece per floor, where every floor is
> perpendicular to the one below it (so we have a kind of 3-dimensional zigzag
> pattern).
>  -- Jonathan Vos Post
> [wondering if my son's earlier intense exposure to Duplo and Lego led him
> towards his double B.S. in Math and Computer Science]
>
>
>
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From davidwwilson at comcast.net  Thu Nov  2 22:49:42 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 2 Nov 2006 16:49:42 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
References: <20061102171720.72610.qmail@web38203.mail.mud.yahoo.com> <002b01c6fea6$a96fcae0$6401a8c0@yourxhtr8hvc4p>
Message-ID: <004e01c6fec8$ce34d100$6401a8c0@yourxhtr8hvc4p>

Regarding A065577:

I altered my counting program to generating a file listing, in increasing 
order, all primes p such that there exists prime q with p < q, p+q = 10^10. 
I wrote a verification program to read this file, check that the p are in 
increasing order, count the p, compute q = 10^10-p, check p < q and that p 
and q are both prime (using trial division).

The verification program has finished, and confirms that a(10) >= 18200488. 
This contradicts Zak Seidov's count a(10) = 18200487, but agrees with Dick 
Mathar's count a(10) = 18200488.

If anyone is interested, I am keeping the text file with the p values for a 
little while. It is 195MB, so e-mailing it is probably not a good option.






From zakseidov at yahoo.com  Fri Nov  3 04:09:56 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Thu, 2 Nov 2006 19:09:56 -0800 (PST)
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
In-Reply-To: <004e01c6fec8$ce34d100$6401a8c0@yourxhtr8hvc4p>
Message-ID: <20061103030956.55909.qmail@web38210.mail.mud.yahoo.com>

David, Dick, Bob, Ivars,
seqfans.

After recalculating (at two different PC's)
and rechecking I could finally 
find source of my error
(silly as usual) in one of codes,
and my final value is 
a(10) = 18200488 (NOT 18200487!)
which coincides now with the Dick-David value.

So, Bob, 
you may wish to give in A065577  the correct value 
a(10) = 18200488 with due refer to David Wilson  and
Richard Mathar (not me).
Also Ivars can extend his table.
Thanks, Zak

BTW1 Still, is it possible to find a(11)... with C++?

BIW2 Not all people even in this closed list
may be interested in Mmca specific problems,
so I''l be happy to have a list of you
particularly Mmca-involved.
Then we may circulate relevant messages
not to all seqfan community.
Please anyone interested write me personally
with subject "Me in Mmca", thanks.
Of course it's not intented to be closed list!
Hope this is not against Gerard' policy.

--- David Wilson <davidwwilson at comcast.net> wrote:

> Regarding A065577:
> 
> I altered my counting program to generating a file
> listing, in increasing 
> order, all primes p such that there exists prime q
> with p < q, p+q = 10^10. 
> I wrote a verification program to read this file,
> check that the p are in 
> increasing order, count the p, compute q = 10^10-p,
> check p < q and that p 
> and q are both prime (using trial division).
> 
> The verification program has finished, and confirms
> that a(10) >= 18200488. 
> This contradicts Zak Seidov's count a(10) =
> 18200487, but agrees with Dick 
> Mathar's count a(10) = 18200488.
> 
> If anyone is interested, I am keeping the text file
> with the p values for a 
> little while. It is 195MB, so e-mailing it is
> probably not a good option.
> 
> 



 
____________________________________________________________________________________
Everyone is raving about the all-new Yahoo! Mail 
(http://advision.webevents.yahoo.com/mailbeta/)






From zakseidov at yahoo.com  Fri Nov  3 05:03:30 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Thu, 2 Nov 2006 20:03:30 -0800 (PST)
Subject: A123998: No primes except 3?
Message-ID: <20061103040330.2777.qmail@web38202.mail.mud.yahoo.com>

Why are there no prime numbers except of 3,
in this sequence?
Zak

A123998
={1,3,9,15,18,39,48,69,78,99,105,114,135,153,165,168,183,189,219,249,273,

288,300,303,309,330,345,363,405,414,438,468,483,498,504,534,585,618,
 639,648,699,714,729,765,804,813,828,879,933}; 
(* %N A123998 Numbers n for which 2n+1 and 4n+1 are
primes. *)



 
__________________________________________________________________________________________
Check out the New Yahoo! Mail - Fire up a more powerful email and get things done faster. 
(http://advision.webevents.yahoo.com/mailbeta) 






From maxale at gmail.com  Fri Nov  3 06:41:20 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 2 Nov 2006 21:41:20 -0800
Subject: A123998: No primes except 3?
In-Reply-To: <20061103040330.2777.qmail@web38202.mail.mud.yahoo.com>
References: <20061103040330.2777.qmail@web38202.mail.mud.yahoo.com>
Message-ID: <d3dac270611022141x4d10f69evdd706982df6e3d28@mail.gmail.com>

Note that if n = 1 mod 3 then 2n+1 is not prime (except n=1);
and if n = 2 mod 3 then 4n+1 is not prime.
Therefore, n must be a multiple of 3, except n=1.

Max

If n>1 and both 2n+1, 4n+1 are primes

On 11/2/06, zak seidov <zakseidov at yahoo.com> wrote:
> Why are there no prime numbers except of 3,
> in this sequence?
> Zak
>
> A123998
> ={1,3,9,15,18,39,48,69,78,99,105,114,135,153,165,168,183,189,219,249,273,
>
> 288,300,303,309,330,345,363,405,414,438,468,483,498,504,534,585,618,
>  639,648,699,714,729,765,804,813,828,879,933};
> (* %N A123998 Numbers n for which 2n+1 and 4n+1 are
> primes. *)
>
>
>
>
> __________________________________________________________________________________________
> Check out the New Yahoo! Mail - Fire up a more powerful email and get things done faster.
> (http://advision.webevents.yahoo.com/mailbeta)
>
>





From zakseidov at yahoo.com  Fri Nov  3 07:01:51 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Thu, 2 Nov 2006 22:01:51 -0800 (PST)
Subject: A123998: No primes except 3?
In-Reply-To: <d3dac270611022141x4d10f69evdd706982df6e3d28@mail.gmail.com>
Message-ID: <20061103060151.3808.qmail@web38207.mail.mud.yahoo.com>

Great!
It'so easy if you are Max, not Zak!
Thanks, 
Zak

--- "Max A." <maxale at gmail.com> wrote:

> Note that if n = 1 mod 3 then 2n+1 is not prime
> (except n=1);
> and if n = 2 mod 3 then 4n+1 is not prime.
> Therefore, n must be a multiple of 3, except n=1.
> 
> Max
> 
> If n>1 and both 2n+1, 4n+1 are primes
> 
> On 11/2/06, zak seidov <zakseidov at yahoo.com> wrote:
> > Why are there no prime numbers except of 3,
> > in this sequence?
> > Zak
> >
> > A123998
> >
>
={1,3,9,15,18,39,48,69,78,99,105,114,135,153,165,168,183,189,219,249,273,
> >
> >
>
288,300,303,309,330,345,363,405,414,438,468,483,498,504,534,585,618,
> >  639,648,699,714,729,765,804,813,828,879,933};
> > (* %N A123998 Numbers n for which 2n+1 and 4n+1
> are
> > primes. *)
> >
> >
> >
> >
> >
>
__________________________________________________________________________________________
> > Check out the New Yahoo! Mail - Fire up a more
> powerful email and get things done faster.
> > (http://advision.webevents.yahoo.com/mailbeta)
> >
> >
> 



 
____________________________________________________________________________________
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From jeremy.gardiner at btinternet.com  Fri Nov  3 09:59:50 2006
From: jeremy.gardiner at btinternet.com (JEREMY GARDINER)
Date: Fri, 3 Nov 2006 08:59:50 +0000 (GMT)
Subject: What is the most famous sequence?
In-Reply-To: <200611020715.AA2085028026@TanyaKhovanova.com>
Message-ID: <20061103085950.5824.qmail@web86610.mail.ird.yahoo.com>

The natural numbers must be the most widely known sequence ...

Tanya Khovanova <tanyakh at TanyaKhovanova.com> wrote:  > What is the most famous sequence?
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From maxale at gmail.com  Fri Nov  3 10:08:28 2006
From: maxale at gmail.com (Max A.)
Date: Fri, 3 Nov 2006 01:08:28 -0800
Subject: querying multidimensional sequences in OEIS
Message-ID: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>

I have a sequence indexed by vectors with non-negative integer components.
Is there a regular way to query such sequence in OEIS ?

As an example, let us consider a sequence of multinomial coefficients.
Any vector with non-negative integer components [k1, k2, ..., kn]
defines an element of this sequence:

a( [k1, k2, ..., kn] ) = (k1 + k2 +  ... + kn)! / (k1! * k2! * ... * kn!)

Assume that I know nothing about these numbers and have only a
black-box computing a(v) for any given vector v with non-negative
integer components. Can I anyhow find this sequence in OEIS and learn
that its elements called "multinomial coefficients" ?

Thanks,
Max





From wouter.meeussen at pandora.be  Fri Nov  3 12:39:40 2006
From: wouter.meeussen at pandora.be (wouter meeussen)
Date: Fri, 3 Nov 2006 12:39:40 +0100
Subject: download seqs resulting from search
Message-ID: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>

dear all,

if I search for all number triangles with row sums equal to the factorials,
I enter
"row sum" "factorial"
and get 9 pages of results in 'short' form.
(1) how can I improve on this query?
(2) how can I download them all in one go?

Wouter





From njas at research.att.com  Fri Nov  3 14:43:47 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Fri, 3 Nov 2006 08:43:47 -0500 (EST)
Subject: download seqs resulting from search
Message-ID: <200611031343.IAA51448@fry.research.att.com>

as i have said before, you can download up to 30 (or is it 50?) sequences
at once by using the old email lookup service

send email to  sequences at research.att.com  saying things like

lookup A000123
lookup A000125
lookup A012345

etc.

Neil





From franktaw at netscape.net  Fri Nov  3 16:18:26 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 10:18:26 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
Message-ID: <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>

I don't know of any standard for functions of an arbitrary finite 
sequence
of positive integers in the OEIS.

In your example, the function is commutative, which makes it 
essentially
a function of the multiset of values - or in other words, of a 
partition.  For
that there is a standard - two of them actually.  One is the 
enumeration
in Abramowitz and Stegun order, as specified in A036036, and the other
is the Mathematica ordering, as specified in A080577.

There are some relevant sequences in the database in both orders, some
only in the A&S order, and a few only in the Mma order.

Since the two orders are the same for partitions of up to 5 objects, 
one
often just search for an early part of the sequence.  If this does not 
suffice,
one should try both before giving up.

For example, the partitions of 5 objects are: [5], [4,1], [3,2], 
[3,1^2],
[2^2,1], [2,1^3], [1^5].  The multinomial coefficients for these are:

1,5,10,20,30,60,120

A search for this turns up A036038, the A&S version, and A078760, the
Mma version.

----
As for arbitrary finite sequences of positive integers, there are two 
main
candidates for an ordering: A066099 and A108244; these are the reverse
of each other.  I would probably go with A066099, as it is closest to 
the
ordering used for the partitions; but you can make an argument for 
either
one.  Possibly both should be used.

Off hand, I can't think of any interesting functions of finite 
sequences of
positive integers that depend on the order of the sequence.

Franklin T. Adams-Watters


-----Original Message-----
From: maxale at gmail.com

I have a sequence indexed by vectors with non-negative integer 
components.?
Is there a regular way to query such sequence in OEIS ??
?
As an example, let us consider a sequence of multinomial coefficients.?
Any vector with non-negative integer components [k1, k2, ..., kn]?
defines an element of this sequence:?
?
a( [k1, k2, ..., kn] ) = (k1 + k2 + ... + kn)! / (k1! * k2! * ... * 
kn!)?
?
Assume that I know nothing about these numbers and have only a?
black-box computing a(v) for any given vector v with non-negative?
integer components. Can I anyhow find this sequence in OEIS and learn?
that its elements called "multinomial coefficients" ??
?
Thanks,?
Max?


________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From rsc at swtch.com  Fri Nov  3 17:06:23 2006
From: rsc at swtch.com (Russ Cox)
Date: Fri, 3 Nov 2006 11:06:23 -0500
Subject: download seqs resulting from search
In-Reply-To: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>
References: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>
Message-ID: <ee9e417a0611030806q1490600ci75384340c6facd67@mail.gmail.com>

> if I search for all number triangles with row sums equal to the factorials,
> I enter
> "row sum" "factorial"
> and get 9 pages of results in 'short' form.
> (1) how can I improve on this query?

Not sure.

> (2) how can I download them all in one go?

Add &n=100 to the URL
and you will get 100 results per page.
Then click on "text" formatting and save the
resulting page.

http://www.research.att.com/~njas/sequences/?q=%22row+sums%22+factorial&n=100&fmt=3

Russ





From franktaw at netscape.net  Fri Nov  3 17:14:55 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 11:14:55 -0500
Subject: download seqs resulting from search
In-Reply-To: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>
References: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>
Message-ID: <8C8CD7A55C060ED-FD0-B6C7@WEBMAIL-MA20.sysops.aol.com>

Try adding "keyword:tabl" to the search.  Then (probably) repeat with 
"keyword:tabf" instead - unless you aren't interested in irregular 
tables.  This won't be all in one go, but two isn't too bad.

Of course, there are undoubtably some tables which sum to factorials 
that don't have the words "row sum" in them.  And maybe some that just 
reference A000142 instead of using the word "factorial".

Franklin T. Adams-Watters


-----Original Message-----
From: wouter.meeussen at pandora.be

dear all,

if I search for all number triangles with row sums equal to the 
factorials,
I enter
"row sum" "factorial"
and get 9 pages of results in 'short' form.
(1) how can I improve on this query?
(2) how can I download them all in one go?

Wouter

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From mlb at well.com  Fri Nov  3 17:26:14 2006
From: mlb at well.com (Marc LeBrun)
Date: Fri, 03 Nov 2006 08:26:14 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <200611031625.kA3GP1mt017249@smtp.well.com>

 >=franktaw at netscape.net

Good advice...maybe it could be made part of the tips-n-tricks section?

 > just search for an early part of the sequence

Or sometimes you can search for a small set of "unusual" later, 
larger, terms that will act as a kind of signature.

 > Off hand, I can't think of any interesting functions of finite sequences of
 > positive integers that depend on the order of the sequence.

Exponents in prime factorizations, say.







From franktaw at netscape.net  Fri Nov  3 19:31:34 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 13:31:34 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <200611031625.kA3GP1mt017249@smtp.well.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com> <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com> <200611031625.kA3GP1mt017249@smtp.well.com>
Message-ID: <8C8CD8D6CD483AD-FD0-C3EF@WEBMAIL-MA20.sysops.aol.com>

Nice suggestion.  That would be A057335.

Franklin T. Adams-Watters


-----Original Message-----
From: mlb at well.com

>=franktaw at netscape.net?
> Off hand, I can't think of any interesting functions of finite 
sequences of?
> positive integers that depend on the order of the sequence.?
?
Exponents in prime factorizations, say.?
?


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From maxale at gmail.com  Fri Nov  3 21:26:20 2006
From: maxale at gmail.com (Max A.)
Date: Fri, 3 Nov 2006 12:26:20 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <200611031625.kA3GP1mt017249@smtp.well.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <200611031625.kA3GP1mt017249@smtp.well.com>
Message-ID: <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com>

On 11/3/06, Marc LeBrun <mlb at well.com> wrote:

>  > Off hand, I can't think of any interesting functions of finite sequences of
>  > positive integers that depend on the order of the sequence.
>
> Exponents in prime factorizations, say.

Oh, that is a very nice suggestion.
Are there any multidimensional sequences represented this way in OEIS?

I have a number of multidimensional sequences in mind that I'd like to
see in OEIS.
E.g.: the coefficients of incomplete Bell polynomials:

b( [k1, k2, ..., kn] ) = (k1 + 2*k2 + ... + n*kn)! / (k1! * k2! * ...
* kn!) / (1!^k1 * 2!^k2 * ... * n!^kn)

and looking similar the number of permutations with given cycle structure:

c( [k1, k2, ..., kn] ) = (k1 + 2*k2 + ... + n*kn)! / (k1! * k2! * ...
* kn!) / (1^k1 * 2^k2 * ... * n^kn)

Max





From mlb at well.com  Fri Nov  3 22:41:00 2006
From: mlb at well.com (Marc LeBrun)
Date: Fri, 03 Nov 2006 13:41:00 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com
 >
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
 <200611031625.kA3GP1mt017249@smtp.well.com>
 <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com>
Message-ID: <200611032139.kA3LdZSV024106@smtp.well.com>

 > I have a number of multidimensional sequences in mind that I'd like to
 > see in OEIS.  E.g.: the coefficients of incomplete Bell polynomials:

Two possible ways to submit sequences of tuples T(1),T(2),...
where T(i)=[T(i,1),T(i,2),...] are:

A. Define a traversal order by a map (i,j) <--> n and set a(n) = T(i,j).
This gives the tabl/tabf style of entry.

B. Encode each tuple as a single number, say by using it as factor
exponents:  T(i) --> 2^T(i,1) * 3^T(i,2) * 5^T(i,3) *...

For example, for Pascal's triangle

A: 1; 1, 1; 1, 2, 1; 1, 3, 3, 1;...

B: 2, 6,    90,      47250,...

However there are a variety of possible ways to do these things--for example
you might subtract 1 from all the Pascal exponents, B': 1, 1, 3, 225,...

Or traverse in the order (1,1), (2,1), (1,2), (3,1), (1,3), (2,2), (1,4),...
which recursively interleaves sequences instead of the usual diagonalizing.

But this sort of thing can also be abused to breed swarms of silly sequences,
so it should be well-motivated.

It's nicest to use familiar schemes (eg as Frank noted for partitions).







From franktaw at netscape.net  Fri Nov  3 22:41:18 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 16:41:18 -0500
Subject: Possibly more errors in OEIS?
Message-ID: <8C8CDA7EE6FDEF2-12C-B6F@WEBMAIL-MA20.sysops.aol.com>

Looking at sequences with keyword:cons, there is a marked shortage of 
sequences with a given duplicate digit (excluding 0), compared with 
those with a given other pair of digits.  For example, restricting our 
attention to sequences not of the form "Decimal expansion of 1/n", and 
ignoring all pairs involving 0 or 1, there are an average of about 1710 
sequences  with any given digit pair, but an average of only about 1627 
with duplicate pairs.  10 of this difference is explained by the 10 
"selvage" sequences, which have no duplicate digits, but the rest is 
unexplained.

I'm guessing that there may be an input error here; that duplicate 
digits sometimes get combined at some point in the process, whether 
before or after submission.  I'm not sure how to test for this, short 
of examining the "cons" sequences and determining which are correct.

Mostly for Neil,

Sequences A081780 to A081796, "continued cotangent" sequences, all have 
the "cons" keyword, but they should not.

Also, sequence A114721 has the "cons" keyword; it should instead have 
the "frac" keyword.

Franklin T. Adams-Watters

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From jawbrey at att.net  Fri Nov  3 22:51:00 2006
From: jawbrey at att.net (Jon Awbrey)
Date: Fri, 03 Nov 2006 16:51:00 -0500
Subject: querying multidimensional sequences in OEIS
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <200611031625.kA3GP1mt017249@smtp.well.com>
	 <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com> <200611032139.kA3LdZSV024106@smtp.well.com>
Message-ID: <454BB9C4.CCCA8C55@att.net>

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o

you might get some ideas from looking at the
tables of "primal codes" or "primal functions"
that I compiled, for instance, here:

http://stderr.org/pipermail/inquiry/2005-July/002846.html

ja

Marc LeBrun wrote:
> 
>  > I have a number of multidimensional sequences in mind that I'd like to
>  > see in OEIS.  E.g.: the coefficients of incomplete Bell polynomials:
> 
> Two possible ways to submit sequences of tuples T(1),T(2),...
> where T(i)=[T(i,1),T(i,2),...] are:
> 
> A. Define a traversal order by a map (i,j) <--> n and set a(n) = T(i,j).
> This gives the tabl/tabf style of entry.
> 
> B. Encode each tuple as a single number, say by using it as factor
> exponents:  T(i) --> 2^T(i,1) * 3^T(i,2) * 5^T(i,3) *...
> 
> For example, for Pascal's triangle
> 
> A: 1; 1, 1; 1, 2, 1; 1, 3, 3, 1;...
> 
> B: 2, 6,    90,      47250,...
> 
> However there are a variety of possible ways to do these things--for example
> you might subtract 1 from all the Pascal exponents, B': 1, 1, 3, 225,...
> 
> Or traverse in the order (1,1), (2,1), (1,2), (3,1), (1,3), (2,2), (1,4),...
> which recursively interleaves sequences instead of the usual diagonalizing.
> 
> But this sort of thing can also be abused to breed swarms of silly sequences,
> so it should be well-motivated.
> 
> It's nicest to use familiar schemes (eg as Frank noted for partitions).

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o
inquiry e-lab: http://stderr.org/pipermail/inquiry/
citizendium development forums: http://smf.citizendium.org
http://www.textop.org/wiki/index.php?title=User:Jon_Awbrey
wikinfo: http://wikinfo.org/wiki.php?title=User:Jon_Awbrey
wp review: http://wikipediareview.com/index.php?showuser=398
o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o






From maxale at gmail.com  Fri Nov  3 23:02:04 2006
From: maxale at gmail.com (Max A.)
Date: Fri, 3 Nov 2006 14:02:04 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <200611032139.kA3LdZSV024106@smtp.well.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <200611031625.kA3GP1mt017249@smtp.well.com>
	 <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com>
	 <200611032139.kA3LdZSV024106@smtp.well.com>
Message-ID: <d3dac270611031402s62ff2746p3a5c749865678cec@mail.gmail.com>

On 11/3/06, Marc LeBrun <mlb at well.com> wrote:
>  > I have a number of multidimensional sequences in mind that I'd like to
>  > see in OEIS.  E.g.: the coefficients of incomplete Bell polynomials:
>
> Two possible ways to submit sequences of tuples T(1),T(2),...
> where T(i)=[T(i,1),T(i,2),...] are:
>
> A. Define a traversal order by a map (i,j) <--> n and set a(n) = T(i,j).
> This gives the tabl/tabf style of entry.
>
> B. Encode each tuple as a single number, say by using it as factor
> exponents:  T(i) --> 2^T(i,1) * 3^T(i,2) * 5^T(i,3) *...

As I understand, A is applicable only to the cases where the dimension
of tuples is fixed while B is perfectly fine for tuples with variable
dimensions (what I'm currently interested in).

>
> For example, for Pascal's triangle
>
> A: 1; 1, 1; 1, 2, 1; 1, 3, 3, 1;...
>
> B: 2, 6,    90,      47250,...

Does B apply to Pascal triangle?
The 5-th term corresponds to the tuple [0,0,5], what it is in Pascal triangle?

Max





From franktaw at netscape.net  Fri Nov  3 23:06:02 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 17:06:02 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com> <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>

<Me>
>Off hand, I can't think of any interesting functions of finite 
sequences of
>positive integers that depend on the order of the sequence.

Of course, I no sooner wrote that than I started thinking of such 
functions.
I'm not sure any of these are tremendously interesting, but they're at 
least
somewhat so.

In the following, b(i), i = 1..k is the finite sequence.  I'm starting 
all these
sequences with the empty sequence, but I searched for them without
that first term.  All are shown here in A066099 order.

Alternating sum
Sum_i (-1)^{i+1) b(i)
0,1,2,0,3,1,-1,1,4,2,0,2,-2,0,2,0

Binomial sum
Sum_i C(k-1,i-1) b(i)
0,1,2,2,3,3,3,4,4,4,4,5,4,6,5,8

Inverse binomial sum
Sum_i (-1)^{i+1) C(k-1,i-1) b(i)
0,1,2,0,3,1,-1,0,4,2,0,1,-2,-2,1,0

Weighted sum
sum_i i b(i)
0,1,2,3,3,4,5,6,4,5,6,7,7,8,9,10
This is A029931

Zero-based weighted sum
sum_i (i-1) b(i)
0,0,0,1,0,1,2,3,0,1,2,3,3,4,5,6

Sum of products of consecutive elements
sum_{i=1}^{k-1} b(i) b(i+1)
0,0,0,1,0,2,2,2,0,3,4,3,3,4,3,3

Number of rises
sum_{b(i)>b(i-1)} 1
0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,0

Number of falls
sum_{b(i)<b(i-1)} 1
0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0

Number of unchanged
sum_{b(i)=b(i-1)} 1
0,0,0,1,0,0,0,2,0,0,1,1,0,0,1,3

Number of non-rises
sum_{b(i)<=b(i-1)} 1
0,0,0,1,0,0,1,2,0,0,1,1,1,1,2,3

Number of non-falls
sum_{b(i)>=b(i-1)} 1
0,0,0,1,0,1,0,2,0,1,1,2,0,1,1,3

Number of monotonically increasing runs
1 + number of falls, but 0 for empty sequence
0,1,1,1,1,2,1,1,1,2,1,2,1,2,1,1

Number of monotonically decreasing runs
1 + number of rises, but 0 for empty sequence
0,1,1,1,1,1,2,1,1,1,1,1,2,2,2,1

Number of runs of equal terms
1 + number of unchanged, but 0 for empty sequence
0,1,1,2,1,1,1,3,1,1,2,2,1,1,2,4

Number of distinct non-empty subsequences
0,1,1,2,1,3,3,3,1,3,2,5,3,5,5,4
E.g., f(1,1) = 2, for the sequences [1], and [1,1].

Number of distinct subsequences
1 + number of distinct non-empty subsequences
1,2,2,3,2,4,4,4,2,4,3,6,4,6,6,5

Number of set partitions
List parts of set partition by their smallest element, and count the 
part sizes.
E.g., 1|2,4|3 (also known as {1,2,3,2}) would count for the sequence 
[1,2,1].
1,1,1,1,1,2,1,1,1,3,3,3,1,2,1,1

Number of permutations
List permutations in cycle form, sorted by the smallest element in the 
cycle,
and count the cycle lengths.
Number of set partitions * Product_i (b(i)-1)!
1,1,1,1,2,2,1,1,6,6,3,3,2,2,1,1

Number of partially ordered sets (unlabelled) by rank
The rank of an element in a poset is the length of the longest chain of 
which it
is the largest element.
1,1,1,1,1,2,1,1,1,3,4,3,1,2,1,1

Number of partially ordered sets (labelled) by rank
1,1,1,2,1,9,3,6,1,28,54,60,4,36,12,24

Number of partially ordered sets (naturally labelled) by rank
Naturally labelled means labels are consistent with the ordering.
1,1,1,1,1,4,1,1,1,11,13,8,1,4,1,1

Number of forests (unlabelled, unordered) with b(i) nodes at height i.
1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,2,2,1,1,1,1,2,1,1,1,1,1

... and, of course, various other forest options.

If there is agreement that this is the right order for these sequences, 
I'll try
to find time to submit them.
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From tbaruchel at free.fr  Fri Nov  3 23:17:29 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Fri, 3 Nov 2006 23:17:29 +0100
Subject: [math-fun] A fast (written in assembly) library for continued
	fractions
Message-ID: <20061103221729.GB2467@>

I wrote for my own purposes the following X86 assembly code, intended
to be called from GFORTRAN. Such a thing should only work if you have
a similar configuration to mine, but anyway, it may interest you.
I wanted a fast library for computing the length in the continued
fraction expansion of sqrt(n) for any integer n that can fit in
a REAL*8 gfortran type for statistical purposes. I also included
a faster function that tells if this length is odd or even. Though
I am not an assembly expert, there should be a significant speed
improvement if you use this code rather than your own fortran
functions, since the whole process is performed on the FPU stack
with no load/store useless travels.

It may help you, it may not. You know about these kinds of software...

Instructions are at the beginning of the file, as comments.
Read them carefully. Code is below.

Regards.

------------------ FILE quad.s BEGINS BELOW --------------------

/*

  Compile your piece of code with something like
    gfortran -fno-underscoring mycode.f90 quad.s
  and add freely your own flags. Enjoy !
  Of course, if you don't want to use the -fno-underscoring
  option for some reason, you can add an underscore at the end
  of the global functions below, like that:
    .globl initfpu_
    initfpu_:
  it should work.

  Remember this piece of code has the following status:
    Seems to work for me, maybe it will also seem for you,
    maybe your computer will explode, maybe it will make the
    earth explode, maybe God will choose not to have created
    the world, think about it before using it.
  (Anyway, if you see some kind of bug, please let me know ;-)
  Remember that
    a) this is x86 assembly code;
    b) it uses the GNU AS (gas) syntax;
    c) it uses the gfortran calling conventions;
  (now you understand why you can't expect it will works ;-)

  Usage. In your program (or subroutine or whatever), declare
  the function you need like:
    
    logical :: oddperiod
    integer*8 :: period

  (the two subroutines don't need to be declared).

  Then use
    
    integer*4 :: oldfpustatus
    call initfpu(oldfpustatus)

  before any computation; if you don't do it, it won't work.
  When you don't need any more the library,
    
    call restorefpu(oldfpustatus)

  will restore the control register of the FPU.

  During all that time, you shouldn't be disturbed too much, 
  for your own mathematical operations, but
  remember the rounding mode of the FPU has been set to
  "round down" (rather than the usual "nearest").

  How does it work? The whole computation is performed on the
  stack of the FPU; only integer values are needed for these
  tasks, but they are stored as REAL*10, 80 bits reals. They
  are not converted to anything during the whole process, thus
  it should work for great integers stored in your own REAL*8,
  64 bits reals. Of course, it may happen (I think it will never,
  but didn't check) that an intermediate result can not fit in
  such a REAL*10 register; in that case, read above what is
  written about the end of the world...
  
*/



/*
  call initfpu(integer*4)
  =====================
    This subroutine should be called before any other call to the
    library; its purpose is to set the rounding control in the FPU
    control register to the "round down" mode.
    The subroutine puts in the integer variable the initial value
    of the FPU control register.

    If you want to use often this subroutine (which you shouldn't:
    speed being an issue, this subroutine is intended to be used
    once, then the functions billions of times, not the contrary ;-),
    maybe removing the "fnclex" can improve the speed, since I don't
    think it plays a great role here.
*/
.globl initfpu
initfpu:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        fnstcw  (%eax)
	fwait
        movw    (%eax), %dx
	andb	$0xf3, %dh
	orb	$0x4, %dh
        movw    %dx, 2(%eax)
	fnclex
	fldcw   2(%eax)
        fwait
        movw    $0, 2(%eax)
	leave
	ret

/*
  call restorefpu(integer*4)
  ========================
    This subroutine should be used for restoring the initial state of
    the FPU control register; the integer variable must contain the
    right value (which should have been previously set by the
    subroutine initfpu).
*/
.globl restorefpu
restorefpu:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
	fclex
	fldcw   (%eax)
	leave
	ret

/*
  sqrtint( real double precision : REAL*8 )
  =========================================
    This function returns the integer part of the square root
    of the given real. The return value has itself the real type.
*/
.global sqrtint
sqrtint:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
	fldl	(%eax)
	fsqrt
        frndint
	leave
	ret

/*
  oddperiod( real double precision : REAL*8 )
  ===========================================
    Despite the type of the real argument, the variable has to
    be some integer value stored as a double precision.
    The function has the logical type. It returns FALSE if
    the argument is a perfect square or if the continued
    fraction expansion of its square root has an even period
    length; it returns TRUE in the remaining case.
      oddperiod(dble(25)) --> FALSE
      oddperiod(dble(13)) --> TRUE
      oddperiod(dble(12)) --> FALSE
    If it may help, you can declare the function as an integer*4
    type rather than a logical one (output will be 0 or 1).

    Though it seems to work with integer*8, I am not absolutely
    sure it *always* will (unless you add the following instruction
      movl $0, %edx
    at the beginning of the code (for instance between the current
    second and third lines of code), but really why would you
    do it? Maybe something in the calling conventions is such that
    the register EDX already contains 0, but, again, who cares?

    If you read the instructions for the period() function, you
    will learn about some 2^32 limitations, but you don't have to
    care about that here: the length is not kept in any register
    during the computation and the code should work as long as the
    values in the FPU stack are the right one.
    
*/
.global oddperiod
oddperiod:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        fldln2
	fldl	(%eax)
        fld     %st
        fsqrt
        frndint
        fld     %st
        fld     %st
        fmul    %st, %st
        fld     %st(3)
        fsubp   %st, %st(1)
        fcom    %st(4)
        fstsw   %ax
        fwait
        sahf
        ja      oddperiod0
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
oddperiod2:
        movl    $0, %eax
        leave
        ret
oddperiod0:
        fld     %st(1)
        fadd    %st(3), %st
        fdiv    %st(1), %st
        frndint
        fld     %st(2)
        fld     %st(1)
        fmul    %st(3), %st
        fsubp   %st, %st(1)
        fxch    %st(3)
        fsub    %st(3), %st
        fabs
        fcomp   %st(6)
        fstsw   %ax
        fwait
        sahf
        ja      oddperiod1
        fld     %st(3)
        fadd    %st, %st
        fsubp   %st, %st(1)
        fcomp   %st(5)
        fstsw   %ax
        fwait
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        sahf
        ja      oddperiod2
        movl    $1, %eax
        leave
        ret
oddperiod1:
        ffree   %st
        fincstp
        fld     %st(1)
        fmul    %st, %st
        fld     %st(4)
        fsubp   %st, %st(1)
        fdivp   %st, %st(1)
        frndint
        jmp oddperiod0

/*
  period( real double precision : REAL*8 )
  ========================================
    Despite the type of the real argument, the variable has to
    be some integer value stored as a double precision.
    The function has integer*8 type, though the output
    can't be greater than 2^32-1 (this choice has been made
    for avoiding sign problem when the result is greater
    (or equal) then 2^31 and smaller then 2^32.
    The result is the length of period in the continued fraction
    expansion of the square root of the given integer.

    If speed is really for you an obsession, remember that it
    is highly unprobable that you use that library for periods
    between the two fatidic values. Maybe it will increase a little
    the speed to use integer*4 which allows to delete twice the line
        movl    $0,%edx
    (not the first time it appears, but the second and third one).
    If you try such a thing and see some speed improvement, please
    let me know.
*/
.global period
period:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        movl    $0, %edx
        fldln2
	fldl	(%eax)
        fld     %st
        fsqrt
        frndint
        fld     %st
        fld     %st
        fmul    %st, %st
        fld     %st(3)
        fsubp   %st, %st(1)
        fcom    %st(4)
        fstsw   %ax
        fwait
        sahf
        ja      period0
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        movl    %edx, %eax
        movl     $0, %edx
        leave
        ret
period0:
        inc     %edx
        fld     %st(1)
        fadd    %st(3), %st
        fdiv    %st(1), %st
        frndint
        fld     %st(2)
        fld     %st(1)
        fmul    %st(3), %st
        fsubp   %st, %st(1)
        fxch    %st(3)
        fsub    %st(3), %st
        fabs
        fcomp   %st(6)
        fstsw   %ax
        fwait
        sahf
        ja      period1
        fld     %st(3)
        fadd    %st, %st
        fsubp   %st, %st(1)
        fcomp   %st(5)
        fstsw   %ax
        fwait
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        sahf
        jb      period2
        add     %edx, %edx
period2:
        movl    %edx, %eax
        movl    $0,%edx
        leave
        ret
period1:
        ffree   %st
        fincstp
        fld     %st(1)
        fmul    %st, %st
        fld     %st(4)
        fsubp   %st, %st(1)
        fdivp   %st, %st(1)
        frndint
        jmp period0

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From tbaruchel at free.fr  Fri Nov  3 23:17:29 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Fri, 3 Nov 2006 23:17:29 +0100
Subject: A fast (written in assembly) library for continued fractions
Message-ID: <20061103221729.GB2467@>

I wrote for my own purposes the following X86 assembly code, intended
to be called from GFORTRAN. Such a thing should only work if you have
a similar configuration to mine, but anyway, it may interest you.
I wanted a fast library for computing the length in the continued
fraction expansion of sqrt(n) for any integer n that can fit in
a REAL*8 gfortran type for statistical purposes. I also included
a faster function that tells if this length is odd or even. Though
I am not an assembly expert, there should be a significant speed
improvement if you use this code rather than your own fortran
functions, since the whole process is performed on the FPU stack
with no load/store useless travels.

It may help you, it may not. You know about these kinds of software...

Instructions are at the beginning of the file, as comments.
Read them carefully. Code is below.

Regards.

------------------ FILE quad.s BEGINS BELOW --------------------

/*

  Compile your piece of code with something like
    gfortran -fno-underscoring mycode.f90 quad.s
  and add freely your own flags. Enjoy !
  Of course, if you don't want to use the -fno-underscoring
  option for some reason, you can add an underscore at the end
  of the global functions below, like that:
    .globl initfpu_
    initfpu_:
  it should work.

  Remember this piece of code has the following status:
    Seems to work for me, maybe it will also seem for you,
    maybe your computer will explode, maybe it will make the
    earth explode, maybe God will choose not to have created
    the world, think about it before using it.
  (Anyway, if you see some kind of bug, please let me know ;-)
  Remember that
    a) this is x86 assembly code;
    b) it uses the GNU AS (gas) syntax;
    c) it uses the gfortran calling conventions;
  (now you understand why you can't expect it will works ;-)

  Usage. In your program (or subroutine or whatever), declare
  the function you need like:
    
    logical :: oddperiod
    integer*8 :: period

  (the two subroutines don't need to be declared).

  Then use
    
    integer*4 :: oldfpustatus
    call initfpu(oldfpustatus)

  before any computation; if you don't do it, it won't work.
  When you don't need any more the library,
    
    call restorefpu(oldfpustatus)

  will restore the control register of the FPU.

  During all that time, you shouldn't be disturbed too much, 
  for your own mathematical operations, but
  remember the rounding mode of the FPU has been set to
  "round down" (rather than the usual "nearest").

  How does it work? The whole computation is performed on the
  stack of the FPU; only integer values are needed for these
  tasks, but they are stored as REAL*10, 80 bits reals. They
  are not converted to anything during the whole process, thus
  it should work for great integers stored in your own REAL*8,
  64 bits reals. Of course, it may happen (I think it will never,
  but didn't check) that an intermediate result can not fit in
  such a REAL*10 register; in that case, read above what is
  written about the end of the world...
  
*/



/*
  call initfpu(integer*4)
  =====================
    This subroutine should be called before any other call to the
    library; its purpose is to set the rounding control in the FPU
    control register to the "round down" mode.
    The subroutine puts in the integer variable the initial value
    of the FPU control register.

    If you want to use often this subroutine (which you shouldn't:
    speed being an issue, this subroutine is intended to be used
    once, then the functions billions of times, not the contrary ;-),
    maybe removing the "fnclex" can improve the speed, since I don't
    think it plays a great role here.
*/
.globl initfpu
initfpu:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        fnstcw  (%eax)
	fwait
        movw    (%eax), %dx
	andb	$0xf3, %dh
	orb	$0x4, %dh
        movw    %dx, 2(%eax)
	fnclex
	fldcw   2(%eax)
        fwait
        movw    $0, 2(%eax)
	leave
	ret

/*
  call restorefpu(integer*4)
  ========================
    This subroutine should be used for restoring the initial state of
    the FPU control register; the integer variable must contain the
    right value (which should have been previously set by the
    subroutine initfpu).
*/
.globl restorefpu
restorefpu:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
	fclex
	fldcw   (%eax)
	leave
	ret

/*
  sqrtint( real double precision : REAL*8 )
  =========================================
    This function returns the integer part of the square root
    of the given real. The return value has itself the real type.
*/
.global sqrtint
sqrtint:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
	fldl	(%eax)
	fsqrt
        frndint
	leave
	ret

/*
  oddperiod( real double precision : REAL*8 )
  ===========================================
    Despite the type of the real argument, the variable has to
    be some integer value stored as a double precision.
    The function has the logical type. It returns FALSE if
    the argument is a perfect square or if the continued
    fraction expansion of its square root has an even period
    length; it returns TRUE in the remaining case.
      oddperiod(dble(25)) --> FALSE
      oddperiod(dble(13)) --> TRUE
      oddperiod(dble(12)) --> FALSE
    If it may help, you can declare the function as an integer*4
    type rather than a logical one (output will be 0 or 1).

    Though it seems to work with integer*8, I am not absolutely
    sure it *always* will (unless you add the following instruction
      movl $0, %edx
    at the beginning of the code (for instance between the current
    second and third lines of code), but really why would you
    do it? Maybe something in the calling conventions is such that
    the register EDX already contains 0, but, again, who cares?

    If you read the instructions for the period() function, you
    will learn about some 2^32 limitations, but you don't have to
    care about that here: the length is not kept in any register
    during the computation and the code should work as long as the
    values in the FPU stack are the right one.
    
*/
.global oddperiod
oddperiod:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        fldln2
	fldl	(%eax)
        fld     %st
        fsqrt
        frndint
        fld     %st
        fld     %st
        fmul    %st, %st
        fld     %st(3)
        fsubp   %st, %st(1)
        fcom    %st(4)
        fstsw   %ax
        fwait
        sahf
        ja      oddperiod0
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
oddperiod2:
        movl    $0, %eax
        leave
        ret
oddperiod0:
        fld     %st(1)
        fadd    %st(3), %st
        fdiv    %st(1), %st
        frndint
        fld     %st(2)
        fld     %st(1)
        fmul    %st(3), %st
        fsubp   %st, %st(1)
        fxch    %st(3)
        fsub    %st(3), %st
        fabs
        fcomp   %st(6)
        fstsw   %ax
        fwait
        sahf
        ja      oddperiod1
        fld     %st(3)
        fadd    %st, %st
        fsubp   %st, %st(1)
        fcomp   %st(5)
        fstsw   %ax
        fwait
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        sahf
        ja      oddperiod2
        movl    $1, %eax
        leave
        ret
oddperiod1:
        ffree   %st
        fincstp
        fld     %st(1)
        fmul    %st, %st
        fld     %st(4)
        fsubp   %st, %st(1)
        fdivp   %st, %st(1)
        frndint
        jmp oddperiod0

/*
  period( real double precision : REAL*8 )
  ========================================
    Despite the type of the real argument, the variable has to
    be some integer value stored as a double precision.
    The function has integer*8 type, though the output
    can't be greater than 2^32-1 (this choice has been made
    for avoiding sign problem when the result is greater
    (or equal) then 2^31 and smaller then 2^32.
    The result is the length of period in the continued fraction
    expansion of the square root of the given integer.

    If speed is really for you an obsession, remember that it
    is highly unprobable that you use that library for periods
    between the two fatidic values. Maybe it will increase a little
    the speed to use integer*4 which allows to delete twice the line
        movl    $0,%edx
    (not the first time it appears, but the second and third one).
    If you try such a thing and see some speed improvement, please
    let me know.
*/
.global period
period:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        movl    $0, %edx
        fldln2
	fldl	(%eax)
        fld     %st
        fsqrt
        frndint
        fld     %st
        fld     %st
        fmul    %st, %st
        fld     %st(3)
        fsubp   %st, %st(1)
        fcom    %st(4)
        fstsw   %ax
        fwait
        sahf
        ja      period0
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        movl    %edx, %eax
        movl     $0, %edx
        leave
        ret
period0:
        inc     %edx
        fld     %st(1)
        fadd    %st(3), %st
        fdiv    %st(1), %st
        frndint
        fld     %st(2)
        fld     %st(1)
        fmul    %st(3), %st
        fsubp   %st, %st(1)
        fxch    %st(3)
        fsub    %st(3), %st
        fabs
        fcomp   %st(6)
        fstsw   %ax
        fwait
        sahf
        ja      period1
        fld     %st(3)
        fadd    %st, %st
        fsubp   %st, %st(1)
        fcomp   %st(5)
        fstsw   %ax
        fwait
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        sahf
        jb      period2
        add     %edx, %edx
period2:
        movl    %edx, %eax
        movl    $0,%edx
        leave
        ret
period1:
        ffree   %st
        fincstp
        fld     %st(1)
        fmul    %st, %st
        fld     %st(4)
        fsubp   %st, %st(1)
        fdivp   %st, %st(1)
        frndint
        jmp period0





From jvospost3 at gmail.com  Fri Nov  3 23:36:07 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 3 Nov 2006 14:36:07 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com>

I think the example givem without A number, is:

 A007188 <http://www.research.att.com/%7Enjas/sequences/A007188>
 Multiplicative encoding of Pascal triangle: Product p(i+1)^C(n,i).
(Formerly M1722)
 +0
2
2, 6, 90, 47250, 66852843750, 2806877704512541816406250,
1216935896582703898519354781702537118597533386230468750
There might be other Godel-coded sequences, and the like, besides Conway's
brilliant "Fractran", above and beyond those with "decimal Godelization" --
but usually these quickly lead to numbers too large for OEIS.

-- Jonathan Vos Post

On 11/3/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
> <Me>
> >Off hand, I can't think of any interesting functions of finite
> sequences of
> >positive integers that depend on the order of the sequence.
>
> Of course, I no sooner wrote that than I started thinking of such
> functions.
> I'm not sure any of these are tremendously interesting, but they're at
> least
> somewhat so.
>
> In the following, b(i), i = 1..k is the finite sequence.  I'm starting
> all these
> sequences with the empty sequence, but I searched for them without
> that first term.  All are shown here in A066099 order.
>
> Alternating sum
> Sum_i (-1)^{i+1) b(i)
> 0,1,2,0,3,1,-1,1,4,2,0,2,-2,0,2,0
>
> Binomial sum
> Sum_i C(k-1,i-1) b(i)
> 0,1,2,2,3,3,3,4,4,4,4,5,4,6,5,8
>
> Inverse binomial sum
> Sum_i (-1)^{i+1) C(k-1,i-1) b(i)
> 0,1,2,0,3,1,-1,0,4,2,0,1,-2,-2,1,0
>
> Weighted sum
> sum_i i b(i)
> 0,1,2,3,3,4,5,6,4,5,6,7,7,8,9,10
> This is A029931
>
> Zero-based weighted sum
> sum_i (i-1) b(i)
> 0,0,0,1,0,1,2,3,0,1,2,3,3,4,5,6
>
> Sum of products of consecutive elements
> sum_{i=1}^{k-1} b(i) b(i+1)
> 0,0,0,1,0,2,2,2,0,3,4,3,3,4,3,3
>
> Number of rises
> sum_{b(i)>b(i-1)} 1
> 0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,0
>
> Number of falls
> sum_{b(i)<b(i-1)} 1
> 0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0
>
> Number of unchanged
> sum_{b(i)=b(i-1)} 1
> 0,0,0,1,0,0,0,2,0,0,1,1,0,0,1,3
>
> Number of non-rises
> sum_{b(i)<=b(i-1)} 1
> 0,0,0,1,0,0,1,2,0,0,1,1,1,1,2,3
>
> Number of non-falls
> sum_{b(i)>=b(i-1)} 1
> 0,0,0,1,0,1,0,2,0,1,1,2,0,1,1,3
>
> Number of monotonically increasing runs
> 1 + number of falls, but 0 for empty sequence
> 0,1,1,1,1,2,1,1,1,2,1,2,1,2,1,1
>
> Number of monotonically decreasing runs
> 1 + number of rises, but 0 for empty sequence
> 0,1,1,1,1,1,2,1,1,1,1,1,2,2,2,1
>
> Number of runs of equal terms
> 1 + number of unchanged, but 0 for empty sequence
> 0,1,1,2,1,1,1,3,1,1,2,2,1,1,2,4
>
> Number of distinct non-empty subsequences
> 0,1,1,2,1,3,3,3,1,3,2,5,3,5,5,4
> E.g., f(1,1) = 2, for the sequences [1], and [1,1].
>
> Number of distinct subsequences
> 1 + number of distinct non-empty subsequences
> 1,2,2,3,2,4,4,4,2,4,3,6,4,6,6,5
>
> Number of set partitions
> List parts of set partition by their smallest element, and count the
> part sizes.
> E.g., 1|2,4|3 (also known as {1,2,3,2}) would count for the sequence
> [1,2,1].
> 1,1,1,1,1,2,1,1,1,3,3,3,1,2,1,1
>
> Number of permutations
> List permutations in cycle form, sorted by the smallest element in the
> cycle,
> and count the cycle lengths.
> Number of set partitions * Product_i (b(i)-1)!
> 1,1,1,1,2,2,1,1,6,6,3,3,2,2,1,1
>
> Number of partially ordered sets (unlabelled) by rank
> The rank of an element in a poset is the length of the longest chain of
> which it
> is the largest element.
> 1,1,1,1,1,2,1,1,1,3,4,3,1,2,1,1
>
> Number of partially ordered sets (labelled) by rank
> 1,1,1,2,1,9,3,6,1,28,54,60,4,36,12,24
>
> Number of partially ordered sets (naturally labelled) by rank
> Naturally labelled means labels are consistent with the ordering.
> 1,1,1,1,1,4,1,1,1,11,13,8,1,4,1,1
>
> Number of forests (unlabelled, unordered) with b(i) nodes at height i.
> 1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,2,2,1,1,1,1,2,1,1,1,1,1
>
> ... and, of course, various other forest options.
>
> If there is agreement that this is the right order for these sequences,
> I'll try
> to find time to submit them.
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and
> industry-leading spam and email virus protection.
>
>
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From jvospost3 at gmail.com  Sat Nov  4 00:59:34 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 3 Nov 2006 15:59:34 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
	 <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com>
Message-ID: <5542af940611031559x7a45e57euad1d783fb83d82b@mail.gmail.com>

To give a specific example:

NEW SEQUENCE FROM Jonathan Vos Post

%I A000001
%S A000001 2, 6, 450, 2836181250, 81492043057751910481759423160156250,
4561157026363824997482074305569280581505536351717093893927260661169357729871499327113563125890139588096951624677718591308593750
%N A000001 Multiplicative encoding of Catalan's triangle: Product
p(i+1)^T(n,i).
%C A000001 This is to A009766 "Catalan's triangle T(n,k) (read by
rows)"  as A007188 "Multiplicative encoding of Pascal triangle: Product
p(i+1)^C(n,i)" is to A007318 "Pascal's triangle read by rows."
%F A000001 a(n) = PRODUCT[i=i..n] p(i+1)^T(n,i), where T(n,i), are
Cataln's triangle as in a009766.
%e A000001 a(1) = p(1)^T(1,1) = 2^1 = 2.
a(2) = p(1)^T(2,1) * p(2)^T(2,2) = 2^1 * 3^1 = 6.
a(3) = p(1)^T(3,1) * p(2)^T(3,2) * p(3)^T(3,3) = 2^1 * 3^2 * 5^2 = 450.
a(4) = 2^1 * 3^3 * 5^5 * 7^5 = 2836181250.
a(5) = 2^1 * 3^4 * 5^9 * 7^14 * 11^14 =
81492043057751910481759423160156250.
a(6) = 2^1 * 3^5 * 5^14 * 7^28 * 11^42 * 13^42.
%Y A000001 Cf. A007188, A007318, A009766.
%O A000001 1
%K A000001 ,easy,nonn,
%A A000001 Jonathan Vos Post (jvospost2 at yahoo.com
<http://us.f551.mail.yahoo.com/ym/Compose?To=jvospost2 at yahoo.com&YY=14244&y5beta=yes&y5beta=yes&order=down&sort=date&pos=0>),
Nov 03 2006
RH
RA 192.20.225.32



On 11/3/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> I think the example givem without A number, is:
>
>  A007188 <http://www.research.att.com/%7Enjas/sequences/A007188>
>  Multiplicative encoding of Pascal triangle: Product p(i+1)^C(n,i).
> (Formerly M1722)
>  +0
> 2
> 2, 6, 90, 47250, 66852843750, 2806877704512541816406250,
> 1216935896582703898519354781702537118597533386230468750
> There might be other Godel-coded sequences, and the like, besides Conway's
> brilliant "Fractran", above and beyond those with "decimal Godelization" --
> but usually these quickly lead to numbers too large for OEIS.
>
> -- Jonathan Vos Post
>
> On 11/3/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
> >
> > <Me>
> > >Off hand, I can't think of any interesting functions of finite
> > sequences of
> > >positive integers that depend on the order of the sequence.
> >
> > Of course, I no sooner wrote that than I started thinking of such
> > functions.
> > I'm not sure any of these are tremendously interesting, but they're at
> > least
> > somewhat so.
> >
> > In the following, b(i), i = 1..k is the finite sequence.  I'm starting
> > all these
> > sequences with the empty sequence, but I searched for them without
> > that first term.  All are shown here in A066099 order.
> >
> > Alternating sum
> > Sum_i (-1)^{i+1) b(i)
> > 0,1,2,0,3,1,-1,1,4,2,0,2,-2,0,2,0
> >
> > Binomial sum
> > Sum_i C(k-1,i-1) b(i)
> > 0,1,2,2,3,3,3,4,4,4,4,5,4,6,5,8
> >
> > Inverse binomial sum
> > Sum_i (-1)^{i+1) C(k-1,i-1) b(i)
> > 0,1,2,0,3,1,-1,0,4,2,0,1,-2,-2,1,0
> >
> > Weighted sum
> > sum_i i b(i)
> > 0,1,2,3,3,4,5,6,4,5,6,7,7,8,9,10
> > This is A029931
> >
> > Zero-based weighted sum
> > sum_i (i-1) b(i)
> > 0,0,0,1,0,1,2,3,0,1,2,3,3,4,5,6
> >
> > Sum of products of consecutive elements
> > sum_{i=1}^{k-1} b(i) b(i+1)
> > 0,0,0,1,0,2,2,2,0,3,4,3,3,4,3,3
> >
> > Number of rises
> > sum_{b(i)>b(i-1)} 1
> > 0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,0
> >
> > Number of falls
> > sum_{b(i)<b(i-1)} 1
> > 0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0
> >
> > Number of unchanged
> > sum_{b(i)=b(i-1)} 1
> > 0,0,0,1,0,0,0,2,0,0,1,1,0,0,1,3
> >
> > Number of non-rises
> > sum_{b(i)<=b(i-1)} 1
> > 0,0,0,1,0,0,1,2,0,0,1,1,1,1,2,3
> >
> > Number of non-falls
> > sum_{b(i)>=b(i-1)} 1
> > 0,0,0,1,0,1,0,2,0,1,1,2,0,1,1,3
> >
> > Number of monotonically increasing runs
> > 1 + number of falls, but 0 for empty sequence
> > 0,1,1,1,1,2,1,1,1,2,1,2,1,2,1,1
> >
> > Number of monotonically decreasing runs
> > 1 + number of rises, but 0 for empty sequence
> > 0,1,1,1,1,1,2,1,1,1,1,1,2,2,2,1
> >
> > Number of runs of equal terms
> > 1 + number of unchanged, but 0 for empty sequence
> > 0,1,1,2,1,1,1,3,1,1,2,2,1,1,2,4
> >
> > Number of distinct non-empty subsequences
> > 0,1,1,2,1,3,3,3,1,3,2,5,3,5,5,4
> > E.g., f(1,1) = 2, for the sequences [1], and [1,1].
> >
> > Number of distinct subsequences
> > 1 + number of distinct non-empty subsequences
> > 1,2,2,3,2,4,4,4,2,4,3,6,4,6,6,5
> >
> > Number of set partitions
> > List parts of set partition by their smallest element, and count the
> > part sizes.
> > E.g., 1|2,4|3 (also known as {1,2,3,2}) would count for the sequence
> > [1,2,1].
> > 1,1,1,1,1,2,1,1,1,3,3,3,1,2,1,1
> >
> > Number of permutations
> > List permutations in cycle form, sorted by the smallest element in the
> > cycle,
> > and count the cycle lengths.
> > Number of set partitions * Product_i (b(i)-1)!
> > 1,1,1,1,2,2,1,1,6,6,3,3,2,2,1,1
> >
> > Number of partially ordered sets (unlabelled) by rank
> > The rank of an element in a poset is the length of the longest chain of
> > which it
> > is the largest element.
> > 1,1,1,1,1,2,1,1,1,3,4,3,1,2,1,1
> >
> > Number of partially ordered sets (labelled) by rank
> > 1,1,1,2,1,9,3,6,1,28,54,60,4,36,12,24
> >
> > Number of partially ordered sets (naturally labelled) by rank
> > Naturally labelled means labels are consistent with the ordering.
> > 1,1,1,1,1,4,1,1,1,11,13,8,1,4,1,1
> >
> > Number of forests (unlabelled, unordered) with b(i) nodes at height i.
> > 1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,2,2,1,1,1,1,2,1,1,1,1,1
> >
> > ... and, of course, various other forest options.
> >
> > If there is agreement that this is the right order for these sequences,
> > I'll try
> > to find time to submit them.
> > ________________________________________________________________________
> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > industry-leading spam and email virus protection.
> >
> >
>
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From davidwwilson at comcast.net  Sat Nov  4 01:40:21 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Fri, 3 Nov 2006 19:40:21 -0500
Subject: A124015 thoughts
Message-ID: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p>

The description of A124015 is

%N A124015 Number of words with n letters in the National Scrabble Association Dictionary.

Some notes:

- It should be noted that the NSAD includes a restricted set of English words (words of 2 to 14 letters, no proper nouns or derivatives, no words with non-alphabetic characters (e.g, contractions, hyphenated words), and it is this restricted set that is being counted.

- The edition of the NSAD used to create A124015 should be specified, since the NSAD is continually edited.  While I do not think that A124015 should change with each new edition of the NSAD, because it may be referenced in other literature. On the other hand, I don't think that a new sequence should be created for each new edition of the NSAD unless the new edition exhibits some interesting statistical departure from the current NSAD (e.g, in some future NSAD, the number of 9-letter words exceeds the number of 8-letter words). This is because I believe that the value of A124015 is mainly to indicate a distribution of English word lengths.

- On the other hand, I have always understood that 4-letter words are most common in literature (where small words appear more often). It might be interesting to create a sequence counting words of length n in the King James Bible, War and Peace, or some other suitably large and stable piece of English literature (we don't have to go overboard on this, just one or two examples indicating word length distributions in literature).

- Another interesting idea: In various languages, what is the distribution of the number of letters in the names of numbers from one to (say) one million?

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From franktaw at netscape.net  Sat Nov  4 02:49:48 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 20:49:48 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <5542af940611031559x7a45e57euad1d783fb83d82b@mail.gmail.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>	 <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com> <5542af940611031559x7a45e57euad1d783fb83d82b@mail.gmail.com>
Message-ID: <8C8CDCAA57554FD-ED8-5165@FWM-M11.sysops.aol.com>

This is OK, but it is the opposite of what we started out talking 
about.  This is a function from integers to sequences; but we are 
talking about functions from sequences to integers.  And not every 
integer encodes a sequence in this way; only those in A055932.

Franklin T. Adams-Watters


-----Original Message-----
From: jvospost3 at gmail.com

  To give a specific example:

 NEW SEQUENCE FROM Jonathan Vos Post

%I A000001
%S A000001 2, 6, 450, 2836181250, 81492043057751910481759423160156250,
4561157026363824997482074305569280581505536351717093893927260661169357729
871499327113563125890139588096951624677718591308593750

%N A000001 Multiplicative encoding of Catalan's triangle: Product
p(i+1)^T(n,i).
%C A000001 This is to A009766 "Catalan's triangle T(n,k) (read by
rows)"  as A007188 "Multiplicative encoding of Pascal triangle: Product

p(i+1)^C(n,i)" is to A007318 "Pascal's triangle read by rows."
%F A000001 a(n) = PRODUCT[i=i..n] p(i+1)^T(n,i), where T(n,i), are
Cataln's triangle as in a009766.
%e A000001 a(1) = p(1)^T(1,1) = 2^1 = 2.

a(2) = p(1)^T(2,1) * p(2)^T(2,2) = 2^1 * 3^1 = 6.
a(3) = p(1)^T(3,1) * p(2)^T(3,2) * p(3)^T(3,3) = 2^1 * 3^2 * 5^2 = 450.
a(4) = 2^1 * 3^3 * 5^5 * 7^5 = 2836181250.
a(5) = 2^1 * 3^4 * 5^9 * 7^14 * 11^14 =

81492043057751910481759423160156250.
a(6) = 2^1 * 3^5 * 5^14 * 7^28 * 11^42 * 13^42.
%Y A000001 Cf. A007188, A007318, A009766.
%O A000001 1
%K A000001 ,easy,nonn,
%A A000001 Jonathan Vos Post (
jvospost2 at yahoo.com), Nov 03 2006
RH
RA 192.20.225.32

On 11/3/06, Jonathan Post <jvospost3 at gmail.com> wrote: I think the 
example givem without A number, is:

   A007188
   Multiplicative encoding of Pascal triangle: Product p(i+1)^C(n,i).
(Formerly M1722)
   +0
2

  2, 6, 90, 47250, 66852843750, 2806877704512541816406250, 
1216935896582703898519354781702537118597533386230468750

  There might be other Godel-coded sequences, and the like, besides 
Conway's brilliant "Fractran", above and beyond those with "decimal 
Godelization" -- but usually these quickly lead to numbers too large 
for OEIS.

 -- Jonathan Vos Post

On 11/3/06, franktaw at netscape.net < franktaw at netscape.net> wrote: <Me>
>Off hand, I can't think of any interesting functions of finite
 sequences of
>positive integers that depend on the order of the sequence.

Of course, I no sooner wrote that than I started thinking of such
functions.
I'm not sure any of these are tremendously interesting, but they're at
least
somewhat so.

In the following, b(i), i = 1..k is the finite sequence. I'm starting
all these
sequences with the empty sequence, but I searched for them without
that first term. All are shown here in A066099 order.

Alternating sum
Sum_i (-1)^{i+1) b(i)
0,1,2,0,3,1,-1,1,4,2,0,2,-2,0,2,0

Binomial sum
Sum_i C(k-1,i-1) b(i)
0,1,2,2,3,3,3,4,4,4,4,5,4,6,5,8

Inverse binomial sum
Sum_i (-1)^{i+1) C(k-1,i-1) b(i)
0,1,2,0,3,1,-1,0,4,2,0,1,-2,-2,1,0

Weighted sum
sum_i i b(i)
0,1,2,3,3,4,5,6,4,5,6,7,7,8,9,10
This is A029931

Zero-based weighted sum
sum_i (i-1) b(i)
0,0,0,1,0,1,2,3,0,1,2,3,3,4,5,6

 Sum of products of consecutive elements
sum_{i=1}^{k-1} b(i) b(i+1)
0,0,0,1,0,2,2,2,0,3,4,3,3,4,3,3

Number of rises
sum_{b(i)>b(i-1)} 1
0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,0

Number of falls
sum_{b(i)<b(i-1)} 1
0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0

Number of unchanged
sum_{b(i)=b(i-1)} 1
0,0,0,1,0,0,0,2,0,0,1,1,0,0,1,3

Number of non-rises
sum_{b(i)<=b(i-1)} 1
0,0,0,1,0,0,1,2,0,0,1,1,1,1,2,3

Number of non-falls
sum_{b(i)>=b(i-1)} 1
0,0,0,1,0,1,0,2,0,1,1,2,0,1,1,3

Number of monotonically increasing runs
1 + number of falls, but 0 for empty sequence
0,1,1,1,1,2,1,1,1,2,1,2,1,2,1,1

Number of monotonically decreasing runs
1 + number of rises, but 0 for empty sequence
0,1,1,1,1,1,2,1,1,1,1,1,2,2,2,1

Number of runs of equal terms
1 + number of unchanged, but 0 for empty sequence
0,1,1,2,1,1,1,3,1,1,2,2,1,1,2,4

Number of distinct non-empty subsequences
0,1,1,2,1,3,3,3,1,3,2,5,3,5,5,4
E.g., f(1,1) = 2, for the sequences [1], and [1,1].

Number of distinct subsequences
1 + number of distinct non-empty subsequences
1,2,2,3,2,4,4,4,2,4,3,6,4,6,6,5

 Number of set partitions
List parts of set partition by their smallest element, and count the
part sizes.
E.g., 1|2,4|3 (also known as {1,2,3,2}) would count for the sequence
[1,2,1].
1,1,1,1,1,2,1,1,1,3,3,3,1,2,1,1

Number of permutations
List permutations in cycle form, sorted by the smallest element in the
cycle,
and count the cycle lengths.
Number of set partitions * Product_i (b(i)-1)!
1,1,1,1,2,2,1,1,6,6,3,3,2,2,1,1

Number of partially ordered sets (unlabelled) by rank
The rank of an element in a poset is the length of the longest chain of
which it
is the largest element.
1,1,1,1,1,2,1,1,1,3,4,3,1,2,1,1

Number of partially ordered sets (labelled) by rank
1,1,1,2,1,9,3,6,1,28,54,60,4,36,12,24

Number of partially ordered sets (naturally labelled) by rank
Naturally labelled means labels are consistent with the ordering.
1,1,1,1,1,4,1,1,1,11,13,8,1,4,1,1

 Number of forests (unlabelled, unordered) with b(i) nodes at height i.
1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,2,2,1,1,1,1,2,1,1,1,1,1

... and, of course, various other forest options.

If there is agreement that this is the right order for these sequences,
I'll try
to find time to submit them.
________________________________________________________________________
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industry-leading spam and email virus protection.








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From maxale at gmail.com  Sat Nov  4 03:09:57 2006
From: maxale at gmail.com (Max A.)
Date: Fri, 3 Nov 2006 18:09:57 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <200611032139.kA3LdZSV024106@smtp.well.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <200611031625.kA3GP1mt017249@smtp.well.com>
	 <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com>
	 <200611032139.kA3LdZSV024106@smtp.well.com>
Message-ID: <d3dac270611031809j6a1f850dj5a1846b40153dc0@mail.gmail.com>

On 11/3/06, Marc LeBrun <mlb at well.com> wrote:

> B. Encode each tuple as a single number, say by using it as factor
> exponents:  T(i) --> 2^T(i,1) * 3^T(i,2) * 5^T(i,3) *...

The sequences I've mentioned encoded this way:

I) Multinomial coefficients:

a( [k1, k2, ..., kn] ) = (k1 + k2 +  ... + kn)! / (k1! * k2! * ... * kn!)

1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 2, 1, 1, 3, 1, 3, 2, 2, 1,
4, 1, 2, 1, 3, 1, 6, 1, 1, 2, 2, 2, 6, 1, 2, 2, 4, 1, 6, 1, 3, 3, 2,
1, 5, 1, 3, 2, 3, 1, 4, 2, 4, 2, 2, 1, 12, 1, 2, 3, 1, 2, 6, 1, 3, 2,
6, 1, 10, 1, 2, 3, 3, 2, 6, 1, 5, 1, 2, 1, 12, 2, 2, 2, 4, 1, 12, 2,
3, 2, 2, 2, 6, 1, 3, 3, 6

II) Coefficients of incomplete Bell polynomials:

b( [k1, k2, ..., kn] ) = (k1 + 2*k2 + ... + n*kn)! / (k1! * k2! * ...
* kn!) / (1!^k1 * 2!^k2 * ... * n!^kn)

1, 1, 1, 1, 1, 3, 1, 1, 3, 4, 1, 6, 1, 5, 10, 1, 1, 15, 1, 10, 15, 6,
1, 10, 10, 7, 15, 15, 1, 60, 1, 1, 21, 8, 35, 45, 1, 9, 28, 20, 1,
105, 1, 21, 105, 10, 1, 15, 35, 70, 36, 28, 1, 105, 56, 35, 45, 11, 1,
210, 1, 12, 210, 1, 84, 168, 1, 36, 55, 280, 1, 105, 1, 13, 280, 45,
126, 252, 1, 35, 105, 14, 1, 420, 120, 15, 66, 56, 1, 840, 210, 55,
78, 16, 165, 21, 1, 315, 378, 280

III) The number of permutations with given cycle structure:

c( [k1, k2, ..., kn] ) = (k1 + 2*k2 + ... + n*kn)! / (k1! * k2! * ...
* kn!) / (1^k1 * 2^k2 * ... * n^kn)

1, 1, 1, 1, 2, 3, 6, 1, 3, 8, 24, 6, 120, 30, 20, 1, 720, 15, 5040,
20, 90, 144, 40320, 10, 40, 840, 15, 90, 362880, 120, 3628800, 1, 504,
5760, 420, 45, 39916800, 45360, 3360, 40, 479001600, 630, 6227020800,
504, 210, 403200, 87178291200, 15, 1260, 280, 25920, 3360,
1307674368000, 105, 2688, 210, 226800, 3991680, 20922789888000, 420,
355687428096000, 43545600, 1260, 1, 20160, 4032, 6402373705728000,
25920, 2217600, 3360, 121645100408832000, 105, 2432902008176640000,
518918400, 1120, 226800, 18144, 30240, 51090942171709440000, 70, 105,
6706022400, 1124000727777607680000, 2520, 172800, 93405312000,
23950080, 1344, 25852016738884976640000, 1680, 151200, 2217600,
283046400, 1394852659200, 1663200, 21, 620448401733239439360000,
11340, 9072, 1120

Max





From jawbrey at att.net  Sat Nov  4 03:26:36 2006
From: jawbrey at att.net (Jon Awbrey)
Date: Fri, 03 Nov 2006 21:26:36 -0500
Subject: querying multidimensional sequences in OEIS
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>	 <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com> <5542af940611031559x7a45e57euad1d783fb83d82b@mail.gmail.com> <8C8CDCAA57554FD-ED8-5165@FWM-M11.sysops.aol.com>
Message-ID: <454BFA5C.ACD4F79B@att.net>

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o

but every positive integer in N = {1, 2, 3, ...}
uniquely codes a finite partial function f : N -> N

http://stderr.org/pipermail/inquiry/2005-July/002846.html

ja

franktaw at netscape.net wrote:
> 
> This is OK, but it is the opposite of what we started out talking
> about.  This is a function from integers to sequences; but we are
> talking about functions from sequences to integers.  And not every
> integer encodes a sequence in this way; only those in A055932.
> 
> Franklin T. Adams-Watters

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o
inquiry e-lab: http://stderr.org/pipermail/inquiry/
citizendium development forums: http://smf.citizendium.org
http://www.textop.org/wiki/index.php?title=User:Jon_Awbrey
wikinfo: http://wikinfo.org/wiki.php?title=User:Jon_Awbrey
wp review: http://wikipediareview.com/index.php?showuser=398
o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o






From jvospost3 at gmail.com  Sat Nov  4 03:49:46 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 3 Nov 2006 18:49:46 -0800
Subject: A124015 thoughts
In-Reply-To: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p>
Message-ID: <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>

I mostly agree.  A large sample of text as actually used in a language, such
as a set of books, newspaper articles, transcripts of phone conversations,
or the like, is often called a "corpus" by those who undertake statistical
analyses.  Various government agencies have done so since well before World
War II, tapping the minds of as many of half the Mathematicians in the
world, with computational resources in advance of corporate and personal
users, and with the help of Mathematicians such as Turing.

There are statistical analyses of various editions of the Bible, of the
works of Shakespeare, and the like, with phalanxes of statistical analysts
on Literature.

Many of these are covered in specialized journals, of questionable value to
seqfans, and unnoticed here except when, for instance, a minor poem is
claimed by such methods to be by Shakespeare.

Hence I second the motion by David Wilson to start with an extremely
well-defined if artificial corpus, such as "In various languages, what is
the distribution of the number of letters in the names of numbers from one
to (say) one million?"

I have commented more extensively on Scrabble in other venues.  Do not ever
agree to use the OED (Oxford English Dictionary) as the official dictionary
when playing Scrabble.  Many misspellings by Tournament rumes are there.
Even though the OED is the definitive dictionary of the largest language on
Earth, and fascinating reading for various purposes, it is just too big and
too weird for game use.  Also, it makes it hard to distinguish a bluff from
a variant.

Now that I think of it, logarithmic trends can be interesting.  How about "what
is the distribution of the number of letters in the names of numbers from
one to one billion?"

-- Jonathan Vos Post

On 11/3/06, David Wilson <davidwwilson at comcast.net> wrote:
>
>  The description of A124015 is
>
> %N A124015 Number of words with n letters in the National Scrabble
> Association Dictionary.
>
> Some notes:
>
> - It should be noted that the NSAD includes a restricted set of English
> words (words of 2 to 14 letters, no proper nouns or derivatives, no words
> with non-alphabetic characters (e.g, contractions, hyphenated words), and
> it is this restricted set that is being counted.
>
> - The edition of the NSAD used to create A124015 should be specified,
> since the NSAD is continually edited.  While I do not think that A124015
> should change with each new edition of the NSAD, because it may be
> referenced in other literature. On the other hand, I don't think that a new
> sequence should be created for each new edition of the NSAD unless the new
> edition exhibits some interesting statistical departure from the current
> NSAD (e.g, in some future NSAD, the number of 9-letter words exceeds the
> number of 8-letter words). This is because I believe that the value of
> A124015 is mainly to indicate a distribution of English word lengths.
>
> - On the other hand, I have always understood that 4-letter words are most
> common in literature (where small words appear more often). It might be
> interesting to create a sequence counting words of length n in the King
> James Bible, War and Peace, or some other suitably large and stable piece of
> English literature (we don't have to go overboard on this, just one or two
> examples indicating word length distributions in literature).
>
> - Another interesting idea: In various languages, what is the distribution
> of the number of letters in the names of numbers from one to (say) one
> million?
>
>
>
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From davidwwilson at comcast.net  Sat Nov  4 17:58:32 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sat, 4 Nov 2006 11:58:32 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
References: <20061103030956.55909.qmail@web38210.mail.mud.yahoo.com>
Message-ID: <001701c70032$75ba7ff0$6401a8c0@yourxhtr8hvc4p>

----- Original Message ----- 
From: "zak seidov" <zakseidov at yahoo.com>
To: <seqfan at ext.jussieu.fr>
Sent: Thursday, November 02, 2006 10:09 PM
Subject: Re: A065577 Number of Goldbach partitions of 10^n? More terms?


> David, Dick, Bob, Ivars,
> seqfans.
>
> After recalculating (at two different PC's)
> and rechecking I could finally
> find source of my error
> (silly as usual) in one of codes,
> and my final value is
> a(10) = 18200488 (NOT 18200487!)
> which coincides now with the Dick-David value.
>
> So, Bob,
> you may wish to give in A065577  the correct value
> a(10) = 18200488 with due refer to David Wilson  and
> Richard Mathar (not me).

I don't remember if Richard Mathar or myself was first to find a(10) = 
18200488. It's not really important to me, since any value would be suspect 
until independently confirmed. The fact that your suspect value led you to a 
bug in your program simply serves as additional confirmation. I have no 
problem with all three of us getting attribution. But then, I have no 
problem with none of us getting attribution, as this computation strikes me 
as on the level of an undergraduate homework problem.

> Also Ivars can extend his table.
> Thanks, Zak
>
> BTW1 Still, is it possible to find a(11)... with C++?

Well, it is possible to find a(11) using a C++ sieving program on an aging 
PC in a few hours, and if you were to do so, you would find that a(11) = 
149091160. Methinks that finding a(12) using the same program on the same PC 
would take about a week, except that said PC is a shared resource.

> BIW2 Not all people even in this closed list
> may be interested in Mmca specific problems,
> so I''l be happy to have a list of you
> particularly Mmca-involved.
> Then we may circulate relevant messages
> not to all seqfan community.
> Please anyone interested write me personally
> with subject "Me in Mmca", thanks.
> Of course it's not intented to be closed list!
> Hope this is not against Gerard' policy.

I truly wish I could afford Mma (or better yet, a math-enabled PDF editor). 
If I'm on anybody's Christmas list...






From mathar at strw.leidenuniv.nl  Sat Nov  4 18:20:07 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Sat, 4 Nov 2006 18:20:07 +0100
Subject: primes in arithmetic progression
Message-ID: <200611041720.kA4HK70T027399@amer.strw.leidenuniv.nl>


Is there an OEIS table that shows the smallest prime p
of k (not necessarily consecutive) primes in arithmetic progression
with common difference d? The table would look similar to this one
below, and contain rather large primes where I am leaving blanks:

    k=1 k=2  k=3  k=4  k=5  k=6
d
0     2   2    2    2    2    2
2     2   3    3
4     2   3    3
6     2   5    5    5    5
8     2   3    3
10    2   3    3
12    2   5    5    5    5
14    2   3    3 
16    2   3
18    2   5    5    5
20    2   3
22    2   7
24    2   5    5   59
26    2   3
28    2   3    3
30    2   7    7    7     7    7
32    2   5
34    2   3    3
36    2   5    7   31

The row d=0 and the column k=1 are degenerate and filled with the
prime 2. All strides d are even. Example for row d=24 and column k=4:
The 4 numbers 59,59+24,59+2*24 and 59+3*24 are all primes.

http://www.research.att.com/~njas/sequences/Sindx_Pri.html#primes_AP

--Richard





From rsc at swtch.com  Sat Nov  4 18:24:53 2006
From: rsc at swtch.com (Russ Cox)
Date: Sat, 4 Nov 2006 12:24:53 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
In-Reply-To: <001701c70032$75ba7ff0$6401a8c0@yourxhtr8hvc4p>
References: <20061103030956.55909.qmail@web38210.mail.mud.yahoo.com>
	 <001701c70032$75ba7ff0$6401a8c0@yourxhtr8hvc4p>
Message-ID: <ee9e417a0611040924h7c62e2a6hc2ba0ec199c7a175@mail.gmail.com>

I have confirmed the values so far
and also computed that a(12) = 1243722370.
Submitted (with a(11)) to OEIS via form.

Russ





From jvospost3 at gmail.com  Sat Nov  4 19:33:14 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 4 Nov 2006 10:33:14 -0800
Subject: primes in arithmetic progression
In-Reply-To: <200611041720.kA4HK70T027399@amer.strw.leidenuniv.nl>
References: <200611041720.kA4HK70T027399@amer.strw.leidenuniv.nl>
Message-ID: <5542af940611041033s7a722e9fjf6ad6643f988477d@mail.gmail.com>

The semiprime analogue of this is a table that shows the smallest semiprime
S
of k (not necessarily consecutive) primes in arithmetic progression
with common difference d (once a 0 appears, the row has ended nonzero
values):

d   k+1  k=2  k=3  k=4  k=5  k=6  k=7 k=8
0   4      4       4      4       4       4       4 ...
1   4      9       33    0
2   4      4       91    213   1383 8129 3091 0
3   4      6       115
4   4      6       6      111
5   4      4       77
6   4      4
7   4      14    51
8   4      6      6       69

-- Fairly easy to extend, fairly easy to prove the first 0 in a row.

The same can be done for 3-almost primes, 4-almost primes, and so forth,
this making a 3-dimensional array.



On 11/4/06, Richard Mathar <mathar at strw.leidenuniv.nl> wrote:
>
>
> Is there an OEIS table that shows the smallest prime p
> of k (not necessarily consecutive) primes in arithmetic progression
> with common difference d? The table would look similar to this one
> below, and contain rather large primes where I am leaving blanks:
>
>     k=1 k=2  k=3  k=4  k=5  k=6
> d
> 0     2   2    2    2    2    2
> 2     2   3    3
> 4     2   3    3
> 6     2   5    5    5    5
> 8     2   3    3
> 10    2   3    3
> 12    2   5    5    5    5
> 14    2   3    3
> 16    2   3
> 18    2   5    5    5
> 20    2   3
> 22    2   7
> 24    2   5    5   59
> 26    2   3
> 28    2   3    3
> 30    2   7    7    7     7    7
> 32    2   5
> 34    2   3    3
> 36    2   5    7   31
>
> The row d=0 and the column k=1 are degenerate and filled with the
> prime 2. All strides d are even. Example for row d=24 and column k=4:
> The 4 numbers 59,59+24,59+2*24 and 59+3*24 are all primes.
>
> http://www.research.att.com/~njas/sequences/Sindx_Pri.html#primes_AP
>
> --Richard
>
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From jvospost3 at gmail.com  Sat Nov  4 19:37:19 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 4 Nov 2006 10:37:19 -0800
Subject: primes in arithmetic progression
In-Reply-To: <5542af940611041033s7a722e9fjf6ad6643f988477d@mail.gmail.com>
References: <200611041720.kA4HK70T027399@amer.strw.leidenuniv.nl>
	 <5542af940611041033s7a722e9fjf6ad6643f988477d@mail.gmail.com>
Message-ID: <5542af940611041037s636536abnf98a7eee7feb349e@mail.gmail.com>

Sorry, I mean: "The semiprime analogue of this is a table that shows the
smallest semiprime S of k (not necessarily consecutive) SEMIprimes in
arithmetic progression
 with common difference d."

On 11/4/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> The semiprime analogue of this is a table that shows the smallest
> semiprime S
> of k (not necessarily consecutive) primes in arithmetic progression
> with common difference d (once a 0 appears, the row has ended nonzero
> values):
>
> d   k+1  k=2  k=3  k=4  k=5  k=6  k=7 k=8
> 0   4      4       4      4       4       4       4 ...
> 1   4      9       33    0
> 2   4      4       91    213   1383 8129 3091 0
> 3   4      6       115
> 4   4      6       6      111
> 5   4      4       77
> 6   4      4
> 7   4      14    51
> 8   4      6      6       69
>
> -- Fairly easy to extend, fairly easy to prove the first 0 in a row.
>
> The same can be done for 3-almost primes, 4-almost primes, and so forth,
> this making a 3-dimensional array.
>
>
>
> On 11/4/06, Richard Mathar <mathar at strw.leidenuniv.nl> wrote:
> >
> >
> > Is there an OEIS table that shows the smallest prime p
> > of k (not necessarily consecutive) primes in arithmetic progression
> > with common difference d? The table would look similar to this one
> > below, and contain rather large primes where I am leaving blanks:
> >
> >     k=1 k=2  k=3  k=4  k=5  k=6
> > d
> > 0     2   2    2    2    2    2
> > 2     2   3    3
> > 4     2   3    3
> > 6     2   5    5    5    5
> > 8     2   3    3
> > 10    2   3    3
> > 12    2   5    5    5    5
> > 14    2   3    3
> > 16    2   3
> > 18    2   5    5    5
> > 20    2   3
> > 22    2   7
> > 24    2   5    5   59
> > 26    2   3
> > 28    2   3    3
> > 30    2   7    7    7     7    7
> > 32    2   5
> > 34    2   3    3
> > 36    2   5    7   31
> >
> > The row d=0 and the column k=1 are degenerate and filled with the
> > prime 2. All strides d are even. Example for row d=24 and column k=4:
> > The 4 numbers 59,59+24,59+2*24 and 59+3*24 are all primes.
> >
> > http://www.research.att.com/~njas/sequences/Sindx_Pri.html#primes_AP<http://www.research.att.com/%7Enjas/sequences/Sindx_Pri.html#primes_AP>
> >
> > --Richard
> >
>
>
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From njas at research.att.com  Sat Nov  4 20:10:10 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sat, 4 Nov 2006 14:10:10 -0500 (EST)
Subject: primes in arithmetic progression
Message-ID: <200611041910.OAA62978@fry.research.att.com>

Just to get the ball rolling, I am adding this entry:

%I A124064
%S A124064 2,2,2,2,3,2,2,3,3,2,2,5,3
%N A124064 Array read by antidiagonals: T(d,k) (k >= 1, d = 2,4,6,8,...) = smallest prime p of k (not necessarily consecutive) primes in arithmetic progression with common difference d.
%C A124064 The row d=0 and the column k=1 are degenerate and are filled with the prime 2.
%O A124064 1,1
%K A124064 nonn,tabl,more
%e A124064 Array begins:
%e A124064 d.\...k=1.k=2..k=3..k=4..k=5..k=6
%e A124064 0..|..2...2....2....2....2....2
%e A124064 2..|..2...3....3
%e A124064 4..|..2...3....3
%e A124064 6..|..2...5....5....5....5
%e A124064 8..|..2...3....3
%e A124064 10.|..2...3....3
%e A124064 12.|..2...5....5....5....5
%e A124064 14.|..2...3....3.
%e A124064 16.|..2...3
%e A124064 18.|..2...5....5....5
%e A124064 20.|..2...3
%e A124064 22.|..2...7
%e A124064 24.|..2...5....5...59
%e A124064 26.|..2...3
%e A124064 28.|..2...3....3
%e A124064 30.|..2...7....7....7.....7....7
%e A124064 32.|..2...5
%e A124064 34.|..2...3....3
%e A124064 36.|..2...5....7...31
%e A124064 Example for row d=24 and column k=4: the 4 numbers 59,59+24,59+2*24 and 59+3*24 are all primes.
%A A124064 Richard Mathar (mathar(AT)strw.leidenuniv.nl), Nov 04 2006




NJAS





From davidwwilson at comcast.net  Sat Nov  4 21:10:45 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sat, 4 Nov 2006 15:10:45 -0500
Subject: Observation on A053669
Message-ID: <000601c7004d$507dc510$6401a8c0@yourxhtr8hvc4p>

It seems to me that

%N A053669 Smallest prime co-prime to n.

could just as easily be described

%N A053669 Smallest number > 1 coprime to n.

and the fact that a(n) is prime falls out.  This definition would also simplify the program.
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From mathar at strw.leidenuniv.nl  Sat Nov  4 21:45:17 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Sat, 4 Nov 2006 21:45:17 +0100
Subject: Observation on A053669
Message-ID: <200611042045.kA4KjHeV011396@amer.strw.leidenuniv.nl>


dww> From seqfan-owner at ext.jussieu.fr  Sat Nov  4 21:12:32 2006
dww> Return-Path: <seqfan-owner at ext.jussieu.fr>
dww> From: "David Wilson" <davidwwilson at comcast.net>
dww> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
dww> Subject: Observation on A053669
dww> ...
dww> It seems to me that
dww> 
dww> %N A053669 Smallest prime co-prime to n.
dww> 
dww> could just as easily be described
dww> 
dww> %N A053669 Smallest number > 1 coprime to n.
dww> 
dww> and the fact that a(n) is prime falls out.  This definition would also =
dww> simplify the program.
dww>...

Yes. If the smallest prime co-prime to n would not be a prime,
it would be built by at least two prime factors, the smaller of which
would be the correct entry (proof by reductio ad absurdum).

It would still be useful to keep in a comment that the numbers are all prime
(this might speed up some searches for the numbers...)

I'd propose to list in a comment the first occurrences of the first primes:
first 5 at a(6)
first 7 is a(30)
first 11 is a(210)
first 13 is a(2310)
first 17 is a(30030)
first 19 is a(510510)
where the indices are 6=2*3, 30=2*3*5, 210=2*3*5*7, 2*3*5*7*11 etc,
(the primorials, I guess, as in A002110)

-- RJM





From jvospost3 at gmail.com  Sat Nov  4 22:55:31 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 4 Nov 2006 13:55:31 -0800
Subject: primes in arithmetic progression
In-Reply-To: <200611041910.OAA62978@fry.research.att.com>
References: <200611041910.OAA62978@fry.research.att.com>
Message-ID: <5542af940611041355vd65aa5cr3a4a9df3cc348c15@mail.gmail.com>

The analogue for 3-almost primes (or, as njas prefers, numbers which are the
product of exactly 3 primes) is left as an exercise for the reader...  I've
just submitted the following.

Subject: PRE-NUMBERED NEW SEQUENCE A124570 FROM Jonathan Vos Post


%I A124570
%S A124570 4, 4, 4, 4, 9, 4, 4, 4, 33, 4, 4, 6, 91, 0, 4, 4, 6, 115,
213, 0, 4, 4, 4, 6, 0, 1383, 0, 4, 4, 4, 77, 111, 0, 8129, 0, 4, 4, 14
%N A124570 Array read by antidiagonals: T(d,k) (k >= 1, d =
1,2,3,4,5,6,...) = smallest semiprime s of k (not necessarily consecutive)
semiprimes in arithmetic progression with common difference d.
%C A124570 Semiprime analogue of A124064 Array read by antidiagonals:
T(d,k) (k >= 1, d = 2,4,6,8,...) = smallest prime p of k (not
necessarily consecutive) primes in arithmetic progression with common
difference d. The row d=0 and the column k=1 are degenerate and are filled with
the semiprime 4. Once a 0 occurs in a row, there are no more nonzero
vales in the row. For example, There cannot be 4 semiprimes in arithmetic
progression with common difference 3, starting with k, because modulo 4
we have {k, k+3, k+6, k+9} == {k+0, k+3, k+2, k+1} and one of these
must be divisible by 4, hence a nonsemiprime (eliminating k = 4 by
inspection).
%e A124570 Array begins:
d.\...k=1.k=2..k=3..k=4..k=5..k=6..k=7..k=8
0..|..4...4....4....4....4....4....4....4
1..|..4...9....33...0....0....0....0....0
2..|..4...4....91...213..1383.8129.3091.0
3..|..4...6....115..0....0....0....0....0
4..|..4...6....6....6....111...
5..|..4...4....77...
6..|..4...4....
7..|..4...14...51...
8..|..4...6....6....69.
Example for row 3: 115 = 5 * 23 is semiprime, 115+3 = 118 = 2 * 59 is
semiprime, and 115+3+3 = 121 = 11^2 is semiprime, so T(3,3) = 115.
%Y A124570 Cf. A000040, A001358, A056809, A070552, A092125, A092126,
A092127, A092128, A092129, A124064.
%O A124570 1,1
%K A124570 ,easy,more,nonn,tabl,
%A A124570 Jonathan Vos Post (jvospost2 at yahoo.com
<http://us.f551.mail.yahoo.com/ym/Compose?To=jvospost2 at yahoo.com&YY=99063&y5beta=yes&y5beta=yes&order=down&sort=date&pos=0&view=a&head=b>),
Nov 04 2006
RH
RA 192.20.225.32
RU
RI



On 11/4/06, N. J. A. Sloane <njas at research.att.com> wrote:
>
> Just to get the ball rolling, I am adding this entry:
>
> %I A124064
> %S A124064 2,2,2,2,3,2,2,3,3,2,2,5,3
> %N A124064 Array read by antidiagonals: T(d,k) (k >= 1, d = 2,4,6,8,...) =
> smallest prime p of k (not necessarily consecutive) primes in arithmetic
> progression with common difference d.
> %C A124064 The row d=0 and the column k=1 are degenerate and are filled
> with the prime 2.
> %O A124064 1,1
> %K A124064 nonn,tabl,more
> %e A124064 Array begins:
> %e A124064 d.\...k=1.k=2..k=3..k=4..k=5..k=6
> %e A124064 0..|..2...2....2....2....2....2
> %e A124064 2..|..2...3....3
> %e A124064 4..|..2...3....3
> %e A124064 6..|..2...5....5....5....5
> %e A124064 8..|..2...3....3
> %e A124064 10.|..2...3....3
> %e A124064 12.|..2...5....5....5....5
> %e A124064 14.|..2...3....3.
> %e A124064 16.|..2...3
> %e A124064 18.|..2...5....5....5
> %e A124064 20.|..2...3
> %e A124064 22.|..2...7
> %e A124064 24.|..2...5....5...59
> %e A124064 26.|..2...3
> %e A124064 28.|..2...3....3
> %e A124064 30.|..2...7....7....7.....7....7
> %e A124064 32.|..2...5
> %e A124064 34.|..2...3....3
> %e A124064 36.|..2...5....7...31
> %e A124064 Example for row d=24 and column k=4: the 4 numbers
> 59,59+24,59+2*24 and 59+3*24 are all primes.
> %A A124064 Richard Mathar (mathar(AT)strw.leidenuniv.nl), Nov 04 2006
>
>
>
>
> NJAS
>
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From jrbibers at indiana.edu  Sat Nov  4 23:17:45 2006
From: jrbibers at indiana.edu (Joseph Biberstine)
Date: Sat, 04 Nov 2006 17:17:45 -0500
Subject: A124015 thoughts
In-Reply-To: <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
Message-ID: <454D1189.8090102@indiana.edu>

	For anyone interested, this Mathematica function, eng[], will (as
usual, barring bugs) give the American English name for integers on 0 to
10^21-1.  Extension of the upper bound should be plain to implement
(just one line of code -- observe the trend).

	All printing is lowercase and without hyphens.  In holding that "and"
should only ever be read for radix points, it is never printed here.
Note this implementation uses a text postprocessor and so is inherently
ugly.  Also note that the helper memoizes for better or worse.  Please,
of course, report all bugs.

	Attached are some diagrams of the predictable distribution of
StringLength[eng[n]] (note this includes spaces).  The third ranges up
to 10^9, sampling every 10^6.

	Here is sample output from random naturals on 0,999999:

 40573, forty thousand five hundred seventy three
108097, one hundred eight thousand ninety seven
539799, five hundred thirty nine thousand seven hundred ninety nine
136868, one hundred thirty six thousand eight hundred sixty eight
597841, five hundred ninety seven thousand eight hundred forty one
272460, two hundred seventy two thousand four hundred sixty
135550, one hundred thirty five thousand five hundred fifty
402424, four hundred two thousand four hundred twenty four
543617, five hundred forty three thousand six hundred seventeen
868945, eight hundred sixty eight thousand nine hundred forty five
951779, nine hundred fifty one thousand seven hundred seventy nine
742845, seven hundred forty two thousand eight hundred forty five
 84140, eighty four thousand one hundred forty
680145, six hundred eighty thousand one hundred forty five
 82734, eighty two thousand seven hundred thirty four
154758, one hundred fifty four thousand seven hundred fifty eight
961084, nine hundred sixty one thousand eighty four
360944, three hundred sixty thousand nine hundred forty four
574416, five hundred seventy four thousand four hundred sixteen
164099, one hundred sixty four thousand ninety nine

	Here is the code:

Clear[eng, engRaw, n];
eng[n_] := (
  pre = engRaw[n];
  If[pre == "", pre = "zero"]; (* fixes "zero" *)
  pre = FixedPoint[StringReplace[#, "  " -> " "] &, pre] ; (* no
consecutive spaces *)
  pre = FixedPoint[StringReplace[#, RegularExpression[" $"] :> ""] &,
pre] ; (* no trailing spaces *)
  pre);
(* This raw function relies heavily on the postprocessor, which fixes
"zero" and deals with whitespace *)
engRaw[n_] := engRaw[n] = (* memo *)
  Which[
    n == 0, "",(* fixed in postprocessing *)
    n == 1, "one",
    n == 2, "two",
    n == 3, "three",
    n == 4, "four",
    n == 5, "five",
    n == 6, "six",
    n == 7, "seven",
    n == 8, "eight",
    n == 9, "nine",
    n == 10, "ten",
    n == 11, "eleven",
    n == 12, "twelve",
    n == 13, "thirteen",
    n == 14, "fourteen",
    n == 15, "fifteen",
    n == 16, "sixteen",
    n == 17, "seventeen",
    n == 18, "eighteen",
    n == 19, "nineteen",
    20 <= n <= 29, "twenty " <> engRaw[Mod[n, 10]],
    30 <= n <= 39, "thirty " <> engRaw[Mod[n, 10]],
    40 <= n <= 49, "forty " <> engRaw[Mod[n, 10]],
    50 <= n <= 59, "fifty " <> engRaw[Mod[n, 10]],
    60 <= n <= 69, "sixty " <> engRaw[Mod[n, 10]],
    70 <= n <= 79, "seventy " <> engRaw[Mod[n, 10]],
    80 <= n <= 89, "eighty " <> engRaw[Mod[n, 10]],
    90 <= n <= 99, "ninety " <> engRaw[Mod[n, 10]],
    10^2 <= n <= 10^3 - 1, engRaw[Quotient[n, 10^2]] <> " hundred " <>
engRaw[Mod[n, 10^2]],
    10^3 <= n <= 10^6 - 1, engRaw[Quotient[n, 10^3]] <> " thousand " <>
engRaw[Mod[n, 10^3]],
    10^6 <= n <= 10^9 - 1, engRaw[Quotient[n, 10^6]] <> " million " <>
engRaw[Mod[n, 10^6]],
    10^9 <= n <= 10^12 - 1, engRaw[Quotient[n, 10^9]] <> " billion " <>
engRaw[Mod[n, 10^9]],
    10^12 <= n <= 10^15 - 1, engRaw[Quotient[n, 10^12]] <> " trillion "
<> engRaw[Mod[n, 10^12]],
    10^15 <= n <= 10^18 - 1, engRaw[Quotient[n, 10^15]] <> " quadrillion
" <> engRaw[Mod[n, 10^15]],
    10^18 <= n <= 10^21 - 1, engRaw[Quotient[n, 10^18]] <> " quintillion
" <> engRaw[Mod[n, 10^18]],
    True, Return["Naturals beyond 10^21-1 are not supported"];];

-JRB

Jonathan Post wrote:
> Now that I think of it, logarithmic trends can be interesting.  How
> about "what is the distribution of the number of letters in the names of
> numbers from one to one billion?"
> 
> -- Jonathan Vos Post
> 
> On 11/3/06, *David Wilson* <davidwwilson at comcast.net
> <mailto:davidwwilson at comcast.net>> wrote:
> 
>     The description of A124015 is
>      
>     %N A124015 Number of words with n letters in the National Scrabble
>     Association Dictionary.
>      
>     Some notes:
>      
>     - It should be noted that the NSAD includes a restricted set of
>     English words (words of 2 to 14 letters, no proper nouns or
>     derivatives, no words with non-alphabetic characters (e.g,
>     contractions, hyphenated words), and it is this restricted set that
>     is being counted.
>      
>     - The edition of the NSAD used to create A124015 should be
>     specified, since the NSAD is continually edited.  While I do not
>     think that A124015 should change with each new edition of the NSAD,
>     because it may be referenced in other literature. On the other hand,
>     I don't think that a new sequence should be created for each new
>     edition of the NSAD unless the new edition exhibits some interesting
>     statistical departure from the current NSAD (e.g, in some future
>     NSAD, the number of 9-letter words exceeds the number of 8-letter
>     words). This is because I believe that the value of A124015 is
>     mainly to indicate a distribution of English word lengths.
>      
>     - On the other hand, I have always understood that 4-letter words
>     are most common in literature (where small words appear more often).
>     It might be interesting to create a sequence counting words of
>     length n in the King James Bible, War and Peace, or some other
>     suitably large and stable piece of English literature (we don't have
>     to go overboard on this, just one or two examples indicating word
>     length distributions in literature).
>      
>     - Another interesting idea: In various languages, what is the
>     distribution of the number of letters in the names of numbers from
>     one to (say) one million?
>      
>      
> 
> 
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From jvospost3 at gmail.com  Sun Nov  5 02:05:28 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 4 Nov 2006 17:05:28 -0800
Subject: A124015 thoughts
In-Reply-To: <454D1189.8090102@indiana.edu>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p>
	 <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
	 <454D1189.8090102@indiana.edu>
Message-ID: <5542af940611041705u66643b27u66c836a95a2b9a1@mail.gmail.com>

Nice code.  And what fascinating graphs! They have both periodicities and a
fractal appearance.  Somehwere in there is the equivalent of Benford's Law,
and Zipf's Law, on lengths names of numbers (as opposed to digit
distribution and power law).

On 11/4/06, Joseph Biberstine <jrbibers at indiana.edu> wrote:
>
>         For anyone interested, this Mathematica function, eng[], will (as
> usual, barring bugs) give the American English name for integers on 0 to
> 10^21-1.  Extension of the upper bound should be plain to implement
> (just one line of code -- observe the trend).
>
>         All printing is lowercase and without hyphens.  In holding that
> "and"
> should only ever be read for radix points, it is never printed here.
> Note this implementation uses a text postprocessor and so is inherently
> ugly.  Also note that the helper memoizes for better or worse.  Please,
> of course, report all bugs.
>
>         Attached are some diagrams of the predictable distribution of
> StringLength[eng[n]] (note this includes spaces).  The third ranges up
> to 10^9, sampling every 10^6.
>
>         Here is sample output from random naturals on 0,999999:
>
> 40573, forty thousand five hundred seventy three
> 108097, one hundred eight thousand ninety seven
> 539799, five hundred thirty nine thousand seven hundred ninety nine
> 136868, one hundred thirty six thousand eight hundred sixty eight
> 597841, five hundred ninety seven thousand eight hundred forty one
> 272460, two hundred seventy two thousand four hundred sixty
> 135550, one hundred thirty five thousand five hundred fifty
> 402424, four hundred two thousand four hundred twenty four
> 543617, five hundred forty three thousand six hundred seventeen
> 868945, eight hundred sixty eight thousand nine hundred forty five
> 951779, nine hundred fifty one thousand seven hundred seventy nine
> 742845, seven hundred forty two thousand eight hundred forty five
> 84140, eighty four thousand one hundred forty
> 680145, six hundred eighty thousand one hundred forty five
> 82734, eighty two thousand seven hundred thirty four
> 154758, one hundred fifty four thousand seven hundred fifty eight
> 961084, nine hundred sixty one thousand eighty four
> 360944, three hundred sixty thousand nine hundred forty four
> 574416, five hundred seventy four thousand four hundred sixteen
> 164099, one hundred sixty four thousand ninety nine
>
>         Here is the code:
>
> Clear[eng, engRaw, n];
> eng[n_] := (
>   pre = engRaw[n];
>   If[pre == "", pre = "zero"]; (* fixes "zero" *)
>   pre = FixedPoint[StringReplace[#, "  " -> " "] &, pre] ; (* no
> consecutive spaces *)
>   pre = FixedPoint[StringReplace[#, RegularExpression[" $"] :> ""] &,
> pre] ; (* no trailing spaces *)
>   pre);
> (* This raw function relies heavily on the postprocessor, which fixes
> "zero" and deals with whitespace *)
> engRaw[n_] := engRaw[n] = (* memo *)
>   Which[
>     n == 0, "",(* fixed in postprocessing *)
>     n == 1, "one",
>     n == 2, "two",
>     n == 3, "three",
>     n == 4, "four",
>     n == 5, "five",
>     n == 6, "six",
>     n == 7, "seven",
>     n == 8, "eight",
>     n == 9, "nine",
>     n == 10, "ten",
>     n == 11, "eleven",
>     n == 12, "twelve",
>     n == 13, "thirteen",
>     n == 14, "fourteen",
>     n == 15, "fifteen",
>     n == 16, "sixteen",
>     n == 17, "seventeen",
>     n == 18, "eighteen",
>     n == 19, "nineteen",
>     20 <= n <= 29, "twenty " <> engRaw[Mod[n, 10]],
>     30 <= n <= 39, "thirty " <> engRaw[Mod[n, 10]],
>     40 <= n <= 49, "forty " <> engRaw[Mod[n, 10]],
>     50 <= n <= 59, "fifty " <> engRaw[Mod[n, 10]],
>     60 <= n <= 69, "sixty " <> engRaw[Mod[n, 10]],
>     70 <= n <= 79, "seventy " <> engRaw[Mod[n, 10]],
>     80 <= n <= 89, "eighty " <> engRaw[Mod[n, 10]],
>     90 <= n <= 99, "ninety " <> engRaw[Mod[n, 10]],
>     10^2 <= n <= 10^3 - 1, engRaw[Quotient[n, 10^2]] <> " hundred " <>
> engRaw[Mod[n, 10^2]],
>     10^3 <= n <= 10^6 - 1, engRaw[Quotient[n, 10^3]] <> " thousand " <>
> engRaw[Mod[n, 10^3]],
>     10^6 <= n <= 10^9 - 1, engRaw[Quotient[n, 10^6]] <> " million " <>
> engRaw[Mod[n, 10^6]],
>     10^9 <= n <= 10^12 - 1, engRaw[Quotient[n, 10^9]] <> " billion " <>
> engRaw[Mod[n, 10^9]],
>     10^12 <= n <= 10^15 - 1, engRaw[Quotient[n, 10^12]] <> " trillion "
> <> engRaw[Mod[n, 10^12]],
>     10^15 <= n <= 10^18 - 1, engRaw[Quotient[n, 10^15]] <> " quadrillion
> " <> engRaw[Mod[n, 10^15]],
>     10^18 <= n <= 10^21 - 1, engRaw[Quotient[n, 10^18]] <> " quintillion
> " <> engRaw[Mod[n, 10^18]],
>     True, Return["Naturals beyond 10^21-1 are not supported"];];
>
> -JRB
>
> Jonathan Post wrote:
> > Now that I think of it, logarithmic trends can be interesting.  How
> > about "what is the distribution of the number of letters in the names of
> > numbers from one to one billion?"
> >
> > -- Jonathan Vos Post
> >
> > On 11/3/06, *David Wilson* <davidwwilson at comcast.net
> > <mailto:davidwwilson at comcast.net>> wrote:
> >
> >     The description of A124015 is
> >
> >     %N A124015 Number of words with n letters in the National Scrabble
> >     Association Dictionary.
> >
> >     Some notes:
> >
> >     - It should be noted that the NSAD includes a restricted set of
> >     English words (words of 2 to 14 letters, no proper nouns or
> >     derivatives, no words with non-alphabetic characters (e.g,
> >     contractions, hyphenated words), and it is this restricted set that
> >     is being counted.
> >
> >     - The edition of the NSAD used to create A124015 should be
> >     specified, since the NSAD is continually edited.  While I do not
> >     think that A124015 should change with each new edition of the NSAD,
> >     because it may be referenced in other literature. On the other hand,
> >     I don't think that a new sequence should be created for each new
> >     edition of the NSAD unless the new edition exhibits some interesting
> >     statistical departure from the current NSAD (e.g, in some future
> >     NSAD, the number of 9-letter words exceeds the number of 8-letter
> >     words). This is because I believe that the value of A124015 is
> >     mainly to indicate a distribution of English word lengths.
> >
> >     - On the other hand, I have always understood that 4-letter words
> >     are most common in literature (where small words appear more often).
> >     It might be interesting to create a sequence counting words of
> >     length n in the King James Bible, War and Peace, or some other
> >     suitably large and stable piece of English literature (we don't have
> >     to go overboard on this, just one or two examples indicating word
> >     length distributions in literature).
> >
> >     - Another interesting idea: In various languages, what is the
> >     distribution of the number of letters in the names of numbers from
> >     one to (say) one million?
> >
> >
> >
> >
>
>
>
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From pxp at rogers.com  Sun Nov  5 02:29:38 2006
From: pxp at rogers.com (Hans Havermann)
Date: Sat, 4 Nov 2006 20:29:38 -0500
Subject: A124015 thoughts
In-Reply-To: <454D1189.8090102@indiana.edu>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com> <454D1189.8090102@indiana.edu>
Message-ID: <3E87B5F0-736C-466D-B4EC-C2D438AA1BA3@rogers.com>

Joseph Biberstine:

> 	For anyone interested, this Mathematica function, eng[], will (as  
> usual, barring bugs) give the American English name for integers on  
> 0 to 10^21-1.  Extension of the upper bound should be plain to  
> implement (just one line of code -- observe the trend).

I forget where I found the following bit of elegant code, but it  
essentially does the same thing as yours (except use a hyphen instead  
of a space between the tens and units for tens > 20):


Ones[n_, pad_:0] := If[Mod[pad, 10] > 0, "-", ""] <> {"", "one",  
"two", "three", "four", "five", "six", "seven", "eight", "nine",  
"ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen",  
"sixteen", "seventeen", "eighteen", "nineteen"}[[n + 1]]

Tens[n_] := {"", "twenty", "thirty", "forty", "fifty", "sixty",  
"seventy", "eighty", "ninety"}[[n]]

LastThree[n_] := LastThree[n] = Module[{h = Quotient[n, 100], m = Mod 
[n, 100]}, If[n > 99, Ones[h, 0] <> " hundred ", ""] <> If[m < 20,  
Ones[m], Tens[Quotient[m, 10]] <> Ones[Mod[m, 10], m]]]

Powers[n_] := Powers[n] = {"", " thousand", " million", " billion", "  
trillion", " quadrillion", " quintillion", " sextillion", "  
septillion", " octillion", "nonillion", " decillion", " undecillion",  
" duodecillion", " tredecillion", " quattuordecillion", "  
quindecillion", " sexdecillion", " septendecillion", "  
octodecillion", " novemdecillion", " vigintillion"}[[n]]

(*NumberName[0] := "zero"*)

NumberName[n_Integer] := Module[{db = Reverse[MapIndexed[If[#1 == 0,  
"", LastThree[#1] <> Powers[#2[[1]]] <> " "] &, FromDigits[#, 10] & / 
@ Reverse /@ Partition[Reverse[IntegerDigits[n, 10]], 3, 3, {1, 1},  
{}]]]}, StringDrop[StringJoin @@ db, -1]]

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From zbi74583 at boat.zero.ad.jp  Sun Nov  5 04:58:03 2006
From: zbi74583 at boat.zero.ad.jp (koh)
Date: Sun, 05 Nov 2006 12:58:03 +0900
Subject: terms lack
Message-ID: <20061105035802.zbi74583@boat.zero.ad.jp>

    Recently I updated more terms of A054572, which are calculated by Richard Mathar.
    And Neil claimed or got angry.
    He said, "You are red card. Don't submit a wrong sequences"
    I am exaggerating a little.
    These words are not exactly what he wrote.



    To Neil
    They are correct, not "wrong". 
    It is only not complete.

    In early time, I didn't do a computer search, so there is a possibility that my old submitted sequences are not complete.
    Now, I calculate both theoretically and with a computer.



    To Seqfans
    I defined many "divisor" functions and calculated many Amicable Numbers and Perfect Numbers using those functions.
    And I submitted them as sequences to  OEIS.

        http://www.research.att.com/~njas/sequences/?q=kohmoto+amicable&sort=0&fmt=0&language=english&go=Search        
        
        http://www.research.att.com/~njas/sequences/?q=kohmoto+perfect&sort=0&fmt=0&language=english&go=Search
   
    I suppose that they are almost complete but some of them might not be complete.

    I wish some one do a computer search about them.
    I think that A054572 is the worst case, because I gave only two terms up to 10^4.
    But sometimes my theory gives a complete set of examples.
    See A099105  , I was sure that these three terms are all, and a computer search also said it is right.

    Yasutoshi
    





From deutsch at duke.poly.edu  Sun Nov  5 05:30:30 2006
From: deutsch at duke.poly.edu (Emeric Deutsch)
Date: Sat, 4 Nov 2006 23:30:30 -0500 (EST)
Subject: No subject
Message-ID: <Pine.GSO.4.62.0611041740160.21855@duke.poly.edu>

Dear Seqfans,
We (Lou Shapiro and myself) would like to submit a few
sequences similar to A124497 (shown at the bottom of this
message). They will refer to 1-2-3-4 trees, 1-2-3-4-5
trees, etc. rather than to 1-2-3 trees. We can describe
the corresponding g.f.'s but we need help in finding the
terms themselves.

Here is the description of all the g.f.'s. Let M[k](z) be
the g.f. of the 1-2-...-k trees with thinning limbs. I will
describe these functions; consequently, we may disregard
the objects counted by them.

Denote by C[k](z) the g.f. of the k-ary trees. C[k]
satisfies C[k]=1+zC[k]^k and the coefficient of z^n
in its series expansion is
(*)      binom(kn,n)/[(k-1)n+1].

I will write Mk for M[k] and Ck for C[k].

Obviously, M1(z)=1/(1-z) (path-trees). The next one is 
M2(z)=M1(z)C2(z^2*M1), i.e. M1 times C2 evaluated at
z^2*M1. This gives no problem since
C2(z)=[1-sqrt(1-4z)]/(2z). We obtain A090344.

Next M3(z)=M2(z)C3(z^3*M2^2); since there is an
explicit expression for C3(z), namely
C3(z)=2/sqrt(3z)*sin[(1/3)*arcsin(sqrt(27z/4))],
this yields easily A124497.

 From here on we need help.
M4(z)=M3(z)C4(z^4*M3^3); 
Most probably sequence starts with
1,1,2,4,10,24,62,160,425,1140,3105 (I used some terms
of the series of C4(z); see (*)).

M5(z)=M4(z)C5(z^5*M4^4);
Most probably sequence starts with
1,1,2,4,10,25,67,180,495,1375,3871 (I used some terms
of the series of C5(z); see (*)).

And so on.

These sequences approach A124344:
1,1,2,4,10,25,68,187,530,1523,4447,13121,39107,...

Many thanks.
Emeric

--------------------------------
A124497  Number of 1-2-3 trees with n edges and with thinning limbs. A 
1-2-3 tree is an ordered tree with vertices of outdegree at most 3. A 
rooted tree with thinning limbs is such that if a node has k children, all 
its children have at most k children.

1, 1, 2, 4, 9, 20, 48, 116, 288, 724, 1849, 4768, 12423, 32628, 86342, 
229952, 616042, 1659012, 4489101, 12199521, 33284546, 91140797, 250396629, 
690043032, 1907022197, 5284167884, 14677681554, 40862469713, 114001697975 
(list; graph; listen)
OFFSET  0,3
FORMULA  G.f.=H*T(H^2*z^3), where 
T=2/sqrt(3*x)*sin((1/3)*arcsin(sqrt(27*x/4))) (solution of T=1+zT^3, 
T(0)=1), H=C(z^2/(1-z))/(1-z), and C(x)=[1-sqrt(1-4x)]/(2x) is the Catalan 
function. More generally, if M[k](z) is the g.f. of the 1-2-...-k trees 
with thinning limbs and C[k](z)=1+z*{C[k](z)}^k is the g.f. of the k-ary 
trees, then M[k](z)=M[k-1](z)C[k](M[k-1]^(k-1)*z^k).
MAPLE  C:=x->(1-sqrt(1-4*x))/2/x: 
T:=x->2/sqrt(3*x)*sin((1/3)*arcsin(sqrt(27*x/4))): M2:=C(z^2/(1-z))/(1-z): 
G:=M2*T(M2^2*z^3): Gser:=series(G, z=0, 40): seq(coeff(Gser, z, n), 
n=0..33);
CROSSREFS  Cf. A090344, A124344.
KEYWORD  nonn,new
AUTHOR  Emeric Deutsch and Louis Shapiro (deutsch(AT)duke.poly.edu, 
lshapiro(AT)Howard.edu), Nov 04 2006





From zakseidov at yahoo.com  Sun Nov  5 07:28:49 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sat, 4 Nov 2006 22:28:49 -0800 (PST)
Subject: Observation on A053669
In-Reply-To: <200611042045.kA4KjHeV011396@amer.strw.leidenuniv.nl>
Message-ID: <20061105062849.51615.qmail@web38208.mail.mud.yahoo.com>

Yes, 
and because a(n_odd)=2
keep only even n's.
Zak

--- Richard Mathar <mathar at strw.leidenuniv.nl> wrote:

> 
> dww> From seqfan-owner at ext.jussieu.fr  Sat Nov  4
> 21:12:32 2006
> dww> Return-Path: <seqfan-owner at ext.jussieu.fr>
> dww> From: "David Wilson" <davidwwilson at comcast.net>
> dww> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
> dww> Subject: Observation on A053669
> dww> ...
> dww> It seems to me that
> dww> 
> dww> %N A053669 Smallest prime co-prime to n.
> dww> 
> dww> could just as easily be described
> dww> 
> dww> %N A053669 Smallest number > 1 coprime to n.
> dww> 
> dww> and the fact that a(n) is prime falls out. 
> This definition would also =
> dww> simplify the program.
> dww>...
> 
> Yes. If the smallest prime co-prime to n would not
> be a prime,
> it would be built by at least two prime factors, the
> smaller of which
> would be the correct entry (proof by reductio ad
> absurdum).
> 
> It would still be useful to keep in a comment that
> the numbers are all prime
> (this might speed up some searches for the
> numbers...)
> 
> I'd propose to list in a comment the first
> occurrences of the first primes:
> first 5 at a(6)
> first 7 is a(30)
> first 11 is a(210)
> first 13 is a(2310)
> first 17 is a(30030)
> first 19 is a(510510)
> where the indices are 6=2*3, 30=2*3*5, 210=2*3*5*7,
> 2*3*5*7*11 etc,
> (the primorials, I guess, as in A002110)
> 
> -- RJM
> 



 
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From davidwwilson at comcast.net  Sun Nov  5 07:38:34 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sun, 5 Nov 2006 01:38:34 -0500
Subject: Notes on A124064 (was Re: primes in arithmetic progression)
References: <200611041910.OAA62978@fry.research.att.com>
Message-ID: <003101c700a5$05184930$6401a8c0@yourxhtr8hvc4p>

This is one of those cases where I think we have jumped the gun, creating a 
sequence without first determining the best structure for that sequence.

Richard Mather proposed a table of T(d,k) = the smallest possible initial 
element of a nondecreasing AP of k primes with difference d. Clearly, T(d,k) 
is defined only for d >= 0 and k >= 1.

For d = 0, it is pretty easy to show T(0, k) = 2 for any k >= 1. For d >= 1, 
however, T(d,k) is defined only for a finite number of k.

Proof: Suppose there exists AP P of primes with difference d >= 1 and 2d+2 
elements. Let c = d+1. Because P is an arithmetic progression of 2c elements 
with difference d, and c is coprime to d, P will have two elements divisible 
by c, call them cx and cy. Whereas P is an increasing sequence of positive 
integers, we can assume that 0 < cx < cy. Hence 0 < x < y, and y >= 2. But 
this makes cy composite, contrary to fact that P is a sequence of primes. 
Hence P does not exist, that is to say, any AP of primes with difference d 
 >= 1 has at most 2d+1 elements, and T(d,k) is undefined for k > 2d+1. QED.

In fact, assuming the k-tuple conjecture is true, we can give the precise 
domain of T for d >= 1: If p is the smallest prime not dividing d, then 
T(d,k) is defined precisely for 1 <= k <= p if the AP of p elements with 
initial element p and difference d is a prime AP, and for 1 <= k <= p-1 
otherwise.

This means that the table of T(d,k) looks as follows:

  \k|  1  2  3  4  5  6  7  8  9 10
  d\|
----+------------------------------
  0 |  2  2  2  2  2  2  2  2  2  2 ...
  1 |  2  2
  2 |  2  3  3
  3 |  2  2
  4 |  2  3  3
  5 |  2  2
  6 |  2  5  5  5  5
  7 |  2
  8 |  2  3  3
  9 |  2  2
 10 |  2  3  3
 11 |  2  2
 12 |  2  5  5  5  5
 13 |  2
 14 |  2  3  3
 15 |  2  2
 16 |  2  3
 17 |  2  2
 18 |  2  5  5  5
 19 |  2
 20 |  2  3  3
 ...

Two important points:

- Sometimes in the OEIS, we are tempted to pare down sequences to remove 
uninteresting values in order to emphasize the interesting values. In such 
cases, I think that the OEIS should include a main sequence that is 
complete, and additional pared-down sequences if they seem to augment our 
understanding of the main sequence. In this case, Mather's sequence included 
T(d,k) only for even d, presumably because T(d,k) is uninteresting for odd 
d. My table above includes T(d,k) for both even and odd d. In fact, it turns 
out that for odd d, T(d,1) = 2 and T(d,2) = 2 if 2+d is prime, not 
scintillating behavior, but not trivial either. My feeling is that the main 
sequence for T(d,k) (presumably A124064) should include all the values of 
T(d,k), and if restricting T(d,k) to even d seems to augment our 
understanding, a second pared-down sequence should be made referring back to 
the complete main sequence.

- For d >= 1 and sufficient k, T(d,k) is undefined. The standard 
diagonalization technique won't work for this irregular table. That is why 
NJAS's proposed A124064 ends where it does, because the next value, T(2,4), 
does not exist (it is not a matter of doing more computation, T(2,4) does 
not exist because there is no AP of 4 primes with difference 2). I can see 
two possible ways to address this problem, the (icky) option of making 
T(d,k) = 0 if undefined, or the (less icky) option of requiring the APs in 
question to be strictly increasing. This would dispense with the infinite d 
= 0 line and leave us with a list of finite sequences which we could 
concatenate to build A124064, vis:

2,2,2,3,3,2,2,2,3,3,2,2,2,5,5,5,5,2,2,3,3,2,2,2,3,3,2,2,2,5,5,5,5,2,2,
3,3,2,2,2,3,2,2,2,5,5,5,2,2,3,3,2,2,2,7,2,2,5,5,59,2,2,3,2,2,2,3,3,2,2,
2,7,7,7,7,7,2,2,5,2,2,3,3,2,2,2,5,7,31,2,2,3,3,2,2,2,3,3,2,2,2,5,5,5,5

We could put the line lengths in another sequence.

2,3,2,3,2,5,1,3,2,3,2,5,1,3,2,2,2,4,1,3,2,2,1,4,1,2,2,3,2,6,1,2,1,3,2,4,
1,3,2,3,2,5,1,2,2,2,1,5,1,3,2,2,1,4,1,2,2,2,2,6,1,2,1,3,2,4,1,3,2,2,2,4,
1,2,1,2,2,4,1,3,2,2,1,4,1,2,2,2,1,6,1,2,1,3,2,5,1,3,2,2,2,4,1,3,2,2,2,4

----- Original Message ----- 
From: "N. J. A. Sloane" <njas at research.att.com>
To: <seqfan at ext.jussieu.fr>; "Richard Mathar" <mathar at strw.leidenuniv.nl>
Sent: Saturday, November 04, 2006 2:10 PM
Subject: Re: primes in arithmetic progression


> Just to get the ball rolling, I am adding this entry:
>
> %I A124064
> %S A124064 2,2,2,2,3,2,2,3,3,2,2,5,3
> %N A124064 Array read by antidiagonals: T(d,k) (k >= 1, d = 2,4,6,8,...) = 
> smallest prime p of k (not necessarily consecutive) primes in arithmetic 
> progression with common difference d.
> %C A124064 The row d=0 and the column k=1 are degenerate and are filled 
> with the prime 2.
> %O A124064 1,1
> %K A124064 nonn,tabl,more
> %e A124064 Array begins:
> %e A124064 d.\...k=1.k=2..k=3..k=4..k=5..k=6
> %e A124064 0..|..2...2....2....2....2....2
> %e A124064 2..|..2...3....3
> %e A124064 4..|..2...3....3
> %e A124064 6..|..2...5....5....5....5
> %e A124064 8..|..2...3....3
> %e A124064 10.|..2...3....3
> %e A124064 12.|..2...5....5....5....5
> %e A124064 14.|..2...3....3.
> %e A124064 16.|..2...3
> %e A124064 18.|..2...5....5....5
> %e A124064 20.|..2...3
> %e A124064 22.|..2...7
> %e A124064 24.|..2...5....5...59
> %e A124064 26.|..2...3
> %e A124064 28.|..2...3....3
> %e A124064 30.|..2...7....7....7.....7....7
> %e A124064 32.|..2...5
> %e A124064 34.|..2...3....3
> %e A124064 36.|..2...5....7...31
> %e A124064 Example for row d=24 and column k=4: the 4 numbers 
> 59,59+24,59+2*24 and 59+3*24 are all primes.
> %A A124064 Richard Mathar (mathar(AT)strw.leidenuniv.nl), Nov 04 2006
>
>
>
>
> NJAS
>
>
> -- 
> No virus found in this incoming message.
> Checked by AVG Free Edition.
> Version: 7.5.430 / Virus Database: 268.13.27/517 - Release Date: 11/3/2006 
> 4:30 PM
> 






From davidwwilson at comcast.net  Sun Nov  5 07:40:23 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sun, 5 Nov 2006 01:40:23 -0500
Subject: Observation on A053669
References: <200611042045.kA4KjHeV011396@amer.strw.leidenuniv.nl>
Message-ID: <003b01c700a5$45a9d3b0$6401a8c0@yourxhtr8hvc4p>

Another possible description of A053669:

Smallest prime non-divisor of n.

----- Original Message ----- 
From: "Richard Mathar" <mathar at strw.leidenuniv.nl>
To: <seqfan at ext.jussieu.fr>
Sent: Saturday, November 04, 2006 3:45 PM
Subject: Re: Observation on A053669


>
> dww> From seqfan-owner at ext.jussieu.fr  Sat Nov  4 21:12:32 2006
> dww> Return-Path: <seqfan-owner at ext.jussieu.fr>
> dww> From: "David Wilson" <davidwwilson at comcast.net>
> dww> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
> dww> Subject: Observation on A053669
> dww> ...
> dww> It seems to me that
> dww>
> dww> %N A053669 Smallest prime co-prime to n.
> dww>
> dww> could just as easily be described
> dww>
> dww> %N A053669 Smallest number > 1 coprime to n.
> dww>
> dww> and the fact that a(n) is prime falls out.  This definition would 
> also =
> dww> simplify the program.
> dww>...
>
> Yes. If the smallest prime co-prime to n would not be a prime,
> it would be built by at least two prime factors, the smaller of which
> would be the correct entry (proof by reductio ad absurdum).
>
> It would still be useful to keep in a comment that the numbers are all 
> prime
> (this might speed up some searches for the numbers...)
>
> I'd propose to list in a comment the first occurrences of the first 
> primes:
> first 5 at a(6)
> first 7 is a(30)
> first 11 is a(210)
> first 13 is a(2310)
> first 17 is a(30030)
> first 19 is a(510510)
> where the indices are 6=2*3, 30=2*3*5, 210=2*3*5*7, 2*3*5*7*11 etc,
> (the primorials, I guess, as in A002110)
>
> -- RJM
>
>
> -- 
> No virus found in this incoming message.
> Checked by AVG Free Edition.
> Version: 7.5.430 / Virus Database: 268.13.27/517 - Release Date: 11/3/2006 
> 4:30 PM
>
> 






From pauldhanna at juno.com  Sun Nov  5 07:49:40 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Sun, 5 Nov 2006 01:49:40 -0500
Subject: A124497: 1-2-3-...-k Trees
Message-ID: <20061105.014940.1648.0.pauldhanna@juno.com>

Emeric (and Seqfans),
     What a fascinating problem.
Although I will not be using this formula now, 
let's extend your formula for k-ary trees: 
(*)  [z^n] C[k](z)^m  =  binomial(kn+m-1,n)*m/[(k-1)n+m] 
to give the coefficient of z^n in the m-th power of C[k](z). 
Note that the logarithm is obtained similarly: 
(**) [z^n] log( C[k](z) ) =  binomial(kn-1,n)/[(k-1)n], for n>0.
No doubt you are already aware of this lovely facts; 
and they may simplify some series representations involved here. 
 
I will not attempt to give a simpler g.f. for your M[n], where:
   M[n+1](z) = M[n](z)*C[n](z^n*M[n-1]^(n-1)); 
 
but using PARI I will list the first 31 terms for n=2..10 (below). 
 
You have some impressive results so far. 
     Paul 
 
(PARI)
{M=1/(1-x+O(x^31));
 for(n=2,10,
   M=M*sum(k=0,30,binomial(n*k,k)/((n-1)*k+1)*(x^n*M^(n-1))^k);
   print("n="n":");print(Vec(M))
 )
}

n=2: 
[1, 1, 2, 3, 6, 11, 23, 47, 102, 221, 493, 1105, 2516, 5763, 13328, 
30995, 72556, 170655, 403351, 957135, 2279948, 5449013, 13063596, 
31406517, 75701508, 182902337, 442885683, 1074604289, 2612341856, 
6361782007, 15518343597]
n=3: 
[1, 1, 2, 4, 9, 20, 48, 116, 288, 724, 1849, 4768, 12423, 32628, 
86342, 229952, 616042, 1659012, 4489101, 12199521, 33284546, 
91140797, 250396629, 690043032, 1907022197, 5284167884, 14677681554, 
40862469713, 114001697975, 318681491431, 892495078849]
n=4: 
[1, 1, 2, 4, 10, 24, 62, 160, 425, 1140, 3105, 8528, 23643, 66008, 
185526, 524384, 1489810, 4251852, 12184745, 35048405, 101156752, 
292865417, 850314803, 2475327088, 7223400899, 21126670372, 
61920289652, 181838859665, 534980035009, 1576653291583, 
4654122552827]
n=5: 
[1, 1, 2, 4, 10, 25, 67, 180, 495, 1375, 3871, 10993, 31493, 90843, 
263686, 769466, 2256135, 6643082, 19634705, 58232350, 173242381, 
516860717, 1546035258, 4635543843, 13929569399, 41943013047, 
126532961332, 382396277940, 1157548532794, 3509416800398, 
10655157352027]
n=6: 
[1, 1, 2, 4, 10, 25, 68, 186, 522, 1479, 4246, 12289, 35872, 105411, 
311662, 926270, 2765778, 8292296, 24953437, 75338686, 228140842, 
692733127, 2108652750, 6433255041, 19668210742, 60247367313, 
184879648441, 568281131800, 1749490180723, 5393738767856, 
16651705792875]
n=7: 
[1, 1, 2, 4, 10, 25, 68, 187, 529, 1514, 4393, 12856, 37944, 112740, 
337017, 1012671, 3056978, 9265765, 28187304, 86028764, 263339173, 
808256040, 2486804090, 7668374150, 23694936466, 73354869098, 
227489892754, 706643494680, 2198340096395, 6848600059320, 
21363953916171]
n=8: 
[1, 1, 2, 4, 10, 25, 68, 187, 530, 1522, 4437, 13056, 38766, 115908, 
348733, 1054775, 3205251, 9780165, 29951916, 92029644, 283606809, 
876333336, 2714448570, 8426790446, 26213887250, 81699432762, 
255072076566, 797641572824, 2498068134817, 7834443329344, 
24602515926899]
n=9: 
[1, 1, 2, 4, 10, 25, 68, 187, 530, 1523, 4446, 13110, 39030, 117060, 
353404, 1072811, 3272526, 10024884, 30825490, 95103360, 294300321, 
913203240, 2840650881, 8856206258, 27667828196, 86601978030, 
271545341505, 852829532176, 2682483667081, 8449321174753, 
26648690288909]
n=10: 
[1, 1, 2, 4, 10, 25, 68, 187, 530, 1523, 4447, 13120, 39095, 117400, 
354974, 1079493, 3299426, 10128934, 31216245, 96538200, 299477771, 
931628670, 2905496841, 9082358938, 28450647731, 89294760932, 
280759352295, 884216374006, 2788990466531, 8809540396813, 
27863492984436]

END. 
----------------------------------------------------------
On Sat, 4 Nov 2006 23:30:30 -0500 (EST) Emeric Deutsch
<deutsch at duke.poly.edu> writes:
> Dear Seqfans,
> We (Lou Shapiro and myself) would like to submit a few
> sequences similar to A124497 (shown at the bottom of this
> message). They will refer to 1-2-3-4 trees, 1-2-3-4-5
> trees, etc. rather than to 1-2-3 trees. We can describe
> the corresponding g.f.'s but we need help in finding the
> terms themselves.
> 
> Here is the description of all the g.f.'s. Let M[k](z) be
> the g.f. of the 1-2-...-k trees with thinning limbs. I will
> describe these functions; consequently, we may disregard
> the objects counted by them.
> 
> Denote by C[k](z) the g.f. of the k-ary trees. C[k]
> satisfies C[k]=1+zC[k]^k and the coefficient of z^n
> in its series expansion is
> (*)      binom(kn,n)/[(k-1)n+1].
> 
> I will write Mk for M[k] and Ck for C[k].
> 
> Obviously, M1(z)=1/(1-z) (path-trees). The next one is 
> M2(z)=M1(z)C2(z^2*M1), i.e. M1 times C2 evaluated at
> z^2*M1. This gives no problem since
> C2(z)=[1-sqrt(1-4z)]/(2z). We obtain A090344.
> 
> Next M3(z)=M2(z)C3(z^3*M2^2); since there is an
> explicit expression for C3(z), namely
> C3(z)=2/sqrt(3z)*sin[(1/3)*arcsin(sqrt(27z/4))],
> this yields easily A124497.
> 
> From here on we need help.
> M4(z)=M3(z)C4(z^4*M3^3); 
> Most probably sequence starts with
> 1,1,2,4,10,24,62,160,425,1140,3105 (I used some terms
> of the series of C4(z); see (*)).
> 
> M5(z)=M4(z)C5(z^5*M4^4);
> Most probably sequence starts with
> 1,1,2,4,10,25,67,180,495,1375,3871 (I used some terms
> of the series of C5(z); see (*)).
> 
> And so on.
> 
> These sequences approach A124344:
> 1,1,2,4,10,25,68,187,530,1523,4447,13121,39107,...
> 
> Many thanks.
> Emeric





From pauldhanna at juno.com  Sun Nov  5 08:04:15 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Sun, 5 Nov 2006 02:04:15 -0500
Subject: A124497: 1-2-3-...-k Trees
Message-ID: <20061105.020415.1648.1.pauldhanna@juno.com>

Emeric (and Seqfans),
     Correction to my typo in prior e-mail:
> I will not attempt to give a simpler g.f. for your M[n], where:
>    M[n+1](z) = M[n](z)*C[n](z^n*M[n-1]^(n-1)); 
 
that should have been:
 
M[1](z) = 1/(1-z);
 
M[n+1](z) = M[n](z)*C[n+1](z^(n+1)*M[n]^n); 
 
This was more clearly given by Emeric, but I correct myself for the sake
of accuracy. 
    Paul





From zakseidov at yahoo.com  Sun Nov  5 09:49:35 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 00:49:35 -0800 (PST)
Subject: A124057 is subset of A045940
Message-ID: <20061105084935.7574.qmail@web38212.mail.mud.yahoo.com>

A124057 is subset of A045940
from which more terms for A124057 
can be retrieved:

602,603,1083,2012,2091,2522,2523,2524,2634,2763,3243,3355,4202,4203,4921,4922,4923,5034,5035,5132,5203,5282,5283,5785,5882,5954,5972,6092,6212,6476,6962,6985,7730,7731,7945,8393,8825,8956,8972


%N A045940 Numbers n such that factorizations of n
through n+3 have same number of primes (including 
               multiplicities).

%N A124057 Numbers n such that n, n+1, n+2 and n+3 are
products of exactly 3 primes.



 
____________________________________________________________________________________
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From jvospost3 at gmail.com  Sun Nov  5 17:53:45 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 5 Nov 2006 08:53:45 -0800
Subject: A124057 is subset of A045940
In-Reply-To: <20061105084935.7574.qmail@web38212.mail.mud.yahoo.com>
References: <20061105084935.7574.qmail@web38212.mail.mud.yahoo.com>
Message-ID: <5542af940611050853h50af8ae5ue5004d32f982bef0@mail.gmail.com>

Dear Zak.

You're right.  I knew that; it's in my handwritten notes.  I should have you
submit that as a comment, and also have A045940 added as a cf to A124057,
and also show your extension. Thank you!

Shalom,

Jonathan

On 11/5/06, zak seidov <zakseidov at yahoo.com> wrote:
>
> A124057 is subset of A045940
> from which more terms for A124057
> can be retrieved:
>
>
> 602,603,1083,2012,2091,2522,2523,2524,2634,2763,3243,3355,4202,4203,4921,4922,4923,5034,5035,5132,5203,5282,5283,5785,5882,5954,5972,6092,6212,6476,6962,6985,7730,7731,7945,8393,8825,8956,8972
>
>
> %N A045940 Numbers n such that factorizations of n
> through n+3 have same number of primes (including
>                multiplicities).
>
> %N A124057 Numbers n such that n, n+1, n+2 and n+3 are
> products of exactly 3 primes.
>
>
>
>
>
> ____________________________________________________________________________________
> Access over 1 million songs - Yahoo! Music Unlimited
> (http://music.yahoo.com/unlimited)
>
>
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From franktaw at netscape.net  Sun Nov  5 18:35:13 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Sun, 05 Nov 2006 12:35:13 -0500
Subject: A124015 thoughts
In-Reply-To: <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
Message-ID: <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com>

Some years ago, researchers at Brown University put together a 
collection of text from a large variety of source to serve as a 
cross-section of English usage, for study of this kind of question.  If 
you're going to add a sequence with statistics of this sort, that would 
be the place to start.

See, e.g., Wikipedia: http://en.wikipedia.org/wiki/Brown_Corpus.

Franklin T. Adams-Watters

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From bdm at cs.anu.edu.au  Sun Nov  5 23:40:24 2006
From: bdm at cs.anu.edu.au (Brendan McKay)
Date: Mon, 6 Nov 2006 09:40:24 +1100
Subject: A124015 thoughts
In-Reply-To: <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com> <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com>
Message-ID: <20061105224024.GC28253@cs.anu.edu.au>

* franktaw at netscape.net <franktaw at netscape.net> [061106 04:35]:
> Some years ago, researchers at Brown University put together a 
> collection of text from a large variety of source to serve as a 
> cross-section of English usage, for study of this kind of question.  If 
> you're going to add a sequence with statistics of this sort, that would 
> be the place to start.
> 
> See, e.g., Wikipedia: http://en.wikipedia.org/wiki/Brown_Corpus.
> 
> Franklin T. Adams-Watters

Yes, but I don't know if statistics based on sampling belong in OEIS.
They are not precise well-defined quantities.  Counts derived from
single texts are more concrete, but even then there are multiple
editions, even changes between printings of the same edition.

Brendan.





From zakseidov at yahoo.com  Sun Nov  5 23:53:40 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 14:53:40 -0800 (PST)
Subject: Don't restrict OEIS sequence domains!
In-Reply-To: <001501c70116$e2fabd20$6401a8c0@yourxhtr8hvc4p>
Message-ID: <20061105225340.19195.qmail@web38215.mail.mud.yahoo.com>

David, 

I simply remember Neil's recommendation
(somewhere in his tip's) to omit trivial cases
e.g. if each second term is zero.

And you may wish to 
put in %C A053699 
some of your observations, 
e.g., a(n_odd)=2.
Zak

--- David Wilson <davidwwilson at comcast.net> wrote:

> ----- Original Message ----- 
> From: "zak seidov" <zakseidov at yahoo.com>
> To: <seqfan at ext.jussieu.fr>
> Sent: Sunday, November 05, 2006 1:28 AM
> Subject: Re: Observation on A053669
> 
> 
> > Yes,
> > and because a(n_odd)=2
> > keep only even n's.
> > Zak
> 
> > [Earlier included message elided]
> ---------------------------
> 
> Let's take a look at Zak's recommendation to
> restrict the domain of A053699 
> to even n:
> 
> A053699(n) currently looks like:
> 
>
2,3,2,3,2,5,2,3,2,3,2,5,2,3,2,3,2,5,2,3,2,3,2,5,2,3,2,3,2,5,2,3,2,3,2,7,...
> 
> But a(k) = 2 for odd k, these values are
> uninteresting, so list only a(2n):
> 
>
3,3,5,3,3,5,3,3,5,3,3,5,3,3,7,3,3,5,3,3,5,3,3,5,3,3,5,3,3,7,3,3,5,3,3,5,...
> 
> But now a(k) = 3 unless 6 divides k, so we should
> list only a(6n):
> 
>
5,5,5,5,7,5,5,5,5,7,5,5,5,5,7,5,5,5,5,7,5,5,5,5,7,5,5,5,5,7,5,5,5,5,11,..
> 
> But again, a(k) = 5 except when 30 divides k, so
> let's eject these boring 
> values and keep only a(30n):
> 
>
7,7,7,7,7,7,11,7,7,7,7,7,7,11,7,7,7,7,7,7,11,7,7,7,7,7,7,11,7,7,7,7,7,...
> 
> But look at all the uninteresting 7's. Maybe we
> should list only a(210n)...
> 
> I think you get the picture. In the case of this
> sequence, we repeatedly 
> eliminate uninteresting domain values, but we never
> flush out the elusive 
> "interesting domain" of the sequence. Indeed, we can
> eliminate the entire 
> domain as uninteresting, whereas I think the
> original sequence is indeed 
> quite interesting.
> 
> Clearly, there are times when restricted domains are
> useful. For example, 
> since the areas of Pythagorean triangles are always
> divisible by 6, it would 
> seem reasonable, in the name of compactness, to
> prefer "Number of 
> Pythagorean triangles with area 6n" over "Number of
> Pythagorean triangles of 
> area n", since the latter would have a(n) = 0 for n
> != 6k. Nevertheless, I 
> would advocate having both sequences in the OEIS,
> because we never know if 
> the user will be searching for the compact or
> inflated version of the 
> sequence.
> 
> Similarly, if we adopt Zak's recommendation to
> restrict A065577 to a domain 
> of even values, and then someone looks up
> 
> 2,3,2,3,2,5,2,3,2,3,2,5
> 
> they will probably miss the most straightforward
> explanation of their 
> sequence.
> 
> In short: Don't restrict domains of OEIS sequences
> unless it makes clear 
> sense to do so, and if you do include a sequence on
> some restricted 
> "interesting" domain, have pity on the OEIS users
> and also include the same 
> sequence on its fullest possible domain.
> 
> ----- Original Message ----- 
> From: "zak seidov" <zakseidov at yahoo.com>
> To: <seqfan at ext.jussieu.fr>
> Sent: Sunday, November 05, 2006 1:28 AM
> Subject: Re: Observation on A053669
> 
> 
> > Yes,
> > and because a(n_odd)=2
> > keep only even n's.
> > Zak 
> 
> 



 
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From franktaw at netscape.net  Mon Nov  6 00:26:46 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Sun, 05 Nov 2006 18:26:46 -0500
Subject: A124015 thoughts
In-Reply-To: <20061105224024.GC28253@cs.anu.edu.au>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com> <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com> <20061105224024.GC28253@cs.anu.edu.au>
Message-ID: <8C8CF48FECF25AB-958-F877@FWM-M05.sysops.aol.com>

I'm not saying one way or the other whether this sort of thing belongs
in the OEIS; just that, if it is going to be there, the Brown Corpus
would be the best source for it.

The Brown Corpus is about as concrete as you can get in this area.  It
is very well defined and a standard research reference.

Franklin T. Adams-Watters


-----Original Message-----
From: bdm at cs.anu.edu.au

  * franktaw at netscape.net <franktaw at netscape.net> [061106 04:35]:
> Some years ago, researchers at Brown University put together a
> collection of text from a large variety of source to serve as a
> cross-section of English usage, for study of this kind of question.  
If
> you're going to add a sequence with statistics of this sort, that 
would
> be the place to start.
>
> See, e.g., Wikipedia: http://en.wikipedia.org/wiki/Brown_Corpus.
>
> Franklin T. Adams-Watters

Yes, but I don't know if statistics based on sampling belong in OEIS.
They are not precise well-defined quantities.  Counts derived from
single texts are more concrete, but even then there are multiple
editions, even changes between printings of the same edition.

Brendan.


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From zakseidov at yahoo.com  Mon Nov  6 00:51:18 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 15:51:18 -0800 (PST)
Subject: A124057 is subset of A045940
Message-ID: <20061105235118.90653.qmail@web38203.mail.mud.yahoo.com>

Just submitted:

My Q to seqfans:
Really, there are no numbers n such that n, n+1, n+2
and n+3 are products of 6 primes?
No for n<7258290.
Thanks, Zak


%I A124728
%S A124728
4023,7314,9162,12122,12123,16674,19434,19940,23874,24723,29094,33234,35124,35125,39234,42182,42183,44163,45175,46988,49147,51793,52854,52855,54584,54585,54663,58375,63594,64074,64075,64323,64491,64712,64713,65943,67507,67962,70262,71124,75123,75446,77283,80792,80793,81025,81963,81975,86823
%N A124728 Numbers n such that n, n+1, n+2 and n+3 are
products of 4 primes.
%C A124728 Subset of A045940  Numbers n such that
factorizations of n through n+3 have same number of
primes (including multiplicities).
Cf. A124057, A124729  Numbers n such that n, n+1, n+2
and n+3 are products of exactly 3,5 primes. 
There are no numbers n such that n, n+1, n+2 and n+3
are products of exactly 6 primes(?)
%e A124728 4023=3^3*149, 4024=2^3*503, 4025=5^2*7*23,
4026=2*3*11*61 (all products of 4 primes).
%Y A124728 A045940, A124057, A124729.
%O A124728 1
%K A124728 ,nonn,
%A A124728 Zak Seidov  (zakseidov at yahoo.com), Nov 05
2006


%I A124729
%S A124729
57967,491875,543303,584647,632148,632149,715374,824523,878875,914823,930123,931623,955448,964143,995874
%N A124729 Numbers n such that n, n+1, n+2 and n+3 are
products of 5 primes.
%C A124729 Subset of A045940  Numbers n such that
factorizations of n through n+3 have same number of
primes (including multiplicities).
Cf. A124057, A124728  Numbers n such that n, n+1, n+2
and n+3 are products of exactly 3,4 primes. 
There are no numbers n such that n, n+1, n+2 and n+3
are products of exactly 6 primes(?)
%e A124729 57967=7^3*13^2, 57968=2^4*3623,
57969=3^3*19*113, 57970=2*5*11*17*31 (all product of 5
primes (including multiplicities)};
632148 is the first number such that n through n+4 are
5-almost primes.
%Y A124729 A045940, A124057, A124728.
%O A124729 1
%K A124729 ,nonn,
%A A124729 Zak Seidov  (zakseidov at yahoo.com), Nov 05
2006



 
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From zakseidov at yahoo.com  Mon Nov  6 02:02:27 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 17:02:27 -0800 (PST)
Subject: n..n+6 are all N-almost primes
Message-ID: <20061106010227.58041.qmail@web38206.mail.mud.yahoo.com>

Just dubmitted.

My Q to gurus:
Are there longer set(s) of successive numbers
n..n+m (m>6)
which are products of the same number (N) of primes
(including multiplicities)?

It seems that for  larger N's, 
the larger m's may be(?).

Thanks, Zak
  
%I A000001
%S A000001
211673,298433,355923,381353,460801,506521,540292,568729,690593,705953,737633,741305,921529
%N A000001 Numbers n such that factorizations of
n..n+6 have same number of primes (including
multiplicities).
%C A000001 Subset of A045940  Numbers n such that
factorizations of n..n+3 have same number of primes
(including multiplicities).
%e A000001 211673=7*11*2749, 211674=2*3*35279,
211675=5^2*8467, 211676=2^2*52919, 211677=3*37*1907,
211677=2*109*971, 211679=13*19*857 all 3-almost
primes;
355923=3^2*71*557, 355924=2^2*101*881,
355925=5^22*23*619, 355926=2*3*137*433,
355927=11*13*19*131, 355928=2^3*44491,
355929=3*7*17*997 all 4-almost primes.
%Y A000001 A045940.
%O A000001 1
%K A000001 ,nonn,
%A A000001 Zak Seidov  (zakseidov at yahoo.com), Nov 05
2006



 
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From zakseidov at yahoo.com  Mon Nov  6 03:49:26 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 18:49:26 -0800 (PST)
Subject: n..n+6 are all N-almost primes
In-Reply-To: <20061106010227.58041.qmail@web38206.mail.mud.yahoo.com>
Message-ID: <20061106024926.64870.qmail@web38203.mail.mud.yahoo.com>

--- zak seidov <zakseidov at yahoo.com> wrote:
> Are there longer set(s) of successive numbers
> n..n+m (m>6)
> which are products of the same number (N) of primes
> (including multiplicities)?

Me: Just submitted:
%I A000001
%S A000001
3405122,3405123,6612470,8360103,8520321,9306710
%N A000001 Numbers n such that n..n+7 are products of
the same number of primes.
%C A000001 Note that because 3405130=2*5*167*2039 is
also 4-almost prime,
3405122 is the first n such that numbers n..n+8 are
products of the same number N of primes (N=4).
%e A000001 3405122=2*7*29*8387, 3405123=3^2*19*19913,
3405124=2^2*127*6703, 3405125=5^3*27241,
3405126=2*3*59*9619, 3405127=11*23*43*313,
3405128=2^3*425641, 3405129=3*7*13*12473 all 4-almost
primes.
%O A000001 1
%K A000001 ,nonn,
%A A000001 Zak Seidov  (zakseidov at yahoo.com), Nov 05
2006

Now we need to find numbers n such that n..n+m (m>8)
are N-almost primes (n'd be >10^7).

Thanks, Zak

BTW The "annoying" ad at the bottom of message is 
from Yahoo!Mail - not me, sorry.

<original message>
> Just dubmitted.
> 
> My Q to gurus:
> Are there longer set(s) of successive numbers
> n..n+m (m>6)
> which are products of the same number (N) of primes
> (including multiplicities)?
> 
> It seems that for  larger N's, 
> the larger m's may be(?).
> 
> Thanks, Zak
>   
> %I A000001
> %S A000001
>
211673,298433,355923,381353,460801,506521,540292,568729,690593,705953,737633,741305,921529
> %N A000001 Numbers n such that factorizations of
> n..n+6 have same number of primes (including
> multiplicities).
> %C A000001 Subset of A045940  Numbers n such that
> factorizations of n..n+3 have same number of primes
> (including multiplicities).
> %e A000001 211673=7*11*2749, 211674=2*3*35279,
> 211675=5^2*8467, 211676=2^2*52919, 211677=3*37*1907,
> 211677=2*109*971, 211679=13*19*857 all 3-almost
> primes;
> 355923=3^2*71*557, 355924=2^2*101*881,
> 355925=5^22*23*619, 355926=2*3*137*433,
> 355927=11*13*19*131, 355928=2^3*44491,
> 355929=3*7*17*997 all 4-almost primes.
> %Y A000001 A045940.
> %O A000001 1
> %K A000001 ,nonn,
> %A A000001 Zak Seidov  (zakseidov at yahoo.com), Nov 05
> 2006
> 




 
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From davidwwilson at comcast.net  Mon Nov  6 04:10:17 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sun, 5 Nov 2006 22:10:17 -0500
Subject: A124015 thoughts
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com> <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com> <20061105224024.GC28253@cs.anu.edu.au> <8C8CF48FECF25AB-958-F877@FWM-M05.sysops.aol.com>
Message-ID: <001901c70151$16106360$6401a8c0@yourxhtr8hvc4p>

Sounds great. Where do I get the Brown Corpus in computer-readable form?

----- Original Message ----- 
From: <franktaw at netscape.net>
To: <seqfan at ext.jussieu.fr>
Sent: Sunday, November 05, 2006 6:26 PM
Subject: Re: A124015 thoughts


> I'm not saying one way or the other whether this sort of thing belongs
> in the OEIS; just that, if it is going to be there, the Brown Corpus
> would be the best source for it.
>
> The Brown Corpus is about as concrete as you can get in this area.  It
> is very well defined and a standard research reference.
>
> Franklin T. Adams-Watters
>
>
> -----Original Message-----
> From: bdm at cs.anu.edu.au
>
>  * franktaw at netscape.net <franktaw at netscape.net> [061106 04:35]:
>> Some years ago, researchers at Brown University put together a
>> collection of text from a large variety of source to serve as a
>> cross-section of English usage, for study of this kind of question.
> If
>> you're going to add a sequence with statistics of this sort, that
> would
>> be the place to start.
>>
>> See, e.g., Wikipedia: http://en.wikipedia.org/wiki/Brown_Corpus.
>>
>> Franklin T. Adams-Watters
>
> Yes, but I don't know if statistics based on sampling belong in OEIS.
> They are not precise well-defined quantities.  Counts derived from
> single texts are more concrete, but even then there are multiple
> editions, even changes between printings of the same edition.
>
> Brendan.
>
>
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading 
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>
> -- 
> No virus found in this incoming message.
> Checked by AVG Free Edition.
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> 5:30 PM
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From dean at math.ucdavis.edu  Mon Nov  6 10:39:09 2006
From: dean at math.ucdavis.edu (Dean Hickerson)
Date: Mon, 6 Nov 2006 01:39:09 -0800
Subject: terms lack
Message-ID: <20061106093909.GA16454@math.ucdavis.edu>

Mostly to Yasutoshi Kohmoto <zbi74583 at boat.zero.ad.jp>:

> Recently I updated more terms of A054572, which are calculated by
> Richard Mathar.
> And Neil claimed or got angry.
> He said, "You are red card. Don't submit a wrong sequences"
> I am exaggerating a little.
> These words are not exactly what he wrote.
>
> To Neil
> They are correct, not "wrong".
> It is only not complete.

The terms that are listed in a sequence should be the start of the sequence.
If there are missing terms in between the listed ones, then the sequence
entry is not just incomplete, it's wrong.

If you've computed some terms of a sequence, but you're not sure if there
are smaller ones that you've missed, then you should only include the ones
that you're sure are at the start of the sequence.  You can add a comment
about the larger terms, stating that it's not known if there are smaller ones.
For example, the "EXTENSIONS" lines for the sequence A000043 (Mersenne
exponents) mention some recently discovered terms.  But there might be
some smaller ones, so these are not included in the sequence.

> I defined many "divisor" functions and calculated many Amicable Numbers and
> Perfect Numbers using those functions.
> And I submitted them as sequences to  OEIS.
>
> http://www.research.att.com/~njas/sequences/?q=kohmoto+amicable&sort=0&fmt=0&language=english&go=Search
>
> http://www.research.att.com/~njas/sequences/?q=kohmoto+perfect&sort=0&fmt=0&language=english&go=Search
>
> I suppose that they are almost complete but some of them might not be
> complete.

I've found missing terms for several of your sequences:


Sequences A045613 and A045614 ("Super Unitary Amicable Number") consist
of pairs of distinct numbers a and b such that

    usigma(usigma(a)) = usigma(usigma(b)) = a+b.

(The definitions don't make this clear, but A045613 contains the smaller
members of the pairs and A045614 contains the larger members.  Also they're
sorted by their smallest members, so 155 comes before 142 in A045614.)  The
first 3 pairs that you've listed are (105, 155), (110, 142) and (2145, 3055).
But you've missed several smaller ones:

    (33, 35), (208, 224), (268, 272), (455, 601), (695, 1033), (812, 956),
    (1609, 1847), (1808, 2512), (1892, 3004), (1913, 2407), (2096, 2224)


A051594 and A051595 are "(-1)sigma amicable numbers".  (Again, the definition
doesn't make it clear that A051595 has the smaller terms and A051594 has the
larger terms.)  The first pair that you've listed is (1969706592, 2236072608).
But there are many smaller ones, starting with:

    (429552, 466200), (497808, 604656), (800496, 1103760).


A036471-A036474 are defined as "Amicable quadruple: 4 different numbers
which satisfy sigma(a)=sigma(b)=sigma(c)=sigma(d)=a+b+c+d".  The first
quadruple that you've listed has smallest element
342151462276356306033089201934180, but here's a much smaller one:

    (3270960, 3361680, 3461040, 3834000)

(I think that's the smallest one, but I could be wrong.)  If someone can
determine the first several terms of this sequence, then it should be
edited; otherwise it should be deleted.


In some of your sequences, I wasn't able to find any smaller terms, but the
ones that you've listed are so large that I'm not at all confident that they
are the smallest ones.  For example, A038362 and A038363 are defined as
"A Rational Amicable Number: two different numbers a, b which satisfies the
following equation: sigma(a)=sigma(b)=(a+b)^3/(a^2+b^2)", and the smallest
pair that you've included is:

    (26403469440047700, 30193441130006700)

Do you have any reason to think that that's the smallest such pair?  If not,
and if you can't determine what the first few terms really are, then the
sequence should be deleted.


I've only looked at a few of your sequences, but that's enough to convince
me that they should all be checked.

Neil, I'll submit corrections for A045613, A045614, A051594, and A051595,
but I'm not going to work on the others.  Maybe they should be deleted or
marked as "probation" until the author or someone else fixes them.

Dean Hickerson
dean at math.ucdavis.edu





From bdm at cs.anu.edu.au  Mon Nov  6 12:45:30 2006
From: bdm at cs.anu.edu.au (Brendan McKay)
Date: Mon, 6 Nov 2006 22:45:30 +1100
Subject: terms lack
In-Reply-To: <20061106093909.GA16454@math.ucdavis.edu>
References: <20061106093909.GA16454@math.ucdavis.edu>
Message-ID: <20061106114530.GB4232@cs.anu.edu.au>

* Dean Hickerson <dean at math.ucdavis.edu> [061106 20:40]:
> The terms that are listed in a sequence should be the start of the sequence.
> If there are missing terms in between the listed ones, then the sequence
> entry is not just incomplete, it's wrong.
> 
> If you've computed some terms of a sequence, but you're not sure if there
> are smaller ones that you've missed, then you should only include the ones
> that you're sure are at the start of the sequence.  You can add a comment
> about the larger terms, stating that it's not known if there are smaller ones.
> For example, the "EXTENSIONS" lines for the sequence A000043 (Mersenne
> exponents) mention some recently discovered terms.  But there might be
> some smaller ones, so these are not included in the sequence.
 
Is there an official rule about this?  I'm not sure it is correct.
Of course we should not submit sequences with possible gaps, but
suppose we can compute a(10) thru a(20) but have trouble with
a(1) thru a(9).  I don't see why we can't submit a(10..20) with
an OFFSET of 10 and a comment explaining why a(1..9) are absent.

Brendan.





From Annette.Warlich at t-online.de  Mon Nov  6 13:05:20 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Mon, 06 Nov 2006 13:05:20 +0100
Subject: Fibbinary sequence
Message-ID: <454F2500.2020107@t-online.de>

A courious result by chance:

A003714	Fibbinary numbers:
       if n = F_i1+F_i2+...+F_ik is the Zeckendorf representation of n
       (i.e. write n in Fibonacci number system)
       then
         a(n) = 2^{i1-2}+2^{i2-2}+...+2^{ik-2}.

	0, 1, 2, 4, 5, 8, 9, 10, 16, 17, 18, 20, 21, 32, 33, 34, 36, 37,
      40, 41, 42, 64, 65, 66, 68, 69, 72, 73, 74, 80, 81, 82, 84, 85,
     128, 129, 130, 132, 133, 136, 137, 138, 144, 145, 146, 148, 149,
     160, 161, 162, 164, 165, 168, 169, 170, 256, 257, 258, 260

I was checking some variants of the Pascal-matrix and got the
following
Let Px =
      1       .    .     .     .    .  .  .
      1      -1    .     .     .    .  .  .
    3/2      -3    1     .     .    .  .  .
      4      -9    6    -1     .    .  .  .
     15     -35   30   -10     1    .  .  .
  137/2  -345/2  165   -75    15   -1  .  .
 1491/4 -2009/2 1050  -560 315/2  -21  1  .
   2374   -6720 7588 -4515  1540 -294 28 -1

then the index r of the rows, r=1..inf(??) , which contain only integers, is A003714
      1, 2, 4, 5, 8, 9, 10, 16, 17, 18, 20, 21, 32, 33, 34, 36, 37,
     40, 41, 42, 64,

(although I checked this only up to N=64)


Explanation: -----------------------------------------------------------

The matrix Px is constructed as a variant of Pj
Pj:
 1  .  .   .  .   . .  .
 1 -1  .   .  .   . .  .
 1 -2  1   .  .   . .  .
 1 -3  3  -1  .   . .  .
 1 -4  6  -4  1   . .  .
 1 -5 10 -10  5  -1 .  .
 1 -6 15 -20 15  -6 1  .
 1 -7 21 -35 35 -21 7 -1

which is the column-signed version of the Pascalmatrix P
   Pj = P*J
where J is diagonal unit with alternating signs (diag(1,-1,1,-1,...))

The matrix S2 of the stirling-numbers of 2'nd kind
 1   .   .    .    .   .  . .
 1   1   .    .    .   .  . .
 1   3   1    .    .   .  . .
 1   7   6    1    .   .  . .
 1  15  25   10    1   .  . .
 1  31  90   65   15   1  . .
 1  63 301  350  140  21  1 .
 1 127 966 1701 1050 266 28 1

and S1 of the stirling numbers of first kind (where also S1 = S2^-1)
     1     .      .    .     .   .   . .
    -1     1      .    .     .   .   . .
     2    -3      1    .     .   .   . .
    -6    11     -6    1     .   .   . .
    24   -50     35  -10     1   .   . .
  -120   274   -225   85   -15   1   . .
   720 -1764   1624 -735   175 -21   1 .
 -5040 13068 -13132 6769 -1960 322 -28 1

with the eigenvalues Z(1) = diag(1,1/2,1/3,1/4) such that

   G = S2 * Z(1) * S2^-1

G:
     1     .    .     .   .    .   .   .
   1/2   1/2    .     .   .    .   .   .
   1/6   1/2  1/3     .   .    .   .   .
     0   1/4  1/2   1/4   .    .   .   .
 -1/30     0  1/3   1/2 1/5    .   .   .
     0 -1/12    0  5/12 1/2  1/6   .   .
  1/42     0 -1/6     0 1/2  1/2 1/7   .
     0  1/12    0 -7/24   0 7/12 1/2 1/8

form an eigensystem of G.
G ist interesting as it contains the bernoulli-
numbers, and coefficients of the integrals of the
bernoulli-functions.



Also G is Eigenmatrix of the signed Pascal-matrix Pj,
with the eigenvalues J = diag(1,-1,1,-1,...)

  Pj = G * J * G^-1

and Pj is used in divergent Euler-summation, Hasse's
zeta-summation and elsewhere.




Now I tried a variant of P by varying G, where the eigenvalues
are not Z(1), but are V(1/2) = diag(1,1/2,1/4,1/8,....) for some
analyses about summing zeta series, since Z(1) and V(1/2) are
closely related in this subject and complementary/mutually supplementary
in certain manners.

Now with the G-variant

  Gx  = S2 * V(1/2) * S2^-1
Gx:
     1      .       .        .      .      .    .     .
   1/2    1/2       .        .      .      .    .     .
     0    3/4     1/4        .      .      .    .     .
  -1/4    3/8     3/4      1/8      .      .    .     .
     0   -5/8   15/16      5/8   1/16      .    .     .
   1/2 -15/16  -15/32    45/32  15/32   1/32    .     .
     0    7/4 -105/32    35/64 105/64  21/64 1/64     .
 -17/8 147/32    7/32 -735/128  35/16 105/64 7/32 1/128

I come to

  Px =         Gx               * J *       Gx^-1
     =  ( S2 * V(1/2) * S2^-1 ) * J * (S2 * V(2) * S2^-1)

(while Pj is
  Pj =         G                * J *       G ^-1
     =  ( S2 * Z(-1) * S2^-1 )  * J * (S2 * Z(1) * S2^-1)
  )

and as shown above:

Px:
      1       .    .     .     .    .  .  .
      1      -1    .     .     .    .  .  .
    3/2      -3    1     .     .    .  .  .
      4      -9    6    -1     .    .  .  .
     15     -35   30   -10     1    .  .  .
  137/2  -345/2  165   -75    15   -1  .  .
 1491/4 -2009/2 1050  -560 315/2  -21  1  .
   2374   -6720 7588 -4515  1540 -294 28 -1

and the rows containing only integers, are uniquely
indicated by A003714, up to r=64, where the rownumber r
starts with 1.

Don't have an idea, how this coincidence could be interesting -
I thought I'd just sentd the observation..

Regards -

Gottfried Helms







From Labos at ana.sote.hu  Mon Nov  6 13:38:39 2006
From: Labos at ana.sote.hu (=?ISO-8859-2?Q?L=DFbos_Elem=DAr?=)
Date: Mon, 6 Nov 2006 13:38:39 MET-1MEST
Subject: terms lack
In-Reply-To: <20061106114530.GB4232@cs.anu.edu.au>
References: <20061106093909.GA16454@math.ucdavis.edu>
Message-ID: <6578DDB2A59@ana.sote.hu>

On 6 Nov 06, at 22:45, Brendan McKay wrote:

> * Dean Hickerson <dean at math.ucdavis.edu> [061106 20:40]:
> > The terms that are listed in a sequence should be the start of the sequence.
> > If there are missing terms in between the listed ones, then the sequence
> > entry is not just incomplete, it's wrong.
> > 
> > If you've computed some terms of a sequence, but you're not sure if there
> > are smaller ones that you've missed, then you should only include the ones
> > that you're sure are at the start of the sequence.  You can add a comment
> > about the larger terms, stating that it's not known if there are smaller ones.
> > For example, the "EXTENSIONS" lines for the sequence A000043 (Mersenne
> > exponents) mention some recently discovered terms.  But there might be
> > some smaller ones, so these are not included in the sequence.
>  
> Is there an official rule about this?  I'm not sure it is correct.
> Of course we should not submit sequences with possible gaps, but
> suppose we can compute a(10) thru a(20) but have trouble with
> a(1) thru a(9).  I don't see why we can't submit a(10..20) with
> an OFFSET of 10 and a comment explaining why a(1..9) are absent.
> 
> Brendan.

There are 2 cases :
 1  -- You cannot even know how many terms are missing.
          This is essentally the case of eg Mersenne primes..
-------------------
 2. --  You can specify the missing terms by list-
          position. In this second case you can introduce 
          a marker signal to point to missing position.

Handling of the cases can follow different standard..
The general case can be worthwhile to publish if 
so many hard computation is behind. 
Ordering terms somehow it can be mentioned that
the sequence is lacunar.
With correct description the situation, the case is
neither wrong, nor good. Perhaps unnecessary...

The empty sequence represents the full lacunarity.
Quite correctly. However the empty sequence is
not unique. 

Labos 






From dean at math.ucdavis.edu  Mon Nov  6 14:22:58 2006
From: dean at math.ucdavis.edu (Dean Hickerson)
Date: Mon, 6 Nov 2006 05:22:58 -0800
Subject: terms lack
Message-ID: <20061106132258.GA19536@math.ucdavis.edu>

Mostly to Brendan McKay

> Is there an official rule about this?  I'm not sure it is correct.
> Of course we should not submit sequences with possible gaps, but
> suppose we can compute a(10) thru a(20) but have trouble with
> a(1) thru a(9).  I don't see why we can't submit a(10..20) with
> an OFFSET of 10 and a comment explaining why a(1..9) are absent.

Sure, that's fine.  In that case the sequence that's in the OEIS starts
with a(10), and it doesn't have any gaps.  It's part of a longer, partially
unknown, one that starts with a(1), but the sequence that it's part of
isn't in the OEIS.

But in cases like Kohmoto's sequences that wouldn't work.  For example,
suppose we know from hand calculation (as Kohmoto did for A045613), that
105, 110, and 2145 are in the sequence.  Presumably we can test that
there are no terms between 105 and 110, but we can't tell if there are
terms below 105 or between 110 and 2145.  So (ignoring the 4-term rule for
submissions) we could submit a sequence with just 105 and 110, and add
comments saying that there might be earlier terms and that 2145 is also
in the sequence, but we shouldn't submit the sequence as starting with
105, 110, 2145.

Dean Hickerson
dean at math.ucdavis.edu





From tbaruchel at free.fr  Mon Nov  6 17:31:32 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Mon, 6 Nov 2006 17:31:32 +0100
Subject: A fast (written in assembly) library for continued fractions
In-Reply-To: <20061103221729.GB2467@>
References: <20061103221729.GB2467@>
Message-ID: <20061106163132.GA30587@>

On Ven, nov 03, 2006 at 11:17:29 +0100, Thomas Baruchel wrote:
> I wrote for my own purposes the following X86 assembly code, intended

The final version of my code can be found at
  http://baruchel.free.fr/~thomas/wp/?p=10
I give also (in the comments of the file) advices for using the functions
from gcc, gfortran, g95. The fastest continued fractions in the west...

Next step will be functions for computing the Pell equation, but
pure FPU stack will not be possible this time (only eight registers), thus
some memory access will be required :-(

Regards,

-- 
Thomas Baruchel
  http://baruchel.free.fr/~thomas/wp/





From Eric.Angelini at kntv.be  Mon Nov  6 18:57:00 2006
From: Eric.Angelini at kntv.be (Eric Angelini)
Date: Mon, 6 Nov 2006 18:57:00 +0100
Subject: Langford/Skolem-like sequence : odd + even terms
Message-ID: <F05775148D000C4B9321A5113D53CB7D191001@KNTV-SERVER01.kntv.local>


Hello SeqFans,
Another Langford/Skolem-like sequence, with a mixture of odd and
even terms -- plus two questions:
http://www.cetteadressecomportecinquantesignes.com/OddEvenSkol.htm
Best,
?.






From noe at sspectra.com  Mon Nov  6 19:04:49 2006
From: noe at sspectra.com (T. D. Noe)
Date: Mon, 6 Nov 2006 10:04:49 -0800
Subject: Cyclotomic polynomial for n=0
Message-ID: <p06230985c175281c5eee@[192.168.1.43]>

For n=1,2,3,... the cyclotomic polynomials are

x-1, x+1, x^2+x+1,...

It seems that there are two conventions for n=0:

x or 1

Mathematica says it is 1.  I think Maple uses x.  Can someone explain this?

Thanks,

Tony





From franktaw at netscape.net  Mon Nov  6 22:27:14 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 06 Nov 2006 16:27:14 -0500
Subject: Cyclotomic polynomial for n=0
In-Reply-To: <p06230985c175281c5eee@[192.168.1.43]>
References: <p06230985c175281c5eee@[192.168.1.43]>
Message-ID: <8C8D001766F0D57-BD0-2F58@WEBMAIL-MA14.sysops.aol.com>

Strictly speaking, the cyclotomic polynomial for n=0 is undefned.  The 
two packages have chosen different ways to extend the concept to n=0; 
it would have been perfectly OK for them to have returned some kind of 
error.

There are reasonable arguments for either choice.  A lot comes down to 
the basic property that the degree is phi(n); the question then is, 
what is phi(0)?  Again, this is strictly speaking undefined, but there 
are two reasonable choices: 0 and 1.  phi(0) = 0 gives 1 as the value 
of the 0th cyclotomic polynomial.  If we take phi(0) = 1, then we need 
to choose a root for the polynomial.  Choosing 0 as the root gives the 
nice property that the cyclotomic polynomials for n >= 0 constitute all 
the primitive monic polynomials all of whose roots lie in the unit 
disk; hence the selection of x as the polynomial.

If you take the formula Phi_n(x) = Product_{0<=k<n, gcd(k,n) = 1} e^{2 
k Pi i / n}, you get Phi_0(x) = 1.  On the other hand, if we regard 
this as a product over distinct residues modulo n that are relatively 
prime to n, we get x - e^{2 Pi i / 0}, which is undefined.

You can also make a case for 0 as the "correct" value.  Phi_n(x) is 
x^n-1, divided by all x - u for primitive dth roots of unity u for d a 
proper divisor of n.  x^0-1 is 0; you can divide it by anything you 
want, and the result is still 0.  (Of course, this becomes an infinite 
product in the denominator - but this is only a plausibility argument.)

Franklin T. Adams-Watters


-----Original Message-----
From: noe at sspectra.com

For n=1,2,3,... the cyclotomic polynomials are

x-1, x+1, x^2+x+1,...

It seems that there are two conventions for n=0:

x or 1

Mathematica says it is 1.  I think Maple uses x.  Can someone explain 
this?

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From zbi74583 at boat.zero.ad.jp  Tue Nov  7 02:34:32 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Tue, 7 Nov 2006 10:34:32 +0900
Subject: terms lack
References: <20061106093909.GA16454@math.ucdavis.edu>
Message-ID: <018901c7020c$e2322a30$ef73d7dc@FMC57937B6BE00>

    To Dean Hicherson

    Thanks for editing my sequences.
    And thanks for a good explanation about "completeness".
    Now, I understand that  the  purpose of OEIS is "Identification of 
sequences", it is necessary.

    Should A038362 be deleted?
    Still, I think that it should exist , because a sequence is a mapping 
from N to N, so a subset of the images is also a good information for 
understanding the mapping.

    Yasutoshi


----- Original Message ----- 
From: "Dean Hickerson" <dean at math.ucdavis.edu>
To: <seqfan at ext.jussieu.fr>
Sent: Monday, November 06, 2006 6:39 PM
Subject: Re: terms lack


> Mostly to Yasutoshi Kohmoto <zbi74583 at boat.zero.ad.jp>:
>
>> Recently I updated more terms of A054572, which are calculated by
>> Richard Mathar.
>> And Neil claimed or got angry.
>> He said, "You are red card. Don't submit a wrong sequences"
>> I am exaggerating a little.
>> These words are not exactly what he wrote.
>>
>> To Neil
>> They are correct, not "wrong".
>> It is only not complete.
>
> The terms that are listed in a sequence should be the start of the 
> sequence.
> If there are missing terms in between the listed ones, then the sequence
> entry is not just incomplete, it's wrong.
>
> If you've computed some terms of a sequence, but you're not sure if there
> are smaller ones that you've missed, then you should only include the ones
> that you're sure are at the start of the sequence.  You can add a comment
> about the larger terms, stating that it's not known if there are smaller 
> ones.
> For example, the "EXTENSIONS" lines for the sequence A000043 (Mersenne
> exponents) mention some recently discovered terms.  But there might be
> some smaller ones, so these are not included in the sequence.
>
>> I defined many "divisor" functions and calculated many Amicable Numbers 
>> and
>> Perfect Numbers using those functions.
>> And I submitted them as sequences to  OEIS.
>>
>> http://www.research.att.com/~njas/sequences/?q=kohmoto+amicable&sort=0&fmt=0&language=english&go=Search
>>
>> http://www.research.att.com/~njas/sequences/?q=kohmoto+perfect&sort=0&fmt=0&language=english&go=Search
>>
>> I suppose that they are almost complete but some of them might not be
>> complete.
>
> I've found missing terms for several of your sequences:
>
>
> Sequences A045613 and A045614 ("Super Unitary Amicable Number") consist
> of pairs of distinct numbers a and b such that
>
>    usigma(usigma(a)) = usigma(usigma(b)) = a+b.
>
> (The definitions don't make this clear, but A045613 contains the smaller
> members of the pairs and A045614 contains the larger members.  Also 
> they're
> sorted by their smallest members, so 155 comes before 142 in A045614.) 
> The
> first 3 pairs that you've listed are (105, 155), (110, 142) and (2145, 
> 3055).
> But you've missed several smaller ones:
>
>    (33, 35), (208, 224), (268, 272), (455, 601), (695, 1033), (812, 956),
>    (1609, 1847), (1808, 2512), (1892, 3004), (1913, 2407), (2096, 2224)
>
>
> A051594 and A051595 are "(-1)sigma amicable numbers".  (Again, the 
> definition
> doesn't make it clear that A051595 has the smaller terms and A051594 has 
> the
> larger terms.)  The first pair that you've listed is (1969706592, 
> 2236072608).
> But there are many smaller ones, starting with:
>
>    (429552, 466200), (497808, 604656), (800496, 1103760).
>
>
> A036471-A036474 are defined as "Amicable quadruple: 4 different numbers
> which satisfy sigma(a)=sigma(b)=sigma(c)=sigma(d)=a+b+c+d".  The first
> quadruple that you've listed has smallest element
> 342151462276356306033089201934180, but here's a much smaller one:
>
>    (3270960, 3361680, 3461040, 3834000)
>
> (I think that's the smallest one, but I could be wrong.)  If someone can
> determine the first several terms of this sequence, then it should be
> edited; otherwise it should be deleted.
>
>
> In some of your sequences, I wasn't able to find any smaller terms, but 
> the
> ones that you've listed are so large that I'm not at all confident that 
> they
> are the smallest ones.  For example, A038362 and A038363 are defined as
> "A Rational Amicable Number: two different numbers a, b which satisfies 
> the
> following equation: sigma(a)=sigma(b)=(a+b)^3/(a^2+b^2)", and the smallest
> pair that you've included is:
>
>    (26403469440047700, 30193441130006700)
>
> Do you have any reason to think that that's the smallest such pair?  If 
> not,
> and if you can't determine what the first few terms really are, then the
> sequence should be deleted.
>
>
> I've only looked at a few of your sequences, but that's enough to convince
> me that they should all be checked.
>
> Neil, I'll submit corrections for A045613, A045614, A051594, and A051595,
> but I'm not going to work on the others.  Maybe they should be deleted or
> marked as "probation" until the author or someone else fixes them.
>
> Dean Hickerson
> dean at math.ucdavis.edu
> 






From maxale at gmail.com  Tue Nov  7 22:14:38 2006
From: maxale at gmail.com (Max A.)
Date: Tue, 7 Nov 2006 13:14:38 -0800
Subject: square arrangements with bit-wise constraints
Message-ID: <d3dac270611071314i6023ba68h4e7e15d63da8c8b3@mail.gmail.com>

There is a problem at MathPages about arrangements of numbers
0,1,...,15 into 4x4 array with certain bit-wise constraints:
http://www.mathpages.com/home/kmath352.htm

This problem (and its variants) can be easily generalized to
(2^n)x(2^n) arrays, e.g.:

In how many distinct ways can the integers 0 through 2^(2n)-1 be arranged
in a (2^n)x(2^n) array such that the bit-wise OR over each row, each column,
and each diagonal is 2^(2n)-1, and the bit-wise AND over each row, column,
and diagonal is 0?

Does anybody wish to compute corresponding sequences and submit them to OEIS?

Regards,
Max





From maxale at gmail.com  Tue Nov  7 23:49:04 2006
From: maxale at gmail.com (Max A.)
Date: Tue, 7 Nov 2006 14:49:04 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>

On 11/3/06, franktaw at netscape.net <franktaw at netscape.net> wrote:

> In the following, b(i), i = 1..k is the finite sequence.  I'm starting
> all these
> sequences with the empty sequence, but I searched for them without
> that first term.  All are shown here in A066099 order.

What is good about this order comparing, say, to the prime
factorization order suggested by Marc?

I would vote for having a standard way to represent such sequences in
OEIS (well documented there, and maybe even having a dedicated
keyword). That is important when one come up with a multidimensional
sequence and want to query it in OEIS. Then it will be enough for
him/her to encode the sequence in the standard way and make a single
query.

Of course, there are many representations possible but one of them
should be called "standard", and every multidimensional sequence
should have this standard representation (among maybe others) in OEIS.

Neil, does that sound reasonable?

Regards,
Max





From mlb at well.com  Wed Nov  8 00:02:52 2006
From: mlb at well.com (Marc LeBrun)
Date: Tue, 07 Nov 2006 15:02:52 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com
 >
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
 <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>
Message-ID: <200611072304.kA7N3xSr010827@smtp.well.com>

 > does that sound reasonable?

It does, but in practice there are already several orderings in wide 
use (as Frank indicated).

Of course it's always possible to submit more than one ordering.

In fact sometimes different ones would be preferred in different contexts.

Or fancier (perhaps special-purpose) lookups could be used.

"That's the great thing about standards--there's so many to choose from!"







From franktaw at netscape.net  Wed Nov  8 00:56:58 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Tue, 07 Nov 2006 18:56:58 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com> <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>
Message-ID: <8C8D0DF8BAA6EAF-124-5E95@FWM-M16.sysops.aol.com>

Exactly what factorization order?  For most numbers, the sequence of
prime power exponents is not a sequence of positive integers.  If 2
represents the sequence (1), what sequence does 3 represent?

If doesn't work any better to use prime power exponents to represent
sequences of non-negative integers.  If 2 is (1) and 3 is (0,1), what 
do
you use to represent (1,0)?

There are ways around these difficulties, but they are pretty 
artificial.

A066099 sorts, first of all, by the number being composed; I think any
reasonable standard candidate should do that.  Second, it is simple,
with a definition tied in to the binary numbers.  Third, it is similar 
to the
Mathematica ordering for partitions (subsort in reverse lexicographic
order), which is one of the two standards for partition ordering in the
OEIS.

(The equivalent of the other, the Abramowitz and Stegun ordering,
was not yet in the OEIS; I did submit it the other day with A-number
A124734.  It is, in any event, not as simple to define as A066099.)

In any event, based on Neil saying to go ahead and submit these
sequences, I  have submitted them with comments stating that
A066099 is the standard ordering.

Franklin T. Adams-Watters


-----Original Message-----
From: maxale at gmail.com

  On 11/3/06, franktaw at netscape.net <franktaw at netscape.net> wrote:

  > In the following, b(i), i = 1..k is the finite sequence. I'm 
starting
 > all these
 > sequences with the empty sequence, but I searched for them without
 > that first term. All are shown here in A066099 order.

 What is good about this order comparing, say, to the prime
 factorization order suggested by Marc?

 I would vote for having a standard way to represent such sequences in
 OEIS (well documented there, and maybe even having a dedicated
 keyword). ...
________________________________________________________________________
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From maxale at gmail.com  Wed Nov  8 01:42:29 2006
From: maxale at gmail.com (Max A.)
Date: Tue, 7 Nov 2006 16:42:29 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8D0DF8BAA6EAF-124-5E95@FWM-M16.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
	 <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>
	 <8C8D0DF8BAA6EAF-124-5E95@FWM-M16.sysops.aol.com>
Message-ID: <d3dac270611071642x66119c57o13f7bf6df6d2436a@mail.gmail.com>

On 11/7/06, franktaw at netscape.net <franktaw at netscape.net> wrote:

> Exactly what factorization order?  For most numbers, the sequence of
> prime power exponents is not a sequence of positive integers.  If 2
> represents the sequence (1), what sequence does 3 represent?
>
> If doesn't work any better to use prime power exponents to represent
> sequences of non-negative integers.  If 2 is (1) and 3 is (0,1), what
> do you use to represent (1,0)?

Actually, the sequences that I mentioned recently does not depend on
trailing zeros, so index (1,0) for them is the same as (1). That's why
the factorization order works well for them.

> A066099 sorts, first of all, by the number being composed; I think any
> reasonable standard candidate should do that.  Second, it is simple,
> with a definition tied in to the binary numbers.  Third, it is similar
> to the
> Mathematica ordering for partitions (subsort in reverse lexicographic
> order), which is one of the two standards for partition ordering in the
> OEIS.

I see now the advantages of the A066099 ordering. And I agree that it
is a good candidate for the "standard".

Max





From arndt at jjj.de  Wed Nov  8 03:54:45 2006
From: arndt at jjj.de (Joerg Arndt)
Date: Wed, 8 Nov 2006 13:54:45 +1100
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com> <20061102011002.GC6041@cochise.anu.edu.au> <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
Message-ID: <20061108025445.GC6252@cochise.anu.edu.au>

* Jonathan Post <jvospost3 at gmail.com> [Nov 08. 2006 13:29]:
> Thank you, Joerg.  That was interesting.  Does your generating function
> hypothesis hold when I extend the sequence by another 5 values, to a(n) = 0, 0,
> 1, 1, 3, 1, 7, 4,
> 14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again, assuming that
> I got the elementary algebra and arithmetic correct.

Alas, the function I gave was spurious, with the
additional terms no OGF is found.

The gp/script can be found here
http://www.jjj.de/pari/ggf.inc.gp
(cannot find Ralf original at the moment).


> If so, I should submit this sequence as by the two us!

Definitely your invention!
But thanks for the kind offer.


> [...]

best regards,    jj





From zbi74583 at boat.zero.ad.jp  Wed Nov  8 06:33:16 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Wed, 8 Nov 2006 14:33:16 +0900
Subject: N(2)=1+1/2^2+2/3^2+5/4^2+4/5^2+2/6^2+6/7^2+....
Message-ID: <02ba01c702f7$69237ea0$ef73d7dc@FMC57937B6BE00>

    Hi, Seqfans

    I asked Mathar to calculate the following
 number.
    N(s)=   Sum_{m=1 to infinity} (-1)Sigma(m)/m^s

     See the end of this mail

     I tried to find an identity like the following

         Zeta(s)Zeta(s-1)=Sum_{m=1 to infinity} Sigma(m)/m^s
         Zeta(s)/Zeta(s+1)=Sum_{m=1 to infinity} Phi(m)/m^s

    But it is too difficult.
    I transformed it a little as follows.

         N(s)=   Sum_{m=1 to infinity} (-1)Sigma(m)/m^s
         =   Product_{p_i} ( Sum_{k=0 to infinity}
(-1)Sigma(p_i^k)/p_i^(k*s) )
         =   Product_{p_i} ( Sum_{k=0 to infinity}
(Sigma(p_i^k)-2)/p_i^(k*s) )
         =   Product_{p_i} ( Sum_{k=0 to infinity}
 Sigma(p_i^k)/p_i^(k*s) -2/(p_i^s-1) )

        Some members of Seqfan might solve it easily.

    Yasutsoshi

----- Original Message ----- 
From: "Richard Mathar" <mathar at strw.leidenuniv.nl>
To: <zbi74583 at boat.zero.ad.jp>
Cc: <mathar at mail.strw.leidenuniv.nl>
Sent: Sunday, November 05, 2006 10:35 PM
Subject: Re: N(2)=1+1/2^2+2/3^2+5/4^2+4/5^2+2/6^2+6/7^2+....


>
> These are 5 tables of partial sums of N(s) up to some maximum
> m of sum_{m=1...max(m)) (-1)Sigma(m)/m^s. The first number
> is s, the second the maximum m up to which the sum has been accumulated,
> and finally the floating point number the partial sum.
> The case s=1 seems not to be a converging series; the partial sums are
> almost
> linear to the upper limit of m. (If your formula with the zeta-function is
> correct, this is expected since zeta(1)=infinty.) I am including some
> values
> for small maximum m so you can check the results for the basic cases.
> N(2) seems to be a converging series. For N(3) you could already get some
> pretty accurate estimate of the series limit.
>
> zeta(2)=1.644934067
> zeta(3)=1.202056903
> zeta(4)=1.082323234
> zeta(3)*zeta(2)=1.977304350
> zeta(4)*zeta(3)=1.301014115
> zeta(5)*zeta(4)=1.122291001
>
>
> PARI:
>
> \\ computes (-1)sigma(n)
> A049060(n)={
>        local(i,resul,rmax,p) ;
>        if(n==1,
>                return(1)
>        ) ;
>        i=factor(n) ;
>        rmax=matsize(i)[1] ;
>        resul=1 ;
>        for(r=1,rmax,
>                p=0 ;
>                for(j=1,i[r,2],
>                        p += i[r,1]^j ;
>                ) ;
>                resul *= p-1 ;
>        ) ;
>        return(resul) ;
> }
>
>
> N(s,mmax)={
> local(resul) ;
> resul=0.0 ;
> for(m=1,mmax,
> resul += A049060(m)/m^s ;
> if(m<10 || (m % 10000 == 0),
> print(s," ",m," ",resul) ;
> ) ;
> ) ;
> }
>
> {
> mmax=1000000 ;
> for(s=1,5,
> N(s,mmax) ;
> ) ;
> }
>
> Tables:
>
> 1 1 1.000000000000000000000000000
> 1 2 1.500000000000000000000000000
> 1 3 2.166666666666666666666666667
> 1 4 3.416666666666666666666666667
> 1 5 4.216666666666666666666666667
> 1 6 4.550000000000000000000000000
> 1 7 5.407142857142857142857142857
> 1 8 7.032142857142857142857142857
> 1 9 8.254365079365079365079365079
> 1 10000 8956.254813409229652481598757
> 1 20000 17913.54053798274101959639246
> 1 30000 26870.58900461199889420226700
> 1 40000 35827.55225952432056619824099
> 1 50000 44784.92216728336777991954546
> 1 60000 53741.30998083683291788901077
> 1 70000 62698.82877172292127401143322
> 1 80000 71656.42458494024747176547617
> 1 90000 80613.31691563743683765060924
> 1 100000 89570.20047616274019702521482
> 1 110000 98527.44018335441754406085900
> 1 120000 107484.2058258699552949703883
> 1 130000 116441.3450587298928159003627
> 1 140000 125399.1600101618061663172293
> 1 150000 134355.0716257765731578804073
> 1 160000 143312.6869188725594888173896
> 1 170000 152270.1691945778107893068833
> 1 180000 161227.3673088893702787242209
> 1 190000 170184.2283671443801174189235
> 1 200000 179141.9177161676803505348327
> 1 210000 188098.1905736334290746346704
> 1 220000 197055.5778178213299226211954
> 1 230000 206012.3243951074382665848232
> 1 240000 214969.0180191522602672455303
> 1 250000 223927.0626654355275622417046
> 1 260000 232884.0479228396547382227555
> 1 270000 241841.6897312605543987962985
> 1 280000 250798.1260711861562596104630
> 1 290000 259755.4115589713808307871029
> 1 300000 268712.2473108133199765189322
> 1 310000 277669.3610844430796369732363
> 1 320000 286626.7360084187087594659096
> 1 330000 295583.0702328420388811375076
> 1 340000 304540.9324662583651406115180
> 1 350000 313497.9488508493337302810705
> 1 360000 322455.2848872162239393519675
> 1 370000 331411.9952274305516780284586
> 1 380000 340369.3878688847801453424847
> 1 390000 349326.5442385613025662494469
> 1 400000 358283.7198151948112652311070
> 1 410000 367240.8596635115166082534173
> 1 420000 376197.2486760902447184900822
> 1 430000 385154.6524560274677028589761
> 1 440000 394112.0480841471143112935114
> 1 450000 403069.2697966051366843904756
> 1 460000 412026.1487890321657369929892
> 1 470000 420983.4928713199267860102478
> 1 480000 429940.0597652682271439943634
> 1 490000 438897.5981659322519223492273
> 1 500000 447855.4516433798944040122164
> 1 510000 456811.4127981064830642426529
> 1 520000 465769.1531758862533175249594
> 1 530000 474726.4930671377953578153300
> 1 540000 483683.7925672078781534825684
> 1 550000 492640.2315981062235193108053
> 1 560000 501597.6949987133986923353009
> 1 570000 510554.3958474732897001847922
> 1 580000 519511.5431567859764988621906
> 1 590000 528468.5920184442509092919859
> 1 600000 537425.2057958757909274862160
> 1 610000 546383.3300255139505983850944
> 1 620000 555340.2697592667964924572269
> 1 630000 564297.7073054137624068690112
> 1 640000 573254.2920689474814450061694
> 1 650000 582211.4873730487914773223047
> 1 660000 591168.5211525597522567544916
> 1 670000 600125.8589826114248021714650
> 1 680000 609083.0330009002048275663287
> 1 690000 618039.4919349372313479054900
> 1 700000 626997.2222632460664089943429
> 1 710000 635954.0917690439691078847726
> 1 720000 644911.6290093784488180074886
> 1 730000 653868.6304875870104467000556
> 1 740000 662825.3447013751015479514900
> 1 750000 671782.5558801465303249818702
> 1 760000 680739.9011662731435592584669
> 1 770000 689696.9475041453715157412648
> 1 780000 698653.1437839886472918561286
> 1 790000 707611.0985532863373721850039
> 1 800000 716568.7987693866776749447598
> 1 810000 725525.6575122445531974385646
> 1 820000 734482.4569919820693082893669
> 1 830000 743439.4779916042359124217226
> 1 840000 752396.3611940500092646796766
> 1 850000 761353.5394438445571430835057
> 1 860000 770310.9434931792736520368474
> 1 870000 779267.1927223886953325647496
> 1 880000 788225.1436674517253585125689
>
>
> 2 1 1.000000000000000000000000000
> 2 2 1.250000000000000000000000000
> 2 3 1.472222222222222222222222222
> 2 4 1.784722222222222222222222222
> 2 5 1.944722222222222222222222222
> 2 6 2.000277777777777777777777778
> 2 7 2.122726757369614512471655329
> 2 8 2.325851757369614512471655329
> 2 9 2.461654226505416981607457798
> 2 10000 8.619813177975507784214344698
> 2 20000 9.240635095871089404115753095
> 2 30000 9.603791897573761358292380061
> 2 40000 9.861458278261452185262848858
> 2 50000 10.06133043757656995196303496
> 2 60000 10.22462154552329778519365923
> 2 70000 10.36270038005636494621123847
> 2 80000 10.48231016697477680498962023
> 2 90000 10.58780522013752853231672862
> 2 100000 10.68217402235127296288785820
> 2 110000 10.76754463607533705360724580
> 2 120000 10.84547789729425765394094918
> 2 130000 10.91717268365027978460835188
> 2 140000 10.98355652885066079792761960
> 2 150000 11.04534545539722125239381859
> 2 160000 11.10315629233238669062836862
> 2 170000 11.15746025373895516747888760
> 2 180000 11.20865775307751148407150182
> 2 190000 11.25708463818456648077093060
> 2 200000 11.30303121795283056763425981
> 2 210000 11.34672871948779384291766276
> 2 220000 11.38839827063844308711114745
> 2 230000 11.42821240662906506074836109
> 2 240000 11.46633165190016188934526342
> 2 250000 11.50290004647923375585461222
> 2 260000 11.53802980091084626273394586
> 2 270000 11.57183602344116902681335304
> 2 280000 11.60440829561909536077171934
> 2 290000 11.63584045378869260629085740
> 2 300000 11.66620539683007648663354504
> 2 310000 11.69557551635064846235471407
> 2 320000 11.72401391041322342493145640
> 2 330000 11.75157395832622563415629569
> 2 340000 11.77831576251751436463182945
> 2 350000 11.80427986574616427809478607
> 2 360000 11.82951337960016320803463309
> 2 370000 11.85405377158818097821428014
> 2 380000 11.87794149361683265940241605
> 2 390000 11.90120806565142413868457075
> 2 400000 11.92388559358277288937420440
> 2 410000 11.94600303970164868968234173
> 2 420000 11.96758569454234998141373716
> 2 430000 11.98866287480559328733313634
> 2 440000 12.00925544388189350449936925
> 2 450000 12.02938482345842327394291528
> 2 460000 12.04907101416111155199518539
> 2 470000 12.06833481184766752219695603
> 2 480000 12.08719139777266186606623586
> 2 490000 12.10566116650605692754670219
> 2 500000 12.12375840131307130932177405
> 2 510000 12.14149351071909408890611230
> 2 520000 12.15888769016783248485026059
> 2 530000 12.17594975401341942285000014
> 2 540000 12.19269280126406301090981599
> 2 550000 12.20912705098196862278557687
> 2 560000 12.22526702472305560991043337
> 2 570000 12.24111996896626519121471994
> 2 580000 12.25669798166764128539680162
> 2 590000 12.27200952384435815319104573
> 2 600000 12.28706299338740013449483108
> 2 610000 12.30187012252190226181814868
> 2 620000 12.31643453756242668558457904
> 2 630000 12.33076670711890682146429120
> 2 640000 12.34487182648867321005497632
> 2 650000 12.35875920337989470739102627
> 2 660000 12.37243430644503889140869087
> 2 670000 12.38590421689421153982279985
> 2 680000 12.39917431914306915108546756
> 2 690000 12.41224965443445935379058168
> 2 700000 12.42513867918653815467126754
> 2 710000 12.43784365088579118428209464
> 2 720000 12.45037185509338005878377820
> 2 730000 12.46272650898502725121177200
> 2 740000 12.47491268376206272216237202
> 2 750000 12.48693594951832440956683550
> 2 760000 12.49880013638169511883229569
> 2 770000 12.51050884124405914256853750
> 2 780000 12.52206537108093533871333277
> 2 790000 12.53347692259745252540675575
> 2 800000 12.54474459716977128685280547
> 2 810000 12.55587125001645305770022096
> 2 820000 12.56686130746829341595469366
> 2 830000 12.57771842011223665163037795
> 2 840000 12.58844534265072464040923243
> 2 850000 12.59904566706465055360427300
> 2 860000 12.60952226922408394518254911
> 2 870000 12.61987642201166975414741555
> 2 880000 12.63011418303496833362065684
> 2 890000 12.64023571583693921823484395
> 2 900000 12.65024408926657285403820450
> 2 910000 12.66014049405210257196085418
> 2 920000 12.66993013960070545251999449
> 2 930000 12.67961329658082585267888159
> 2 940000 12.68919341076487592672709130
> 2 950000 12.69867217785123447166660908
> 2 960000 12.70805040456023104531710849
> 2 970000 12.71733330696500792660758365
> 2 980000 12.72652037909829933006161382
> 2 990000 12.73561385879130711822105048
> 2 1000000 12.74461639944508312563225494
>
>
> 3 1 1.000000000000000000000000000
> 3 2 1.125000000000000000000000000
> 3 3 1.199074074074074074074074074
> 3 4 1.277199074074074074074074074
> 3 5 1.309199074074074074074074074
> 3 6 1.318458333333333333333333333
> 3 7 1.335951044703595724003887269
> 3 8 1.361341669703595724003887269
> 3 9 1.376430832940907109463420877
> 3 10000 1.463913982110741753040591678
> 3 20000 1.463958761182127253891639226
> 3 30000 1.463973687899719174955107484
> 3 40000 1.463981151485566123839035784
> 3 50000 1.463985629911768534082474807
> 3 60000 1.463988615257603046255509048
> 3 70000 1.463990747941060935932260523
> 3 80000 1.463992347452072574407883095
> 3 90000 1.463993591418911296987275373
> 3 100000 1.463994586598910692322299337
> 3 110000 1.463995400874372718599240224
> 3 120000 1.463996079403452056486658683
> 3 130000 1.463996653569731083112186861
> 3 140000 1.463997145746936259101012298
> 3 150000 1.463997572212239449568643565
> 3 160000 1.463997945442508018944894819
> 3 170000 1.463998274756427172013405468
> 3 180000 1.463998567470471281896562070
> 3 190000 1.463998829362945706567178177
> 3 200000 1.463999065087923587578410681
> 3 210000 1.463999278329572278322382185
> 3 220000 1.463999472210290336272529678
> 3 230000 1.463999649219477985324795694
> 3 240000 1.463999811477588227637144335
> 3 250000 1.463999960777271571969049099
> 3 260000 1.464000098575628251036076097
> 3 270000 1.464000226175768714632044135
> 3 280000 1.464000344645861174695942201
> 3 290000 1.464000454956284414358238477
> 3 300000 1.464000557907605374141443674
> 3 310000 1.464000654220015163939180059
> 3 320000 1.464000744515464470055773720
> 3 330000 1.464000829328753859822237621
> 3 340000 1.464000909166687500211857215
> 3 350000 1.464000984435262908416401907
> 3 360000 1.464001055524739843653406744
> 3 370000 1.464001122766895437258623701
> 3 380000 1.464001186474841361410807209
> 3 390000 1.464001246914091200041356493
> 3 400000 1.464001304331492624240843891
> 3 410000 1.464001358947841911631088786
> 3 420000 1.464001410959152183436079651
> 3 430000 1.464001460557008749462202957
> 3 440000 1.464001507900298795165105625
> 3 450000 1.464001553138542176224706874
> 3 460000 1.464001596408268761316169011
> 3 470000 1.464001637838876141306278665
> 3 480000 1.464001677539798346550656570
> 3 490000 1.464001715624418180233792011
> 3 500000 1.464001752186865804286414230
> 3 510000 1.464001787308096983826068464
> 3 520000 1.464001821085252219783666269
> 3 530000 1.464001853586288170141300728
> 3 540000 1.464001884883421221311137792
> 3 550000 1.464001915039618326560450277
> 3 560000 1.464001944122159842425557477
> 3 570000 1.464001972181866862157721242
> 3 580000 1.464001999275380923272745020
> 3 590000 1.464002025450188768071703730
> 3 600000 1.464002050751299694154935270
> 3 610000 1.464002075226966770079630464
> 3 620000 1.464002098909928297454328876
> 3 630000 1.464002121842320025653578842
> 3 640000 1.464002144055977272720855659
> 3 650000 1.464002165587616796774276693
> 3 660000 1.464002186466412269783936185
> 3 670000 1.464002206722643544544226821
> 3 680000 1.464002226382732127695424983
> 3 690000 1.464002245471462412103956612
> 3 700000 1.464002264017435494080639785
> 3 710000 1.464002282039246813434225341
> 3 720000 1.464002299561748289079351576
> 3 730000 1.464002316603153605988980695
> 3 740000 1.464002333183468992396940641
> 3 750000 1.464002349322542550155865090
> 3 760000 1.464002365037130225625255935
> 3 770000 1.464002380343037616331619950
> 3 780000 1.464002395255086253184010138
> 3 790000 1.464002409792474848650673824
> 3 800000 1.464002423965997236326475471
> 3 810000 1.464002437788253358795504488
> 3 820000 1.464002451273302166423360831
> 3 830000 1.464002464433742258170070393
> 3 840000 1.464002477280648906020788669
> 3 850000 1.464002489825692224210435982
> 3 860000 1.464002502079291985535094261
> 3 870000 1.464002514049668308408128899
> 3 880000 1.464002525750211478039722139
> 3 890000 1.464002537187197706443946991
> 3 900000 1.464002548369947911638829284
> 3 910000 1.464002559305408714470469907
> 3 920000 1.464002570004676399632165939
> 3 930000 1.464002580473145840547517414
> 3 940000 1.464002590719441673840433620
> 3 950000 1.464002600750056430793286939
> 3 960000 1.464002610570359301053681776
> 3 970000 1.464002620190111512591642127
> 3 980000 1.464002629612902400536905838
> 3 990000 1.464002638845007329954067381
> 3 1000000 1.464002647892926094713436024
>
>
> 4 1 1.000000000000000000000000000
> 4 2 1.062500000000000000000000000
> 4 3 1.087191358024691358024691358
> 4 4 1.106722608024691358024691358
> 4 5 1.113122608024691358024691358
> 4 6 1.114665817901234567901234568
> 4 7 1.117164776668414909425599416
> 4 8 1.120338604793414909425599416
> 4 9 1.122015178486449507809992039
> 4 10000 1.126342859836933119635341727
> 4 20000 1.126342863195068461987333491
> 4 30000 1.126342863816977724856173635
> 4 40000 1.126342864034657367890755384
> 4 50000 1.126342864135419045299551969
> 4 60000 1.126342864190149270391012376
> 4 70000 1.126342864223154637804140480
> 4 80000 1.126342864244576264194082203
> 4 90000 1.126342864259261709833237965
> 4 100000 1.126342864269766231843710036
> 4 110000 1.126342864277538770311620766
> 4 120000 1.126342864283450137689488953
> 4 130000 1.126342864288050791700191891
> 4 140000 1.126342864291701518257416350
> 4 150000 1.126342864294646135946293287
> 4 160000 1.126342864297056569571605552
> 4 170000 1.126342864299054229967212718
> 4 180000 1.126342864300728234359584700
> 4 190000 1.126342864302144893188134103
> 4 200000 1.126342864303354525085913547
> 4 210000 1.126342864304395340180474300
> 4 220000 1.126342864305297594007409955
> 4 230000 1.126342864306084686590084820
> 4 240000 1.126342864306775457553279160
> 4 250000 1.126342864307385095784161975
> 4 260000 1.126342864307925686556878209
> 4 270000 1.126342864308407364997509003
> 4 280000 1.126342864308838305443671084
> 4 290000 1.126342864309225477016399704
> 4 300000 1.126342864309574563445429650
> 4 310000 1.126342864309890425673710622
> 4 320000 1.126342864310177149073405791
> 4 330000 1.126342864310438174074903581
> 4 340000 1.126342864310676548757732228
> 4 350000 1.126342864310894763694630232
> 4 360000 1.126342864311095054783294446
> 4 370000 1.126342864311279313873956099
> 4 380000 1.126342864311449231412772564
> 4 390000 1.126342864311606242436985681
> 4 400000 1.126342864311751625747966668
> 4 410000 1.126342864311886501047539562
> 4 420000 1.126342864312011847410167188
> 4 430000 1.126342864312128564203286214
> 4 440000 1.126342864312237413459871362
> 4 450000 1.126342864312339084988654298
> 4 460000 1.126342864312434194531858928
> 4 470000 1.126342864312523302682800842
> 4 480000 1.126342864312606892652019146
> 4 490000 1.126342864312685425831403550
> 4 500000 1.126342864312759296677644308
> 4 510000 1.126342864312828850304390296
> 4 520000 1.126342864312894443054768726
> 4 530000 1.126342864312956355217879326
> 4 540000 1.126342864313014859446213372
> 4 550000 1.126342864313070196425470463
> 4 560000 1.126342864313122601535952824
> 4 570000 1.126342864313172268506265819
> 4 580000 1.126342864313219391141266622
> 4 590000 1.126342864313264137597311253
> 4 600000 1.126342864313306663426890322
> 4 610000 1.126342864313347121770685419
> 4 620000 1.126342864313385633102777798
> 4 630000 1.126342864313422327185152818
> 4 640000 1.126342864313457311416964776
> 4 650000 1.126342864313490695748042037
> 4 660000 1.126342864313522573574899694
> 4 670000 1.126342864313553035740126664
> 4 680000 1.126342864313582163334291352
> 4 690000 1.126342864313610031536993753
> 4 700000 1.126342864313636717733889822
> 4 710000 1.126342864313662281830870319
> 4 720000 1.126342864313686789977325667
> 4 730000 1.126342864313710296430980716
> 4 740000 1.126342864313732855692110296
> 4 750000 1.126342864313754519823275656
> 4 760000 1.126342864313775334724375007
> 4 770000 1.126342864313795343268423855
> 4 780000 1.126342864313814585400999315
> 4 790000 1.126342864313833105102330919
> 4 800000 1.126342864313850934103761246
> 4 810000 1.126342864313868105235437633
> 4 820000 1.126342864313884651902605718
> 4 830000 1.126342864313900604514476466
> 4 840000 1.126342864313915990566003172
> 4 850000 1.126342864313930837276585851
> 4 860000 1.126342864313945169444268968
> 4 870000 1.126342864313959008481394737
> 4 880000 1.126342864313972380956143569
> 4 890000 1.126342864313985304497313295
> 4 900000 1.126342864313997799554982470
> 4 910000 1.126342864314009883288627525
> 4 920000 1.126342864314021576813991216
> 4 930000 1.126342864314032894395637362
> 4 940000 1.126342864314043853302500574
> 4 950000 1.126342864314054467992200910
> 4 960000 1.126342864314064751304010854
> 4 970000 1.126342864314074720218563997
> 4 980000 1.126342864314084384860927016
> 4 990000 1.126342864314093757784864329
> 4 1000000 1.126342864314102851387025112
>
>
> 5 1 1.000000000000000000000000000
> 5 2 1.031250000000000000000000000
> 5 3 1.039480452674897119341563786
> 5 4 1.044363265174897119341563786
> 5 5 1.045643265174897119341563786
> 5 6 1.045900466820987654320987654
> 5 7 1.046257460930584845967325490
> 5 8 1.046654189446209845967325490
> 5 9 1.046840475412102579121146892
> 5 10000 1.047127646962981639236350714
> 5 20000 1.047127646963242803683634380
> 5 30000 1.047127646963269060444667980
> 5 40000 1.047127646963275452376929668
> 5 50000 1.047127646963277728778610486
> 5 60000 1.047127646963278734909173763
> 5 70000 1.047127646963279246707166974
> 5 80000 1.047127646963279534023699805
> 5 90000 1.047127646963279707590251386
> 5 100000 1.047127646963279818572046343
> 5 110000 1.047127646963279892819669335
> 5 120000 1.047127646963279944352143863
> 5 130000 1.047127646963279981235720216
> 5 140000 1.047127646963280008327383488
> 5 150000 1.047127646963280028667159600
> 5 160000 1.047127646963280044239866946
> 5 170000 1.047127646963280056361640181
> 5 180000 1.047127646963280065937720739
> 5 190000 1.047127646963280073602739981
> 5 200000 1.047127646963280079811373591
> 5 210000 1.047127646963280084892514359
> 5 220000 1.047127646963280089092048700
> 5 230000 1.047127646963280092592526331
> 5 240000 1.047127646963280095533744603
> 5 250000 1.047127646963280098023437624
> 5 260000 1.047127646963280100144477685
> 5 270000 1.047127646963280101962983953
> 5 280000 1.047127646963280103530722491
> 5 290000 1.047127646963280104889770905
> 5 300000 1.047127646963280106073563423
> 5 310000 1.047127646963280107109545028
> 5 320000 1.047127646963280108020080532
> 5 330000 1.047127646963280108823485592
> 5 340000 1.047127646963280109535261396
> 5 350000 1.047127646963280110167943658
> 5 360000 1.047127646963280110732291221
> 5 370000 1.047127646963280111237235413
> 5 380000 1.047127646963280111690454876
> 5 390000 1.047127646963280112098366088
> 5 400000 1.047127646963280112466502499
> 5 410000 1.047127646963280112799594529
> 5 420000 1.047127646963280113101691720
> 5 430000 1.047127646963280113376369686
> 5 440000 1.047127646963280113626641318
> 5 450000 1.047127646963280113855154497
> 5 460000 1.047127646963280114064219587
> 5 470000 1.047127646963280114255879066
> 5 480000 1.047127646963280114431883567
> 5 490000 1.047127646963280114593830274
> 5 500000 1.047127646963280114743084164
> 5 510000 1.047127646963280114880831757
> 5 520000 1.047127646963280115008212055
> 5 530000 1.047127646963280115126153891
> 5 540000 1.047127646963280115235519959
> 5 550000 1.047127646963280115337066821
> 5 560000 1.047127646963280115431500449
> 5 570000 1.047127646963280115519415563
> 5 580000 1.047127646963280115601376090
> 5 590000 1.047127646963280115677873082
> 5 600000 1.047127646963280115749351712
> 5 610000 1.047127646963280115816230987
> 5 620000 1.047127646963280115878856405
> 5 630000 1.047127646963280115937571799
> 5 640000 1.047127646963280115992669514
> 5 650000 1.047127646963280116044432219
> 5 660000 1.047127646963280116093104355
> 5 670000 1.047127646963280116138915493
> 5 680000 1.047127646963280116182070550
> 5 690000 1.047127646963280116222756887
> 5 700000 1.047127646963280116261156889
> 5 710000 1.047127646963280116297420392
> 5 720000 1.047127646963280116331699689
> 5 730000 1.047127646963280116364124370
> 5 740000 1.047127646963280116394819088
> 5 750000 1.047127646963280116423900164
> 5 760000 1.047127646963280116451471135
> 5 770000 1.047127646963280116477627542
> 5 780000 1.047127646963280116502457450
> 5 790000 1.047127646963280116526050682
> 5 800000 1.047127646963280116548478239
> 5 810000 1.047127646963280116569809891
> 5 820000 1.047127646963280116590113535
> 5 830000 1.047127646963280116609450953
> 5 840000 1.047127646963280116627878223
> 5 850000 1.047127646963280116645449098
> 5 860000 1.047127646963280116662212610
> 5 870000 1.047127646963280116678212198
> 5 880000 1.047127646963280116693495678
> 5 890000 1.047127646963280116708099149
> 5 900000 1.047127646963280116722060660
> 5 910000 1.047127646963280116735413376
> 5 920000 1.047127646963280116748193681
> 5 930000 1.047127646963280116760429366
> 5 940000 1.047127646963280116772150557
> 5 950000 1.047127646963280116783383436
> 5 960000 1.047127646963280116794151686
> 5 970000 1.047127646963280116804482528
> 5 980000 1.047127646963280116814395317
> 5 990000 1.047127646963280116823911289
> 5 1000000 1.047127646963280116833050882
>
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From ralf at ark.in-berlin.de  Wed Nov  8 09:26:05 2006
From: ralf at ark.in-berlin.de (Ralf Stephan)
Date: Wed, 8 Nov 2006 09:26:05 +0100
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <20061108025445.GC6252@cochise.anu.edu.au>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com> <20061102011002.GC6041@cochise.anu.edu.au> <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com> <20061108025445.GC6252@cochise.anu.edu.au>
Message-ID: <20061108082605.GA21119@ark.in-berlin.de>

You wrote 
> * Jonathan Post <jvospost3 at gmail.com> [Nov 08. 2006 13:29]:
> > Thank you, Joerg.  That was interesting.  Does your generating function
> > hypothesis hold when I extend the sequence by another 5 values, to a(n) = 0, 0,
> > 1, 1, 3, 1, 7, 4,
> > 14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again, assuming that
> > I got the elementary algebra and arithmetic correct.
> 
> Alas, the function I gave was spurious, with the
> additional terms no OGF is found.

Quite possible. Always treat them as hypotheses, as Jonathan said.


ralf






From maxale at gmail.com  Wed Nov  8 11:27:40 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 8 Nov 2006 02:27:40 -0800
Subject: A038379 is all wrong
Message-ID: <d3dac270611080227lc6e0ad2qde71bd834c86d0f2@mail.gmail.com>

A038379 is defined as the "number of positive semi-definite real {0,1}
n X n matrices".
But the sequence and comments have nothing to do with the description.

In particular, all given values do not match the definition.
It lists 1, 3, 27, 729, 52649 while the correct values (that match the
definition) are 2, 7, 59, 1468, 102751 (I will be grateful if somebody
check these values).
For the simplest case n=1, it is clear that there two positive
semi-definite real {0,1}-matrices:
(0) and (1).

There is also a comment:
%C A038379 For n <= 4 a(n) = the upper bound 3^C(n,2).
Why the upper bound is 3^C(n,2) ? We have only two options for each
element of the matrix.
There would be an upper bound of this type if:
1) The main diagonal of a matrix is fixed, and the matrix is
symmetric. Then there are C(n,2) distinct non-diagonal elements whose
values should be assigned.
2) There are three (not two!) possible values for each non-diagonal
element of the matrix.

It seems that the definition of A038379 does not match the rest. Can
anybody find a definition of A038379 that is consistent with its
current content? If so, I will submit the real "number of positive
semi-definite real {0,1} n X n matrices" as a separate sequence.

Regards,
Max





From njas at research.att.com  Wed Nov  8 15:54:53 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 8 Nov 2006 09:54:53 -0500 (EST)
Subject: A038379 is all wrong
Message-ID: <200611081454.JAA83283@fry.research.att.com>

Max,   maybe I meant to say "Number of positive semi-definite real
matrices with entries {-1, 0, +1}." - can you check?

Neil





From maxale at gmail.com  Wed Nov  8 22:59:13 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 8 Nov 2006 13:59:13 -0800
Subject: A038379 is all wrong
In-Reply-To: <200611081454.JAA83283@fry.research.att.com>
References: <200611081454.JAA83283@fry.research.att.com>
Message-ID: <d3dac270611081359t117aab4ag1545e9c3130fa53a@mail.gmail.com>

On 11/8/06, N. J. A. Sloane <njas at research.att.com> wrote:
> Max,   maybe I meant to say "Number of positive semi-definite real
> matrices with entries {-1, 0, +1}." - can you check?

For the comment to make sense, the main diagonal should be fixed as well.
I will check most obvious variants.

Max





From jvospost3 at gmail.com  Thu Nov  9 02:51:54 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 8 Nov 2006 17:51:54 -0800
Subject: primes in arithmetic progression
In-Reply-To: <335cf8920611042302v7db1ca58xb8b67f872981b204@mail.gmail.com>
References: <200611041910.OAA62978@fry.research.att.com>
	 <5542af940611041355vd65aa5cr3a4a9df3cc348c15@mail.gmail.com>
	 <335cf8920611042302v7db1ca58xb8b67f872981b204@mail.gmail.com>
Message-ID: <5542af940611081751y69ee576as6473da25e91a2175@mail.gmail.com>

Sorry, Olivier.  I hope not to repeat the mistake.

Thank you again, Richard J. Mathar, for your corrections and extensions of 6
Nov 2006, which appear online today for A124570.  I'm surprised that nobody
looked in this direction, before you started it.  NJAS was right to see the
value, and 'kick things off" -- and I hope that it does not stop with my
clumsy shot at the semiprime case, which you so gallantly repaired.

I'm mostly satisfied with the results of myself and tens of millions of
others voting in the U.S. elections Tuesday 7 Nove 2006.  I feel as if the
whole country is being corrected and extended, to benefits both domestic and
global. But this is not the venue to discuss that.



On 11/4/06, Olivier Gerard <olivier.gerard at gmail.com> wrote:
>
> Hello Jonathan,
>
> Thanks for this message.
>
> Please remember NOT to send your posts to seqfan as HTML,
> even with a pure text version.
>
> regards,
>
> Olivier
>
>
> On 11/4/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> >
> > The analogue for 3-almost primes (or, as njas prefers, numbers which are
> > the product of exactly 3 primes) is left as an exercise for the reader...
> > I've just submitted the following.
> >
> > Subject: PRE-NUMBERED NEW SEQUENCE A124570 FROM Jonathan Vos Post
> >
> >
> >
>
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From jvospost3 at gmail.com  Thu Nov  9 04:11:11 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 8 Nov 2006 19:11:11 -0800
Subject: Scrabblomics; COMMENT FROM Jonathan Vos Post RE A113172
In-Reply-To: <20061108213820.28640.qmail@web55109.mail.re4.yahoo.com>
References: <20061108213820.28640.qmail@web55109.mail.re4.yahoo.com>
Message-ID: <5542af940611081911h2ff3a639v3a4b53b5c2927de2@mail.gmail.com>

Here's more fun, with a silly if plausibly motivated name.

Subject: NEW SEQUENCE FROM Jonathan Vos Post


%I A000001
%S A000001 1,
3, 3,
8, 10, 8,
9, 12, 12, 9,
4, 15, 19, 15, 4,
7, 15, 19, 19, 15, 7,
8, 18, 19, 21, 19, 18, 8,
9, 22, 20, 11, 11, 20, 22, 9,
4, 15, 17, 15, 18, 15, 17, 15, 4,
7, 15, 18, 18, 20, 20, 18, 18, 15, 7,
8, 18, 20, 22, 21, 11, 21, 22, 20, 18, 8,
9, 22, 21, 17, 19, 18, 18, 19, 17, 21, 22, 9
%N A000001 Pascrabble Triangle, by rows.
%C A000001 The apex of the triangle is 1.  Any other value is the
Scrabble value of English name for the number which is the sum of the
numbers above. This is generated the same way as A007318 Pascal's
triangle read by rows, except apply A113172 to each sum. The first
column of this triangle is 1, 3, 8, 9, 4, 7, 8, 9, 4, 7, 8, 9, 4, 7...
= iterations 1, A113172(1), A113172(A113172(1)),
A113172(A113172(A113172(1))).
The central pascrabble numbers T(2n+1,n) = 1, 10, 19, 21, 18, 11, 22, ...
%F A000001 T(1,1) = 1; for i > 1, T(i,j) = A113172(T(i-1, j-1)+T(i-1, j).
%e A000001 Triangle begins:
row.|.values in row
.1..|01
.2..|03.03
.3..|08.10.08
.4..|09.12.12.09
.5..|04.15.19.15.04
.6..|07.15.19.19.15.07
.7..|08.18.19.21.19.18.08
.8..|09.22.20.11.11.20.22.09
.9..|04.15.17.15.18.15.17.15.04
10..|07.15.18.18.20.20.18.18.15.07
11..|08.18.20.22.21.11.21.22.20.18.8
12..|09.22.21.17.19.18.18.19.17.21.22.09
%Y A000001 cf. A007318 Pascal's triangle read by rows, A113172.
%O A000001 1
%K A000001 ,easy,nonn,tabl,word,
%A A000001 Jonathan Vos Post (jvospost2 at yahoo.com), Nov 08 2006

I think that this cut & paste is pure text, and not HTML (as Olivier asked)
but I am not sure.

On 11/8/06, jonathan post <jvospost2 at yahoo.com> wrote:
>
> My backup copy (as no autoreply yet).
>
> Subject: COMMENT FROM Jonathan Vos Post RE A113172
>
>
> %I A113172
> %S A113172
> 13,3,6,8,7,10,10,8,9,4,3,9,12,11,11,13,14,12,12,8,12,15,18,20,19,22,22
> %N A113172 Scrabble value of English word for the
> number n.
> %C A113172 Is 12 the unique fixed point of Scrabble
> value of English word for the number n, where
> A113172(n) = n? 12 --> "TWELVE" --> 1+4+1+1+4+1 -->
> 12. For what n are A113172(n) prime? For what n are
> there cycles of what length for iterations n,
> A113172(n), A113172(A113172(n)),
> A113172(A113172(A113172(n))), ...? For what n does
> A113172(n) | n? These are key questions in general
> Scrabblomics.
> %F A113172 Scrabble value of English word for the
> number n.
> %O A113172 1
> %K A113172 ,easy,nonn,word,
> %A A113172 Jonathan Vos Post (jvospost2 at yahoo.com),
> Nov 08 2006
>
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From rkg at cpsc.ucalgary.ca  Thu Nov  9 20:26:26 2006
From: rkg at cpsc.ucalgary.ca (Richard Guy)
Date: Thu, 9 Nov 2006 12:26:26 -0700 (MST)
Subject: Small notation clean-up
In-Reply-To: <5542af940611081751y69ee576as6473da25e91a2175@mail.gmail.com>
References: <200611041910.OAA62978@fry.research.att.com> 
 <5542af940611041355vd65aa5cr3a4a9df3cc348c15@mail.gmail.com> 
 <335cf8920611042302v7db1ca58xb8b67f872981b204@mail.gmail.com>
 <5542af940611081751y69ee576as6473da25e91a2175@mail.gmail.com>
Message-ID: <Pine.LNX.4.63.0611091221220.23740@csl>

Would someone more competent & energetic than I
clean up the (usually unexplained)   S_n
notation that appears sporadically in OEIS ?

Evidently it means a ``star'' with  n  vertices,
and so is in fact the complete bipartite graph
K_(1,n-1), a notation which is more widely
known.   Best to all,   R.





From maxale at gmail.com  Thu Nov  9 20:35:39 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 9 Nov 2006 11:35:39 -0800
Subject: Small notation clean-up
In-Reply-To: <Pine.LNX.4.63.0611091221220.23740@csl>
References: <200611041910.OAA62978@fry.research.att.com>
	 <5542af940611041355vd65aa5cr3a4a9df3cc348c15@mail.gmail.com>
	 <335cf8920611042302v7db1ca58xb8b67f872981b204@mail.gmail.com>
	 <5542af940611081751y69ee576as6473da25e91a2175@mail.gmail.com>
	 <Pine.LNX.4.63.0611091221220.23740@csl>
Message-ID: <d3dac270611091135u58ec268au1f85498bf1ea978b@mail.gmail.com>

On 11/9/06, Richard Guy <rkg at cpsc.ucalgary.ca> wrote:
> Would someone more competent & energetic than I
> clean up the (usually unexplained)   S_n
> notation that appears sporadically in OEIS ?
>
> Evidently it means a ``star'' with  n  vertices,

It is also often used for the symmetric group of order n.

Max





From maxale at gmail.com  Thu Nov  9 23:11:00 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 9 Nov 2006 14:11:00 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <20061108025445.GC6252@cochise.anu.edu.au>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <20061102011002.GC6041@cochise.anu.edu.au>
	 <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
	 <20061108025445.GC6252@cochise.anu.edu.au>
Message-ID: <d3dac270611091411t71b55c5fl1087e070509b8fee@mail.gmail.com>

On 11/7/06, Joerg Arndt <arndt at jjj.de> wrote:
> * Jonathan Post <jvospost3 at gmail.com> [Nov 08. 2006 13:29]:
> > Thank you, Joerg.  That was interesting.  Does your generating function
> > hypothesis hold when I extend the sequence by another 5 values, to a(n) = 0, 0,
> > 1, 1, 3, 1, 7, 4,
> > 14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again, assuming that
> > I got the elementary algebra and arithmetic correct.
>
> Alas, the function I gave was spurious, with the
> additional terms no OGF is found.

Not surprising at all!
This sequence is not well-defined (see above in this discussion), so
one may view its terms as random numbers. How can they have an OGF? ;)

Max





From jvospost3 at gmail.com  Thu Nov  9 23:15:12 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 9 Nov 2006 14:15:12 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <d3dac270611091411t71b55c5fl1087e070509b8fee@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <20061102011002.GC6041@cochise.anu.edu.au>
	 <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
	 <20061108025445.GC6252@cochise.anu.edu.au>
	 <d3dac270611091411t71b55c5fl1087e070509b8fee@mail.gmail.com>
Message-ID: <5542af940611091415m4db107dei8659adb11bdcbffa@mail.gmail.com>

Max's position is well taken.  I think that "random" overstates it
somewhat.  I think that even careful definition leaves a residuum of chaos
in the proper sense.  I have not yet posted any seq based on this, out of
caution.  When I do, I'll probably post just one and see what happens.

On 11/9/06, Max A. <maxale at gmail.com> wrote:
>
> On 11/7/06, Joerg Arndt <arndt at jjj.de> wrote:
> > * Jonathan Post <jvospost3 at gmail.com> [Nov 08. 2006 13:29]:
> > > Thank you, Joerg.  That was interesting.  Does your generating
> function
> > > hypothesis hold when I extend the sequence by another 5 values, to
> a(n) = 0, 0,
> > > 1, 1, 3, 1, 7, 4,
> > > 14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again,
> assuming that
> > > I got the elementary algebra and arithmetic correct.
> >
> > Alas, the function I gave was spurious, with the
> > additional terms no OGF is found.
>
> Not surprising at all!
> This sequence is not well-defined (see above in this discussion), so
> one may view its terms as random numbers. How can they have an OGF? ;)
>
> Max
>
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From maxale at gmail.com  Thu Nov  9 23:41:15 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 9 Nov 2006 14:41:15 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <5542af940611091415m4db107dei8659adb11bdcbffa@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <20061102011002.GC6041@cochise.anu.edu.au>
	 <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
	 <20061108025445.GC6252@cochise.anu.edu.au>
	 <d3dac270611091411t71b55c5fl1087e070509b8fee@mail.gmail.com>
	 <5542af940611091415m4db107dei8659adb11bdcbffa@mail.gmail.com>
Message-ID: <d3dac270611091441la822b41v6a324c49ec2a1a76@mail.gmail.com>

On 11/9/06, Jonathan Post <jvospost3 at gmail.com> wrote:

> Max's position is well taken.  I think that "random" overstates it somewhat.

OK, it may not be random from your point of view since you followed
some rules (I believe) defining the terms of this sequence. The
problem is that you did not let us know these rules!
Remember the example I gave above? It was about different
representations of b(1) as 2-nd degree polynomials of w:
b(1) = w^2 = 2w^2+w+1 = 3w^2+2w+2 = ...
They all are equal (since w=(-1 + i*sqrt(3))/2) but at the same time
they all have different coefficients as polynomials of w.
For some (unknown to us) reason, you chose the first representation
that gave the 1st term of your sequence equal 0 (as the coefficient of
w^0 in b(1)=w^2). But you chose b(1)=2w^2+w+1, the first term would be
1; if you chose b(1)=3w^2+2w+2, it would be 2, etc.

There is yet another way to define your sequence precisely. This time
let's forget the exact value of w and keep in mind only that w^3 = 1.
Then every element of the sequence b(0) = w, b(1) = w^2, b(n) =
w*b(n-1) + b(n-2) have unique representation as a 2-nd degree
polynomial of w:
b(2) = -x + 1
b(3) = -2*x^2 + x
b(4) = x^2 + x - 3
b(5) = 3*x^2 - 4*x + 1
b(6) = -5*x^2 + 6
b(7) = -3*x^2 + 10*x - 6
b(8) = 15*x^2 - 6*x - 9
...
This gives us the following sequence of coefficients of w^0:
0, 0, 1, 0, -3, 1, 6, -6, -9
But it is different from your original sequence (again).

Max





From franktaw at netscape.net  Fri Nov 10 00:40:10 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 09 Nov 2006 18:40:10 -0500
Subject: Incorrect conjecture
Message-ID: <8C8D26F876C6EFF-76C-A121@WEBMAIL-MA17.sysops.aol.com>

The conjecture in http://www.research.att.com/~njas/sequences/A092903 
is wrong.  This conjecture is that A092903 is not the same as A005153.  
Translated, the opposite of this conjecture is that if every number up 
to m can be expressed as the sum of distinct divisors of m, then every 
number up to sigma(m) can be.  (The latter condition is known as m is 
practical.)

Following the link from A005153 to 
http://planetmath.org/encyclopedia/PracticalNumber.html, we find the 
lemma (paraphrased for non-graphical presentation): an integer m >= 2 
with factorization Product_{i=1}^k p_i^e_i with the p_i in ascending 
order is practical if and only if p_1 = 2 and, for 1 < i <= k, p_i <= 
sigma(Product_{j < i} p_j^e_j) + 1.

Proof: Let m =  Product_{i=1}^k p_i^e_i  be a number not of this form.  
If p_1 != 2, 2 cannot be represented as a sum.  If p_i > 
sigma(Product_{j<i} p_j^e_j) + 1, then sigma(Product_{j<i} p_j^e_j) + 1 
cannot be represented as the sum, and since this number is less than 
p_i, it is less than m.  Hence not every number up to m can be 
expressed as the sum of distinct divisors of m.   Thus if every number 
up to m is expressible as the sum of distinct divisors, so is every 
number up to sigma(m).  The converse is trivial, so the conditions are 
equivalent.

Hence A092903 is a duplicate of A005153.

Franklin T. Adams-Watters

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From davidwwilson at comcast.net  Fri Nov 10 03:06:32 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 9 Nov 2006 21:06:32 -0500
Subject: Explanation of A123937
Message-ID: <002f01c7046c$d7f7f880$6401a8c0@yourxhtr8hvc4p>

This is a hurried attempt to provide background on the genesis of A123937.  I am sure that a most of what I write here is a poor rehash of basic theory, and this whole post probably equates to a few lines of generating function theory.

For d and n in Z+, let A_d(n) be the set of integer points on the d-sphere of radius sqrt(n) centered at the origin, that is:

    [1] a_d(n) = #({P in Z^d : |P|^2 = n}).

A table of a_d(n) for a few small values of d and n:

Table of a_d(n)

 \d|  0    1    2    3    4    5     6     7      8      9      10
 n\|
---+--------------------------------------------------------------
 0 |  1    1    1    1    1    1     1     1      1      1       1
 1 |  0    2    4    6    8   10    12    14     16     18      20
 2 |  0    0    4   12   24   40    60    84    112    144     180
 3 |  0    0    0    8   32   80   160   280    448    672     960
 4 |  0    2    4    6   24   90   252   574   1136   2034    3380
 5 |  0    0    8   24   48  112   312   840   2016   4320    8424
 6 |  0    0    0   24   96  240   544  1288   3136   7392   16320
 7 |  0    0    0    0   64  320   960  2368   5504  12672   28800
 8 |  0    0    4   12   24  200  1020  3444   9328  22608   52020
 9 |  0    2    4   30  104  250   876  3542  12112  34802   88660
10 |  0    0    8   24  144  560  1560  4424  14112  44640  129064

Each sequence a_d is a column of this table. All of these column sequences already exist in the OEIS.

It turns out that a_d can be gotten by convolving the sequence j = (1,2,0,0,2,...) = A000122 d times with the convolutional identity sequence e = (1,0,0,0,0,...) = A000007. We could say that a_d = j^d as a "convolutional power".

The generating function of a_d is t(x)^d where t is the generating function of j. t turns out to be the Jacobi theta function theta_3, as noted on A000122. If we apply the finite difference transform to the sequence of generating functions {t(x)^d}, we obtain the sequence {(t(x)-1)^d}. Expanding these functions back to sequences gives the sequence of sequences {b_d} where b_d = (j-e)^d as a convolutional power. This allows us to constuct a table of b_d:

Table of b_d(n)

 \d|  0    1    2    3    4    5    6    7    8    9    10
 n\|
---+------------------------------------------------------
 0 |  1    0    0    0    0    0    0    0    0    0     0
 1 |  0    2    0    0    0    0    0    0    0    0     0
 2 |  0    0    4    0    0    0    0    0    0    0     0
 3 |  0    0    0    8    0    0    0    0    0    0     0
 4 |  0    2    0    0   16    0    0    0    0    0     0
 5 |  0    0    8    0    0   32    0    0    0    0     0
 6 |  0    0    0   24    0    0   64    0    0    0     0
 7 |  0    0    0    0   64    0    0  128    0    0     0
 8 |  0    0    4    0    0  160    0    0  256    0     0
 9 |  0    2    0   24    0    0  384    0    0  512     0
10 |  0    0    8    0   96    0    0  896    0    0  1024

It turns out that the process we used to get from the a_d table to the b_d table, namely, converting columns of a_d to generating functions, applying the finite difference transform to the gf's, and expanding the resulting functions back to columns of b_d, is the same as applying the finite difference transform to the rows of the a_d table to obtain rows of the b_d table.

An inductive proof starting with b_d = (j-e)^d where j-e = (0,2,0,0,2,...) shows that b_d(d) = 2^d and b_d(n) = 0 for n < d. This means that the nth element of the nth row of the b_d table is the last nonzero element in that row, which in turn implies that the nth row of the a_d table, that is, the number of integer points P with |P|^2 = n, is a polynomial of degree n in d.

If, instead of [1], we had started with

    [1] a_d'(n) = #({P in Z^d : |P|^2 <= n})

which counts points inside as well as on the d-sphere, the nth row of a_d' table would be gotten by adding termwise the rows 0 through n of the a_d table. The sequence a_d' has generating function t(x)^d/(1-x). Similarly, the nth row of the corresponding b_d' table is the sum of rows 0 through n of the b_d table, and sequence b_d has generating function (t(x)-1)^d)/(1-x). The table of b_d', omitting the superdiagonal zeroes, is sequence A123937.

The recurrence given in the title of A123937 follows from the relation

    b_d = (1,1,1,1,1,...) = A000012 if d = 0; (j-e) * b_{d-1}' if d >= 1.

I know this is a spotty exposition, but I'm sure the loose ends could be tied together by a real number theorist.

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From zakseidov at yahoo.com  Sat Nov 11 14:53:18 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sat, 11 Nov 2006 05:53:18 -0800 (PST)
Subject: A052530: to be edited
Message-ID: <20061111135318.71618.qmail@web38206.mail.mud.yahoo.com>

Both %F's and %t don't accord with entries %S 
(mainly with first term of %S),
and note this misprint(?):
"a(n)-4*a(n+1)+a(n+2)})"

%S A052530
1,2,8,30,112,418,1560,5822,21728,81090,302632,

%F A052530 a(0) = c, a(1) = p*c^3; a(n+2) =
p*c^2*a(n+1) - a(n), for p = 1, c = 2.

%F A052530 Recurrence: {a(0)=1, a(1)=2, a(2)=8, 
a(n)-4*a(n+1)+a(n+2)}

%t A052530 a[0] = c; a[1] = p*c^3; a[n_] := a[n] =
p*c^2*a[n - 1] - a[n - 2]; p = 1; c = 2; 
Table[ a[n], {n, 0, 20} ]

Anyone may wish to fix it? 
Zak


 
____________________________________________________________________________________
Want to start your own business?
Learn how on Yahoo! Small Business.
http://smallbusiness.yahoo.com/r-index





From davidwwilson at comcast.net  Sat Nov 11 16:34:08 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sat, 11 Nov 2006 10:34:08 -0500
Subject: A124057 is subset of A045940
References: <20061105235118.90653.qmail@web38203.mail.mud.yahoo.com>
Message-ID: <004501c705a6$d4c0bea0$6501a8c0@yourxhtr8hvc4p>

Nah, there's lots of them:

Yah, lots:

8706123 24463374 32442848 32942943 36782289 48580623 55486248 57476573
59600365 59757774 62481222 62664810 62884590 63262374 63728124 64724373
65159575 65450824 69362487 70302087 70370223 70785924 71494773 72060272
72503682 73256910 73638422 74066874 74361858 75356070 75414702 78676623
83349123 84405123 88094895 92349423 93803982 96192782 97757955 97822374
98474661 99104550 100708374 100997575 101302623 101905622 101905623
102956103 105596334 110476143 111673374 112529493 112801974 113067422
114811332 114940375 115338894 115466874 116079774 118269150 119076174
120919623 123052110 123821970 123952542 124045207 124232373 124465624
124581534 126115623 126134874 126619623 127151127 128095749 129079248
129764789 131404623 131422575 132568225 132617814 133891623 133918134
134350623 134421174 135691143 138468069 142680174 142904430 142904431
145863150 148286886 151757655 152779624 153101583 153129248 153930510
154322523 155740023 155826423 158252994


----- Original Message ----- 
From: "zak seidov" <zakseidov at yahoo.com>
To: <seqfan at ext.jussieu.fr>
Sent: Sunday, November 05, 2006 6:51 PM
Subject: Re: A124057 is subset of A045940


> Just submitted:
> 
> My Q to seqfans:
> Really, there are no numbers n such that n, n+1, n+2
> and n+3 are products of 6 primes?
> No for n<7258290.
> Thanks, Zak






From njas at research.att.com  Sat Nov 11 22:52:30 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sat, 11 Nov 2006 16:52:30 -0500 (EST)
Subject: querying multidimensional sequences in OEIS
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
    <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
    <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <200611112152.QAA93259@fry.research.att.com>

Max, the other day you said:


Of course, there are many representations possible but one of them
should be called "standard", and every multidimensional sequence
should have this standard representation (among maybe others) in OEIS.

Neil, does that sound reasonable?


Me:  it is very hard to force a standard representation - there are
too many people who use the OEIS and don't spend much time
reading rules

even such a simple thing as the "offset" and "keywords" gives
a lot of people trouble.

in short, it would not work

Neil





From njas at research.att.com  Sun Nov 12 04:40:14 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sat, 11 Nov 2006 22:40:14 -0500 (EST)
Subject: Thanks to Tony Noe.
Message-ID: <200611120340.WAA18989@fry.research.att.com>


Seqfans:

in the past few weeks Tony Noe has corrected several hundred entries 
in the OEIS.  the kinds of errors he has found include numbers
broken by commas, numbers run together, dropped digits, duplicated 
numbers, and just plain wrong entries.

It is astonishing to me that there were so many errors.

And it suggests that there must be thousands of other errors
waiting to be found.

I would like to thank him for all his work.   Incidentally he
has also added hundreds of b-files.

Right now I am in the middle of a large batch of updates - there
will be a new version of the OEIS by dawn.

Tony:  thanks!


Neil





From neoneye at gmail.com  Sun Nov 12 08:27:28 2006
From: neoneye at gmail.com (Simon Strandgaard)
Date: Sun, 12 Nov 2006 08:27:28 +0100
Subject: growing binary trees
Message-ID: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>

I made up an interesting sequence:

1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 26, 33, 41,
50, 62, 77, 94, 115, 142, 174, 212, 260, 319,
389, 475, 582, 711, 867, 1060, 1296


Its a binary tree where leafs is getting
added when a nodes internal counter reaches 3.





below is my slow ruby program:

class Node
  def initialize
    @n = 1
    @childs = []
  end
  def count
    @childs.inject(@n){|n,child| n + child.count}
  end
  def grow
    return @n += 1 if @n < 3
    @childs.each{|child| child.grow }
    @childs << Node.new if @childs.size < 2
  end
end

result = []
node = Node.new
30.times do |i|
  result << node.count
  node.grow
end
p result







# btw: the grow method below outputs A000071: Fibonacci numbers - 1.
def grow
  return @n += 1 if @n < 1
  @childs.each{|child| child.grow }
  @childs << Node.new if @childs.size < 2
end


# btw: the grow method below outputs A054405: Row sums of array T as in A055215.
def grow
  return @n += 1 if @n < 2
  @childs.each{|child| child.grow }
  @childs << Node.new if @childs.size < 2
end




--
Simon Strandgaard





From ralf at ark.in-berlin.de  Sun Nov 12 10:09:02 2006
From: ralf at ark.in-berlin.de (Ralf Stephan)
Date: Sun, 12 Nov 2006 10:09:02 +0100
Subject: growing binary trees
In-Reply-To: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
Message-ID: <20061112090902.GA1205@ark.in-berlin.de>

> I made up an interesting sequence:
> 
> 1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 26, 33, 41,
> 50, 62, 77, 94, 115, 142, 174, 212, 260, 319,
> 389, 475, 582, 711, 867, 1060, 1296

According to superseeker:

                                     2      3      4
                        6 + 2 x + 2 x  - 3 x  - 4 x
                     [- ----------------------------, ogf]
                                5            3
                              -x  - 1 + x + x

Second cumulative sums of:

%I A017817
%S A017817 1,0,0,1,1,0,1,2,1,1,3,3,2,4,6,5,6,10,11,11,16,21,22,27,37,43,49,64,80,
%T A017817 92,113,144,172,205,257,316,377,462,573,693,839,1035,1266,1532,1874,
%U A017817 2301,2798,3406,4175,5099,6204,7581,9274,11303,13785,16855,20577,25088
%N A017817 a(0)=1, a(1)=a(2)=0, a(3)=1; a(n)=a(n-3)+a(n-4).
%F A017817 G.f.: 1/(1-x^3-x^4).
%F A017817 a(n)/a(n-1) tends to r = 1.2207440846..., a real root of x^4 - x - 1 = 0. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Oct 22 2006

First sums of:

Transformation T018 gave a match with:
%I A122511
%S A122511 1,1,1,2,2,2,3,4,4,5,7,8,9,12,15,17,21,27,32,38,48,59,70,86,107,129,156,
%T A122511 193,236,285,349,429,521,634,778,950,1155,1412,1728,2105,2567,3140,3833,
%U A122511 4672,5707,6973,8505,10379,12680,15478
%N A122511 Alternative method for A079398 using vector matrirx Markov.
%K A122511 nonn,uned
%A A122511 Roger Bagula (rlbagula(AT)sbcglobal.net), Sep 16 2006

which is IDENTICAL with

%I A079398
%S A079398 0,1,1,1,1,2,2,2,3,4,4,5,7,8,9,12,15,17,21,27,32,38,48,59,70,86,107,129,
%T A079398 156,193,236,285,349,429,521,634,778,950,1155,1412,1728,2105,2567,3140,
%U A079398 3833,4672,5707,6973,8505,10379,12680,15478,18884,23059,28158,34362
%N A079398 Conjectured values of d(n), the dimension of a Z-module in MZV(conv).
%C A079398 From the conjecture of Zagier, Drinfeld, Kontsevich and Goncharov (see link).
%C A079398 P(0)=P(1)=P(2)=P(3)=1, for m>3: P(m) = P(m-3) + P(m-4) is the 3rd sequence in the series: Fibonacci sequence, Padovan sequence, ... The Padovan sequence (whose ratio of successive terms approaches the plastic constant) is similar to the Perrin sequence. - Jonathan Vos Post (jvospost2(AT)yahoo.com), Jan 23 2005
%D A079398 Michel Waldschmidt, "Multiple Zeta values and Euler-Zagier numbers", in Number theory and discrete mathematics, International conference in honour of Srinivasa Ramanujan, Center for Advanced Study in Mathematics, Panjab University, Chandigarh, (Oct 02, 2000).
%H A079398 Michel Waldschmidt, <a href="http://www.institut.math.jussieu.fr/~miw/articles/pdf/Transparents.pdf">Multiple Zeta values and Euler-Zagier numbers</a>
%H A079398 Eric W. Weisstein, <a href="http://mathworld.wolfram.com/PadovanSequence.html">Padovan Sequence</a>.
%F A079398 a(1)=0 a(2)=a(3)=a(4)=1 for n>=4 a(n)=a(n-2)+a(n-3)
%F A079398 a(n)=sum{k=0..floor((n-1)/2), binomial(floor((n-k-1)/3), k)} (offset 0); a(n)=sum{k=0..floor(n/2), binomial(floor((n-k-1)/3), k)}-0^n. (offset 0). - Paul Barry (pbarry(AT)wit.ie), Jul 06 2004
%F A079398 For n>1, a(n) = P(n-2) where P(n) is defined by: P(0)=P(1)=P(2)=P(3)=1, for m>3: P(m) = P(m-3) + P(m-4). - Jonathan Vos Post (jvospost2(AT)yahoo.com), Jan 23 2005
%Y A079398 Cf. A000931.
%K A079398 nonn
%O A079398 1,6
%A A079398 Benoit Cloitre (abmt(AT)wanadoo.fr), Feb 16 2003


ralf





From neoneye at gmail.com  Sun Nov 12 16:54:56 2006
From: neoneye at gmail.com (Simon Strandgaard)
Date: Sun, 12 Nov 2006 16:54:56 +0100
Subject: growing binary trees
In-Reply-To: <20061112090902.GA1205@ark.in-berlin.de>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
	 <20061112090902.GA1205@ark.in-berlin.de>
Message-ID: <df1390cc0611120754n68a94a44k76bc07615c3bdb98@mail.gmail.com>

On 11/12/06, Ralf Stephan <ralf at ark.in-berlin.de> wrote:
> > I made up an interesting sequence:
> >
> > 1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 26, 33, 41,
> > 50, 62, 77, 94, 115, 142, 174, 212, 260, 319,
> > 389, 475, 582, 711, 867, 1060, 1296
>
> According to superseeker:
[snip]

I don't see how its already in oeis?

How is A017817, A122511, A079398 related to it?
There is no match between my numbers and their numbers.
There doesn't seem to be any constant I can add to
make them match. Im an amateur.


--
Simon Strandgaard





From ralf at ark.in-berlin.de  Sun Nov 12 17:18:14 2006
From: ralf at ark.in-berlin.de (Ralf Stephan)
Date: Sun, 12 Nov 2006 17:18:14 +0100
Subject: growing binary trees
In-Reply-To: <20061112090902.GA1205@ark.in-berlin.de>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com> <20061112090902.GA1205@ark.in-berlin.de>
Message-ID: <20061112161814.GA11423@ark.in-berlin.de>

Simon, sorry for the cryptic reply.

In other words:

Let n>=0, and your sequence a(n) = 1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 
26, 33, 41, 50, 62, 77, 94, 115, 142, 174, 212, 260, 319, ...

then 

A079398(n) = a(n-2) - a(n-1) 
A122511(n) = a(n+1) - a(n)      
A017817(n) = A122511(n+3) - A122511(n+2)

All of these sequences satisfy the recurrence
a(n) = a(n-2) + a(n-3) + c

and are thus related to the Padovan sequence A000931.
With your sequence, c appears to be 3 but this has
to be proved, and all of the above is only a conjecture,
based on solid evidence, though.


ralf







From neoneye at gmail.com  Sun Nov 12 17:30:14 2006
From: neoneye at gmail.com (Simon Strandgaard)
Date: Sun, 12 Nov 2006 17:30:14 +0100
Subject: growing binary trees
In-Reply-To: <20061112161814.GA11423@ark.in-berlin.de>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
	 <20061112090902.GA1205@ark.in-berlin.de>
	 <20061112161814.GA11423@ark.in-berlin.de>
Message-ID: <df1390cc0611120830u79249ff5ue1f9c588cde122d2@mail.gmail.com>

On 11/12/06, Ralf Stephan <ralf at ark.in-berlin.de> wrote:
[snip]
> All of these sequences satisfy the recurrence
> a(n) = a(n-2) + a(n-3) + c
>
> and are thus related to the Padovan sequence A000931.
> With your sequence, c appears to be 3 but this has
> to be proved, and all of the above is only a conjecture,
> based on solid evidence, though.
[snip]

I get it. Thanks for the explaination :-)

--
Simon Strandgaard





From njas at research.att.com  Sun Nov 12 17:38:06 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sun, 12 Nov 2006 11:38:06 -0500 (EST)
Subject: growing binary trees
Message-ID: <200611121638.LAA08650@fry.research.att.com>

I will add Simon S.'s sequence to the OEIS - it will be A123015 
in a few minutes

Can Ruby lines be combined onto one line?

NJAS





From njas at research.att.com  Sun Nov 12 18:22:45 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sun, 12 Nov 2006 12:22:45 -0500 (EST)
Subject: growing binary trees
Message-ID: <200611121722.MAA17909@fry.research.att.com>

About this sequence (it will be A123015):


1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 26, 33, 41,
50, 62, 77, 94, 115, 142, 174, 212, 260, 319,
389, 475, 582, 711, 867, 1060, 1296


Its a binary tree where leafs is getting
added when a nodes internal counter reaches 3.

- I really don't understand the definition.  What exactly is the
n-th term?

Neil Sloane





From neoneye at gmail.com  Sun Nov 12 19:10:23 2006
From: neoneye at gmail.com (Simon Strandgaard)
Date: Sun, 12 Nov 2006 19:10:23 +0100
Subject: growing binary trees
In-Reply-To: <200611121638.LAA08650@fry.research.att.com>
References: <200611121638.LAA08650@fry.research.att.com>
Message-ID: <df1390cc0611121010r2cf66135gfa52aff18c944540@mail.gmail.com>

On 11/12/06, N. J. A. Sloane <njas at research.att.com> wrote:
> I will add Simon S.'s sequence to the OEIS - it will be A123015
> in a few minutes
>
> Can Ruby lines be combined onto one line?


I cannot express this thing as math, and its difficult for me to
compact the code without loss of readability.


below is a 5 line version:

class Node; def initialize; @n = 1; @c = [] end
def count; @c.inject(@n){|n,c| n + c.count} end
def grow; return @n += 1 if @n < 3; @c.each{|c| c.grow }
@c << Node.new if @c.size < 2; end; end; r = []; node = Node.new
30.times { r << node.count; node.grow }; p r


here is the oneliner (less readable):

class C;def initialize;@n=1;@c=[] end;def n;@c.inject(@n){|n,c|n+c.n}
end;def g;return @n+=1 if @n<3; @c.each{|c|c.g};@c<<C.new if
@c.size<2;end;end;r=[];c=C.new;30.times{r<<c.n;c.g};p r



the n-th term is:
a(0)=1, a(1)=2, a(2)=3, a(3)=4, a(4)=6, a(5)=8, a(6)=10, a(7)=13, ...


code for n'th term is below:

class Node
 def initialize
   @n = 1
   @childs = []
 end
 def count
   @childs.inject(@n){|n,child| n + child.count}
 end
 def grow
   return @n += 1 if @n < 3
   @childs.each{|child| child.grow }
   @childs << Node.new if @childs.size < 2
 end
end

def a(n)
  node = Node.new
  n.times{ node.grow }
  node.count
end

0.upto(30) {|n| p a(n) }



visually it works like this:

step#0
  1


step#1
  2


step#2
  3


step#3
  3
 /
1


step#4
  3
 / \
2   1


step#5
  3
 / \
3   2


step#6
    3
   / \
  3   3
 /
1


step#7
      3
    /   \
  3       3
 / \     /
2   1   1


step#8
      3
    /   \
  3       3
 / \     / \
3   2   2   1



--
Simon Strandgaard





From maxale at gmail.com  Mon Nov 13 07:53:16 2006
From: maxale at gmail.com (Max A.)
Date: Sun, 12 Nov 2006 22:53:16 -0800
Subject: growing binary trees
In-Reply-To: <20061112090902.GA1205@ark.in-berlin.de>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
	 <20061112090902.GA1205@ark.in-berlin.de>
Message-ID: <d3dac270611122253m4930fb58qdf5d06ce4fbf8026@mail.gmail.com>

On 11/12/06, Ralf Stephan <ralf at ark.in-berlin.de> wrote:

> %I A122511
> %S A122511 1,1,1,2,2,2,3,4,4,5,7,8,9,12,15,17,21,27,32,38,48,59,70,86,107,129,156,
> %T A122511 193,236,285,349,429,521,634,778,950,1155,1412,1728,2105,2567,3140,3833,
> %U A122511 4672,5707,6973,8505,10379,12680,15478
> %N A122511 Alternative method for A079398 using vector matrirx Markov.
> %K A122511 nonn,uned
> %A A122511 Roger Bagula (rlbagula(AT)sbcglobal.net), Sep 16 2006
>
> which is IDENTICAL with
>
> %I A079398
> %S A079398 0,1,1,1,1,2,2,2,3,4,4,5,7,8,9,12,15,17,21,27,32,38,48,59,70,86,107,129,
> %T A079398 156,193,236,285,349,429,521,634,778,950,1155,1412,1728,2105,2567,3140,
> %U A079398 3833,4672,5707,6973,8505,10379,12680,15478,18884,23059,28158,34362
> %N A079398 Conjectured values of d(n), the dimension of a Z-module in MZV(conv).
> %C A079398 From the conjecture of Zagier, Drinfeld, Kontsevich and Goncharov (see link).
> %C A079398 P(0)=P(1)=P(2)=P(3)=1, for m>3: P(m) = P(m-3) + P(m-4) is the 3rd sequence in the series: Fibonacci sequence, Padovan sequence, ... The Padovan sequence (whose ratio of successive terms approaches the plastic constant) is similar to the Perrin sequence. - Jonathan Vos Post (jvospost2(AT)yahoo.com), Jan 23 2005
> %D A079398 Michel Waldschmidt, "Multiple Zeta values and Euler-Zagier numbers", in Number theory and discrete mathematics, International conference in honour of Srinivasa Ramanujan, Center for Advanced Study in Mathematics, Panjab University, Chandigarh, (Oct 02, 2000).
> %H A079398 Michel Waldschmidt, <a href="http://www.institut.math.jussieu.fr/~miw/articles/pdf/Transparents.pdf">Multiple Zeta values and Euler-Zagier numbers</a>
> %H A079398 Eric W. Weisstein, <a href="http://mathworld.wolfram.com/PadovanSequence.html">Padovan Sequence</a>.
> %F A079398 a(1)=0 a(2)=a(3)=a(4)=1 for n>=4 a(n)=a(n-2)+a(n-3)
> %F A079398 a(n)=sum{k=0..floor((n-1)/2), binomial(floor((n-k-1)/3), k)} (offset 0); a(n)=sum{k=0..floor(n/2), binomial(floor((n-k-1)/3), k)}-0^n. (offset 0). - Paul Barry (pbarry(AT)wit.ie), Jul 06 2004
> %F A079398 For n>1, a(n) = P(n-2) where P(n) is defined by: P(0)=P(1)=P(2)=P(3)=1, for m>3: P(m) = P(m-3) + P(m-4). - Jonathan Vos Post (jvospost2(AT)yahoo.com), Jan 23 2005
> %Y A079398 Cf. A000931.
> %K A079398 nonn
> %O A079398 1,6
> %A A079398 Benoit Cloitre (abmt(AT)wanadoo.fr), Feb 16 2003

Should not then A122511 be marked as a duplicate of A079398 with an
appropriate comment?
I do not see much sense in keeping them separate.

Max





From davidwwilson at comcast.net  Mon Nov 13 18:50:32 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Mon, 13 Nov 2006 12:50:32 -0500
Subject: A102567 = A106498
Message-ID: <000601c7074c$375f4d50$6501a8c0@yourxhtr8hvc4p>

A102567 and A106498 are duplicates and should be merged. The latter sequence has an associated b-file.
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From maxale at gmail.com  Mon Nov 13 19:24:29 2006
From: maxale at gmail.com (Max A.)
Date: Mon, 13 Nov 2006 10:24:29 -0800
Subject: A050259 and a new solution to 2^n = 3 (mod n)
Message-ID: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>

SeqFans,

In course of exploring all solutions of the congruence 2^n = 3 (mod n)
below 10^16, I have found a new solution:

n = 3468371109448915

This came as a big surprise to me since my original goal was to prove
that 8365386194032363 is the second term of A050259. But now it
appears to be at least the third term.

So far I've submitted the new solution as a comment to A050259, but
hopefully it will take its true place in this sequence soon.

Regards,
Max

P.S. See also http://mathworld.wolfram.com/2.html





From rkg at cpsc.ucalgary.ca  Mon Nov 13 20:15:20 2006
From: rkg at cpsc.ucalgary.ca (Richard Guy)
Date: Mon, 13 Nov 2006 12:15:20 -0700 (MST)
Subject: A050259 and a new solution to 2^n = 3 (mod n)
In-Reply-To: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>
References: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>
Message-ID: <Pine.LNX.4.63.0611131211270.24385@csl>

Congratulations!  May I know Max's full name,
so that he can earn undying fame in UPINT as
well as in OEIS ?   Thanks,   R.

On Mon, 13 Nov 2006, Max A. wrote:

> SeqFans,
>
> In course of exploring all solutions of the congruence 2^n = 3 
> (mod n)
> below 10^16, I have found a new solution:
>
> n = 3468371109448915
>
> This came as a big surprise to me since my original goal was 
> to prove
> that 8365386194032363 is the second term of A050259. But now 
> it
> appears to be at least the third term.
>
> So far I've submitted the new solution as a comment to 
> A050259, but
> hopefully it will take its true place in this sequence soon.
>
> Regards,
> Max
>
> P.S. See also http://mathworld.wolfram.com/2.html






From jvospost3 at gmail.com  Mon Nov 13 21:34:20 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Mon, 13 Nov 2006 12:34:20 -0800
Subject: A050259 and a new solution to 2^n = 3 (mod n)
In-Reply-To: <Pine.LNX.4.63.0611131211270.24385@csl>
References: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>
	 <Pine.LNX.4.63.0611131211270.24385@csl>
Message-ID: <5542af940611131234l54af8bcar76fcba8e87d0fb1b@mail.gmail.com>

Is this a hierarchy of fame, or a graph, or directed graph, or a lattice, or
what? What percentage of "nice" sequences are referenced in each successive
edition of UPINT (Unsolved problems in Number Theory)? How many OEIS seqs
per edition of UPINT, now that the two are more tightly interlinked? What
estimates does one make to extrapolate for future editions? Should there be
a "very nice" keyword for seqs that pose problems in UPINT, when they get
solved in later editions?

Only hypothetical for me, as I've had over 1400 seqs in OEIS and not one is
yet "nice", although some are comments on "nice" seqs. So if there is a
hierarchy of fame, I'm just above the lowest  level (or highest, depending
on indexing) in quality, which is most assuredly not correlated well with
quantity.  In Science Fiction, for instance, there are indeed a hierarchies
of fame, based on awards (Hugo, Nebula), coauthorships, pay, anthologies,
and the like.  I'm also just somewhat above the bottom, in part for my
coauthorships with Asimov, Bradbury, Clarke, and Heinlein.  I've done
research on Asimov Number as the Science Fiction equivalent of Erdos Number,
but that's another story.

By the way, I have the penultimate edition (so far) of UPINT as a gift from
the wonderful Tony Noe, when he bought the ultimate edition (so far). I echo
the applause by njas for all the hundreds of errors in OEIS found and
corrected by Tony, and the many many b-files.

Best,

Jonathan Vos Post

On 11/13/06, Richard Guy <rkg at cpsc.ucalgary.ca> wrote:
>
> Congratulations!  May I know Max's full name,
> so that he can earn undying fame in UPINT as
> well as in OEIS ?   Thanks,   R.
>
> On Mon, 13 Nov 2006, Max A. wrote:
>
> > SeqFans,
> >
> > In course of exploring all solutions of the congruence 2^n = 3
> > (mod n)
> > below 10^16, I have found a new solution:
> >
> > n = 3468371109448915
> >
> > This came as a big surprise to me since my original goal was
> > to prove
> > that 8365386194032363 is the second term of A050259. But now
> > it
> > appears to be at least the third term.
> >
> > So far I've submitted the new solution as a comment to
> > A050259, but
> > hopefully it will take its true place in this sequence soon.
> >
> > Regards,
> > Max
> >
> > P.S. See also http://mathworld.wolfram.com/2.html
>
>
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From r.rosenthal at web.de  Mon Nov 13 23:12:57 2006
From: r.rosenthal at web.de (Rainer Rosenthal)
Date: Mon, 13 Nov 2006 23:12:57 +0100
Subject: A050259 and a new solution to 2^n = 3 (mod n)
In-Reply-To: <Pine.LNX.4.63.0611131211270.24385@csl>
References: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com> <Pine.LNX.4.63.0611131211270.24385@csl>
Message-ID: <4558EDE9.2080404@web.de>

Richard Guy wrote:
> Congratulations!  May I know Max's full name,
> so that he can earn undying fame in UPINT as
> well as in OEIS ?   Thanks,   R.
> 
> On Mon, 13 Nov 2006, Max A. wrote:
>>In course of exploring all solutions of the congruence 2^n = 3 
>>(mod n) below 10^16, I have found a new solution:
>>
>>n = 3468371109448915
>>P.S. See also http://mathworld.wolfram.com/2.html

This is a lovely finding, which I will have to tell
immediately in the German Newsgroup de.sci.mathematik.

Congratulations from a friend of OEIS and UPINT and
of things that are not too difficult to comprehend :-)
Not only congratulations to Max for finding, but to all
of us for being witnesses.

Best regards,
Rainer Rosenthal
r.rosenthal at web.de






From franktaw at netscape.net  Tue Nov 14 00:01:52 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 13 Nov 2006 18:01:52 -0500
Subject: A102567 = A106498
In-Reply-To: <000601c7074c$375f4d50$6501a8c0@yourxhtr8hvc4p>
References: <000601c7074c$375f4d50$6501a8c0@yourxhtr8hvc4p>
Message-ID: <8C8D58ED7DCDA34-128-9CD7@FWM-R16.sysops.aol.com>

A116148 (n concatenated with n-1 gives the product of two numbers which 
differ by 2) is also a duplicate of these two.  The only possible 
exceptions would be numbers of the form 10^n-1; but then the 
concatenation is 10^{2n}+10^n-1, and 10^{2n}+10^n is never a square.

Franklin T. Adams-Watters


-----Original Message-----
From: davidwwilson at comcast.net

A102567 and A106498 are duplicates and should be merged.?The latter 
sequence has an associated b-file.




________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From maxale at gmail.com  Tue Nov 14 00:52:58 2006
From: maxale at gmail.com (Max A.)
Date: Mon, 13 Nov 2006 15:52:58 -0800
Subject: A050259 and a new solution to 2^n = 3 (mod n)
In-Reply-To: <Pine.LNX.4.63.0611131211270.24385@csl>
References: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>
	 <Pine.LNX.4.63.0611131211270.24385@csl>
Message-ID: <d3dac270611131552v64b33034n8ed43b34faf905a4@mail.gmail.com>

I answered Richard privately.
But if other people also want to refer to my full name, it is
Max Alekseyev

Regards,
Max

P.S. It is not a conspiracy but rather my way to beat spamming. So
far, spam emails never addressed me by my full name, and I can easily
recognize important emails that do contain my full name in the To:
field.

On 11/13/06, Richard Guy <rkg at cpsc.ucalgary.ca> wrote:
> Congratulations!  May I know Max's full name,
> so that he can earn undying fame in UPINT as
> well as in OEIS ?   Thanks,   R.
>
> On Mon, 13 Nov 2006, Max A. wrote:
>
> > SeqFans,
> >
> > In course of exploring all solutions of the congruence 2^n = 3
> > (mod n)
> > below 10^16, I have found a new solution:
> >
> > n = 3468371109448915
> >
> > This came as a big surprise to me since my original goal was
> > to prove
> > that 8365386194032363 is the second term of A050259. But now
> > it
> > appears to be at least the third term.
> >
> > So far I've submitted the new solution as a comment to
> > A050259, but
> > hopefully it will take its true place in this sequence soon.
> >
> > Regards,
> > Max
> >
> > P.S. See also http://mathworld.wolfram.com/2.html
>
>





From bowerc at usa.net  Tue Nov 14 02:10:04 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Mon, 13 Nov 2006 17:10:04 -0800
Subject: growing binary trees
Message-ID: <397kkNBJe5074S28.1163466604@cmsweb28.cms.usa.net>

 From Simon's diagrams, I can see this is analogous to Fibonacci's
rabbits. Basically, the behavior of the node is given by its age.
A node at age 0 or 1 grows and one at age 2 or 3 produces a new node.

 From this its easy to get a g.f. of

(1+x+x^2)/(1-x-x^3+x^5)

1 2 3 4 6 8 10 13 17 21 26 33 41 50 62 77 94 115 142 174 212 260 319 389
475 582 711 867 1060 1296 1581 1930 2359 2880 3514 4292 5242 6397 7809
9537

Also, the number of nodes has g.f.

1/(1-x-x^3+x^5)
1 1 1 2 3 3 4 6 7 8 11 14 16 20 26 31 37 47 58 69 85 106 128 155 192 235
284 348 428 520 633 777 949 1154 1411 1727 2104 2566 3139 3832

which isn't in the EIS either.

------ Original Message ------
From: "Simon Strandgaard" <neoneye at gmail.com>
To: njas at research.att.comCc: seqfan at ext.jussieu.fr
Subject: Re: growing binary trees

...

> visually it works like this:
> 
> step#0
>   1
> 
> 
> step#1
>   2
> 
> 
> step#2
>   3
> 
> 
> step#3
>   3
>  /
> 1
> 
> 
> step#4
>   3
>  / \
> 2   1
> 
> 
> step#5
>   3
>  / \
> 3   2
> 
> 
> step#6
>     3
>    / \
>   3   3
>  /
> 1
> 
> 
> step#7
>       3
>     /   \
>   3       3
>  / \     /
> 2   1   1
> 
> 
> step#8
>       3
>     /   \
>   3       3
>  / \     / \
> 3   2   2   1
> 
> 
> 
> --
> Simon Strandgaard
> 









From zbi74583 at boat.zero.ad.jp  Tue Nov 14 06:50:02 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Tue, 14 Nov 2006 14:50:02 +0900
Subject: A036471
Message-ID: <013d01c707b0$bedd9660$d967d7dc@FMC57937B6BE00>

    Hi, Seqfans

    An exhaustive search of A036471  with a computer up to 10^7 is almost 
impossible.

    So, using my algorithm, I searched small examples which are ?regular 
type?.



    See this site, there is an explanation of "type".



              http://amicable.homepage.dk/apstat.htm#typesys







    I found these examples.



    2^9*3^2*13*31*(5*11,71)*(7*23,191) 
11 digits

    2^15*3*5^2*11*31*257*(1439,19*71,23*59,29*47) 
16 digits

    2^3*3^4*5^2*(7*23,191)*(17*19,359) 
9 digits

    (2^3*19*41,2^5*199)*(3*5*7*13,3^2*5*7*139) 
8 digits

    (2^3*19*41,2^5*199)*(3^3*5*7*71,3^3*5*17*31) 
9 digits

    (2^3*19*41,2^5*199)*(3^2*7*13*5*17,3^2*7*13*107) 
9 digits

    2^7*3^2*13*(5*11,71)*(7*59,479) 
9 digits

    2^5*3^4*5^2*(17*19,359)*(23*29,719) 
11 digits







    Where, x*(y,z) means (x*z,x*y). (x,y)*(z,u)=(x*z,x*u,y*z,y*z)





    I think that if a smaller one than 3270960  exists, then it must be 
?irregular".



    Regurer type Amicable Quadruple are generated from ?seeds? which are 
two pairs or one quadruple of primes or almost primes, {x,y} and {z,u} or 
{x,y,z,u}.

    The numbers x,y,z,u of small Amicable Quadruple must satisfy the 
following conditions.

             .

    o Sigma(x)=Sigma(y) and Sigma(z)=Sigma(u), or 
Sigma(x)=Sigma(y)=Sigma(z)=Sigma(u)

    o all prime factors of x,y,z,u are small

    o all prime factors of x+y and z+u are small, or all  prime factors of 
x+y+z+u are small



    For example, if {x,y}={7*23,191} and {z,u}={17*19,359} then

    o Sigma(7*23)=Sigma(191), Sigma(17*19)=Sigma(359)

    o all prime factors of 7*23,191,17*19,359  are small

    o x+y=2^5*11, z+u=2*11*31

    So, (7*23,191)*(17*19,359) generates 2^3*3^4*5^2*(7*23,191)*(17*19,359)





    If we obtain all small seeds then we will calculate all small regular 
terms of A036471.



    Yasutoshi


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From g.resta at iit.cnr.it  Tue Nov 14 14:40:49 2006
From: g.resta at iit.cnr.it (Giovanni Resta)
Date: Tue, 14 Nov 2006 14:40:49 +0100
Subject: A036471
In-Reply-To: <013d01c707b0$bedd9660$d967d7dc@FMC57937B6BE00>
References: <013d01c707b0$bedd9660$d967d7dc@FMC57937B6BE00>
Message-ID: <4559C761.70305@iit.cnr.it>

kohmoto wrote:
>
> Hi, Seqfans
>
> An exhaustive search of A036471 with a computer up to 10^7 is almost
> impossible.
>
Here it is the result of an exhaustive search
with a computer up to 10^7, in the format
sigma -> {a,b,c,d} (with a,b,c,d<10^7)

13927680 -> {3270960, 3361680, 3461040, 3834000}
16248960 -> {3767400, 4090320, 4150440, 4240800}
19498752 -> {4651920, 4839120, 4918320, 5089392}
21228480 -> {4969440, 5116320, 5475960, 5666760}
24373440 -> {5682600, 5831280, 5920200, 6939360}
24998400 -> {5405400, 6029100, 6421800, 7142100}
27855360 -> {6514200, 6640200, 6768720, 7932240}
28304640 -> {6126120, 6541920, 7559640, 8076960}
28304640 -> {6126120, 6922080, 7280280, 7976160}
29030400 -> {6320160, 6902280, 7685496, 8122464}
29030400 -> {6977880, 7087080, 7227360, 7738080}
29877120 -> {7013160, 7436520, 7688520, 7738920}
29998080 -> {6819120, 7327320, 7715400, 8136240}
29998080 -> {6966960, 7054320, 7840560, 8136240}
32497920 -> {7706160, 7722000, 7948080, 9121680}
33022080 -> {7731360, 7852320, 8125920, 9312480}
33868800 -> {7469280, 8157240, 8873760, 9368520}
33868800 -> {7469280, 8157240, 9098460, 9143820}

Note that there are values which admit more than one quadruple,
and there are also quadruples which share 1 or 2 common values.

Here are some larger quadruples, but now there can be some
missing among these, because here I used an heuristic.

{8353800, 8920800, 10063200, 10159800}
{8288280, 9360540, 9830520, 10260180}
{8316000, 8759520, 10260180, 10403820}
{9258480, 9621360, 9661680, 10455984}
{9009000, 10174500, 10521000, 11179980}
{10048500, 10174500, 10188360, 10473120}
{9840600, 10098000, 10914120, 10930320}
{9923760, 10316880, 10747440, 10794960}
{9563400, 10085040, 11241720, 11964240}
{9646560, 10667160, 11502540, 11729340}
{9767520, 11202840, 11217960, 11357280}
{9828000, 10796940, 11362680, 12041820}
{10417680, 10538640, 10979280, 13061520}
{10450440, 11614680, 11802420, 13306860}
{10450440, 12179160, 12204360, 12340440}

etc.etc. (program is running).

Giovanni Resta






From davidwwilson at comcast.net  Tue Nov 14 18:00:27 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Tue, 14 Nov 2006 12:00:27 -0500
Subject: A023153 through A023161
References: <p06230918c17f1b479bcf@[192.168.1.43]>
Message-ID: <006d01c7080e$632ff480$6501a8c0@yourxhtr8hvc4p>

Tony Noe has found that A023153 through A023161 are not multiplicative as I 
had marked them.

At the time, I was convinced that in sequence {s_k(n) = k^(2^n)}, a cycle of 
period x (mod a) and a cycle of period y (mod b) implied a unique cycle of 
period xy (mod ab). However, if gcd(x, y) > 1, the cycles (mod a) and (mod 
b) can align in more than one way, producing more than one cycle (mod ab). 
For example, the cycle (2,4) (mod 7) and the cycle (4,7) (mod 9) produce two 
cycles (4,16) and (25,58) (mod 63). This means that sometimes f(ab) > 
f(a)f(b).

This problem affects A023153 through A023161, and Tony Noe has shown that 
indeed none of these are multiplicative.

NJAS: Please remove the "mult" keyword and any comments related to 
multiplicativity from A023153-A023161.

Tony: Since these sequences are almost multiplicative, but are not, and you 
seem to be investigating these sequences, could you please add lines such as

%C A023153 Not multiplicative: a(63) = 10, but a(7)*a(9) = 3*3 = 9.

Thanks for spotting my error, Tony. Seems I have been bitten by the dragon I 
released.

----- Original Message ----- 
From: "T. D. Noe" <noe at sspectra.com>
To: <davidwwilson at comcast.net>
Sent: Tuesday, November 14, 2006 2:06 AM
Subject: Question about A023161


> David, according to my calculations, this sequence fails to be
> multiplicative at a(667).
>
> Best regards,
>
> Tony
>
>
> -- 
> No virus found in this incoming message.
> Checked by AVG Free Edition.
> Version: 7.5.430 / Virus Database: 268.14.5/533 - Release Date: 11/13/2006 
> 8:56 PM
>
> 






From jvospost3 at gmail.com  Tue Nov 14 20:46:00 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Tue, 14 Nov 2006 11:46:00 -0800
Subject: A023153 through A023161
In-Reply-To: <006d01c7080e$632ff480$6501a8c0@yourxhtr8hvc4p>
References: <p06230918c17f1b479bcf@192.168.1.43>
	 <006d01c7080e$632ff480$6501a8c0@yourxhtr8hvc4p>
Message-ID: <5542af940611141146j1a3212b0u2d31815c6bee8387@mail.gmail.com>

"almost multiplicative" -- interesting notion.  What other "almost
multiplicative" are there in OEIS masquerading as multiplicative (i'm
guessing that the amazing Tony Noe is searching), and what are but with no
keyword on that? Can "almost multiplicative" be axiomatizeed, or is it just
a subjective interpretation of "almost"? What pattens are there in "almost
multiplicative" -- new sequences on where multiplicativity breaks down?

-- Jonathan Vos Post

On 11/14/06, David Wilson <davidwwilson at comcast.net> wrote:
>
> Tony Noe has found that A023153 through A023161 are not multiplicative as
> I
> had marked them.
>
> At the time, I was convinced that in sequence {s_k(n) = k^(2^n)}, a cycle
> of
> period x (mod a) and a cycle of period y (mod b) implied a unique cycle of
> period xy (mod ab). However, if gcd(x, y) > 1, the cycles (mod a) and (mod
> b) can align in more than one way, producing more than one cycle (mod ab).
> For example, the cycle (2,4) (mod 7) and the cycle (4,7) (mod 9) produce
> two
> cycles (4,16) and (25,58) (mod 63). This means that sometimes f(ab) >
> f(a)f(b).
>
> This problem affects A023153 through A023161, and Tony Noe has shown that
> indeed none of these are multiplicative.
>
> NJAS: Please remove the "mult" keyword and any comments related to
> multiplicativity from A023153-A023161.
>
> Tony: Since these sequences are almost multiplicative, but are not, and
> you
> seem to be investigating these sequences, could you please add lines such
> as
>
> %C A023153 Not multiplicative: a(63) = 10, but a(7)*a(9) = 3*3 = 9.
>
> Thanks for spotting my error, Tony. Seems I have been bitten by the dragon
> I
> released.
>
> ----- Original Message -----
> From: "T. D. Noe" <noe at sspectra.com>
> To: <davidwwilson at comcast.net>
> Sent: Tuesday, November 14, 2006 2:06 AM
> Subject: Question about A023161
>
>
> > David, according to my calculations, this sequence fails to be
> > multiplicative at a(667).
> >
> > Best regards,
> >
> > Tony
> >
> >
> > --
> > No virus found in this incoming message.
> > Checked by AVG Free Edition.
> > Version: 7.5.430 / Virus Database: 268.14.5/533 - Release Date:
> 11/13/2006
> > 8:56 PM
> >
> >
>
>
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From pauldhanna at juno.com  Wed Nov 15 05:04:51 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Tue, 14 Nov 2006 23:04:51 -0500
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
Message-ID: <20061114.230452.536.0.pauldhanna@juno.com>

Seqfans, 
    Could someone compute the initial terms of the following. 
I do not believe that it is in the OEIS (?). 
  
Number of labelled nodes in generation n of a rooted tree 
where a node with label k has k child nodes such that 
the label of each child node is the least unused label in 
the path from the root to that child node, 
and where root is labeled '1'. 
  
Thus every distinct path in the tree eventually forms 
a permutation of the positive integers. 
 
The wording needs to be made more precise and concise. 
 
Here are the initial nodes of the tree for generations 0..4: 
Gen 0: [1];
Gen 1: (1)->[2];
Gen 2: (2)->[3,4];
Gen 3: (3)->[4,5,6], (4)->[3,5,6,7];
Gen 4: (4)->[5,6,7,8], (5)->[4,6,7,8,9], (6)->[4,5,7,8,9,10], 
(3)->[5,6,7], (5)->[3,6,7,8,9], (6)->[3,5,7,8,9,10], 
(7)->[3,5,6,8,9,10,11]; ...
 
Thus, the number of nodes in generation n begins:
[1, 1, 2, 7, 36, 248, ...]
 
The maximum node in generation n begins:
[1, 2, 4, 7, 11, ...]
 
Would appreciate it if someone could compute more terms.
Thanks,
      Paul





From franktaw at netscape.net  Wed Nov 15 05:13:17 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Tue, 14 Nov 2006 23:13:17 -0500
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
In-Reply-To: <20061114.230452.536.0.pauldhanna@juno.com>
References: <20061114.230452.536.0.pauldhanna@juno.com>
Message-ID: <8C8D68382FC1E11-FEC-53E1@FWM-M20.sysops.aol.com>

You need to reword that a bit.  Your example makes clear that the
children of a given node must all have distinct labels, but you don't
say so.

Without that condition, the problem is pretty easy.  :-)

Franklin T. Adams-Watters


-----Original Message-----
From: pauldhanna at juno.com

...

Number of labelled nodes in generation n of a rooted tree
where a node with label k has k child nodes such that
the label of each child node is the least unused label in
the path from the root to that child node,
and where root is labeled '1'.

...
________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From njas at research.att.com  Wed Nov 15 06:13:02 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 15 Nov 2006 00:13:02 -0500 (EST)
Subject: king chickens
Message-ID: <200611150513.AAA96507@fry.research.att.com>

can anyone extend this?

%I A123553
%S A123553 1,2,12,104
%N A123553 A "king chicken" in a tournament graph (a directed labeled graph on n nodes) is a player A who for any other player B either beats B directly or beats someone who beats B.  Sequence gives total number of king chickens in all 2^(n(n-1)/2) tourna
%C A123553 H. G. Landau showed in 1951 that a player is a king chicken if and only if he has the highest score. There may be several king chickens in a tournament.
%D A123553 S. B. Maurer, The king chicken theorems, Math. Mag., 53 (1980), 67-80.
%e A123553 For n = 3 there are 8 tournaments: six of the form A beats B and C, and B beats C, with one king chicken (A), and two of the form A beats B beats C beats A, with three king chickens each (A or B or C), for a total of 6*1 + 2*3 = 12.
%K A123553 nonn,more,new
%O A123553 1,2
%A A123553 njas, Nov 14 2006

Neil





From bdm at cs.anu.edu.au  Wed Nov 15 06:37:18 2006
From: bdm at cs.anu.edu.au (Brendan McKay)
Date: Wed, 15 Nov 2006 16:37:18 +1100
Subject: king chickens
In-Reply-To: <200611150513.AAA96507@fry.research.att.com>
References: <200611150513.AAA96507@fry.research.att.com>
Message-ID: <20061115053717.GA15829@cs.anu.edu.au>


I don't have time right now, so if someone who knows how to use nauty
wants to jump into this, here is how to do it:
1. Fetch the unlabelled tournaments from
      http://cs.anu.edu.au/~bdm/data/digraphs.html
2. For each unlabelled tournament compute the automorphism group size A
   and the number of king chickens K.
3. The answer is the sum of K*(n!/A) over the unlabelled tournaments.

That will get the answer up to n=10 fairly easily.  The next case n=11
is plausible but more effort as there are nearly a billion tournaments.

Incidentally, the definition of "tournament" in the entry is incomplete.
It has to add that exactly one of the directed edges u->v, v->u is
present for each pair u,v of distinct vertices.

Brendan.


* N. J. A. Sloane <njas at research.att.com> [061115 16:13]:
> can anyone extend this?
> 
> %I A123553
> %S A123553 1,2,12,104
> %N A123553 A "king chicken" in a tournament graph (a directed labeled graph on n nodes) is a player A who for any other player B either beats B directly or beats someone who beats B.  Sequence gives total number of king chickens in all 2^(n(n-1)/2) tourna
> %C A123553 H. G. Landau showed in 1951 that a player is a king chicken if and only if he has the highest score. There may be several king chickens in a tournament.
> %D A123553 S. B. Maurer, The king chicken theorems, Math. Mag., 53 (1980), 67-80.
> %e A123553 For n = 3 there are 8 tournaments: six of the form A beats B and C, and B beats C, with one king chicken (A), and two of the form A beats B beats C beats A, with three king chickens each (A or B or C), for a total of 6*1 + 2*3 = 12.
> %K A123553 nonn,more,new
> %O A123553 1,2
> %A A123553 njas, Nov 14 2006
> 
> Neil





From g.resta at iit.cnr.it  Wed Nov 15 11:32:25 2006
From: g.resta at iit.cnr.it (Giovanni Resta)
Date: Wed, 15 Nov 2006 11:32:25 +0100
Subject: A question on A068480 
Message-ID: <455AECB9.5050308@iit.cnr.it>

A068480 is the sequence on numbers n such that gcd(n!-1,2^n+1)>1.

I noticed that for all the numbers listed gcd(n!-1,2^n+1) = 2n+1.

Is this a rule (proof ?) or a coincidence (counterexample ?) .

thanks,
giovanni.







From mathar at strw.leidenuniv.nl  Wed Nov 15 13:40:44 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Wed, 15 Nov 2006 13:40:44 +0100
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
Message-ID: <200611151240.kAFCeiKc009825@amer.strw.leidenuniv.nl>


pdh> From seqfan-owner at ext.jussieu.fr  Wed Nov 15 05:08:51 2006
pdh> To: seqfan at ext.jussieu.fr
pdh> Date: Tue, 14 Nov 2006 23:04:51 -0500
pdh> Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
pdh> From: "Paul D. Hanna" <pauldhanna at juno.com>
pdh> ...
pdh> Number of labelled nodes in generation n of a rooted tree 
pdh> where a node with label k has k child nodes such that 
pdh> the label of each child node is the least unused label in 
pdh> the path from the root to that child node, 
pdh> and where root is labeled '1'. 
pdh>   
pdh> Thus every distinct path in the tree eventually forms 
pdh> a permutation of the positive integers. 
pdh>  
pdh> The wording needs to be made more precise and concise. 
pdh>  
pdh> Here are the initial nodes of the tree for generations 0..4: 
pdh> Gen 0: [1];
pdh> Gen 1: (1)->[2];
pdh> Gen 2: (2)->[3,4];
pdh> Gen 3: (3)->[4,5,6], (4)->[3,5,6,7];
pdh> Gen 4: (4)->[5,6,7,8], (5)->[4,6,7,8,9], (6)->[4,5,7,8,9,10], 
pdh> (3)->[5,6,7], (5)->[3,6,7,8,9], (6)->[3,5,7,8,9,10], 
pdh> (7)->[3,5,6,8,9,10,11]; ...
pdh>  
pdh> Thus, the number of nodes in generation n begins:
pdh> [1, 1, 2, 7, 36, 248, ...]
pdh>  
pdh> The maximum node in generation n begins:
pdh> [1, 2, 4, 7, 11, ...]

My count of node numbers and maximum nodes is
Gen 0 1 1
Gen 1 1 2
Gen 2 2 4
Gen 3 7 7
Gen 4 36 11
Gen 5 248 16
Gen 6 2157 22
Gen 7 22761 29
Gen 8 283220 37
Gen 9 4068411 46

This list is limited by the simplicity of my program which keeps all the
trees in memory at the same time and runs into allocation problems if the
tenth generations are attempted.





From pauldhanna at juno.com  Wed Nov 15 14:34:15 2006
From: pauldhanna at juno.com (Paul D Hanna)
Date: Wed, 15 Nov 2006 13:34:15 GMT
Subject: Trees with Labeled Nodes that form Permutation of Positive Integer
	s
Message-ID: <20061115.053511.1712.764552@webmail42.nyc.untd.com>

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From franktaw at netscape.net  Wed Nov 15 18:34:02 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 15 Nov 2006 12:34:02 -0500
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
In-Reply-To: <20061114.230452.536.0.pauldhanna@juno.com>
References: <20061114.230452.536.0.pauldhanna@juno.com>
Message-ID: <8C8D6F3606C258E-3A4-7467@FWM-R09.sysops.aol.com>

From: pauldhanna at juno.com

>    Could someone compute the initial terms of the following.
>I do not believe that it is in the OEIS (?).
>
>Number of labelled nodes in generation n of a rooted tree
>where a node with label k has k child nodes such that
>the label of each child node is the least unused label in
>the path from the root to that child node,
excluding those used by earlier children of the same parent,
>and where root is labeled '1'.
>
>Thus every distinct path in the tree eventually forms
>a permutation of the positive integers.
>
>...

This isn't really true.  There are many paths in this tree
that do not form a permutation of the positive integers.
You get such a permutation iff, infinitely often, you choose
the first child from the current node.

It is true, of course, that any finite initial path can be
extended to a path that is a permuatation of the positive
integers.

It is an interesting problem to construct an interesting tree
where every path from the root is a permutation of the
positive integers.  I think the following construction works:
start with node 1, which arbitrarily has node 2 as a child
(by the construction that follows, 1 would have no children).
For each node with label m, add BigOmega(m) children, where
BigOmega(m) is the number of prime factors of m with
repetition; label these chilidren with the first positive integers
not occurring in the path from the root to the current node.

The tree starts:

......1
......|
......2
......|
......3
......|
......4
...../.\
....5...6
.../....|\
..6.....5.7
./.\....|.|
7...8...7.5
|../|\..|.|
8.7.9.A.8.8

Any time you hit a prime, there is only one child in the next
generation.  If you keep choosing non-prime children, the
primes start accumulating in the list of positive integers not
occurring in the path, until eventually all children of a node will
be prime.

One could of course use Omega(m), the number of distinct
prime divisors, instead of BigOmega(m).  I think tau(m)-1 would
also work.  A001511(m) (the power of 2 dividing 2m) works, too
- in this case, the single children are under the odd nodes.
________________________________________________________________________
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From qq-quet at mindspring.com  Wed Nov 15 19:23:30 2006
From: qq-quet at mindspring.com (Leroy Quet)
Date: Wed, 15 Nov 06 11:23:30 -0700
Subject: Max Score For A Game
Message-ID: <E1GkPUS-0005H1-00@pop04.mail.atl.earthlink.net>

Below is a game I invented (as posted to sci.math and rec.puzzles -- it 
is just one of many stupid silly simple games I have posted there).

I post it here to seq.fans because I wonder what the sequence is where 
the nth term is the maximum possible score for an n-by-n grid. (We could 
also have a table where the element in the mth column and nth row is the 
maximum score for an m-by-n grid.)
Of course, if there is a way to always get the score n^2 -1 for each 
n-by-n case, or m*n -1 for each m-by-n case (n or m equal 1 excluded), 
then there is be no need to calculate the sequence.

Thanks,
Leroy Quet

=====

Here is a solitaire game played on a square grid.
(I hope this game is more fun than my last game.)

Start by drawing an r-by-r grid on paper.
(I suggest an r of 5 or so for beginners.)
 
Put a "1" in each square of the left-most column
and in each square of the top-most row of the grid.

On each move the player chooses any one of the grid's
empty squares. This square is (m,n), which is the
square in the mth column from the left side of the
grid and in the nth row from the top.

The player then sums up all the integers already
written in the mth column and nth row.
So, if the grid looks like this, where the * is
the square the player has chosen to fill in next
with an integer, 
1 1 1 1
1 3
1 * 6
1 6

then the sum of the column 2's terms and row 3's
terms is 1+3+6 +1+6 = 17. Let this sum be s.

Next the player counts the number of integers in the
squares to the left of (m,n) and above (m,n)
(ie, the squares with coordinates (j,k), 1 <=j <=m,
1 <=k <=n) which are coprime to s.
If the count is c integers in the given rectangle
which are coprime to s, then the player writes c
in square (m,n).

Play continues until there is an integer in every
square of the grid.
The player's score is the number in the last square
he/she fills in.

Example game:
(r=4)
(View with fixed-width font.)

1 1 1 1   1 1 1 1   1 1 1 1   1 1 1 1 
1         1 3       1 3       1 3 5  
1         1         1   6     1   6
1         1         1         1

1 1 1 1   1 1 1 1   1 1 1 1
1 3 5 6   1 3 5 6   1 3 5 6
1   6     1   6     1 5 6
1         1 6       1 6

1 1 1 1   1 1 1 1   1 1 1 1
1 3 5 6   1 3 5 6   1 3 5 6
1 5 6     1 5 6     1 5 6 11
1 6   10  1 6 1110  1 6 1110

So the player gets 11 points.

(It seems that a good strategy is to
{unlike in my example} finish in the
lower right square, and to {as in my
example, 1+6+10+1+5+6=29} have a row
and column sum on the last move which
is a prime.)

Question: Is it always possible to get,
for an r-by-r grid (r >= 2), r^2 -1 points?

Thanks,
Leroy Quet





From davidwwilson at comcast.net  Wed Nov 15 19:48:18 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Wed, 15 Nov 2006 13:48:18 -0500
Subject: Domino questions
Message-ID: <001501c708e6$9e07a570$6501a8c0@yourxhtr8hvc4p>

1. On an n x n (m x n) checkerboard, what is the smallest number of dominoes that can be placed so that no further dominoes can be placed?

2. Suppose two players, A and B, alternately place dominoes on an n x n (m x n) checkerboard until no further play is possible. A plays to maximize the final number of dominoes on the board, B to minimize. What is the number of dominoes on the board at the end of the game if A moves first? If B moves first?
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From franktaw at netscape.net  Wed Nov 15 21:01:19 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 15 Nov 2006 15:01:19 -0500
Subject: Domino questions
In-Reply-To: <001501c708e6$9e07a570$6501a8c0@yourxhtr8hvc4p>
References: <001501c708e6$9e07a570$6501a8c0@yourxhtr8hvc4p>
Message-ID: <8C8D707F39EB04F-934-6F47@FWM-R01.sysops.aol.com>

From: davidwwilson at comcast.net
>1. On an n x n (m x n) checkerboard, what is the smallest number
>of dominoes that can be placed so that no further dominoes can
>be placed?

This might be ceiling(n^2/3), A008810.  Can someone please
check the reference there: "J. H. Conway and R. K. Guy, The Book
of Numbers, Copernicus Press, NY, 1996, number of red blocks in
Fig 2.5."
?
>2. Suppose two players, A and B, alternately place dominoes on an
>n x n (m x n) checkerboard until no further play is possible. A plays
>to maximize the final number of dominoes on the board, B to
>minimize. What is the number of dominoes on the board at the end
>of the game if A moves first? If B moves first?

You don't want much, do you?

(Note for anyone who may be confused - that comment means that
I think this is likely to be a hard problem.)

Franklin T. Adams-Watters
________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From njas at research.att.com  Wed Nov 15 23:40:20 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 15 Nov 2006 17:40:20 -0500 (EST)
Subject: request for a b-file for a core file
Message-ID: <200611152240.RAA60703@fry.research.att.com>


Could someone help produce a b-file for A000014?
(and A001678 along the way)

Here is the sequence:

%I A000014 M0320 N0118
%S A000014 0,1,1,0,1,1,2,2,4,5,10,14,26,42,78,132,249,445,842,1561,2988,5671,10981,
%T A000014 21209,41472,81181,160176,316749,629933,1256070,2515169,5049816,10172638,
%U A000014 20543579,41602425,84440886,171794492,350238175,715497037,1464407113
%N A000014 Number of series-reduced trees with n nodes.


I could use 200, or 500 or so terms

Notes on computing A14 and A1678:

looking at the H'book of Graph Theory, p. 525:

they define f(x) (essentially A1678 but with a different offset):
f(x) = 1*x+1*x^3+1*x^4+2*x^5+3*x^6+6*x^7+10*x^8+19*x^9+35*x^10+...

by

f = (x/(x+1))* exp( sum( f(x^i) / i, i=1..infinity ) )

then A14 is given by

(1+x)f(x) - ((1+x)/2) f(x)^2 + ((1-x)/2) f(x^2)

= x + x^2 + x^4 + x^5 + 2 x^6 + ...

= A14

I actually haven't checked these formulae - there are
slightly different versions in the entry for A14 itself.

Thanks!

Neil






From bowerc at usa.net  Thu Nov 16 01:35:58 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 15 Nov 2006 16:35:58 -0800
Subject: request for a b-file for a core file
Message-ID: <964kkPaI79512S23.1163637358@cmsweb23.cms.usa.net>

I just sent a b-file of terms 0-500 to Neil for A001678.
I'm still working on a version for A000014

------ Original Message ------
From: "N. J. A. Sloane" <njas at research.att.com>
To: seqfans at seqfan.net, seqfan at ext.jussieu.frCc: njas at research.att.com
Subject: request for a b-file for a core file

> 
> Could someone help produce a b-file for A000014?
> (and A001678 along the way)
> 
> Here is the sequence:
> 
> %I A000014 M0320 N0118
> %S A000014
0,1,1,0,1,1,2,2,4,5,10,14,26,42,78,132,249,445,842,1561,2988,5671,10981,
> %T A000014
21209,41472,81181,160176,316749,629933,1256070,2515169,5049816,10172638,
> %U A000014
20543579,41602425,84440886,171794492,350238175,715497037,1464407113
> %N A000014 Number of series-reduced trees with n nodes.
> 
> 
> I could use 200, or 500 or so terms
> 
> Notes on computing A14 and A1678:
> 
> looking at the H'book of Graph Theory, p. 525:
> 
> they define f(x) (essentially A1678 but with a different offset):
> f(x) = 1*x+1*x^3+1*x^4+2*x^5+3*x^6+6*x^7+10*x^8+19*x^9+35*x^10+...
> 
> by
> 
> f = (x/(x+1))* exp( sum( f(x^i) / i, i=1..infinity ) )
> 
> then A14 is given by
> 
> (1+x)f(x) - ((1+x)/2) f(x)^2 + ((1-x)/2) f(x^2)
> 
> = x + x^2 + x^4 + x^5 + 2 x^6 + ...
> 
> = A14
> 
> I actually haven't checked these formulae - there are
> slightly different versions in the entry for A14 itself.
> 
> Thanks!
> 
> Neil
> 
> 









From rkg at cpsc.ucalgary.ca  Thu Nov 16 01:50:35 2006
From: rkg at cpsc.ucalgary.ca (Richard Guy)
Date: Wed, 15 Nov 2006 17:50:35 -0700 (MST)
Subject: Domino questions
In-Reply-To: <8C8D707F39EB04F-934-6F47@FWM-R01.sysops.aol.com>
References: <001501c708e6$9e07a570$6501a8c0@yourxhtr8hvc4p>
 <8C8D707F39EB04F-934-6F47@FWM-R01.sysops.aol.com>
Message-ID: <Pine.LNX.4.63.0611151749420.25970@csl>

1.  Checked!  Unfortunately, in the printing,
red and orange are barely distinguishable.   R.

On Wed, 15 Nov 2006, franktaw at netscape.net wrote:

> From: davidwwilson at comcast.net
>> 1. On an n x n (m x n) checkerboard, what is the smallest number
>> of dominoes that can be placed so that no further dominoes can
>> be placed?
>
> This might be ceiling(n^2/3), A008810.  Can someone please
> check the reference there: "J. H. Conway and R. K. Guy, The Book
> of Numbers, Copernicus Press, NY, 1996, number of red blocks in
> Fig 2.5."
> ?
>> 2. Suppose two players, A and B, alternately place dominoes on an
>> n x n (m x n) checkerboard until no further play is possible. A plays
>> to maximize the final number of dominoes on the board, B to
>> minimize. What is the number of dominoes on the board at the end
>> of the game if A moves first? If B moves first?
>
> You don't want much, do you?
>
> (Note for anyone who may be confused - that comment means that
> I think this is likely to be a hard problem.)
>
> Franklin T. Adams-Watters
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading 
> spam and email virus protection.
>

From bowerc at usa.net  Thu Nov 16 02:01:10 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 15 Nov 2006 17:01:10 -0800
Subject: request for a b-file for a core file
Message-ID: <888kkPBak9480S24.1163638870@cmsweb24.cms.usa.net>

I just sent the b-file for A000014

Christian

------ Original Message ------
From: "Christian G. Bower" <bowerc at usa.net>
To: "seqfan" <seqfan at ext.jussieu.fr>
Subject: Re: request for a b-file for a core file

> I just sent a b-file of terms 0-500 to Neil for A001678.
> I'm still working on a version for A000014
> 








From zbi74583 at boat.zero.ad.jp  Thu Nov 16 04:45:46 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Thu, 16 Nov 2006 12:45:46 +0900
Subject: A036471
Message-ID: <011301c70931$b92d7500$ca73d7dc@FMC57937B6BE00>

    Thanks, Giovanni Resta.



    The smallest term!!

    It is great!

       .

    Now human being became to know much about A036471.       

    The first term is 3270960 and known last term is       

   

    



    

    

    n=32583657,    19.6million digits.



    Yasutoshi

    


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From maxale at gmail.com  Thu Nov 16 10:48:46 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 16 Nov 2006 01:48:46 -0800
Subject: A question on A068480
In-Reply-To: <455AECB9.5050308@iit.cnr.it>
References: <455AECB9.5050308@iit.cnr.it>
Message-ID: <d3dac270611160148q5f52517gbc511cc7b9fd042c@mail.gmail.com>

It is unlikely for n!-1 and 2^n+1 to have a common prime factor
(unless in a special case described below), especially since
1) all prime factors of n!-1 are greater than n;
2) all prime factors of 2^n+1 are of the form p=2kq+1 where q is a
divisor of n, and the multiplicative of 2 modulo p divides 2q.

The special case is when n is even and p=2n+1 is prime. In this case,
p divides both n!-1 and 2^n+1, and thus gcd(n!-1,2^n+1). And it is
very likely that gcd(n!-1,2^n+1) = p.

Overall, the statement "if gcd(n!-1,2^n+1)>1 then
gcd(n!-1,2^n+1)=2n+1" sounds plausible and I verified it for n up to
45,000. But it may be very hard to give a proof.

Max

On 11/15/06, Giovanni Resta <g.resta at iit.cnr.it> wrote:
> A068480 is the sequence on numbers n such that gcd(n!-1,2^n+1)>1.
>
> I noticed that for all the numbers listed gcd(n!-1,2^n+1) = 2n+1.
>
> Is this a rule (proof ?) or a coincidence (counterexample ?) .
>
> thanks,
> giovanni.
>
>
>





From maxale at gmail.com  Thu Nov 16 11:49:25 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 16 Nov 2006 02:49:25 -0800
Subject: A question on A068480
In-Reply-To: <d3dac270611160148q5f52517gbc511cc7b9fd042c@mail.gmail.com>
References: <455AECB9.5050308@iit.cnr.it>
	 <d3dac270611160148q5f52517gbc511cc7b9fd042c@mail.gmail.com>
Message-ID: <d3dac270611160249k774410b5p3b07705ed642b8ae@mail.gmail.com>

On 11/16/06, Max A. <maxale at gmail.com> wrote:

> The special case is when n is even and p=2n+1 is prime. In this case,
> p divides both n!-1 and 2^n+1, and thus gcd(n!-1,2^n+1). And it is
> very likely that gcd(n!-1,2^n+1) = p.

Oh, my bad! Correct statement:

The special case is when n is of the form 4k+1 and p=2n+1 is prime
(implying that 2 and -1 are non-squares modulo p), and p also belongs
to A058302.
In this case, p divides both n!-1 and 2^n+1, and thus gcd(n!-1,2^n+1).
And it is very likely that gcd(n!-1,2^n+1) = p.

The hypothesis about gcd(n!-1,2^n+1) implies that:
8k+3 belong to A058302 if and only if 4k+1 belong to A068480.

Max





From davidwwilson at comcast.net  Thu Nov 16 17:47:04 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 16 Nov 2006 11:47:04 -0500
Subject: b-files for power series expansions
Message-ID: <000601c7099e$d8c5e880$6501a8c0@yourxhtr8hvc4p>

There is a seqload of OEIS sequences that are expansions of real functions, which I guess means power series coefficients. An example is A000810.  Would Mma or some other math power tool be able to quickly generate b-files for these monsters?
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From mathar at strw.leidenuniv.nl  Thu Nov 16 18:44:15 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Thu, 16 Nov 2006 18:44:15 +0100
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
Message-ID: <200611161744.kAGHiFGX005987@amer.strw.leidenuniv.nl>


pdh> From seqfan-owner at ext.jussieu.fr  Wed Nov 15 05:08:51 2006
pdh> Return-Path: <seqfan-owner at ext.jussieu.fr>
pdh> To: seqfan at ext.jussieu.fr
pdh> Date: Tue, 14 Nov 2006 23:04:51 -0500
pdh> Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
pdh> From: "Paul D. Hanna" <pauldhanna at juno.com>
pdh> ...
pdh> Number of labelled nodes in generation n of a rooted tree 
pdh> where a node with label k has k child nodes such that 
pdh> the label of each child node is the least unused label in 
pdh> the path from the root to that child node, 
pdh> and where root is labeled '1'. 
pdh>   ...
pdh> Here are the initial nodes of the tree for generations 0..4: 
pdh> Gen 0: [1];
pdh> Gen 1: (1)->[2];
pdh> Gen 2: (2)->[3,4];
pdh> Gen 3: (3)->[4,5,6], (4)->[3,5,6,7];
pdh> Gen 4: (4)->[5,6,7,8], (5)->[4,6,7,8,9], (6)->[4,5,7,8,9,10], 
pdh> (3)->[5,6,7], (5)->[3,6,7,8,9], (6)->[3,5,7,8,9,10], 
pdh> (7)->[3,5,6,8,9,10,11]; ...
pdh>  
pdh> Thus, the number of nodes in generation n begins:
pdh> [1, 1, 2, 7, 36, 248, ...]
pdh>  
pdh> The maximum node in generation n begins:
pdh> [1, 2, 4, 7, 11, ...]
pdh>  
pdh> Would appreciate it if someone could compute more terms.
pdh> Thanks,
pdh>       Paul

rjm> My count of node numbers and maximum nodes is
rjm> Gen 0 1 1
rjm> Gen 1 1 2
rjm> Gen 2 2 4
rjm> Gen 3 7 7
rjm> Gen 4 36 11
rjm> Gen 5 248 16
rjm> Gen 6 2157 22
rjm> Gen 7 22761 29
rjm> Gen 8 283220 37
rjm> Gen 9 4068411 46

The maximum node label is trivially A000124 because the maximum node will be
generated from the child with itself the highest label, and this adds 'n' to
the maximum label for each new generation. This produces the arithmetic sum
sequence that is shown above as the last number in each row.

For those who have some spare computer cycles and Maple: below is a program
that runs recursively ("memory friendly") through the entire tree up to 
some maximum generation 'maxgen' (to be set/edited in the fourth but last line
below, maxgen=10 equivalent to Gen up to 9 above). It prints a film of
progresses made to count the nodes found so far at all levels, and should show
all numbers in the list monotonically increasing slowly towards
[1, 1, 2, 7, 36, 248, 2157, 22761, 283220, 4068411, 66367834, ...]
                                                    ^^^ 66367834 is Gen 10
I can also provide my older (but non-recursive, memory-limited) C++ version.

Maple V:

gen := proc(parents,maxgen,ocounts,lvl)
	local thislbl,lbl,childlbl,counts,npar ;
	counts := ocounts ;
	counts[lvl] := counts[lvl]+1 ;
	if nops(parents) < maxgen then
		thislbl := op(-1,parents) ;
		childlbl := 1 ;
		for lbl from 1 to thislbl do
			while ( childlbl in parents ) or ( childlbl = thislbl ) do
				childlbl := childlbl+1 ;
			od ;
			npar := [op(parents),childlbl] ;
			# this would generate a full list of all paths:
			# print("route",npar) ;
			if nops(counts) < lvl+1 then
				counts := [op(counts),0] ;
			fi ;
			counts := gen(npar,maxgen,counts,lvl+1) ;
			childlbl := childlbl+1 ;
		od ;
	fi ;
	# this produces a snapshot of the counts accumulated so far:
	# if the current node creation level is not too high, print it...
	if lvl <= maxgen -3 then
		print(counts) ;
	fi ;
	RETURN(counts) ;
end:

# Edit the maximum number below...total run time probably grows exponentially
# with the number chosen here.
maxgen := 8 ;
parents := [1,2] ;
n := [1,0] ;
gen(parents,maxgen,n,2) ;
print(%) ;





From franktaw at netscape.net  Thu Nov 16 18:59:28 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 16 Nov 2006 12:59:28 -0500
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
In-Reply-To: <200611161744.kAGHiFGX005987@amer.strw.leidenuniv.nl>
References: <200611161744.kAGHiFGX005987@amer.strw.leidenuniv.nl>
Message-ID: <8C8D7C018493BAF-87C-1422@FWM-M21.sysops.aol.com>

I don't know Maple well enough - or have enough time - to tell: does 
this keep each path separately, or does it keep a count of the number 
of paths with a given set of integers on the path?

The latter approach should be much faster and less memory intensive, 
once you get past the first few generations.  (Probably still 
exponential, though.)

Franklin T. Adams-Watters

-----Original Message-----
From: mathar at strw.leidenuniv.nl

...

Maple V:

gen := proc(parents,maxgen,ocounts,lvl)
    local thislbl,lbl,childlbl,counts,npar ;
    counts := ocounts ;
    counts[lvl] := counts[lvl]+1 ;
    if nops(parents) < maxgen then
        thislbl := op(-1,parents) ;
        childlbl := 1 ;
        for lbl from 1 to thislbl do
            while ( childlbl in parents ) or ( childlbl = thislbl ) do
                childlbl := childlbl+1 ;
            od ;
            npar := [op(parents),childlbl] ;
            # this would generate a full list of all paths:
            # print("route",npar) ;
            if nops(counts) < lvl+1 then
                counts := [op(counts),0] ;
            fi ;
            counts := gen(npar,maxgen,counts,lvl+1) ;
            childlbl := childlbl+1 ;
        od ;
    fi ;
    # this produces a snapshot of the counts accumulated so far:
    # if the current node creation level is not too high, print it...
    if lvl <= maxgen -3 then
        print(counts) ;
    fi ;
    RETURN(counts) ;
end:

# Edit the maximum number below...total run time probably grows 
exponentially
# with the number chosen here.
maxgen := 8 ;
parents := [1,2] ;
n := [1,0] ;
gen(parents,maxgen,n,2) ;
print(%) ;

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From zakseidov at yahoo.com  Thu Nov 16 19:33:20 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Thu, 16 Nov 2006 10:33:20 -0800 (PST)
Subject: b-files for power series expansions
In-Reply-To: <000601c7099e$d8c5e880$6501a8c0@yourxhtr8hvc4p>
Message-ID: <20061116183320.50052.qmail@web38201.mail.mud.yahoo.com>

Coefficients of the series
1, 1, 4, 13/3, 44/3, 242/15, 2408/45, 18581/315
differ from A000810 entries (?!),
Zak

A000810
1,1,8,26,352,1936,38528,297296,7869952,78098176
%N A000810 Expansion of (sin x + cos x)/cos 3x.


--- David Wilson <davidwwilson at comcast.net> wrote:

> There is a seqload of OEIS sequences that are
> expansions of real functions, which I guess means
> power series coefficients. An example is A000810. 
> Would Mma or some other math power tool be able to
> quickly generate b-files for these monsters?



 
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From franktaw at netscape.net  Thu Nov 16 20:25:13 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 16 Nov 2006 14:25:13 -0500
Subject: b-files for power series expansions
In-Reply-To: <20061116183320.50052.qmail@web38201.mail.mud.yahoo.com>
Message-ID: <8C8D7CC12CFC338-87C-19DE@FWM-M21.sysops.aol.com>

A000810 is the exponential generating function:

1/0!, 1/1!, 8/2!, 26/3!, 352/4!, ...

Franklin T. Adams-Watters


-----Original Message-----
From: zakseidov at yahoo.com

Coefficients of the series
1, 1, 4, 13/3, 44/3, 242/15, 2408/45, 18581/315
differ from A000810 entries (?!),
Zak

A000810
1,1,8,26,352,1936,38528,297296,7869952,78098176
%N A000810 Expansion of (sin x + cos x)/cos 3x.


--- David Wilson <davidwwilson at comcast.net> wrote:

> There is a seqload of OEIS sequences that are
> expansions of real functions, which I guess means
> power series coefficients. An example is A000810.
> Would Mma or some other math power tool be able to
> quickly generate b-files for these monsters?




_________________________________________________________________________
___________
Sponsored Link

Mortgage rates near 39yr lows.
$420k for $1,399/mo. Calculate new payment!
www.LowerMyBills.com/lre

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From mathar at strw.leidenuniv.nl  Thu Nov 16 20:31:50 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Thu, 16 Nov 2006 20:31:50 +0100
Subject: b-files for power series expansions
Message-ID: <200611161931.kAGJVoHb015119@amer.strw.leidenuniv.nl>


dww> From seqfan-owner at ext.jussieu.fr  Thu Nov 16 17:49:20 2006
dww> From: "David Wilson" <davidwwilson at comcast.net>
dww> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
dww> Subject: b-files for power series expansions
dww> Date: Thu, 16 Nov 2006 11:47:04 -0500
dww> ...
dww> There is a seqload of OEIS sequences that are expansions of real =
dww> functions, which I guess means power series coefficients. An example is =
dww> A000810.  Would Mma or some other math power tool be able to quickly =
dww> generate b-files for these monsters?
dww>...

These are usually taylor series coefficients multiplied by n!
(the E.g.f. given in the definition). Where the functions are even,
the odd terms (coefficients equal to 0) are omitted.

I put some of them into
http://www.strw.leidenuniv.nl/~mathar/progs/b000810.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000813.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000816.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000819.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000822.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000825.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000828.txt

Unrelated are
http://www.strw.leidenuniv.nl/~mathar/progs/b007504.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b049060.txt





From mathar at strw.leidenuniv.nl  Thu Nov 16 22:05:41 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Thu, 16 Nov 2006 22:05:41 +0100
Subject: Re^2: Trees with Labeled Nodes that form Permutation of Positive Integers
Message-ID: <200611162105.kAGL5fkI022493@amer.strw.leidenuniv.nl>


ftaw> From seqfan-owner at ext.jussieu.fr  Thu Nov 16 19:01:33 2006
ftaw> Return-Path: <seqfan-owner at ext.jussieu.fr>
ftaw> To: seqfan at ext.jussieu.fr
ftaw> Subject: Re: Trees with Labeled Nodes that form Permutation of Positive Integers
ftaw> 
ftaw> I don't know Maple well enough - or have enough time - to tell: does 
ftaw> this keep each path separately, or does it keep a count of the number 
ftaw> of paths with a given set of integers on the path?
ftaw> 
ftaw> The latter approach should be much faster and less memory intensive, 
ftaw> once you get past the first few generations.  (Probably still 
ftaw> exponential, though.)
ftaw> 
ftaw> Franklin T. Adams-Watters
ftaw> 

There is no such optimization in the program shown below. For each
node, it descends into all branches/leafs independent of whether the
set of labels in the current node and higher up in the trunk is the same
or not the same as any set seen earlier. The state vector is
only the vector of counts (one number per generation) which is updated
at each level of recursion, and the single list of labels of the
ancestors of the current node.
So yes, there is room for optimization by introducing some memory
in here, certainly at the cost of increasing the machine memory requirement
to remember all the sub-counts of partial tree in some hash area.

The first action to become faster is to return to a compiled language.
Since this here does not involve any type of functionality which needs a number
theoretic package, one would try to exploit the fact that C would typically be a
factor of 5 to 10 faster than Maple. I don't want to appear selfish and
will therefore leave all this as an exercise to others :-)

RJM

rjm> -----Original Message-----
rjm> From: mathar at strw.leidenuniv.nl
rjm> 
rjm> ...
rjm> 
rjm> Maple V:
rjm> 
rjm> gen := proc(parents,maxgen,ocounts,lvl)
rjm>     local thislbl,lbl,childlbl,counts,npar ;
rjm>     counts := ocounts ;
rjm>     counts[lvl] := counts[lvl]+1 ;
rjm>     if nops(parents) < maxgen then
rjm>         thislbl := op(-1,parents) ;
rjm>         childlbl := 1 ;
rjm>         for lbl from 1 to thislbl do
rjm>             while ( childlbl in parents ) or ( childlbl = thislbl ) do
rjm>                 childlbl := childlbl+1 ;
rjm>             od ;
rjm>             npar := [op(parents),childlbl] ;
rjm>             # this would generate a full list of all paths:
rjm>             # print("route",npar) ;
rjm>             if nops(counts) < lvl+1 then
rjm>                 counts := [op(counts),0] ;
rjm>             fi ;
rjm>             counts := gen(npar,maxgen,counts,lvl+1) ;
rjm>             childlbl := childlbl+1 ;
rjm>         od ;
rjm>     fi ;
rjm>     # this produces a snapshot of the counts accumulated so far:
rjm>     # if the current node creation level is not too high, print it...
rjm>     if lvl <= maxgen -3 then
rjm>         print(counts) ;
rjm>     fi ;
rjm>     RETURN(counts) ;
rjm> end:
rjm> 
rjm> # Edit the maximum number below...total run time probably grows 
rjm> exponentially
rjm> # with the number chosen here.
rjm> maxgen := 8 ;
rjm> parents := [1,2] ;
rjm> n := [1,0] ;
rjm> gen(parents,maxgen,n,2) ;
rjm> print(%) ;





From franktaw at netscape.net  Fri Nov 17 00:46:06 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 16 Nov 2006 18:46:06 -0500
Subject: Re^2: Trees with Labeled Nodes that form Permutation of Positive
 Integers
In-Reply-To: <200611162105.kAGL5fkI022493@amer.strw.leidenuniv.nl>
References: <200611162105.kAGL5fkI022493@amer.strw.leidenuniv.nl>
Message-ID: <8C8D7F084CDAF74-554-265F@FWM-M18.sysops.aol.com>

No hashing is necessary.  There are (n-1)! possible sets of labels.  
Each
sorted sequence of labels has an increase of at most k between the kth
and k+1st label.  So you can take the first differences, subtract 1,
reverse, and treat as a base factorial number.  E.g., for 1,2,4,5,9, 
the
differences are 1,2,1,4; subtract one and reverse to get 3010,
3*3!+0*2!+1*1! = 19, so this will be at (zero-based) index 19.  (This 
is
easily reversible to get the label set from the index and level.)

For each label set, you'll need to keep a count of how many nodes
there are with each possible final label, to tell what to generate in
the next level.  Since 1 and 2 can't be labels after level 2, you only
need to keep n-2 such counters for each set.

We can tell that this algorithm is on the order of n! to compute level 
n.
The naive approach is proportional to the size of the nth level, which 
is
certainly greater; since the largest term at level n is C(n,2)+1 ~ 
n^2/2,
it is certainly no more than (n^2/2)!.  I'm guessing that is it more 
like
c^n n!, for some c between 1.6 and 2 (probably closer to 2, from the
data, though convergence is slow; k 2^n n!/n^2 gives a fairly good
fit to the available data).

Based on this, the eventual savings should be much larger than merely
switching to a compiled language.  I'll also leave this as an exercise,
though. :-)

Franklin T. Adams-Watters


-----Original Message-----
From: mathar at strw.leidenuniv.nl

ftaw> From seqfan-owner at ext.jussieu.fr  Thu Nov 16 19:01:33 2006
ftaw>
ftaw> I don't know Maple well enough - or have enough time - to tell: 
does
ftaw> this keep each path separately, or does it keep a count of the 
number
ftaw> of paths with a given set of integers on the path?
ftaw>
ftaw> The latter approach should be much faster and less memory 
intensive,
ftaw> once you get past the first few generations.  (Probably still
ftaw> exponential, though.)
ftaw>
ftaw> Franklin T. Adams-Watters
ftaw>

There is no such optimization in the program shown below. For each
node, it descends into all branches/leafs independent of whether the
set of labels in the current node and higher up in the trunk is the same
or not the same as any set seen earlier. The state vector is
only the vector of counts (one number per generation) which is updated
at each level of recursion, and the single list of labels of the
ancestors of the current node.
So yes, there is room for optimization by introducing some memory
in here, certainly at the cost of increasing the machine memory 
requirement
to remember all the sub-counts of partial tree in some hash area.

The first action to become faster is to return to a compiled language.
Since this here does not involve any type of functionality which needs 
a number
theoretic package, one would try to exploit the fact that C would 
typically be a
factor of 5 to 10 faster than Maple. I don't want to appear selfish and
will therefore leave all this as an exercise to others :-)

RJM

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From pauldhanna at juno.com  Fri Nov 17 07:13:46 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Fri, 17 Nov 2006 01:13:46 -0500
Subject: Sub-Fibonacci Tree
Message-ID: <20061117.011346.692.1.pauldhanna@juno.com>

Seqfans, 
     Consider the following "sub-Fibonacci tree" and in particular 
its associated sequence defined by: 
 
Sum of labels of nodes in generation n of a rooted tree in which a node 
with label k has child nodes assigned labels of successive integers 
beginning with k+1 such that the number of child nodes for each node 
is given by the label of the parent node, starting at generation n=0 
with a root node with label '1' followed by a child node with label '2'. 
 
(Perhaps someone could make this description more concise?!)
 
EXAMPLE.
The initial nodes of the tree for generations 0..5 are: 
gen 0: [1];
gen 1: [2];
gen 2: [3];
gen 3: [4,5];
gen 4: (4)->[5,6,7],(5)->[6,7,8];
gen 5: (5)->[6,7,8,9],(6)->[7,8,9,10],(7)->[8,9,10,11], 
(6)->[7,8,9,10,11],(7)->[8,9,10,11,12],(8)->[9,10,11,12,13];
  
By definition, there are 2 child nodes for node [3] since 
the parent of node [3] is [2]; likewise, 
there are 3 child nodes for nodes [4] and [5] (in gen.3) 
since the parent of both nodes has label [3] (in gen.2). 
 
Notice the minimum label in generation n is n+1, and 
the maximum label in generation n is Fibonacci(n+2).  
  
 From this tree I am interested in gleaning 2 sequences. 
  
SEQUENCE 1. 
I compute the number of nodes in generation n to begin: 
[1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368, ...]
 
which is equal (with offset) to A005270:
"Sub-Fibonacci sequences of length n."
1, 1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368 
 
SEQUENCE 2. 
But the following related sequence is NOT found in the OEIS: 
 
Sum of labels of nodes in generation n:
[1, 2, 3, 9, 39, 252, 2361, 32077, 631058, ...]
 
EXTENSION?
Could someone compute more terms of sequences 1 and 2? 
 
Thanks,
    Paul





From pauldhanna at juno.com  Fri Nov 17 08:55:47 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Fri, 17 Nov 2006 02:55:47 -0500
Subject: Sub-Fibonacci Tree
Message-ID: <20061117.025548.692.2.pauldhanna@juno.com>

Seqfans,
    Here I extend A005270 by one more term and also extend the sequence 
a(n) = sum of labels of nodes in generation n in the sub-Fibonacci tree.
I demonstrate a slightly faster method to compute these sequences. 
 
First, I am changing the description to be (hopefully) more clear: 
 
"Sum of labels of nodes in generation n of the sub-Fibonacci tree in
which 
a node with label k and parent node with label g, has g child nodes that 
are given labels beginning with k+1 through k+g; the tree starts at 
generation n=0 with a root node labeled '1' and a child node labeled
'2'."
 
Would appreciate any suggestions to change in wording.
  
 
EXTENDING A005270.
There is a faster way to compute b(n)=A005270(n+1): 
b(n) = "number of nodes in generation n of the sub-Fibonacci tree"
 
by employing the following statistic:
 
b(n+1) = sum of (parent label)*(label) over all nodes in generation n. 
 
For example, 
b(4) = 27 = 3*(4+5);
b(5) = 177 = 4*(5+6+7) + 5*(6+7+8);
b(6) = 1680 = 5*(6+7+8+9) + 6*(7+8+9+10) + 7*(8+9+10+11)+ 
 6*(7+8+9+10+11) + 7*(8+9+10+11+12) + 8*(9+10+11+12+13); ...

Using this method, I extend sequence A005270 by one more term:
1, 1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368, 13067353,
 

SUM OF LABELS SEQUENCE. 
In like manner, the computation of:
a(n) = "sum of labels of nodes in generation n of the sub-Fibonacci tree"

can be accomplished by the statistic:
 
a(n+1) = sum of (parent label)*(label) over all nodes in generation n 
+ sum of (parent label)*[label*(label+1)/2] over all nodes in generation
n-1. 
 
For example, 
a(2) = 3 = 1*2  + 1*( 1*2/2 ) ;
a(3) = 9 = 2*3  + 1*( 2*3/2 ) ;
a(4) = 39 = 3*(4+5)  + 2*( 3*4/2 ) ;
a(5) = 252 = 4*(5+6+7) + 5*(6+7+8)  + 3*( 4*5/2 + 5*6/2 ) ;
a(6) = 2361 = 5*(6+7+8+9) + 6*(7+8+9+10) + 7*(8+9+10+11) + 
 6*(7+8+9+10+11) + 7*(8+9+10+11+12) + 8*(9+10+11+12+13)  +
 4*( 5*6/2 + 6*7/2 + 7*8/2 ) + 5*( 6*7/2 + 7*8/2 + 8*9/2) ;
...
 
With this method, I extend the sum of labels in generation n to: 
1, 2, 3, 9, 39, 252, 2361, 32077, 631058, 18035534, 
 

Would someone please confirm my calculation of the terms: 
b(10) = A005270(11) = 13067353  and 
a(10) = 18035534
 
and maybe compute a few more terms? 
Thanks,
    Paul
----------------------------------------------------
> EXAMPLE.
> The initial nodes of the tree for generations 0..5 are: 
> gen 0: [1];
> gen 1: [2];
> gen 2: [3];
> gen 3: [4,5];
> gen 4: (4)->[5,6,7],(5)->[6,7,8];
> gen 5: (5)->[6,7,8,9],(6)->[7,8,9,10],(7)->[8,9,10,11], 
> (6)->[7,8,9,10,11],(7)->[8,9,10,11,12],(8)->[9,10,11,12,13];
>   
> By definition, there are 2 child nodes for node [3] since 
> the parent of node [3] is [2]; likewise, 
> there are 3 child nodes for nodes [4] and [5] (in gen.3) 
> since the parent of both nodes has label [3] (in gen.2). 
>  
> Notice the minimum label in generation n is n+1, and 
> the maximum label in generation n is Fibonacci(n+2).  
>   
> From this tree I am interested in gleaning 2 sequences. 
>   
> SEQUENCE 1. 
> I compute the number of nodes in generation n to begin: 
> [1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368, ...]
>  
> which is equal (with offset) to A005270:
> "Sub-Fibonacci sequences of length n."
> 1, 1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368 
>  
> SEQUENCE 2. 
> But the following related sequence is NOT found in the OEIS: 
>  
> Sum of labels of nodes in generation n:
> [1, 2, 3, 9, 39, 252, 2361, 32077, 631058, ...]
>  
> EXTENSION?
> Could someone compute more terms of sequences 1 and 2? 
>  
> Thanks,
>     Paul





From zbi74583 at boat.zero.ad.jp  Sat Nov 18 02:27:22 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Sat, 18 Nov 2006 10:27:22 +0900
Subject: A085058
Message-ID: <019201c70ab0$b44b3ce0$ca73d7dc@FMC57937B6BE00>

    Hi, Seqfans

    If 2^k(n) is the largest power of two dividing tangent number T(n) then 
numbers such that 2*n-k(n) are as follows.



    S : 2,3,2,4,2,3,2,5,2,....



    It looks like A085058.

    Are they the same seqences ?



    Yasutoshi


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From alec at mihailovs.com  Sun Nov 19 11:38:43 2006
From: alec at mihailovs.com (Alec Mihailovs)
Date: Sun, 19 Nov 2006 04:38:43 -0600
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
References: <200611161744.kAGHiFGX005987@amer.strw.leidenuniv.nl>
Message-ID: <000c01c70bc6$e36ead60$03fea8c0@Media>

From: "Richard Mathar" <mathar at strw.leidenuniv.nl>
Sent: Thursday, November 16, 2006 11:44 AM
>
> pdh> Date: Tue, 14 Nov 2006 23:04:51 -0500
> pdh> From: "Paul D. Hanna" <pauldhanna at juno.com>
> pdh> ...
> pdh> Number of labelled nodes in generation n of a rooted tree
> pdh> where a node with label k has k child nodes such that
> pdh> the label of each child node is the least unused label in
> pdh> the path from the root to that child node,
> pdh> and where root is labeled '1'.
> pdh>   ...
> pdh> Thus, the number of nodes in generation n begins:
> pdh> [1, 1, 2, 7, 36, 248, ...]
>
> [1, 1, 2, 7, 36, 248, 2157, 22761, 283220, 4068411, 66367834, ...]
>                                                    ^^^ 66367834 is Gen 10

Maple 10 can compile (some) procedures in C. I used the following 
back-tracking procedure,

f:=proc(n::integer[4],A::Array(datatype=integer[4]),
B::Array(datatype=integer[4]))::integer[4]; local c::integer[4],
i::integer[4],len::integer[4],m::integer[4]; c,len,m:=0,3,3;
while len>1 do if len=n then c:=c+1;m:=A[len];B[m]:=0;len:=len-1;
B[A[len]]:=B[A[len]]+1 elif B[A[len]]<=A[len] then for i from m+1
do if B[i]=0 then break fi od; len:=len+1;A[len]:=i;B[i]:=1;m:=2
else m:=A[len];B[m]:=0;len:=len-1;B[A[len]]:=B[A[len]]+1 fi od; c end:

cf:=Compiler:-Compile(f):

F:=proc(n::posint) local A,B;
if n<3 then 1 elif n=3 then 2 else
A:=Array([$1..3,0$(n-3)],datatype=integer[4]);
B:=Array([1$3,0$((n-2)*(n+1)/2)],datatype=integer[4]);
cf(n,A,B) fi end:

seq(F(n),n=1..12);

  1, 1, 2, 7, 36, 248, 2157, 22761, 283220, 4068411, 66367834,

        1213504295

That confirms Richard Mathar's and Paul D. Hanna's calculations and gives 
one more element of the sequence. Further elements could not be calculated 
on my 32-bit computer using this procedure, because they are greater than 
2^32. Changing the types of c and f from integer[4] to float[8] allows 
further calculations. I did that and after about an hour got the next 
element,

F(13);

                      24606397802


Alec Mihailovs
http://mihailovs.com/Alec/








From deutsch at duke.poly.edu  Mon Nov 20 17:33:59 2006
From: deutsch at duke.poly.edu (Emeric Deutsch)
Date: Mon, 20 Nov 2006 11:33:59 -0500 (EST)
Subject: help with a new sequence 
Message-ID: <Pine.GSO.4.62.0611201046020.12396@duke.poly.edu>

Seqfans,

I intend to submit to OEIS the following new sequence (triangle):

1,1,1,1,3,1,1,4,2,6,1,1,5,5,10,10,10,1,

Triangle read by rows: T(n,k) is the number of Dyck paths of
semilength n whose ascent lengths form the k-th partition of
the integer n; the partitions of n are ordered in the way
exemplified by [5], [4,1], [3,2], [3,1,1], [2,2,1], [2,1,1,1],
[1,1,1,1,1] (the "Mathematica" ordering).

Equivalently, T(n,k) is the number of ordered trees with n
edges whose node degrees form the k-th partition of the
integer n.

Example: T(5,3)=5 because the 3rd partition of 5 is [3,2] and
we have (UU)DD(UUU)DDD, (UUU)DDD(UU)DD, (UU)D(UUU)DDDD,
(UUU)D(UU)DDDD, and (UUU)DD(UU)DDD; here U=(1,1), D=(1,-1) and
the ascents are shown between parentheses.

Triangle starts:
1;
1,1;
1,3,1;
1,4,2,6,1;
1,5,5,10,10,10,1;

Row n has A000041(n) terms (=number of partitions of n).

------------

So far this is all I have; terms have been found by
straightforward counting. I am sure you can find more
terms, more facts, etc. I'd appreciate any collaboration.

Thanks, Emeric





From pauldhanna at juno.com  Mon Nov 20 19:39:48 2006
From: pauldhanna at juno.com (Paul D Hanna)
Date: Mon, 20 Nov 2006 18:39:48 GMT
Subject: help with a new sequence 
Message-ID: <20061120.104022.757.551642@webmail37.nyc.untd.com>

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From franktaw at netscape.net  Mon Nov 20 19:49:27 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 20 Nov 2006 13:49:27 -0500
Subject: help with a new sequence 
In-Reply-To: <20061120.104022.757.551642@webmail37.nyc.untd.com>
References: <20061120.104022.757.551642@webmail37.nyc.untd.com>
Message-ID: <8C8DAEBBD6AFC69-91C-B5F3@FWM-M20.sysops.aol.com>

This is A008284, a famous function.

Franklin T. Adams-Watters


-----Original Message-----
From: pauldhanna at juno.com

Seqfans,
??????Row n of a?related triangle?could?count?the number of terms in 
row n
of?Emerics triangle that?sum to?form the respective?terms of the 
Narayana triangle.
?
This triangle P?would begin:
1;
1, 1;
1, 1, 1;
1, 2, 1, 1;
1, 2, 2, 1, 1; ...

where row sums equal the partition numbers.
?
For example, row 3 of P?is [1, 2, 1, 1]
since Narayana row 3 is [1, 4+2, 6, 1]
formed from row 3 of?Emerics triangle: [1, 4, 2, 6, 1].
?
Is the resulting partition-related triangle P?already in the OEIS?

It?might have an interesting matrix inverse.
??? Paul
?
--?Emeric?Deutsch?<deutsch at duke.poly.edu>?wrote:
...
Triangle?starts:
1;
1,1;
1,3,1;
1,4,2,6,1;
1,5,5,10,10,10,1;

Row?n?has?A000041(n)?terms?(=number?of?partitions?of?n).
...
Thanks,?Emeric




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From franktaw at netscape.net  Mon Nov 20 20:55:23 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 20 Nov 2006 14:55:23 -0500
Subject: help with a new sequence 
In-Reply-To: <Pine.GSO.4.62.0611201046020.12396@duke.poly.edu>
References: <Pine.GSO.4.62.0611201046020.12396@duke.poly.edu>
Message-ID: <8C8DAF4F355FB9E-B7C-B33B@FWM-M19.sysops.aol.com>

How about if you look at the sequence (composition) of ascents, instead 
of the partition?

I think this starts (using the A066099 ordering):

1
1,1
1,2,1,1
1,3,2,3,1,2,1,1
1,4,3,6,2,5,3,1,1,3,2,2,1,3,4,1

Amazingly, this corresponds to Stern's diatomic array (version 3, 
A070879) for the first 18 terms.
Makes me wonder if A070879 can be interpreted as a function on 
compositions.

Franklin T. Adams-Watters


-----Original Message-----
From: deutsch at duke.poly.edu

Seqfans,?
?
I intend to submit to OEIS the following new sequence (triangle):?
?
1,1,1,1,3,1,1,4,2,6,1,1,5,5,10,10,10,1,?
?
Triangle read by rows: T(n,k) is the number of Dyck paths of?
semilength n whose ascent lengths form the k-th partition of?
the integer n; the partitions of n are ordered in the way?
exemplified by [5], [4,1], [3,2], [3,1,1], [2,2,1], [2,1,1,1],?
[1,1,1,1,1] (the "Mathematica" ordering).?
?
Equivalently, T(n,k) is the number of ordered trees with n?
edges whose node degrees form the k-th partition of the?
integer n.?
?
Example: T(5,3)=5 because the 3rd partition of 5 is [3,2] and?
we have (UU)DD(UUU)DDD, (UUU)DDD(UU)DD, (UU)D(UUU)DDDD,?
(UUU)D(UU)DDDD, and (UUU)DD(UU)DDD; here U=(1,1), D=(1,-1) and?
the ascents are shown between parentheses.?
?
Triangle starts:?
1;?
1,1;?
1,3,1;?
1,4,2,6,1;?
1,5,5,10,10,10,1;?
?
Row n has A000041(n) terms (=number of partitions of n).?
?
------------?
?
So far this is all I have; terms have been found by?
straightforward counting. I am sure you can find more?
terms, more facts, etc. I'd appreciate any collaboration.?
?
Thanks, Emeric?


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From franktaw at netscape.net  Mon Nov 20 22:54:07 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 20 Nov 2006 16:54:07 -0500
Subject: help with a new sequence 
In-Reply-To: <8C8DAF4F355FB9E-B7C-B33B@FWM-M19.sysops.aol.com>
References: <Pine.GSO.4.62.0611201046020.12396@duke.poly.edu> <8C8DAF4F355FB9E-B7C-B33B@FWM-M19.sysops.aol.com>
Message-ID: <8C8DB058980377A-B7C-BB3C@FWM-M19.sysops.aol.com>

Let f(a_0, a_1, ..., a_m) be the number of Dyck paths associated with 
the composition
[a_0, a_1, ..., a_m].   We have f(a_0) = 1; and for m > 0,

f(a_0, a_1, ..., a_m) = Sum_{i=0}^{a_0-1} f(a_1+i, a_2, ..., a_m).

Using this, I get the following table (including row 0, the empty 
composition):

1,
1,
1,1,
1,2,1,1,
1,3,2,3,1,2,1,1,
1,4,3,6,2,5,3,4,1,3,2,3,1,2,1,1,
1,5,4,10,3,9,6,10,2,7,5,9,3,7,4,5,1,4,3,6,2,5,3,4,1,3,2,3,1,2,1,1,
1,6,5,15,4,14,10,20,3,12,9,19,6,16,10,15,2,9,7,16,5,14,9,14,3,10,7,12,4,9
,5,6,1,5,4,
   10,3,9,6,10,2,7,5,9,3,7,4,5,1,4,3,6,2,5,3,4,1,3,2,3,1,2,1,1

And accumulating by partition, for Emeric's original table (but also 
including row 0):

1,
1,
1,1,
1,3,1,
1,4,2,6,1,
1,5,5,10,10,10,1,
1,6,6,15,3,30,20,5,30,15,1,
1,7,7,21,7,42,35,21,21,105,35,35,70,21,1,
1,8,8,28,8,56,56,4,56,28,168,70,28,84,168,280,56,14,140,140,28,1,
1,9,9,36,9,72,84,9,72,36,252,126,36,72,252,252,504,126,12,252,252,84,756,
630,84,
   126,420,252,36,1

Given a partition of n into k parts, [a_1^e_1,...,a_k^e_k], the number 
of
associated Dyck paths appears to be n!/(n-k+1)!/(Product (e_i)!).

Now, maybe somebody can prove that.

Franklin T. Adams-Watters


-----Original Message-----
From: franktaw at netscape.net

How about if you look at the sequence (composition) of ascents, instead 
of the partition??
?
I think this starts (using the A066099 ordering):?
?
1?
1,1?
1,2,1,1?
1,3,2,3,1,2,1,1?
1,4,3,6,2,5,3,1,1,3,2,2,1,3,4,1?
?
Amazingly, this corresponds to Stern's diatomic array (version 3, 
A070879) for the first 18 terms.?
Makes me wonder if A070879 can be interpreted as a function on 
compositions.?
?
Franklin T. Adams-Watters?
?
-----Original Message-----?
From: deutsch at duke.poly.edu?
?
Seqfans,??
??
I intend to submit to OEIS the following new sequence (triangle):??
??
1,1,1,1,3,1,1,4,2,6,1,1,5,5,10,10,10,1,??
??
Triangle read by rows: T(n,k) is the number of Dyck paths of??
semilength n whose ascent lengths form the k-th partition of??
the integer n; the partitions of n are ordered in the way??
exemplified by [5], [4,1], [3,2], [3,1,1], [2,2,1], [2,1,1,1],??
[1,1,1,1,1] (the "Mathematica" ordering).??
??
Equivalently, T(n,k) is the number of ordered trees with n??
edges whose node degrees form the k-th partition of the??
integer n.??
??
Example: T(5,3)=5 because the 3rd partition of 5 is [3,2] and??
we have (UU)DD(UUU)DDD, (UUU)DDD(UU)DD, (UU)D(UUU)DDDD,??
(UUU)D(UU)DDDD, and (UUU)DD(UU)DDD; here U=(1,1), D=(1,-1) and??
the ascents are shown between parentheses.??
??
Triangle starts:??
1;??
1,1;??
1,3,1;??
1,4,2,6,1;??
1,5,5,10,10,10,1;??
??
Row n has A000041(n) terms (=number of partitions of n).??
??
------------??
??
So far this is all I have; terms have been found by??
straightforward counting. I am sure you can find more??
terms, more facts, etc. I'd appreciate any collaboration.??
??
Thanks, Emeric??
?
________________________________________________________________________?

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From njas at research.att.com  Mon Nov 20 23:25:04 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Mon, 20 Nov 2006 17:25:04 -0500 (EST)
Subject: two "find the next term" puzzles from Knuth Vol 4
Message-ID: <200611202225.RAA93655@fry.research.att.com>


the two puzzles (and their solutions) are

%I A123896
%S A123896 0,1,1,1,12,12,12,12,12,12,100,121,122,123,123
%N A123896 A123895(n^2).
...

%I A123902
%S A123902 0,1,12,100,112,121,122,123
%N A123902 A123896 sorted and uniqued.
...

which are based on this:

%I A123895
%S A123895 0,1,1,1,1,1,1,1,1,1,10,11,12,12,12,12,12,12,12,12,10,12,11,12,12,
%T A123895 12,12,12,12,12,10,12,12,11,12,12,12,12,12,12,10,12,12,12,11,12,12,
%U A123895 12,12,12,10,12,12,12,12,11,12,12,12,12,10,12,12,12,12,12,11,12,12
%N A123895 Restricted growth string for the (decimal expansion of the) number n.
%C A123895 Write n in base 10 prefixed with a 0. Read this string from left to r
ight. Write a 0 each time you see the first distinct digit (which is 0), write a
1 each time you see the second distinct digit, write a 2 each time you see the t
hird distinct digit, and so on. Finally, delete the leading zeros.
%D A123895 D. E. Knuth, TAOCP, Vol. 4, Section 7.2.1.5, Problems 4 and 5.
%e A123895 To find a(66041171): 066041171 -> 011023343 -> 11023343.
%Y A123895 Cf. A123896, A123902.
%K A123895 nonn,base,new
%O A123895 0,11
%A A123895 njas, Nov 20 2006

It would be nice if someone could extend the first two, and provide
maple or mma or other code for the third one

Neil






From franktaw at netscape.net  Mon Nov 20 23:59:41 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 20 Nov 2006 17:59:41 -0500
Subject: two "find the next term" puzzles from Knuth Vol 4
In-Reply-To: <200611202225.RAA93655@fry.research.att.com>
References: <200611202225.RAA93655@fry.research.att.com>
Message-ID: <8C8DB0EB222DED3-B7C-BF60@FWM-M19.sysops.aol.com>

A123896:

0 1 1 1 12 12 12 12 12 12 100 121 122 123 123 112 123 123 123 123 100 
112 121 123 123 123 121 123 123 123 100 123 1023 1023 1123 1223 1234 
1234 1222 1231 1200 1231 1234 1234 1234 1012 1223 1102 1203 1203 1200 
1203 1203 1203 1234 1023 1213 1234 1123 1234 1200 1234 1233 1232 1023 
1223 1234 1123 1231 1234 1200 1023 1234 1234 1234 1231 1223 1232 1023 
1234 1200 1213 1234 1223 1023 1223 1234 1234 1122 1234 1200

A123902:

0 1 12 100 112 121 122 123 1012 1023 1102 1122 1123 1200 1201 1203 1213 
1222 1223 1231 1232 1233 1234 10000 10012 10023 10102 10123 10201 10202 
10203 10213 10223 10231 10232 10234 11023 11102 11200 11203 11211 11213 
11221 11223 11232 11234 12003 12013 12023 12032 12033 12034 12100 12113 
12121 12131 12133 12134 12200 12203 12212 12213 12231 12232 12233 12234 
12300 12301 12304 12312 12313 12314 12321 12323 12324 12331 12332 12334 
12341 12342

Sorry, I didn't use Mathematica, Maple, or even PARI; I used Excel and 
VBA.

Here's a VBA program:

Public Function RestrictedGrowthString(ByVal x As String) As String
    Dim i As Long
    Dim dig As Integer
    Dim pos As Long

    For i = 1 To Len(x)
        If Mid(x, i, 1) = "0" Then
            RestrictedGrowthString = RestrictedGrowthString & "0"
        Else
            pos = InStr(x, Mid(x, i, 1))
            If pos = i Then
                dig = dig + 1
                 RestrictedGrowthString = RestrictedGrowthString & 
Format(dig)
            Else
                 RestrictedGrowthString = RestrictedGrowthString & 
Mid(RestrictedGrowthString, pos, 1)
            End If
        End If
    Next i
End Function

Franklin T. Adams-Watters


-----Original Message-----
From: njas at research.att.com

the two puzzles (and their solutions) are

%I A123896
%S A123896 0,1,1,1,12,12,12,12,12,12,100,121,122,123,123
%N A123896 A123895(n^2).
...

%I A123902
%S A123902 0,1,12,100,112,121,122,123
%N A123902 A123896 sorted and uniqued.
...

which are based on this:

%I A123895
%S A123895 
0,1,1,1,1,1,1,1,1,1,10,11,12,12,12,12,12,12,12,12,10,12,11,12,12,
%T A123895 
12,12,12,12,12,10,12,12,11,12,12,12,12,12,12,10,12,12,12,11,12,12,
%U A123895 
12,12,12,10,12,12,12,12,11,12,12,12,12,10,12,12,12,12,12,11,12,12
%N A123895 Restricted growth string for the (decimal expansion of the) 
number n.
%C A123895 Write n in base 10 prefixed with a 0. Read this string from 
left to r
ight. Write a 0 each time you see the first distinct digit (which is 
0), write a
1 each time you see the second distinct digit, write a 2 each time you 
see the t
hird distinct digit, and so on. Finally, delete the leading zeros.
%D A123895 D. E. Knuth, TAOCP, Vol. 4, Section 7.2.1.5, Problems 4 and 
5.
%e A123895 To find a(66041171): 066041171 -> 011023343 -> 11023343.
%Y A123895 Cf. A123896, A123902.
%K A123895 nonn,base,new
%O A123895 0,11
%A A123895 njas, Nov 20 2006

It would be nice if someone could extend the first two, and provide
maple or mma or other code for the third one

Neil


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From njas at research.att.com  Tue Nov 21 00:43:03 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Mon, 20 Nov 2006 18:43:03 -0500 (EST)
Subject: King chickens - I was wrong!
Message-ID: <200611202343.SAA88240@fry.research.att.com>


Martin Fuller pointed out that I was wrong.
A highest scorer in a tournament is a king chicken,
but not conversely.

So there are really two sequences, A123553 (K.C.'s),
now corrected by Martin, and A125031, the total number
of highest scorers, computed by Martin Fuller to n=11
(this is what Gordon Royle computed out to n=10).

Maurer also defines "Emperors", chickens
who peck everybody else, which is n*2^((n-1)(n-2)/2)
(it will be A123903).

Neil





From zakseidov at yahoo.com  Tue Nov 21 06:19:11 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Mon, 20 Nov 2006 21:19:11 -0800 (PST)
Subject: A125097.gif: (Very) nice patterns
Message-ID: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>

Neil, 
seqfans,

1. Have a look at this full graph -
aint these patterns (very) nice?!
It may be known, but still it's nice...

2. Is it possible to 
put "the extended graph" into OEIS?
In the standard graph no patterns are seen at all...

3. Neil removed  keyword "nice" from A125097
(as he did regularly with ny sequences)-
as if I insist that it's me who is "nice"
not it is the seq which is nice.

Thanks, Zak


 
____________________________________________________________________________________
Sponsored Link

$200,000 mortgage for $660/ mo - 
30/15 yr fixed, reduce debt - 
http://yahoo.ratemarketplace.com
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From maxale at gmail.com  Tue Nov 21 07:03:55 2006
From: maxale at gmail.com (Max A.)
Date: Mon, 20 Nov 2006 22:03:55 -0800
Subject: A125097.gif: (Very) nice patterns
In-Reply-To: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>
References: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>
Message-ID: <d3dac270611202203p2b9fff6s2bc522bc6662e735@mail.gmail.com>

Zak,

I do not see anything special about this graph.
Yes, it includes some points from the line y=2x+1, and, yes, it is
symmetric the region [0,500]x[0,1000] (i.e., a(n)=k iff
a((k-1)/2)=2n+1). But that does not make it any nicer.

Regards,
Max

On 11/20/06, zak seidov <zakseidov at yahoo.com> wrote:
> Neil,
> seqfans,
>
> 1. Have a look at this full graph -
> aint these patterns (very) nice?!
> It may be known, but still it's nice...
>
> 2. Is it possible to
> put "the extended graph" into OEIS?
> In the standard graph no patterns are seen at all...
>
> 3. Neil removed  keyword "nice" from A125097
> (as he did regularly with ny sequences)-
> as if I insist that it's me who is "nice"
> not it is the seq which is nice.
>
> Thanks, Zak
>
>
>
> ____________________________________________________________________________________
> Sponsored Link
>
> $200,000 mortgage for $660/ mo -
> 30/15 yr fixed, reduce debt -
> http://yahoo.ratemarketplace.com
>
>





From pauldhanna at juno.com  Tue Nov 21 11:07:54 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Tue, 21 Nov 2006 05:07:54 -0500
Subject: Formula for New Sequence A125054 ?
Message-ID: <20061121.050754.1448.0.pauldhanna@juno.com>

Seqfans,
     Could someone find a formula for this new sequence? 
Sequence: A125054  begins: 
 
1,3,21,327,9129,396363,24615741,2068052367,225742096209,
31048132997523,5252064083753061,1071525520294178007,
259439870666594250489,73542221109962636293083,
24125551094579137082039181,9068240688454120376775401247,
3871645204706420218816959159969,
(I can supply many more terms if needed).
 
The sequence forms the Central terms of a new triangle A125053 
(a variant of triangle A008301 - enumeration of binary trees). 
 
Triangle A125053 is nice since the first column (and row sums) 
form the Secant numbers A000364 (an unexpected result!). 
  
Below I define the triangle.
 
I would be very interested in a formula, perhaps an E.g.f.? 
(I know I am asking a lot, but Seqfans surprise me all the time 
with amazing formulas). 
  
Thanks, 
      Paul 
---------------------------------------------------------
  
If we write triangle A125053 like this: 
.......................... ...1; 
.................... ...1, ...3, ...1; 
.............. ...5, ..15, ..21, ..15, ...5; 
........ ..61, .183, .285, .327, .285, .183, ..61; 
.. 1385, 4155, 6681, 8475, 9129, 8475, 6681, 4155, 1385; 

then the first nonzero term is the sum of the previous row: 
1385 = 61 + 183 + 285 + 327 + 285 + 183 + 61,

the next term is 3 times the first:
4155 = 3*1385,

and the remaining terms in each row are obtained by the rule 
illustrated by: 
6681 = 2*4155 - 1385 - 4*61 ;
8475 = 2*6681 - 4155 - 4*183 ;
9129 = 2*8475 - 6681 - 4*285 ;
8475 = 2*9129 - 8475 - 4*327 ;
6681 = 2*8475 - 9129 - 4*285 ;
4155 = 2*6681 - 8475 - 4*183 ;
1385 = 2*4155 - 6681 - 4*61 .
 
An alternate recurrence is illustrated by:
4155 = 1385 + 2*(61 + 183 + 285 + 327 + 285 + 183 + 61);
6681 = 4155 + 2*(183 + 285 + 327 + 285 + 183);
8475 = 6681 + 2*(285 + 327 + 285);
9129 = 8475 + 2*(327); 
 
and then for k>n, k<=2n, T(n,k) = T(n,2*n-k).
 
END.





From pauldhanna at juno.com  Tue Nov 21 14:58:26 2006
From: pauldhanna at juno.com (Paul D Hanna)
Date: Tue, 21 Nov 2006 13:58:26 GMT
Subject: Formula for New Sequence A125054 ?
Message-ID: <20061121.055901.20413.609605@webmail43.nyc.untd.com>

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From antti.karttunen at gmail.com  Tue Nov 21 15:27:22 2006
From: antti.karttunen at gmail.com (Antti Karttunen)
Date: Tue, 21 Nov 2006 16:27:22 +0200
Subject: Binary tree terminology, again. Re: Formula for New Sequence A125054
 ?
In-Reply-To: <20061121.055901.20413.609605@webmail43.nyc.untd.com>
References: <20061121.055901.20413.609605@webmail43.nyc.untd.com>
Message-ID: <45630CCA.8000103@gmail.com>

Paul D Hanna wrote:

> Seqfans,
>
>      Could someone find a formula for this new sequence?
> Sequence: A125054  begins:
>  
> 1,3,21,327,9129,396363,24615741,2068052367,225742096209,
> 31048132997523,5252064083753061,1071525520294178007,
> 259439870666594250489,73542221109962636293083,
> 24125551094579137082039181,9068240688454120376775401247,
> 3871645204706420218816959159969,
> (I can supply many more terms if needed).
>  
> The sequence forms the Central terms of a new triangle A125053
> (a variant of triangle A008301 - enumeration of binary trees).
>
Is there any more helpful description for A008301 and A125053 ?
E.g. do they relate to non-planar binary trees
(as http://www.research.att.com/~njas/sequences/A001190 )
or to (rooted, unlabeled) planar binary trees, as Catalan numbers, 
A000108, do?
Labeled or unlabeled? Rooted or non-rooted?


Cumprimentos,

Antti Karttunen
writing from Porto, Portugal.

>  
> Triangle A125053 is nice since the first column (and row sums)
> form the Euler numbers A000364 (an unexpected result!).
>  
>
> Seqfans,
>     One should always try Superseeker first!
> Superseeker says:
>
> A125054 = Binomial transform of A000182 (e.g.f. tan(x))
>
> and it holds true.
>
> I did not expect such a simple answer! 
>     Paul
>

>  
>






From tbaruchel at free.fr  Tue Nov 21 15:47:28 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Tue, 21 Nov 2006 15:47:28 +0100 (CET)
Subject: Almost the number of divisors of n+1
Message-ID: <20061121154018.X1354@localhost.my.domain>

Hi, I have some number-theory related empirically collected datas to study.
As long as I can see, I have a link between each integer and the number
of divisors of its successor. But though the rule seems to work very
well in some cases: predecessors of prime numbers seem to appear 0 or 1 times,
predecessors of numbers like 12 or 24 seem to appear much more, etc. I am not
sure this is the very exact rule. Of course, the trouble may come from
the fact that my datas is quite hard to "extract", but maybe you know some
variants of
   a(n) is the number of divisors of (n+1)
Have you ever seen closely-related sequences like that one?

Regards,

-- 
Thomas Baruchel





From davidwwilson at comcast.net  Tue Nov 21 18:18:36 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Tue, 21 Nov 2006 12:18:36 -0500
Subject: Confirmation
Message-ID: <004601c70d91$14d8f3d0$6501a8c0@yourxhtr8hvc4p>

Define

    f(1) = 1
    f(n+1) = f(n) + (f(n) mod n+3364).

I wrote a computer program to find the smallest x with

    x+3365 divides f(x)

The value of x found is just under 8,350,000,000 and the corresponding f(x) just under 2^64. Although my program included an overflow check, I am still not comfortable because this value of f(x) is close to my 64-bit unsigned integer size.

Could someone with a fast computer, preferably a UNIX box, please confirm or refute my values?
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From maxale at gmail.com  Tue Nov 21 20:47:20 2006
From: maxale at gmail.com (Max A.)
Date: Tue, 21 Nov 2006 11:47:20 -0800
Subject: Confirmation
In-Reply-To: <004601c70d91$14d8f3d0$6501a8c0@yourxhtr8hvc4p>
References: <004601c70d91$14d8f3d0$6501a8c0@yourxhtr8hvc4p>
Message-ID: <d3dac270611211147i7b7ef94ev729014f094d24dc4@mail.gmail.com>

I've got

x = 8334201245
f(x) = 17364786429187398690

I've also checked that the value f(n) was non-decreasing during
computations. Since f(n+1) < f(n)+n+3364 and f(x)+x+3364 < 2^64, that
means no overflow in 64-bit unsigned integer happened.

Max

On 11/21/06, David Wilson <davidwwilson at comcast.net> wrote:
>
>
> Define
>
>     f(1) = 1
>     f(n+1) = f(n) + (f(n) mod n+3364).
>
> I wrote a computer program to find the smallest x with
>
>     x+3365 divides f(x)
>
> The value of x found is just under 8,350,000,000 and the corresponding f(x)
> just under 2^64. Although my program included an overflow check, I am still
> not comfortable because this value of f(x) is close to my 64-bit unsigned
> integer size.
>
> Could someone with a fast computer, preferably a UNIX box, please confirm or
> refute my values?





From all at abouthugo.de  Tue Nov 21 22:22:29 2006
From: all at abouthugo.de (Hugo Pfoertner)
Date: Tue, 21 Nov 2006 22:22:29 +0100
Subject: Wrong new terms in A085000
Message-ID: <45636E15.FD6E8D3F@abouthugo.de>

Seqfans, Neil,

the sequence A085000 "Maximal determinant of an n X n matrix using the
integers 1 to n^2."

that had previously the terms a(1)...a(6)
1,10,412,40800,6839492,1865999570,

has recently been extendend as follows:

762074188050,440477621828418, 345964972957674150

The comment says:
a(7) - a(9) were found by Richard Mueller
(richard.mueller(AT)informatik.hu-berlin.de) and Johannes Dewender
(johannes.dewender(AT)informatik.hu-berlin.de) in cooperation, Oct 22
2006 
Entry updated by Richard Mueller  Oct 22 2006

Unfortunately bigger counterexamples exist for all 3 terms:

d7:=det mat
((28, 9,34,41, 1,40,22),
 (47,25, 7,36,35,16, 8),
 (3, 27, 6,39,31,26,43),
 (38,21,32,14,18, 4,48),
 (19, 5,37,13,49,33,20),
 (29,42,15, 2,17,45,24),
 (11,46,44,30,23,12,10));
 
 Reduce: d7 := 762139309293

d8:= det mat
((13, 53, 40, 25,  2, 45, 58, 23),
 (21, 22,  8, 54, 24, 56, 18, 57),
 (60, 11, 36, 44, 12,  7, 49, 41),
 (15, 19, 48,  4, 52, 26, 38, 59),
 (20, 62, 34, 51, 42,  1, 17, 33),
 (37, 29,  3, 31, 61, 35, 55, 10),
 (30, 14, 63, 46, 39, 47, 16,  5),
 (64, 50, 28,  6, 27, 43,  9, 32));

Reduce: d8 := 440857916120379

d9:= det mat
((40, 48, 72, 26,  6, 50, 39, 78, 10),
 (23, 77, 42,  7, 37, 41, 74, 12, 56),
 (52, 47,  8, 54,  5, 64, 19, 44, 75),
 (59, 25, 13, 43, 58, 76, 63, 30,  3),
 (79, 36, 71, 60, 31, 18, 34,  2, 38),
 (32, 81, 22, 61, 67, 15, 17, 53, 21),
 (20, 24, 66, 27, 73, 68,  4, 33, 55),
 ( 1, 14, 45, 80, 35, 29, 69, 46, 49),
 (65, 16, 28, 11, 57,  9, 51, 70, 62));

Reduce: d9 := 346202123011922653

(Thanks to Rainer Rosenthal for X-checking the determinants with Maple).

None of the counterexamples is known to be the biggest possible
solution. The best value I've found in 2 CPU years for
a(7)=762140212575, but currently I don't have access to the computer
where I've found this solution and the corresponding arrangement of
matrix elements.

My suggestion is to remove the 3 new terms and the new Mathematica
program and to restore the previous state of the sequence. The
determination of a(8) and a(9) is currently beyond our available
computational capabilities. I will try to confirm a(7) and to find
improved lower bounds for a(8) and a(9) by running the slightly improved
program given at the link for some months on a cluster of Intel Itanium
2 CPUs.

In the meantime I suggest to add a comment:

Known lower bounds for a(7)...a(9) are
762140212575, 440857916120379, 346202123011922653,

Hugo Pfoertner





From mlb at well.com  Wed Nov 22 00:25:34 2006
From: mlb at well.com (Marc LeBrun)
Date: Tue, 21 Nov 2006 15:25:34 -0800
Subject: Almost the number of divisors of n+1
In-Reply-To: <20061121154018.X1354@localhost.my.domain>
References: <20061121154018.X1354@localhost.my.domain>
Message-ID: <200611212325.kALNPQh3019729@smtp.well.com>

 >  a(n) is the number of divisors of (n+1)
 > Have you ever seen closely-related sequences like that one?

Some random and rather vague ideas:

1. I faintly recall seeing something about algorithms for
factoring n involving properties of n-1 (I mean besides
the obvious connection to "Fermat's little theorem").
Perhaps in Knuth's TAOCP?

2. It sounds like you might have a case where the relationship
you're seeing, a(n) ~ d(n), is actually a first approximation
to some identity like

   a(n) = d(n) + F1(n) + F2(n) + ...

where the Fk depend on the factor structure of n or something.
So perhaps you might try to "take out" the d(n) and then look
further into the "error" and see what you can guess about how
the Fk's specifically vary with primes, highly composites, etc.
For example you might try "fitting" a(n) with Moebius-style sums.

3. There's some rather surprising known relationships involving
divisors--my favorite is one of Legendre's that is basically an
identity found by expanding elliptic functions in power series.
This suggests that you might want to express your relationships
as generating functions and see what emerges.

4. Sometimes there is simple underlying "combinatorial machinery".
For example I was once puzzled by some coefficients that turned
out to be formed by taking each divisor of n, *adding 1*, and then
multiplying them all together!  It took me a while to see that's
algebraically identical to summing the products of every possible
subset of the divisors.  But the "add 1" way of expressing this
was totally opaque, whereas the "every subset" explained them.

Happy hunting!







From zbi74583 at boat.zero.ad.jp  Wed Nov 22 02:15:26 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Wed, 22 Nov 2006 10:15:26 +0900
Subject: N(2)
Message-ID: <00a201c70dd3$b538af40$977fd7dc@FMC57937B6BE00>

    I wonder if N(2) goes infinity.

    Where N(s)=   Sum_{m=1 to infinity} (-1)Sigma(m)/m^s

    Mathar computed up to m=1000000



  s m N(s)

> 2 1 1.000000000000000000000000000
> 2 2 1.250000000000000000000000000
> 2 3 1.472222222222222222222222222
> 2 4 1.784722222222222222222222222
> 2 5 1.944722222222222222222222222
> 2 6 2.000277777777777777777777778
> 2 7 2.122726757369614512471655329
> 2 8 2.325851757369614512471655329
> 2 9 2.461654226505416981607457798
> 2 10000 8.619813177975507784214344698
> 2 20000 9.240635095871089404115753095
> 2 30000 9.603791897573761358292380061
> 2 40000 9.861458278261452185262848858
> 2 50000 10.06133043757656995196303496
> 2 60000 10.22462154552329778519365923
> 2 70000 10.36270038005636494621123847
> 2 80000 10.48231016697477680498962023
> 2 90000 10.58780522013752853231672862
> 2 100000 10.68217402235127296288785820
> 2 110000 10.76754463607533705360724580
> 2 120000 10.84547789729425765394094918
> 2 130000 10.91717268365027978460835188
> 2 140000 10.98355652885066079792761960
> 2 150000 11.04534545539722125239381859
> 2 160000 11.10315629233238669062836862
> 2 170000 11.15746025373895516747888760
> 2 180000 11.20865775307751148407150182
> 2 190000 11.25708463818456648077093060
> 2 200000 11.30303121795283056763425981
> 2 210000 11.34672871948779384291766276
> 2 220000 11.38839827063844308711114745
> 2 230000 11.42821240662906506074836109
> 2 240000 11.46633165190016188934526342
> 2 250000 11.50290004647923375585461222
> 2 260000 11.53802980091084626273394586
> 2 270000 11.57183602344116902681335304
> 2 280000 11.60440829561909536077171934
> 2 290000 11.63584045378869260629085740
> 2 300000 11.66620539683007648663354504
> 2 310000 11.69557551635064846235471407
> 2 320000 11.72401391041322342493145640
> 2 330000 11.75157395832622563415629569
> 2 340000 11.77831576251751436463182945
> 2 350000 11.80427986574616427809478607
> 2 360000 11.82951337960016320803463309
> 2 370000 11.85405377158818097821428014
> 2 380000 11.87794149361683265940241605
> 2 390000 11.90120806565142413868457075
> 2 400000 11.92388559358277288937420440
> 2 410000 11.94600303970164868968234173
> 2 420000 11.96758569454234998141373716
> 2 430000 11.98866287480559328733313634
> 2 440000 12.00925544388189350449936925
> 2 450000 12.02938482345842327394291528
> 2 460000 12.04907101416111155199518539
> 2 470000 12.06833481184766752219695603
> 2 480000 12.08719139777266186606623586
> 2 490000 12.10566116650605692754670219
> 2 500000 12.12375840131307130932177405
> 2 510000 12.14149351071909408890611230
> 2 520000 12.15888769016783248485026059
> 2 530000 12.17594975401341942285000014
> 2 540000 12.19269280126406301090981599
> 2 550000 12.20912705098196862278557687
> 2 560000 12.22526702472305560991043337
> 2 570000 12.24111996896626519121471994
> 2 580000 12.25669798166764128539680162
> 2 590000 12.27200952384435815319104573
> 2 600000 12.28706299338740013449483108
> 2 610000 12.30187012252190226181814868
> 2 620000 12.31643453756242668558457904
> 2 630000 12.33076670711890682146429120
> 2 640000 12.34487182648867321005497632
> 2 650000 12.35875920337989470739102627
> 2 660000 12.37243430644503889140869087
> 2 670000 12.38590421689421153982279985
> 2 680000 12.39917431914306915108546756
> 2 690000 12.41224965443445935379058168
> 2 700000 12.42513867918653815467126754
> 2 710000 12.43784365088579118428209464
> 2 720000 12.45037185509338005878377820
> 2 730000 12.46272650898502725121177200
> 2 740000 12.47491268376206272216237202
> 2 750000 12.48693594951832440956683550
> 2 760000 12.49880013638169511883229569
> 2 770000 12.51050884124405914256853750
> 2 780000 12.52206537108093533871333277
> 2 790000 12.53347692259745252540675575
> 2 800000 12.54474459716977128685280547
> 2 810000 12.55587125001645305770022096
> 2 820000 12.56686130746829341595469366
> 2 830000 12.57771842011223665163037795
> 2 840000 12.58844534265072464040923243
> 2 850000 12.59904566706465055360427300
> 2 860000 12.60952226922408394518254911
> 2 870000 12.61987642201166975414741555
> 2 880000 12.63011418303496833362065684
> 2 890000 12.64023571583693921823484395
> 2 900000 12.65024408926657285403820450
> 2 910000 12.66014049405210257196085418
> 2 920000 12.66993013960070545251999449
> 2 930000 12.67961329658082585267888159
> 2 940000 12.68919341076487592672709130
> 2 950000 12.69867217785123447166660908
> 2 960000 12.70805040456023104531710849
> 2 970000 12.71733330696500792660758365
> 2 980000 12.72652037909829933006161382
> 2 990000 12.73561385879130711822105048
> 2 1000000 12.74461639944508312563225494



         UnitaryPhi(m) < (-1)Sigma(m) < Sigma(m)

         So, U(2) < N(2) < S(2)=Zeta(1):*Zeta(2)

         Where,U(s) = Sum_{m=1 to infinity} UnitaryPhi(m)/m^s , S(s) = 
Sum_{m=1 to infinity} Sigma(m)/m^s



    I suppose that U(2) is already known.

    Does anyone know it?



    Yasutoshi
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From zbi74583 at boat.zero.ad.jp  Wed Nov 22 02:15:24 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Wed, 22 Nov 2006 10:15:24 +0900
Subject: A085058
References: <019201c70ab0$b44b3ce0$ca73d7dc@FMC57937B6BE00> <007501c70ac2$a6b28900$6303f0d5@speedy>
Message-ID: <00a101c70dd3$b38cd400$977fd7dc@FMC57937B6BE00>

   Is it an easy well known fact?

   I would like to know the reason why they are the same.



   Yasutoshi



  ----- Original Message ----- 
  From: Vladeta Jovovic
  To: kohmoto
  Sent: Saturday, November 18, 2006 12:35 PM
  Subject: Re: A085058



  Yes, they are the same sequence: see A101921.

  Best wishes,
  Vladeta Jovovic

    ----- Original Message ----- 
    From: kohmoto
    To: seqfan at ext.jussieu.fr
    Sent: Saturday, November 18, 2006 2:27 AM
    Subject: A085058


        Hi, Seqfans

        If 2^k(n) is the largest power of two dividing tangent number T(n) 
then numbers such that 2*n-k(n) are as follows.



        S : 2,3,2,4,2,3,2,5,2,....



        It looks like A085058.

        Are they the same seqences ?



        Yasutoshi




    __________ NOD32 1.1454 (20060321) Information __________

    This message was checked by NOD32 antivirus system.
    http://www.nod32.com
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From zbi74583 at boat.zero.ad.jp  Wed Nov 22 02:15:22 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Wed, 22 Nov 2006 10:15:22 +0900
Subject: A020955
Message-ID: <009801c70dd3$b0eb0000$977fd7dc@FMC57937B6BE00>

    Hi, Seafans.

    I suppose that no one understand what A020955 means.



    So, I explain it a little.

    First of all, see my home page and click ?Free Class?, recently I 
rewrote it.



         http://boat.zero.ad.jp/~zbi74583/another02.htm



    The idea ?ma??.an abbreviation of ?make??.is represented by a 
mapping from subsets of a set to members of the set..

    Because the idea ?ma? has the same property as a mapping.



    Number of subsets of a set which has N members is 2^N, and the 
possibility of images of each subsets is N, so number of the mapping 
subsets ->members is N^(2^N).



    But if the set is a model of Free Class then the mapping must satisfy 
some conditions.         .

    They are as follows.



    If the set is {0,1,2,3,?.n} then { } ma 0, {0} ma1, {0,1} ma 2, {0,1,2} 
ma 3,?.,{0,1,2,3,?.n} ma n .



    So, images of n+1 subsets are already decided.

    Hence the number of the mapping or the number of models which has N 
members is N^(2^N-n-1) .



    Yasutoshi


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From aplewe at sbcglobal.net  Wed Nov 22 02:59:20 2006
From: aplewe at sbcglobal.net (Andrew Plewe)
Date: Tue, 21 Nov 2006 17:59:20 -0800
Subject: Almost the number of divisors of n+1
In-Reply-To: <200611212325.kALNPQh3019729@smtp.well.com>
Message-ID: <200611220159.kAM1xMZG084485@shiva.jussieu.fr>

Marc said:

Some random and rather vague ideas:

1. I faintly recall seeing something about algorithms for
factoring n involving properties of n-1 (I mean besides
the obvious connection to "Fermat's little theorem").
Perhaps in Knuth's TAOCP? 



I have that book in front of me right now; he demonstrates how to use the
properties of n-1 to verify primality (pg. 375, in vol. 2, 2nd edition)
after applying Fermat's theorm. 

One a similar note, one could also try to solve:

n - 1 = a mod (a + 1) (or, more generally, n - x = a mod (a + x))

in which case a + 1 (or x) will be a factor of n, but I don't think that's a
particularly practical method -- finding a "mod difference" for n - x seems
to be just as hard as factoring n, even if you know the factors of n - x.

	-Andrew Plewe-

-----Original Message-----
From: Marc LeBrun [mailto:mlb at well.com] 
Sent: Tuesday, November 21, 2006 3:26 PM
To: seqfan at ext.jussieu.fr
Subject: Re: Almost the number of divisors of n+1

 >  a(n) is the number of divisors of (n+1)
 > Have you ever seen closely-related sequences like that one?

Some random and rather vague ideas:

1. I faintly recall seeing something about algorithms for
factoring n involving properties of n-1 (I mean besides
the obvious connection to "Fermat's little theorem").
Perhaps in Knuth's TAOCP?

2. It sounds like you might have a case where the relationship
you're seeing, a(n) ~ d(n), is actually a first approximation
to some identity like

   a(n) = d(n) + F1(n) + F2(n) + ...

where the Fk depend on the factor structure of n or something.
So perhaps you might try to "take out" the d(n) and then look
further into the "error" and see what you can guess about how
the Fk's specifically vary with primes, highly composites, etc.
For example you might try "fitting" a(n) with Moebius-style sums.

3. There's some rather surprising known relationships involving
divisors--my favorite is one of Legendre's that is basically an
identity found by expanding elliptic functions in power series.
This suggests that you might want to express your relationships
as generating functions and see what emerges.

4. Sometimes there is simple underlying "combinatorial machinery".
For example I was once puzzled by some coefficients that turned
out to be formed by taking each divisor of n, *adding 1*, and then
multiplying them all together!  It took me a while to see that's
algebraically identical to summing the products of every possible
subset of the divisors.  But the "add 1" way of expressing this
was totally opaque, whereas the "every subset" explained them.

Happy hunting!







From bowerc at usa.net  Wed Nov 22 03:31:46 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Tue, 21 Nov 2006 18:31:46 -0800
Subject: Assembly of nothing
Message-ID: <129kkVcEU2922S01.1164162706@cmsweb01.cms.usa.net>

This is about a problem with sequence transforms I've been thinking
about for a couple of years now. I call it the "assembly of nothing"
problem.

Assembly refers to creating a set of combinatorial structures
classified by number of nodes of the form A(B) where B is a
preexisting set of structures and A is a rule for how to place
the B-structures into the new structure.

One type of assembly is combinatorial substitution (or composition).
In this instance the "A" rule is itself a collection of structures.
The assembly is to take copies of "B" structures and use them as
the points of the "A" structure.

A variant of this assembly is one where certain pairs of points in
the "A" structure are set aside to be filled with the same "B"
structure (i.e. 2 copies of the same structure) or different
"B" structures.

What's particularly notable about these assemblies is that they
can be described by index series which are formal polynomial
series of the indexed variables x_1, x_2, x_3, ....

The "nothing" part of the title refers to "B" structures that have
a size (or number of points) of zero. The formulas detailing the
counting of these structures frequently forbid the "B" structure
from containing any zero sized members. In many cases this seems
like a necessary restriction, but in others, it can get in the
way of some useful combinatorial questions.

Let's consider 2 types of assemblies.

Sets of structures, allowing the same structure multiple times
(as implemented by the Euler transform for unlabeled structures
and the Exponential transform for labeled structures.)

Sets of structures, requiring all structures to be different
(as implemented by the Weigh transform for unlabeled structures.)

http://www.research.att.com/~njas/sequences/A000041
The partition sequence A000041 counts unlabeled sets of nonempty sets
hence it has g.f. EULER(x/(1-x)) if we consider EULER as a g.f.
transformation.

http://www.research.att.com/~njas/sequences/A000009
A000009 counts sets of nonempty sets all different from one another
with g.f. WEIGH(x/(1-x)).

in the first case, it makes sense that the "B" structure (the
nonempty sets) are not allowed to have 0 sized elements (i.e.
be empty). If we tried to count partitions of n and allowed n
to be partitioned into 0s, we would have infinitely many
possibilities. 1, 1+0, 1+0+0, ....

In the second case, we can allow empty "B" structures, since
each structure can only be used once giving us 4 partitions
of 3 into distinct parts: 3, 3+0, 1+2, 1+2+0, or in general
2*A000009(n) partitions of n into distinct parts allowing 0
as a part. Hence we would like to say that
WEIGH(1/(1-x)) = 2*WEIGH(x/(1-x)).

The commonly used formulas appear to support these observations.

The Euler transform as applied to a sequence is often rendered:

(1-x)^(-a1)*(1-x^2)^(-a2)*(1-x^3)^(-a3)*(1-x^4)^(-a4)*...

with the Weigh transform as

(1+x)^a1*(1+x^2)^a2*(1+x^3)^a3*(1+x^4)^a4*...

both of these formulas ignore the a0 term, but if we were to extend
these one term backward we would have Euler start

(1-1)^(-a0)*...

and Weigh

(1+1)^a0*...

and we can see that in Euler's case given any positive a0 we begin
with 0 raised to a negative power giving an infinite value and for
Weigh's case we start the product with 2^a0.

A similar effect is seen from the index series.

An index series is applied to a g.f. as follows:

If s(x_1, x_2, x_3, ...) is the index series, then it maps the
g.f. A(x) to s(A(x), A(x^2), A(x^3), ...) thus x_2^2 + x_3*x_4
maps as A(x^2)^2+A(x^3)*A(x^4).

Euler's i.s. is exp(x_1/1 + x_2/2 + x_3/3 + ...)

Weigh's i.s. is exp(x_1/1 - x_2/2 + x_3/3 - ...)

We can see that if A(x) has a positive constant term, then
that term will be multiplied by the divergent harmonic series
1+1/2+1/3+... and blow up. With Weigh we have the alternating
harmonic series 1-1/2+1/3-...=ln 2 giving it the 2^a0 character.

So in these 2 examples, the problem seems manageable, but if one
is looking for a general theory to cover all such assemblies or
all transforms defined by i.s., things are not so easy.

Consider 2 other assembly rules:

Ordered lists of structures, allowing the same structure multiple times
(as implemented by the Invert transform for unlabeled structures.)

Ordered lists of structures, requiring a structure not have a copy
of itself as a neighbor
(I call the unlabeled version of this the Carlitz transform. It's
not in the EIS canon yet.)

Invert's i.s. is 1/(1-x_1)
Carlitz's i.s. is 1/(1-x_1+x_2-x_3+...)

Invert creates compositions (or ordered partitions) the way Euler
creates partitions.

Hence we have INVERT(x/(1-x)) = (1-x)/(1-2x) = 1,1,2,4,8,16,...

Just as in the unordered partitions, we would not allow one to
compose n into 0s, there would be infinitely many choices.
Note that unlike the Euler transform which blows up for a0>0,
The Invert transform can be calculated for any a0 != 1, although
it's difficult to give a combinatorial interpretation to any
nonzero value.

http://www.research.att.com/~njas/sequences/A003242
Carlitz(x/(1-x)) gives us A003242 Carlitz compositions, those
where no two adjacent parts are equal. We can allow Carlitz
compositions to contain 0s since the separation of the 0s by other
numbers keeps it finite: Carlitz(1/(1-x)) = A114900.
http://www.research.att.com/~njas/sequences/A114900
However, the i.s. does not tell us how to do it. If A(x)=1 what
does one make of Carlitz(A(x)) = 1/(1-1+1-1+...)

I recently had an idea about how to deal with issues like this:
use an analytic interpretation of the g.f. of a sequence.

If a sequence represents structures contains only finitely many
structures, then the number of structures is A(1) where A(x) is
the g.f. My idea is to take the case where A(x) has a radius of
convergence (ROC) >0 and <=1 so that it cannot be evaluated by
adding the terms, and associating its size with A(1) of the
analytic extension of the g.f. Here in the Carlitz case we have
1-x+x^2-x^3+... = 1/(1+x) hence A(1) = 1/2 and 1/(1/2)=2, the
expected value for Carlitz compositions of 0 allowing 0 as a part.
(The 2 are the empty composition and zero.)

In this way I can evaluate a transform of a sequence whose only
nonzero term is a_0 as T(a_0*x)(1) and for more general sequences
I can apply the transform to a 2 dimensional sequence as follows.

If sequence has g.f. A(x) then let C(x,y) = y*A(x).

Apply T to C(x,y) getting D(x,y) as output:

Let d_n = [x^n] D(x,1)

I suspect this will be an adequate solution to many problems of
this sort, but I'm not sure how far this method can be trusted.
Part of the problem is I don't know my analytic function theory
as well as I should. (If anyone out there does, please chime in.)
More importantly, I've run across some examples that challenge
the method.

My first example involves the Euler transform. It's basically only
a problem because I would expect that if f(1)=g(1) then
(Tf)(1) = (Tg)(1) for any transform with both f and g in its
domain. It seems reasonable as 1^n=1 for all n so that i.s.'s
behave like g.f.'s when evaluated at 1. More importantly, the
choice to use the linear term in "C(x,y) = y*A(x)" as opposed
to "C(x,y) = y^2*A(x)" was arnitrary and it seemingly should
not matter which term I chose, so I would like some assurance
that I'll get the same answer following any such path.

Consider

EULER(0) = 1
EULER(x-x^2) = 1+x
EULER(x-x^3) = 1+x+x^2

For all of these, the inputs have f(1)=0, but the outputs have
g(x)=1,2 or 3

I think the reason for this anamoly may be in the instability
Euler has at 0. Just as Weigh behaves like 2^x, Euler behaves
like 0^(-x) giving 0 for negative values of x, infinity for
positive values and doing what we see here at x=0. 0^0 is
generally defined as 1, but analytically it can be anything and
that may be the limitation I'm running into.

My other example concerns me, because I'm interested in formulas
that can be applied to arbitrary sequences, not just integer
sequences. Most of our familiar transform formulas work fine for
arbitrary sequences, and it would be nice if all of them did,
however...


Ordered lists of structures, requiring all structures to be different
(as implemented by the AGK transform for unlabeled structures
found on
http://www.research.att.com/~njas/sequences/transforms2.html )

Obviously, this assembly makes sense for empty structures since
each structure can appear at most once. If I allow there to be
n types of empty structures in the input then I have
A003149(n) empty structures in the output.
http://www.research.att.com/~njas/sequences/A003149

While it seems quite plausible that this sequence can be
interpolated into a nice analytic function, the method I illustrated
above offers no help as for n anything other that a
nonnegative integer gives me a sequence with ROC of zero.

So the idea in posting this is to present some new ideas to the
sequence community, get some opinions from people who know some
things that I don't know and get some ideas about how to handle
difficult cases.

Christian








From davidwwilson at comcast.net  Wed Nov 22 04:22:55 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Tue, 21 Nov 2006 22:22:55 -0500
Subject: Confirmation
References: <200611211857.kALIvmai029436@amer.strw.leidenuniv.nl>
Message-ID: <001001c70de5$812f2190$6501a8c0@yourxhtr8hvc4p>

Thanks for your help, your results agree with my values.





From jawbrey at att.net  Wed Nov 22 04:28:31 2006
From: jawbrey at att.net (Jon Awbrey)
Date: Tue, 21 Nov 2006 22:28:31 -0500
Subject: Assembly of nothing
References: <129kkVcEU2922S01.1164162706@cmsweb01.cms.usa.net>
Message-ID: <4563C3DE.FB5134DC@att.net>

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o

This is one of the impulses that led to the construction of riffs and rotes.
It was probably in my freshman chemistry class, or later when I worked as a
hospital orderly and used to run errands for the nuclear medicine unit that
I was thinking about the techniques of molecular tagging, which is like the
business of labelling graphs, only you have to label molecules with bits of
physical structure, an extra neutron or the like, that alters the resulting
structure.

So the question arises whether you can have a class of structures that is
closed under self-labeling.  Somewhat roughly trying to use your notation,
if A(B) is the set of all structures where those of type A are labeled by
those of type B, can you have the doamin equation A = A(A) for any type A?

Or something like that ...

Jon Awbrey

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o
inquiry e-lab: http://stderr.org/pipermail/inquiry/
wikinfo: http://wikinfo.org/wiki.php?title=User:Jon_Awbrey
wp review: http://wikipediareview.com/index.php?showuser=398
o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o






From Annette.Warlich at t-online.de  Wed Nov 22 13:37:30 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Wed, 22 Nov 2006 13:37:30 +0100
Subject: Eigensequences
Message-ID: <4564448A.5010200@t-online.de>

In  a recent treatise of the pascal matrix and its
connections to the bernoulli-numbers I restated
their connection as an eigenvector/eigensequence-
relation.

Say P is the lower triangular matrix of binomial-
coefficients, Pj the matrix, where the column-signs
alternate and B the vector of bernoulli-numbers
(b_1 = +1/2)
then
   Pj * B = B

Now in connection with the subject of divergent
summmation I investigated powerseries, say
  V(x) = [1,x,x^2,x^3,... ]
in connection with Pj, like

  ??? = Pj * V(x)      // binomial-transform
  ??? = V(x)~ * Pj

(using ~ for the transpose)

Now since

   Pj * B = B

also

   V(x) * Pj * B = V(x) * B

and the effect of the coefficients of Pj
disappear, if I consider powerseries with
the bernoulli-numbers as coefficients:

 f(x) = b0 + b1*x + b2*x^2 + b3*x^3 + ...
      =    b0
         + (1*b0 - 1*b1)*x
         + (1*b0 - 2*b1 + 1*b2)*x^2
         + ...

---------
Now incidently I also found, that, using J as a diagonal-
unit-matrix with units with alternating signs
 J = diag(1,-1,1,-1,...)

such that also

 Pj  = P * J

then a matrix G, containing the bernoulli-numbers
in the first column (and shifted and scaled
in a certain manner in the follwing columns)

 Pj = G * J * G^-1

and thus, for instance, the above eigensequence-
property of the bernoulli-numbers can be extended
to an infinite set of coefficients arising from

   V(x) * Pj * G = V(x) * G * J

in the infinite number of columns of the result.

The second column, for instance adds to the above
f(x) the function g(x)

 g(x) =    1*b0*x + 2*b1*x^2 + 3*b2*x^3 + 4*b3*x^4 + ...
      = -  (    1* 0
             + (1* 0 - 1*1*b0)*x
             + (1* 0 - 2*1*b0 + 1*2*b1)*x^2
             + (1* 0 - 3*1*b0 + 3*2*b1 -  1*3*b2)*x^3
             + ...
      = -  (
               ( - 1*b0)*x
             + ( - 2*b0 + 2*b1)*x^2
             + ( - 3*b0 + 6*b1 -  3*b2)*x^3
             + ...

or, using x=1, the said eigensequence-operation, here
with the associated eigenvalue -1.

-------------------------------------
Even more interesting seems to me, that the said matrix
G can be decomposed into an eigensystem, where

  G = St2 * Z(1) * St2^-1

with St2 = matrix of stirling-numbers of 2'nd kind
     Z(s)= diag(1,  1/2^s,  1/3^s, 1/4^s,... )
the diagonal-matrix of the zeta-summands of power s.

That means, that even the bernoulli-numbers have
eigensequences related to them (even infinite many),
namely the columns of St2 , associated with the
eigenvalues (1,1/2,1/3,...) such that

  V(x) * G * St2*Z(1) = V(x)*St2

and for instance,

  f(x) = 1 + 1*x + 1*x^2 + 1*x^3 + ....  // taken from V(x)*St2

       =       1*b0                     // taken from V(x) * G * St2 * Z(1)
          + (1/2*b0 + b1)*m
          + (1/3*b0 + b1 +     b2 )*m^2
          + (1/4*b0 + b1 + 3/2*b2 +      b3 )*m^3
          + (1/5*b0 + b1 +   2*b2 +    2*b3 +     b4)*m^4
          + (1/6*b0 + b1 + 5/2*b2 + 10/3*b3 + 5/2*b4 +      b5)*m^5
           ...

which may look kind trivial, since the involved Stirling-
coefficients are all 1
Other coefficients occur in the next columns 1,2,3,..., for instance
in column 1:

   g(x)  =.  x + S2_1*x^2 + S3_1*x^3 + S4_1*x^4 + S5_1*x^5 + ....
         =              b0 * x
             ( 2/3*S2_1*b0 +      2*b1 ) * x^2
             ( 1/2*S3_1*b0 + 2*S2_1*b1 + 3     *b2 )*m^3
             ( 2/5*S4_1*b0 + 2*S3_1*b1 + 4*S2_1*b2 +    4     *b3 ) *m^4
             ( 1/3*S5_1*b0 + 2*S4_1*b1 + 5*S3_1*b2 + 20/3*S2_1*b3  +  5*b4) *m^5

and so on for the next columns.
With the involved Stirling-numbers evaluated we have then for this 2'nd
column:

   g(x)  =   .    2*x + 6*x^2 + 14*x^3 + 30*x^4  +  62*x^5
         =          2*b0 * x
            +   (   4*b0 +  4*b1)*x^2
            +   (   7*b0 + 12*b1 +  6*b2) *x^3
            +   (  12*b0 + 28*b1 + 24*b2 +  8*b3) *m^4
            +   (62/3*b0 + 60*b1 + 70*b2 + 40*b3 + 10*b4))))*m^5
              ...
which, if the bernoulli-numbers are inserted is obvious.
Using x=1 and written as sequences, we have another
eigensequence-expression and implicitely infinitely many.

This description was animated by Neil's article in the
integer-sequences-journal, which I came across recently -
though I do not know, if the above is actually of interest
related to the article (my current context is originally
the study of divergent summation, where these identities
popped up)

Regards-

Gottfried Helms







From Annette.Warlich at t-online.de  Wed Nov 22 13:46:11 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Wed, 22 Nov 2006 13:46:11 +0100
Subject: Eigensequences /editing corrections
Message-ID: <45644693.1000503@t-online.de>

// upps. In the previous article I've overlooked some
// needed editings, replacing the m,m^2,... by x,x^2 ...
// sorry. Repost here:


In  a recent treatise of the pascal matrix and its
connections to the bernoulli-numbers I restated
their connection as an eigenvector/eigensequence-
relation.

Say P is the lower triangular matrix of binomial-
coefficients, Pj the matrix, where the column-signs
alternate and B the vector of bernoulli-numbers
(b_1 = +1/2)
then
   Pj * B = B

Now in connection with the subject of divergent
summmation I investigated powerseries, say
  V(x) = [1,x,x^2,x^3,... ]
in connection with Pj, like

  ??? = Pj * V(x)      // binomial-transform
  ??? = V(x)~ * Pj

(using ~ for the transpose)

Now since

   Pj * B = B

also

   V(x) * Pj * B = V(x) * B

and the effect of the coefficients of Pj
disappear, if I consider powerseries with
the bernoulli-numbers as coefficients:

 f(x) = b0 + b1*x + b2*x^2 + b3*x^3 + ...
      =    b0
         + (1*b0 - 1*b1)*x
         + (1*b0 - 2*b1 + 1*b2)*x^2
         + ...

---------
Now incidently I also found, that, using J as a diagonal-
unit-matrix with units with alternating signs
 J = diag(1,-1,1,-1,...)

such that also

 Pj  = P * J

then a matrix G, containing the bernoulli-numbers
in the first column (and shifted and scaled
in a certain manner in the follwing columns)

 Pj = G * J * G^-1

and thus, for instance, the above eigensequence-
property of the bernoulli-numbers can be extended
to an infinite set of coefficients arising from

   V(x) * Pj * G = V(x) * G * J

in the infinite number of columns of the result.

The second column, for instance adds to the above
f(x) the function g(x)

 g(x) =    1*b0*x + 2*b1*x^2 + 3*b2*x^3 + 4*b3*x^4 + ...
      = -  (    1* 0
             + (1* 0 - 1*1*b0)*x
             + (1* 0 - 2*1*b0 + 1*2*b1)*x^2
             + (1* 0 - 3*1*b0 + 3*2*b1 -  1*3*b2)*x^3
             + ...
      = -  (
               ( - 1*b0)*x
             + ( - 2*b0 + 2*b1)*x^2
             + ( - 3*b0 + 6*b1 -  3*b2)*x^3
             + ...

or, using x=1, the said eigensequence-operation, here
with the associated eigenvalue -1.

-------------------------------------
Even more interesting seems to me, that the said matrix
G can be decomposed into an eigensystem, where

  G = St2 * Z(1) * St2^-1

with St2 = matrix of stirling-numbers of 2'nd kind
     Z(s)= diag(1,  1/2^s,  1/3^s, 1/4^s,... )
the diagonal-matrix of the zeta-summands of power s.

That means, that even the bernoulli-numbers have
eigensequences related to them (even infinite many),
namely the columns of St2 , associated with the
eigenvalues (1,1/2,1/3,...) such that

  V(x) * G * St2*Z(1) = V(x)*St2

and for instance,

  f(x) = 1 + 1*x + 1*x^2 + 1*x^3 + ....  // taken from V(x)*St2

       =       1*b0                     // taken from V(x) * G * St2 * Z(1)
          + (1/2*b0 + b1)*x
          + (1/3*b0 + b1 +     b2 )*x^2
          + (1/4*b0 + b1 + 3/2*b2 +      b3 )*x^3
          + (1/5*b0 + b1 +   2*b2 +    2*b3 +     b4)*x^4
          + (1/6*b0 + b1 + 5/2*b2 + 10/3*b3 + 5/2*b4 +      b5)*x^5
           ...

which may look kind trivial, since the involved Stirling-
coefficients are all 1
Other coefficients occur in the next columns 1,2,3,..., for instance
in column 1:

   g(x)  =.  x + S2_1*x^2 + S3_1*x^3 + S4_1*x^4 + S5_1*x^5 + ....
         =              b0 * x
             ( 2/3*S2_1*b0 +      2*b1 ) * x^2
             ( 1/2*S3_1*b0 + 2*S2_1*b1 + 3     *b2 )*x^3
             ( 2/5*S4_1*b0 + 2*S3_1*b1 + 4*S2_1*b2 +    4     *b3 ) *x^4
             ( 1/3*S5_1*b0 + 2*S4_1*b1 + 5*S3_1*b2 + 20/3*S2_1*b3  +  5*b4) *x^5

and so on for the next columns.
With the involved Stirling-numbers evaluated we have then for this 2'nd
column:

   g(x)  =   .    2*x + 6*x^2 + 14*x^3 + 30*x^4  +  62*x^5
         =          2*b0 * x
            +   (   4*b0 +  4*b1)*x^2
            +   (   7*b0 + 12*b1 +  6*b2) *x^3
            +   (  12*b0 + 28*b1 + 24*b2 +  8*b3) *x^4
            +   (62/3*b0 + 60*b1 + 70*b2 + 40*b3 + 10*b4) *x^5
              ...
which, if the bernoulli-numbers are inserted is obvious.
Using x=1 and written as sequences, we have another
eigensequence-expression and implicitely infinitely many.

This description was animated by Neil's article in the
integer-sequences-journal, which I came across recently -
though I do not know, if the above is actually of interest
related to the article (my current context is originally
the study of divergent summation, where these identities
popped up)

Regards-

Gottfried Helms








From njas at research.att.com  Wed Nov 22 17:58:32 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 22 Nov 2006 11:58:32 -0500 (EST)
Subject: A125097.gif: (Very) nice patterns
References: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>
Message-ID: <200611221658.LAA41364@fry.research.att.com>

The other day Zak said:
> Neil,
> seqfans,
>
> 1. Have a look at this full graph -
> aint these patterns (very) nice?!
> It may be known, but still it's nice...
>
> 2. Is it possible to
> put "the extended graph" into OEIS?
> In the standard graph no patterns are seen at all...
>
> 3. Neil removed  keyword "nice" from A125097
> (as he did regularly with ny sequences)-
> as if I insist that it's me who is "nice"
> not it is the seq which is nice.
>

Here is my reply:
1. i will use the b-file
2. we don't need the gif since the graph button will use the b-file
3. the definition of the sequence is
     a(n) is k such that k*(2n+1) mod 1000 = 123, or 2n+1 if no such k exists.
Now there is rule that a sequence should not depend on an arbitrary 
parameter. Here we have two of them. Why not:
a(n) is k such that k*(2n+1) mod 44445555 = 321345, or 2n+1 if no such k exists. ?
4. the sequence is not finite (a(1000) = 123)
5. although i know you are a nice guy, i do not think this sequence is nice.
Best regards
Neil





From grafix at csl.pl  Wed Nov 22 18:58:30 2006
From: grafix at csl.pl (Artur)
Date: Wed, 22 Nov 2006 18:58:30 +0100
Subject: All primes
In-Reply-To: <200611221658.LAA41364@fry.research.att.com>
References: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com> <200611221658.LAA41364@fry.research.att.com>
Message-ID: <op.tjfivsouat030q@tata>

Dear Members
I'm looking for members which have Mathematica Programme and can help in
finding the first number such that
(PrimeQ[2(n) + 1] == True) && (PrimeQ[4(
       n) + 1] == True) && (PrimeQ[6(n) + 1] == True) && (PrimeQ[8(
       n) + 1] == True) && (PrimeQ[10(n) + 1] == True) && (PrimeQ[12(
       n) + 1] == True) && (PrimeQ[14(n) + 1] == True) && (PrimeQ[16(
       n) + 1] == True) && (PrimeQ[18(n) + 1] == True) && (PrimeQ[20(
       n) + 1] == True) && (PrimeQ[22(n) + 1] == True)
these n is bigger than 4850000000000
I have Mathematica procedure which is 4000 much more quickest as above but  
for as big numbers PrimeQ working slow and will be good divided range of  
searching on few subranges and run each other on different computer.
If somebody can help let me know. I need information how long time  
computer can run without reset.
BEST REGARDS
ADRTUR JASINSKI






From rgwv at rgwv.com  Wed Nov 22 19:22:17 2006
From: rgwv at rgwv.com (Robert G. Wilson v)
Date: Wed, 22 Nov 2006 13:22:17 -0500
Subject: Whereabouts of Joe K. Crump & www.spacefire.com
Message-ID: <45649559.9030304@rgwv.com>

,

Does anyone on this list know the whereabouts of Joe K. Crump or the
status of the website www.spacefire.com ? Thanks in advance for any
information.

Bob.





From noe at sspectra.com  Wed Nov 22 19:27:50 2006
From: noe at sspectra.com (T. D. Noe)
Date: Wed, 22 Nov 2006 10:27:50 -0800
Subject: All primes
In-Reply-To: <op.tjfivsouat030q@tata>
References: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>
 <200611221658.LAA41364@fry.research.att.com> <op.tjfivsouat030q@tata>
Message-ID: <p06230905c18a46fef974@[192.168.1.43]>

At 7:04 PM +0100 11/22/06, Artur wrote:
>Dear Members
>I'm looking for members which have Mathematica Programme and can help in
>finding the first number such that
>(PrimeQ[2(n) + 1] == True) && (PrimeQ[4(
>       n) + 1] == True) && (PrimeQ[6(n) + 1] == True) && (PrimeQ[8(
>       n) + 1] == True) && (PrimeQ[10(n) + 1] == True) && (PrimeQ[12(
>       n) + 1] == True) && (PrimeQ[14(n) + 1] == True) && (PrimeQ[16(
>       n) + 1] == True) && (PrimeQ[18(n) + 1] == True) && (PrimeQ[20(
>       n) + 1] == True) && (PrimeQ[22(n) + 1] == True)
>these n is bigger than 4850000000000

See A071576.

Tony






From mathar at strw.leidenuniv.nl  Wed Nov 22 19:39:07 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Wed, 22 Nov 2006 19:39:07 +0100
Subject: Whereabouts of Joe K. Crump & www.spacefire.com
Message-ID: <200611221839.kAMId7db031776@amer.strw.leidenuniv.nl>


> From seqfan-owner at ext.jussieu.fr  Wed Nov 22 19:09:21 2006
> Date: Wed, 22 Nov 2006 13:22:17 -0500
> From: "Robert G. Wilson v" <rgwv at rgwv.com>
> To: "seqfan at ext.jussieu.fr" <SeqFan at ext.jussieu.fr>
> Subject: Whereabouts of Joe K. Crump & www.spacefire.com
> 
> Does anyone on this list know the whereabouts of Joe K. Crump or the
> status of the website www.spacefire.com ? Thanks in advance for any
> information.
..
According to Max (whom we now know as Alekseyev), there is a 
archive.org backup of his web page:

http://groups-beta.google.com/group/sci.math/browse_thread/thread/8d061b676bd7e32b/071ae01f53f9a1f5?hl=en#071ae01f53f9a1f5
(sci.math of Nov 14)
http://web.archive.org/web/20040609202646/www.spacefire.com/numbertheory/

An e-mail address of 2002 is shown in
http://www.primepuzzles.net/thepuzzlers/Crump.htm
and becomes
ssaywer1 at rr.carolina.com
in the archived web page of 2000.





From franktaw at netscape.net  Wed Nov 22 22:20:32 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 22 Nov 2006 16:20:32 -0500
Subject: Categories
Message-ID: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>

How many categories are there?

First, how many categories are there with n morphisms and k objects?  
This table starts:

 1
 2  1
 7  3 1
35 16 3 1

The first column is A058129, the number of monoids; the main diagonal 
is all 1's.  I am not
100% certain of the 16 in the final row.

Taking the row sums, we get:

1,3,11,55

the number of categories with n morphisms.  This is probably not in the 
OEIS (only
A001776 is possible - other matches become less than A058129).  The 
inverse Euler
transform,

1,2,8,41

is the number of connected categories with n morphisms; this is 
likewise probably not
in the OEIS (only A052447 is possible).

Can somebody generate more data?

Franklin T. Adams-Watters

A category is a collection of objects and morphisms; each morphism is 
from one object
to another (not necessarily different) object.  Where the destination 
of one morphism
is the source of a second, their composition is defined; composition is 
associative where
it is defined.  Each object has an identity morphism, which connects it 
to itself; this
is an identity when composed with morphisms coming in and with 
morphisms going
out.
________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From jvospost3 at gmail.com  Wed Nov 22 23:55:44 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 22 Nov 2006 14:55:44 -0800
Subject: Categories
In-Reply-To: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
Message-ID: <5542af940611221455r6e693b45kae8a9af38c6ee9eb@mail.gmail.com>

This actually relates to the enumeration results I've made on "prime"
endofunctions, namely endofunctions which are not unions of other prime
endofunctions nor categorical products of endofunctions. My partial results
here are in A125024, A124023, and a comment on A048888 that I submitted a
day or two ago about the enumeration of prime endofunctions in terms of
numbrals.

I have been working on enumerating not just endofunctions, but
(categorically) endomorphisms of endofunctions, (2-categorically)
endomorphisms of endomorphisms of endofunctions.  I've been posting partial
results on John Baez's "n-category cafe" and apparently just annoyed John
Baez, who didn't yet see that I was finding a new combinatorial result at
the base level (endofunctions) and working on the n-category hierarchy above
that, so those comments may already have been deleted from his blog.

That is, in counting all categories on n objects, most are trivially the
union of smaller disjoint categories (that is, the sagittal digraph is not
connected), or categorical products of smaller categories.  They key to
enumeration involves thus both a tree-enumeration, and the numbral part
which, per a recent paper in the Journal on Integer Sequences, relate to
balanced ordered trees.

Hard to summarize less cryptically, here, but I stand by in support of what
you're doing, and think my approach is genuinely related.

-- Jonathan Vos Post

On 11/22/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
> How many categories are there?
>
> First, how many categories are there with n morphisms and k objects?
> This table starts:
>
> 1
> 2  1
> 7  3 1
> 35 16 3 1
>
> The first column is A058129, the number of monoids; the main diagonal
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
>
> Taking the row sums, we get:
>
> 1,3,11,55
>
> the number of categories with n morphisms.  This is probably not in the
> OEIS (only
> A001776 is possible - other matches become less than A058129).  The
> inverse Euler
> transform,
>
> 1,2,8,41
>
> is the number of connected categories with n morphisms; this is
> likewise probably not
> in the OEIS (only A052447 is possible).
>
> Can somebody generate more data?
>
> Franklin T. Adams-Watters
>
> A category is a collection of objects and morphisms; each morphism is
> from one object
> to another (not necessarily different) object.  Where the destination
> of one morphism
> is the source of a second, their composition is defined; composition is
> associative where
> it is defined.  Each object has an identity morphism, which connects it
> to itself; this
> is an identity when composed with morphisms coming in and with
> morphisms going
> out.
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and
> industry-leading spam and email virus protection.
>
>
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From jvospost3 at gmail.com  Thu Nov 23 01:39:16 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 22 Nov 2006 16:39:16 -0800
Subject: Categories
In-Reply-To: <5542af940611221455r6e693b45kae8a9af38c6ee9eb@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <5542af940611221455r6e693b45kae8a9af38c6ee9eb@mail.gmail.com>
Message-ID: <5542af940611221639n2fde8e43j93dc69594ae9392b@mail.gmail.com>

Next question is thus: "How many 2-categories (bicategories) are there whose
objects are the categories on n objects?"

See:
Basic Bicategories
Tom Leinster (1998-10-04)  arXiv.org:math/9810017 A concise guide to very
basic bicategory theory, from the definition of a bicategory to the
coherence theorem. Comment: 11 pages; LaTeX 2e with Paul Taylor's diagram
macros

This is even harder to answer, because there seem to be at least 7 different
definitions of n-categories, of which categories = 1-categories, and I've
just raised the enumeration probem for n=2.

www.ima.umn.edu/categories/abstracts.html
Abstracts for the IMA 2004 Summer Program: n-Categories.
Ross Street has proposed two related definitions of n-category, one in 1987
and one in 2002.

See also:
Tom Leinster (2003-05-02) arXiv.org:math/0305049 Higher-dimensional category
theory is the study of n-categories, operads, braided monoidal categories,
and other such exotic structures. It draws its inspiration from areas as
diverse as topology, quantum algebra, mathematical physics, logic, and
theoretical computer science. This is the first book on ... Comment: Book,
410 pages

http://www.lepp.cornell.edu/spr/2002-04/msg0041086.html

In the beginning, there was nothing.
But with nothing came nothingness, the vacuity, which was something.
So now there was nothingness and somethingness,
vacuity and triviality, falsehood and truth,
which were 2 things.
So now there was a set of nothingness and somethingness,
and the whole realm of sets of elements
sprang up out of the set of nothingness and somethingness,
and sprouted functions between them
to relate them back to the set of nothingness and somethingness.
So now there was a category of sets and functions,
and the whole realm of categories of objects and morphisms
sprang up out of the category of sets and functions,
and sprouted functors between them,
to relate them back to the category of sets and functions,
which sprouted natural transformations between *them*,
to relate the relationships.
So now there was ...


-- Toby
   toby at math.ucr.edu

But your question is fundamental, and I am eager to see an answer.  I hope
it is not rude for me to be anticipating follow-ups.

Happy Thanksgiving,

Jonathan Vos Post

On 11/22/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> This actually relates to the enumeration results I've made on "prime"
> endofunctions, namely endofunctions which are not unions of other prime
> endofunctions nor categorical products of endofunctions. My partial results
> here are in A125024, A124023, and a comment on A048888 that I submitted a
> day or two ago about the enumeration of prime endofunctions in terms of
> numbrals.
>
> I have been working on enumerating not just endofunctions, but
> (categorically) endomorphisms of endofunctions, (2-categorically)
> endomorphisms of endomorphisms of endofunctions.  I've been posting partial
> results on John Baez's "n-category cafe" and apparently just annoyed John
> Baez, who didn't yet see that I was finding a new combinatorial result at
> the base level (endofunctions) and working on the n-category hierarchy above
> that, so those comments may already have been deleted from his blog.
>
> That is, in counting all categories on n objects, most are trivially the
> union of smaller disjoint categories (that is, the sagittal digraph is not
> connected), or categorical products of smaller categories.  They key to
> enumeration involves thus both a tree-enumeration, and the numbral part
> which, per a recent paper in the Journal on Integer Sequences, relate to
> balanced ordered trees.
>
> Hard to summarize less cryptically, here, but I stand by in support of
> what you're doing, and think my approach is genuinely related.
>
> -- Jonathan Vos Post
>
> On 11/22/06, franktaw at netscape.net <franktaw at netscape.net > wrote:
> >
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> > 1
> > 2  1
> > 7  3 1
> > 35 16 3 1
> >
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
> > Taking the row sums, we get:
> >
> > 1,3,11,55
> >
> > the number of categories with n morphisms.  This is probably not in the
> > OEIS (only
> > A001776 is possible - other matches become less than A058129).  The
> > inverse Euler
> > transform,
> >
> > 1,2,8,41
> >
> > is the number of connected categories with n morphisms; this is
> > likewise probably not
> > in the OEIS (only A052447 is possible).
> >
> > Can somebody generate more data?
> >
> > Franklin T. Adams-Watters
> >
> > A category is a collection of objects and morphisms; each morphism is
> > from one object
> > to another (not necessarily different) object.  Where the destination
> > of one morphism
> > is the source of a second, their composition is defined; composition is
> > associative where
> > it is defined.  Each object has an identity morphism, which connects it
> > to itself; this
> > is an identity when composed with morphisms coming in and with
> > morphisms going
> > out.
> > ________________________________________________________________________
> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > industry-leading spam and email virus protection.
> >
> >
>
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From maxale at gmail.com  Thu Nov 23 04:52:55 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 22 Nov 2006 19:52:55 -0800
Subject: Categories
In-Reply-To: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
Message-ID: <d3dac270611221952h2853749ev8fcb5ced9e2f80a7@mail.gmail.com>

On 11/22/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
> How many categories are there?
>
> First, how many categories are there with n morphisms and k objects?

Does that correspond to transitively closed digraphs with k (labeled?)
vertices and n edges?

Max





From maxale at gmail.com  Thu Nov 23 05:10:51 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 22 Nov 2006 20:10:51 -0800
Subject: Whereabouts of Joe K. Crump & www.spacefire.com
In-Reply-To: <200611221839.kAMId7db031776@amer.strw.leidenuniv.nl>
References: <200611221839.kAMId7db031776@amer.strw.leidenuniv.nl>
Message-ID: <d3dac270611222010m17a14a53x75b9de0cf2ffbe3a@mail.gmail.com>

On 11/22/06, Richard Mathar <mathar at strw.leidenuniv.nl> wrote:

> http://web.archive.org/web/20040609202646/www.spacefire.com/numbertheory/

I've put an updated version of Joe's Database of 2^x-3 factorizations at
http://www-cse.ucsd.edu/users/maxal/2nm3_db.txt

Please contribute if you like.

Max





From pauldhanna at juno.com  Thu Nov 23 07:19:41 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Thu, 23 Nov 2006 01:19:41 -0500
Subject: Formula for A112319? - Diagonals in Self-Compositions of (x+x^2)
Message-ID: <20061123.011941.272.1.pauldhanna@juno.com>

Seqfans, 
     Would someone like to try to find a formula for A112319:  
"Coefficients of x^n in the (n-1)-th self-composition of (x + x^2) for
n>=1."
1, 1, 2, 9, 64, 630, 7916, 121023, 2179556, 45179508, 1059312264, 
 
The related triangles A122888 and A122890 seem to indicate that 
A112319 may have a formula involving the Catalan numbers: 
http://www.research.att.com/~njas/sequences/A122890
 
Below I copy A112319 and triangle A122888 for your convenience. 
Other diagonals in self-compositions of (x+x^2) are A112317, A112320.
 
Be it a recurrence, series reversion, etc., any formula for these
diagonals 
could lead to other discoveries involving self-compositions of functions.

     Paul 
----------------------------------------------------------------

A122888
Triangle, read by rows, where row n lists the coefficients of x^k, 
k=1..2^n, in the n-th self-composition of (x + x^2) for n>=0. 

EXAMPLE  
Triangle begins: 
1; 
1, 1; 
1, 2, 2, 1; 
1, 3, 6, 9, 10, 8, 4, 1; 
1, 4, 12, 30, 64, 118, 188, 258, 302, 298, 244, 162, 84, 32, 8, 1; 

Multiplying the g.f. of column k by (1-x)^k, k>=1, with leading zeros, 
yields the g.f. of row k in the triangle A122890: 
1; 
0, 1; 
0, 0, 2; 
0, 0, 1, 5; 
0, 0, 0, 10, 14; 
0, 0, 0, 8, 70, 42; 
0, 0, 0, 4, 160, 424, 132; 
0, 0, 0, 1, 250, 1978, 2382, 429; 
0, 0, 0, 0, 302, 6276, 19508, 12804, 1430; ... 

in which the main diagonal is the Catalan numbers 
and the row sums form the factorials. 

----------------------------------------------------------------

A112319  
Coefficients of x^n in the (n-1)-th self-composition of (x + x^2) for
n>=1. 

1, 1, 2, 9, 64, 630, 7916, 121023, 2179556, 45179508, 1059312264, 
27715541568, 800423573676, 25289923553700, 867723362137464, 
32128443862364255, 1276818947065793736, 54208515369076658640, 
2448636361058495090816
 
FORMULA  
a(n) = [x^n] F_{n-1}(x) where F_n(x) = F_{n-1}(x+x^2) 
with F_1(x) = x+x^2 and F_0(x)=x for n>=1. 
 
EXAMPLE  
Initial terms in self-compositions of (x+x^2) are: 
F(x) = x + (1)*x^2 
F(F(x)) = x + 2*x^2 + (2)*x^3 + x^4 
F(F(F(x))) = x + 3*x^2 + 6*x^3+ (9)*x^4 +... 
F(F(F(F(x)))) = x + 4*x^2 + 12*x^3 + 30*x^4 + (64)*x^5 +... 
F(F(F(F(F(x))))) = x + 5*x^2 + 20*x^3 + 70*x^4 + 220*x^5 + (630)*x^6 +...
 
----------------------------------------------------------------
END.
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From jvospost3 at gmail.com  Thu Nov 23 07:34:45 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 22 Nov 2006 22:34:45 -0800
Subject: Categories
In-Reply-To: <d3dac270611221952h2853749ev8fcb5ced9e2f80a7@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <d3dac270611221952h2853749ev8fcb5ced9e2f80a7@mail.gmail.com>
Message-ID: <5542af940611222234r3ae28d48h1c3e6541143896a7@mail.gmail.com>

First, John Baez agrees that an earlier post of his partially answered my
follow-up question about 2-categories (some assembly required):

http://golem.ph.utexas.edu/category/2006/10/klein_2geometry_vi.html#c005157

 Re: Klein 2-Geometry VI

Tim writes:

So I'll wrap up and say what I think I've discovered. Just as
finite-dimensional vector spaces and their subspaces are intimately related
to finite sets and their subsets, so finite-dimensional 2-vector spaces and
their 2-subspaces are intimately related to finite graphs and their
subgraphs. These are obviously much more complicated and I'm not at all
confident my analyses here are correct, particularly the identification of
the various 2-Grassmannians associated to various graph maps. The basic idea
seems sound though.

Yeah, that's a great idea!

The way I see it, your fundamental idea is this. There's a 2-functor from
the 2-category of directed graphs to the 2-category of vector 2-spaces
(?2-term chain complexes). And, it goes like this:

   - Any directed graph gives a 2-term chain complex.
   - Any map between directed graphs gives a chain map between 2-term
   chain complexes.
   - And, there's also a kind of "homotopy between maps between directed
   graphs", which gives a chain homotopy between chain maps between 2-term
   chain complexes.

(Do graph theorists ever think about those homotopies? They should!)

All this generalizes further, to an (n+1)-functor from n-dimensional cell
complexes to (n+1)-term chain complexes.

I'm not sure how to get some amazing new insights into projective n-geometry
from this way of thinking - but that's not surprising, since I just read
your post 2 minutes ago. We should mull on it.

Thanks!
Posted by: John Baez <http://math.ucr.edu/home/baez/> on October 8, 2006
7:25 PM |

Second, I took Category theory in grad school 1976 or 1977, and forgot what
little I knew, and have relearned only some. So anyone in seqfans who
learned more and remebered more should jump in here, and don't worry about
making me look foolish, as I take the risk of exposing my ignorance every
time I post here.

If one asks: "how many categories are there with n morphisms and k objects?"
one is asking up to isomorphism. I'm making a first cut here off the top of
my head, which may be no more than an ill-informed guess.

The sagittal graph of a general morphism a.k.a. homomorphism (not restricted
to a monomorphism, epimorphism, isomorphism, endomorphism, or automorphism)
leaves us wondering what the structure of the objects are.  I think, by the
way that the question is asked, that they have no structure, i.e. that they
are just points. If they have structure, that affects everything.
unstructrured objects (points), then we still get different counts if the
vertices and arcs are labeled or unlabeled, and whether one allows loops or
not. if the edges are unlabeled, and loops are allowed, then any of the k
objects can be mapped to any of the k objects including itself, giving k^k,
if n = k.  If n>k then we can 't use more than k of the n because the
mapping must be functional, that is, only one arrow out of each point (but
multiple arrows to a point is allowed).

 If n<k then the first point can be mapped to any of the k points, and we
run out of arrows when we reach k^n.  Now we see why labeling matters, as if
they are unlabeled, we count only once in equivalence classes, but if
labeled each such mapping is distinct.  Each loop and arc is de facto
labeled by its source point and target point, as an ordered pair.  If you
want to allow other labels, you have to multiply by the proper combinational
multiplier.

So we're taking a summation of n=0 (just the k points and none map to any),
n=1 (one of k points maps to one of k), n=2 and so forth.

That starts the answer.  For k=n and any point can map to any, we have the
count of endomorphisms being the same as the count of endofunctions already
in OEIS.

Category theory has differing approaches depending on whether we have
digraphs, multgraphs,  reversible digraphs, reflexive digraphs, irreflexive
digraphs, and various tricks to represent undirected graphs with directed
graphs.

So exact definitions are necessary.

Sorry, just a first cut.  Not enough sleep.  Some time tomorrow before gests
and Thanksgiving dinner kicks in, but no promises.

Thanks for the dialog.

Best,

Jonathan Vos Post
<http://golem.ph.utexas.edu/cgi-bin/MT-3.0/sxp-comments.pl?entry_id=955;parent_id=5157>


On 11/22/06, Max A. <maxale at gmail.com> wrote:
>
> On 11/22/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
>
> Does that correspond to transitively closed digraphs with k (labeled?)
> vertices and n edges?
>
> Max
>
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From Roland.Bacher at ujf-grenoble.fr  Thu Nov 23 07:54:39 2006
From: Roland.Bacher at ujf-grenoble.fr (Roland Bacher)
Date: Thu, 23 Nov 2006 07:54:39 +0100
Subject: Categories
In-Reply-To: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
Message-ID: <20061123065439.GA28473@fourier.ujf-grenoble.fr>


This numbers can also be obtained as the numbers
of the following quotients of quivers:

Associate to a category with n morphisms and k objects 
the quiver with k vertices corresponding to the objects 
and a(X,Y) arrows directed from the object X to the object
Y if there are a(X,Y) morphisms from X into Y.

Given two morphism g:X-->Y, f:Y-->Z with composition
h=f o g:X-->Z, put the relation gf=h on the quiver algebra

The resulting quotient algebra has dimension n and a basis
given by the (simple) arrows.

This leads to the following "algorithm" for enumerating
all categories with n morphisms and k objects:

(a) enumerate all quivers (directed graphs) with k vertices and
n oriented edges.

(b) Associate the following "weight" to such a quiver as follows :

(b1)given a triplet of vertices X,Y,Z,
set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
(this counts the number of ways which associate a morphism
X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the 
number of oriented arrows starting at U and ending at W.

(b2) associate to a given quiver the weight 
\prod_{X,Y,Z} w(X,Y,Z)
where the product is over all triplets of vertices.

The total sum of such weighted quivers yields then the solution.

Roland Bacher



   


On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.net wrote:
> How many categories are there?
> 
> First, how many categories are there with n morphisms and k objects?  
> This table starts:
> 
> 1
> 2  1
> 7  3 1
> 35 16 3 1
> 
> The first column is A058129, the number of monoids; the main diagonal 
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
> 
> Taking the row sums, we get:
> 
> 1,3,11,55
> 
> the number of categories with n morphisms.  This is probably not in the 
> OEIS (only
> A001776 is possible - other matches become less than A058129).  The 
> inverse Euler
> transform,
> 
> 1,2,8,41
> 
> is the number of connected categories with n morphisms; this is 
> likewise probably not
> in the OEIS (only A052447 is possible).
> 
> Can somebody generate more data?
> 
> Franklin T. Adams-Watters
> 
> A category is a collection of objects and morphisms; each morphism is 
> from one object
> to another (not necessarily different) object.  Where the destination 
> of one morphism
> is the source of a second, their composition is defined; composition is 
> associative where
> it is defined.  Each object has an identity morphism, which connects it 
> to itself; this
> is an identity when composed with morphisms coming in and with 
> morphisms going
> out.
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and 
> industry-leading spam and email virus protection.
> 





From jvospost3 at gmail.com  Thu Nov 23 18:14:47 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 23 Nov 2006 09:14:47 -0800
Subject: Categories
In-Reply-To: <20061123065439.GA28473@fourier.ujf-grenoble.fr>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
Message-ID: <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>

Roland Bacher correctly refers to quivers as summarized in wikipedia at

http://en.wikipedia.org/wiki/Quiver_%28mathematics%29

On 11/22/06, Roland Bacher <Roland.Bacher at ujf-grenoble.fr> wrote:
>
>
> This numbers can also be obtained as the numbers
> of the following quotients of quivers:
>
> Associate to a category with n morphisms and k objects
> the quiver with k vertices corresponding to the objects
> and a(X,Y) arrows directed from the object X to the object
> Y if there are a(X,Y) morphisms from X into Y.
>
> Given two morphism g:X-->Y, f:Y-->Z with composition
> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
>
> The resulting quotient algebra has dimension n and a basis
> given by the (simple) arrows.
>
> This leads to the following "algorithm" for enumerating
> all categories with n morphisms and k objects:
>
> (a) enumerate all quivers (directed graphs) with k vertices and
> n oriented edges.
>
> (b) Associate the following "weight" to such a quiver as follows :
>
> (b1)given a triplet of vertices X,Y,Z,
> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> (this counts the number of ways which associate a morphism
> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> number of oriented arrows starting at U and ending at W.
>
> (b2) associate to a given quiver the weight
> \prod_{X,Y,Z} w(X,Y,Z)
> where the product is over all triplets of vertices.
>
> The total sum of such weighted quivers yields then the solution.
>
> Roland Bacher
>
>
>
>
>
>
> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.net wrote:
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> > 1
> > 2  1
> > 7  3 1
> > 35 16 3 1
> >
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
> > Taking the row sums, we get:
> >
> > 1,3,11,55
> >
> > the number of categories with n morphisms.  This is probably not in the
> > OEIS (only
> > A001776 is possible - other matches become less than A058129).  The
> > inverse Euler
> > transform,
> >
> > 1,2,8,41
> >
> > is the number of connected categories with n morphisms; this is
> > likewise probably not
> > in the OEIS (only A052447 is possible).
> >
> > Can somebody generate more data?
> >
> > Franklin T. Adams-Watters
> >
> > A category is a collection of objects and morphisms; each morphism is
> > from one object
> > to another (not necessarily different) object.  Where the destination
> > of one morphism
> > is the source of a second, their composition is defined; composition is
> > associative where
> > it is defined.  Each object has an identity morphism, which connects it
> > to itself; this
> > is an identity when composed with morphisms coming in and with
> > morphisms going
> > out.
> > ________________________________________________________________________
> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > industry-leading spam and email virus protection.
> >
>
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From jvospost3 at gmail.com  Thu Nov 23 21:10:51 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 23 Nov 2006 12:10:51 -0800
Subject: Categories
In-Reply-To: <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
Message-ID: <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>

*"Formalized Proof, Computation, and the Construction Problem in Algebraic
geometry", by Carlos Simpson.

[PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
2004<http://arxiv.org/pdf/math/0410224> File
Format: PDF/Adobe Acrobat - View as
HTML<http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5>
finite integer N, *how many categories* are there with N morphisms? What *
...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
(1965), 450-464.



On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> Roland Bacher correctly refers to quivers as summarized in wikipedia at
>
> http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
>
> On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> >
> >
> > This numbers can also be obtained as the numbers
> > of the following quotients of quivers:
> >
> > Associate to a category with n morphisms and k objects
> > the quiver with k vertices corresponding to the objects
> > and a(X,Y) arrows directed from the object X to the object
> > Y if there are a(X,Y) morphisms from X into Y.
> >
> > Given two morphism g:X-->Y, f:Y-->Z with composition
> > h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> >
> > The resulting quotient algebra has dimension n and a basis
> > given by the (simple) arrows.
> >
> > This leads to the following "algorithm" for enumerating
> > all categories with n morphisms and k objects:
> >
> > (a) enumerate all quivers (directed graphs) with k vertices and
> > n oriented edges.
> >
> > (b) Associate the following "weight" to such a quiver as follows :
> >
> > (b1)given a triplet of vertices X,Y,Z,
> > set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> > (this counts the number of ways which associate a morphism
> > X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> > number of oriented arrows starting at U and ending at W.
> >
> > (b2) associate to a given quiver the weight
> > \prod_{X,Y,Z} w(X,Y,Z)
> > where the product is over all triplets of vertices.
> >
> > The total sum of such weighted quivers yields then the solution.
> >
> > Roland Bacher
> >
> >
> >
> >
> >
> >
> > On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.net wrote:
> > > How many categories are there?
> > >
> > > First, how many categories are there with n morphisms and k objects?
> > > This table starts:
> > >
> > > 1
> > > 2  1
> > > 7  3 1
> > > 35 16 3 1
> > >
> > > The first column is A058129, the number of monoids; the main diagonal
> > > is all 1's.  I am not
> > > 100% certain of the 16 in the final row.
> > >
> > > Taking the row sums, we get:
> > >
> > > 1,3,11,55
> > >
> > > the number of categories with n morphisms.  This is probably not in
> > the
> > > OEIS (only
> > > A001776 is possible - other matches become less than A058129).  The
> > > inverse Euler
> > > transform,
> > >
> > > 1,2,8,41
> > >
> > > is the number of connected categories with n morphisms; this is
> > > likewise probably not
> > > in the OEIS (only A052447 is possible).
> > >
> > > Can somebody generate more data?
> > >
> > > Franklin T. Adams-Watters
> > >
> > > A category is a collection of objects and morphisms; each morphism is
> > > from one object
> > > to another (not necessarily different) object.  Where the destination
> > > of one morphism
> > > is the source of a second, their composition is defined; composition
> > is
> > > associative where
> > > it is defined.  Each object has an identity morphism, which connects
> > it
> > > to itself; this
> > > is an identity when composed with morphisms coming in and with
> > > morphisms going
> > > out.
> > >
> > ________________________________________________________________________
> > > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > > industry-leading spam and email virus protection.
> > >
> >
>
>
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From jvospost3 at gmail.com  Thu Nov 23 21:20:58 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 23 Nov 2006 12:20:58 -0800
Subject: Categories
In-Reply-To: <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
	 <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
Message-ID: <5542af940611231220oe45dce2rbee869c1dea94b9e@mail.gmail.com>

FOM: Set theory vs category theory: Representability of categories *Vaughan
Pratt* pratt at cs.Stanford.EDU
<fom%40cs.nyu.edu?Subject=FOM%3A%20Set%20theory%20vs%20category%20theory%3A%20Representability%20of%20categories&In-Reply-To=199801271058.LAA14876%40brics.dk>
*Tue Jan 27 13:37:42 EST 1998

http://cs.nyu.edu/pipermail/fom/1998-January/001021.html
*



A number of representation theorems were found by Czech algebraists and
category theorists during the latter half of the 1960's, representing
arbitrary small categories as categories of familiar concrete structures
and their homomorphisms.  Two notable representations are in terms
of semigroups, and of directed graphs.  In each case the theorem is
that every small category is representable as a category of semigroups
(directed graphs, etc.) and their homomorphisms (of the standard kind
in each case).

The criteria used here for representability of objects  a  and morphisms
f:a-&gt;b of a category C by respectively structures F(a) and homomorphisms
F(f): F(a)->F(b) of the concrete representing category D are as follows.

(i) F:C->D is a *functor*, that is, a graph homomorphism (viewing the
morphisms of C and D as edges of their respective underlying graphs)
that preserves composition and identities (a condition exactly analogous
to those of monoid homomorphism and group homomorphism).

(ii) F is *faithful*, i.e. for any pair a,b of objects of C, distinct
morphisms f:a-&gt;b of C are represented by distinct homomorphisms
F(f):F(a)->F(b) of D.

(iii) F is *full*, i.e. every homomorphism from F(a) to F(b) in D is
the representation under F of some morphism from a to b in C.

The intuitive meaning of full and faithful is that the representing
object is just as stiff as the object it represents when you wiggle both.
Fullness means it is at least as stiff (D offers no new transformations),
while faithfulness means it is at most as stiff (no transformations are
lost by identification).

Unfortunately the project of embedding arbitrary categories fully and
faithfully in familiar concrete categories has turned out not to have
the impact of its counterparts in group theory (permutations), Boolean
algebras (fields of sets), distributive lattices (rings of sets), etc.
These results have in consequence not been widely publicized in the
introductory categorical literature.

The most objectionable feature of these embeddings in my view (and I
am aware of no other serious objection) is their lack of respect for
concreteness itself.  If one takes for C a small *concrete* category,
such as all quotient groups of the additive group of natural numbers
and their group homomorphisms, the above-mentioned embedding represents
the finite group Z_2 as an infinite semigroup!  And if &quot;natural
numbers&quot;

is replaced by "reals", the corresponding embedding represents Z_2 as an
uncountable semigroup.  Such a lack of respect for concreteness would
seem to undermine the very point of representing abstract objects as
concrete structures.

In contrast a finite group of order n is representable by permutations of
n things, a finite Boolean algebra as a field of subsets of a finite set,
and so on.  In these better-known representation theorems, cardinality
is respected, typically to within at most two exponentials (which in
the infinite case means to within two beth numbers) and often much better.

This unsatisfactory situation with the concrete representation of
objects of arbitrary categories prompts the speculation that the notion
of category is too broad to admit of any improvement in the situation.
After all, any random graph G is made a category by adding to its
edges all paths in G, with composition defined as path concatenation
(the free category generated by G).  How could there possibly be a
cardinality-respecting representation of the vertices of a random graph
in terms of the objects of a familiar concrete category?  When one can
do any violence one wishes to the structure by adding an edge at random
between any two vertices, it would seem that any concrete representation
of each object would inevitably have a cardinality on the order of the
whole category.

But in fact there *is* a representation that does the job.  Let us make
two easily understood modifications to the usual notion of topological
space.

First, drop the requirement that the open sets be closed under arbitrary
union and finite intersection.  (The role of this traditional restriction
is in effect to limit the &quot;signature&quot; of a topological space
to just that
needed to express the notion of limit, or even less with the more discrete
spaces; dropping it greatly broadens the range of possible signatures.)

Second, in place of the customary two degrees of membership in the open
sets (ordinary sets), permit a set K of possible degrees of membership,
call such open sets fuzzy (Zadeh's fuzzy sets are the case K the reals).

Define continuity as usual for functions between topological spaces,
wording the definition in such a way however that its extension to
fuzzy sets is made clear.  Specifically, f:X-&gt;Y is continuous when for
every open set g:Y-&gt;K (here expressed as a characteristic function),
the composite

		   f      g
		X ---> Y ---> K

is an open set of X.  When K = {0,1} it is easily verified that this is
the standard notion of continuity.

Call such generalized topological spaces Chu spaces over K.  An even more
general notion was first studied in detail by Peter Chu in the 1970's,
a master's student of Michael Barr.  The more elementary version above
was first studied by Lafont and Streicher in 1991 (LICS) under the rubric
of games but the term &quot;Chu construction&quot; was already in the computer
science air in the late 1980's making it too late to change the name.

Definitions.

1.  A concrete category is a pair (C,U) where C is a category and U:C-&gt;Set
is a faithful functor, the "forgetful" functor giving the *underlying
set* of each object along with the realization of each morphism of C as
a function (a morphism of Set).

2.  A concrete functor F:(C,U)->(D,V) between two concrete categories
is one satisfying VF = U.  That is, the underlying set VF(a) of the
representation F(a) of a is the same as the underlying set U(a) of
a itself.

Remark.  The elements of all the representing sets of a small concrete
category (C,U) themselves form a single set, namely the disjoint or
marked union of all the sets in the category.  Call this set Elt(C,U),
the set of elements of (C,U).

A very minor technicality requires the notion of an *honest* concrete
category, namely one having, for all objects a and b, a morphism from
a to b if U(a) is empty (necessarily at most one, by concreteness).
Dishonest concrete categories cannot be represented as below because
the empty topological space has nowhere to store the information as to
which other spaces it does or does not have functions to.  (Thanks to
Peter Freyd for pointing out the lacuna in my proof that necessitated
this condition.)

Theorem.  Every small honest concrete category (C,U) embeds fully,
faithfully, and concretely in the category of Chu spaces over Elt(C,U).

(This theorem enjoys all the advantages of the previous embeddings, but
in addition satisfies the strongest possible concreteness requirement:
the representing object has the *same* underlying set as the object it
represents, and the representing morphisms viewed concretely are exactly
the same functions.)

Proof.  Let b be an object of C with underlying set X = U(b).  Represent
b by a generalized topological space X, having one open set Y_h for
each morphism h:b-&gt;c in C for some c.  Form Y_h by taking the degree of
membership of each point x in Y_h to be U(h)(x), i.e. the element of U(c)
which is the image of x under the underlying function U(h) of h.

It is then straightforward to show that this representation is faithful,
full, and concrete.

Obviously the burden of the obstacle that we have overcome has merely
been shifted to the open sets, the number of which must now be on the
order of the cardinality of the category.  But since when has the number
of open sets been an obstacle to topology?  The points of a space are
clearly visible, but our embedding has left these untouched.  Who has
ever seen an open set or cares about how many there are?

For more information on Chu spaces consult

	<A HREF="http://boole.stanford.edu/chuguide.html">http://boole.stanford.edu/chuguide.html</A>

My original interest in Chu spaces was as a model of concurrent
behavior.  More recently I have become interested in them as a primarily
set-theoretic foundation for mathematics that more directly emulates
what category theory has to offer than the extant litany of single-sorted
and and multisorted relational structures and algebras with and without
(ordinary) topology and their standard homomorphisms, all of which are
fully, faithfully, and concretely representable by Chu spaces.

---
Name: Vaughan Pratt
Position: Professor of Computer Science
Institution: Stanford University
Research interests: Foundations of computation and mathematics
For more information: <A
HREF="http://boole.stanford.edu">http://boole.stanford.edu</A>
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From maxale at gmail.com  Thu Nov 23 23:13:45 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 23 Nov 2006 14:13:45 -0800
Subject: connected components in subgraphs of the complete graph: A125205 - A125207
Message-ID: <d3dac270611231413v4ea7e950ifc9405a7ed04e057@mail.gmail.com>

SeqFans,

I've just sent to Neil the following three sequences that you may find
interesting.

It would be also interesting to compute unlabeled variants of these
sequences. If you like to contribute, please do.

Regards,
Max


%I A125205
%S A125205 1,2,1,3,6,3,1,4,18,30,24,15,6,1,5,40,135,250,295,282,215,120,45,10,1,6,
%T A125205 75,420,1385,3015,4800,6365,7170,6705,5065,3009,1365,455,105,15,1,7,126,
%U A125205 1050,5355,18690,47880,96796,166890,251370,329945,373947,362292,297115
%N A125205 Triangular array T(n,k) (n>=1, 0<=k<=n(n-1)/2) giving the
total number
           of connected components in all subgraphs (V,E') with |E'|=k of
           the complete labeled graph K_n=(V,E).
%F A125205 g.f.: Sum_{n,k}
T(n,k)*x^n/n!*y^k=(F(x,y)-1)*exp(F(x,y)-1)=G(x,y)*ln(G(x,y))
           where G(x,y)=Sum_{n=0..oo} (1+y)^(n(n-1)/2)*x^n/n!, and
F(x,y)=1+ln(G(x,y))
           is g.f. of A062734.
%e A125205 The array starts with
           1
           2, 1
           3, 6, 3, 1
           4, 18, 30, 24, 15, 6, 1
           5, 40, 135, 250, 295, 282, 215, 120, 45, 10, 1
           ...
%e A125205 T(3,1)=6 since there are three different subgraphs of K_3
with one edge,
           and each subgraph has two connected components.
%o A125205 (PARI) { reverse(v)=vector(length(v),i,v[length(v)+1-i]) }
           G=sum(n=0,6,(1+y)^(n*(n-1)/2)*x^n/n!); K=G*log(G);
           for(n=1,6,print(reverse(Vec(n!*polcoeff(K,n,x)))))
%Y A125205 Cf. A062734
%Y A125205 Cf. A125206 (row-reversed version), A125207 (row sums)
%K A125205 nonn,tabf
%O A125205 1,2
%A A125205 Max Alekseyev (maxal(AT)cs.ucsd.edu), Nov 23 2006



%I A125206
%S A125206 1,1,2,1,3,6,3,1,6,15,24,30,18,4,1,10,45,120,215,282,295,250,135,40,5,1,
%T A125206 15,105,455,1365,3009,5065,6705,7170,6365,4800,3015,1385,420,75,6,1,21,210,
%U A125206 1330,5985,20349,54271,116385,204225,297115,362292,373947,329945,251370
%N A125206 Triangular array T(n,k) (n>=1, 0<=k<=n(n-1)/2) giving the
total number
           of connected components in all subgraphs obtained from the complete
           labeled graph K_n by removing k edges.
%C A125206 Row-reversed version of A125205, see A125205 for further details
%e A125206 The array starts with
           1
           1, 2
           1, 3, 6, 3
           1, 6, 15, 24, 30, 18, 4
           1, 10, 45, 120, 215, 282, 295, 250, 135, 40, 5
           ...
%Y A125206 Cf. A125205 (row-reversed version), A125207 (row sums)
%K A125206 nonn,tabf
%O A125206 1,3
%A A125206 Max Alekseyev (maxal(AT)cs.ucsd.edu), Nov 23 2006




%I A125207
%S A125207 1,3,13,98,1398,39956,2354240,286394544,71225744048,35884971729760,36419817759267072,
%T A125207 74221711070826087424,303193538300703211111936,2480118087478081928075065344,
%U A125207 40601989279034990139321984265216,1329877330680067685563700135615633408
%N A125207 The total number of connected components in all subgraphs
obtained from
           the complete labeled graph K_n by removing zero or more edges.
%F A125207 E.g.f.: (F(x)-1)*exp(F(x)-1) = G(x)*ln(G(x)) where
           G(x)=Sum_{n=0..oo} 2^(n(n-1)/2)*x^n/n! and F(x)=1+ln(G(x))
is e.g.f. of A001187
%e A125207 For n=2, we have two graph on two vertices: complete and
empty, the former has
           one connected component while the latter has two connected
components. The total
           number of connected components is 3, which is a(2).
%o A125207 (PARI) G=sum(n=0,30,2^(n*(n-1)/2)*x^n/n!) + O(x31); v=Vec(G*log(G));
           for(i=1,length(v),v[i]*=i!); print(v)
%Y A125207 Cf. A001187, A125205, A125206
%K A125207 nonn
%O A125207 1,2
%A A125207 Max Alekseyev (maxal(AT)cs.ucsd.edu), Nov 23 2006





From davidwwilson at comcast.net  Fri Nov 24 01:39:43 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 23 Nov 2006 19:39:43 -0500
Subject: b-files for power series expansions
Message-ID: <01ad01c70f61$09624cf0$6501a8c0@yourxhtr8hvc4p>

I guess my point was, if someone has access to Mma or some other symbolic math package, it would be nice if they could write a program to generate b-files for sequences that are e.g.f. expansions of real functions. We could then generate a whole slew of b-files fairly easily.
 
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From maxale at gmail.com  Fri Nov 24 02:04:09 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 23 Nov 2006 17:04:09 -0800
Subject: Chestnut
In-Reply-To: <000601c6f538$1ca30fe0$6401a8c0@yourxhtr8hvc4p>
References: <000601c6f538$1ca30fe0$6401a8c0@yourxhtr8hvc4p>
Message-ID: <d3dac270611231704ua5487a7s547879c8e77174b7@mail.gmail.com>

On 10/21/06, David Wilson <davidwwilson at comcast.net> wrote:
>
>
>
>
> This is a chestnut that I am convinced it is true, but I cannot prove it. I
> have posed it on math-fun occasionally, and never got an answer. I would
> like to see it posed to some better minds that some of you might know, e.g,
> JHC or other number theorist.
>
> Let S and T be sets of real numbers. Call T a divider of S if some element
> of T lies strictly between any two elements of S.
>
> For integer n >= 1, let Fence(n) be the set of all rationals with
> denominator n, that is, { k/n : k in Z }.
>
> For real set S, let f(S) be the least n such that Fence(n) is a divider of
> S, if such an n exists.
>
> Let Recip(n) be the set of all integer reciprocals on [0,1] with denominator
> <= n, that is, { 1/b : 1 <= b <= n }
>
> Let Farey(n) be the set of all rationals on [0,1] with denominator <= n,
> that is, { a/b : 0 <= a <= b, 1 <= b <= n }
>
> Is f(Farey(n)) = f(Recip(n)) for every n?
>
> The apparently common sequences a(n) = f(Recip(n)) =? f(Farey(n)) is in the
> OEIS, I cannot find it in the short time I have.





From Roland.Bacher at ujf-grenoble.fr  Fri Nov 24 08:53:08 2006
From: Roland.Bacher at ujf-grenoble.fr (Roland Bacher)
Date: Fri, 24 Nov 2006 08:53:08 +0100
Subject: Categories
In-Reply-To: <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com> <20061123065439.GA28473@fourier.ujf-grenoble.fr> <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com> <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
Message-ID: <20061124075307.GA17864@fourier.ujf-grenoble.fr>


I realised that the formula below for enumerating categories
using quivers is completely wrong. It gives only an upper bound
since one has to require moreover associativity for the composition of 
morphisms. The quiver approach works however in principle but
is much more complex (and I guess this approach has been 
used in some of the previous approaches).   Roland Bacher

On Thu, Nov 23, 2006 at 12:10:51PM -0800, Jonathan Post wrote:
> *"Formalized Proof, Computation, and the Construction Problem in Algebraic
> geometry", by Carlos Simpson.
> 
> [PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
> 2004<http://arxiv.org/pdf/math/0410224> File
> Format: PDF/Adobe Acrobat - View as
> HTML<http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5>
> finite integer N, *how many categories* are there with N morphisms? What *
> ...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
> (1965), 450-464.
> 
> 
> 
> On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> >
> >Roland Bacher correctly refers to quivers as summarized in wikipedia at
> >
> >http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
> >
> >On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> >>
> >>
> >> This numbers can also be obtained as the numbers
> >> of the following quotients of quivers:
> >>
> >> Associate to a category with n morphisms and k objects
> >> the quiver with k vertices corresponding to the objects
> >> and a(X,Y) arrows directed from the object X to the object
> >> Y if there are a(X,Y) morphisms from X into Y.
> >>
> >> Given two morphism g:X-->Y, f:Y-->Z with composition
> >> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> >>
> >> The resulting quotient algebra has dimension n and a basis
> >> given by the (simple) arrows.
> >>
> >> This leads to the following "algorithm" for enumerating
> >> all categories with n morphisms and k objects:
> >>
> >> (a) enumerate all quivers (directed graphs) with k vertices and
> >> n oriented edges.
> >>
> >> (b) Associate the following "weight" to such a quiver as follows :
> >>
> >> (b1)given a triplet of vertices X,Y,Z,
> >> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> >> (this counts the number of ways which associate a morphism
> >> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> >> number of oriented arrows starting at U and ending at W.
> >>
> >> (b2) associate to a given quiver the weight
> >> \prod_{X,Y,Z} w(X,Y,Z)
> >> where the product is over all triplets of vertices.
> >>
> >> The total sum of such weighted quivers yields then the solution.
> >>
> >> Roland Bacher
> >>
> >>
> >>
> >>
> >>
> >>
> >> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.net wrote:
> >> > How many categories are there?
> >> >
> >> > First, how many categories are there with n morphisms and k objects?
> >> > This table starts:
> >> >
> >> > 1
> >> > 2  1
> >> > 7  3 1
> >> > 35 16 3 1
> >> >
> >> > The first column is A058129, the number of monoids; the main diagonal
> >> > is all 1's.  I am not
> >> > 100% certain of the 16 in the final row.
> >> >
> >> > Taking the row sums, we get:
> >> >
> >> > 1,3,11,55
> >> >
> >> > the number of categories with n morphisms.  This is probably not in
> >> the
> >> > OEIS (only
> >> > A001776 is possible - other matches become less than A058129).  The
> >> > inverse Euler
> >> > transform,
> >> >
> >> > 1,2,8,41
> >> >
> >> > is the number of connected categories with n morphisms; this is
> >> > likewise probably not
> >> > in the OEIS (only A052447 is possible).
> >> >
> >> > Can somebody generate more data?
> >> >
> >> > Franklin T. Adams-Watters
> >> >
> >> > A category is a collection of objects and morphisms; each morphism is
> >> > from one object
> >> > to another (not necessarily different) object.  Where the destination
> >> > of one morphism
> >> > is the source of a second, their composition is defined; composition
> >> is
> >> > associative where
> >> > it is defined.  Each object has an identity morphism, which connects
> >> it
> >> > to itself; this
> >> > is an identity when composed with morphisms coming in and with
> >> > morphisms going
> >> > out.
> >> >
> >> ________________________________________________________________________
> >> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> >> > industry-leading spam and email virus protection.
> >> >
> >>
> >
> >





From tbaruchel at free.fr  Fri Nov 24 15:12:13 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Fri, 24 Nov 2006 15:12:13 +0100 (CET)
Subject: Quadratic residues
Message-ID: <20061124150520.F1510@localhost.my.domain>


I asked recently about
>  a(n) is the number of divisors of (n+1)
> Have you ever seen closely-related sequences like that one?

After some hints and a
> Happy hunting!
I finally noticed this :
   a) prime numbers 4n+3 lead to the set of their quadratic residues
   b) prime numbers 4n+1 lead to the set of their quadratic non-residues
   c) odd composite numbers :
      15 --> quadratic residues
      21 --> difficult to understand at first sight
Since my datas are empirical and need much work it is difficult to get
much information. Even numbers are difficult to study also.
Have you ever heard about such a behaviour? In that case, what would
you predict for 21?

WARNING. When I speak of "quadratic residues", I mean the residue has
a GCD = 1 with the number (convention may vary).

Thank you for everything,

-- 
Thomas Baruchel





From tbaruchel at free.fr  Fri Nov 24 15:59:29 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Fri, 24 Nov 2006 15:59:29 +0100 (CET)
Subject: Quadratic residues
In-Reply-To: <20061124150520.F1510@localhost.my.domain>
References: <20061124150520.F1510@localhost.my.domain>
Message-ID: <20061124155641.B1631@localhost.my.domain>

On Fri, 24 Nov 2006, Thomas Baruchel wrote:
> Have you ever heard about such a behaviour? In that case, what would
> you predict for 21?

Forget it, I just found : Jacobi symbol + quadratic reciprocity theorem

Thanks.





From eclark at math.usf.edu  Fri Nov 24 16:28:29 2006
From: eclark at math.usf.edu (Edwin Clark)
Date: Fri, 24 Nov 2006 10:28:29 -0500 (EST)
Subject: Categories
In-Reply-To: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
Message-ID: <Pine.LNX.4.58.0611240958550.32273@shell.cas.usf.edu>

On Wed, 22 Nov 2006 franktaw at netscape.net wrote:

> How many categories are there?
> 
> First, how many categories are there with n morphisms and k objects?  
> This table starts:
> 
>  1
>  2  1
>  7  3 1
> 35 16 3 1
> 
> The first column is A058129, the number of monoids; the main diagonal 
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
> 

As you say, the number of one object categories on n objects is M(n)=the 
number of monoids with n elements. But this is known only up to n = 7
according to the OEIS. This seems to make the number of TWO object
categories with n morphisms a very difficult problem...Let alone THREE
object categories...

MathSciNet gives only one hit on "two object categor*":
------------------------------------------------------------------
Rosick\'y, Ji\v r?
Codensity and binding categories.
Comment. Math. Univ. Carolinae 16 (1975), no. 3, 515--529.

If there is no measurable cardinal, the author constructs a two-object 
category all of whose well-powered complete extensions are binding. 
-------------------------------------------------------------------
which is clearly not related to the enumeration problem. 

A search on "categor* with two objects" give a few more hits
but as far as I can see still no papers related to enumeration..







From jvospost3 at gmail.com  Fri Nov 24 17:51:12 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 24 Nov 2006 08:51:12 -0800
Subject: Categories
In-Reply-To: <Pine.LNX.4.58.0611240958550.32273@shell.cas.usf.edu>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <Pine.LNX.4.58.0611240958550.32273@shell.cas.usf.edu>
Message-ID: <5542af940611240851l23b0d23di56bf7bf28fe71f82@mail.gmail.com>

One should also count loops, i.e. an arrow from an object to itself.

Then there are 2 categories on one object, with and without the loop:
Notating V(C) = the vertices in category C (which we label in this
illustrartion but enumerate up to isomorphism, i.e. on forgetting labels;
and L(C) is the set of loops of C, and A(C) is the set of (proper) arcs.
That is, I've separated the loops out from their being a mere subset of
arcs, to aid in enumeration. Then each category is, for this enumeration, a
triple: C= V(C) U L(C) U A(C).

C_1,1 = (1,null,null);
C_1,2 = (1, 1->1,null).

There are 6 categories on 2 objects:
C_2,1 = ({1,2},null,null);
C_2,2 = ({1,2},null,(1->1))
note that the above is isomorphic to C_2,2 = ({1,2},null,(2->2))
C_2,3 = ({1,2},{(1->1),(2->2)},null);
C_2,4 = ({1,2},null,(1->2)) which is isomorphic to C_2,4 =
({1,2},null,(2->1));
C_2,5 = ({1,2},(1>1),(1->2)) which is isomorphic to ({1,2},(2>2),(2->1));
C_2,6 = ({1,2},null,(1>1),(2->1)).

There do seem to be 35 categories on 3 objects, which are essentially the
same as the 7 endomorphisms on 3 objects, with each loop either left in, or
thrown away.

The combinatorics are now straightforward.  We have the endofunction
enumeration, and binary choices on each loop.

I have many, many pages of drawings of these, but not wanting to either scan
them (not having a scanner) nor write out the explicit sets as above.

Is this clear enough, though, to proceed?

-- Jonathan
C_2,2 = ({1,2},(1->1),null)


On 11/24/06, Edwin Clark <eclark at math.usf.edu> wrote:
>
> On Wed, 22 Nov 2006 franktaw at netscape.net wrote:
>
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> >  1
> >  2  1
> >  7  3 1
> > 35 16 3 1
> >
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
>
> As you say, the number of one object categories on n objects is M(n)=the
> number of monoids with n elements. But this is known only up to n = 7
> according to the OEIS. This seems to make the number of TWO object
> categories with n morphisms a very difficult problem...Let alone THREE
> object categories...
>
> MathSciNet gives only one hit on "two object categor*":
> ------------------------------------------------------------------
> Rosick\'y, Ji\v r?
> Codensity and binding categories.
> Comment. Math. Univ. Carolinae 16 (1975), no. 3, 515--529.
>
> If there is no measurable cardinal, the author constructs a two-object
> category all of whose well-powered complete extensions are binding.
> -------------------------------------------------------------------
> which is clearly not related to the enumeration problem.
>
> A search on "categor* with two objects" give a few more hits
> but as far as I can see still no papers related to enumeration..
>
>
>
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From jvospost3 at gmail.com  Fri Nov 24 20:39:20 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 24 Nov 2006 11:39:20 -0800
Subject: Categories
In-Reply-To: <20061124075307.GA17864@fourier.ujf-grenoble.fr>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
	 <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
	 <20061124075307.GA17864@fourier.ujf-grenoble.fr>
Message-ID: <5542af940611241139sa6472eoc31f5980db42a918@mail.gmail.com>

Offline, I'm coordinating with Edwin Clark on definitions.

But I'm counting as lower bounds  (many pages of paper with drawings, too
lengthy for me to turn into ascii, don't have scanner):

2 categories on 1 point;
6 categories on 2 points;
16 categories on 3 points;
46 categories on 4 points;
116 categories on 2 points;

in each case where the underlying digraph is of an endofunction. The full
count on categories on n points is a weighted partial sum on these, related
to my enumeration of "prime" endofunctions being the sum of Fobonacci(n),
tribonacci(n), teranacci(n), ...

-- I can give examples on request.  Of course, I can always be off by 1 or
more on each count, through sheer blunder.  But I am being self-consistent
in methodology, and it can be done by software using the same algorithm,
carefully translated.

-- Jonathan


On 11/23/06, Roland Bacher <Roland.Bacher at ujf-grenoble.fr> wrote:
>
>
> I realised that the formula below for enumerating categories
> using quivers is completely wrong. It gives only an upper bound
> since one has to require moreover associativity for the composition of
> morphisms. The quiver approach works however in principle but
> is much more complex (and I guess this approach has been
> used in some of the previous approaches).   Roland Bacher
>
> On Thu, Nov 23, 2006 at 12:10:51PM -0800, Jonathan Post wrote:
> > *"Formalized Proof, Computation, and the Construction Problem in
> Algebraic
> > geometry", by Carlos Simpson.
> >
> > [PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
> > 2004<http://arxiv.org/pdf/math/0410224> File
> > Format: PDF/Adobe Acrobat - View as
> > HTML<
> http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5
> >
> > finite integer N, *how many categories* are there with N morphisms? What
> *
> > ...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
> > (1965), 450-464.
> >
> >
> >
> > On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> > >
> > >Roland Bacher correctly refers to quivers as summarized in wikipedia at
> > >
> > >http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
> > >
> > >On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> > >>
> > >>
> > >> This numbers can also be obtained as the numbers
> > >> of the following quotients of quivers:
> > >>
> > >> Associate to a category with n morphisms and k objects
> > >> the quiver with k vertices corresponding to the objects
> > >> and a(X,Y) arrows directed from the object X to the object
> > >> Y if there are a(X,Y) morphisms from X into Y.
> > >>
> > >> Given two morphism g:X-->Y, f:Y-->Z with composition
> > >> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> > >>
> > >> The resulting quotient algebra has dimension n and a basis
> > >> given by the (simple) arrows.
> > >>
> > >> This leads to the following "algorithm" for enumerating
> > >> all categories with n morphisms and k objects:
> > >>
> > >> (a) enumerate all quivers (directed graphs) with k vertices and
> > >> n oriented edges.
> > >>
> > >> (b) Associate the following "weight" to such a quiver as follows :
> > >>
> > >> (b1)given a triplet of vertices X,Y,Z,
> > >> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> > >> (this counts the number of ways which associate a morphism
> > >> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> > >> number of oriented arrows starting at U and ending at W.
> > >>
> > >> (b2) associate to a given quiver the weight
> > >> \prod_{X,Y,Z} w(X,Y,Z)
> > >> where the product is over all triplets of vertices.
> > >>
> > >> The total sum of such weighted quivers yields then the solution.
> > >>
> > >> Roland Bacher
> > >>
> > >>
> > >>
> > >>
> > >>
> > >>
> > >> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.netwrote:
> > >> > How many categories are there?
> > >> >
> > >> > First, how many categories are there with n morphisms and k
> objects?
> > >> > This table starts:
> > >> >
> > >> > 1
> > >> > 2  1
> > >> > 7  3 1
> > >> > 35 16 3 1
> > >> >
> > >> > The first column is A058129, the number of monoids; the main
> diagonal
> > >> > is all 1's.  I am not
> > >> > 100% certain of the 16 in the final row.
> > >> >
> > >> > Taking the row sums, we get:
> > >> >
> > >> > 1,3,11,55
> > >> >
> > >> > the number of categories with n morphisms.  This is probably not in
> > >> the
> > >> > OEIS (only
> > >> > A001776 is possible - other matches become less than A058129).  The
> > >> > inverse Euler
> > >> > transform,
> > >> >
> > >> > 1,2,8,41
> > >> >
> > >> > is the number of connected categories with n morphisms; this is
> > >> > likewise probably not
> > >> > in the OEIS (only A052447 is possible).
> > >> >
> > >> > Can somebody generate more data?
> > >> >
> > >> > Franklin T. Adams-Watters
> > >> >
> > >> > A category is a collection of objects and morphisms; each morphism
> is
> > >> > from one object
> > >> > to another (not necessarily different) object.  Where the
> destination
> > >> > of one morphism
> > >> > is the source of a second, their composition is defined;
> composition
> > >> is
> > >> > associative where
> > >> > it is defined.  Each object has an identity morphism, which
> connects
> > >> it
> > >> > to itself; this
> > >> > is an identity when composed with morphisms coming in and with
> > >> > morphisms going
> > >> > out.
> > >> >
> > >>
> ________________________________________________________________________
> > >> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > >> > industry-leading spam and email virus protection.
> > >> >
> > >>
> > >
> > >
>
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From jvospost3 at gmail.com  Fri Nov 24 23:41:12 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 24 Nov 2006 14:41:12 -0800
Subject: Categories
In-Reply-To: <5542af940611241139sa6472eoc31f5980db42a918@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
	 <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
	 <20061124075307.GA17864@fourier.ujf-grenoble.fr>
	 <5542af940611241139sa6472eoc31f5980db42a918@mail.gmail.com>
Message-ID: <5542af940611241441w68d51ab5pc6ba0832f740d4e8@mail.gmail.com>

Somehow the reply to Frank went astray.

The wikipedia entry on directed graph makes some of the definitional
distinctions I've made, and then some.  Further, it makes the useful
statement:

In Category theory,
a category can be considered a directed multigraph
with the objects as vertices and the morphisms as directed edges.
The functors between categories induce then some,
but not necessarily all, of the digraph morphisms.


-- Jonathan


On 11/24/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> Offline, I'm coordinating with Edwin Clark on definitions.
>
> But I'm counting as lower bounds  (many pages of paper with drawings, too
> lengthy for me to turn into ascii, don't have scanner):
>
> 2 categories on 1 point;
> 6 categories on 2 points;
> 16 categories on 3 points;
> 46 categories on 4 points;
> 116 categories on 2 points;
>
> in each case where the underlying digraph is of an endofunction. The full
> count on categories on n points is a weighted partial sum on these, related
> to my enumeration of "prime" endofunctions being the sum of Fobonacci(n),
> tribonacci(n), teranacci(n), ...
>
> -- I can give examples on request.  Of course, I can always be off by 1 or
> more on each count, through sheer blunder.  But I am being self-consistent
> in methodology, and it can be done by software using the same algorithm,
> carefully translated.
>
> -- Jonathan
>
>
> On 11/23/06, Roland Bacher <Roland.Bacher at ujf-grenoble.fr> wrote:
> >
> >
> > I realised that the formula below for enumerating categories
> > using quivers is completely wrong. It gives only an upper bound
> > since one has to require moreover associativity for the composition of
> > morphisms. The quiver approach works however in principle but
> > is much more complex (and I guess this approach has been
> > used in some of the previous approaches).   Roland Bacher
> >
> > On Thu, Nov 23, 2006 at 12:10:51PM -0800, Jonathan Post wrote:
> > > *"Formalized Proof, Computation, and the Construction Problem in
> > Algebraic
> > > geometry", by Carlos Simpson.
> > >
> > > [PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
> > > 2004<http://arxiv.org/pdf/math/0410224> File
> > > Format: PDF/Adobe Acrobat - View as
> > > HTML<http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5
> > >
> > > finite integer N, *how many categories* are there with N morphisms?
> > What *
> > > ...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
> > > (1965), 450-464.
> > >
> > >
> > >
> > > On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> > > >
> > > >Roland Bacher correctly refers to quivers as summarized in wikipedia
> > at
> > > >
> > > >http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
> > > >
> > > >On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> > > >>
> > > >>
> > > >> This numbers can also be obtained as the numbers
> > > >> of the following quotients of quivers:
> > > >>
> > > >> Associate to a category with n morphisms and k objects
> > > >> the quiver with k vertices corresponding to the objects
> > > >> and a(X,Y) arrows directed from the object X to the object
> > > >> Y if there are a(X,Y) morphisms from X into Y.
> > > >>
> > > >> Given two morphism g:X-->Y, f:Y-->Z with composition
> > > >> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> > > >>
> > > >> The resulting quotient algebra has dimension n and a basis
> > > >> given by the (simple) arrows.
> > > >>
> > > >> This leads to the following "algorithm" for enumerating
> > > >> all categories with n morphisms and k objects:
> > > >>
> > > >> (a) enumerate all quivers (directed graphs) with k vertices and
> > > >> n oriented edges.
> > > >>
> > > >> (b) Associate the following "weight" to such a quiver as follows :
> > > >>
> > > >> (b1)given a triplet of vertices X,Y,Z,
> > > >> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> > > >> (this counts the number of ways which associate a morphism
> > > >> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> > > >> number of oriented arrows starting at U and ending at W.
> > > >>
> > > >> (b2) associate to a given quiver the weight
> > > >> \prod_{X,Y,Z} w(X,Y,Z)
> > > >> where the product is over all triplets of vertices.
> > > >>
> > > >> The total sum of such weighted quivers yields then the solution.
> > > >>
> > > >> Roland Bacher
> > > >>
> > > >>
> > > >>
> > > >>
> > > >>
> > > >>
> > > >> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.netwrote:
> > > >> > How many categories are there?
> > > >> >
> > > >> > First, how many categories are there with n morphisms and k
> > objects?
> > > >> > This table starts:
> > > >> >
> > > >> > 1
> > > >> > 2  1
> > > >> > 7  3 1
> > > >> > 35 16 3 1
> > > >> >
> > > >> > The first column is A058129, the number of monoids; the main
> > diagonal
> > > >> > is all 1's.  I am not
> > > >> > 100% certain of the 16 in the final row.
> > > >> >
> > > >> > Taking the row sums, we get:
> > > >> >
> > > >> > 1,3,11,55
> > > >> >
> > > >> > the number of categories with n morphisms.  This is probably not
> > in
> > > >> the
> > > >> > OEIS (only
> > > >> > A001776 is possible - other matches become less than
> > A058129).  The
> > > >> > inverse Euler
> > > >> > transform,
> > > >> >
> > > >> > 1,2,8,41
> > > >> >
> > > >> > is the number of connected categories with n morphisms; this is
> > > >> > likewise probably not
> > > >> > in the OEIS (only A052447 is possible).
> > > >> >
> > > >> > Can somebody generate more data?
> > > >> >
> > > >> > Franklin T. Adams-Watters
> > > >> >
> > > >> > A category is a collection of objects and morphisms; each
> > morphism is
> > > >> > from one object
> > > >> > to another (not necessarily different) object.  Where the
> > destination
> > > >> > of one morphism
> > > >> > is the source of a second, their composition is defined;
> > composition
> > > >> is
> > > >> > associative where
> > > >> > it is defined.  Each object has an identity morphism, which
> > connects
> > > >> it
> > > >> > to itself; this
> > > >> > is an identity when composed with morphisms coming in and with
> > > >> > morphisms going
> > > >> > out.
> > > >> >
> > > >>
> > ________________________________________________________________________
> > > >> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > > >> > industry-leading spam and email virus protection.
> > > >> >
> > > >>
> > > >
> > > >
> >
>
>
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From njas at research.att.com  Sat Nov 25 11:51:53 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sat, 25 Nov 2006 05:51:53 -0500 (EST)
Subject: request for new member
Message-ID: <200611251051.FAA87094@fry.research.att.com>


Olivier:

Richard Guy sent me this:

>From rkg at cpsc.ucalgary.ca  Fri Nov 24 17:07:00 2006
Delivered-To: njas at research.att.com
Date: Fri, 24 Nov 2006 15:02:50 -0700 (MST)
From: Richard Guy <rkg at cpsc.ucalgary.ca>
X-X-Sender: rkg at csl
To: "Neil J. A. Sloane" <njas at research.att.com>
cc: "Saff, Kevin" <kevin.saff at gmail.com>
Subject: New Subscriber
X-Virus-Scanned: by amavisd-new at cpsc.ucalgary.ca
X-Spam-Checker-Version: SpamAssassin 3.0.6 (2005-12-07) on 
        mail-purple.research.att.com
X-Spam-Level: 
X-Spam-Status: No, score=-2.9 required=4.0 tests=AWL,BAYES_00 autolearn=ham 
        version=3.0.6

Neil,
      May Kevin Saff join the seq-fan list?

      His email address is as above.

      Thanks & best wishes,    R.

Neil







From Annette.Warlich at t-online.de  Sat Nov 25 20:26:49 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Sat, 25 Nov 2006 20:26:49 +0100
Subject: Curious binomial-identity /A002720
Message-ID: <456898F9.8020608@t-online.de>


By chance I came across this curious identity
involving the Pascal-triangle.

Assume a row n, say n=4 and the column n, combined
each weighted with the running factorial as in the example:


           1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
 ratio =   -----------------------------------------  -------------------
           1/0! + 4/1! + 6/2! + 4/3! + 1/4!           weighted row-sum


then

  ratio = e  (=exp(1))

The actual sums are the entries of A002720
 http://www.research.att.com/~njas/sequences/A002720

Regards -

Gottfried Helms






From Annette.Warlich at t-online.de  Sat Nov 25 20:41:53 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Sat, 25 Nov 2006 20:41:53 +0100
Subject: Curious binomial-identity /A002720 (small correction of prev.
 post)
In-Reply-To: <456898F9.8020608@t-online.de>
References: <456898F9.8020608@t-online.de>
Message-ID: <45689C81.2020302@t-online.de>

Am 25.11.2006 20:26 schrieb Gottfried Helms:
> By chance I came across this curious identity
> involving the Pascal-triangle.
> 
> Assume a row n, say n=4 and the column n, combined
> each weighted with the running factorial as in the example:
> 
> 
>            1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
>  ratio =   -----------------------------------------  -------------------
>            1/0! + 4/1! + 6/2! + 4/3! + 1/4!           weighted row-sum
> 
> 
> then
> 
>   ratio = e  (=exp(1))
> 

------------------------------------

> The actual sums are the entries of A002720
>  http://www.research.att.com/~njas/sequences/A002720
> 
that should be corrected; the entries in A002720 are

 A(n)  =  weighted-rowsum(n) * n!
       =  weighted-colsum(n) * n! / exp(1)

I forgot to mention the additional n!, since in numerator and
denominator of the above fraction they cancel out, sorry.

Gottfried Helms






From jvospost3 at gmail.com  Sat Nov 25 21:10:27 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 25 Nov 2006 12:10:27 -0800
Subject: Curious binomial-identity /A002720 (small correction of prev. post)
In-Reply-To: <45689C81.2020302@t-online.de>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de>
Message-ID: <5542af940611251210g5e4916bcvee22981abd609ca1@mail.gmail.com>

There are several formulae for A002720, including a(n) = Sum k!C(n, k)^2,
k=0..n.

What you write might be true in the asymptotic limit, but not for any term,
as each term is rational and dividing by e would make each term
transcendental.  I'm sure that you meant the limit, right?

On 11/25/06, Gottfried Helms <Annette.Warlich at t-online.de> wrote:
>
> Am 25.11.2006 20:26 schrieb Gottfried Helms:
> > By chance I came across this curious identity
> > involving the Pascal-triangle.
> >
> > Assume a row n, say n=4 and the column n, combined
> > each weighted with the running factorial as in the example:
> >
> >
> >            1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
> >  ratio =
> -----------------------------------------  -------------------
> >            1/0! + 4/1! + 6/2! + 4/3! + 1/4!           weighted row-sum
> >
> >
> > then
> >
> >   ratio = e  (=exp(1))
> >
>
> ------------------------------------
>
> > The actual sums are the entries of A002720
> >  http://www.research.att.com/~njas/sequences/A002720
> >
> that should be corrected; the entries in A002720 are
>
> A(n)  =  weighted-rowsum(n) * n!
>        =  weighted-colsum(n) * n! / exp(1)
>
> I forgot to mention the additional n!, since in numerator and
> denominator of the above fraction they cancel out, sorry.
>
> Gottfried Helms
>
>
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From Annette.Warlich at t-online.de  Sat Nov 25 22:41:35 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Sat, 25 Nov 2006 22:41:35 +0100
Subject: Curious binomial-identity /A002720 (small correction of prev.
 post)
In-Reply-To: <5542af940611251210g5e4916bcvee22981abd609ca1@mail.gmail.com>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de> <5542af940611251210g5e4916bcvee22981abd609ca1@mail.gmail.com>
Message-ID: <4568B88F.1020800@t-online.de>

Am 25.11.2006 21:10 schrieb Jonathan Post:
> There are several formulae for A002720, including  a(n) = Sum k!C(n,
> k)^2, k=0..n.
> 
> What you write might be true in the asymptotic limit, but not for any
> term, as each term is rational and dividing by e would make each term
> transcendental.  I'm sure that you meant the limit, right?

Yes; as the +... in the numerator should indicate. The sum of columns
have infinitely many terms.

I've seen the generation-formula involving the exp()-function; maybe
this translates in an obvious way (... only with gf's I don't have
*any* experience... sigh...) and the asymptotic formula, which looks
a bit more complicated, but additionally relates it to a pi-expression.

But that we have

   1/3! +  3/2! + 3/1! + 1/0!  = A(3) /3!
                       + 3/1!
                       + 6/2!
                       +10/3!
                       + ...
                        -----
    limit k->oo         A(3) /3! * exp(1)

 or

   1/4! +  4/3! + 6/2! + 4/1! + 1/0!  = A(4)/4!
                              + 4/1!
                              +10/2!
                              +20/3!
                              + ...
                              -----
    limit k->oo               A(4)/4! *exp(1)

has something of a certain beauty... :-)


Gottfried Helms







From jrbibers at indiana.edu  Sat Nov 25 22:53:40 2006
From: jrbibers at indiana.edu (Joseph Biberstine)
Date: Sat, 25 Nov 2006 16:53:40 -0500
Subject: Curious binomial-identity /A002720 (small correction of prev.
 post)
In-Reply-To: <45689C81.2020302@t-online.de>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de>
Message-ID: <4568BB64.9050703@indiana.edu>

The example you gave has an off-by-one row index as follows:

Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n+1)

Note that the row index is incremented.  Mathematica simplifies this
fraction to LaguerreL[-(n+1), 1]/LaguerreL[n+1, -1].  This is not equal
to Exp[1], but interestingly seems to approach Exp[1] monotonically from
below as n approaches infinity.

I'm sure you meant:

Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n)

Mathematica simplifies this to LaguerreL[-(n+1), 1]/LaguerreL[n, -1].
Although I can't get it to explicitly resolve that this is exactly
Exp[1], the values do seem to bear this truth out.

So, I believe your example should read: e.g. for n=3,

    1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
e = -----------------------------------------  -------------------
           1/0! + 3/1! + 3/2! + 1/3!           weighted row-sum

This does seem to hold for every col/row pair in the triangle.

Nice find :)
-JRB

Gottfried Helms wrote:
> Am 25.11.2006 20:26 schrieb Gottfried Helms:
>> By chance I came across this curious identity
>> involving the Pascal-triangle.
>>
>> Assume a row n, say n=4 and the column n, combined
>> each weighted with the running factorial as in the example:
>>
>>
>>            1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
>>  ratio =   -----------------------------------------  -------------------
>>            1/0! + 4/1! + 6/2! + 4/3! + 1/4!           weighted row-sum
>>
>>
>> then
>>
>>   ratio = e  (=exp(1))
>>
> 
> ------------------------------------
> 
>> The actual sums are the entries of A002720
>>  http://www.research.att.com/~njas/sequences/A002720
>>
> that should be corrected; the entries in A002720 are
> 
>  A(n)  =  weighted-rowsum(n) * n!
>        =  weighted-colsum(n) * n! / exp(1)
> 
> I forgot to mention the additional n!, since in numerator and
> denominator of the above fraction they cancel out, sorry.
> 
> Gottfried Helms
> 





From Annette.Warlich at t-online.de  Sun Nov 26 00:43:25 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Sun, 26 Nov 2006 00:43:25 +0100
Subject: Curious binomial-identity /A002720 (small correction of prev.
 post)
In-Reply-To: <4568BB64.9050703@indiana.edu>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de> <4568BB64.9050703@indiana.edu>
Message-ID: <4568D51D.8000305@uni-kassel.de>

Am 25.11.2006 22:53 schrieb Joseph Biberstine:
> The example you gave has an off-by-one row index as follows:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n+1)
> 
> Note that the row index is incremented.  Mathematica simplifies this
> fraction to LaguerreL[-(n+1), 1]/LaguerreL[n+1, -1].  This is not equal
> to Exp[1], but interestingly seems to approach Exp[1] monotonically from
> below as n approaches infinity.
> 
> I'm sure you meant:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n)
> 
> Mathematica simplifies this to LaguerreL[-(n+1), 1]/LaguerreL[n, -1].
> Although I can't get it to explicitly resolve that this is exactly
> Exp[1], the values do seem to bear this truth out.
> 
> So, I believe your example should read: e.g. for n=3,
> 
>     1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
> e = -----------------------------------------  -------------------
>            1/0! + 3/1! + 3/2! + 1/3!           weighted row-sum
> 
> This does seem to hold for every col/row pair in the triangle.

Yes, your correction is right - I didn't look right at the
matrix.


The connection to the Laguerre-polynomials is not too
much surprising; I'm dealing with a ceratin classes
of lower-triangular matrices, which are derived from the
logic of the computing of the Pascal-matrix by matrix-
exponentiation.

While

  Pascalmatrix     = Pk(1)=  matrixexp( subdiagonal(1; [1,2,3,4,5,...]))

the Laguerre-matrix occurs from

  Laguerrematrix(*) = Pk(2) = matrixexp( subdiagonal(1; [1,2^2,3^2,4^2,5^2,...]))
           ((*) means: in a signed and simply scaled form)


I also have
                      Pk(0) = matrixexp( subdiagonal(1; [1,2^0,3^0,4^0,5^0,...]))

(where the vectors are filled in the first principal subdiagonal of an
infinite square-matrix)




Now the current matrix was created by a somehow routinely
"completion of my toolbox" of pascal-like-matrices, as a
hadamard-product of

      Pf =    Pk(1) (x) Pk(0)

         =    1/0!
              1/1!  1/0!
              1/2!  2/1!  1/0!
              1/3!  3/2!  3/1!  1/0!
              1/4!  4/3!  6/2!  4/1!   1/0!
                ....


Then with a powerseries-column-vector

      V(x) = [1,x,x^2,x^3,....] ~

and a diagonal factorial-matrix (used for scaling)

     dF(m) = diag(0! , 1!^m, 2!^m .... )

I got

   dF(1)  *   Pf  * V(1) = A002720       // A002720 read as column-vector
                                         // Due to the entry of Paul Berry

and
   dF(1) *    Pf~ * V(1) = A002720  *exp(1)  // added today



As a table it looks like:

                                                 Row-Sum
         =    1/0!                              =  1/0!    =  A(0)/0!
              1/1!  1/0!                        =  2/1!    =  A(1)/1!
              1/2!  2/1!  1/0!                  =  7/2!    =  A(2)/2!
              1/3!  3/2!  3/1!  1/0!            = 34/3!    =  A(3)/3!
              1/4!  4/3!  6/2!  4/1!   1/0!     =209/4!    =  A(4)/4!
                ....
             --------------------------------------------------------
Col-sum   e*[  1    2/1!  7/2! 34/3! 209/4! ]

Note, that using

   dF(1)  *   Pf  * V(2) =          ???
   dF(1) *    Pf~ * V(2) = exp(2) * ???

  in the second an exponentiation of 2 occurs, and generally for an s

   dF(1) *    Pf~ * V(s) = exp(s) * ???

 with unknown, but -heuristically- integer sequences of coefficients in ???
 for integer s>=0 .


------------------------------------------------------------------------


The Pascal- and (unsigned,scaled) Laguerre-matrix are also
related by

  Pk(0)                              // "base"-matrix
  Pk(1) =   dF(1)* Pk(0) * dF(-1)    // Pascal-matrix
  Pk(2) =   dF(1)* Pk(1) * dF(-1)    // unsigned,scaled Laguerrematrix
 ...
(which is due to the construction via matrix-exponentiation
of the related subdiagonal-matrices)
so finally the occurence of the relation to the Laguerre-matrix
using Mathematica is not too surprising...

Regards -

Gottfried Helms







From zakseidov at yahoo.com  Sun Nov 26 20:58:31 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 26 Nov 2006 11:58:31 -0800 (PST)
Subject: A124080 10 times triangular numbers
Message-ID: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>

Neil, Zerinvary, seqfans,
Why this sequence?!
Thanks, Zak 

%I A124080
%S A124080
0,10,30,60,100,150,210,280,360,450,550,660,780,910,1050,1200,1360,1530,
%T A124080
1710,1900,2100,2310,2530,2760,3000,3250,3510,3780,4060,4350,4650,4960,
%U A124080
5280,5610,5950,6300,6660,7030,7410,7800,8200,8610,9030,9460,9900,10350
%N A124080 10 times triangular numbers.
%F A124080 a(n)=10*C(n,2), n>=1
%p A124080 [seq(10*binomial(n,2) ,n=1..51)];
%Y A124080 Cf. A028895, A046092, A045943, A002378,
A028896, A024966, A033996, A027468.
%K A124080 easy,nonn,new
%O A124080 0,2
%A A124080 Zerinvary Lajos
(zerinvarylajos(AT)yahoo.com), Nov 24 2006



 
____________________________________________________________________________________
Do you Yahoo!?
Everyone is raving about the all-new Yahoo! Mail beta.
http://new.mail.yahoo.com





From jvospost3 at gmail.com  Sun Nov 26 22:23:25 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 26 Nov 2006 13:23:25 -0800
Subject: Categories
In-Reply-To: <20061124075307.GA17864@fourier.ujf-grenoble.fr>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
	 <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
	 <20061124075307.GA17864@fourier.ujf-grenoble.fr>
Message-ID: <5542af940611261323y13ada5f4p9e8f6e5d9051f93d@mail.gmail.com>

I can, with Edwin Clark's permission, forward to Frank my complete
correspondance with Edwin Clark about category enumeration via careful
definition.  I am asked not to cut & paste out of context.  But my emails to
Franklin T. Adams-Watters alone went astray, for whatever reason. so I am
not sure what to do.  I've also asked some Haskell programming experts about
Frank's enumeration question (properly citing Frank), as Haskell has monads
and functors and other categorical stuff built in.  I'll share any useful
answers that I get.

On 11/23/06, Roland Bacher <Roland.Bacher at ujf-grenoble.fr> wrote:
>
>
> I realised that the formula below for enumerating categories
> using quivers is completely wrong. It gives only an upper bound
> since one has to require moreover associativity for the composition of
> morphisms. The quiver approach works however in principle but
> is much more complex (and I guess this approach has been
> used in some of the previous approaches).   Roland Bacher
>
> On Thu, Nov 23, 2006 at 12:10:51PM -0800, Jonathan Post wrote:
> > *"Formalized Proof, Computation, and the Construction Problem in
> Algebraic
> > geometry", by Carlos Simpson.
> >
> > [PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
> > 2004<http://arxiv.org/pdf/math/0410224> File
> > Format: PDF/Adobe Acrobat - View as
> > HTML<
> http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5
> >
> > finite integer N, *how many categories* are there with N morphisms? What
> *
> > ...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
> > (1965), 450-464.
> >
> >
> >
> > On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> > >
> > >Roland Bacher correctly refers to quivers as summarized in wikipedia at
> > >
> > >http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
> > >
> > >On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> > >>
> > >>
> > >> This numbers can also be obtained as the numbers
> > >> of the following quotients of quivers:
> > >>
> > >> Associate to a category with n morphisms and k objects
> > >> the quiver with k vertices corresponding to the objects
> > >> and a(X,Y) arrows directed from the object X to the object
> > >> Y if there are a(X,Y) morphisms from X into Y.
> > >>
> > >> Given two morphism g:X-->Y, f:Y-->Z with composition
> > >> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> > >>
> > >> The resulting quotient algebra has dimension n and a basis
> > >> given by the (simple) arrows.
> > >>
> > >> This leads to the following "algorithm" for enumerating
> > >> all categories with n morphisms and k objects:
> > >>
> > >> (a) enumerate all quivers (directed graphs) with k vertices and
> > >> n oriented edges.
> > >>
> > >> (b) Associate the following "weight" to such a quiver as follows :
> > >>
> > >> (b1)given a triplet of vertices X,Y,Z,
> > >> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> > >> (this counts the number of ways which associate a morphism
> > >> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> > >> number of oriented arrows starting at U and ending at W.
> > >>
> > >> (b2) associate to a given quiver the weight
> > >> \prod_{X,Y,Z} w(X,Y,Z)
> > >> where the product is over all triplets of vertices.
> > >>
> > >> The total sum of such weighted quivers yields then the solution.
> > >>
> > >> Roland Bacher
> > >>
> > >>
> > >>
> > >>
> > >>
> > >>
> > >> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.netwrote:
> > >> > How many categories are there?
> > >> >
> > >> > First, how many categories are there with n morphisms and k
> objects?
> > >> > This table starts:
> > >> >
> > >> > 1
> > >> > 2  1
> > >> > 7  3 1
> > >> > 35 16 3 1
> > >> >
> > >> > The first column is A058129, the number of monoids; the main
> diagonal
> > >> > is all 1's.  I am not
> > >> > 100% certain of the 16 in the final row.
> > >> >
> > >> > Taking the row sums, we get:
> > >> >
> > >> > 1,3,11,55
> > >> >
> > >> > the number of categories with n morphisms.  This is probably not in
> > >> the
> > >> > OEIS (only
> > >> > A001776 is possible - other matches become less than A058129).  The
> > >> > inverse Euler
> > >> > transform,
> > >> >
> > >> > 1,2,8,41
> > >> >
> > >> > is the number of connected categories with n morphisms; this is
> > >> > likewise probably not
> > >> > in the OEIS (only A052447 is possible).
> > >> >
> > >> > Can somebody generate more data?
> > >> >
> > >> > Franklin T. Adams-Watters
> > >> >
> > >> > A category is a collection of objects and morphisms; each morphism
> is
> > >> > from one object
> > >> > to another (not necessarily different) object.  Where the
> destination
> > >> > of one morphism
> > >> > is the source of a second, their composition is defined;
> composition
> > >> is
> > >> > associative where
> > >> > it is defined.  Each object has an identity morphism, which
> connects
> > >> it
> > >> > to itself; this
> > >> > is an identity when composed with morphisms coming in and with
> > >> > morphisms going
> > >> > out.
> > >> >
> > >>
> ________________________________________________________________________
> > >> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > >> > industry-leading spam and email virus protection.
> > >> >
> > >>
> > >
> > >
>
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From jvospost3 at gmail.com  Sun Nov 26 22:27:11 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 26 Nov 2006 13:27:11 -0800
Subject: A124080 10 times triangular numbers
In-Reply-To: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
References: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
Message-ID: <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>

I wondered too.  It intersects the much more interesting
 A000537 <http://www.research.att.com/%7Enjas/sequences/A000537>
 Sum of first n cubes; or n-th triangular number squared.  at 100.

Not that all of my roughly 1 kilophi sequences and comments are interesting,
either.

-- Jonathan

On 11/26/06, zak seidov <zakseidov at yahoo.com> wrote:
>
> Neil, Zerinvary, seqfans,
> Why this sequence?!
> Thanks, Zak
>
> %I A124080
> %S A124080
> 0,10,30,60,100,150,210,280,360,450,550,660,780,910,1050,1200,1360,1530,
> %T A124080
> 1710,1900,2100,2310,2530,2760,3000,3250,3510,3780,4060,4350,4650,4960,
> %U A124080
> 5280,5610,5950,6300,6660,7030,7410,7800,8200,8610,9030,9460,9900,10350
> %N A124080 10 times triangular numbers.
> %F A124080 a(n)=10*C(n,2), n>=1
> %p A124080 [seq(10*binomial(n,2) ,n=1..51)];
> %Y A124080 Cf. A028895, A046092, A045943, A002378,
> A028896, A024966, A033996, A027468.
> %K A124080 easy,nonn,new
> %O A124080 0,2
> %A A124080 Zerinvary Lajos
> (zerinvarylajos(AT)yahoo.com), Nov 24 2006
>
>
>
>
>
> ____________________________________________________________________________________
> Do you Yahoo!?
> Everyone is raving about the all-new Yahoo! Mail beta.
> http://new.mail.yahoo.com
>
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From jvospost3 at gmail.com  Sun Nov 26 23:43:48 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 26 Nov 2006 14:43:48 -0800
Subject: A124080 10 times triangular numbers
In-Reply-To: <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>
References: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
	 <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>
Message-ID: <5542af940611261443g32cd3bf3lf2174cfc2d42e716@mail.gmail.com>

On the other hand, every sequence is related to some interesting sequence...
I just submitted by form the following:

 NEW SEQUENCE FROM Jonathan Vos Post


%I A000001
%S A000001 11, 29, 31, 59, 61, 101, 149, 151, 211, 281, 359, 449, 659,
661, 911, 1049, 1051, 1201, 1361, 1531, 1709, 1901, 2099, 2309, 2311,
2531, 2999, 3001
%N A000001 Primes of the form A124080 (10 times triangular numbers), +1
or -1.
%C A000001 The j, such that A124080(j)-1 is prime or A124080(j)+1 is
prime, where repetition means a twin prime, are 1, 2, 2, 3, 3, 4, 5, 5,
6, 7, 8, 9, 11, 11, 13, 14, 14, 15, 16, 17, 18, 19, 20, 21, 22, 24, 24.
%F A000001 {A124080(j)-1 when prime} U {A124080(j)+1 when prime} =
{i = 10*T(j)-1 such that i is prime} U {i = 10*T(j)+1 such that i is
prime} where T(j) = A000217(j) = j*(j+1)/2.
%e A000001 a(1) = A124080(1)+1 = (10*T(1)) - 1 = 10*(1*(1+1)/2) + 1 =
10+1 = 11 is prime.
a(2) = A124080(2)-1 = (10*T(2))-1 = 10*(2*(2+1)/2) - 1 = 30-1 = 29 is
prime.
a(3) = A124080(2)+1 = (10*T(2))+1 = 10*(2*(2+1)/2) + 1 = 30+1 = 31 is
prime.
%Y A000001 Cf. A000040, A000217, A028895, A046092, A045943, A002378,
A028896, A024966, A033996, A027468.
%O A000001 1,1
%K A000001 ,easy,nonn,
%A A000001 Jonathan Vos Post (jvospost2 at yahoo.com
<http://us.f551.mail.yahoo.com/ym/Compose?To=jvospost2 at yahoo.com&YY=86457&y5beta=yes&y5beta=yes&order=down&sort=date&pos=0&view=a&head=b>),
Nov 26 2006
RH
RA 192.20.225.32
RU
RI



On 11/26/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> I wondered too.  It intersects the much more interesting
>  A000537 <http://www.research.att.com/%7Enjas/sequences/A000537>
>  Sum of first n cubes; or n-th triangular number squared.  at 100.
>
> Not that all of my roughly 1 kilophi sequences and comments are
> interesting, either.
>
> -- Jonathan
>
> On 11/26/06, zak seidov <zakseidov at yahoo.com> wrote:
> >
> > Neil, Zerinvary, seqfans,
> > Why this sequence?!
> > Thanks, Zak
> >
> > %I A124080
> > %S A124080
> > 0,10,30,60,100,150,210,280,360,450,550,660,780,910,1050,1200,1360,1530,
> > %T A124080
> > 1710,1900,2100,2310,2530,2760,3000,3250,3510,3780,4060,4350,4650,4960,
> > %U A124080
> > 5280,5610,5950,6300,6660,7030,7410,7800,8200,8610,9030,9460,9900,10350
> > %N A124080 10 times triangular numbers.
> > %F A124080 a(n)=10*C(n,2), n>=1
> > %p A124080 [seq(10*binomial(n,2) ,n=1..51)];
> > %Y A124080 Cf. A028895, A046092, A045943, A002378,
> > A028896, A024966, A033996, A027468.
> > %K A124080 easy,nonn,new
> > %O A124080 0,2
> > %A A124080 Zerinvary Lajos
> > (zerinvarylajos(AT)yahoo.com), Nov 24 2006
> >
> >
> >
> >
> >
> > ____________________________________________________________________________________
> > Do you Yahoo!?
> > Everyone is raving about the all-new Yahoo! Mail beta.
> > http://new.mail.yahoo.com
> >
>
>
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From njas at research.att.com  Mon Nov 27 03:23:49 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sun, 26 Nov 2006 21:23:49 -0500 (EST)
Subject: A124080 10 times triangular numbers
References: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
    <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>
Message-ID: <200611270223.VAA41276@fry.research.att.com>

JVP, you submitted this sequence:
%N A000001 Primes of the form A124080 (10 times triangular numbers), +1
or -1.
I will use it, but I do not find it interesting.  Why?  Because it 
is derivative.  It is based on a sequence of little interest - acceptable
only because it was sent in by someone who perhaps does not know
what is worth being included in the database - and is a step further 
down the ladder towards the basement.

I seem to have to play the role of the singing teacher
at times:  so be it - this song was off-key!

Neil





From jvospost3 at gmail.com  Mon Nov 27 04:10:47 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 26 Nov 2006 19:10:47 -0800
Subject: A124080 10 times triangular numbers
In-Reply-To: <200611270223.VAA41276@fry.research.att.com>
References: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
	 <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>
	 <200611270223.VAA41276@fry.research.att.com>
Message-ID: <5542af940611261910x10195634teeac87bdf4bfd1e7@mail.gmail.com>

As usual, you're right, Neil.

I was trying to be nice to the newcomer, by showing him the collaborative
nature of the endeavor. I'm sure that his enthusiasm can be made fruitful.
I've certainly erred in judgment, many times, and you and your associate
editors have been splendid in guiding me towards better work.

Sometimes, as they sing: "you've got to be cruel to be kind."

Hence the "probation" keyword and, one of my useful suggestions to you and
your board, the "less" keyword.

Sorry for where that road leads, which I help pave with good intentions.

-- Jonathan

On 11/26/06, N. J. A. Sloane <njas at research.att.com> wrote:
>
> JVP, you submitted this sequence:
> %N A000001 Primes of the form A124080 (10 times triangular numbers), +1
> or -1.
> I will use it, but I do not find it interesting.  Why?  Because it
> is derivative.  It is based on a sequence of little interest - acceptable
> only because it was sent in by someone who perhaps does not know
> what is worth being included in the database - and is a step further
> down the ladder towards the basement.
>
> I seem to have to play the role of the singing teacher
> at times:  so be it - this song was off-key!
>
> Neil
>
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From pauldhanna at juno.com  Mon Nov 27 04:58:25 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Sun, 26 Nov 2006 22:58:25 -0500
Subject: Missing Table?
Message-ID: <20061126.225826.332.1.pauldhanna@juno.com>

Seqfans,
     Where is the rectangular table referred to by these sequences?
 
A040027  Main diagonal of an array of binomial recurrence coefficients.  
 0, 1, 1, 3, 9, 31, 121, 523, 2469, 12611, 69161, 404663,  
A045501  Second diagonal in table of binomial recurrence coefficients.  
 1, 1, 4, 14, 54, 233, 1101, 5625, 30846, 
A045499  Third diagonal in table of binomial recurrence coefficients. 
 1, 1, 5, 20, 85, 400, 2046, 11226, 65676, 407787, 2675410, 
A045500  Fourth diagonal in table of binomial recurrence coefficients. 
 1, 1, 6, 27, 125, 635, 3488, 20425, 126817, 831915, 5744784, 41618459 
(AUTHOR of these sequences:  H. W. Gould) 
 
The closest thing I can find is Emeric Deutsch's recent triangle A124496,

in which these sequences appear as columns. 
 
The lower half of the table (I guess) must look like this:
1, ...
1, 1, ...
1, 1, 3, ...
1, 1, 4, 9, ...
1, 1, 5, 14, 31, ...
1, 1, 6, 20, 54, 121, ...
1, 1, 7, 27, 85, 233, 523, ...
1, 1, 8, 35, 125, 400, 1101, 2469, ...
1, 1, 9, 44, 175, 635, 2046, 5625, 12611, ...
 
Is the complete table in the OEIS?  
If not, it certainly should be. 
     Paul
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From mathar at strw.leidenuniv.nl  Mon Nov 27 10:47:35 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Mon, 27 Nov 2006 10:47:35 +0100
Subject: Missing Table?
Message-ID: <200611270947.kAR9lZBJ001619@amer.strw.leidenuniv.nl>


The master table of A040027 could also be A046936..but this does not fit the
2nd, 3rd and 4th diagonal as well as A124496 does.

RJM

> From seqfan-owner at ext.jussieu.fr  Mon Nov 27 05:03:23 2006
> Return-Path: <seqfan-owner at ext.jussieu.fr>
> To: Seqfan at ext.jussieu.fr
> Cc: gould at math.wvu.edu
> Date: Sun, 26 Nov 2006 22:58:25 -0500
> Subject: Missing Table?
> From: "Paul D. Hanna" <pauldhanna at juno.com>
> 
> Seqfans,
>      Where is the rectangular table referred to by these sequences?
>  
> A040027  Main diagonal of an array of binomial recurrence coefficients.  
>  0, 1, 1, 3, 9, 31, 121, 523, 2469, 12611, 69161, 404663,  
> A045501  Second diagonal in table of binomial recurrence coefficients.  
>  1, 1, 4, 14, 54, 233, 1101, 5625, 30846, 
> A045499  Third diagonal in table of binomial recurrence coefficients. 
>  1, 1, 5, 20, 85, 400, 2046, 11226, 65676, 407787, 2675410, 
> A045500  Fourth diagonal in table of binomial recurrence coefficients. 
>  1, 1, 6, 27, 125, 635, 3488, 20425, 126817, 831915, 5744784, 41618459 
> (AUTHOR of these sequences:  H. W. Gould) 
>  
> The closest thing I can find is Emeric Deutsch's recent triangle A124496,
> 
> in which these sequences appear as columns. 
>  
> The lower half of the table (I guess) must look like this:
> 1, ...
> 1, 1, ...
> 1, 1, 3, ...
> 1, 1, 4, 9, ...
> 1, 1, 5, 14, 31, ...
> 1, 1, 6, 20, 54, 121, ...
> 1, 1, 7, 27, 85, 233, 523, ...
> 1, 1, 8, 35, 125, 400, 1101, 2469, ...
> 1, 1, 9, 44, 175, 635, 2046, 5625, 12611, ...
>  
> Is the complete table in the OEIS?  
> If not, it certainly should be. 
>      Paul





From tanyakh at TanyaKhovanova.com  Mon Nov 27 17:05:13 2006
From: tanyakh at TanyaKhovanova.com (Tanya Khovanova)
Date: Mon, 27 Nov 2006 08:05:13 -0800
Subject: practical numbers
Message-ID: <200611270805.AA74449328@TanyaKhovanova.com>


Hello seqfans,

There is some confusion with practical numbers. Wikipedia article on practical numbers: http://en.wikipedia.org/wiki/Practical_number
gives the following definition of a practical number:
A practical number or panarithmic number is a positive integer n such that all preceding positive integers are a sum of distinct divisors of n.

At the same time it refers to the sequence A005153 which gives a slightly different definition:
Practical numbers (first definition): all k <= sigma(n) are sums of distinct divisors of n. Also called panarithmic numbers.

Browsing other websites (including my own :-) ) I've found that the wiki definition is used more often. Moreover, my calculation with the my definition exactly matches the sequence A005153, though they are supposed to be different.

Please, clarify my confusion.

Tanya Khovanova






From mathar at strw.leidenuniv.nl  Mon Nov 27 17:54:41 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Mon, 27 Nov 2006 17:54:41 +0100
Subject: practical numbers
Message-ID: <200611271654.kARGsfOX022242@amer.strw.leidenuniv.nl>

> From seqfan-owner at ext.jussieu.fr  Mon Nov 27 17:07:28 2006
> Date: Mon, 27 Nov 2006 08:05:13 -0800
> From: "Tanya Khovanova" <tanyakh at tanyakhovanova.com>
> Reply-To: <tanyakh at tanyakhovanova.com>
> To: <seqfan at ext.jussieu.fr>
> Subject: practical numbers
> ...
> There is some confusion with practical numbers. Wikipedia article on practical numbers: http://en.wikipedia.org/wiki/Practical_numbe
> r
> gives the following definition of a practical number:
> A practical number or panarithmic number is a positive integer n such that all preceding positive integers are a sum of distinct div
> isors of n.
> 
> At the same time it refers to the sequence A005153 which gives a slightly different definition:
> Practical numbers (first definition): all k <= sigma(n) are sums of distinct divisors of n. Also called panarithmic numbers.
> 
> Browsing other websites (including my own :-) ) I've found that the wiki definition is used more often. Moreover, my calculation wit
> h the my definition exactly matches the sequence A005153, though they are supposed to be different.
> 
> Please, clarify my confusion.
> ...

The A005153 definition [with reference to the sigma(n)] is also used
in
http://www.dm.unipi.it/gauss-pages/melfi/public_html/articoli/jnt.ps
but the same author uses in his web page
http://members.unine.ch/giuseppe.melfi/pratica.html
just the other definition.
see also http://citeseer.ist.psu.edu/285.html
http://arxiv.org/abs/math.NT/0404555

B. M Stewart in Am J. Math 76 (4) (1954) p 779, available from
http://www.jstor.org, quotes Srinivasan [Current Science (1948), pp 179] with
the definition without the sigma.

E J Scourfield in J. Number Theory 62 (1) (1997) p 163 uses the definition
including the sigma:
http://www.emis.de/projects/EULER/detail?ide=1997saiaentidividens&matchno=7&matchtotal=18&q=cr%3AE*+Saias+

I have to stop here. The recent IAU resolution on what a useful definition of a
"planet" might be, in particular the status of Pluto, be has already caused
enough confusion, and we're still recovering from this.

Richard





From franktaw at netscape.net  Mon Nov 27 18:04:03 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 27 Nov 2006 12:04:03 -0500
Subject: practical numbers
In-Reply-To: <200611270805.AA74449328@TanyaKhovanova.com>
References: <200611270805.AA74449328@TanyaKhovanova.com>
Message-ID: <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com>

The definitions are equivalent.  The condition cited in the Wikipedia 
article, to make everything up to n a sum of the divisors of n, is 
sufficient to make everything up to sigma(n) be such a sum.

Franklin T. Adams-Watters


-----Original Message-----
From: tanyakh at TanyaKhovanova.com

Hello seqfans,

There is some confusion with practical numbers. Wikipedia article on 
practical
numbers: http://en.wikipedia.org/wiki/Practical_number
gives the following definition of a practical number:
A practical number or panarithmic number is a positive integer n such 
that all
preceding positive integers are a sum of distinct divisors of n.

At the same time it refers to the sequence A005153 which gives a 
slightly
different definition:
Practical numbers (first definition): all k <= sigma(n) are sums of 
distinct
divisors of n. Also called panarithmic numbers.

Browsing other websites (including my own :-) ) I've found that the 
wiki
definition is used more often. Moreover, my calculation with the my 
definition
exactly matches the sequence A005153, though they are supposed to be 
different.

Please, clarify my confusion.

Tanya Khovanova


________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From davidwwilson at comcast.net  Mon Nov 27 18:30:31 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Mon, 27 Nov 2006 12:30:31 -0500
Subject: practical numbers
References: <200611270805.AA74449328@TanyaKhovanova.com> <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com>
Message-ID: <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p>

Assuming Franklin's observation to be correct, why not

%C A005153 Also, n such that all k <= n are sums of distinct divisors of n.

Some poorly-thought-out questions before I head off to work:

Are there not-quite-practical numbers, that is, numbers such that n-1 is the 
only number < n which is not the sum of distinct divisors of n?

If not, is there some interesting bound f(n) < n such that if all k <= f(n) 
are sums of distinct divisors of n, then n is also a sum of distinct 
divisors of n?

----- Original Message ----- 
From: <franktaw at netscape.net>
To: <tanyakh at TanyaKhovanova.com>; <seqfan at ext.jussieu.fr>
Sent: Monday, November 27, 2006 12:04 PM
Subject: Re: practical numbers


> The definitions are equivalent.  The condition cited in the Wikipedia 
> article, to make everything up to n a sum of the divisors of n, is 
> sufficient to make everything up to sigma(n) be such a sum.
>
> Franklin T. Adams-Watters






From davidwwilson at comcast.net  Mon Nov 27 18:42:56 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Mon, 27 Nov 2006 12:42:56 -0500
Subject: practical numbers
References: <200611270805.AA74449328@TanyaKhovanova.com> <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com> <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p>
Message-ID: <000601c7124b$7c8b0040$6501a8c0@yourxhtr8hvc4p>

Simce any number n is a repeated sum of any of its divisors, "sum of 
distinct divisors" seems like overkill.

Anyway, another question: What is the smallest number a(n) that is a sum of 
its distinct divisors in n ways? 






From franktaw at netscape.net  Mon Nov 27 20:12:37 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 27 Nov 2006 14:12:37 -0500
Subject: practical numbers
In-Reply-To: <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p>
References: <200611270805.AA74449328@TanyaKhovanova.com> <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com> <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p>
Message-ID: <8C8E06F230775CD-7BC-A6B4@FWM-M18.sysops.aol.com>

To answer the second question first, it suffices to show that every 
number up to and including ceiling(n/2) is a sum of the divisors of n.  
Except for n=3, that can be floor(n/2) instead.  (If n is odd, 2 is not 
a sum; if n is even, every m with n/2 <= m < n is a sum n/2 plus a 
number less than n/2.)  This answers the first question, essentially in 
the negative: 3 is the only such number.

More interesting, perhaps: what about numbers such that every number up 
to n, with only 1 exception, are the sum of divisors of n?  (Note that 
this has to be n in this case, not sigma(n), since the possible sums 
are symmetric with respect to sigma(n) - if a set S of divisors sums to 
m, then the complement of S, with respect to the set of divisors of n, 
sums to sigma(n) - m.)

70 is one such number, skipping only 4; I think 70p for any prime 5 <= 
p < 70 does the same.  945 is another, skipping 2.  (945 is the 
smallest odd abundant number.)  I think that these values (70, 350, 
490, 770, 910, 945) are the only ones up to 1000.

The number skipped is always even; in fact, it always one less than a 
prime divisor p of n (and 1 more than the sum of the divisors of a 
practical divisor of n - which implies that that practical divisor is a 
square or twice a square).  Except for n=3, there must be another prime 
divisor q of n with p < q < 2p.

Franklin T. Adams-Watters


-----Original Message-----
From: davidwwilson at comcast.net

Assuming Franklin's observation to be correct, why not?
?
%C A005153 Also, n such that all k <= n are sums of distinct divisors 
of n.?
?
Some poorly-thought-out questions before I head off to work:?
?
Are there not-quite-practical numbers, that is, numbers such that n-1 
is the only number < n which is not the sum of distinct divisors of n??
?
If not, is there some interesting bound f(n) < n such that if all k <= 
f(n) are sums of distinct divisors of n, then n is also a sum of 
distinct divisors of n??
?
----- Original Message ----- From: <franktaw at netscape.net>?
To: <tanyakh at TanyaKhovanova.com>; <seqfan at ext.jussieu.fr>?
Sent: Monday, November 27, 2006 12:04 PM?
Subject: Re: practical numbers?
?
> The definitions are equivalent. The condition cited in the Wikipedia 
 > article, to make everything up to n a sum of the divisors of n, is > 
sufficient to make everything up to sigma(n) be such a sum.?
>?
> Franklin T. Adams-Watters?
?


________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From franktaw at netscape.net  Mon Nov 27 20:25:04 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 27 Nov 2006 14:25:04 -0500
Subject: practical numbers
In-Reply-To: <000601c7124b$7c8b0040$6501a8c0@yourxhtr8hvc4p>
References: <200611270805.AA74449328@TanyaKhovanova.com> <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com> <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p> <000601c7124b$7c8b0040$6501a8c0@yourxhtr8hvc4p>
Message-ID: <8C8E070DFC523C7-7BC-A7AB@FWM-M18.sysops.aol.com>

I don't think "sum of distinct divisors" is really overkill.  If you 
leave out the "distinct", you'll get people complaining "but every 
number is a sum of its divisors" - or worse, misunderstanding.

A096356.  (Which should really be edited to say "proper divisors" 
instead of just "divisors".  Adding an initial 1 at offset 0 would also 
make sense to me.)

Franklin T. Adams-Watters


-----Original Message-----
From: davidwwilson at comcast.net

Simce any number n is a repeated sum of any of its divisors, "sum of 
distinct divisors" seems like overkill.?
?
Anyway, another question: What is the smallest number a(n) that is a 
sum of its distinct divisors in n ways? ?


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From mathar at strw.leidenuniv.nl  Mon Nov 27 21:46:49 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Mon, 27 Nov 2006 21:46:49 +0100
Subject: Re^2: practical numbers
Message-ID: <200611272046.kARKknOV015522@amer.strw.leidenuniv.nl>


> From seqfan-owner at ext.jussieu.fr  Mon Nov 27 18:44:49 2006
> From: "David Wilson" <davidwwilson at comcast.net>
> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
> Subject: Re: practical numbers
> ...
> Anyway, another question: What is the smallest number a(n) that is a sum of 
> its distinct divisors in n ways? 

For each n we get the "trivial" solution where the sum consists of n itself
as the only term. Further solutions, where the sum contains more than one term
(all of which are distinct divisors), are listed below in the format
n [list of terms that sum to n]:

6 [1, 2, 3]
12 [1, 2, 3, 6]
12 [2, 4, 6]
18 [1, 2, 6, 9]
18 [3, 6, 9]
20 [1, 4, 5, 10]
24 [1, 2, 3, 4, 6, 8]
24 [1, 2, 3, 6, 12]
24 [2, 4, 6, 12]
24 [1, 3, 8, 12]
24 [4, 8, 12]
28 [1, 2, 4, 7, 14]
30 [1, 3, 5, 6, 15]
30 [2, 3, 10, 15]
30 [5, 10, 15]
36 [2, 3, 4, 6, 9, 12]
36 [2, 3, 4, 9, 18]
36 [1, 2, 6, 9, 18]
36 [3, 6, 9, 18]
36 [1, 2, 3, 12, 18]
36 [2, 4, 12, 18]
36 [6, 12, 18]
40 [1, 2, 4, 5, 8, 20]
40 [1, 4, 5, 10, 20]
40 [2, 8, 10, 20]
42 [1, 6, 14, 21]
42 [7, 14, 21]


n=6 has 1 extra solution, n=12 or 18 have 2 extra solutions, n=20 has one,
n=24 has five etc.

Maple program to create that one (not checked!):

a := proc(n)
	local findx,d,f,found,i,count ;
	count :=0 :
	# decompose n into all sums that are candidates for the decomposition
	d := combinat[partition](n) ;
	# take each sum indivdually
	for findx from 1 to nops(d) do
		f := op(findx,d) ;
		# start with the assumption that it works
		found := true ;
		for i from 2 to nops(f) do
			# if two terms are equal in the Maple list, 
			# we flag this case as "not admitted"
			if op(i,f) = op(i-1,f) then
				found := false ;
				break ;
			fi ;
		od ;
		for i from 1 to nops(f) do
			# if any term is not a divisor,
			# we flag this case as "not admitted"
			if n mod op(i,f) <> 0 then
				found := false ;
				break ;
			fi ;
		od ;
		# If this sum seems to work and has more than one term, we list it
		if found and nops(f) > 1 then
			printf("%a %a\n",n,f) ;
			count := count+1 ;
		fi ;
	od ;
	RETURN(count) ;
end:

# loop over the most fundamental cases..
for n from 1 to 47 do
	a(n) ;
od ;

# Richard Mathar





From zbi74583 at boat.zero.ad.jp  Tue Nov 28 07:18:47 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Tue, 28 Nov 2006 15:18:47 +0900
Subject: A038260
Message-ID: <025d01c712b5$16134290$977fd7dc@FMC57937B6BE00>

    Neil wrote on comment line of A038260.

    >Surely there are smaller members! njas, Nov 07 2006



    Why do you think that?

    I searched the ?seeds? for ?regular? Rational Amicable Number again.

       .

    And {11*19, 239 } is the smallest one.           .

    So, I think that at least it is the first terms of regular Rational 
Amicable Pair.

            .

    Such a Diophantine equation which is high degree and is involved with 
Sigma function has many conditions  , so the solution must have many prime 
factors for having good property, hence it becomes to have many digits.



    I guess that 2^2*7*3*5^2*13*31*139*277*3877*11*19 is the smallest and if 
an irregular example exists then it has around 16 digits.



    Yasutoshi


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From Annette.Warlich at t-online.de  Wed Nov 29 08:38:17 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Wed, 29 Nov 2006 08:38:17 +0100
Subject: Curious binomial-identity /A002720
In-Reply-To: <4568BB64.9050703@indiana.edu>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de> <4568BB64.9050703@indiana.edu>
Message-ID: <456D38E9.8030900@t-online.de>

Am 25.11.2006 22:53 schrieb Joseph Biberstine:
> The example you gave has an off-by-one row index as follows:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n+1)
> 
> Note that the row index is incremented.  Mathematica simplifies this
> fraction to LaguerreL[-(n+1), 1]/LaguerreL[n+1, -1].  This is not equal
> to Exp[1], but interestingly seems to approach Exp[1] monotonically from
> below as n approaches infinity.
> 
> I'm sure you meant:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n)
> 
> Mathematica simplifies this to LaguerreL[-(n+1), 1]/LaguerreL[n, -1].
> Although I can't get it to explicitly resolve that this is exactly
> Exp[1], the values do seem to bear this truth out.


I think I got it. As usual I like to write that problem in
matrix-mode.
So with

     V(x)~ * Pf = B

defined as the (infinite) matrix product as set of formal powerseries
in x:

                     1/0!   .     .     .
                     1/1!  1/0!
                     1/2!  2/1!  1/0!
                     1/3!  3/2!  3/1!  1/0!
                 .....
                   -------------------------------
[1 x x^2 x^3...]  = [ b0   b1    b2     b3 .... ]

we have, using the apostroph ' for derivative wrt x:

   b0  =            exp(x)      /0!
   b1  =  x*  [ x  *exp(x) ]'   /1!
   b2 =   x^2*[ x^2*exp(x) ]''  /2!
   b3 =   x^3*[ x^3*exp(x) ]''' /3!
 ...
   bk =   x^k*[ x^k*exp(x) ]'(k)' /k!

Expanding the derivatives gives

(1):

   b0 *0! / exp(x) =   1
   b1 *1! / exp(x) =   1 + x

   b2 *2! / exp(x) =   (2 + 4x + x^2)
                   =   2!/0! + 2*2!/1!  + 1*2!/2! x^2

   b3 *3! / exp(x) =   (6 + 18x + 9x^2 + x^3)
                        3! + 3*3!/1! x + 3*3!/2! x^2 + 1 *3!/3! x^3

   b4 *4! / exp(x) =   (24   + 96x  + 72x^2   + 16x^3 + 1x^4)
                       (1*4! + 4*4!/1! x + 6*4!/2!x^2 + 4*4!/3!x^3 + 1*4!/4!*x^4)



and the rhs written as matrix, where also the k! of the k'th row is cancelled , gives

(2):
   b0 / exp(x)      =   1                              *  [1,  x,  x^2,  x^3, ....]~
   b1 / exp(x)      =   1/0!  1/1!
   b2 / exp(x)      =   1/0!  2/1!  1/2!
   b3 / exp(x)      =   1/0!  3/1!  3/2!   1/3!
     ...            =     ....

If we now consider the result when letting x->1 we can further use
the horizontal symmetry of the binomial-coefficients to reorder
the rows and just get the matrix Pf again as coefficients

(3):

 [b0 b1 b2 b3...]~ /exp(1) =    1/0!                            * [1, 1, 1, 1, 1, ...]~
                                1/1!   1/0!
                                1/2!   2/1!   1/0!
                                1/3!   3/2!   3/1!   1/0!


 so

   V(1)~ * Pf = B~
 or
           Pf~ * V(1) = B
 and
           Pf  * V(1) = B / exp(x)

The systematic of the derivation of matrix Pf as resultant
of the derivative-proceeding in (1) and (2) needs now a better
systematic proof, but I think, that could be done...


--------------------------------------------------

One could proceed one further step and state

 for a current row and column m of Pf we have

(4):
    V(x)~ * Pf[0..inf,m] = exp(x) * Pf[m,0..m] * x^(2m) * V(1/x)

or (Pari:)

   suminf( k=0, x^k*Pf[k+1,m+1]) = exp(x) * x^(2*m) *  sum(k=0,m,Pf[m+1,k+1]/x^k)

where for a row r and col c the entry Pf[r+1,c+1] may be replaced by

      Pf[r+1,c+1] = ch(r,c) / (r-c)!          // for r>=c
                  = 0                         // for r<c

where ch(a,b) is the binomial-coefficient.

----------------

The common notation for the above observation may be given as:

    oo            x^k               m    m              x^(m-k)
   sum   ch(k,m)*------   = exp(x) x  * sum   ch(m,k)* ---------
    k=m          (k-m)!                 k=0              (m-k)!

                where ch(a,b) is the binomial-coefficient.


Gottfried Helms






From bowerc at usa.net  Wed Nov 29 10:49:58 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 29 Nov 2006 01:49:58 -0800
Subject: Categories
Message-ID: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>


I have one more row to add

------ Original Message ------
From: franktaw at netscape.net
To: seqfan at ext.jussieu.fr
Subject: Categories

> How many categories are there?
> 
> First, how many categories are there with n morphisms and k objects?  
> This table starts:
> 
>  1
>  2  1
>  7  3 1
> 35 16 3 1
> 
228 77 20 3 1

> The first column is A058129, the number of monoids; the main diagonal 
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
> 
> Taking the row sums, we get:
> 
> 1,3,11,55
329
> 
> the number of categories with n morphisms.  This is probably not in the 
> OEIS (only
> A001776 is possible - other matches become less than A058129).

While the 329 is eerily close to A001776's 330, EULER(A058129) is a
lower limit for this sequence and we have 2982 vs 2345 for #6 there.
>  The 
> inverse Euler
> transform,
> 
> 1,2,8,41
258
> 
> is the number of connected categories with n morphisms; this is 
> likewise probably not
> in the OEIS (only A052447 is possible).
No longer possible
> 
> Can somebody generate more data?
> 
> Franklin T. Adams-Watters
> 

Christian

PS

I don't know if it will be feasible to collect enough data to do
this, but the columns of the triangle converge and that convergence
would make an interesting sequence in and of itself.








From zakseidov at yahoo.com  Wed Nov 29 13:05:28 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 29 Nov 2006 04:05:28 -0800 (PST)
Subject: Difference between neighbor square and cube
Message-ID: <566064.14049.qm@web38215.mail.mud.yahoo.com>

Just submitted (with no autoreply).
My Q is:
Is the Conjectured list (see %C).. 
old hat or what?
Thanks, Zak

%I A125643
%S A125643
0,0,1,1,4,8,9,16,25,27,36,49,64,64,81,100,121,125,144,169,196,216,225,256,289,324,343,361,400,441,484,512,529,576,625,676,729,729,784,841,900,961,1000,1024,1089,1156,1225,1296,1331,1369,1444,1521,1600,1681,1728,1764,1849,1936,2025,2116,2197
%N A125643 Squares and cubes (with repetitions)
%C A125643 Cf. A002760 Squares and cubes (without
repetitions).
Conjectured list of numbers not appeared as difference
between neighbor terms:
5,6,10,14,16,22,23,29,32,34,42,44,46,50,52,54,58,62,64,66,70,72,78,82,84,86,88,90,92,93,94,96,98,102,105,110,112,114,117,118,120,122,124,126,130,132,134,136,140,141,142,144,153,156,158,160,162,164,165,166,172,176,178,179,182,188,190,192,194
%Y A125643 A002760
%O A125643 0
%K A125643 ,nonn,
%A A125643 Zak Seidov  (zakseidov at gmail.com), Nov 29 2006


 
____________________________________________________________________________________
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Everyone is raving about the all-new Yahoo! Mail beta.
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From maxale at gmail.com  Wed Nov 29 13:37:32 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 29 Nov 2006 04:37:32 -0800
Subject: Difference between neighbor square and cube
In-Reply-To: <566064.14049.qm@web38215.mail.mud.yahoo.com>
References: <566064.14049.qm@web38215.mail.mud.yahoo.com>
Message-ID: <d3dac270611290437x23c81f2fq6fb37e5d662b801d@mail.gmail.com>

Zak,

Your conjecture is related to Mordell curves of the form y^2 = x^3 + n, see
http://mathworld.wolfram.com/MordellCurve.html
http://tnt.math.metro-u.ac.jp/simath/MORDELL/

The latter link can help to resolve the most difficult case of whether
a given number is the difference of a square and a cube. In
particular, for n=5 and n=-5 the corresponding equation y^2 = x^3 + n
has no positive solutions, meaning that 5 is not a difference of
positive square and cube (not necessary neighbors, so this is even
stronger than needed by conjecture).
It is also easy to see that 5 is not the difference of two cubes. And
5 can be represented in an unique way as the difference of two
squares: 5 = 3^2 - 2^2 but 2^2 and 3^2 are not neighbors in A125643
since they are separated by 2^3.
Therefore, 5 indeed never appears as the difference between neighbor
terms of A125643.

Max


On 11/29/06, zak seidov <zakseidov at yahoo.com> wrote:
> Just submitted (with no autoreply).
> My Q is:
> Is the Conjectured list (see %C)..
> old hat or what?
> Thanks, Zak
>
> %I A125643
> %S A125643
> 0,0,1,1,4,8,9,16,25,27,36,49,64,64,81,100,121,125,144,169,196,216,225,256,289,324,343,361,400,441,484,512,529,576,625,676,729,729,784,841,900,961,1000,1024,1089,1156,1225,1296,1331,1369,1444,1521,1600,1681,1728,1764,1849,1936,2025,2116,2197
> %N A125643 Squares and cubes (with repetitions)
> %C A125643 Cf. A002760 Squares and cubes (without
> repetitions).
> Conjectured list of numbers not appeared as difference
> between neighbor terms:
> 5,6,10,14,16,22,23,29,32,34,42,44,46,50,52,54,58,62,64,66,70,72,78,82,84,86,88,90,92,93,94,96,98,102,105,110,112,114,117,118,120,122,124,126,130,132,134,136,140,141,142,144,153,156,158,160,162,164,165,166,172,176,178,179,182,188,190,192,194
> %Y A125643 A002760
> %O A125643 0
> %K A125643 ,nonn,
> %A A125643 Zak Seidov  (zakseidov at gmail.com), Nov 29 2006
>
>
>
> ____________________________________________________________________________________
> Do you Yahoo!?
> Everyone is raving about the all-new Yahoo! Mail beta.
> http://new.mail.yahoo.com
>





From zakseidov at yahoo.com  Wed Nov 29 14:44:52 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 29 Nov 2006 05:44:52 -0800 (PST)
Subject: Difference between neighbor square and cube
In-Reply-To: <d3dac270611290437x23c81f2fq6fb37e5d662b801d@mail.gmail.com>
Message-ID: <948080.43313.qm@web38204.mail.mud.yahoo.com>

Max,
thanks a lot for your (excellent as usual) reply!!

Using it I found:
A054504(*
Numbers n such that Mordell's equation y^2 = x^3 + n
has no integral solutions. *)=
{6, 7, 11, 13, 14, 20, 21, 23, 29, 32, 34, 39, 42, 45,
46, 47, 51, 53, 58, 59, 60, 61, 62, 66, 67, 69, 70,
74, 75, 77, 78, 83, 84, 85, 86, 87, 88, 90, 93, 95,
96, 102, 103, 104, 109, 110, 111, 114, 115, 116, 118,
123, 124, 130, 133, 135, 137, 139, 140, 146, 147, 149,
153, 155};

A081121=(*
Numbers n such that Mordell's equation y^2 = x^3 - n
has no integral solutions. *){
3, 5, 6, 9, 10, 12, 14, 16, 17, 21, 22, 24, 29, 30,
31, 32, 33, 34, 36, 37, 38, 41, 42, 43, 46, 50, 51,
52, 57, 58, 59, 62, 65, 66, 68, 69, 70, 73, 75, 77,
78, 80, 82, 84, 85, 86, 88, 90, 91, 92, 93, 94, 96,
97, 98, 99};

Intersection[A081121,A054504]
ND={6,14,21,29,32,34,42,46,51,58,59,62,66,69,70,75,77,78,84,85,86,88,90,93,96}

And ND list gives 
(Absolute values of) Non-difference between square and
cube (not only neighbor as in %C A125643).

Hence any number in ID should be in %C A125643.
But 21 is in ND and absent in %C A125643!

Indeed in A125643,
we have:
..,64,81,100,121,125,...,
and 121-100=11^2-10^2=21!!

21 is not difference between neighbor square and cube,
but it is difference between neighbor terms in
A125643.

As a result: my subject 'Difference between neighbor
square and cube' is not correct,
and in %C A125643 we have 
'Difference between neighbor square/cube and
square/cube'.

Sorry and thanks to Max & all seqfans,
Zak  


--- "Max A." <maxale at gmail.com> wrote:

> Zak,
> 
> Your conjecture is related to Mordell curves of the
> form y^2 = x^3 + n, see
> http://mathworld.wolfram.com/MordellCurve.html
> http://tnt.math.metro-u.ac.jp/simath/MORDELL/
> 
> The latter link can help to resolve the most
> difficult case of whether
> a given number is the difference of a square and a
> cube. In
> particular, for n=5 and n=-5 the corresponding
> equation y^2 = x^3 + n
> has no positive solutions, meaning that 5 is not a
> difference of
> positive square and cube (not necessary neighbors,
> so this is even
> stronger than needed by conjecture).
> It is also easy to see that 5 is not the difference
> of two cubes. And
> 5 can be represented in an unique way as the
> difference of two
> squares: 5 = 3^2 - 2^2 but 2^2 and 3^2 are not
> neighbors in A125643
> since they are separated by 2^3.
> Therefore, 5 indeed never appears as the difference
> between neighbor
> terms of A125643.
> 
> Max
> 
> 
> On 11/29/06, zak seidov <zakseidov at yahoo.com> wrote:
> > Just submitted (with no autoreply).
> > My Q is:
> > Is the Conjectured list (see %C)..
> > old hat or what?
> > Thanks, Zak
> >
> > %I A125643
> > %S A125643
> >
>
0,0,1,1,4,8,9,16,25,27,36,49,64,64,81,100,121,125,144,169,196,216,225,256,289,324,343,361,400,441,484,512,529,576,625,676,729,729,784,841,900,961,1000,1024,1089,1156,1225,1296,1331,1369,1444,1521,1600,1681,1728,1764,1849,1936,2025,2116,2197
> > %N A125643 Squares and cubes (with repetitions)
> > %C A125643 Cf. A002760 Squares and cubes (without
> > repetitions).
> > Conjectured list of numbers not appeared as
> difference
> > between neighbor terms:
> >
>
5,6,10,14,16,22,23,29,32,34,42,44,46,50,52,54,58,62,64,66,70,72,78,82,84,86,88,90,92,93,94,96,98,102,105,110,112,114,117,118,120,122,124,126,130,132,134,136,140,141,142,144,153,156,158,160,162,164,165,166,172,176,178,179,182,188,190,192,194
> > %Y A125643 A002760
> > %O A125643 0
> > %K A125643 ,nonn,
> > %A A125643 Zak Seidov  (zakseidov at gmail.com), Nov
> 29 2006
> >
> >
> >
> >
>
____________________________________________________________________________________
> > Do you Yahoo!?
> > Everyone is raving about the all-new Yahoo! Mail
> beta.
> > http://new.mail.yahoo.com
> >
> 



 
____________________________________________________________________________________
Do you Yahoo!?
Everyone is raving about the all-new Yahoo! Mail beta.
http://new.mail.yahoo.com





From jvospost3 at gmail.com  Wed Nov 29 19:52:44 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 29 Nov 2006 10:52:44 -0800
Subject: Categories
In-Reply-To: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
References: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
Message-ID: <5542af940611291052l1a3b2654k7467715f3073dfde@mail.gmail.com>

Thank you, Christian.  That looks right.

Note that the partial sums of the first column are A118601 Number of monoids
(semigroups with identity) of order <=n.  This first column also equals the
row sums of A058137, which partitions the monoids of order n by the number k
of idempotents.






On 11/29/06, Christian G. Bower <bowerc at usa.net> wrote:
>
>
> I have one more row to add
>
> ------ Original Message ------
> Received: Wed, 22 Nov 2006 01:21:01 PM PST
> From: franktaw at netscape.net
> To: seqfan at ext.jussieu.fr
> Subject: Categories
>
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> >  1
> >  2  1
> >  7  3 1
> > 35 16 3 1
> >
> 228 77 20 3 1
>
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
> > Taking the row sums, we get:
> >
> > 1,3,11,55
> 329
> >
> > the number of categories with n morphisms.  This is probably not in the
> > OEIS (only
> > A001776 is possible - other matches become less than A058129).
>
> While the 329 is eerily close to A001776's 330, EULER(A058129) is a
> lower limit for this sequence and we have 2982 vs 2345 for #6 there.
> >  The
> > inverse Euler
> > transform,
> >
> > 1,2,8,41
> 258
> >
> > is the number of connected categories with n morphisms; this is
> > likewise probably not
> > in the OEIS (only A052447 is possible).
> No longer possible
> >
> > Can somebody generate more data?
> >
> > Franklin T. Adams-Watters
> >
>
> Christian
>
> PS
>
> I don't know if it will be feasible to collect enough data to do
> this, but the columns of the triangle converge and that convergence
> would make an interesting sequence in and of itself.
>
>
>
>
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From franktaw at netscape.net  Wed Nov 29 20:09:00 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 29 Nov 2006 14:09:00 -0500
Subject: Categories
In-Reply-To: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
References: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
Message-ID: <8C8E200F6031B3A-794-8706@FWM-M14.sysops.aol.com>

Thanks, Christian.

I think we need to look at one more table:  the number of connected
categories with n morphisms and k objects.  This starts:

  1
  2
  7 1
 35 6
228 28 2

(Row n has length ceiling(n/2).)

The table of the number of categories that I started with is a two-
dimensional Euler transform of this table.

The downward-sloping diagonal sums of this table are 1,3,15,???.
Shifting this left and taking the Euler transform gives the limiting
values Christian is referring to; starting with b(0):

1,3,21,???

For the table above, a(2n-1,n) has a purely algebraic or graph-
theoretic interpretation.  It is the number of connected
anti-transitive relations on n objects (meaning that if a R b and
b R c, then NOT (a R c)); equivalently, the number of bipartite
oriented trees, where each edge origin is in the same part.

If I haven't made any mistakes, this sequence starts:

1,1,2,3,6,10

This is not enough data to determine whether it is in the OEIS.

Franklin T. Adams-Watters


-----Original Message-----
From: bowerc at usa.net

I have one more row to add

------ Original Message ------

> How many categories are there?
>
> First, how many categories are there with n morphisms and k objects?
> This table starts:
>
>  1
>  2  1
>  7  3 1
> 35 16 3 1
>
228 77 20 3 1

> The first column is A058129, the number of monoids; the main diagonal
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
>
> Taking the row sums, we get:
>
> 1,3,11,55
329
>
> the number of categories with n morphisms.  This is probably not in 
the
> OEIS (only
> A001776 is possible - other matches become less than A058129).

While the 329 is eerily close to A001776's 330, EULER(A058129) is a
lower limit for this sequence and we have 2982 vs 2345 for #6 there.
>  The
> inverse Euler
> transform,
>
> 1,2,8,41
258
>
> is the number of connected categories with n morphisms; this is
> likewise probably not
> in the OEIS (only A052447 is possible).
No longer possible
>
> Can somebody generate more data?
>
> Franklin T. Adams-Watters
>

Christian

PS

I don't know if it will be feasible to collect enough data to do
this, but the columns of the triangle converge and that convergence
would make an interesting sequence in and of itself.

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From jvospost3 at gmail.com  Wed Nov 29 20:23:47 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 29 Nov 2006 11:23:47 -0800
Subject: Categories
In-Reply-To: <8C8E200F6031B3A-794-8706@FWM-M14.sysops.aol.com>
References: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
	 <8C8E200F6031B3A-794-8706@FWM-M14.sysops.aol.com>
Message-ID: <5542af940611291123l4f9c3385l283f3b2243dab1c2@mail.gmail.com>

The issue of "connected" in the underlying functional graph, as a subset of
the set of all digraphs, is further revealed by the distinction (definitions
at MathWorld) of Weakly Conncted versus Strongly Conncted Digraphs.

Every functional digraph is the disjoint union of connected digraphs.
That's why there is a recursion in the enumeration of F(n) = functional
graphs on n points. F(n) is equal or greater to the number of functional
graphs on n-1 points, unioned with an isolated point looping to itself, plus
the number of functional graphs on n-2 points unioned with any of the 3
functional graphs on 3 points, ..., throwing out any double or multiple
counts by equivalence class on isomorphisms of the unlabelled digraphs.

Note that MathWorld's correct definition of functional graph, and its
Mathematica, is identical to what I'd called endofunctions in earlier
posting of this thread.  Hence the enumeration problem that I solved for
"prime" endofunctions is identical to a subset of your category enumeration
problem, relating to the number of categories that are not unions nor
categorical products of smaller categories.

However, when I referred to 2-categories before, I should more correctly
have said Bicategories, the difference being based on whether associativity
is strict or up to 2-isomorphism.  Your enumeration of categories is more
formally explained in terms of Cat, the category of categories and functors,
but this is not the right venue for me to go into excruciating details.

On 11/29/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
> Thanks, Christian.
>
> I think we need to look at one more table:  the number of connected
> categories with n morphisms and k objects.  This starts:
>
>   1
>   2
>   7 1
> 35 6
> 228 28 2
>
> (Row n has length ceiling(n/2).)
>
> The table of the number of categories that I started with is a two-
> dimensional Euler transform of this table.
>
> The downward-sloping diagonal sums of this table are 1,3,15,???.
> Shifting this left and taking the Euler transform gives the limiting
> values Christian is referring to; starting with b(0):
>
> 1,3,21,???
>
> For the table above, a(2n-1,n) has a purely algebraic or graph-
> theoretic interpretation.  It is the number of connected
> anti-transitive relations on n objects (meaning that if a R b and
> b R c, then NOT (a R c)); equivalently, the number of bipartite
> oriented trees, where each edge origin is in the same part.
>
> If I haven't made any mistakes, this sequence starts:
>
> 1,1,2,3,6,10
>
> This is not enough data to determine whether it is in the OEIS.
>
> Franklin T. Adams-Watters
>
>
> -----Original Message-----
> From: bowerc at usa.net
>
> I have one more row to add
>
> ------ Original Message ------
> Received: Wed, 22 Nov 2006 01:21:01 PM PST
>
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> >  1
> >  2  1
> >  7  3 1
> > 35 16 3 1
> >
> 228 77 20 3 1
>
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
> > Taking the row sums, we get:
> >
> > 1,3,11,55
> 329
> >
> > the number of categories with n morphisms.  This is probably not in
> the
> > OEIS (only
> > A001776 is possible - other matches become less than A058129).
>
> While the 329 is eerily close to A001776's 330, EULER(A058129) is a
> lower limit for this sequence and we have 2982 vs 2345 for #6 there.
> >  The
> > inverse Euler
> > transform,
> >
> > 1,2,8,41
> 258
> >
> > is the number of connected categories with n morphisms; this is
> > likewise probably not
> > in the OEIS (only A052447 is possible).
> No longer possible
> >
> > Can somebody generate more data?
> >
> > Franklin T. Adams-Watters
> >
>
> Christian
>
> PS
>
> I don't know if it will be feasible to collect enough data to do
> this, but the columns of the triangle converge and that convergence
> would make an interesting sequence in and of itself.
>
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and
> industry-leading spam and email virus protection.
>
>
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From njas at research.att.com  Wed Nov 29 22:34:56 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 29 Nov 2006 16:34:56 -0500 (EST)
Subject: OEIS on vacation in December
Message-ID: <200611292134.QAA53261@fry.research.att.com>


This means:
- do send in sequences in published works or 
   that are important for your work
- do send important corrections, extensions, updates
- do send EDITED versions

- don't send in sequences that you made up (unless
   they are really beautiful - and if they
   are base dependent or involve primes 
   they are probably not)
- don't send very minor corrections

I will be doing occasional updates, but I won't be
working on the OEIS 24/7 !

Thanks!

Neil





From maxale at gmail.com  Wed Nov 29 22:45:38 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 29 Nov 2006 13:45:38 -0800
Subject: Difference between neighbor square and cube
In-Reply-To: <948080.43313.qm@web38204.mail.mud.yahoo.com>
References: <d3dac270611290437x23c81f2fq6fb37e5d662b801d@mail.gmail.com>
	 <948080.43313.qm@web38204.mail.mud.yahoo.com>
Message-ID: <d3dac270611291345n2ad5dae4id319f3bce8fca71c@mail.gmail.com>

On 11/29/06, zak seidov <zakseidov at yahoo.com> wrote:

> Intersection[A081121,A054504]
> ND={6,14,21,29,32,34,42,46,51,58,59,62,66,69,70,75,77,78,84,85,86,88,90,93,96}

That's A110223. But Robert G. Wilson posed as "conjectured".
I believe this sequence can be confirmed/verified using the
information about Mordell curves from
http://tnt.math.metro-u.ac.jp/simath/MORDELL/

Max





From bowerc at usa.net  Thu Nov 30 00:42:31 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 29 Nov 2006 15:42:31 -0800
Subject: Categories
Message-ID: <623kkCXpf3086S27.1164843751@cmsweb27.cms.usa.net>

ftaw
> I think we need to look at one more table:  the number of connected
> categories with n morphisms and k objects.  This starts:

>   1
>   2
>   7 1
>  35 6
> 228 28 2

> (Row n has length ceiling(n/2).)
...
> For the table above, a(2n-1,n) has a purely algebraic or graph-
> theoretic interpretation.  It is the number of connected
> anti-transitive relations on n objects (meaning that if a R b and
> b R c, then NOT (a R c)); equivalently, the number of bipartite
> oriented trees, where each edge origin is in the same part.
> 
> If I haven't made any mistakes, this sequence starts:
> 
> 1,1,2,3,6,10
> 
> This is not enough data to determine whether it is in the OEIS.

For n>1 it's A122086
http://www.research.att.com/~njas/sequences/A122086
A sequence which could use a nicer description and formula such as
a(n) = A000055(n) - A000081(n/2) where a sequence evaluated at a
noninteger is 0.
http://www.research.att.com/~njas/sequences/A000055
http://www.research.att.com/~njas/sequences/A000081

Christian








From bowerc at usa.net  Thu Nov 30 01:36:23 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 29 Nov 2006 16:36:23 -0800
Subject: Categories
Message-ID: <209kkdajX8954S10.1164846983@cmsweb10.cms.usa.net>

I have row 6 now:

> How many categories are there?
>
> First, how many categories are there with n morphisms and k objects?
> This table starts:
>
>  1
>  2  1
>  7  3 1
> 35 16 3 1
>
228 77 20 3 1
2237 485 111 21 3 1

> Taking the row sums, we get:
>
> 1,3,11,55
329
2858

>  The
> inverse Euler
> transform,
>
> 1,2,8,41
258
2407

> I think we need to look at one more table:  the number of connected
> categories with n morphisms and k objects.  This starts:

>   1
>   2
>   7 1
>  35 6
> 228 28 2
2237 159 11
31559 ?   ? 3

> (Row n has length ceiling(n/2).)
> 
> The table of the number of categories that I started with is a two-
> dimensional Euler transform of this table.
> 
> The downward-sloping diagonal sums of this table are 1,3,15,???.
77
> Shifting this left and taking the Euler transform gives the limiting
> values Christian is referring to; starting with b(0):
> 
> 1,3,21,???
132








From franktaw at netscape.net  Thu Nov 30 14:53:38 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 30 Nov 2006 08:53:38 -0500
Subject: Categories
In-Reply-To: <623kkCXpf3086S27.1164843751@cmsweb27.cms.usa.net>
References: <623kkCXpf3086S27.1164843751@cmsweb27.cms.usa.net>
Message-ID: <8C8E29E12708888-5D8-4EB@WEBMAIL-RA17.sysops.aol.com>

A typo there; that should be a(n) = 2*A000055(n) - A000081(n/2).

Franklin T. Adams-Watters


-----Original Message-----
From: bowerc at usa.net

...
For n>1 it's A122086
http://www.research.att.com/~njas/sequences/A122086
A sequence which could use a nicer description and formula such as
a(n) = A000055(n) - A000081(n/2) where a sequence evaluated at a
noninteger is 0.
http://www.research.att.com/~njas/sequences/A000055
http://www.research.att.com/~njas/sequences/A000081

Christian





________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From jvospost3 at gmail.com  Thu Nov 30 19:34:51 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 30 Nov 2006 10:34:51 -0800
Subject: Categories
In-Reply-To: <8C8E29E12708888-5D8-4EB@WEBMAIL-RA17.sysops.aol.com>
References: <623kkCXpf3086S27.1164843751@cmsweb27.cms.usa.net>
	 <8C8E29E12708888-5D8-4EB@WEBMAIL-RA17.sysops.aol.com>
Message-ID: <5542af940611301034t3bba13fdnadc86411facc535d@mail.gmail.com>

a(n) = 2*A000055(n) - A000081(n/2) = (2* Number of trees with n unlabeled
nodes) - Number of rooted trees with n/2 nodes (or connected functions with
a fixed point)

where a(n) is downwards sloping diagonal sums of main category table, or
what?

It isn't until we look at bicategories that we are forced to look at the
symmetries and structure of the individual categories being enumerated here,
as objects in the higher categories, and have to enumerate the morphisms of
the functions. A (finite, concrete) category whose underlying directed graph
has no symmetry (i.e. automorphism group of order 1) can only map to itself
or an identical copy of itself, point to point, loop to loop, arc to arc.
But any cycle of length >1 in that directed graph allows morphisms to other
directed graphs.  The directed graph C_n, a cycle of length n, can thus map
to the same graph but mapping each point to the next, and each arc to the
next, and so forth. That requires then analysis of cycles in directed
graphs, as by de Bruijn in 1952 and more deeply understood since then. This
is another enumeration problem, unnecessary to completing the answer to
Frank's original question, and more tied to my endofunctions analysis. Sorry
if my interruptions distracted you from the straightforward question.  Good
work, guys!

Funny how Category Theory folks tend to be addicted to theory for its own
sake, and less willing to "get their hands dirty" with anything as old
fashioned as enumeration.  OEIS helps us focus here, by demanding the
integer sequences!

On 11/30/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
> A typo there; that should be a(n) = 2*A000055(n) - A000081(n/2).
>
> Franklin T. Adams-Watters
>
>
> -----Original Message-----
> From: bowerc at usa.net
>
> ...
> For n>1 it's A122086
> http://www.research.att.com/~njas/sequences/A122086
> A sequence which could use a nicer description and formula such as
> a(n) = A000055(n) - A000081(n/2) where a sequence evaluated at a
> noninteger is 0.
> http://www.research.att.com/~njas/sequences/A000055
> http://www.research.att.com/~njas/sequences/A000081
>
> Christian
>
>
>
>
>
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and
> industry-leading spam and email virus protection.
>
>
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From all at abouthugo.de  Thu Nov 30 21:28:06 2006
From: all at abouthugo.de (Hugo Pfoertner)
Date: Thu, 30 Nov 2006 21:28:06 +0100
Subject: Difference between neighbor square and cube
References: <948080.43313.qm@web38204.mail.mud.yahoo.com>
Message-ID: <456F3ED6.D0619322@abouthugo.de>

zak seidov wrote:
> 
> Max,
> thanks a lot for your (excellent as usual) reply!!
> 
> Using it I found:
> A054504(*
> Numbers n such that Mordell's equation y^2 = x^3 + n
> has no integral solutions. *)=
> {6, 7, 11, 13, 14, 20, 21, 23, 29, 32, 34, 39, 42, 45,
> 46, 47, 51, 53, 58, 59, 60, 61, 62, 66, 67, 69, 70,
> 74, 75, 77, 78, 83, 84, 85, 86, 87, 88, 90, 93, 95,
> 96, 102, 103, 104, 109, 110, 111, 114, 115, 116, 118,
> 123, 124, 130, 133, 135, 137, 139, 140, 146, 147, 149,
> 153, 155};
> 
> A081121=(*
> Numbers n such that Mordell's equation y^2 = x^3 - n
> has no integral solutions. *){
> 3, 5, 6, 9, 10, 12, 14, 16, 17, 21, 22, 24, 29, 30,
> 31, 32, 33, 34, 36, 37, 38, 41, 42, 43, 46, 50, 51,
> 52, 57, 58, 59, 62, 65, 66, 68, 69, 70, 73, 75, 77,
> 78, 80, 82, 84, 85, 86, 88, 90, 91, 92, 93, 94, 96,
> 97, 98, 99};
> 
> Intersection[A081121,A054504]
> ND={6,14,21,29,32,34,42,46,51,58,59,62,66,69,70,75,77,78,84,85,86,88,90,93,96}
> 
> And ND list gives
> (Absolute values of) Non-difference between square and
> cube (not only neighbor as in %C A125643).
> 
> Hence any number in ID should be in %C A125643.
> But 21 is in ND and absent in %C A125643!
> 
> Indeed in A125643,
> we have:
> ..,64,81,100,121,125,...,
> and 121-100=11^2-10^2=21!!
> 
> 21 is not difference between neighbor square and cube,
> but it is difference between neighbor terms in
> A125643.
> 
> As a result: my subject 'Difference between neighbor
> square and cube' is not correct,
> and in %C A125643 we have
> 'Difference between neighbor square/cube and
> square/cube'.
> 
> Sorry and thanks to Max & all seqfans,
> Zak

Also closely related:

http://www.research.att.com/~njas/sequences/A087285
http://www.research.att.com/~njas/sequences/A087286
http://www.research.att.com/~njas/sequences/A088017

Hugo





From bowerc at usa.net  Thu Nov 30 22:29:20 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Thu, 30 Nov 2006 13:29:20 -0800
Subject: Categories
Message-ID: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>



------ Original Message ------
From: "Jonathan Post" <jvospost3 at gmail.com>
To: "franktaw at netscape.net" <franktaw at netscape.net>Cc: bowerc at usa.net,
seqfan at ext.jussieu.fr, jvospost2 at yahoo.com
Subject: Re: Categories

> a(n) = 2*A000055(n) - A000081(n/2) = (2* Number of trees with n unlabeled
> nodes) - Number of rooted trees with n/2 nodes (or connected functions with
> a fixed point)
> 
> where a(n) is downwards sloping diagonal sums of main category table, or
> what?

Actually it's the "corner" values from this table

>   *1*
>   2
>   7 *1*
>  35 6
> 228 28 *2*
> 2237 159 11
> 31559 ?   ? *3*

The downward sloping sums were an aide in calculating the convergent
sequence of the columns

The interpretation of these corners comes from the fact we have a
saturated category. It has 2n-1 elements, n of which are object
identities. It's connected, so it has a minimum of n-1 "crossover"
morphisms, it also has a maximum of n-1 cm's since there are only
n-1 elements left. Any connected graph with n points and n-1 edges is
a tree. The associativity of categories states that if we have an
a-b morphism (an edge on the graph) and a b-c we must have an a-c.
But if we have all 3, it's no longer a tree, hence we never have the
a-b and the b-c, hence each point has all edges either pointing in or
pointing out, hence the biparite graph analogy, hence a little
combinatorial magic.

...
> Funny how Category Theory folks tend to be addicted to theory for its > own
> sake, and less willing to "get their hands dirty" with anything as old
> fashioned as enumeration.  OEIS helps us focus here, by demanding the
> integer sequences!
> 

I try to stay out of philosophical stuff here, but sometimes it's too
hard to resist. I figure I'm guilty of spending too much time counting
stuff and not taking the theory in. Most of the world would not give a
hoot about either, so I'm glad there are folk spending time on what I
would not give any time to.

Still, I find it jarring any time I read a math paper and they write
about several things that can be represented as sequences, yet all they
give is a mathematical description, rarely do they cite the OEIS
reference or even give me the first n terms leaving me to calculate them
myself and hope I didn't make a mistake. If I wrote a paper, I'd
include that stuff, not just because I'm an OEIS nerd, but also because
I would want people to be able to find my paper who didn't know they
were looking for it.

Just a little more self indulgence here with my list of semigroup pipe
dreams. After doing sequences on semigroups
http://www.research.att.com/~njas/sequences/A027851
and monoids
http://www.research.att.com/~njas/sequences/A058129
I knew categories was one of the next logical steps, but I held off
because I knew it would be difficult. Perhaps I can spur someone else's
interest in related problems that may prove even more difficult.

There's transformation semigroups, the semigroup version of permutation
groups
http://www.research.att.com/~njas/sequences/A000638
I've seen TSs written about, but no attempts to enumerate them.
(Note, a TS is any set of endofunctions closed under composition)
There TS's close cousins which could be called:
transformation monoids
relation semigroups
relation monoids
(Composition of relations is such that (a,c) is in R1(R2) iff there
is b such that (a,b) in R1 and (b,c) in R2)
Then there are the generalizations of categories to semigroups and
groupoids. Note I mean the groupoid that is a set and closed operation
http://www.research.att.com/~njas/sequences/A001329
Not the group like category groupoid which might also be interesting
to count and not too difficult; oh terminology.
Anyway, with the semigroup and groupoid categories, there comes the
issue of treating objects and morphims. The ordinary (monoid like)
categories avoid this problem by having object identities among the
morphisms that can be associatied with the objects themselves. Without
that restriction, it's possible to have objects untouched by any
morphism or two structures with identical (or isomorphic) multiplication
tables, but not isomorphic when you take the objects into account.
Basically many of the same counting problems one has with multigraphs
(and probably the same solutions too.)








From jvospost3 at gmail.com  Thu Nov 30 22:53:42 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 30 Nov 2006 13:53:42 -0800
Subject: Categories
In-Reply-To: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>
References: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>
Message-ID: <5542af940611301353m486a4788wff2e4eb539b76943@mail.gmail.com>

I strongly agree with Christian.  My doctoral dissertation involved taking
the Krohn-Rhodes decomposition of the semigroup of differential operators of
a system of nonlinear differential equations that had been in the
mathematical Biology literature for decades, and routinely dismissed as not
capable of closed-form solution. Marvin Minsky (who had been the PhD advisor
of the eponymous John Rhodes) thought my work was the best use yet of his
student's theorem.  The great Stan Ulam (coinventor of cellular automata,
the H-bomb, and the nuclear pulse rocket) also thought it was cool. But only
half a dozen biologists in theb world knew that much about semigroups, and
some of them wrote to me from Edinburgh, Scotland, and places in the USSR.
Ulam died before we could coauthor; Semigroup Forum accepted my paper with
delight for a special issue, but wanted it reformatted, and I was laid off
from the company where I worked, and they erased on only machine-readable
version of the paper.  Equations were harder to typeset those decades
ago...  But yes, lots of so-called professionals go out of their way to
avoid citing the OEIS, or give any useful examples.  Also yes: semigroup
enumerations are worthwhile.  Also, you nailed the differences between those
and categories.

See for instance:

  A118581 <http://www.research.att.com/~njas/sequences/A118581> Number of
nonisomorphic *semigroups* of order <= n.

  A118099 <http://www.research.att.com/~njas/sequences/A118099> Number of
inverse *semigroups* of order <= n.

  A118100 <http://www.research.att.com/~njas/sequences/A118100>

Number of commutative *semigroups* of order <= n.

  A118601 <http://www.research.att.com/~njas/sequences/A118601> Number of
monoids (*semigroups* with identity) of order <= n.
  A113534 <http://www.research.att.com/~njas/sequences/A113534> Ascending
descending base exponent transform of the flipped tribonacci substitution (
A092782 <http://www.research.att.com/~njas/sequences/A092782>).
with its semigroup reference, and some others.

-- Jonathan


On 11/30/06, Christian G. Bower <bowerc at usa.net> wrote:
>
>
>
> ------ Original Message ------
> Received: Thu, 30 Nov 2006 10:34:55 AM PST
> From: "Jonathan Post" <jvospost3 at gmail.com>
> To: "franktaw at netscape.net" <franktaw at netscape.net>Cc: bowerc at usa.net,
> seqfan at ext.jussieu.fr, jvospost2 at yahoo.com
> Subject: Re: Categories
>
> > a(n) = 2*A000055(n) - A000081(n/2) = (2* Number of trees with n
> unlabeled
> > nodes) - Number of rooted trees with n/2 nodes (or connected functions
> with
> > a fixed point)
> >
> > where a(n) is downwards sloping diagonal sums of main category table, or
> > what?
>
> Actually it's the "corner" values from this table
>
> >   *1*
> >   2
> >   7 *1*
> >  35 6
> > 228 28 *2*
> > 2237 159 11
> > 31559 ?   ? *3*
>
> The downward sloping sums were an aide in calculating the convergent
> sequence of the columns
>
> The interpretation of these corners comes from the fact we have a
> saturated category. It has 2n-1 elements, n of which are object
> identities. It's connected, so it has a minimum of n-1 "crossover"
> morphisms, it also has a maximum of n-1 cm's since there are only
> n-1 elements left. Any connected graph with n points and n-1 edges is
> a tree. The associativity of categories states that if we have an
> a-b morphism (an edge on the graph) and a b-c we must have an a-c.
> But if we have all 3, it's no longer a tree, hence we never have the
> a-b and the b-c, hence each point has all edges either pointing in or
> pointing out, hence the biparite graph analogy, hence a little
> combinatorial magic.
>
> ...
> > Funny how Category Theory folks tend to be addicted to theory for its >
> own
> > sake, and less willing to "get their hands dirty" with anything as old
> > fashioned as enumeration.  OEIS helps us focus here, by demanding the
> > integer sequences!
> >
>
> I try to stay out of philosophical stuff here, but sometimes it's too
> hard to resist. I figure I'm guilty of spending too much time counting
> stuff and not taking the theory in. Most of the world would not give a
> hoot about either, so I'm glad there are folk spending time on what I
> would not give any time to.
>
> Still, I find it jarring any time I read a math paper and they write
> about several things that can be represented as sequences, yet all they
> give is a mathematical description, rarely do they cite the OEIS
> reference or even give me the first n terms leaving me to calculate them
> myself and hope I didn't make a mistake. If I wrote a paper, I'd
> include that stuff, not just because I'm an OEIS nerd, but also because
> I would want people to be able to find my paper who didn't know they
> were looking for it.
>
> Just a little more self indulgence here with my list of semigroup pipe
> dreams. After doing sequences on semigroups
> http://www.research.att.com/~njas/sequences/A027851
> and monoids
> http://www.research.att.com/~njas/sequences/A058129
> I knew categories was one of the next logical steps, but I held off
> because I knew it would be difficult. Perhaps I can spur someone else's
> interest in related problems that may prove even more difficult.
>
> There's transformation semigroups, the semigroup version of permutation
> groups
> http://www.research.att.com/~njas/sequences/A000638
> I've seen TSs written about, but no attempts to enumerate them.
> (Note, a TS is any set of endofunctions closed under composition)
> There TS's close cousins which could be called:
> transformation monoids
> relation semigroups
> relation monoids
> (Composition of relations is such that (a,c) is in R1(R2) iff there
> is b such that (a,b) in R1 and (b,c) in R2)
> Then there are the generalizations of categories to semigroups and
> groupoids. Note I mean the groupoid that is a set and closed operation
> http://www.research.att.com/~njas/sequences/A001329
> Not the group like category groupoid which might also be interesting
> to count and not too difficult; oh terminology.
> Anyway, with the semigroup and groupoid categories, there comes the
> issue of treating objects and morphims. The ordinary (monoid like)
> categories avoid this problem by having object identities among the
> morphisms that can be associatied with the objects themselves. Without
> that restriction, it's possible to have objects untouched by any
> morphism or two structures with identical (or isomorphic) multiplication
> tables, but not isomorphic when you take the objects into account.
> Basically many of the same counting problems one has with multigraphs
> (and probably the same solutions too.)
>
>
>
>
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From franktaw at netscape.net  Thu Nov 30 22:54:39 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 30 Nov 2006 16:54:39 -0500
Subject: Categories
In-Reply-To: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>
References: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>
Message-ID: <8C8E2E14509A6B2-934-A97D@FWM-M10.sysops.aol.com>

I'd like to put my plug in for more application in math.

I heard of a case some years ago where a paper was published about some 
kind of mathematical object (I think it was a kind of topological 
space, but it really doesn't matter).  Three or four more papers were 
published, establishing more and more properties for this type of 
object - until finally it was proved that they don't exist!  This 
wouldn't have happened if somebody had asked for an example at an early 
stage.

There is a strong tendency in mathematics to start at the end.  The 
researcher pursues a line of thought, which eventually leads to a 
spiffy proof.  The proof is then published, with no hint of the process 
by which it was reached.  This is a disservice to anybody who might use 
a similar approach to solve some other problem.  It is especially a 
disservice when presented to students.

On a more personal level, I find when looking a math paper, I want to 
know how this relates to problems that I am already interested in or at 
least familiar with.  If I can't get an answer to that, I have a hard 
time maintaining any interest.

Franklin T. Adams-Watters

________________________________________________________________________
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From zakseidov at yahoo.com  Wed Nov  1 06:43:12 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Tue, 31 Oct 2006 21:43:12 -0800 (PST)
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
Message-ID: <20061101054313.25333.qmail@web38210.mail.mud.yahoo.com>

Dear seqfans,
Bob, Ivars,

In A065577, 
a(1)=2 because 10=3+7=5+5?
Also a(2)=6 because
100=3+97=11+89=17+83=29+71=41+59=47+53?

The point is that in Ivars Peterson's
"Goldbach's Prime Pairs"
http://www.maa.org/mathland/mathtrek_8_21_00.html
we read:

" 
Note that Goldbach partitions take into account the
order of the primes that are summed. For example, 10
has two Goldbach partitions: 3 + 7 and 7 + 3.

Integer No. of Goldbach partitions 
10 2 
100 6 
1,000 28 
10,000 127 
100,000 810 
1,000,000 5,402 
10,000,000 38,807 
100,000,000 291,400
" 

Me: But in A065577 a(1)=2 and a(2)=6 
(and also other terms as I checked them)
without taking into account 
the order of the primes 
that are summed(?!)

Can anyoone (Bob? Ivars?) clarify this?
Thanks, Zak

BTW Two next terms are calculated by me
(for n=9, 10) as 2274205, 18200487.
Can anyone check these?
Thanks, Zak


%I A065577
%S A065577 2,6,28,127,810,5402,38807,291400
%N A065577 Number of Goldbach partitions of 10^n.
%H A065577 Ivars Peterson's MathTrek, <a
href="http://www.maa.org/mathland/mathtrek_8_21_00.html">Goldbach's
Prime Pairs</a>
%H A065577 Science News Online, week of Aug. 19, 2000;
Vol. 158, No. 8 <a href="http://www.sciencenews.org/
20000819/mathtrek.asp">Goldbach's Prime Pairs</a>
<skip>
%A A065577 Robert G. Wilson v (rgwv(AT)rgwv.com), Dec
01 2001




 
____________________________________________________________________________________
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From ralf at ark.in-berlin.de  Wed Nov  1 10:35:28 2006
From: ralf at ark.in-berlin.de (Ralf Stephan)
Date: Wed, 1 Nov 2006 10:35:28 +0100
Subject: Re^2: Error-finding program
In-Reply-To: <200610312150.k9VLoKfn018613@amer.strw.leidenuniv.nl>
References: <200610312150.k9VLoKfn018613@amer.strw.leidenuniv.nl>
Message-ID: <20061101093528.GA19017@ark.in-berlin.de>

> The definitions of A026468 and A026469 are
>  a(n) = least positive integer > a(n-1) and not a(i)^2 + a(j)^2 for 1<=i<=j<=n.
> and
>  a(n) = least positive integer > a(n-1) and not a(i)^2 + a(j)^2 for 1<=i<=j<=n.  	
> which looks quite identical. So why can A026468/9 be two different
> sequences? Is there a "not" too much or too little, or rather one of the 
> "less-equal" signs to be corrected to "less" or a starting condition on a(2)
> to be added?

My guess is that Clark started the same algorithm with 1, 2 and then
tried it with starting values 1, 3, so the definition of A026469 should 
be amended with ", starting with 1, 3".


ralf






From njas at research.att.com  Wed Nov  1 14:16:47 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 1 Nov 2006 08:16:47 -0500 (EST)
Subject: Re^2: Error-finding program
References: <200610312150.k9VLoKfn018613@amer.strw.leidenuniv.nl>
Message-ID: <200611011316.IAA53531@fry.research.att.com>

Ralf,  I think you've solved it!   Thanks!   Neil


My guess is that Clark started the same algorithm with 1, 2 and then
tried it with starting values 1, 3, so the definition of A026469 should 
be amended with ", starting with 1, 3".





From mathar at strw.leidenuniv.nl  Wed Nov  1 14:17:58 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Wed, 1 Nov 2006 14:17:58 +0100
Subject: A065577 Number of Goldbach partitions of 10^n?
Message-ID: <200611011317.kA1DHwmp018846@amer.strw.leidenuniv.nl>


zs> From seqfan-owner at ext.jussieu.fr  Wed Nov  1 06:44:53 2006
zs> Return-Path: <seqfan-owner at ext.jussieu.fr>
zs> Date: Tue, 31 Oct 2006 21:43:12 -0800 (PST)
zs> From: zak seidov <zakseidov at yahoo.com>
zs> Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
zs> To: seqfan at ext.jussieu.fr, ip at sciserv.org, rgwv at rgwv.com
zs> 
zs> Dear seqfans,
zs> Bob, Ivars,
zs> 
zs> In A065577, 
zs> a(1)=2 because 10=3+7=5+5?
zs> Also a(2)=6 because
zs> 100=3+97=11+89=17+83=29+71=41+59=47+53?
zs> ....

One might call A065577 explicitly the partitions without regard to order. The
conversion between the two counts is simple, see
http://mathworld.wolfram.com/GoldbachPartition.html

The table of n, ordered, and non-ordered partitions is

                                    1, 3, 2
                                   2, 12, 6
                                   3, 56, 28
                                  4, 254, 127
                                 5, 1620, 810
                                6, 10804, 5402
                                7, 77614, 38807

with the ordered partitions in the OEIS as A073610, the non-ordered in A061358 (?)
(Cf from A065577 to these two might also help to show the difference,
A065577(n)=A061358(10^n).)
With the exception of the n=1 leading term, the count of the
ordered partitions is twice that of the non-ordered partitions,
because 10^n/2 is not a prime then.

The Maple program to print the small table above:

A065577 := proc(n,orderd)
	local N,a,i,ip;
	N := 10^n ;
	a := 0 ;
	i := 1 ;
	ip := 2 ;
	while 2*ip <= N do
		if isprime(N-ip) then
			if orderd and ip <> N-ip then
				a := a+ 2;
			else
				a := a+ 1;
			fi ;
		fi ;
		i := i+1 ;
		ip := ithprime(i) ;
	od ;
	RETURN(a) ;
end:

for n from 1 to 20 do
	print(n,A065577(n,true),A065577(n,false)) ;
od ;

In PARI

A065577(n)={
	local(N,a,i,ip);
	N = 10^n ;
	a = 0 ;
	i = 1 ;
	ip = 2 ;
	while(2*ip <= N,
		if(isprime(N-ip),
			a++ ;
		) ;
		i++ ;
		\\ip = prime(i) ; \\ slower and needs -p switch
		ip = nextprime(ip+1) ; \\ pseudoprimes only
	) ;
	return(a) ;
}

{
	for(n=1,20,
		print(n," ",A065577(n)) ;
	)
}

The PARI output confirms the n=9 case quoted by Zak (I've not gone to n=10):

1 2
2 6
3 28
4 127
5 810
6 5402
7 38807
8 291400
9 2274205

-- Richard





From zakseidov at yahoo.com  Wed Nov  1 15:29:37 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 1 Nov 2006 06:29:37 -0800 (PST)
Subject: A065577 Number of Goldbach partitions of 10^n?
In-Reply-To: <200611011317.kA1DHwmp018846@amer.strw.leidenuniv.nl>
Message-ID: <20061101142937.98860.qmail@web38214.mail.mud.yahoo.com>

Richard,
thanks for your reply.
It seems this'd concern mainly Ivars!
Also could you please calculate cases n>=10) -
I guess this'd not take for you hours(days!)
as with my Mathematica.
thanks, Zak


--- Richard Mathar <mathar at strw.leidenuniv.nl> wrote:

> 
> zs> From seqfan-owner at ext.jussieu.fr  Wed Nov  1
> 06:44:53 2006
> zs> Return-Path: <seqfan-owner at ext.jussieu.fr>
> zs> Date: Tue, 31 Oct 2006 21:43:12 -0800 (PST)
> zs> From: zak seidov <zakseidov at yahoo.com>
> zs> Subject: A065577 Number of Goldbach partitions
> of 10^n? More terms?
> zs> To: seqfan at ext.jussieu.fr, ip at sciserv.org,
> rgwv at rgwv.com
> zs> 
> zs> Dear seqfans,
> zs> Bob, Ivars,
> zs> 
> zs> In A065577, 
> zs> a(1)=2 because 10=3+7=5+5?
> zs> Also a(2)=6 because
> zs> 100=3+97=11+89=17+83=29+71=41+59=47+53?
> zs> ....
> 
> One might call A065577 explicitly the partitions
> without regard to order. The
> conversion between the two counts is simple, see
> http://mathworld.wolfram.com/GoldbachPartition.html
> 
> The table of n, ordered, and non-ordered partitions
> is
> 
>                                     1, 3, 2
>                                    2, 12, 6
>                                    3, 56, 28
>                                   4, 254, 127
>                                  5, 1620, 810
>                                 6, 10804, 5402
>                                 7, 77614, 38807
> 
> with the ordered partitions in the OEIS as A073610,
> the non-ordered in A061358 (?)
> (Cf from A065577 to these two might also help to
> show the difference,
> A065577(n)=A061358(10^n).)
> With the exception of the n=1 leading term, the
> count of the
> ordered partitions is twice that of the non-ordered
> partitions,
> because 10^n/2 is not a prime then.
> 
> The Maple program to print the small table above:
> 
> A065577 := proc(n,orderd)
> 	local N,a,i,ip;
> 	N := 10^n ;
> 	a := 0 ;
> 	i := 1 ;
> 	ip := 2 ;
> 	while 2*ip <= N do
> 		if isprime(N-ip) then
> 			if orderd and ip <> N-ip then
> 				a := a+ 2;
> 			else
> 				a := a+ 1;
> 			fi ;
> 		fi ;
> 		i := i+1 ;
> 		ip := ithprime(i) ;
> 	od ;
> 	RETURN(a) ;
> end:
> 
> for n from 1 to 20 do
> 	print(n,A065577(n,true),A065577(n,false)) ;
> od ;
> 
> In PARI
> 
> A065577(n)={
> 	local(N,a,i,ip);
> 	N = 10^n ;
> 	a = 0 ;
> 	i = 1 ;
> 	ip = 2 ;
> 	while(2*ip <= N,
> 		if(isprime(N-ip),
> 			a++ ;
> 		) ;
> 		i++ ;
> 		\\ip = prime(i) ; \\ slower and needs -p switch
> 		ip = nextprime(ip+1) ; \\ pseudoprimes only
> 	) ;
> 	return(a) ;
> }
> 
> {
> 	for(n=1,20,
> 		print(n," ",A065577(n)) ;
> 	)
> }
> 
> The PARI output confirms the n=9 case quoted by Zak
> (I've not gone to n=10):
> 
> 1 2
> 2 6
> 3 28
> 4 127
> 5 810
> 6 5402
> 7 38807
> 8 291400
> 9 2274205
> 
> -- Richard
> 



 
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From tanyakh at TanyaKhovanova.com  Wed Nov  1 16:46:50 2006
From: tanyakh at TanyaKhovanova.com (Tanya Khovanova)
Date: Wed,  1 Nov 2006 07:46:50 -0800
Subject: What is the most famous sequence?
Message-ID: <200611010746.AA12650352@TanyaKhovanova.com>

What is the most famous sequence? The most basic (natural numbers?)? The most simple (0,0,0, ... or 1,1,1, ...?)? The most versatile (Catalan numbers?)?

I would be curious to know your opinion.

I tried to check the number of Google hits for some sequence numbers. Here is what I've got:
A000040 (prime numbers) 68,800
A000045 (Fibonacci)     66,000
A000108 (Catalan)       62,500
A000027 (natural)       47,500

Can someone write a program to find top ten? What is the most referenced sequence inside Encyclopedia? Can we find the most hits outside Encyclopedia? Is it easy to do?

Tanya


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From jvospost3 at gmail.com  Wed Nov  1 18:58:55 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 1 Nov 2006 09:58:55 -0800
Subject: New & old LEGO seqs; production cuts
Message-ID: <5542af940611010958me17f347w3bf2d2f2d6c9dec0@mail.gmail.com>

I like the new seqs about Lego, such as
 A123762 <http://www.research.att.com/%7Enjas/sequences/A123762>
 Number of ways, counted up to symmetry, to build a contiguous building with
n *LEGO* blocks of size 1x2.

Ironically timed, as *"Recent restructuring and production cuts have left
Lego unable to fill
orders<http://www.cbc.ca/money/story/2006/10/31/legoproduction.html>for
the upcoming holiday season. Affected products include Duplo bricks,
Lego City sets, and (horror of horrors!) Star Wars and Lego Technik
sets."*According to the article Lego stands to lose $127 million in
holiday sales.

http://www.cbc.ca/money/story/2006/10/31/legoproduction.html

It might be useful to add crossrefs to earlier seqs such as:

 A007576 <http://www.research.att.com/%7Enjas/sequences/A007576>
 Number of maximally stable towers of 2 X 2 *LEGO* blocks.

 A082679 <http://www.research.att.com/%7Enjas/sequences/A082679>
 Number of *Lego* towers, one piece per floor, where every floor is
perpendicular to the one below it (so we have a kind of 3-dimensional zigzag
pattern).
 -- Jonathan Vos Post
[wondering if my son's earlier intense exposure to Duplo and Lego led him
towards his double B.S. in Math and Computer Science]
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From dean at math.ucdavis.edu  Wed Nov  1 21:12:15 2006
From: dean at math.ucdavis.edu (Dean Hickerson)
Date: Wed, 1 Nov 2006 12:12:15 -0800
Subject: Re^2: Error-finding program
Message-ID: <20061101201215.GA2388@math.ucdavis.edu>

Ralf Stephan wrote:

> My guess is that Clark started the same algorithm with 1, 2 and then
> tried it with starting values 1, 3, so the definition of A026469 should 
> be amended with ", starting with 1, 3".

Actually, A026469 should just say "starting with 1", since the 3 follows
from the rest of the definition.  A026468 should say "starting with 1,2",
since the 2 is artificially added to the sequence.

Dean Hickerson
dean at math.ucdavis.edu





From franktaw at netscape.net  Thu Nov  2 02:04:26 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 01 Nov 2006 20:04:26 -0500
Subject: What is the most famous sequence?
In-Reply-To: <200611010746.AA12650352@TanyaKhovanova.com>
References: <200611010746.AA12650352@TanyaKhovanova.com>
Message-ID: <8C8CC31F9FF2FAD-370-91EC@WEBMAIL-MA05.sysops.aol.com>

I would have to say that the most famous sequence is the primes.

To get the most referenced inside the OEIS, just do a search that 
matches
everything (or almost everything) - I tend to use "-keyword:dead".  Be 
sure
you're sorting by references (for many searches, such as this one, the 
default
sort by relevance works, too).  You'll see the sequences sorted by how 
many
references there are.

The top ten are (with references as of 11/1/06 6 PM CST):

1666 A000040 Primes
1305 A000045 Fibonacci
1177 A000108 Catalan
1009 A000010 Totient function
 883 A000217 Triangular
 870 A000005 Number of Divisors
 808 A000203 Sum of Divisors
 674 A002275 Repunits
 616 A001358 Semiprimes
 594 A000142 Factorials

This tends to under-estimate the importance of sequences like the 
squares
and the powers of 2.  If somebody enters a formula for a sequence that
uses the totient function, they are likely to include a reference to 
A000010.
Adding a formula which squares an integer will not cause people to add 
a
reference to the squares, however.

Incidently, here are some searches, intended to identify how popular 
certain
topics are in the OEIS, with the number of sequences matched:

25267 Prime
17853 keyword:base
10904 Digit -maple:digit
 7087 Square (but this covers a lot more than just the squares)
 6384 Binomial
 5548 Power
 4904 Divisor
 4051 Polynomial
 3901 Keyword:cons
 3722 Transform
 3698 Partition
 3638 Factor
 3526 Keyword:cofr
 3448 Permutation
 3326 Pi
 3325 Fibonacci
 3308 Matrix OR Matrices (search each separately, then subtract number
        where both occur)
 3138 Group
 2601 Phi (includes both golden ratio and totient function)
 2238 Sigma
 2189 Graph
 2116 Composite
 1971 Tree
 1553 Keyword:frac
 1540 Vector
 1549 Catalan
 1511 Reverse
 1197 Factorial (but I can't search for "!" operator)
 1165 Modulo OR Modulus
 1159 Palindrome
 1051 Perfect
 1034 Semiprime

Obviously, many of these searches do not perfectly identify the 
intended
subject.

Franklin T. Adams-Watters


-----Original Message-----
From: tanyakh at TanyaKhovanova.com

What is the most famous sequence? The most basic (natural numbers?)? 
The most
simple (0,0,0, ... or 1,1,1, ...?)? The most versatile (Catalan 
numbers?)?

I would be curious to know your opinion.

I tried to check the number of Google hits for some sequence numbers. 
Here is
what I've got:
A000040 (prime numbers) 68,800
A000045 (Fibonacci)     66,000
A000108 (Catalan)       62,500
A000027 (natural)       47,500

Can someone write a program to find top ten? What is the most 
referenced
sequence inside Encyclopedia? Can we find the most hits outside 
Encyclopedia? Is
it easy to do?

Tanya

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From arndt at jjj.de  Thu Nov  2 02:10:02 2006
From: arndt at jjj.de (Joerg Arndt)
Date: Thu, 2 Nov 2006 12:10:02 +1100
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
Message-ID: <20061102011002.GC6041@cochise.anu.edu.au>

* Jonathan Post <jvospost3 at gmail.com> [Nov 01. 2006 10:14]:
> Eisenstein-Fibonacci sequences.  Cube root of unity
> analogues of square-root of unity A014291  Imaginary
> Rabbits.
> 
> Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
> where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
> = (-1 - i*sqrt(3))/2.
> 
> [...]
> The coefficients of 1 = a(n) = 0, 0, 1, 1, 3, 1, 7, 4,
> 14, 17, 28, 54, 66, 143, 182, 350, ...
> 
> which does not seem to be in OEIS.

Ralf Stephan's ggf() says that the OGF may be:

? g=ggf([1, 1, 3, 1, 7, 4,14, 17, 28, 54, 66, 143, 182, 350])
(x^5 - 3*x^3 + x + 1)/(-x^6 + 3*x^4 - x^3 - 3*x^2 + 1)
? factor(g)

[x - 1 1]

[x^4 + x^3 - 2*x^2 - 2*x - 1 1]

[x^2 + x - 1 -1]

[x^4 - x^3 - x^2 + x + 1 -1]


> [...]







From eclark at math.usf.edu  Thu Nov  2 03:28:27 2006
From: eclark at math.usf.edu (Edwin Clark)
Date: Wed, 1 Nov 2006 21:28:27 -0500 (EST)
Subject: What is the most famous sequence?
In-Reply-To: <200611010746.AA12650352@TanyaKhovanova.com>
References: <200611010746.AA12650352@TanyaKhovanova.com>
Message-ID: <Pine.LNX.4.58.0611012122360.14271@shell.cas.usf.edu>


> What is the most famous sequence?

Since the novel and movie "The Da Vince Code"-- not to mention 
Knuth's "Art of Computer Programming" it would have to be the 
Fibonacci numbers.







From zakseidov at yahoo.com  Thu Nov  2 07:12:28 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 1 Nov 2006 22:12:28 -0800 (PST)
Subject: A124450: no autoreply from OEIS
Message-ID: <20061102061228.7046.qmail@web38206.mail.mud.yahoo.com>

Just submitted:
A124450 (no autoreply from OEIS).
Hope someone may wish to add more terms:
Mathematica isn't the best in such calcs.
Thanks, Zak

%I A124450
%S A124450
5,47,491,4919,49877,499943,4999913,49999757,499999931,4999999937,49999999811,499999999769,4999999998431,49999999999619,499999999999769,4999999999998557,49999999999998887,499999999999999679,4999999999999999661,49999999999999998647
%N A124450 Lesser of pair of closest primes summed to
10^n.
%C A124450 Cf. A065577 Number of Goldbach partitions
of 10^n.
%F A124450 10^n - a(n) is prime.
%e A124450 10^1=5+5; 10^2=47+53; 10^3=491+509;
10^4=4919+5081; 10^5=49877=50123; 10^6=499943+500057;
10^7=4999913+5000087; 10^8=49999757+50000243;
10^9=499999931+500000069;
10^10=4999999937+5000000063}, etc.
%Y A124450 A065577
%O A124450 1
%K A124450 ,nonn,
%A A124450 Zak Seidov  (zakseidov at yahoo.com), Nov 02 2006


 
____________________________________________________________________________________
Everyone is raving about the all-new Yahoo! Mail 
(http://advision.webevents.yahoo.com/mailbeta/)






From zakseidov at yahoo.com  Thu Nov  2 08:12:42 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 1 Nov 2006 23:12:42 -0800 (PST)
Subject: SEQ FROM Zak Seidov 
In-Reply-To: <200611020706.kA276utj016855@exspider.research.att.com>
Message-ID: <20061102071242.66437.qmail@web38208.mail.mud.yahoo.com>

Neil,
in contrast,
in this case I got immediate autoreply from OEIS.
Does submitting with ascribed or non-ascribed 
sequence number matter?
Zak 
--- The On-Line Encyclopedia of Integer Sequences
<oeis at research.att.com> wrote:

> The following is a copy of the email message that
> was sent to njas
> containing the sequence you submitted.
> 
> All greater than and less than signs have been
> replaced by their html
> equivalents.  They will be changed back when the
> message is processed.
> 
> This copy is just for your records.  No reply is
> expected.
>  Subject: NEW SEQUENCE FROM Zak Seidov 
> 
> 
> %I A000001
> %S A000001
>
5,,3,59,11,17,29,11,71,71,23,11,29,29,11,83,3,11,281,11,101,71,23,257,401,293,107,293,53,11,113,251,47,587,23,179,389,59,173,17,1427,83,431,53,563,593,41,47,239,383,431,1181,701,971,149,593,569,149,191,1973
> %N A000001 Lesser of pair of farsest primes summed
> to 10^n.
> %C A000001 Cf. A065577 Number of Goldbach partitions
> of 10^n, A124450 Lesser of pair of closest primes
> summed to 10^n.
> %F A000001 10^n - a(n) is prime.
> %e A000001 10^1 = 5 + 5, 10^2 = 3 + 97, 10^3 = 3 +
> 997, 10^4 = 59 + 9941, 10^5 = 11 + 99989, 10^6 = 17
> + 999983, 10^7 = 29 + 9999971, 10^8 = 11 + 99999989,
> 10^9 = 
> 71 + 999999929, 10^10 = 23 + 9999999977, etc.
> %Y A000001 A065577, A124450.
> %O A000001 1
> %K A000001 ,nonn,
> %A A000001 Zak Seidov  (zakseidov at yahoo.com), Nov 02
> 2006
> RH 
> RA 192.20.225.32
> RU 
> RI 
> 



 
____________________________________________________________________________________
Access over 1 million songs - Yahoo! Music Unlimited 
(http://music.yahoo.com/unlimited)






From jvospost3 at gmail.com  Thu Nov  2 08:44:39 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 1 Nov 2006 23:44:39 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <20061102011002.GC6041@cochise.anu.edu.au>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <20061102011002.GC6041@cochise.anu.edu.au>
Message-ID: <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>

Thank you, Joerg.  That was interesting.  Does your generating function
hypothesis hold when I extend the sequence by another 5 values, to a(n) = 0,
0, 1, 1, 3, 1, 7, 4,
14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again, assuming
that I got the elementary algebra and arithmetic correct.

If so, I should submit this sequence as by the two us!

Eisenstein-Fibonacci sequences.  Cube root of unity analogues of square-root
of unity A014291  Imaginary Rabbits.

[update of 1 Nov 06]

Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
= (-1 - i*sqrt(3))/2.

There are various integer sequences derived from this
complex sequence (such as the real part, the imaginary
part, the polynomial coefficient is of 1, w, and
w^2)Unless my arithmetic by hand is in error again,
is:

n   b(n)
0   w
1   w^2
2   w(w^2) + w = w^3 + w = 1 + w.
3   w(1 + w) + w^2 = w + w^2 + w^2 = 1 + 2w^2.
4   w(1 + 2w^2) + 1 + w = w + 2w^3 + 1 + w = 3 + 2w.
5   w(3 + 2w) + 1 + 2w^2 = 3w + 2w^2 + 1 + 2w^2.
   = 1 + 3w + 4w^2.
6  w(1 + 3w + 4w^2) + 3 + 2w = w + 3w^2 + 4w^3 + 3 +
2w
   = 7 + 3w + 3w^2.
7  w(7 + 3w + 3w^2) + 1 + 3w + 4w^2 = 7w + 3w^2 + 3w^3

  + 1 + 3w + 4w^2 = 4 + 10w + 7w^2.
8  w(4 + 10w + 7w^2) + 7 + 3w + 3w^2 = 4w + 10w^2 +
  7w^3 + 7 + 3w + 3w^2 = 14 + 7w + 13w^2.
9  w(14 + 7w + 13w^2) + 4 + 10w + 7w^2 = 14w + 7w^2 +
  13w^3 + 4 + 10w + 7w^2 = 17 + 24w + 14w^2.
10 w(17 + 24w + 14w^2) + 14 + 7w + 13w^2 = 17w + 24w^2
  + 14w^3 + 14 + 7w + 13w^2 = 28 + 24w + 37w^2.
11 w(28 + 24w + 37w^2) + 17 + 24w + 14w^2 = 28w +
  24w^2 + 37w^3 + 17 + 24w + 14w^2 = 54 + 52w +
38w^2.
12 w(54 + 52w + 38w^2) + 28 + 24w + 37w^2 = 54w +
  52w^2 + 38w^3 + 28 + 24w + 37w^2 = 66 + 78w +
89w^2.
13 w(66 + 78w + 89w^2) + 54 + 52w + 38w^2 =
  66w + 78w^2 + 89w^3 + 54 + 52w + 38w^2 =
  143 + 118w + 116w^2.
14 w(143 + 118w + 116w^2) + 66 + 78w + 89w^2 =
  143w + 118w^2 + 116w^3 + 66 + 78w + 89w^2 =
  182 + 221w + 207w^2.
15 w(182 + 221w + 207w^2) + 143 + 118w + 116w^2 =
  182w + 221w^2 + 207w^3 + 143 + 118w + 116w^2 =
  350 + 300w + 337w^2.

16 w(350 + 300w + 337w^2) + 350 + 300w + 337w^2 =
    350w + 300w^2 + 337w^3 + 350 + 300w + 337w^2 =
    687 +  650w + 637w^2.

17 w(687 +  650w + 637w^2) + 350 + 300w + 337w^2 =
    687w +  650w^2 + 637w^3 + 350 + 300w + 337w^2 =
    987 + 987w + 987w^2.

18  w(987 + 987w + 987w^2) + 687 +  650w + 637w^2 =
    987w + 987w^2 + 987w^3 + 687 +  650w + 637w^2 =
    1674 + 1637w + 1624w^2.

19  w(1674 + 1637w + 1624w^2) + 987 + 987w + 987w^2 =
     1674w + 1637w^2 + 1624w^3 + 987 + 987w + 987w^2
=
     2611 + 2661w + 2624w^2.

20  w(2611 + 2661w + 2624w^2) + 1674 + 1637w + 1624w^2
=
     2611w + 2661w^2 + 2624w^3 + 1674 + 1637w +
1624w^2 =
     4298 + 4248w + 4285w^2.


The coefficients of 1 = a(n) = 0, 0, 1, 1, 3, 1, 7, 4,
14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611,
4298

which does not seem to be in OEIS.

To excerpt my comment in A107890 (and some other
seqs):

"Eisenstein integers are of the form a + b*omega,
where a and b are ordinary integers, and omega = (-1 +
i*sqrt(3))/2 is a cube root of 1, the other cube roots
of 1 being 1 and omega^2 = (-1 - i*sqrt(3))/2.
Eisenstein integers are complex numbers that are also
members of the imaginary quadratic field Q(sqrt -3) =
Z[omega]. The sums, differences, and products of
Eisenstein integers are other Eisenstein integers."

Other sequences come from:
b(0) = w, b(1) = w^2, b(n) = b(n-1) + w*b(n-2),
and from the O, J notation of D?rrie (1965) in
Weisstein, Eric W. "Eisenstein Integer." From
MathWorld--A Wolfram Web Resource.
http://mathworld.wolfram.com/EisensteinInteger.html


Eisenstein-Fib.doc

=== end 2nd email ===

On 11/1/06, Joerg Arndt <arndt at jjj.de> wrote:
>
> * Jonathan Post <jvospost3 at gmail.com> [Nov 01. 2006 10:14]:
> > Eisenstein-Fibonacci sequences.  Cube root of unity
> > analogues of square-root of unity A014291  Imaginary
> > Rabbits.
> >
> > Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
> > where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
> > = (-1 - i*sqrt(3))/2.
> >
> > [...]
> > The coefficients of 1 = a(n) = 0, 0, 1, 1, 3, 1, 7, 4,
> > 14, 17, 28, 54, 66, 143, 182, 350, ...
> >
> > which does not seem to be in OEIS.
>
> Ralf Stephan's ggf() says that the OGF may be:
>
> ? g=ggf([1, 1, 3, 1, 7, 4,14, 17, 28, 54, 66, 143, 182, 350])
> (x^5 - 3*x^3 + x + 1)/(-x^6 + 3*x^4 - x^3 - 3*x^2 + 1)
> ? factor(g)
>
> [x - 1 1]
>
> [x^4 + x^3 - 2*x^2 - 2*x - 1 1]
>
> [x^2 + x - 1 -1]
>
> [x^4 - x^3 - x^2 + x + 1 -1]
>
>
> > [...]
>
>
>
-------------- next part --------------
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From maxale at gmail.com  Thu Nov  2 09:23:51 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 2 Nov 2006 00:23:51 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
Message-ID: <d3dac270611020023t17f583aep351f8e1e187148cc@mail.gmail.com>

On 10/22/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> Eisenstein-Fibonacci sequences.  Cube root of unity
>  analogues of square-root of unity A014291  Imaginary
>  Rabbits.
>
>  Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
>  where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
>  = (-1 - i*sqrt(3))/2.

First off, using the identity w^2 = -w-1, we can uniquely represent
b(n) as u(n)+v(n)*w where u(n) and v(n) are integers.
In particular, u(0)=0, v(0)=1, u(1)=-1, v(1)=-1.

Now the recurrence relation b(n) = w*b(n-1) + b(n-2) implies that
u(n) = u(n-2) - v(n-1)
v(n) = v(n-2) + u(n-1) - v(n-1)
that in turn implies that
[u(n-1),v(n-1),u(n),v(n)] = M*[u(n-2),v(n-2),u(n-1),v(n-1)]
where vectors are column-vectors and the matrix M is defined as
[ 0, 0, 1, 0 ]
[ 0, 0, 0, 1 ]
[ 1, 0, 0, -1 ]
[ 0, 1, 1, -1 ]

The characteristic polynomial of the matrix M is
x^4 + x^3 - x^2 - x + 1
meaning that u(n) and v(n) satisfies the recurrence
u(n) = - u(n-1) + u(n-2) + u(n-3) - u(n-4)
v(n) = - v(n-1) + v(n-2) + v(n-3) - v(n-4)
with u(0)=0, u(1)=-1, u(2)=1, u(3)=-2
and v(0)=1, v(1)=-1, v(2)=1, v(3)=-1.

>  There are various integer sequences derived from this
>  complex sequence (such as the real part, the imaginary
>  part, the polynomial coefficient is of 1, w, and
>  w^2)Unless my arithmetic by hand is in error again,
>  is:
>
>  n   b(n)
>  0   w
>  1   w^2
>  2   w(w^2) + w = w^3 + w = 1 + w.
>  3   w(1 + w) + w^2 = w + w^2 + w^2 = 1 + 2w^2.

It is not clear why you don't reduce the "polynomials" of w to the first degree.
Higher degree "polynomials" are not uniquely defined.
E.g., b(1)=w^2=2w^2+w+1=3w^2+2w+2=...
but if you reduce to the first degree "polynomial", there is an unique
representation:
b(1) = -w-1.

[...]

>  The coefficients of 1 = a(n) = 0, 0, 1, 1, 3, 1, 7, 4,
>  14, 17, 28, 54, 66, 143, 182, 350, ...

This sequence is not well-defined (see above).
Probably, what you are interested in is u(n):
0, -1, 1, -2, 2, -2, 1, 1, -4, 8, -12, 15, -15, 10, 2, -22, 49, -79, 104, -112

Max





From maxale at gmail.com  Thu Nov  2 13:06:51 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 2 Nov 2006 04:06:51 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <d3dac270611020023t17f583aep351f8e1e187148cc@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <d3dac270611020023t17f583aep351f8e1e187148cc@mail.gmail.com>
Message-ID: <d3dac270611020406l10f3deb8jb69c41cb2b267651@mail.gmail.com>

On 11/2/06, Max A. <maxale at gmail.com> wrote:

> >  Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
> >  where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
> >  = (-1 - i*sqrt(3))/2.
>
> First off, using the identity w^2 = -w-1, we can uniquely represent
> b(n) as u(n)+v(n)*w where u(n) and v(n) are integers.

[...]

> u(n) and v(n) satisfies the recurrence
> u(n) = - u(n-1) + u(n-2) + u(n-3) - u(n-4)
> v(n) = - v(n-1) + v(n-2) + v(n-3) - v(n-4)
> with u(0)=0, u(1)=-1, u(2)=1, u(3)=-2
> and v(0)=1, v(1)=-1, v(2)=1, v(3)=-1.

o.g.f. for u(n) is
-x/(1 + x - x^2 - x^3 + x^4)
and
o.g.f. for v(n) is
(1-x^2)/(1 + x - x^2 - x^3 + x^4)

Hence, o.g.f. for b(n) is
(-x + w*(1-x^2)) / (1 + x - x^2 - x^3 + x^4)

Max





From njas at research.att.com  Thu Nov  2 14:24:15 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Thu, 2 Nov 2006 08:24:15 -0500 (EST)
Subject: SEQ FROM Zak Seidov
Message-ID: <200611021324.IAA74292@fry.research.att.com>


Does submitting with ascribed or non-ascribed 
sequence number matter?
Zak 


Me:  I don't know.  But I see you are still using yahoo!
Why not switch to gmail?  Then you won't have these problems.

Best

Neil





From mathar at strw.leidenuniv.nl  Thu Nov  2 14:40:51 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Thu, 2 Nov 2006 14:40:51 +0100
Subject: Re^2:  Eisenstein-Fibonacci sequences
Message-ID: <200611021340.kA2DepPv014946@amer.strw.leidenuniv.nl>


ma> From seqfan-owner at ext.jussieu.fr  Thu Nov  2 13:12:31 2006
ma> Return-Path: <seqfan-owner at ext.jussieu.fr>
ma> Date: Thu, 2 Nov 2006 04:06:51 -0800
ma> From: "Max A." <maxale at gmail.com>
ma> To: "Jonathan Post" <jvospost3 at gmail.com>
ma> Subject: Re: Eisenstein-Fibonacci sequences
ma> Cc: "Sequence Fans Mailing List" <seqfan at ext.jussieu.fr>
ma> 
ma> On 11/2/06, Max A. <maxale at gmail.com> wrote:
ma> 
ma> > >  Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
ma> > >  where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
ma> > >  = (-1 - i*sqrt(3))/2.
ma> >
ma> > First off, using the identity w^2 = -w-1, we can uniquely represent
ma> > b(n) as u(n)+v(n)*w where u(n) and v(n) are integers.
ma> 
ma> [...]
ma> 
ma> > u(n) and v(n) satisfies the recurrence
ma> > u(n) = - u(n-1) + u(n-2) + u(n-3) - u(n-4)
ma> > v(n) = - v(n-1) + v(n-2) + v(n-3) - v(n-4)
ma> > with u(0)=0, u(1)=-1, u(2)=1, u(3)=-2
ma> > and v(0)=1, v(1)=-1, v(2)=1, v(3)=-1.
ma> 
ma> o.g.f. for u(n) is
ma> -x/(1 + x - x^2 - x^3 + x^4)
ma> and
ma> o.g.f. for v(n) is
ma> (1-x^2)/(1 + x - x^2 - x^3 + x^4)
ma> 
ma> Hence, o.g.f. for b(n) is
ma> (-x + w*(1-x^2)) / (1 + x - x^2 - x^3 + x^4)
ma> 
ma> Max

The g.f. above [-x+w(1-x^2)]/[1+x-x^2-x^3+x^4] generates
the reduced  polynomial
                          2             3      4            5               6
w + (-1 - w) x + (1 + w) x  + (-2 - w) x  + 2 x  + (w - 2) x  + (-3 w + 1) x

                  7               8              9      10
     + (1 + 5 w) x  + (-4 - 7 w) x  + (8 + 8 w) x  + O(x  )

If one doesn't reduce the polynomial, one gets the simpler
g(x)=w/[1-w*x-x^2] which generates the non-reduced and more generic
b(0)+b(1) x+ b(2) x^2 + b(3) x^3+ ...
=

     2       3       2     4      2   3     5      3       4
w + w  x + (w  + w) x  + (w  + 2 w ) x  + (w  + 3 w  + w) x  +

      6      4      2   5     7      5      3       6
    (w  + 4 w  + 3 w ) x  + (w  + 5 w  + 6 w  + w) x  +

      8      6       4      2   7     9      7       5       3       8
    (w  + 6 w  + 10 w  + 4 w ) x  + (w  + 7 w  + 15 w  + 10 w  + w) x  +

      10      8       6       4      2   9      10
    (w   + 8 w  + 21 w  + 20 w  + 5 w ) x  + O(x  )

and which becomes (apart from an offset) A000045 for w=1, A052542 for w=2,
A052906 for w=3, A000007 for w=0, and signed variations of these for w=-1,-2 or -3..
The coefficients table of the non-reduced g.f. is in A102426, which is
essentially the same as A098925 and A092865.

R. Mathar





From tanyakh at TanyaKhovanova.com  Thu Nov  2 16:15:08 2006
From: tanyakh at TanyaKhovanova.com (Tanya Khovanova)
Date: Thu,  2 Nov 2006 07:15:08 -0800
Subject: What is the most famous sequence?
Message-ID: <200611020715.AA2085028026@TanyaKhovanova.com>

Would this mean that Fibonacci sequence is the most popularized?

---------- Original Message ----------------------------------
From: Edwin Clark <eclark at math.usf.edu>
Date:  Wed, 1 Nov 2006 21:28:27 -0500 (EST)

>
>> What is the most famous sequence?
>
>Since the novel and movie "The Da Vince Code"-- not to mention 
>Knuth's "Art of Computer Programming" it would have to be the 
>Fibonacci numbers.
>
>
>


___________________________________________________________
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From davidwwilson at comcast.net  Thu Nov  2 16:46:38 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 2 Nov 2006 10:46:38 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
References: <20061101054313.25333.qmail@web38210.mail.mud.yahoo.com>
Message-ID: <008e01c6fe96$19f1bbe0$6401a8c0@yourxhtr8hvc4p>

I wrote a C++ program using sieve method and calculated

a(7) = 38807
a(8) = 291400
a(9) = 2274205
a(10) = 18200488

a(7), a(8) and a(9) agree with previous calculations.

However, Zak Seidov gave

However, my a(10) = 18200487.

So now I have to go root out the discrepancy. I am going to have my computer 
generate a file of all the partitions of 10^10 and do an independent check 
of the results, and hopefully uncover a problem in my program. 






From rsc at swtch.com  Thu Nov  2 17:26:57 2006
From: rsc at swtch.com (Russ Cox)
Date: Thu, 2 Nov 2006 11:26:57 -0500
Subject: SEQ FROM Zak Seidov
In-Reply-To: <20061102071242.66437.qmail@web38208.mail.mud.yahoo.com>
References: <200611020706.kA276utj016855@exspider.research.att.com>
	 <20061102071242.66437.qmail@web38208.mail.mud.yahoo.com>
Message-ID: <ee9e417a0611020826h1daaac8xa85b9f6cc4b0ebdf@mail.gmail.com>

As I said in my earlier mail, the semantics of the
Yahoo "spam protection" are such that if you
submit two sequences within some time period
(maybe 12 hours to a day), the first might get
delayed many hours while the second will go through
immediately.

Out of curiosity, do you have oeis at research.att.com
in your address book?

Russ





From davidwwilson at comcast.net  Thu Nov  2 17:29:52 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 2 Nov 2006 11:29:52 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
Message-ID: <001101c6fe9c$205d0c40$6401a8c0@yourxhtr8hvc4p>

Ok, that last message was incoherent.

For A065577 Zak Seidov got a(10) = 18200488, I got a(10) = 18200487.  I am 
checking my program. 






From pxp at rogers.com  Thu Nov  2 18:36:24 2006
From: pxp at rogers.com (Hans Havermann)
Date: Thu, 2 Nov 2006 12:36:24 -0500
Subject: A124450
In-Reply-To: <20061102061228.7046.qmail@web38206.mail.mud.yahoo.com>
References: <20061102061228.7046.qmail@web38206.mail.mud.yahoo.com>
Message-ID: <D7CB7013-474C-4B78-B336-E5DBA449F167@rogers.com>

zak seidov:

> Hope someone may wish to add more terms:
> Mathematica isn't the best in such calcs.

> %S A124450
> 5,47,491,4919,49877,499943,4999913,49999757,499999931,4999999937,49999 
> 999811,499999999769,4999999998431,49999999999619,499999999999769,49999 
> 99999998557,49999999999998887,499999999999999679,4999999999999999661,4 
> 9999999999999998647
> %N A124450 Lesser of pair of closest primes summed to 10^n.

I think you overstate the Mathematica difficulty. With no effort at  
any sort of efficiency in the programming, it still only takes me 8  
seconds to generate 50 terms.

Table[h =10^n/2; c=0; While[PrimeQ[h-c]==False || PrimeQ[h+c]==False,  
c++]; h-c, {n, 1, 50}]

{5, 47, 491, 4919, 49877, 499943, 4999913, 49999757, 499999931,  
4999999937, 49999999811, 499999999769, 4999999998431, 49999999999619,  
499999999999769, 4999999999998557, 49999999999998887,  
499999999999999679, 4999999999999999661, 49999999999999998647,  
499999999999999999637, 4999999999999999999481,  
49999999999999999998677, 499999999999999999998497,  
4999999999999999999999259, 49999999999999999999998779,  
499999999999999999999999043, 4999999999999999999999998947,  
49999999999999999999999999661, 499999999999999999999999994069,  
4999999999999999999999999997879, 49999999999999999999999999997699,  
499999999999999999999999999997969,  
4999999999999999999999999999995227,  
49999999999999999999999999999995263,  
499999999999999999999999999999989719,  
4999999999999999999999999999999998683,  
49999999999999999999999999999999999871,  
499999999999999999999999999999999996127,  
4999999999999999999999999999999999998557,  
49999999999999999999999999999999999999613,  
499999999999999999999999999999999999988231,  
4999999999999999999999999999999999999991729,  
49999999999999999999999999999999999999994663,  
499999999999999999999999999999999999999997117,  
4999999999999999999999999999999999999999992863,  
49999999999999999999999999999999999999999991807,  
499999999999999999999999999999999999999999991507,  
4999999999999999999999999999999999999999999993199,  
49999999999999999999999999999999999999999999997081}

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From davidwwilson at comcast.net  Thu Nov  2 18:45:17 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 2 Nov 2006 12:45:17 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
References: <20061102171720.72610.qmail@web38203.mail.mud.yahoo.com>
Message-ID: <002b01c6fea6$a96fcae0$6401a8c0@yourxhtr8hvc4p>

Well let's get it right this time

I got a(10) = 18200488, in agreement with Dick Mathar's value.





From jvospost3 at gmail.com  Thu Nov  2 19:01:55 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 2 Nov 2006 10:01:55 -0800
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
In-Reply-To: <002b01c6fea6$a96fcae0$6401a8c0@yourxhtr8hvc4p>
References: <20061102171720.72610.qmail@web38203.mail.mud.yahoo.com>
	 <002b01c6fea6$a96fcae0$6401a8c0@yourxhtr8hvc4p>
Message-ID: <5542af940611021001x27e8951akc9b931928d84333e@mail.gmail.com>

Amusing coincidence that the digits of a(10) = 18200488 consist entirely of
0 and powers of 2.  [base].

-- Jonathan Vos Post

On 11/2/06, David Wilson <davidwwilson at comcast.net> wrote:
>
> Well let's get it right this time
>
> I got a(10) = 18200488, in agreement with Dick Mathar's value.
>
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From jvospost3 at gmail.com  Thu Nov  2 21:54:44 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 2 Nov 2006 12:54:44 -0800
Subject: New & old LEGO seqs; production cuts
In-Reply-To: <C16FD8A8.20890%eilers@math.ku.dk>
References: <5542af940611010958me17f347w3bf2d2f2d6c9dec0@mail.gmail.com>
	 <C16FD8A8.20890%eilers@math.ku.dk>
Message-ID: <5542af940611021254w14b12a38y14021854e772de77@mail.gmail.com>

Dear Soren,

Thank you for a clear and thoughful answer.

My son (in the room with me now) agrees that baby/childhood play with Duplo
and Lego guided him towards his Math/Computer Science expertise today.

He and I agree, however, that it's a shame that the manufacturer doesn't
make 4-D Lego blocks, or, more properly, the ones that they do make are of
extension along the t-axis only in the trivial way, and cannot be freely
rotated in Minkowski space.

So, I wonder, how well does your approach to counting under symmetry work if
extended to an additional spacial dimension?

I suppose that we can consider an AxBxC 4-D Lego to be actually an AxBxCx1
Lego in Z^4. I think that adjacency is well-defined, all rotations are about
planes, and there is a relationship to 4-D analogues of polyominoes, i.e. to
polyhypercubes.

Best,

Jonathan Vos Post

On 11/2/06, S?ren Eilers <eilers at math.ku.dk> wrote:
>
>  Dear Jonathan
>
> Thank you, and yes, I think that LEGO inspires kids to learn about
> fundamental mathematical concepts such as symmetry.
>
> I've also considered adding crossrefs to the sequences you mentioned but
> decided against it.
>
> Regarding A007576: Watson definitely counts a subset of what we count in
> A123818, but as you can see in his paper he only looks at "planar" buildings
> where all blocks could be placed in, say, [0,2]xRxR. This means that one of
> the dimensions of the blocks used is irrelevant, and Watson could just have
> well have counted what we call FLAT structures with 1x2 blocks. Thus, IF one
> was to crossref I would suggest to do it to A123764, but perhaps this is a
> little hard to explain in the limited space of an OEIS entry.
>
> But Watson has even more restrictions; he looks only at MAX HEIGHT
> buildings with one block at each level. This is exactly what LEGO did wrong
> when they claimed that A123828(6)=102981500 [the number they wanted was
> (46^5+2^5)/2, but they got the last digit wrong too].
>
> Of course counting FLAT and MAX HEIGHT structures with 2x2 or 1x2 blocks
> is easy, the formula is 3^(n-1) or (3^(n-1)+1)/2 depending on whether or not
> you want to identify up to symmetry. Watson knows this and tries to count
> how many of these structures would tip over if placed on a flat surface,
> essentially computing the x-coordinate of the center of gravity. So I think
> the kind of LEGO counting done by Watson is quite different from the kind of
> LEGO counting we do.
>
>     {Btw, A007576 is indeed the same as A086821}
>     {If somebody was interested I could probably quite easily compute how
> many of "our" LEGO buildings are stable in the sense of Watson}
>
> Regarding Zeilberger's LEGO counts, they all relate, as far as I recall
> from a relatively hard look some time ago, to buildings with blocks of
> varying size, and hence are quite different in nature from what we do. In
> that case, n is not a number of blocks, but the total volume/weight of the
> structure, so that's a very different ball game.
>
> Best regards,
> Soren
>
>
>
>
>
> On 01/11/06 18:58, "Jonathan Post" <jvospost3 at gmail.com> wrote:
>
> I like the new seqs about Lego, such as
> A123762 <http://www.research.att.com/%7Enjas/sequences/A123762><http://www.research.att.com/%7Enjas/sequences/A123762>
>   Number of ways, counted up to symmetry, to build a contiguous building
> with n *LEGO* blocks of size 1x2.
>
> Ironically timed, as *"Recent restructuring and production cuts have left
> Lego unable to fill orders
> <http://www.cbc.ca/money/story/2006/10/31/legoproduction.html><http://www.cbc.ca/money/story/2006/10/31/legoproduction.html>for the upcoming holiday season. Affected products include Duplo bricks,
> Lego City sets, and (horror of horrors!) Star Wars and Lego Technik sets."
> * According to the article Lego stands to lose $127 million in holiday
> sales.
>
> http://www.cbc.ca/money/story/2006/10/31/legoproduction.html
>
> It might be useful to add crossrefs to earlier seqs such as:
>
> A007576 <http://www.research.att.com/%7Enjas/sequences/A007576><http://www.research.att.com/%7Enjas/sequences/A007576>
>   Number of maximally stable towers of 2 X 2 *LEGO* blocks.
>
>
> A082679 <http://www.research.att.com/%7Enjas/sequences/A082679><http://www.research.att.com/%7Enjas/sequences/A082679>
>   Number of *Lego* towers, one piece per floor, where every floor is
> perpendicular to the one below it (so we have a kind of 3-dimensional zigzag
> pattern).
>  -- Jonathan Vos Post
> [wondering if my son's earlier intense exposure to Duplo and Lego led him
> towards his double B.S. in Math and Computer Science]
>
>
>
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From davidwwilson at comcast.net  Thu Nov  2 22:49:42 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 2 Nov 2006 16:49:42 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
References: <20061102171720.72610.qmail@web38203.mail.mud.yahoo.com> <002b01c6fea6$a96fcae0$6401a8c0@yourxhtr8hvc4p>
Message-ID: <004e01c6fec8$ce34d100$6401a8c0@yourxhtr8hvc4p>

Regarding A065577:

I altered my counting program to generating a file listing, in increasing 
order, all primes p such that there exists prime q with p < q, p+q = 10^10. 
I wrote a verification program to read this file, check that the p are in 
increasing order, count the p, compute q = 10^10-p, check p < q and that p 
and q are both prime (using trial division).

The verification program has finished, and confirms that a(10) >= 18200488. 
This contradicts Zak Seidov's count a(10) = 18200487, but agrees with Dick 
Mathar's count a(10) = 18200488.

If anyone is interested, I am keeping the text file with the p values for a 
little while. It is 195MB, so e-mailing it is probably not a good option.






From zakseidov at yahoo.com  Fri Nov  3 04:09:56 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Thu, 2 Nov 2006 19:09:56 -0800 (PST)
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
In-Reply-To: <004e01c6fec8$ce34d100$6401a8c0@yourxhtr8hvc4p>
Message-ID: <20061103030956.55909.qmail@web38210.mail.mud.yahoo.com>

David, Dick, Bob, Ivars,
seqfans.

After recalculating (at two different PC's)
and rechecking I could finally 
find source of my error
(silly as usual) in one of codes,
and my final value is 
a(10) = 18200488 (NOT 18200487!)
which coincides now with the Dick-David value.

So, Bob, 
you may wish to give in A065577  the correct value 
a(10) = 18200488 with due refer to David Wilson  and
Richard Mathar (not me).
Also Ivars can extend his table.
Thanks, Zak

BTW1 Still, is it possible to find a(11)... with C++?

BIW2 Not all people even in this closed list
may be interested in Mmca specific problems,
so I''l be happy to have a list of you
particularly Mmca-involved.
Then we may circulate relevant messages
not to all seqfan community.
Please anyone interested write me personally
with subject "Me in Mmca", thanks.
Of course it's not intented to be closed list!
Hope this is not against Gerard' policy.

--- David Wilson <davidwwilson at comcast.net> wrote:

> Regarding A065577:
> 
> I altered my counting program to generating a file
> listing, in increasing 
> order, all primes p such that there exists prime q
> with p < q, p+q = 10^10. 
> I wrote a verification program to read this file,
> check that the p are in 
> increasing order, count the p, compute q = 10^10-p,
> check p < q and that p 
> and q are both prime (using trial division).
> 
> The verification program has finished, and confirms
> that a(10) >= 18200488. 
> This contradicts Zak Seidov's count a(10) =
> 18200487, but agrees with Dick 
> Mathar's count a(10) = 18200488.
> 
> If anyone is interested, I am keeping the text file
> with the p values for a 
> little while. It is 195MB, so e-mailing it is
> probably not a good option.
> 
> 



 
____________________________________________________________________________________
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From zakseidov at yahoo.com  Fri Nov  3 05:03:30 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Thu, 2 Nov 2006 20:03:30 -0800 (PST)
Subject: A123998: No primes except 3?
Message-ID: <20061103040330.2777.qmail@web38202.mail.mud.yahoo.com>

Why are there no prime numbers except of 3,
in this sequence?
Zak

A123998
={1,3,9,15,18,39,48,69,78,99,105,114,135,153,165,168,183,189,219,249,273,

288,300,303,309,330,345,363,405,414,438,468,483,498,504,534,585,618,
 639,648,699,714,729,765,804,813,828,879,933}; 
(* %N A123998 Numbers n for which 2n+1 and 4n+1 are
primes. *)



 
__________________________________________________________________________________________
Check out the New Yahoo! Mail - Fire up a more powerful email and get things done faster. 
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From maxale at gmail.com  Fri Nov  3 06:41:20 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 2 Nov 2006 21:41:20 -0800
Subject: A123998: No primes except 3?
In-Reply-To: <20061103040330.2777.qmail@web38202.mail.mud.yahoo.com>
References: <20061103040330.2777.qmail@web38202.mail.mud.yahoo.com>
Message-ID: <d3dac270611022141x4d10f69evdd706982df6e3d28@mail.gmail.com>

Note that if n = 1 mod 3 then 2n+1 is not prime (except n=1);
and if n = 2 mod 3 then 4n+1 is not prime.
Therefore, n must be a multiple of 3, except n=1.

Max

If n>1 and both 2n+1, 4n+1 are primes

On 11/2/06, zak seidov <zakseidov at yahoo.com> wrote:
> Why are there no prime numbers except of 3,
> in this sequence?
> Zak
>
> A123998
> ={1,3,9,15,18,39,48,69,78,99,105,114,135,153,165,168,183,189,219,249,273,
>
> 288,300,303,309,330,345,363,405,414,438,468,483,498,504,534,585,618,
>  639,648,699,714,729,765,804,813,828,879,933};
> (* %N A123998 Numbers n for which 2n+1 and 4n+1 are
> primes. *)
>
>
>
>
> __________________________________________________________________________________________
> Check out the New Yahoo! Mail - Fire up a more powerful email and get things done faster.
> (http://advision.webevents.yahoo.com/mailbeta)
>
>





From zakseidov at yahoo.com  Fri Nov  3 07:01:51 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Thu, 2 Nov 2006 22:01:51 -0800 (PST)
Subject: A123998: No primes except 3?
In-Reply-To: <d3dac270611022141x4d10f69evdd706982df6e3d28@mail.gmail.com>
Message-ID: <20061103060151.3808.qmail@web38207.mail.mud.yahoo.com>

Great!
It'so easy if you are Max, not Zak!
Thanks, 
Zak

--- "Max A." <maxale at gmail.com> wrote:

> Note that if n = 1 mod 3 then 2n+1 is not prime
> (except n=1);
> and if n = 2 mod 3 then 4n+1 is not prime.
> Therefore, n must be a multiple of 3, except n=1.
> 
> Max
> 
> If n>1 and both 2n+1, 4n+1 are primes
> 
> On 11/2/06, zak seidov <zakseidov at yahoo.com> wrote:
> > Why are there no prime numbers except of 3,
> > in this sequence?
> > Zak
> >
> > A123998
> >
>
={1,3,9,15,18,39,48,69,78,99,105,114,135,153,165,168,183,189,219,249,273,
> >
> >
>
288,300,303,309,330,345,363,405,414,438,468,483,498,504,534,585,618,
> >  639,648,699,714,729,765,804,813,828,879,933};
> > (* %N A123998 Numbers n for which 2n+1 and 4n+1
> are
> > primes. *)
> >
> >
> >
> >
> >
>
__________________________________________________________________________________________
> > Check out the New Yahoo! Mail - Fire up a more
> powerful email and get things done faster.
> > (http://advision.webevents.yahoo.com/mailbeta)
> >
> >
> 



 
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From jeremy.gardiner at btinternet.com  Fri Nov  3 09:59:50 2006
From: jeremy.gardiner at btinternet.com (JEREMY GARDINER)
Date: Fri, 3 Nov 2006 08:59:50 +0000 (GMT)
Subject: What is the most famous sequence?
In-Reply-To: <200611020715.AA2085028026@TanyaKhovanova.com>
Message-ID: <20061103085950.5824.qmail@web86610.mail.ird.yahoo.com>

The natural numbers must be the most widely known sequence ...

Tanya Khovanova <tanyakh at TanyaKhovanova.com> wrote:  > What is the most famous sequence?
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From maxale at gmail.com  Fri Nov  3 10:08:28 2006
From: maxale at gmail.com (Max A.)
Date: Fri, 3 Nov 2006 01:08:28 -0800
Subject: querying multidimensional sequences in OEIS
Message-ID: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>

I have a sequence indexed by vectors with non-negative integer components.
Is there a regular way to query such sequence in OEIS ?

As an example, let us consider a sequence of multinomial coefficients.
Any vector with non-negative integer components [k1, k2, ..., kn]
defines an element of this sequence:

a( [k1, k2, ..., kn] ) = (k1 + k2 +  ... + kn)! / (k1! * k2! * ... * kn!)

Assume that I know nothing about these numbers and have only a
black-box computing a(v) for any given vector v with non-negative
integer components. Can I anyhow find this sequence in OEIS and learn
that its elements called "multinomial coefficients" ?

Thanks,
Max





From wouter.meeussen at pandora.be  Fri Nov  3 12:39:40 2006
From: wouter.meeussen at pandora.be (wouter meeussen)
Date: Fri, 3 Nov 2006 12:39:40 +0100
Subject: download seqs resulting from search
Message-ID: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>

dear all,

if I search for all number triangles with row sums equal to the factorials,
I enter
"row sum" "factorial"
and get 9 pages of results in 'short' form.
(1) how can I improve on this query?
(2) how can I download them all in one go?

Wouter





From njas at research.att.com  Fri Nov  3 14:43:47 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Fri, 3 Nov 2006 08:43:47 -0500 (EST)
Subject: download seqs resulting from search
Message-ID: <200611031343.IAA51448@fry.research.att.com>

as i have said before, you can download up to 30 (or is it 50?) sequences
at once by using the old email lookup service

send email to  sequences at research.att.com  saying things like

lookup A000123
lookup A000125
lookup A012345

etc.

Neil





From franktaw at netscape.net  Fri Nov  3 16:18:26 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 10:18:26 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
Message-ID: <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>

I don't know of any standard for functions of an arbitrary finite 
sequence
of positive integers in the OEIS.

In your example, the function is commutative, which makes it 
essentially
a function of the multiset of values - or in other words, of a 
partition.  For
that there is a standard - two of them actually.  One is the 
enumeration
in Abramowitz and Stegun order, as specified in A036036, and the other
is the Mathematica ordering, as specified in A080577.

There are some relevant sequences in the database in both orders, some
only in the A&S order, and a few only in the Mma order.

Since the two orders are the same for partitions of up to 5 objects, 
one
often just search for an early part of the sequence.  If this does not 
suffice,
one should try both before giving up.

For example, the partitions of 5 objects are: [5], [4,1], [3,2], 
[3,1^2],
[2^2,1], [2,1^3], [1^5].  The multinomial coefficients for these are:

1,5,10,20,30,60,120

A search for this turns up A036038, the A&S version, and A078760, the
Mma version.

----
As for arbitrary finite sequences of positive integers, there are two 
main
candidates for an ordering: A066099 and A108244; these are the reverse
of each other.  I would probably go with A066099, as it is closest to 
the
ordering used for the partitions; but you can make an argument for 
either
one.  Possibly both should be used.

Off hand, I can't think of any interesting functions of finite 
sequences of
positive integers that depend on the order of the sequence.

Franklin T. Adams-Watters


-----Original Message-----
From: maxale at gmail.com

I have a sequence indexed by vectors with non-negative integer 
components.?
Is there a regular way to query such sequence in OEIS ??
?
As an example, let us consider a sequence of multinomial coefficients.?
Any vector with non-negative integer components [k1, k2, ..., kn]?
defines an element of this sequence:?
?
a( [k1, k2, ..., kn] ) = (k1 + k2 + ... + kn)! / (k1! * k2! * ... * 
kn!)?
?
Assume that I know nothing about these numbers and have only a?
black-box computing a(v) for any given vector v with non-negative?
integer components. Can I anyhow find this sequence in OEIS and learn?
that its elements called "multinomial coefficients" ??
?
Thanks,?
Max?


________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From rsc at swtch.com  Fri Nov  3 17:06:23 2006
From: rsc at swtch.com (Russ Cox)
Date: Fri, 3 Nov 2006 11:06:23 -0500
Subject: download seqs resulting from search
In-Reply-To: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>
References: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>
Message-ID: <ee9e417a0611030806q1490600ci75384340c6facd67@mail.gmail.com>

> if I search for all number triangles with row sums equal to the factorials,
> I enter
> "row sum" "factorial"
> and get 9 pages of results in 'short' form.
> (1) how can I improve on this query?

Not sure.

> (2) how can I download them all in one go?

Add &n=100 to the URL
and you will get 100 results per page.
Then click on "text" formatting and save the
resulting page.

http://www.research.att.com/~njas/sequences/?q=%22row+sums%22+factorial&n=100&fmt=3

Russ





From franktaw at netscape.net  Fri Nov  3 17:14:55 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 11:14:55 -0500
Subject: download seqs resulting from search
In-Reply-To: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>
References: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>
Message-ID: <8C8CD7A55C060ED-FD0-B6C7@WEBMAIL-MA20.sysops.aol.com>

Try adding "keyword:tabl" to the search.  Then (probably) repeat with 
"keyword:tabf" instead - unless you aren't interested in irregular 
tables.  This won't be all in one go, but two isn't too bad.

Of course, there are undoubtably some tables which sum to factorials 
that don't have the words "row sum" in them.  And maybe some that just 
reference A000142 instead of using the word "factorial".

Franklin T. Adams-Watters


-----Original Message-----
From: wouter.meeussen at pandora.be

dear all,

if I search for all number triangles with row sums equal to the 
factorials,
I enter
"row sum" "factorial"
and get 9 pages of results in 'short' form.
(1) how can I improve on this query?
(2) how can I download them all in one go?

Wouter

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From mlb at well.com  Fri Nov  3 17:26:14 2006
From: mlb at well.com (Marc LeBrun)
Date: Fri, 03 Nov 2006 08:26:14 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <200611031625.kA3GP1mt017249@smtp.well.com>

 >=franktaw at netscape.net

Good advice...maybe it could be made part of the tips-n-tricks section?

 > just search for an early part of the sequence

Or sometimes you can search for a small set of "unusual" later, 
larger, terms that will act as a kind of signature.

 > Off hand, I can't think of any interesting functions of finite sequences of
 > positive integers that depend on the order of the sequence.

Exponents in prime factorizations, say.







From franktaw at netscape.net  Fri Nov  3 19:31:34 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 13:31:34 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <200611031625.kA3GP1mt017249@smtp.well.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com> <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com> <200611031625.kA3GP1mt017249@smtp.well.com>
Message-ID: <8C8CD8D6CD483AD-FD0-C3EF@WEBMAIL-MA20.sysops.aol.com>

Nice suggestion.  That would be A057335.

Franklin T. Adams-Watters


-----Original Message-----
From: mlb at well.com

>=franktaw at netscape.net?
> Off hand, I can't think of any interesting functions of finite 
sequences of?
> positive integers that depend on the order of the sequence.?
?
Exponents in prime factorizations, say.?
?


________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From maxale at gmail.com  Fri Nov  3 21:26:20 2006
From: maxale at gmail.com (Max A.)
Date: Fri, 3 Nov 2006 12:26:20 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <200611031625.kA3GP1mt017249@smtp.well.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <200611031625.kA3GP1mt017249@smtp.well.com>
Message-ID: <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com>

On 11/3/06, Marc LeBrun <mlb at well.com> wrote:

>  > Off hand, I can't think of any interesting functions of finite sequences of
>  > positive integers that depend on the order of the sequence.
>
> Exponents in prime factorizations, say.

Oh, that is a very nice suggestion.
Are there any multidimensional sequences represented this way in OEIS?

I have a number of multidimensional sequences in mind that I'd like to
see in OEIS.
E.g.: the coefficients of incomplete Bell polynomials:

b( [k1, k2, ..., kn] ) = (k1 + 2*k2 + ... + n*kn)! / (k1! * k2! * ...
* kn!) / (1!^k1 * 2!^k2 * ... * n!^kn)

and looking similar the number of permutations with given cycle structure:

c( [k1, k2, ..., kn] ) = (k1 + 2*k2 + ... + n*kn)! / (k1! * k2! * ...
* kn!) / (1^k1 * 2^k2 * ... * n^kn)

Max





From mlb at well.com  Fri Nov  3 22:41:00 2006
From: mlb at well.com (Marc LeBrun)
Date: Fri, 03 Nov 2006 13:41:00 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com
 >
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
 <200611031625.kA3GP1mt017249@smtp.well.com>
 <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com>
Message-ID: <200611032139.kA3LdZSV024106@smtp.well.com>

 > I have a number of multidimensional sequences in mind that I'd like to
 > see in OEIS.  E.g.: the coefficients of incomplete Bell polynomials:

Two possible ways to submit sequences of tuples T(1),T(2),...
where T(i)=[T(i,1),T(i,2),...] are:

A. Define a traversal order by a map (i,j) <--> n and set a(n) = T(i,j).
This gives the tabl/tabf style of entry.

B. Encode each tuple as a single number, say by using it as factor
exponents:  T(i) --> 2^T(i,1) * 3^T(i,2) * 5^T(i,3) *...

For example, for Pascal's triangle

A: 1; 1, 1; 1, 2, 1; 1, 3, 3, 1;...

B: 2, 6,    90,      47250,...

However there are a variety of possible ways to do these things--for example
you might subtract 1 from all the Pascal exponents, B': 1, 1, 3, 225,...

Or traverse in the order (1,1), (2,1), (1,2), (3,1), (1,3), (2,2), (1,4),...
which recursively interleaves sequences instead of the usual diagonalizing.

But this sort of thing can also be abused to breed swarms of silly sequences,
so it should be well-motivated.

It's nicest to use familiar schemes (eg as Frank noted for partitions).







From franktaw at netscape.net  Fri Nov  3 22:41:18 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 16:41:18 -0500
Subject: Possibly more errors in OEIS?
Message-ID: <8C8CDA7EE6FDEF2-12C-B6F@WEBMAIL-MA20.sysops.aol.com>

Looking at sequences with keyword:cons, there is a marked shortage of 
sequences with a given duplicate digit (excluding 0), compared with 
those with a given other pair of digits.  For example, restricting our 
attention to sequences not of the form "Decimal expansion of 1/n", and 
ignoring all pairs involving 0 or 1, there are an average of about 1710 
sequences  with any given digit pair, but an average of only about 1627 
with duplicate pairs.  10 of this difference is explained by the 10 
"selvage" sequences, which have no duplicate digits, but the rest is 
unexplained.

I'm guessing that there may be an input error here; that duplicate 
digits sometimes get combined at some point in the process, whether 
before or after submission.  I'm not sure how to test for this, short 
of examining the "cons" sequences and determining which are correct.

Mostly for Neil,

Sequences A081780 to A081796, "continued cotangent" sequences, all have 
the "cons" keyword, but they should not.

Also, sequence A114721 has the "cons" keyword; it should instead have 
the "frac" keyword.

Franklin T. Adams-Watters

________________________________________________________________________
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From jawbrey at att.net  Fri Nov  3 22:51:00 2006
From: jawbrey at att.net (Jon Awbrey)
Date: Fri, 03 Nov 2006 16:51:00 -0500
Subject: querying multidimensional sequences in OEIS
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <200611031625.kA3GP1mt017249@smtp.well.com>
	 <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com> <200611032139.kA3LdZSV024106@smtp.well.com>
Message-ID: <454BB9C4.CCCA8C55@att.net>

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o

you might get some ideas from looking at the
tables of "primal codes" or "primal functions"
that I compiled, for instance, here:

http://stderr.org/pipermail/inquiry/2005-July/002846.html

ja

Marc LeBrun wrote:
> 
>  > I have a number of multidimensional sequences in mind that I'd like to
>  > see in OEIS.  E.g.: the coefficients of incomplete Bell polynomials:
> 
> Two possible ways to submit sequences of tuples T(1),T(2),...
> where T(i)=[T(i,1),T(i,2),...] are:
> 
> A. Define a traversal order by a map (i,j) <--> n and set a(n) = T(i,j).
> This gives the tabl/tabf style of entry.
> 
> B. Encode each tuple as a single number, say by using it as factor
> exponents:  T(i) --> 2^T(i,1) * 3^T(i,2) * 5^T(i,3) *...
> 
> For example, for Pascal's triangle
> 
> A: 1; 1, 1; 1, 2, 1; 1, 3, 3, 1;...
> 
> B: 2, 6,    90,      47250,...
> 
> However there are a variety of possible ways to do these things--for example
> you might subtract 1 from all the Pascal exponents, B': 1, 1, 3, 225,...
> 
> Or traverse in the order (1,1), (2,1), (1,2), (3,1), (1,3), (2,2), (1,4),...
> which recursively interleaves sequences instead of the usual diagonalizing.
> 
> But this sort of thing can also be abused to breed swarms of silly sequences,
> so it should be well-motivated.
> 
> It's nicest to use familiar schemes (eg as Frank noted for partitions).

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o
inquiry e-lab: http://stderr.org/pipermail/inquiry/
citizendium development forums: http://smf.citizendium.org
http://www.textop.org/wiki/index.php?title=User:Jon_Awbrey
wikinfo: http://wikinfo.org/wiki.php?title=User:Jon_Awbrey
wp review: http://wikipediareview.com/index.php?showuser=398
o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o






From maxale at gmail.com  Fri Nov  3 23:02:04 2006
From: maxale at gmail.com (Max A.)
Date: Fri, 3 Nov 2006 14:02:04 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <200611032139.kA3LdZSV024106@smtp.well.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <200611031625.kA3GP1mt017249@smtp.well.com>
	 <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com>
	 <200611032139.kA3LdZSV024106@smtp.well.com>
Message-ID: <d3dac270611031402s62ff2746p3a5c749865678cec@mail.gmail.com>

On 11/3/06, Marc LeBrun <mlb at well.com> wrote:
>  > I have a number of multidimensional sequences in mind that I'd like to
>  > see in OEIS.  E.g.: the coefficients of incomplete Bell polynomials:
>
> Two possible ways to submit sequences of tuples T(1),T(2),...
> where T(i)=[T(i,1),T(i,2),...] are:
>
> A. Define a traversal order by a map (i,j) <--> n and set a(n) = T(i,j).
> This gives the tabl/tabf style of entry.
>
> B. Encode each tuple as a single number, say by using it as factor
> exponents:  T(i) --> 2^T(i,1) * 3^T(i,2) * 5^T(i,3) *...

As I understand, A is applicable only to the cases where the dimension
of tuples is fixed while B is perfectly fine for tuples with variable
dimensions (what I'm currently interested in).

>
> For example, for Pascal's triangle
>
> A: 1; 1, 1; 1, 2, 1; 1, 3, 3, 1;...
>
> B: 2, 6,    90,      47250,...

Does B apply to Pascal triangle?
The 5-th term corresponds to the tuple [0,0,5], what it is in Pascal triangle?

Max





From franktaw at netscape.net  Fri Nov  3 23:06:02 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 17:06:02 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com> <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>

<Me>
>Off hand, I can't think of any interesting functions of finite 
sequences of
>positive integers that depend on the order of the sequence.

Of course, I no sooner wrote that than I started thinking of such 
functions.
I'm not sure any of these are tremendously interesting, but they're at 
least
somewhat so.

In the following, b(i), i = 1..k is the finite sequence.  I'm starting 
all these
sequences with the empty sequence, but I searched for them without
that first term.  All are shown here in A066099 order.

Alternating sum
Sum_i (-1)^{i+1) b(i)
0,1,2,0,3,1,-1,1,4,2,0,2,-2,0,2,0

Binomial sum
Sum_i C(k-1,i-1) b(i)
0,1,2,2,3,3,3,4,4,4,4,5,4,6,5,8

Inverse binomial sum
Sum_i (-1)^{i+1) C(k-1,i-1) b(i)
0,1,2,0,3,1,-1,0,4,2,0,1,-2,-2,1,0

Weighted sum
sum_i i b(i)
0,1,2,3,3,4,5,6,4,5,6,7,7,8,9,10
This is A029931

Zero-based weighted sum
sum_i (i-1) b(i)
0,0,0,1,0,1,2,3,0,1,2,3,3,4,5,6

Sum of products of consecutive elements
sum_{i=1}^{k-1} b(i) b(i+1)
0,0,0,1,0,2,2,2,0,3,4,3,3,4,3,3

Number of rises
sum_{b(i)>b(i-1)} 1
0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,0

Number of falls
sum_{b(i)<b(i-1)} 1
0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0

Number of unchanged
sum_{b(i)=b(i-1)} 1
0,0,0,1,0,0,0,2,0,0,1,1,0,0,1,3

Number of non-rises
sum_{b(i)<=b(i-1)} 1
0,0,0,1,0,0,1,2,0,0,1,1,1,1,2,3

Number of non-falls
sum_{b(i)>=b(i-1)} 1
0,0,0,1,0,1,0,2,0,1,1,2,0,1,1,3

Number of monotonically increasing runs
1 + number of falls, but 0 for empty sequence
0,1,1,1,1,2,1,1,1,2,1,2,1,2,1,1

Number of monotonically decreasing runs
1 + number of rises, but 0 for empty sequence
0,1,1,1,1,1,2,1,1,1,1,1,2,2,2,1

Number of runs of equal terms
1 + number of unchanged, but 0 for empty sequence
0,1,1,2,1,1,1,3,1,1,2,2,1,1,2,4

Number of distinct non-empty subsequences
0,1,1,2,1,3,3,3,1,3,2,5,3,5,5,4
E.g., f(1,1) = 2, for the sequences [1], and [1,1].

Number of distinct subsequences
1 + number of distinct non-empty subsequences
1,2,2,3,2,4,4,4,2,4,3,6,4,6,6,5

Number of set partitions
List parts of set partition by their smallest element, and count the 
part sizes.
E.g., 1|2,4|3 (also known as {1,2,3,2}) would count for the sequence 
[1,2,1].
1,1,1,1,1,2,1,1,1,3,3,3,1,2,1,1

Number of permutations
List permutations in cycle form, sorted by the smallest element in the 
cycle,
and count the cycle lengths.
Number of set partitions * Product_i (b(i)-1)!
1,1,1,1,2,2,1,1,6,6,3,3,2,2,1,1

Number of partially ordered sets (unlabelled) by rank
The rank of an element in a poset is the length of the longest chain of 
which it
is the largest element.
1,1,1,1,1,2,1,1,1,3,4,3,1,2,1,1

Number of partially ordered sets (labelled) by rank
1,1,1,2,1,9,3,6,1,28,54,60,4,36,12,24

Number of partially ordered sets (naturally labelled) by rank
Naturally labelled means labels are consistent with the ordering.
1,1,1,1,1,4,1,1,1,11,13,8,1,4,1,1

Number of forests (unlabelled, unordered) with b(i) nodes at height i.
1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,2,2,1,1,1,1,2,1,1,1,1,1

... and, of course, various other forest options.

If there is agreement that this is the right order for these sequences, 
I'll try
to find time to submit them.
________________________________________________________________________
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From tbaruchel at free.fr  Fri Nov  3 23:17:29 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Fri, 3 Nov 2006 23:17:29 +0100
Subject: [math-fun] A fast (written in assembly) library for continued
	fractions
Message-ID: <20061103221729.GB2467@>

I wrote for my own purposes the following X86 assembly code, intended
to be called from GFORTRAN. Such a thing should only work if you have
a similar configuration to mine, but anyway, it may interest you.
I wanted a fast library for computing the length in the continued
fraction expansion of sqrt(n) for any integer n that can fit in
a REAL*8 gfortran type for statistical purposes. I also included
a faster function that tells if this length is odd or even. Though
I am not an assembly expert, there should be a significant speed
improvement if you use this code rather than your own fortran
functions, since the whole process is performed on the FPU stack
with no load/store useless travels.

It may help you, it may not. You know about these kinds of software...

Instructions are at the beginning of the file, as comments.
Read them carefully. Code is below.

Regards.

------------------ FILE quad.s BEGINS BELOW --------------------

/*

  Compile your piece of code with something like
    gfortran -fno-underscoring mycode.f90 quad.s
  and add freely your own flags. Enjoy !
  Of course, if you don't want to use the -fno-underscoring
  option for some reason, you can add an underscore at the end
  of the global functions below, like that:
    .globl initfpu_
    initfpu_:
  it should work.

  Remember this piece of code has the following status:
    Seems to work for me, maybe it will also seem for you,
    maybe your computer will explode, maybe it will make the
    earth explode, maybe God will choose not to have created
    the world, think about it before using it.
  (Anyway, if you see some kind of bug, please let me know ;-)
  Remember that
    a) this is x86 assembly code;
    b) it uses the GNU AS (gas) syntax;
    c) it uses the gfortran calling conventions;
  (now you understand why you can't expect it will works ;-)

  Usage. In your program (or subroutine or whatever), declare
  the function you need like:
    
    logical :: oddperiod
    integer*8 :: period

  (the two subroutines don't need to be declared).

  Then use
    
    integer*4 :: oldfpustatus
    call initfpu(oldfpustatus)

  before any computation; if you don't do it, it won't work.
  When you don't need any more the library,
    
    call restorefpu(oldfpustatus)

  will restore the control register of the FPU.

  During all that time, you shouldn't be disturbed too much, 
  for your own mathematical operations, but
  remember the rounding mode of the FPU has been set to
  "round down" (rather than the usual "nearest").

  How does it work? The whole computation is performed on the
  stack of the FPU; only integer values are needed for these
  tasks, but they are stored as REAL*10, 80 bits reals. They
  are not converted to anything during the whole process, thus
  it should work for great integers stored in your own REAL*8,
  64 bits reals. Of course, it may happen (I think it will never,
  but didn't check) that an intermediate result can not fit in
  such a REAL*10 register; in that case, read above what is
  written about the end of the world...
  
*/



/*
  call initfpu(integer*4)
  =====================
    This subroutine should be called before any other call to the
    library; its purpose is to set the rounding control in the FPU
    control register to the "round down" mode.
    The subroutine puts in the integer variable the initial value
    of the FPU control register.

    If you want to use often this subroutine (which you shouldn't:
    speed being an issue, this subroutine is intended to be used
    once, then the functions billions of times, not the contrary ;-),
    maybe removing the "fnclex" can improve the speed, since I don't
    think it plays a great role here.
*/
.globl initfpu
initfpu:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        fnstcw  (%eax)
	fwait
        movw    (%eax), %dx
	andb	$0xf3, %dh
	orb	$0x4, %dh
        movw    %dx, 2(%eax)
	fnclex
	fldcw   2(%eax)
        fwait
        movw    $0, 2(%eax)
	leave
	ret

/*
  call restorefpu(integer*4)
  ========================
    This subroutine should be used for restoring the initial state of
    the FPU control register; the integer variable must contain the
    right value (which should have been previously set by the
    subroutine initfpu).
*/
.globl restorefpu
restorefpu:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
	fclex
	fldcw   (%eax)
	leave
	ret

/*
  sqrtint( real double precision : REAL*8 )
  =========================================
    This function returns the integer part of the square root
    of the given real. The return value has itself the real type.
*/
.global sqrtint
sqrtint:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
	fldl	(%eax)
	fsqrt
        frndint
	leave
	ret

/*
  oddperiod( real double precision : REAL*8 )
  ===========================================
    Despite the type of the real argument, the variable has to
    be some integer value stored as a double precision.
    The function has the logical type. It returns FALSE if
    the argument is a perfect square or if the continued
    fraction expansion of its square root has an even period
    length; it returns TRUE in the remaining case.
      oddperiod(dble(25)) --> FALSE
      oddperiod(dble(13)) --> TRUE
      oddperiod(dble(12)) --> FALSE
    If it may help, you can declare the function as an integer*4
    type rather than a logical one (output will be 0 or 1).

    Though it seems to work with integer*8, I am not absolutely
    sure it *always* will (unless you add the following instruction
      movl $0, %edx
    at the beginning of the code (for instance between the current
    second and third lines of code), but really why would you
    do it? Maybe something in the calling conventions is such that
    the register EDX already contains 0, but, again, who cares?

    If you read the instructions for the period() function, you
    will learn about some 2^32 limitations, but you don't have to
    care about that here: the length is not kept in any register
    during the computation and the code should work as long as the
    values in the FPU stack are the right one.
    
*/
.global oddperiod
oddperiod:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        fldln2
	fldl	(%eax)
        fld     %st
        fsqrt
        frndint
        fld     %st
        fld     %st
        fmul    %st, %st
        fld     %st(3)
        fsubp   %st, %st(1)
        fcom    %st(4)
        fstsw   %ax
        fwait
        sahf
        ja      oddperiod0
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
oddperiod2:
        movl    $0, %eax
        leave
        ret
oddperiod0:
        fld     %st(1)
        fadd    %st(3), %st
        fdiv    %st(1), %st
        frndint
        fld     %st(2)
        fld     %st(1)
        fmul    %st(3), %st
        fsubp   %st, %st(1)
        fxch    %st(3)
        fsub    %st(3), %st
        fabs
        fcomp   %st(6)
        fstsw   %ax
        fwait
        sahf
        ja      oddperiod1
        fld     %st(3)
        fadd    %st, %st
        fsubp   %st, %st(1)
        fcomp   %st(5)
        fstsw   %ax
        fwait
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        sahf
        ja      oddperiod2
        movl    $1, %eax
        leave
        ret
oddperiod1:
        ffree   %st
        fincstp
        fld     %st(1)
        fmul    %st, %st
        fld     %st(4)
        fsubp   %st, %st(1)
        fdivp   %st, %st(1)
        frndint
        jmp oddperiod0

/*
  period( real double precision : REAL*8 )
  ========================================
    Despite the type of the real argument, the variable has to
    be some integer value stored as a double precision.
    The function has integer*8 type, though the output
    can't be greater than 2^32-1 (this choice has been made
    for avoiding sign problem when the result is greater
    (or equal) then 2^31 and smaller then 2^32.
    The result is the length of period in the continued fraction
    expansion of the square root of the given integer.

    If speed is really for you an obsession, remember that it
    is highly unprobable that you use that library for periods
    between the two fatidic values. Maybe it will increase a little
    the speed to use integer*4 which allows to delete twice the line
        movl    $0,%edx
    (not the first time it appears, but the second and third one).
    If you try such a thing and see some speed improvement, please
    let me know.
*/
.global period
period:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        movl    $0, %edx
        fldln2
	fldl	(%eax)
        fld     %st
        fsqrt
        frndint
        fld     %st
        fld     %st
        fmul    %st, %st
        fld     %st(3)
        fsubp   %st, %st(1)
        fcom    %st(4)
        fstsw   %ax
        fwait
        sahf
        ja      period0
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        movl    %edx, %eax
        movl     $0, %edx
        leave
        ret
period0:
        inc     %edx
        fld     %st(1)
        fadd    %st(3), %st
        fdiv    %st(1), %st
        frndint
        fld     %st(2)
        fld     %st(1)
        fmul    %st(3), %st
        fsubp   %st, %st(1)
        fxch    %st(3)
        fsub    %st(3), %st
        fabs
        fcomp   %st(6)
        fstsw   %ax
        fwait
        sahf
        ja      period1
        fld     %st(3)
        fadd    %st, %st
        fsubp   %st, %st(1)
        fcomp   %st(5)
        fstsw   %ax
        fwait
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        sahf
        jb      period2
        add     %edx, %edx
period2:
        movl    %edx, %eax
        movl    $0,%edx
        leave
        ret
period1:
        ffree   %st
        fincstp
        fld     %st(1)
        fmul    %st, %st
        fld     %st(4)
        fsubp   %st, %st(1)
        fdivp   %st, %st(1)
        frndint
        jmp period0

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From tbaruchel at free.fr  Fri Nov  3 23:17:29 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Fri, 3 Nov 2006 23:17:29 +0100
Subject: A fast (written in assembly) library for continued fractions
Message-ID: <20061103221729.GB2467@>

I wrote for my own purposes the following X86 assembly code, intended
to be called from GFORTRAN. Such a thing should only work if you have
a similar configuration to mine, but anyway, it may interest you.
I wanted a fast library for computing the length in the continued
fraction expansion of sqrt(n) for any integer n that can fit in
a REAL*8 gfortran type for statistical purposes. I also included
a faster function that tells if this length is odd or even. Though
I am not an assembly expert, there should be a significant speed
improvement if you use this code rather than your own fortran
functions, since the whole process is performed on the FPU stack
with no load/store useless travels.

It may help you, it may not. You know about these kinds of software...

Instructions are at the beginning of the file, as comments.
Read them carefully. Code is below.

Regards.

------------------ FILE quad.s BEGINS BELOW --------------------

/*

  Compile your piece of code with something like
    gfortran -fno-underscoring mycode.f90 quad.s
  and add freely your own flags. Enjoy !
  Of course, if you don't want to use the -fno-underscoring
  option for some reason, you can add an underscore at the end
  of the global functions below, like that:
    .globl initfpu_
    initfpu_:
  it should work.

  Remember this piece of code has the following status:
    Seems to work for me, maybe it will also seem for you,
    maybe your computer will explode, maybe it will make the
    earth explode, maybe God will choose not to have created
    the world, think about it before using it.
  (Anyway, if you see some kind of bug, please let me know ;-)
  Remember that
    a) this is x86 assembly code;
    b) it uses the GNU AS (gas) syntax;
    c) it uses the gfortran calling conventions;
  (now you understand why you can't expect it will works ;-)

  Usage. In your program (or subroutine or whatever), declare
  the function you need like:
    
    logical :: oddperiod
    integer*8 :: period

  (the two subroutines don't need to be declared).

  Then use
    
    integer*4 :: oldfpustatus
    call initfpu(oldfpustatus)

  before any computation; if you don't do it, it won't work.
  When you don't need any more the library,
    
    call restorefpu(oldfpustatus)

  will restore the control register of the FPU.

  During all that time, you shouldn't be disturbed too much, 
  for your own mathematical operations, but
  remember the rounding mode of the FPU has been set to
  "round down" (rather than the usual "nearest").

  How does it work? The whole computation is performed on the
  stack of the FPU; only integer values are needed for these
  tasks, but they are stored as REAL*10, 80 bits reals. They
  are not converted to anything during the whole process, thus
  it should work for great integers stored in your own REAL*8,
  64 bits reals. Of course, it may happen (I think it will never,
  but didn't check) that an intermediate result can not fit in
  such a REAL*10 register; in that case, read above what is
  written about the end of the world...
  
*/



/*
  call initfpu(integer*4)
  =====================
    This subroutine should be called before any other call to the
    library; its purpose is to set the rounding control in the FPU
    control register to the "round down" mode.
    The subroutine puts in the integer variable the initial value
    of the FPU control register.

    If you want to use often this subroutine (which you shouldn't:
    speed being an issue, this subroutine is intended to be used
    once, then the functions billions of times, not the contrary ;-),
    maybe removing the "fnclex" can improve the speed, since I don't
    think it plays a great role here.
*/
.globl initfpu
initfpu:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        fnstcw  (%eax)
	fwait
        movw    (%eax), %dx
	andb	$0xf3, %dh
	orb	$0x4, %dh
        movw    %dx, 2(%eax)
	fnclex
	fldcw   2(%eax)
        fwait
        movw    $0, 2(%eax)
	leave
	ret

/*
  call restorefpu(integer*4)
  ========================
    This subroutine should be used for restoring the initial state of
    the FPU control register; the integer variable must contain the
    right value (which should have been previously set by the
    subroutine initfpu).
*/
.globl restorefpu
restorefpu:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
	fclex
	fldcw   (%eax)
	leave
	ret

/*
  sqrtint( real double precision : REAL*8 )
  =========================================
    This function returns the integer part of the square root
    of the given real. The return value has itself the real type.
*/
.global sqrtint
sqrtint:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
	fldl	(%eax)
	fsqrt
        frndint
	leave
	ret

/*
  oddperiod( real double precision : REAL*8 )
  ===========================================
    Despite the type of the real argument, the variable has to
    be some integer value stored as a double precision.
    The function has the logical type. It returns FALSE if
    the argument is a perfect square or if the continued
    fraction expansion of its square root has an even period
    length; it returns TRUE in the remaining case.
      oddperiod(dble(25)) --> FALSE
      oddperiod(dble(13)) --> TRUE
      oddperiod(dble(12)) --> FALSE
    If it may help, you can declare the function as an integer*4
    type rather than a logical one (output will be 0 or 1).

    Though it seems to work with integer*8, I am not absolutely
    sure it *always* will (unless you add the following instruction
      movl $0, %edx
    at the beginning of the code (for instance between the current
    second and third lines of code), but really why would you
    do it? Maybe something in the calling conventions is such that
    the register EDX already contains 0, but, again, who cares?

    If you read the instructions for the period() function, you
    will learn about some 2^32 limitations, but you don't have to
    care about that here: the length is not kept in any register
    during the computation and the code should work as long as the
    values in the FPU stack are the right one.
    
*/
.global oddperiod
oddperiod:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        fldln2
	fldl	(%eax)
        fld     %st
        fsqrt
        frndint
        fld     %st
        fld     %st
        fmul    %st, %st
        fld     %st(3)
        fsubp   %st, %st(1)
        fcom    %st(4)
        fstsw   %ax
        fwait
        sahf
        ja      oddperiod0
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
oddperiod2:
        movl    $0, %eax
        leave
        ret
oddperiod0:
        fld     %st(1)
        fadd    %st(3), %st
        fdiv    %st(1), %st
        frndint
        fld     %st(2)
        fld     %st(1)
        fmul    %st(3), %st
        fsubp   %st, %st(1)
        fxch    %st(3)
        fsub    %st(3), %st
        fabs
        fcomp   %st(6)
        fstsw   %ax
        fwait
        sahf
        ja      oddperiod1
        fld     %st(3)
        fadd    %st, %st
        fsubp   %st, %st(1)
        fcomp   %st(5)
        fstsw   %ax
        fwait
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        sahf
        ja      oddperiod2
        movl    $1, %eax
        leave
        ret
oddperiod1:
        ffree   %st
        fincstp
        fld     %st(1)
        fmul    %st, %st
        fld     %st(4)
        fsubp   %st, %st(1)
        fdivp   %st, %st(1)
        frndint
        jmp oddperiod0

/*
  period( real double precision : REAL*8 )
  ========================================
    Despite the type of the real argument, the variable has to
    be some integer value stored as a double precision.
    The function has integer*8 type, though the output
    can't be greater than 2^32-1 (this choice has been made
    for avoiding sign problem when the result is greater
    (or equal) then 2^31 and smaller then 2^32.
    The result is the length of period in the continued fraction
    expansion of the square root of the given integer.

    If speed is really for you an obsession, remember that it
    is highly unprobable that you use that library for periods
    between the two fatidic values. Maybe it will increase a little
    the speed to use integer*4 which allows to delete twice the line
        movl    $0,%edx
    (not the first time it appears, but the second and third one).
    If you try such a thing and see some speed improvement, please
    let me know.
*/
.global period
period:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        movl    $0, %edx
        fldln2
	fldl	(%eax)
        fld     %st
        fsqrt
        frndint
        fld     %st
        fld     %st
        fmul    %st, %st
        fld     %st(3)
        fsubp   %st, %st(1)
        fcom    %st(4)
        fstsw   %ax
        fwait
        sahf
        ja      period0
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        movl    %edx, %eax
        movl     $0, %edx
        leave
        ret
period0:
        inc     %edx
        fld     %st(1)
        fadd    %st(3), %st
        fdiv    %st(1), %st
        frndint
        fld     %st(2)
        fld     %st(1)
        fmul    %st(3), %st
        fsubp   %st, %st(1)
        fxch    %st(3)
        fsub    %st(3), %st
        fabs
        fcomp   %st(6)
        fstsw   %ax
        fwait
        sahf
        ja      period1
        fld     %st(3)
        fadd    %st, %st
        fsubp   %st, %st(1)
        fcomp   %st(5)
        fstsw   %ax
        fwait
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        sahf
        jb      period2
        add     %edx, %edx
period2:
        movl    %edx, %eax
        movl    $0,%edx
        leave
        ret
period1:
        ffree   %st
        fincstp
        fld     %st(1)
        fmul    %st, %st
        fld     %st(4)
        fsubp   %st, %st(1)
        fdivp   %st, %st(1)
        frndint
        jmp period0





From jvospost3 at gmail.com  Fri Nov  3 23:36:07 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 3 Nov 2006 14:36:07 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com>

I think the example givem without A number, is:

 A007188 <http://www.research.att.com/%7Enjas/sequences/A007188>
 Multiplicative encoding of Pascal triangle: Product p(i+1)^C(n,i).
(Formerly M1722)
 +0
2
2, 6, 90, 47250, 66852843750, 2806877704512541816406250,
1216935896582703898519354781702537118597533386230468750
There might be other Godel-coded sequences, and the like, besides Conway's
brilliant "Fractran", above and beyond those with "decimal Godelization" --
but usually these quickly lead to numbers too large for OEIS.

-- Jonathan Vos Post

On 11/3/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
> <Me>
> >Off hand, I can't think of any interesting functions of finite
> sequences of
> >positive integers that depend on the order of the sequence.
>
> Of course, I no sooner wrote that than I started thinking of such
> functions.
> I'm not sure any of these are tremendously interesting, but they're at
> least
> somewhat so.
>
> In the following, b(i), i = 1..k is the finite sequence.  I'm starting
> all these
> sequences with the empty sequence, but I searched for them without
> that first term.  All are shown here in A066099 order.
>
> Alternating sum
> Sum_i (-1)^{i+1) b(i)
> 0,1,2,0,3,1,-1,1,4,2,0,2,-2,0,2,0
>
> Binomial sum
> Sum_i C(k-1,i-1) b(i)
> 0,1,2,2,3,3,3,4,4,4,4,5,4,6,5,8
>
> Inverse binomial sum
> Sum_i (-1)^{i+1) C(k-1,i-1) b(i)
> 0,1,2,0,3,1,-1,0,4,2,0,1,-2,-2,1,0
>
> Weighted sum
> sum_i i b(i)
> 0,1,2,3,3,4,5,6,4,5,6,7,7,8,9,10
> This is A029931
>
> Zero-based weighted sum
> sum_i (i-1) b(i)
> 0,0,0,1,0,1,2,3,0,1,2,3,3,4,5,6
>
> Sum of products of consecutive elements
> sum_{i=1}^{k-1} b(i) b(i+1)
> 0,0,0,1,0,2,2,2,0,3,4,3,3,4,3,3
>
> Number of rises
> sum_{b(i)>b(i-1)} 1
> 0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,0
>
> Number of falls
> sum_{b(i)<b(i-1)} 1
> 0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0
>
> Number of unchanged
> sum_{b(i)=b(i-1)} 1
> 0,0,0,1,0,0,0,2,0,0,1,1,0,0,1,3
>
> Number of non-rises
> sum_{b(i)<=b(i-1)} 1
> 0,0,0,1,0,0,1,2,0,0,1,1,1,1,2,3
>
> Number of non-falls
> sum_{b(i)>=b(i-1)} 1
> 0,0,0,1,0,1,0,2,0,1,1,2,0,1,1,3
>
> Number of monotonically increasing runs
> 1 + number of falls, but 0 for empty sequence
> 0,1,1,1,1,2,1,1,1,2,1,2,1,2,1,1
>
> Number of monotonically decreasing runs
> 1 + number of rises, but 0 for empty sequence
> 0,1,1,1,1,1,2,1,1,1,1,1,2,2,2,1
>
> Number of runs of equal terms
> 1 + number of unchanged, but 0 for empty sequence
> 0,1,1,2,1,1,1,3,1,1,2,2,1,1,2,4
>
> Number of distinct non-empty subsequences
> 0,1,1,2,1,3,3,3,1,3,2,5,3,5,5,4
> E.g., f(1,1) = 2, for the sequences [1], and [1,1].
>
> Number of distinct subsequences
> 1 + number of distinct non-empty subsequences
> 1,2,2,3,2,4,4,4,2,4,3,6,4,6,6,5
>
> Number of set partitions
> List parts of set partition by their smallest element, and count the
> part sizes.
> E.g., 1|2,4|3 (also known as {1,2,3,2}) would count for the sequence
> [1,2,1].
> 1,1,1,1,1,2,1,1,1,3,3,3,1,2,1,1
>
> Number of permutations
> List permutations in cycle form, sorted by the smallest element in the
> cycle,
> and count the cycle lengths.
> Number of set partitions * Product_i (b(i)-1)!
> 1,1,1,1,2,2,1,1,6,6,3,3,2,2,1,1
>
> Number of partially ordered sets (unlabelled) by rank
> The rank of an element in a poset is the length of the longest chain of
> which it
> is the largest element.
> 1,1,1,1,1,2,1,1,1,3,4,3,1,2,1,1
>
> Number of partially ordered sets (labelled) by rank
> 1,1,1,2,1,9,3,6,1,28,54,60,4,36,12,24
>
> Number of partially ordered sets (naturally labelled) by rank
> Naturally labelled means labels are consistent with the ordering.
> 1,1,1,1,1,4,1,1,1,11,13,8,1,4,1,1
>
> Number of forests (unlabelled, unordered) with b(i) nodes at height i.
> 1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,2,2,1,1,1,1,2,1,1,1,1,1
>
> ... and, of course, various other forest options.
>
> If there is agreement that this is the right order for these sequences,
> I'll try
> to find time to submit them.
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and
> industry-leading spam and email virus protection.
>
>
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From jvospost3 at gmail.com  Sat Nov  4 00:59:34 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 3 Nov 2006 15:59:34 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
	 <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com>
Message-ID: <5542af940611031559x7a45e57euad1d783fb83d82b@mail.gmail.com>

To give a specific example:

NEW SEQUENCE FROM Jonathan Vos Post

%I A000001
%S A000001 2, 6, 450, 2836181250, 81492043057751910481759423160156250,
4561157026363824997482074305569280581505536351717093893927260661169357729871499327113563125890139588096951624677718591308593750
%N A000001 Multiplicative encoding of Catalan's triangle: Product
p(i+1)^T(n,i).
%C A000001 This is to A009766 "Catalan's triangle T(n,k) (read by
rows)"  as A007188 "Multiplicative encoding of Pascal triangle: Product
p(i+1)^C(n,i)" is to A007318 "Pascal's triangle read by rows."
%F A000001 a(n) = PRODUCT[i=i..n] p(i+1)^T(n,i), where T(n,i), are
Cataln's triangle as in a009766.
%e A000001 a(1) = p(1)^T(1,1) = 2^1 = 2.
a(2) = p(1)^T(2,1) * p(2)^T(2,2) = 2^1 * 3^1 = 6.
a(3) = p(1)^T(3,1) * p(2)^T(3,2) * p(3)^T(3,3) = 2^1 * 3^2 * 5^2 = 450.
a(4) = 2^1 * 3^3 * 5^5 * 7^5 = 2836181250.
a(5) = 2^1 * 3^4 * 5^9 * 7^14 * 11^14 =
81492043057751910481759423160156250.
a(6) = 2^1 * 3^5 * 5^14 * 7^28 * 11^42 * 13^42.
%Y A000001 Cf. A007188, A007318, A009766.
%O A000001 1
%K A000001 ,easy,nonn,
%A A000001 Jonathan Vos Post (jvospost2 at yahoo.com
<http://us.f551.mail.yahoo.com/ym/Compose?To=jvospost2 at yahoo.com&YY=14244&y5beta=yes&y5beta=yes&order=down&sort=date&pos=0>),
Nov 03 2006
RH
RA 192.20.225.32



On 11/3/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> I think the example givem without A number, is:
>
>  A007188 <http://www.research.att.com/%7Enjas/sequences/A007188>
>  Multiplicative encoding of Pascal triangle: Product p(i+1)^C(n,i).
> (Formerly M1722)
>  +0
> 2
> 2, 6, 90, 47250, 66852843750, 2806877704512541816406250,
> 1216935896582703898519354781702537118597533386230468750
> There might be other Godel-coded sequences, and the like, besides Conway's
> brilliant "Fractran", above and beyond those with "decimal Godelization" --
> but usually these quickly lead to numbers too large for OEIS.
>
> -- Jonathan Vos Post
>
> On 11/3/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
> >
> > <Me>
> > >Off hand, I can't think of any interesting functions of finite
> > sequences of
> > >positive integers that depend on the order of the sequence.
> >
> > Of course, I no sooner wrote that than I started thinking of such
> > functions.
> > I'm not sure any of these are tremendously interesting, but they're at
> > least
> > somewhat so.
> >
> > In the following, b(i), i = 1..k is the finite sequence.  I'm starting
> > all these
> > sequences with the empty sequence, but I searched for them without
> > that first term.  All are shown here in A066099 order.
> >
> > Alternating sum
> > Sum_i (-1)^{i+1) b(i)
> > 0,1,2,0,3,1,-1,1,4,2,0,2,-2,0,2,0
> >
> > Binomial sum
> > Sum_i C(k-1,i-1) b(i)
> > 0,1,2,2,3,3,3,4,4,4,4,5,4,6,5,8
> >
> > Inverse binomial sum
> > Sum_i (-1)^{i+1) C(k-1,i-1) b(i)
> > 0,1,2,0,3,1,-1,0,4,2,0,1,-2,-2,1,0
> >
> > Weighted sum
> > sum_i i b(i)
> > 0,1,2,3,3,4,5,6,4,5,6,7,7,8,9,10
> > This is A029931
> >
> > Zero-based weighted sum
> > sum_i (i-1) b(i)
> > 0,0,0,1,0,1,2,3,0,1,2,3,3,4,5,6
> >
> > Sum of products of consecutive elements
> > sum_{i=1}^{k-1} b(i) b(i+1)
> > 0,0,0,1,0,2,2,2,0,3,4,3,3,4,3,3
> >
> > Number of rises
> > sum_{b(i)>b(i-1)} 1
> > 0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,0
> >
> > Number of falls
> > sum_{b(i)<b(i-1)} 1
> > 0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0
> >
> > Number of unchanged
> > sum_{b(i)=b(i-1)} 1
> > 0,0,0,1,0,0,0,2,0,0,1,1,0,0,1,3
> >
> > Number of non-rises
> > sum_{b(i)<=b(i-1)} 1
> > 0,0,0,1,0,0,1,2,0,0,1,1,1,1,2,3
> >
> > Number of non-falls
> > sum_{b(i)>=b(i-1)} 1
> > 0,0,0,1,0,1,0,2,0,1,1,2,0,1,1,3
> >
> > Number of monotonically increasing runs
> > 1 + number of falls, but 0 for empty sequence
> > 0,1,1,1,1,2,1,1,1,2,1,2,1,2,1,1
> >
> > Number of monotonically decreasing runs
> > 1 + number of rises, but 0 for empty sequence
> > 0,1,1,1,1,1,2,1,1,1,1,1,2,2,2,1
> >
> > Number of runs of equal terms
> > 1 + number of unchanged, but 0 for empty sequence
> > 0,1,1,2,1,1,1,3,1,1,2,2,1,1,2,4
> >
> > Number of distinct non-empty subsequences
> > 0,1,1,2,1,3,3,3,1,3,2,5,3,5,5,4
> > E.g., f(1,1) = 2, for the sequences [1], and [1,1].
> >
> > Number of distinct subsequences
> > 1 + number of distinct non-empty subsequences
> > 1,2,2,3,2,4,4,4,2,4,3,6,4,6,6,5
> >
> > Number of set partitions
> > List parts of set partition by their smallest element, and count the
> > part sizes.
> > E.g., 1|2,4|3 (also known as {1,2,3,2}) would count for the sequence
> > [1,2,1].
> > 1,1,1,1,1,2,1,1,1,3,3,3,1,2,1,1
> >
> > Number of permutations
> > List permutations in cycle form, sorted by the smallest element in the
> > cycle,
> > and count the cycle lengths.
> > Number of set partitions * Product_i (b(i)-1)!
> > 1,1,1,1,2,2,1,1,6,6,3,3,2,2,1,1
> >
> > Number of partially ordered sets (unlabelled) by rank
> > The rank of an element in a poset is the length of the longest chain of
> > which it
> > is the largest element.
> > 1,1,1,1,1,2,1,1,1,3,4,3,1,2,1,1
> >
> > Number of partially ordered sets (labelled) by rank
> > 1,1,1,2,1,9,3,6,1,28,54,60,4,36,12,24
> >
> > Number of partially ordered sets (naturally labelled) by rank
> > Naturally labelled means labels are consistent with the ordering.
> > 1,1,1,1,1,4,1,1,1,11,13,8,1,4,1,1
> >
> > Number of forests (unlabelled, unordered) with b(i) nodes at height i.
> > 1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,2,2,1,1,1,1,2,1,1,1,1,1
> >
> > ... and, of course, various other forest options.
> >
> > If there is agreement that this is the right order for these sequences,
> > I'll try
> > to find time to submit them.
> > ________________________________________________________________________
> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > industry-leading spam and email virus protection.
> >
> >
>
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From davidwwilson at comcast.net  Sat Nov  4 01:40:21 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Fri, 3 Nov 2006 19:40:21 -0500
Subject: A124015 thoughts
Message-ID: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p>

The description of A124015 is

%N A124015 Number of words with n letters in the National Scrabble Association Dictionary.

Some notes:

- It should be noted that the NSAD includes a restricted set of English words (words of 2 to 14 letters, no proper nouns or derivatives, no words with non-alphabetic characters (e.g, contractions, hyphenated words), and it is this restricted set that is being counted.

- The edition of the NSAD used to create A124015 should be specified, since the NSAD is continually edited.  While I do not think that A124015 should change with each new edition of the NSAD, because it may be referenced in other literature. On the other hand, I don't think that a new sequence should be created for each new edition of the NSAD unless the new edition exhibits some interesting statistical departure from the current NSAD (e.g, in some future NSAD, the number of 9-letter words exceeds the number of 8-letter words). This is because I believe that the value of A124015 is mainly to indicate a distribution of English word lengths.

- On the other hand, I have always understood that 4-letter words are most common in literature (where small words appear more often). It might be interesting to create a sequence counting words of length n in the King James Bible, War and Peace, or some other suitably large and stable piece of English literature (we don't have to go overboard on this, just one or two examples indicating word length distributions in literature).

- Another interesting idea: In various languages, what is the distribution of the number of letters in the names of numbers from one to (say) one million?

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From franktaw at netscape.net  Sat Nov  4 02:49:48 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 20:49:48 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <5542af940611031559x7a45e57euad1d783fb83d82b@mail.gmail.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>	 <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com> <5542af940611031559x7a45e57euad1d783fb83d82b@mail.gmail.com>
Message-ID: <8C8CDCAA57554FD-ED8-5165@FWM-M11.sysops.aol.com>

This is OK, but it is the opposite of what we started out talking 
about.  This is a function from integers to sequences; but we are 
talking about functions from sequences to integers.  And not every 
integer encodes a sequence in this way; only those in A055932.

Franklin T. Adams-Watters


-----Original Message-----
From: jvospost3 at gmail.com

  To give a specific example:

 NEW SEQUENCE FROM Jonathan Vos Post

%I A000001
%S A000001 2, 6, 450, 2836181250, 81492043057751910481759423160156250,
4561157026363824997482074305569280581505536351717093893927260661169357729
871499327113563125890139588096951624677718591308593750

%N A000001 Multiplicative encoding of Catalan's triangle: Product
p(i+1)^T(n,i).
%C A000001 This is to A009766 "Catalan's triangle T(n,k) (read by
rows)"  as A007188 "Multiplicative encoding of Pascal triangle: Product

p(i+1)^C(n,i)" is to A007318 "Pascal's triangle read by rows."
%F A000001 a(n) = PRODUCT[i=i..n] p(i+1)^T(n,i), where T(n,i), are
Cataln's triangle as in a009766.
%e A000001 a(1) = p(1)^T(1,1) = 2^1 = 2.

a(2) = p(1)^T(2,1) * p(2)^T(2,2) = 2^1 * 3^1 = 6.
a(3) = p(1)^T(3,1) * p(2)^T(3,2) * p(3)^T(3,3) = 2^1 * 3^2 * 5^2 = 450.
a(4) = 2^1 * 3^3 * 5^5 * 7^5 = 2836181250.
a(5) = 2^1 * 3^4 * 5^9 * 7^14 * 11^14 =

81492043057751910481759423160156250.
a(6) = 2^1 * 3^5 * 5^14 * 7^28 * 11^42 * 13^42.
%Y A000001 Cf. A007188, A007318, A009766.
%O A000001 1
%K A000001 ,easy,nonn,
%A A000001 Jonathan Vos Post (
jvospost2 at yahoo.com), Nov 03 2006
RH
RA 192.20.225.32

On 11/3/06, Jonathan Post <jvospost3 at gmail.com> wrote: I think the 
example givem without A number, is:

   A007188
   Multiplicative encoding of Pascal triangle: Product p(i+1)^C(n,i).
(Formerly M1722)
   +0
2

  2, 6, 90, 47250, 66852843750, 2806877704512541816406250, 
1216935896582703898519354781702537118597533386230468750

  There might be other Godel-coded sequences, and the like, besides 
Conway's brilliant "Fractran", above and beyond those with "decimal 
Godelization" -- but usually these quickly lead to numbers too large 
for OEIS.

 -- Jonathan Vos Post

On 11/3/06, franktaw at netscape.net < franktaw at netscape.net> wrote: <Me>
>Off hand, I can't think of any interesting functions of finite
 sequences of
>positive integers that depend on the order of the sequence.

Of course, I no sooner wrote that than I started thinking of such
functions.
I'm not sure any of these are tremendously interesting, but they're at
least
somewhat so.

In the following, b(i), i = 1..k is the finite sequence. I'm starting
all these
sequences with the empty sequence, but I searched for them without
that first term. All are shown here in A066099 order.

Alternating sum
Sum_i (-1)^{i+1) b(i)
0,1,2,0,3,1,-1,1,4,2,0,2,-2,0,2,0

Binomial sum
Sum_i C(k-1,i-1) b(i)
0,1,2,2,3,3,3,4,4,4,4,5,4,6,5,8

Inverse binomial sum
Sum_i (-1)^{i+1) C(k-1,i-1) b(i)
0,1,2,0,3,1,-1,0,4,2,0,1,-2,-2,1,0

Weighted sum
sum_i i b(i)
0,1,2,3,3,4,5,6,4,5,6,7,7,8,9,10
This is A029931

Zero-based weighted sum
sum_i (i-1) b(i)
0,0,0,1,0,1,2,3,0,1,2,3,3,4,5,6

 Sum of products of consecutive elements
sum_{i=1}^{k-1} b(i) b(i+1)
0,0,0,1,0,2,2,2,0,3,4,3,3,4,3,3

Number of rises
sum_{b(i)>b(i-1)} 1
0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,0

Number of falls
sum_{b(i)<b(i-1)} 1
0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0

Number of unchanged
sum_{b(i)=b(i-1)} 1
0,0,0,1,0,0,0,2,0,0,1,1,0,0,1,3

Number of non-rises
sum_{b(i)<=b(i-1)} 1
0,0,0,1,0,0,1,2,0,0,1,1,1,1,2,3

Number of non-falls
sum_{b(i)>=b(i-1)} 1
0,0,0,1,0,1,0,2,0,1,1,2,0,1,1,3

Number of monotonically increasing runs
1 + number of falls, but 0 for empty sequence
0,1,1,1,1,2,1,1,1,2,1,2,1,2,1,1

Number of monotonically decreasing runs
1 + number of rises, but 0 for empty sequence
0,1,1,1,1,1,2,1,1,1,1,1,2,2,2,1

Number of runs of equal terms
1 + number of unchanged, but 0 for empty sequence
0,1,1,2,1,1,1,3,1,1,2,2,1,1,2,4

Number of distinct non-empty subsequences
0,1,1,2,1,3,3,3,1,3,2,5,3,5,5,4
E.g., f(1,1) = 2, for the sequences [1], and [1,1].

Number of distinct subsequences
1 + number of distinct non-empty subsequences
1,2,2,3,2,4,4,4,2,4,3,6,4,6,6,5

 Number of set partitions
List parts of set partition by their smallest element, and count the
part sizes.
E.g., 1|2,4|3 (also known as {1,2,3,2}) would count for the sequence
[1,2,1].
1,1,1,1,1,2,1,1,1,3,3,3,1,2,1,1

Number of permutations
List permutations in cycle form, sorted by the smallest element in the
cycle,
and count the cycle lengths.
Number of set partitions * Product_i (b(i)-1)!
1,1,1,1,2,2,1,1,6,6,3,3,2,2,1,1

Number of partially ordered sets (unlabelled) by rank
The rank of an element in a poset is the length of the longest chain of
which it
is the largest element.
1,1,1,1,1,2,1,1,1,3,4,3,1,2,1,1

Number of partially ordered sets (labelled) by rank
1,1,1,2,1,9,3,6,1,28,54,60,4,36,12,24

Number of partially ordered sets (naturally labelled) by rank
Naturally labelled means labels are consistent with the ordering.
1,1,1,1,1,4,1,1,1,11,13,8,1,4,1,1

 Number of forests (unlabelled, unordered) with b(i) nodes at height i.
1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,2,2,1,1,1,1,2,1,1,1,1,1

... and, of course, various other forest options.

If there is agreement that this is the right order for these sequences,
I'll try
to find time to submit them.
________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and
industry-leading spam and email virus protection.








________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From maxale at gmail.com  Sat Nov  4 03:09:57 2006
From: maxale at gmail.com (Max A.)
Date: Fri, 3 Nov 2006 18:09:57 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <200611032139.kA3LdZSV024106@smtp.well.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <200611031625.kA3GP1mt017249@smtp.well.com>
	 <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com>
	 <200611032139.kA3LdZSV024106@smtp.well.com>
Message-ID: <d3dac270611031809j6a1f850dj5a1846b40153dc0@mail.gmail.com>

On 11/3/06, Marc LeBrun <mlb at well.com> wrote:

> B. Encode each tuple as a single number, say by using it as factor
> exponents:  T(i) --> 2^T(i,1) * 3^T(i,2) * 5^T(i,3) *...

The sequences I've mentioned encoded this way:

I) Multinomial coefficients:

a( [k1, k2, ..., kn] ) = (k1 + k2 +  ... + kn)! / (k1! * k2! * ... * kn!)

1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 2, 1, 1, 3, 1, 3, 2, 2, 1,
4, 1, 2, 1, 3, 1, 6, 1, 1, 2, 2, 2, 6, 1, 2, 2, 4, 1, 6, 1, 3, 3, 2,
1, 5, 1, 3, 2, 3, 1, 4, 2, 4, 2, 2, 1, 12, 1, 2, 3, 1, 2, 6, 1, 3, 2,
6, 1, 10, 1, 2, 3, 3, 2, 6, 1, 5, 1, 2, 1, 12, 2, 2, 2, 4, 1, 12, 2,
3, 2, 2, 2, 6, 1, 3, 3, 6

II) Coefficients of incomplete Bell polynomials:

b( [k1, k2, ..., kn] ) = (k1 + 2*k2 + ... + n*kn)! / (k1! * k2! * ...
* kn!) / (1!^k1 * 2!^k2 * ... * n!^kn)

1, 1, 1, 1, 1, 3, 1, 1, 3, 4, 1, 6, 1, 5, 10, 1, 1, 15, 1, 10, 15, 6,
1, 10, 10, 7, 15, 15, 1, 60, 1, 1, 21, 8, 35, 45, 1, 9, 28, 20, 1,
105, 1, 21, 105, 10, 1, 15, 35, 70, 36, 28, 1, 105, 56, 35, 45, 11, 1,
210, 1, 12, 210, 1, 84, 168, 1, 36, 55, 280, 1, 105, 1, 13, 280, 45,
126, 252, 1, 35, 105, 14, 1, 420, 120, 15, 66, 56, 1, 840, 210, 55,
78, 16, 165, 21, 1, 315, 378, 280

III) The number of permutations with given cycle structure:

c( [k1, k2, ..., kn] ) = (k1 + 2*k2 + ... + n*kn)! / (k1! * k2! * ...
* kn!) / (1^k1 * 2^k2 * ... * n^kn)

1, 1, 1, 1, 2, 3, 6, 1, 3, 8, 24, 6, 120, 30, 20, 1, 720, 15, 5040,
20, 90, 144, 40320, 10, 40, 840, 15, 90, 362880, 120, 3628800, 1, 504,
5760, 420, 45, 39916800, 45360, 3360, 40, 479001600, 630, 6227020800,
504, 210, 403200, 87178291200, 15, 1260, 280, 25920, 3360,
1307674368000, 105, 2688, 210, 226800, 3991680, 20922789888000, 420,
355687428096000, 43545600, 1260, 1, 20160, 4032, 6402373705728000,
25920, 2217600, 3360, 121645100408832000, 105, 2432902008176640000,
518918400, 1120, 226800, 18144, 30240, 51090942171709440000, 70, 105,
6706022400, 1124000727777607680000, 2520, 172800, 93405312000,
23950080, 1344, 25852016738884976640000, 1680, 151200, 2217600,
283046400, 1394852659200, 1663200, 21, 620448401733239439360000,
11340, 9072, 1120

Max





From jawbrey at att.net  Sat Nov  4 03:26:36 2006
From: jawbrey at att.net (Jon Awbrey)
Date: Fri, 03 Nov 2006 21:26:36 -0500
Subject: querying multidimensional sequences in OEIS
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>	 <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com> <5542af940611031559x7a45e57euad1d783fb83d82b@mail.gmail.com> <8C8CDCAA57554FD-ED8-5165@FWM-M11.sysops.aol.com>
Message-ID: <454BFA5C.ACD4F79B@att.net>

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o

but every positive integer in N = {1, 2, 3, ...}
uniquely codes a finite partial function f : N -> N

http://stderr.org/pipermail/inquiry/2005-July/002846.html

ja

franktaw at netscape.net wrote:
> 
> This is OK, but it is the opposite of what we started out talking
> about.  This is a function from integers to sequences; but we are
> talking about functions from sequences to integers.  And not every
> integer encodes a sequence in this way; only those in A055932.
> 
> Franklin T. Adams-Watters

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o
inquiry e-lab: http://stderr.org/pipermail/inquiry/
citizendium development forums: http://smf.citizendium.org
http://www.textop.org/wiki/index.php?title=User:Jon_Awbrey
wikinfo: http://wikinfo.org/wiki.php?title=User:Jon_Awbrey
wp review: http://wikipediareview.com/index.php?showuser=398
o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o






From jvospost3 at gmail.com  Sat Nov  4 03:49:46 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 3 Nov 2006 18:49:46 -0800
Subject: A124015 thoughts
In-Reply-To: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p>
Message-ID: <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>

I mostly agree.  A large sample of text as actually used in a language, such
as a set of books, newspaper articles, transcripts of phone conversations,
or the like, is often called a "corpus" by those who undertake statistical
analyses.  Various government agencies have done so since well before World
War II, tapping the minds of as many of half the Mathematicians in the
world, with computational resources in advance of corporate and personal
users, and with the help of Mathematicians such as Turing.

There are statistical analyses of various editions of the Bible, of the
works of Shakespeare, and the like, with phalanxes of statistical analysts
on Literature.

Many of these are covered in specialized journals, of questionable value to
seqfans, and unnoticed here except when, for instance, a minor poem is
claimed by such methods to be by Shakespeare.

Hence I second the motion by David Wilson to start with an extremely
well-defined if artificial corpus, such as "In various languages, what is
the distribution of the number of letters in the names of numbers from one
to (say) one million?"

I have commented more extensively on Scrabble in other venues.  Do not ever
agree to use the OED (Oxford English Dictionary) as the official dictionary
when playing Scrabble.  Many misspellings by Tournament rumes are there.
Even though the OED is the definitive dictionary of the largest language on
Earth, and fascinating reading for various purposes, it is just too big and
too weird for game use.  Also, it makes it hard to distinguish a bluff from
a variant.

Now that I think of it, logarithmic trends can be interesting.  How about "what
is the distribution of the number of letters in the names of numbers from
one to one billion?"

-- Jonathan Vos Post

On 11/3/06, David Wilson <davidwwilson at comcast.net> wrote:
>
>  The description of A124015 is
>
> %N A124015 Number of words with n letters in the National Scrabble
> Association Dictionary.
>
> Some notes:
>
> - It should be noted that the NSAD includes a restricted set of English
> words (words of 2 to 14 letters, no proper nouns or derivatives, no words
> with non-alphabetic characters (e.g, contractions, hyphenated words), and
> it is this restricted set that is being counted.
>
> - The edition of the NSAD used to create A124015 should be specified,
> since the NSAD is continually edited.  While I do not think that A124015
> should change with each new edition of the NSAD, because it may be
> referenced in other literature. On the other hand, I don't think that a new
> sequence should be created for each new edition of the NSAD unless the new
> edition exhibits some interesting statistical departure from the current
> NSAD (e.g, in some future NSAD, the number of 9-letter words exceeds the
> number of 8-letter words). This is because I believe that the value of
> A124015 is mainly to indicate a distribution of English word lengths.
>
> - On the other hand, I have always understood that 4-letter words are most
> common in literature (where small words appear more often). It might be
> interesting to create a sequence counting words of length n in the King
> James Bible, War and Peace, or some other suitably large and stable piece of
> English literature (we don't have to go overboard on this, just one or two
> examples indicating word length distributions in literature).
>
> - Another interesting idea: In various languages, what is the distribution
> of the number of letters in the names of numbers from one to (say) one
> million?
>
>
>
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From davidwwilson at comcast.net  Sat Nov  4 17:58:32 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sat, 4 Nov 2006 11:58:32 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
References: <20061103030956.55909.qmail@web38210.mail.mud.yahoo.com>
Message-ID: <001701c70032$75ba7ff0$6401a8c0@yourxhtr8hvc4p>

----- Original Message ----- 
From: "zak seidov" <zakseidov at yahoo.com>
To: <seqfan at ext.jussieu.fr>
Sent: Thursday, November 02, 2006 10:09 PM
Subject: Re: A065577 Number of Goldbach partitions of 10^n? More terms?


> David, Dick, Bob, Ivars,
> seqfans.
>
> After recalculating (at two different PC's)
> and rechecking I could finally
> find source of my error
> (silly as usual) in one of codes,
> and my final value is
> a(10) = 18200488 (NOT 18200487!)
> which coincides now with the Dick-David value.
>
> So, Bob,
> you may wish to give in A065577  the correct value
> a(10) = 18200488 with due refer to David Wilson  and
> Richard Mathar (not me).

I don't remember if Richard Mathar or myself was first to find a(10) = 
18200488. It's not really important to me, since any value would be suspect 
until independently confirmed. The fact that your suspect value led you to a 
bug in your program simply serves as additional confirmation. I have no 
problem with all three of us getting attribution. But then, I have no 
problem with none of us getting attribution, as this computation strikes me 
as on the level of an undergraduate homework problem.

> Also Ivars can extend his table.
> Thanks, Zak
>
> BTW1 Still, is it possible to find a(11)... with C++?

Well, it is possible to find a(11) using a C++ sieving program on an aging 
PC in a few hours, and if you were to do so, you would find that a(11) = 
149091160. Methinks that finding a(12) using the same program on the same PC 
would take about a week, except that said PC is a shared resource.

> BIW2 Not all people even in this closed list
> may be interested in Mmca specific problems,
> so I''l be happy to have a list of you
> particularly Mmca-involved.
> Then we may circulate relevant messages
> not to all seqfan community.
> Please anyone interested write me personally
> with subject "Me in Mmca", thanks.
> Of course it's not intented to be closed list!
> Hope this is not against Gerard' policy.

I truly wish I could afford Mma (or better yet, a math-enabled PDF editor). 
If I'm on anybody's Christmas list...






From mathar at strw.leidenuniv.nl  Sat Nov  4 18:20:07 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Sat, 4 Nov 2006 18:20:07 +0100
Subject: primes in arithmetic progression
Message-ID: <200611041720.kA4HK70T027399@amer.strw.leidenuniv.nl>


Is there an OEIS table that shows the smallest prime p
of k (not necessarily consecutive) primes in arithmetic progression
with common difference d? The table would look similar to this one
below, and contain rather large primes where I am leaving blanks:

    k=1 k=2  k=3  k=4  k=5  k=6
d
0     2   2    2    2    2    2
2     2   3    3
4     2   3    3
6     2   5    5    5    5
8     2   3    3
10    2   3    3
12    2   5    5    5    5
14    2   3    3 
16    2   3
18    2   5    5    5
20    2   3
22    2   7
24    2   5    5   59
26    2   3
28    2   3    3
30    2   7    7    7     7    7
32    2   5
34    2   3    3
36    2   5    7   31

The row d=0 and the column k=1 are degenerate and filled with the
prime 2. All strides d are even. Example for row d=24 and column k=4:
The 4 numbers 59,59+24,59+2*24 and 59+3*24 are all primes.

http://www.research.att.com/~njas/sequences/Sindx_Pri.html#primes_AP

--Richard





From rsc at swtch.com  Sat Nov  4 18:24:53 2006
From: rsc at swtch.com (Russ Cox)
Date: Sat, 4 Nov 2006 12:24:53 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
In-Reply-To: <001701c70032$75ba7ff0$6401a8c0@yourxhtr8hvc4p>
References: <20061103030956.55909.qmail@web38210.mail.mud.yahoo.com>
	 <001701c70032$75ba7ff0$6401a8c0@yourxhtr8hvc4p>
Message-ID: <ee9e417a0611040924h7c62e2a6hc2ba0ec199c7a175@mail.gmail.com>

I have confirmed the values so far
and also computed that a(12) = 1243722370.
Submitted (with a(11)) to OEIS via form.

Russ





From jvospost3 at gmail.com  Sat Nov  4 19:33:14 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 4 Nov 2006 10:33:14 -0800
Subject: primes in arithmetic progression
In-Reply-To: <200611041720.kA4HK70T027399@amer.strw.leidenuniv.nl>
References: <200611041720.kA4HK70T027399@amer.strw.leidenuniv.nl>
Message-ID: <5542af940611041033s7a722e9fjf6ad6643f988477d@mail.gmail.com>

The semiprime analogue of this is a table that shows the smallest semiprime
S
of k (not necessarily consecutive) primes in arithmetic progression
with common difference d (once a 0 appears, the row has ended nonzero
values):

d   k+1  k=2  k=3  k=4  k=5  k=6  k=7 k=8
0   4      4       4      4       4       4       4 ...
1   4      9       33    0
2   4      4       91    213   1383 8129 3091 0
3   4      6       115
4   4      6       6      111
5   4      4       77
6   4      4
7   4      14    51
8   4      6      6       69

-- Fairly easy to extend, fairly easy to prove the first 0 in a row.

The same can be done for 3-almost primes, 4-almost primes, and so forth,
this making a 3-dimensional array.



On 11/4/06, Richard Mathar <mathar at strw.leidenuniv.nl> wrote:
>
>
> Is there an OEIS table that shows the smallest prime p
> of k (not necessarily consecutive) primes in arithmetic progression
> with common difference d? The table would look similar to this one
> below, and contain rather large primes where I am leaving blanks:
>
>     k=1 k=2  k=3  k=4  k=5  k=6
> d
> 0     2   2    2    2    2    2
> 2     2   3    3
> 4     2   3    3
> 6     2   5    5    5    5
> 8     2   3    3
> 10    2   3    3
> 12    2   5    5    5    5
> 14    2   3    3
> 16    2   3
> 18    2   5    5    5
> 20    2   3
> 22    2   7
> 24    2   5    5   59
> 26    2   3
> 28    2   3    3
> 30    2   7    7    7     7    7
> 32    2   5
> 34    2   3    3
> 36    2   5    7   31
>
> The row d=0 and the column k=1 are degenerate and filled with the
> prime 2. All strides d are even. Example for row d=24 and column k=4:
> The 4 numbers 59,59+24,59+2*24 and 59+3*24 are all primes.
>
> http://www.research.att.com/~njas/sequences/Sindx_Pri.html#primes_AP
>
> --Richard
>
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From jvospost3 at gmail.com  Sat Nov  4 19:37:19 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 4 Nov 2006 10:37:19 -0800
Subject: primes in arithmetic progression
In-Reply-To: <5542af940611041033s7a722e9fjf6ad6643f988477d@mail.gmail.com>
References: <200611041720.kA4HK70T027399@amer.strw.leidenuniv.nl>
	 <5542af940611041033s7a722e9fjf6ad6643f988477d@mail.gmail.com>
Message-ID: <5542af940611041037s636536abnf98a7eee7feb349e@mail.gmail.com>

Sorry, I mean: "The semiprime analogue of this is a table that shows the
smallest semiprime S of k (not necessarily consecutive) SEMIprimes in
arithmetic progression
 with common difference d."

On 11/4/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> The semiprime analogue of this is a table that shows the smallest
> semiprime S
> of k (not necessarily consecutive) primes in arithmetic progression
> with common difference d (once a 0 appears, the row has ended nonzero
> values):
>
> d   k+1  k=2  k=3  k=4  k=5  k=6  k=7 k=8
> 0   4      4       4      4       4       4       4 ...
> 1   4      9       33    0
> 2   4      4       91    213   1383 8129 3091 0
> 3   4      6       115
> 4   4      6       6      111
> 5   4      4       77
> 6   4      4
> 7   4      14    51
> 8   4      6      6       69
>
> -- Fairly easy to extend, fairly easy to prove the first 0 in a row.
>
> The same can be done for 3-almost primes, 4-almost primes, and so forth,
> this making a 3-dimensional array.
>
>
>
> On 11/4/06, Richard Mathar <mathar at strw.leidenuniv.nl> wrote:
> >
> >
> > Is there an OEIS table that shows the smallest prime p
> > of k (not necessarily consecutive) primes in arithmetic progression
> > with common difference d? The table would look similar to this one
> > below, and contain rather large primes where I am leaving blanks:
> >
> >     k=1 k=2  k=3  k=4  k=5  k=6
> > d
> > 0     2   2    2    2    2    2
> > 2     2   3    3
> > 4     2   3    3
> > 6     2   5    5    5    5
> > 8     2   3    3
> > 10    2   3    3
> > 12    2   5    5    5    5
> > 14    2   3    3
> > 16    2   3
> > 18    2   5    5    5
> > 20    2   3
> > 22    2   7
> > 24    2   5    5   59
> > 26    2   3
> > 28    2   3    3
> > 30    2   7    7    7     7    7
> > 32    2   5
> > 34    2   3    3
> > 36    2   5    7   31
> >
> > The row d=0 and the column k=1 are degenerate and filled with the
> > prime 2. All strides d are even. Example for row d=24 and column k=4:
> > The 4 numbers 59,59+24,59+2*24 and 59+3*24 are all primes.
> >
> > http://www.research.att.com/~njas/sequences/Sindx_Pri.html#primes_AP<http://www.research.att.com/%7Enjas/sequences/Sindx_Pri.html#primes_AP>
> >
> > --Richard
> >
>
>
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From njas at research.att.com  Sat Nov  4 20:10:10 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sat, 4 Nov 2006 14:10:10 -0500 (EST)
Subject: primes in arithmetic progression
Message-ID: <200611041910.OAA62978@fry.research.att.com>

Just to get the ball rolling, I am adding this entry:

%I A124064
%S A124064 2,2,2,2,3,2,2,3,3,2,2,5,3
%N A124064 Array read by antidiagonals: T(d,k) (k >= 1, d = 2,4,6,8,...) = smallest prime p of k (not necessarily consecutive) primes in arithmetic progression with common difference d.
%C A124064 The row d=0 and the column k=1 are degenerate and are filled with the prime 2.
%O A124064 1,1
%K A124064 nonn,tabl,more
%e A124064 Array begins:
%e A124064 d.\...k=1.k=2..k=3..k=4..k=5..k=6
%e A124064 0..|..2...2....2....2....2....2
%e A124064 2..|..2...3....3
%e A124064 4..|..2...3....3
%e A124064 6..|..2...5....5....5....5
%e A124064 8..|..2...3....3
%e A124064 10.|..2...3....3
%e A124064 12.|..2...5....5....5....5
%e A124064 14.|..2...3....3.
%e A124064 16.|..2...3
%e A124064 18.|..2...5....5....5
%e A124064 20.|..2...3
%e A124064 22.|..2...7
%e A124064 24.|..2...5....5...59
%e A124064 26.|..2...3
%e A124064 28.|..2...3....3
%e A124064 30.|..2...7....7....7.....7....7
%e A124064 32.|..2...5
%e A124064 34.|..2...3....3
%e A124064 36.|..2...5....7...31
%e A124064 Example for row d=24 and column k=4: the 4 numbers 59,59+24,59+2*24 and 59+3*24 are all primes.
%A A124064 Richard Mathar (mathar(AT)strw.leidenuniv.nl), Nov 04 2006




NJAS





From davidwwilson at comcast.net  Sat Nov  4 21:10:45 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sat, 4 Nov 2006 15:10:45 -0500
Subject: Observation on A053669
Message-ID: <000601c7004d$507dc510$6401a8c0@yourxhtr8hvc4p>

It seems to me that

%N A053669 Smallest prime co-prime to n.

could just as easily be described

%N A053669 Smallest number > 1 coprime to n.

and the fact that a(n) is prime falls out.  This definition would also simplify the program.
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From mathar at strw.leidenuniv.nl  Sat Nov  4 21:45:17 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Sat, 4 Nov 2006 21:45:17 +0100
Subject: Observation on A053669
Message-ID: <200611042045.kA4KjHeV011396@amer.strw.leidenuniv.nl>


dww> From seqfan-owner at ext.jussieu.fr  Sat Nov  4 21:12:32 2006
dww> Return-Path: <seqfan-owner at ext.jussieu.fr>
dww> From: "David Wilson" <davidwwilson at comcast.net>
dww> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
dww> Subject: Observation on A053669
dww> ...
dww> It seems to me that
dww> 
dww> %N A053669 Smallest prime co-prime to n.
dww> 
dww> could just as easily be described
dww> 
dww> %N A053669 Smallest number > 1 coprime to n.
dww> 
dww> and the fact that a(n) is prime falls out.  This definition would also =
dww> simplify the program.
dww>...

Yes. If the smallest prime co-prime to n would not be a prime,
it would be built by at least two prime factors, the smaller of which
would be the correct entry (proof by reductio ad absurdum).

It would still be useful to keep in a comment that the numbers are all prime
(this might speed up some searches for the numbers...)

I'd propose to list in a comment the first occurrences of the first primes:
first 5 at a(6)
first 7 is a(30)
first 11 is a(210)
first 13 is a(2310)
first 17 is a(30030)
first 19 is a(510510)
where the indices are 6=2*3, 30=2*3*5, 210=2*3*5*7, 2*3*5*7*11 etc,
(the primorials, I guess, as in A002110)

-- RJM





From jvospost3 at gmail.com  Sat Nov  4 22:55:31 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 4 Nov 2006 13:55:31 -0800
Subject: primes in arithmetic progression
In-Reply-To: <200611041910.OAA62978@fry.research.att.com>
References: <200611041910.OAA62978@fry.research.att.com>
Message-ID: <5542af940611041355vd65aa5cr3a4a9df3cc348c15@mail.gmail.com>

The analogue for 3-almost primes (or, as njas prefers, numbers which are the
product of exactly 3 primes) is left as an exercise for the reader...  I've
just submitted the following.

Subject: PRE-NUMBERED NEW SEQUENCE A124570 FROM Jonathan Vos Post


%I A124570
%S A124570 4, 4, 4, 4, 9, 4, 4, 4, 33, 4, 4, 6, 91, 0, 4, 4, 6, 115,
213, 0, 4, 4, 4, 6, 0, 1383, 0, 4, 4, 4, 77, 111, 0, 8129, 0, 4, 4, 14
%N A124570 Array read by antidiagonals: T(d,k) (k >= 1, d =
1,2,3,4,5,6,...) = smallest semiprime s of k (not necessarily consecutive)
semiprimes in arithmetic progression with common difference d.
%C A124570 Semiprime analogue of A124064 Array read by antidiagonals:
T(d,k) (k >= 1, d = 2,4,6,8,...) = smallest prime p of k (not
necessarily consecutive) primes in arithmetic progression with common
difference d. The row d=0 and the column k=1 are degenerate and are filled with
the semiprime 4. Once a 0 occurs in a row, there are no more nonzero
vales in the row. For example, There cannot be 4 semiprimes in arithmetic
progression with common difference 3, starting with k, because modulo 4
we have {k, k+3, k+6, k+9} == {k+0, k+3, k+2, k+1} and one of these
must be divisible by 4, hence a nonsemiprime (eliminating k = 4 by
inspection).
%e A124570 Array begins:
d.\...k=1.k=2..k=3..k=4..k=5..k=6..k=7..k=8
0..|..4...4....4....4....4....4....4....4
1..|..4...9....33...0....0....0....0....0
2..|..4...4....91...213..1383.8129.3091.0
3..|..4...6....115..0....0....0....0....0
4..|..4...6....6....6....111...
5..|..4...4....77...
6..|..4...4....
7..|..4...14...51...
8..|..4...6....6....69.
Example for row 3: 115 = 5 * 23 is semiprime, 115+3 = 118 = 2 * 59 is
semiprime, and 115+3+3 = 121 = 11^2 is semiprime, so T(3,3) = 115.
%Y A124570 Cf. A000040, A001358, A056809, A070552, A092125, A092126,
A092127, A092128, A092129, A124064.
%O A124570 1,1
%K A124570 ,easy,more,nonn,tabl,
%A A124570 Jonathan Vos Post (jvospost2 at yahoo.com
<http://us.f551.mail.yahoo.com/ym/Compose?To=jvospost2 at yahoo.com&YY=99063&y5beta=yes&y5beta=yes&order=down&sort=date&pos=0&view=a&head=b>),
Nov 04 2006
RH
RA 192.20.225.32
RU
RI



On 11/4/06, N. J. A. Sloane <njas at research.att.com> wrote:
>
> Just to get the ball rolling, I am adding this entry:
>
> %I A124064
> %S A124064 2,2,2,2,3,2,2,3,3,2,2,5,3
> %N A124064 Array read by antidiagonals: T(d,k) (k >= 1, d = 2,4,6,8,...) =
> smallest prime p of k (not necessarily consecutive) primes in arithmetic
> progression with common difference d.
> %C A124064 The row d=0 and the column k=1 are degenerate and are filled
> with the prime 2.
> %O A124064 1,1
> %K A124064 nonn,tabl,more
> %e A124064 Array begins:
> %e A124064 d.\...k=1.k=2..k=3..k=4..k=5..k=6
> %e A124064 0..|..2...2....2....2....2....2
> %e A124064 2..|..2...3....3
> %e A124064 4..|..2...3....3
> %e A124064 6..|..2...5....5....5....5
> %e A124064 8..|..2...3....3
> %e A124064 10.|..2...3....3
> %e A124064 12.|..2...5....5....5....5
> %e A124064 14.|..2...3....3.
> %e A124064 16.|..2...3
> %e A124064 18.|..2...5....5....5
> %e A124064 20.|..2...3
> %e A124064 22.|..2...7
> %e A124064 24.|..2...5....5...59
> %e A124064 26.|..2...3
> %e A124064 28.|..2...3....3
> %e A124064 30.|..2...7....7....7.....7....7
> %e A124064 32.|..2...5
> %e A124064 34.|..2...3....3
> %e A124064 36.|..2...5....7...31
> %e A124064 Example for row d=24 and column k=4: the 4 numbers
> 59,59+24,59+2*24 and 59+3*24 are all primes.
> %A A124064 Richard Mathar (mathar(AT)strw.leidenuniv.nl), Nov 04 2006
>
>
>
>
> NJAS
>
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From jrbibers at indiana.edu  Sat Nov  4 23:17:45 2006
From: jrbibers at indiana.edu (Joseph Biberstine)
Date: Sat, 04 Nov 2006 17:17:45 -0500
Subject: A124015 thoughts
In-Reply-To: <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
Message-ID: <454D1189.8090102@indiana.edu>

	For anyone interested, this Mathematica function, eng[], will (as
usual, barring bugs) give the American English name for integers on 0 to
10^21-1.  Extension of the upper bound should be plain to implement
(just one line of code -- observe the trend).

	All printing is lowercase and without hyphens.  In holding that "and"
should only ever be read for radix points, it is never printed here.
Note this implementation uses a text postprocessor and so is inherently
ugly.  Also note that the helper memoizes for better or worse.  Please,
of course, report all bugs.

	Attached are some diagrams of the predictable distribution of
StringLength[eng[n]] (note this includes spaces).  The third ranges up
to 10^9, sampling every 10^6.

	Here is sample output from random naturals on 0,999999:

 40573, forty thousand five hundred seventy three
108097, one hundred eight thousand ninety seven
539799, five hundred thirty nine thousand seven hundred ninety nine
136868, one hundred thirty six thousand eight hundred sixty eight
597841, five hundred ninety seven thousand eight hundred forty one
272460, two hundred seventy two thousand four hundred sixty
135550, one hundred thirty five thousand five hundred fifty
402424, four hundred two thousand four hundred twenty four
543617, five hundred forty three thousand six hundred seventeen
868945, eight hundred sixty eight thousand nine hundred forty five
951779, nine hundred fifty one thousand seven hundred seventy nine
742845, seven hundred forty two thousand eight hundred forty five
 84140, eighty four thousand one hundred forty
680145, six hundred eighty thousand one hundred forty five
 82734, eighty two thousand seven hundred thirty four
154758, one hundred fifty four thousand seven hundred fifty eight
961084, nine hundred sixty one thousand eighty four
360944, three hundred sixty thousand nine hundred forty four
574416, five hundred seventy four thousand four hundred sixteen
164099, one hundred sixty four thousand ninety nine

	Here is the code:

Clear[eng, engRaw, n];
eng[n_] := (
  pre = engRaw[n];
  If[pre == "", pre = "zero"]; (* fixes "zero" *)
  pre = FixedPoint[StringReplace[#, "  " -> " "] &, pre] ; (* no
consecutive spaces *)
  pre = FixedPoint[StringReplace[#, RegularExpression[" $"] :> ""] &,
pre] ; (* no trailing spaces *)
  pre);
(* This raw function relies heavily on the postprocessor, which fixes
"zero" and deals with whitespace *)
engRaw[n_] := engRaw[n] = (* memo *)
  Which[
    n == 0, "",(* fixed in postprocessing *)
    n == 1, "one",
    n == 2, "two",
    n == 3, "three",
    n == 4, "four",
    n == 5, "five",
    n == 6, "six",
    n == 7, "seven",
    n == 8, "eight",
    n == 9, "nine",
    n == 10, "ten",
    n == 11, "eleven",
    n == 12, "twelve",
    n == 13, "thirteen",
    n == 14, "fourteen",
    n == 15, "fifteen",
    n == 16, "sixteen",
    n == 17, "seventeen",
    n == 18, "eighteen",
    n == 19, "nineteen",
    20 <= n <= 29, "twenty " <> engRaw[Mod[n, 10]],
    30 <= n <= 39, "thirty " <> engRaw[Mod[n, 10]],
    40 <= n <= 49, "forty " <> engRaw[Mod[n, 10]],
    50 <= n <= 59, "fifty " <> engRaw[Mod[n, 10]],
    60 <= n <= 69, "sixty " <> engRaw[Mod[n, 10]],
    70 <= n <= 79, "seventy " <> engRaw[Mod[n, 10]],
    80 <= n <= 89, "eighty " <> engRaw[Mod[n, 10]],
    90 <= n <= 99, "ninety " <> engRaw[Mod[n, 10]],
    10^2 <= n <= 10^3 - 1, engRaw[Quotient[n, 10^2]] <> " hundred " <>
engRaw[Mod[n, 10^2]],
    10^3 <= n <= 10^6 - 1, engRaw[Quotient[n, 10^3]] <> " thousand " <>
engRaw[Mod[n, 10^3]],
    10^6 <= n <= 10^9 - 1, engRaw[Quotient[n, 10^6]] <> " million " <>
engRaw[Mod[n, 10^6]],
    10^9 <= n <= 10^12 - 1, engRaw[Quotient[n, 10^9]] <> " billion " <>
engRaw[Mod[n, 10^9]],
    10^12 <= n <= 10^15 - 1, engRaw[Quotient[n, 10^12]] <> " trillion "
<> engRaw[Mod[n, 10^12]],
    10^15 <= n <= 10^18 - 1, engRaw[Quotient[n, 10^15]] <> " quadrillion
" <> engRaw[Mod[n, 10^15]],
    10^18 <= n <= 10^21 - 1, engRaw[Quotient[n, 10^18]] <> " quintillion
" <> engRaw[Mod[n, 10^18]],
    True, Return["Naturals beyond 10^21-1 are not supported"];];

-JRB

Jonathan Post wrote:
> Now that I think of it, logarithmic trends can be interesting.  How
> about "what is the distribution of the number of letters in the names of
> numbers from one to one billion?"
> 
> -- Jonathan Vos Post
> 
> On 11/3/06, *David Wilson* <davidwwilson at comcast.net
> <mailto:davidwwilson at comcast.net>> wrote:
> 
>     The description of A124015 is
>      
>     %N A124015 Number of words with n letters in the National Scrabble
>     Association Dictionary.
>      
>     Some notes:
>      
>     - It should be noted that the NSAD includes a restricted set of
>     English words (words of 2 to 14 letters, no proper nouns or
>     derivatives, no words with non-alphabetic characters (e.g,
>     contractions, hyphenated words), and it is this restricted set that
>     is being counted.
>      
>     - The edition of the NSAD used to create A124015 should be
>     specified, since the NSAD is continually edited.  While I do not
>     think that A124015 should change with each new edition of the NSAD,
>     because it may be referenced in other literature. On the other hand,
>     I don't think that a new sequence should be created for each new
>     edition of the NSAD unless the new edition exhibits some interesting
>     statistical departure from the current NSAD (e.g, in some future
>     NSAD, the number of 9-letter words exceeds the number of 8-letter
>     words). This is because I believe that the value of A124015 is
>     mainly to indicate a distribution of English word lengths.
>      
>     - On the other hand, I have always understood that 4-letter words
>     are most common in literature (where small words appear more often).
>     It might be interesting to create a sequence counting words of
>     length n in the King James Bible, War and Peace, or some other
>     suitably large and stable piece of English literature (we don't have
>     to go overboard on this, just one or two examples indicating word
>     length distributions in literature).
>      
>     - Another interesting idea: In various languages, what is the
>     distribution of the number of letters in the names of numbers from
>     one to (say) one million?
>      
>      
> 
> 
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From jvospost3 at gmail.com  Sun Nov  5 02:05:28 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 4 Nov 2006 17:05:28 -0800
Subject: A124015 thoughts
In-Reply-To: <454D1189.8090102@indiana.edu>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p>
	 <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
	 <454D1189.8090102@indiana.edu>
Message-ID: <5542af940611041705u66643b27u66c836a95a2b9a1@mail.gmail.com>

Nice code.  And what fascinating graphs! They have both periodicities and a
fractal appearance.  Somehwere in there is the equivalent of Benford's Law,
and Zipf's Law, on lengths names of numbers (as opposed to digit
distribution and power law).

On 11/4/06, Joseph Biberstine <jrbibers at indiana.edu> wrote:
>
>         For anyone interested, this Mathematica function, eng[], will (as
> usual, barring bugs) give the American English name for integers on 0 to
> 10^21-1.  Extension of the upper bound should be plain to implement
> (just one line of code -- observe the trend).
>
>         All printing is lowercase and without hyphens.  In holding that
> "and"
> should only ever be read for radix points, it is never printed here.
> Note this implementation uses a text postprocessor and so is inherently
> ugly.  Also note that the helper memoizes for better or worse.  Please,
> of course, report all bugs.
>
>         Attached are some diagrams of the predictable distribution of
> StringLength[eng[n]] (note this includes spaces).  The third ranges up
> to 10^9, sampling every 10^6.
>
>         Here is sample output from random naturals on 0,999999:
>
> 40573, forty thousand five hundred seventy three
> 108097, one hundred eight thousand ninety seven
> 539799, five hundred thirty nine thousand seven hundred ninety nine
> 136868, one hundred thirty six thousand eight hundred sixty eight
> 597841, five hundred ninety seven thousand eight hundred forty one
> 272460, two hundred seventy two thousand four hundred sixty
> 135550, one hundred thirty five thousand five hundred fifty
> 402424, four hundred two thousand four hundred twenty four
> 543617, five hundred forty three thousand six hundred seventeen
> 868945, eight hundred sixty eight thousand nine hundred forty five
> 951779, nine hundred fifty one thousand seven hundred seventy nine
> 742845, seven hundred forty two thousand eight hundred forty five
> 84140, eighty four thousand one hundred forty
> 680145, six hundred eighty thousand one hundred forty five
> 82734, eighty two thousand seven hundred thirty four
> 154758, one hundred fifty four thousand seven hundred fifty eight
> 961084, nine hundred sixty one thousand eighty four
> 360944, three hundred sixty thousand nine hundred forty four
> 574416, five hundred seventy four thousand four hundred sixteen
> 164099, one hundred sixty four thousand ninety nine
>
>         Here is the code:
>
> Clear[eng, engRaw, n];
> eng[n_] := (
>   pre = engRaw[n];
>   If[pre == "", pre = "zero"]; (* fixes "zero" *)
>   pre = FixedPoint[StringReplace[#, "  " -> " "] &, pre] ; (* no
> consecutive spaces *)
>   pre = FixedPoint[StringReplace[#, RegularExpression[" $"] :> ""] &,
> pre] ; (* no trailing spaces *)
>   pre);
> (* This raw function relies heavily on the postprocessor, which fixes
> "zero" and deals with whitespace *)
> engRaw[n_] := engRaw[n] = (* memo *)
>   Which[
>     n == 0, "",(* fixed in postprocessing *)
>     n == 1, "one",
>     n == 2, "two",
>     n == 3, "three",
>     n == 4, "four",
>     n == 5, "five",
>     n == 6, "six",
>     n == 7, "seven",
>     n == 8, "eight",
>     n == 9, "nine",
>     n == 10, "ten",
>     n == 11, "eleven",
>     n == 12, "twelve",
>     n == 13, "thirteen",
>     n == 14, "fourteen",
>     n == 15, "fifteen",
>     n == 16, "sixteen",
>     n == 17, "seventeen",
>     n == 18, "eighteen",
>     n == 19, "nineteen",
>     20 <= n <= 29, "twenty " <> engRaw[Mod[n, 10]],
>     30 <= n <= 39, "thirty " <> engRaw[Mod[n, 10]],
>     40 <= n <= 49, "forty " <> engRaw[Mod[n, 10]],
>     50 <= n <= 59, "fifty " <> engRaw[Mod[n, 10]],
>     60 <= n <= 69, "sixty " <> engRaw[Mod[n, 10]],
>     70 <= n <= 79, "seventy " <> engRaw[Mod[n, 10]],
>     80 <= n <= 89, "eighty " <> engRaw[Mod[n, 10]],
>     90 <= n <= 99, "ninety " <> engRaw[Mod[n, 10]],
>     10^2 <= n <= 10^3 - 1, engRaw[Quotient[n, 10^2]] <> " hundred " <>
> engRaw[Mod[n, 10^2]],
>     10^3 <= n <= 10^6 - 1, engRaw[Quotient[n, 10^3]] <> " thousand " <>
> engRaw[Mod[n, 10^3]],
>     10^6 <= n <= 10^9 - 1, engRaw[Quotient[n, 10^6]] <> " million " <>
> engRaw[Mod[n, 10^6]],
>     10^9 <= n <= 10^12 - 1, engRaw[Quotient[n, 10^9]] <> " billion " <>
> engRaw[Mod[n, 10^9]],
>     10^12 <= n <= 10^15 - 1, engRaw[Quotient[n, 10^12]] <> " trillion "
> <> engRaw[Mod[n, 10^12]],
>     10^15 <= n <= 10^18 - 1, engRaw[Quotient[n, 10^15]] <> " quadrillion
> " <> engRaw[Mod[n, 10^15]],
>     10^18 <= n <= 10^21 - 1, engRaw[Quotient[n, 10^18]] <> " quintillion
> " <> engRaw[Mod[n, 10^18]],
>     True, Return["Naturals beyond 10^21-1 are not supported"];];
>
> -JRB
>
> Jonathan Post wrote:
> > Now that I think of it, logarithmic trends can be interesting.  How
> > about "what is the distribution of the number of letters in the names of
> > numbers from one to one billion?"
> >
> > -- Jonathan Vos Post
> >
> > On 11/3/06, *David Wilson* <davidwwilson at comcast.net
> > <mailto:davidwwilson at comcast.net>> wrote:
> >
> >     The description of A124015 is
> >
> >     %N A124015 Number of words with n letters in the National Scrabble
> >     Association Dictionary.
> >
> >     Some notes:
> >
> >     - It should be noted that the NSAD includes a restricted set of
> >     English words (words of 2 to 14 letters, no proper nouns or
> >     derivatives, no words with non-alphabetic characters (e.g,
> >     contractions, hyphenated words), and it is this restricted set that
> >     is being counted.
> >
> >     - The edition of the NSAD used to create A124015 should be
> >     specified, since the NSAD is continually edited.  While I do not
> >     think that A124015 should change with each new edition of the NSAD,
> >     because it may be referenced in other literature. On the other hand,
> >     I don't think that a new sequence should be created for each new
> >     edition of the NSAD unless the new edition exhibits some interesting
> >     statistical departure from the current NSAD (e.g, in some future
> >     NSAD, the number of 9-letter words exceeds the number of 8-letter
> >     words). This is because I believe that the value of A124015 is
> >     mainly to indicate a distribution of English word lengths.
> >
> >     - On the other hand, I have always understood that 4-letter words
> >     are most common in literature (where small words appear more often).
> >     It might be interesting to create a sequence counting words of
> >     length n in the King James Bible, War and Peace, or some other
> >     suitably large and stable piece of English literature (we don't have
> >     to go overboard on this, just one or two examples indicating word
> >     length distributions in literature).
> >
> >     - Another interesting idea: In various languages, what is the
> >     distribution of the number of letters in the names of numbers from
> >     one to (say) one million?
> >
> >
> >
> >
>
>
>
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From pxp at rogers.com  Sun Nov  5 02:29:38 2006
From: pxp at rogers.com (Hans Havermann)
Date: Sat, 4 Nov 2006 20:29:38 -0500
Subject: A124015 thoughts
In-Reply-To: <454D1189.8090102@indiana.edu>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com> <454D1189.8090102@indiana.edu>
Message-ID: <3E87B5F0-736C-466D-B4EC-C2D438AA1BA3@rogers.com>

Joseph Biberstine:

> 	For anyone interested, this Mathematica function, eng[], will (as  
> usual, barring bugs) give the American English name for integers on  
> 0 to 10^21-1.  Extension of the upper bound should be plain to  
> implement (just one line of code -- observe the trend).

I forget where I found the following bit of elegant code, but it  
essentially does the same thing as yours (except use a hyphen instead  
of a space between the tens and units for tens > 20):


Ones[n_, pad_:0] := If[Mod[pad, 10] > 0, "-", ""] <> {"", "one",  
"two", "three", "four", "five", "six", "seven", "eight", "nine",  
"ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen",  
"sixteen", "seventeen", "eighteen", "nineteen"}[[n + 1]]

Tens[n_] := {"", "twenty", "thirty", "forty", "fifty", "sixty",  
"seventy", "eighty", "ninety"}[[n]]

LastThree[n_] := LastThree[n] = Module[{h = Quotient[n, 100], m = Mod 
[n, 100]}, If[n > 99, Ones[h, 0] <> " hundred ", ""] <> If[m < 20,  
Ones[m], Tens[Quotient[m, 10]] <> Ones[Mod[m, 10], m]]]

Powers[n_] := Powers[n] = {"", " thousand", " million", " billion", "  
trillion", " quadrillion", " quintillion", " sextillion", "  
septillion", " octillion", "nonillion", " decillion", " undecillion",  
" duodecillion", " tredecillion", " quattuordecillion", "  
quindecillion", " sexdecillion", " septendecillion", "  
octodecillion", " novemdecillion", " vigintillion"}[[n]]

(*NumberName[0] := "zero"*)

NumberName[n_Integer] := Module[{db = Reverse[MapIndexed[If[#1 == 0,  
"", LastThree[#1] <> Powers[#2[[1]]] <> " "] &, FromDigits[#, 10] & / 
@ Reverse /@ Partition[Reverse[IntegerDigits[n, 10]], 3, 3, {1, 1},  
{}]]]}, StringDrop[StringJoin @@ db, -1]]

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From zbi74583 at boat.zero.ad.jp  Sun Nov  5 04:58:03 2006
From: zbi74583 at boat.zero.ad.jp (koh)
Date: Sun, 05 Nov 2006 12:58:03 +0900
Subject: terms lack
Message-ID: <20061105035802.zbi74583@boat.zero.ad.jp>

    Recently I updated more terms of A054572, which are calculated by Richard Mathar.
    And Neil claimed or got angry.
    He said, "You are red card. Don't submit a wrong sequences"
    I am exaggerating a little.
    These words are not exactly what he wrote.



    To Neil
    They are correct, not "wrong". 
    It is only not complete.

    In early time, I didn't do a computer search, so there is a possibility that my old submitted sequences are not complete.
    Now, I calculate both theoretically and with a computer.



    To Seqfans
    I defined many "divisor" functions and calculated many Amicable Numbers and Perfect Numbers using those functions.
    And I submitted them as sequences to  OEIS.

        http://www.research.att.com/~njas/sequences/?q=kohmoto+amicable&sort=0&fmt=0&language=english&go=Search        
        
        http://www.research.att.com/~njas/sequences/?q=kohmoto+perfect&sort=0&fmt=0&language=english&go=Search
   
    I suppose that they are almost complete but some of them might not be complete.

    I wish some one do a computer search about them.
    I think that A054572 is the worst case, because I gave only two terms up to 10^4.
    But sometimes my theory gives a complete set of examples.
    See A099105  , I was sure that these three terms are all, and a computer search also said it is right.

    Yasutoshi
    





From deutsch at duke.poly.edu  Sun Nov  5 05:30:30 2006
From: deutsch at duke.poly.edu (Emeric Deutsch)
Date: Sat, 4 Nov 2006 23:30:30 -0500 (EST)
Subject: No subject
Message-ID: <Pine.GSO.4.62.0611041740160.21855@duke.poly.edu>

Dear Seqfans,
We (Lou Shapiro and myself) would like to submit a few
sequences similar to A124497 (shown at the bottom of this
message). They will refer to 1-2-3-4 trees, 1-2-3-4-5
trees, etc. rather than to 1-2-3 trees. We can describe
the corresponding g.f.'s but we need help in finding the
terms themselves.

Here is the description of all the g.f.'s. Let M[k](z) be
the g.f. of the 1-2-...-k trees with thinning limbs. I will
describe these functions; consequently, we may disregard
the objects counted by them.

Denote by C[k](z) the g.f. of the k-ary trees. C[k]
satisfies C[k]=1+zC[k]^k and the coefficient of z^n
in its series expansion is
(*)      binom(kn,n)/[(k-1)n+1].

I will write Mk for M[k] and Ck for C[k].

Obviously, M1(z)=1/(1-z) (path-trees). The next one is 
M2(z)=M1(z)C2(z^2*M1), i.e. M1 times C2 evaluated at
z^2*M1. This gives no problem since
C2(z)=[1-sqrt(1-4z)]/(2z). We obtain A090344.

Next M3(z)=M2(z)C3(z^3*M2^2); since there is an
explicit expression for C3(z), namely
C3(z)=2/sqrt(3z)*sin[(1/3)*arcsin(sqrt(27z/4))],
this yields easily A124497.

 From here on we need help.
M4(z)=M3(z)C4(z^4*M3^3); 
Most probably sequence starts with
1,1,2,4,10,24,62,160,425,1140,3105 (I used some terms
of the series of C4(z); see (*)).

M5(z)=M4(z)C5(z^5*M4^4);
Most probably sequence starts with
1,1,2,4,10,25,67,180,495,1375,3871 (I used some terms
of the series of C5(z); see (*)).

And so on.

These sequences approach A124344:
1,1,2,4,10,25,68,187,530,1523,4447,13121,39107,...

Many thanks.
Emeric

--------------------------------
A124497  Number of 1-2-3 trees with n edges and with thinning limbs. A 
1-2-3 tree is an ordered tree with vertices of outdegree at most 3. A 
rooted tree with thinning limbs is such that if a node has k children, all 
its children have at most k children.

1, 1, 2, 4, 9, 20, 48, 116, 288, 724, 1849, 4768, 12423, 32628, 86342, 
229952, 616042, 1659012, 4489101, 12199521, 33284546, 91140797, 250396629, 
690043032, 1907022197, 5284167884, 14677681554, 40862469713, 114001697975 
(list; graph; listen)
OFFSET  0,3
FORMULA  G.f.=H*T(H^2*z^3), where 
T=2/sqrt(3*x)*sin((1/3)*arcsin(sqrt(27*x/4))) (solution of T=1+zT^3, 
T(0)=1), H=C(z^2/(1-z))/(1-z), and C(x)=[1-sqrt(1-4x)]/(2x) is the Catalan 
function. More generally, if M[k](z) is the g.f. of the 1-2-...-k trees 
with thinning limbs and C[k](z)=1+z*{C[k](z)}^k is the g.f. of the k-ary 
trees, then M[k](z)=M[k-1](z)C[k](M[k-1]^(k-1)*z^k).
MAPLE  C:=x->(1-sqrt(1-4*x))/2/x: 
T:=x->2/sqrt(3*x)*sin((1/3)*arcsin(sqrt(27*x/4))): M2:=C(z^2/(1-z))/(1-z): 
G:=M2*T(M2^2*z^3): Gser:=series(G, z=0, 40): seq(coeff(Gser, z, n), 
n=0..33);
CROSSREFS  Cf. A090344, A124344.
KEYWORD  nonn,new
AUTHOR  Emeric Deutsch and Louis Shapiro (deutsch(AT)duke.poly.edu, 
lshapiro(AT)Howard.edu), Nov 04 2006





From zakseidov at yahoo.com  Sun Nov  5 07:28:49 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sat, 4 Nov 2006 22:28:49 -0800 (PST)
Subject: Observation on A053669
In-Reply-To: <200611042045.kA4KjHeV011396@amer.strw.leidenuniv.nl>
Message-ID: <20061105062849.51615.qmail@web38208.mail.mud.yahoo.com>

Yes, 
and because a(n_odd)=2
keep only even n's.
Zak

--- Richard Mathar <mathar at strw.leidenuniv.nl> wrote:

> 
> dww> From seqfan-owner at ext.jussieu.fr  Sat Nov  4
> 21:12:32 2006
> dww> Return-Path: <seqfan-owner at ext.jussieu.fr>
> dww> From: "David Wilson" <davidwwilson at comcast.net>
> dww> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
> dww> Subject: Observation on A053669
> dww> ...
> dww> It seems to me that
> dww> 
> dww> %N A053669 Smallest prime co-prime to n.
> dww> 
> dww> could just as easily be described
> dww> 
> dww> %N A053669 Smallest number > 1 coprime to n.
> dww> 
> dww> and the fact that a(n) is prime falls out. 
> This definition would also =
> dww> simplify the program.
> dww>...
> 
> Yes. If the smallest prime co-prime to n would not
> be a prime,
> it would be built by at least two prime factors, the
> smaller of which
> would be the correct entry (proof by reductio ad
> absurdum).
> 
> It would still be useful to keep in a comment that
> the numbers are all prime
> (this might speed up some searches for the
> numbers...)
> 
> I'd propose to list in a comment the first
> occurrences of the first primes:
> first 5 at a(6)
> first 7 is a(30)
> first 11 is a(210)
> first 13 is a(2310)
> first 17 is a(30030)
> first 19 is a(510510)
> where the indices are 6=2*3, 30=2*3*5, 210=2*3*5*7,
> 2*3*5*7*11 etc,
> (the primorials, I guess, as in A002110)
> 
> -- RJM
> 



 
__________________________________________________________________________________________
Check out the New Yahoo! Mail - Fire up a more powerful email and get things done faster. 
(http://advision.webevents.yahoo.com/mailbeta) 






From davidwwilson at comcast.net  Sun Nov  5 07:38:34 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sun, 5 Nov 2006 01:38:34 -0500
Subject: Notes on A124064 (was Re: primes in arithmetic progression)
References: <200611041910.OAA62978@fry.research.att.com>
Message-ID: <003101c700a5$05184930$6401a8c0@yourxhtr8hvc4p>

This is one of those cases where I think we have jumped the gun, creating a 
sequence without first determining the best structure for that sequence.

Richard Mather proposed a table of T(d,k) = the smallest possible initial 
element of a nondecreasing AP of k primes with difference d. Clearly, T(d,k) 
is defined only for d >= 0 and k >= 1.

For d = 0, it is pretty easy to show T(0, k) = 2 for any k >= 1. For d >= 1, 
however, T(d,k) is defined only for a finite number of k.

Proof: Suppose there exists AP P of primes with difference d >= 1 and 2d+2 
elements. Let c = d+1. Because P is an arithmetic progression of 2c elements 
with difference d, and c is coprime to d, P will have two elements divisible 
by c, call them cx and cy. Whereas P is an increasing sequence of positive 
integers, we can assume that 0 < cx < cy. Hence 0 < x < y, and y >= 2. But 
this makes cy composite, contrary to fact that P is a sequence of primes. 
Hence P does not exist, that is to say, any AP of primes with difference d 
 >= 1 has at most 2d+1 elements, and T(d,k) is undefined for k > 2d+1. QED.

In fact, assuming the k-tuple conjecture is true, we can give the precise 
domain of T for d >= 1: If p is the smallest prime not dividing d, then 
T(d,k) is defined precisely for 1 <= k <= p if the AP of p elements with 
initial element p and difference d is a prime AP, and for 1 <= k <= p-1 
otherwise.

This means that the table of T(d,k) looks as follows:

  \k|  1  2  3  4  5  6  7  8  9 10
  d\|
----+------------------------------
  0 |  2  2  2  2  2  2  2  2  2  2 ...
  1 |  2  2
  2 |  2  3  3
  3 |  2  2
  4 |  2  3  3
  5 |  2  2
  6 |  2  5  5  5  5
  7 |  2
  8 |  2  3  3
  9 |  2  2
 10 |  2  3  3
 11 |  2  2
 12 |  2  5  5  5  5
 13 |  2
 14 |  2  3  3
 15 |  2  2
 16 |  2  3
 17 |  2  2
 18 |  2  5  5  5
 19 |  2
 20 |  2  3  3
 ...

Two important points:

- Sometimes in the OEIS, we are tempted to pare down sequences to remove 
uninteresting values in order to emphasize the interesting values. In such 
cases, I think that the OEIS should include a main sequence that is 
complete, and additional pared-down sequences if they seem to augment our 
understanding of the main sequence. In this case, Mather's sequence included 
T(d,k) only for even d, presumably because T(d,k) is uninteresting for odd 
d. My table above includes T(d,k) for both even and odd d. In fact, it turns 
out that for odd d, T(d,1) = 2 and T(d,2) = 2 if 2+d is prime, not 
scintillating behavior, but not trivial either. My feeling is that the main 
sequence for T(d,k) (presumably A124064) should include all the values of 
T(d,k), and if restricting T(d,k) to even d seems to augment our 
understanding, a second pared-down sequence should be made referring back to 
the complete main sequence.

- For d >= 1 and sufficient k, T(d,k) is undefined. The standard 
diagonalization technique won't work for this irregular table. That is why 
NJAS's proposed A124064 ends where it does, because the next value, T(2,4), 
does not exist (it is not a matter of doing more computation, T(2,4) does 
not exist because there is no AP of 4 primes with difference 2). I can see 
two possible ways to address this problem, the (icky) option of making 
T(d,k) = 0 if undefined, or the (less icky) option of requiring the APs in 
question to be strictly increasing. This would dispense with the infinite d 
= 0 line and leave us with a list of finite sequences which we could 
concatenate to build A124064, vis:

2,2,2,3,3,2,2,2,3,3,2,2,2,5,5,5,5,2,2,3,3,2,2,2,3,3,2,2,2,5,5,5,5,2,2,
3,3,2,2,2,3,2,2,2,5,5,5,2,2,3,3,2,2,2,7,2,2,5,5,59,2,2,3,2,2,2,3,3,2,2,
2,7,7,7,7,7,2,2,5,2,2,3,3,2,2,2,5,7,31,2,2,3,3,2,2,2,3,3,2,2,2,5,5,5,5

We could put the line lengths in another sequence.

2,3,2,3,2,5,1,3,2,3,2,5,1,3,2,2,2,4,1,3,2,2,1,4,1,2,2,3,2,6,1,2,1,3,2,4,
1,3,2,3,2,5,1,2,2,2,1,5,1,3,2,2,1,4,1,2,2,2,2,6,1,2,1,3,2,4,1,3,2,2,2,4,
1,2,1,2,2,4,1,3,2,2,1,4,1,2,2,2,1,6,1,2,1,3,2,5,1,3,2,2,2,4,1,3,2,2,2,4

----- Original Message ----- 
From: "N. J. A. Sloane" <njas at research.att.com>
To: <seqfan at ext.jussieu.fr>; "Richard Mathar" <mathar at strw.leidenuniv.nl>
Sent: Saturday, November 04, 2006 2:10 PM
Subject: Re: primes in arithmetic progression


> Just to get the ball rolling, I am adding this entry:
>
> %I A124064
> %S A124064 2,2,2,2,3,2,2,3,3,2,2,5,3
> %N A124064 Array read by antidiagonals: T(d,k) (k >= 1, d = 2,4,6,8,...) = 
> smallest prime p of k (not necessarily consecutive) primes in arithmetic 
> progression with common difference d.
> %C A124064 The row d=0 and the column k=1 are degenerate and are filled 
> with the prime 2.
> %O A124064 1,1
> %K A124064 nonn,tabl,more
> %e A124064 Array begins:
> %e A124064 d.\...k=1.k=2..k=3..k=4..k=5..k=6
> %e A124064 0..|..2...2....2....2....2....2
> %e A124064 2..|..2...3....3
> %e A124064 4..|..2...3....3
> %e A124064 6..|..2...5....5....5....5
> %e A124064 8..|..2...3....3
> %e A124064 10.|..2...3....3
> %e A124064 12.|..2...5....5....5....5
> %e A124064 14.|..2...3....3.
> %e A124064 16.|..2...3
> %e A124064 18.|..2...5....5....5
> %e A124064 20.|..2...3
> %e A124064 22.|..2...7
> %e A124064 24.|..2...5....5...59
> %e A124064 26.|..2...3
> %e A124064 28.|..2...3....3
> %e A124064 30.|..2...7....7....7.....7....7
> %e A124064 32.|..2...5
> %e A124064 34.|..2...3....3
> %e A124064 36.|..2...5....7...31
> %e A124064 Example for row d=24 and column k=4: the 4 numbers 
> 59,59+24,59+2*24 and 59+3*24 are all primes.
> %A A124064 Richard Mathar (mathar(AT)strw.leidenuniv.nl), Nov 04 2006
>
>
>
>
> NJAS
>
>
> -- 
> No virus found in this incoming message.
> Checked by AVG Free Edition.
> Version: 7.5.430 / Virus Database: 268.13.27/517 - Release Date: 11/3/2006 
> 4:30 PM
> 






From davidwwilson at comcast.net  Sun Nov  5 07:40:23 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sun, 5 Nov 2006 01:40:23 -0500
Subject: Observation on A053669
References: <200611042045.kA4KjHeV011396@amer.strw.leidenuniv.nl>
Message-ID: <003b01c700a5$45a9d3b0$6401a8c0@yourxhtr8hvc4p>

Another possible description of A053669:

Smallest prime non-divisor of n.

----- Original Message ----- 
From: "Richard Mathar" <mathar at strw.leidenuniv.nl>
To: <seqfan at ext.jussieu.fr>
Sent: Saturday, November 04, 2006 3:45 PM
Subject: Re: Observation on A053669


>
> dww> From seqfan-owner at ext.jussieu.fr  Sat Nov  4 21:12:32 2006
> dww> Return-Path: <seqfan-owner at ext.jussieu.fr>
> dww> From: "David Wilson" <davidwwilson at comcast.net>
> dww> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
> dww> Subject: Observation on A053669
> dww> ...
> dww> It seems to me that
> dww>
> dww> %N A053669 Smallest prime co-prime to n.
> dww>
> dww> could just as easily be described
> dww>
> dww> %N A053669 Smallest number > 1 coprime to n.
> dww>
> dww> and the fact that a(n) is prime falls out.  This definition would 
> also =
> dww> simplify the program.
> dww>...
>
> Yes. If the smallest prime co-prime to n would not be a prime,
> it would be built by at least two prime factors, the smaller of which
> would be the correct entry (proof by reductio ad absurdum).
>
> It would still be useful to keep in a comment that the numbers are all 
> prime
> (this might speed up some searches for the numbers...)
>
> I'd propose to list in a comment the first occurrences of the first 
> primes:
> first 5 at a(6)
> first 7 is a(30)
> first 11 is a(210)
> first 13 is a(2310)
> first 17 is a(30030)
> first 19 is a(510510)
> where the indices are 6=2*3, 30=2*3*5, 210=2*3*5*7, 2*3*5*7*11 etc,
> (the primorials, I guess, as in A002110)
>
> -- RJM
>
>
> -- 
> No virus found in this incoming message.
> Checked by AVG Free Edition.
> Version: 7.5.430 / Virus Database: 268.13.27/517 - Release Date: 11/3/2006 
> 4:30 PM
>
> 






From pauldhanna at juno.com  Sun Nov  5 07:49:40 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Sun, 5 Nov 2006 01:49:40 -0500
Subject: A124497: 1-2-3-...-k Trees
Message-ID: <20061105.014940.1648.0.pauldhanna@juno.com>

Emeric (and Seqfans),
     What a fascinating problem.
Although I will not be using this formula now, 
let's extend your formula for k-ary trees: 
(*)  [z^n] C[k](z)^m  =  binomial(kn+m-1,n)*m/[(k-1)n+m] 
to give the coefficient of z^n in the m-th power of C[k](z). 
Note that the logarithm is obtained similarly: 
(**) [z^n] log( C[k](z) ) =  binomial(kn-1,n)/[(k-1)n], for n>0.
No doubt you are already aware of this lovely facts; 
and they may simplify some series representations involved here. 
 
I will not attempt to give a simpler g.f. for your M[n], where:
   M[n+1](z) = M[n](z)*C[n](z^n*M[n-1]^(n-1)); 
 
but using PARI I will list the first 31 terms for n=2..10 (below). 
 
You have some impressive results so far. 
     Paul 
 
(PARI)
{M=1/(1-x+O(x^31));
 for(n=2,10,
   M=M*sum(k=0,30,binomial(n*k,k)/((n-1)*k+1)*(x^n*M^(n-1))^k);
   print("n="n":");print(Vec(M))
 )
}

n=2: 
[1, 1, 2, 3, 6, 11, 23, 47, 102, 221, 493, 1105, 2516, 5763, 13328, 
30995, 72556, 170655, 403351, 957135, 2279948, 5449013, 13063596, 
31406517, 75701508, 182902337, 442885683, 1074604289, 2612341856, 
6361782007, 15518343597]
n=3: 
[1, 1, 2, 4, 9, 20, 48, 116, 288, 724, 1849, 4768, 12423, 32628, 
86342, 229952, 616042, 1659012, 4489101, 12199521, 33284546, 
91140797, 250396629, 690043032, 1907022197, 5284167884, 14677681554, 
40862469713, 114001697975, 318681491431, 892495078849]
n=4: 
[1, 1, 2, 4, 10, 24, 62, 160, 425, 1140, 3105, 8528, 23643, 66008, 
185526, 524384, 1489810, 4251852, 12184745, 35048405, 101156752, 
292865417, 850314803, 2475327088, 7223400899, 21126670372, 
61920289652, 181838859665, 534980035009, 1576653291583, 
4654122552827]
n=5: 
[1, 1, 2, 4, 10, 25, 67, 180, 495, 1375, 3871, 10993, 31493, 90843, 
263686, 769466, 2256135, 6643082, 19634705, 58232350, 173242381, 
516860717, 1546035258, 4635543843, 13929569399, 41943013047, 
126532961332, 382396277940, 1157548532794, 3509416800398, 
10655157352027]
n=6: 
[1, 1, 2, 4, 10, 25, 68, 186, 522, 1479, 4246, 12289, 35872, 105411, 
311662, 926270, 2765778, 8292296, 24953437, 75338686, 228140842, 
692733127, 2108652750, 6433255041, 19668210742, 60247367313, 
184879648441, 568281131800, 1749490180723, 5393738767856, 
16651705792875]
n=7: 
[1, 1, 2, 4, 10, 25, 68, 187, 529, 1514, 4393, 12856, 37944, 112740, 
337017, 1012671, 3056978, 9265765, 28187304, 86028764, 263339173, 
808256040, 2486804090, 7668374150, 23694936466, 73354869098, 
227489892754, 706643494680, 2198340096395, 6848600059320, 
21363953916171]
n=8: 
[1, 1, 2, 4, 10, 25, 68, 187, 530, 1522, 4437, 13056, 38766, 115908, 
348733, 1054775, 3205251, 9780165, 29951916, 92029644, 283606809, 
876333336, 2714448570, 8426790446, 26213887250, 81699432762, 
255072076566, 797641572824, 2498068134817, 7834443329344, 
24602515926899]
n=9: 
[1, 1, 2, 4, 10, 25, 68, 187, 530, 1523, 4446, 13110, 39030, 117060, 
353404, 1072811, 3272526, 10024884, 30825490, 95103360, 294300321, 
913203240, 2840650881, 8856206258, 27667828196, 86601978030, 
271545341505, 852829532176, 2682483667081, 8449321174753, 
26648690288909]
n=10: 
[1, 1, 2, 4, 10, 25, 68, 187, 530, 1523, 4447, 13120, 39095, 117400, 
354974, 1079493, 3299426, 10128934, 31216245, 96538200, 299477771, 
931628670, 2905496841, 9082358938, 28450647731, 89294760932, 
280759352295, 884216374006, 2788990466531, 8809540396813, 
27863492984436]

END. 
----------------------------------------------------------
On Sat, 4 Nov 2006 23:30:30 -0500 (EST) Emeric Deutsch
<deutsch at duke.poly.edu> writes:
> Dear Seqfans,
> We (Lou Shapiro and myself) would like to submit a few
> sequences similar to A124497 (shown at the bottom of this
> message). They will refer to 1-2-3-4 trees, 1-2-3-4-5
> trees, etc. rather than to 1-2-3 trees. We can describe
> the corresponding g.f.'s but we need help in finding the
> terms themselves.
> 
> Here is the description of all the g.f.'s. Let M[k](z) be
> the g.f. of the 1-2-...-k trees with thinning limbs. I will
> describe these functions; consequently, we may disregard
> the objects counted by them.
> 
> Denote by C[k](z) the g.f. of the k-ary trees. C[k]
> satisfies C[k]=1+zC[k]^k and the coefficient of z^n
> in its series expansion is
> (*)      binom(kn,n)/[(k-1)n+1].
> 
> I will write Mk for M[k] and Ck for C[k].
> 
> Obviously, M1(z)=1/(1-z) (path-trees). The next one is 
> M2(z)=M1(z)C2(z^2*M1), i.e. M1 times C2 evaluated at
> z^2*M1. This gives no problem since
> C2(z)=[1-sqrt(1-4z)]/(2z). We obtain A090344.
> 
> Next M3(z)=M2(z)C3(z^3*M2^2); since there is an
> explicit expression for C3(z), namely
> C3(z)=2/sqrt(3z)*sin[(1/3)*arcsin(sqrt(27z/4))],
> this yields easily A124497.
> 
> From here on we need help.
> M4(z)=M3(z)C4(z^4*M3^3); 
> Most probably sequence starts with
> 1,1,2,4,10,24,62,160,425,1140,3105 (I used some terms
> of the series of C4(z); see (*)).
> 
> M5(z)=M4(z)C5(z^5*M4^4);
> Most probably sequence starts with
> 1,1,2,4,10,25,67,180,495,1375,3871 (I used some terms
> of the series of C5(z); see (*)).
> 
> And so on.
> 
> These sequences approach A124344:
> 1,1,2,4,10,25,68,187,530,1523,4447,13121,39107,...
> 
> Many thanks.
> Emeric





From pauldhanna at juno.com  Sun Nov  5 08:04:15 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Sun, 5 Nov 2006 02:04:15 -0500
Subject: A124497: 1-2-3-...-k Trees
Message-ID: <20061105.020415.1648.1.pauldhanna@juno.com>

Emeric (and Seqfans),
     Correction to my typo in prior e-mail:
> I will not attempt to give a simpler g.f. for your M[n], where:
>    M[n+1](z) = M[n](z)*C[n](z^n*M[n-1]^(n-1)); 
 
that should have been:
 
M[1](z) = 1/(1-z);
 
M[n+1](z) = M[n](z)*C[n+1](z^(n+1)*M[n]^n); 
 
This was more clearly given by Emeric, but I correct myself for the sake
of accuracy. 
    Paul





From zakseidov at yahoo.com  Sun Nov  5 09:49:35 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 00:49:35 -0800 (PST)
Subject: A124057 is subset of A045940
Message-ID: <20061105084935.7574.qmail@web38212.mail.mud.yahoo.com>

A124057 is subset of A045940
from which more terms for A124057 
can be retrieved:

602,603,1083,2012,2091,2522,2523,2524,2634,2763,3243,3355,4202,4203,4921,4922,4923,5034,5035,5132,5203,5282,5283,5785,5882,5954,5972,6092,6212,6476,6962,6985,7730,7731,7945,8393,8825,8956,8972


%N A045940 Numbers n such that factorizations of n
through n+3 have same number of primes (including 
               multiplicities).

%N A124057 Numbers n such that n, n+1, n+2 and n+3 are
products of exactly 3 primes.



 
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From jvospost3 at gmail.com  Sun Nov  5 17:53:45 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 5 Nov 2006 08:53:45 -0800
Subject: A124057 is subset of A045940
In-Reply-To: <20061105084935.7574.qmail@web38212.mail.mud.yahoo.com>
References: <20061105084935.7574.qmail@web38212.mail.mud.yahoo.com>
Message-ID: <5542af940611050853h50af8ae5ue5004d32f982bef0@mail.gmail.com>

Dear Zak.

You're right.  I knew that; it's in my handwritten notes.  I should have you
submit that as a comment, and also have A045940 added as a cf to A124057,
and also show your extension. Thank you!

Shalom,

Jonathan

On 11/5/06, zak seidov <zakseidov at yahoo.com> wrote:
>
> A124057 is subset of A045940
> from which more terms for A124057
> can be retrieved:
>
>
> 602,603,1083,2012,2091,2522,2523,2524,2634,2763,3243,3355,4202,4203,4921,4922,4923,5034,5035,5132,5203,5282,5283,5785,5882,5954,5972,6092,6212,6476,6962,6985,7730,7731,7945,8393,8825,8956,8972
>
>
> %N A045940 Numbers n such that factorizations of n
> through n+3 have same number of primes (including
>                multiplicities).
>
> %N A124057 Numbers n such that n, n+1, n+2 and n+3 are
> products of exactly 3 primes.
>
>
>
>
>
> ____________________________________________________________________________________
> Access over 1 million songs - Yahoo! Music Unlimited
> (http://music.yahoo.com/unlimited)
>
>
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From franktaw at netscape.net  Sun Nov  5 18:35:13 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Sun, 05 Nov 2006 12:35:13 -0500
Subject: A124015 thoughts
In-Reply-To: <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
Message-ID: <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com>

Some years ago, researchers at Brown University put together a 
collection of text from a large variety of source to serve as a 
cross-section of English usage, for study of this kind of question.  If 
you're going to add a sequence with statistics of this sort, that would 
be the place to start.

See, e.g., Wikipedia: http://en.wikipedia.org/wiki/Brown_Corpus.

Franklin T. Adams-Watters

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From bdm at cs.anu.edu.au  Sun Nov  5 23:40:24 2006
From: bdm at cs.anu.edu.au (Brendan McKay)
Date: Mon, 6 Nov 2006 09:40:24 +1100
Subject: A124015 thoughts
In-Reply-To: <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com> <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com>
Message-ID: <20061105224024.GC28253@cs.anu.edu.au>

* franktaw at netscape.net <franktaw at netscape.net> [061106 04:35]:
> Some years ago, researchers at Brown University put together a 
> collection of text from a large variety of source to serve as a 
> cross-section of English usage, for study of this kind of question.  If 
> you're going to add a sequence with statistics of this sort, that would 
> be the place to start.
> 
> See, e.g., Wikipedia: http://en.wikipedia.org/wiki/Brown_Corpus.
> 
> Franklin T. Adams-Watters

Yes, but I don't know if statistics based on sampling belong in OEIS.
They are not precise well-defined quantities.  Counts derived from
single texts are more concrete, but even then there are multiple
editions, even changes between printings of the same edition.

Brendan.





From zakseidov at yahoo.com  Sun Nov  5 23:53:40 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 14:53:40 -0800 (PST)
Subject: Don't restrict OEIS sequence domains!
In-Reply-To: <001501c70116$e2fabd20$6401a8c0@yourxhtr8hvc4p>
Message-ID: <20061105225340.19195.qmail@web38215.mail.mud.yahoo.com>

David, 

I simply remember Neil's recommendation
(somewhere in his tip's) to omit trivial cases
e.g. if each second term is zero.

And you may wish to 
put in %C A053699 
some of your observations, 
e.g., a(n_odd)=2.
Zak

--- David Wilson <davidwwilson at comcast.net> wrote:

> ----- Original Message ----- 
> From: "zak seidov" <zakseidov at yahoo.com>
> To: <seqfan at ext.jussieu.fr>
> Sent: Sunday, November 05, 2006 1:28 AM
> Subject: Re: Observation on A053669
> 
> 
> > Yes,
> > and because a(n_odd)=2
> > keep only even n's.
> > Zak
> 
> > [Earlier included message elided]
> ---------------------------
> 
> Let's take a look at Zak's recommendation to
> restrict the domain of A053699 
> to even n:
> 
> A053699(n) currently looks like:
> 
>
2,3,2,3,2,5,2,3,2,3,2,5,2,3,2,3,2,5,2,3,2,3,2,5,2,3,2,3,2,5,2,3,2,3,2,7,...
> 
> But a(k) = 2 for odd k, these values are
> uninteresting, so list only a(2n):
> 
>
3,3,5,3,3,5,3,3,5,3,3,5,3,3,7,3,3,5,3,3,5,3,3,5,3,3,5,3,3,7,3,3,5,3,3,5,...
> 
> But now a(k) = 3 unless 6 divides k, so we should
> list only a(6n):
> 
>
5,5,5,5,7,5,5,5,5,7,5,5,5,5,7,5,5,5,5,7,5,5,5,5,7,5,5,5,5,7,5,5,5,5,11,..
> 
> But again, a(k) = 5 except when 30 divides k, so
> let's eject these boring 
> values and keep only a(30n):
> 
>
7,7,7,7,7,7,11,7,7,7,7,7,7,11,7,7,7,7,7,7,11,7,7,7,7,7,7,11,7,7,7,7,7,...
> 
> But look at all the uninteresting 7's. Maybe we
> should list only a(210n)...
> 
> I think you get the picture. In the case of this
> sequence, we repeatedly 
> eliminate uninteresting domain values, but we never
> flush out the elusive 
> "interesting domain" of the sequence. Indeed, we can
> eliminate the entire 
> domain as uninteresting, whereas I think the
> original sequence is indeed 
> quite interesting.
> 
> Clearly, there are times when restricted domains are
> useful. For example, 
> since the areas of Pythagorean triangles are always
> divisible by 6, it would 
> seem reasonable, in the name of compactness, to
> prefer "Number of 
> Pythagorean triangles with area 6n" over "Number of
> Pythagorean triangles of 
> area n", since the latter would have a(n) = 0 for n
> != 6k. Nevertheless, I 
> would advocate having both sequences in the OEIS,
> because we never know if 
> the user will be searching for the compact or
> inflated version of the 
> sequence.
> 
> Similarly, if we adopt Zak's recommendation to
> restrict A065577 to a domain 
> of even values, and then someone looks up
> 
> 2,3,2,3,2,5,2,3,2,3,2,5
> 
> they will probably miss the most straightforward
> explanation of their 
> sequence.
> 
> In short: Don't restrict domains of OEIS sequences
> unless it makes clear 
> sense to do so, and if you do include a sequence on
> some restricted 
> "interesting" domain, have pity on the OEIS users
> and also include the same 
> sequence on its fullest possible domain.
> 
> ----- Original Message ----- 
> From: "zak seidov" <zakseidov at yahoo.com>
> To: <seqfan at ext.jussieu.fr>
> Sent: Sunday, November 05, 2006 1:28 AM
> Subject: Re: Observation on A053669
> 
> 
> > Yes,
> > and because a(n_odd)=2
> > keep only even n's.
> > Zak 
> 
> 



 
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From franktaw at netscape.net  Mon Nov  6 00:26:46 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Sun, 05 Nov 2006 18:26:46 -0500
Subject: A124015 thoughts
In-Reply-To: <20061105224024.GC28253@cs.anu.edu.au>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com> <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com> <20061105224024.GC28253@cs.anu.edu.au>
Message-ID: <8C8CF48FECF25AB-958-F877@FWM-M05.sysops.aol.com>

I'm not saying one way or the other whether this sort of thing belongs
in the OEIS; just that, if it is going to be there, the Brown Corpus
would be the best source for it.

The Brown Corpus is about as concrete as you can get in this area.  It
is very well defined and a standard research reference.

Franklin T. Adams-Watters


-----Original Message-----
From: bdm at cs.anu.edu.au

  * franktaw at netscape.net <franktaw at netscape.net> [061106 04:35]:
> Some years ago, researchers at Brown University put together a
> collection of text from a large variety of source to serve as a
> cross-section of English usage, for study of this kind of question.  
If
> you're going to add a sequence with statistics of this sort, that 
would
> be the place to start.
>
> See, e.g., Wikipedia: http://en.wikipedia.org/wiki/Brown_Corpus.
>
> Franklin T. Adams-Watters

Yes, but I don't know if statistics based on sampling belong in OEIS.
They are not precise well-defined quantities.  Counts derived from
single texts are more concrete, but even then there are multiple
editions, even changes between printings of the same edition.

Brendan.


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From zakseidov at yahoo.com  Mon Nov  6 00:51:18 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 15:51:18 -0800 (PST)
Subject: A124057 is subset of A045940
Message-ID: <20061105235118.90653.qmail@web38203.mail.mud.yahoo.com>

Just submitted:

My Q to seqfans:
Really, there are no numbers n such that n, n+1, n+2
and n+3 are products of 6 primes?
No for n<7258290.
Thanks, Zak


%I A124728
%S A124728
4023,7314,9162,12122,12123,16674,19434,19940,23874,24723,29094,33234,35124,35125,39234,42182,42183,44163,45175,46988,49147,51793,52854,52855,54584,54585,54663,58375,63594,64074,64075,64323,64491,64712,64713,65943,67507,67962,70262,71124,75123,75446,77283,80792,80793,81025,81963,81975,86823
%N A124728 Numbers n such that n, n+1, n+2 and n+3 are
products of 4 primes.
%C A124728 Subset of A045940  Numbers n such that
factorizations of n through n+3 have same number of
primes (including multiplicities).
Cf. A124057, A124729  Numbers n such that n, n+1, n+2
and n+3 are products of exactly 3,5 primes. 
There are no numbers n such that n, n+1, n+2 and n+3
are products of exactly 6 primes(?)
%e A124728 4023=3^3*149, 4024=2^3*503, 4025=5^2*7*23,
4026=2*3*11*61 (all products of 4 primes).
%Y A124728 A045940, A124057, A124729.
%O A124728 1
%K A124728 ,nonn,
%A A124728 Zak Seidov  (zakseidov at yahoo.com), Nov 05
2006


%I A124729
%S A124729
57967,491875,543303,584647,632148,632149,715374,824523,878875,914823,930123,931623,955448,964143,995874
%N A124729 Numbers n such that n, n+1, n+2 and n+3 are
products of 5 primes.
%C A124729 Subset of A045940  Numbers n such that
factorizations of n through n+3 have same number of
primes (including multiplicities).
Cf. A124057, A124728  Numbers n such that n, n+1, n+2
and n+3 are products of exactly 3,4 primes. 
There are no numbers n such that n, n+1, n+2 and n+3
are products of exactly 6 primes(?)
%e A124729 57967=7^3*13^2, 57968=2^4*3623,
57969=3^3*19*113, 57970=2*5*11*17*31 (all product of 5
primes (including multiplicities)};
632148 is the first number such that n through n+4 are
5-almost primes.
%Y A124729 A045940, A124057, A124728.
%O A124729 1
%K A124729 ,nonn,
%A A124729 Zak Seidov  (zakseidov at yahoo.com), Nov 05
2006



 
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From zakseidov at yahoo.com  Mon Nov  6 02:02:27 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 17:02:27 -0800 (PST)
Subject: n..n+6 are all N-almost primes
Message-ID: <20061106010227.58041.qmail@web38206.mail.mud.yahoo.com>

Just dubmitted.

My Q to gurus:
Are there longer set(s) of successive numbers
n..n+m (m>6)
which are products of the same number (N) of primes
(including multiplicities)?

It seems that for  larger N's, 
the larger m's may be(?).

Thanks, Zak
  
%I A000001
%S A000001
211673,298433,355923,381353,460801,506521,540292,568729,690593,705953,737633,741305,921529
%N A000001 Numbers n such that factorizations of
n..n+6 have same number of primes (including
multiplicities).
%C A000001 Subset of A045940  Numbers n such that
factorizations of n..n+3 have same number of primes
(including multiplicities).
%e A000001 211673=7*11*2749, 211674=2*3*35279,
211675=5^2*8467, 211676=2^2*52919, 211677=3*37*1907,
211677=2*109*971, 211679=13*19*857 all 3-almost
primes;
355923=3^2*71*557, 355924=2^2*101*881,
355925=5^22*23*619, 355926=2*3*137*433,
355927=11*13*19*131, 355928=2^3*44491,
355929=3*7*17*997 all 4-almost primes.
%Y A000001 A045940.
%O A000001 1
%K A000001 ,nonn,
%A A000001 Zak Seidov  (zakseidov at yahoo.com), Nov 05
2006



 
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From zakseidov at yahoo.com  Mon Nov  6 03:49:26 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 18:49:26 -0800 (PST)
Subject: n..n+6 are all N-almost primes
In-Reply-To: <20061106010227.58041.qmail@web38206.mail.mud.yahoo.com>
Message-ID: <20061106024926.64870.qmail@web38203.mail.mud.yahoo.com>

--- zak seidov <zakseidov at yahoo.com> wrote:
> Are there longer set(s) of successive numbers
> n..n+m (m>6)
> which are products of the same number (N) of primes
> (including multiplicities)?

Me: Just submitted:
%I A000001
%S A000001
3405122,3405123,6612470,8360103,8520321,9306710
%N A000001 Numbers n such that n..n+7 are products of
the same number of primes.
%C A000001 Note that because 3405130=2*5*167*2039 is
also 4-almost prime,
3405122 is the first n such that numbers n..n+8 are
products of the same number N of primes (N=4).
%e A000001 3405122=2*7*29*8387, 3405123=3^2*19*19913,
3405124=2^2*127*6703, 3405125=5^3*27241,
3405126=2*3*59*9619, 3405127=11*23*43*313,
3405128=2^3*425641, 3405129=3*7*13*12473 all 4-almost
primes.
%O A000001 1
%K A000001 ,nonn,
%A A000001 Zak Seidov  (zakseidov at yahoo.com), Nov 05
2006

Now we need to find numbers n such that n..n+m (m>8)
are N-almost primes (n'd be >10^7).

Thanks, Zak

BTW The "annoying" ad at the bottom of message is 
from Yahoo!Mail - not me, sorry.

<original message>
> Just dubmitted.
> 
> My Q to gurus:
> Are there longer set(s) of successive numbers
> n..n+m (m>6)
> which are products of the same number (N) of primes
> (including multiplicities)?
> 
> It seems that for  larger N's, 
> the larger m's may be(?).
> 
> Thanks, Zak
>   
> %I A000001
> %S A000001
>
211673,298433,355923,381353,460801,506521,540292,568729,690593,705953,737633,741305,921529
> %N A000001 Numbers n such that factorizations of
> n..n+6 have same number of primes (including
> multiplicities).
> %C A000001 Subset of A045940  Numbers n such that
> factorizations of n..n+3 have same number of primes
> (including multiplicities).
> %e A000001 211673=7*11*2749, 211674=2*3*35279,
> 211675=5^2*8467, 211676=2^2*52919, 211677=3*37*1907,
> 211677=2*109*971, 211679=13*19*857 all 3-almost
> primes;
> 355923=3^2*71*557, 355924=2^2*101*881,
> 355925=5^22*23*619, 355926=2*3*137*433,
> 355927=11*13*19*131, 355928=2^3*44491,
> 355929=3*7*17*997 all 4-almost primes.
> %Y A000001 A045940.
> %O A000001 1
> %K A000001 ,nonn,
> %A A000001 Zak Seidov  (zakseidov at yahoo.com), Nov 05
> 2006
> 




 
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From davidwwilson at comcast.net  Mon Nov  6 04:10:17 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sun, 5 Nov 2006 22:10:17 -0500
Subject: A124015 thoughts
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com> <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com> <20061105224024.GC28253@cs.anu.edu.au> <8C8CF48FECF25AB-958-F877@FWM-M05.sysops.aol.com>
Message-ID: <001901c70151$16106360$6401a8c0@yourxhtr8hvc4p>

Sounds great. Where do I get the Brown Corpus in computer-readable form?

----- Original Message ----- 
From: <franktaw at netscape.net>
To: <seqfan at ext.jussieu.fr>
Sent: Sunday, November 05, 2006 6:26 PM
Subject: Re: A124015 thoughts


> I'm not saying one way or the other whether this sort of thing belongs
> in the OEIS; just that, if it is going to be there, the Brown Corpus
> would be the best source for it.
>
> The Brown Corpus is about as concrete as you can get in this area.  It
> is very well defined and a standard research reference.
>
> Franklin T. Adams-Watters
>
>
> -----Original Message-----
> From: bdm at cs.anu.edu.au
>
>  * franktaw at netscape.net <franktaw at netscape.net> [061106 04:35]:
>> Some years ago, researchers at Brown University put together a
>> collection of text from a large variety of source to serve as a
>> cross-section of English usage, for study of this kind of question.
> If
>> you're going to add a sequence with statistics of this sort, that
> would
>> be the place to start.
>>
>> See, e.g., Wikipedia: http://en.wikipedia.org/wiki/Brown_Corpus.
>>
>> Franklin T. Adams-Watters
>
> Yes, but I don't know if statistics based on sampling belong in OEIS.
> They are not precise well-defined quantities.  Counts derived from
> single texts are more concrete, but even then there are multiple
> editions, even changes between printings of the same edition.
>
> Brendan.
>
>
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading 
> spam and email virus protection.
>
>
>
> -- 
> No virus found in this incoming message.
> Checked by AVG Free Edition.
> Version: 7.5.430 / Virus Database: 268.13.28/518 - Release Date: 11/4/2006 
> 5:30 PM
>
> 






From dean at math.ucdavis.edu  Mon Nov  6 10:39:09 2006
From: dean at math.ucdavis.edu (Dean Hickerson)
Date: Mon, 6 Nov 2006 01:39:09 -0800
Subject: terms lack
Message-ID: <20061106093909.GA16454@math.ucdavis.edu>

Mostly to Yasutoshi Kohmoto <zbi74583 at boat.zero.ad.jp>:

> Recently I updated more terms of A054572, which are calculated by
> Richard Mathar.
> And Neil claimed or got angry.
> He said, "You are red card. Don't submit a wrong sequences"
> I am exaggerating a little.
> These words are not exactly what he wrote.
>
> To Neil
> They are correct, not "wrong".
> It is only not complete.

The terms that are listed in a sequence should be the start of the sequence.
If there are missing terms in between the listed ones, then the sequence
entry is not just incomplete, it's wrong.

If you've computed some terms of a sequence, but you're not sure if there
are smaller ones that you've missed, then you should only include the ones
that you're sure are at the start of the sequence.  You can add a comment
about the larger terms, stating that it's not known if there are smaller ones.
For example, the "EXTENSIONS" lines for the sequence A000043 (Mersenne
exponents) mention some recently discovered terms.  But there might be
some smaller ones, so these are not included in the sequence.

> I defined many "divisor" functions and calculated many Amicable Numbers and
> Perfect Numbers using those functions.
> And I submitted them as sequences to  OEIS.
>
> http://www.research.att.com/~njas/sequences/?q=kohmoto+amicable&sort=0&fmt=0&language=english&go=Search
>
> http://www.research.att.com/~njas/sequences/?q=kohmoto+perfect&sort=0&fmt=0&language=english&go=Search
>
> I suppose that they are almost complete but some of them might not be
> complete.

I've found missing terms for several of your sequences:


Sequences A045613 and A045614 ("Super Unitary Amicable Number") consist
of pairs of distinct numbers a and b such that

    usigma(usigma(a)) = usigma(usigma(b)) = a+b.

(The definitions don't make this clear, but A045613 contains the smaller
members of the pairs and A045614 contains the larger members.  Also they're
sorted by their smallest members, so 155 comes before 142 in A045614.)  The
first 3 pairs that you've listed are (105, 155), (110, 142) and (2145, 3055).
But you've missed several smaller ones:

    (33, 35), (208, 224), (268, 272), (455, 601), (695, 1033), (812, 956),
    (1609, 1847), (1808, 2512), (1892, 3004), (1913, 2407), (2096, 2224)


A051594 and A051595 are "(-1)sigma amicable numbers".  (Again, the definition
doesn't make it clear that A051595 has the smaller terms and A051594 has the
larger terms.)  The first pair that you've listed is (1969706592, 2236072608).
But there are many smaller ones, starting with:

    (429552, 466200), (497808, 604656), (800496, 1103760).


A036471-A036474 are defined as "Amicable quadruple: 4 different numbers
which satisfy sigma(a)=sigma(b)=sigma(c)=sigma(d)=a+b+c+d".  The first
quadruple that you've listed has smallest element
342151462276356306033089201934180, but here's a much smaller one:

    (3270960, 3361680, 3461040, 3834000)

(I think that's the smallest one, but I could be wrong.)  If someone can
determine the first several terms of this sequence, then it should be
edited; otherwise it should be deleted.


In some of your sequences, I wasn't able to find any smaller terms, but the
ones that you've listed are so large that I'm not at all confident that they
are the smallest ones.  For example, A038362 and A038363 are defined as
"A Rational Amicable Number: two different numbers a, b which satisfies the
following equation: sigma(a)=sigma(b)=(a+b)^3/(a^2+b^2)", and the smallest
pair that you've included is:

    (26403469440047700, 30193441130006700)

Do you have any reason to think that that's the smallest such pair?  If not,
and if you can't determine what the first few terms really are, then the
sequence should be deleted.


I've only looked at a few of your sequences, but that's enough to convince
me that they should all be checked.

Neil, I'll submit corrections for A045613, A045614, A051594, and A051595,
but I'm not going to work on the others.  Maybe they should be deleted or
marked as "probation" until the author or someone else fixes them.

Dean Hickerson
dean at math.ucdavis.edu





From bdm at cs.anu.edu.au  Mon Nov  6 12:45:30 2006
From: bdm at cs.anu.edu.au (Brendan McKay)
Date: Mon, 6 Nov 2006 22:45:30 +1100
Subject: terms lack
In-Reply-To: <20061106093909.GA16454@math.ucdavis.edu>
References: <20061106093909.GA16454@math.ucdavis.edu>
Message-ID: <20061106114530.GB4232@cs.anu.edu.au>

* Dean Hickerson <dean at math.ucdavis.edu> [061106 20:40]:
> The terms that are listed in a sequence should be the start of the sequence.
> If there are missing terms in between the listed ones, then the sequence
> entry is not just incomplete, it's wrong.
> 
> If you've computed some terms of a sequence, but you're not sure if there
> are smaller ones that you've missed, then you should only include the ones
> that you're sure are at the start of the sequence.  You can add a comment
> about the larger terms, stating that it's not known if there are smaller ones.
> For example, the "EXTENSIONS" lines for the sequence A000043 (Mersenne
> exponents) mention some recently discovered terms.  But there might be
> some smaller ones, so these are not included in the sequence.
 
Is there an official rule about this?  I'm not sure it is correct.
Of course we should not submit sequences with possible gaps, but
suppose we can compute a(10) thru a(20) but have trouble with
a(1) thru a(9).  I don't see why we can't submit a(10..20) with
an OFFSET of 10 and a comment explaining why a(1..9) are absent.

Brendan.





From Annette.Warlich at t-online.de  Mon Nov  6 13:05:20 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Mon, 06 Nov 2006 13:05:20 +0100
Subject: Fibbinary sequence
Message-ID: <454F2500.2020107@t-online.de>

A courious result by chance:

A003714	Fibbinary numbers:
       if n = F_i1+F_i2+...+F_ik is the Zeckendorf representation of n
       (i.e. write n in Fibonacci number system)
       then
         a(n) = 2^{i1-2}+2^{i2-2}+...+2^{ik-2}.

	0, 1, 2, 4, 5, 8, 9, 10, 16, 17, 18, 20, 21, 32, 33, 34, 36, 37,
      40, 41, 42, 64, 65, 66, 68, 69, 72, 73, 74, 80, 81, 82, 84, 85,
     128, 129, 130, 132, 133, 136, 137, 138, 144, 145, 146, 148, 149,
     160, 161, 162, 164, 165, 168, 169, 170, 256, 257, 258, 260

I was checking some variants of the Pascal-matrix and got the
following
Let Px =
      1       .    .     .     .    .  .  .
      1      -1    .     .     .    .  .  .
    3/2      -3    1     .     .    .  .  .
      4      -9    6    -1     .    .  .  .
     15     -35   30   -10     1    .  .  .
  137/2  -345/2  165   -75    15   -1  .  .
 1491/4 -2009/2 1050  -560 315/2  -21  1  .
   2374   -6720 7588 -4515  1540 -294 28 -1

then the index r of the rows, r=1..inf(??) , which contain only integers, is A003714
      1, 2, 4, 5, 8, 9, 10, 16, 17, 18, 20, 21, 32, 33, 34, 36, 37,
     40, 41, 42, 64,

(although I checked this only up to N=64)


Explanation: -----------------------------------------------------------

The matrix Px is constructed as a variant of Pj
Pj:
 1  .  .   .  .   . .  .
 1 -1  .   .  .   . .  .
 1 -2  1   .  .   . .  .
 1 -3  3  -1  .   . .  .
 1 -4  6  -4  1   . .  .
 1 -5 10 -10  5  -1 .  .
 1 -6 15 -20 15  -6 1  .
 1 -7 21 -35 35 -21 7 -1

which is the column-signed version of the Pascalmatrix P
   Pj = P*J
where J is diagonal unit with alternating signs (diag(1,-1,1,-1,...))

The matrix S2 of the stirling-numbers of 2'nd kind
 1   .   .    .    .   .  . .
 1   1   .    .    .   .  . .
 1   3   1    .    .   .  . .
 1   7   6    1    .   .  . .
 1  15  25   10    1   .  . .
 1  31  90   65   15   1  . .
 1  63 301  350  140  21  1 .
 1 127 966 1701 1050 266 28 1

and S1 of the stirling numbers of first kind (where also S1 = S2^-1)
     1     .      .    .     .   .   . .
    -1     1      .    .     .   .   . .
     2    -3      1    .     .   .   . .
    -6    11     -6    1     .   .   . .
    24   -50     35  -10     1   .   . .
  -120   274   -225   85   -15   1   . .
   720 -1764   1624 -735   175 -21   1 .
 -5040 13068 -13132 6769 -1960 322 -28 1

with the eigenvalues Z(1) = diag(1,1/2,1/3,1/4) such that

   G = S2 * Z(1) * S2^-1

G:
     1     .    .     .   .    .   .   .
   1/2   1/2    .     .   .    .   .   .
   1/6   1/2  1/3     .   .    .   .   .
     0   1/4  1/2   1/4   .    .   .   .
 -1/30     0  1/3   1/2 1/5    .   .   .
     0 -1/12    0  5/12 1/2  1/6   .   .
  1/42     0 -1/6     0 1/2  1/2 1/7   .
     0  1/12    0 -7/24   0 7/12 1/2 1/8

form an eigensystem of G.
G ist interesting as it contains the bernoulli-
numbers, and coefficients of the integrals of the
bernoulli-functions.



Also G is Eigenmatrix of the signed Pascal-matrix Pj,
with the eigenvalues J = diag(1,-1,1,-1,...)

  Pj = G * J * G^-1

and Pj is used in divergent Euler-summation, Hasse's
zeta-summation and elsewhere.




Now I tried a variant of P by varying G, where the eigenvalues
are not Z(1), but are V(1/2) = diag(1,1/2,1/4,1/8,....) for some
analyses about summing zeta series, since Z(1) and V(1/2) are
closely related in this subject and complementary/mutually supplementary
in certain manners.

Now with the G-variant

  Gx  = S2 * V(1/2) * S2^-1
Gx:
     1      .       .        .      .      .    .     .
   1/2    1/2       .        .      .      .    .     .
     0    3/4     1/4        .      .      .    .     .
  -1/4    3/8     3/4      1/8      .      .    .     .
     0   -5/8   15/16      5/8   1/16      .    .     .
   1/2 -15/16  -15/32    45/32  15/32   1/32    .     .
     0    7/4 -105/32    35/64 105/64  21/64 1/64     .
 -17/8 147/32    7/32 -735/128  35/16 105/64 7/32 1/128

I come to

  Px =         Gx               * J *       Gx^-1
     =  ( S2 * V(1/2) * S2^-1 ) * J * (S2 * V(2) * S2^-1)

(while Pj is
  Pj =         G                * J *       G ^-1
     =  ( S2 * Z(-1) * S2^-1 )  * J * (S2 * Z(1) * S2^-1)
  )

and as shown above:

Px:
      1       .    .     .     .    .  .  .
      1      -1    .     .     .    .  .  .
    3/2      -3    1     .     .    .  .  .
      4      -9    6    -1     .    .  .  .
     15     -35   30   -10     1    .  .  .
  137/2  -345/2  165   -75    15   -1  .  .
 1491/4 -2009/2 1050  -560 315/2  -21  1  .
   2374   -6720 7588 -4515  1540 -294 28 -1

and the rows containing only integers, are uniquely
indicated by A003714, up to r=64, where the rownumber r
starts with 1.

Don't have an idea, how this coincidence could be interesting -
I thought I'd just sentd the observation..

Regards -

Gottfried Helms







From Labos at ana.sote.hu  Mon Nov  6 13:38:39 2006
From: Labos at ana.sote.hu (=?ISO-8859-2?Q?L=DFbos_Elem=DAr?=)
Date: Mon, 6 Nov 2006 13:38:39 MET-1MEST
Subject: terms lack
In-Reply-To: <20061106114530.GB4232@cs.anu.edu.au>
References: <20061106093909.GA16454@math.ucdavis.edu>
Message-ID: <6578DDB2A59@ana.sote.hu>

On 6 Nov 06, at 22:45, Brendan McKay wrote:

> * Dean Hickerson <dean at math.ucdavis.edu> [061106 20:40]:
> > The terms that are listed in a sequence should be the start of the sequence.
> > If there are missing terms in between the listed ones, then the sequence
> > entry is not just incomplete, it's wrong.
> > 
> > If you've computed some terms of a sequence, but you're not sure if there
> > are smaller ones that you've missed, then you should only include the ones
> > that you're sure are at the start of the sequence.  You can add a comment
> > about the larger terms, stating that it's not known if there are smaller ones.
> > For example, the "EXTENSIONS" lines for the sequence A000043 (Mersenne
> > exponents) mention some recently discovered terms.  But there might be
> > some smaller ones, so these are not included in the sequence.
>  
> Is there an official rule about this?  I'm not sure it is correct.
> Of course we should not submit sequences with possible gaps, but
> suppose we can compute a(10) thru a(20) but have trouble with
> a(1) thru a(9).  I don't see why we can't submit a(10..20) with
> an OFFSET of 10 and a comment explaining why a(1..9) are absent.
> 
> Brendan.

There are 2 cases :
 1  -- You cannot even know how many terms are missing.
          This is essentally the case of eg Mersenne primes..
-------------------
 2. --  You can specify the missing terms by list-
          position. In this second case you can introduce 
          a marker signal to point to missing position.

Handling of the cases can follow different standard..
The general case can be worthwhile to publish if 
so many hard computation is behind. 
Ordering terms somehow it can be mentioned that
the sequence is lacunar.
With correct description the situation, the case is
neither wrong, nor good. Perhaps unnecessary...

The empty sequence represents the full lacunarity.
Quite correctly. However the empty sequence is
not unique. 

Labos 






From dean at math.ucdavis.edu  Mon Nov  6 14:22:58 2006
From: dean at math.ucdavis.edu (Dean Hickerson)
Date: Mon, 6 Nov 2006 05:22:58 -0800
Subject: terms lack
Message-ID: <20061106132258.GA19536@math.ucdavis.edu>

Mostly to Brendan McKay

> Is there an official rule about this?  I'm not sure it is correct.
> Of course we should not submit sequences with possible gaps, but
> suppose we can compute a(10) thru a(20) but have trouble with
> a(1) thru a(9).  I don't see why we can't submit a(10..20) with
> an OFFSET of 10 and a comment explaining why a(1..9) are absent.

Sure, that's fine.  In that case the sequence that's in the OEIS starts
with a(10), and it doesn't have any gaps.  It's part of a longer, partially
unknown, one that starts with a(1), but the sequence that it's part of
isn't in the OEIS.

But in cases like Kohmoto's sequences that wouldn't work.  For example,
suppose we know from hand calculation (as Kohmoto did for A045613), that
105, 110, and 2145 are in the sequence.  Presumably we can test that
there are no terms between 105 and 110, but we can't tell if there are
terms below 105 or between 110 and 2145.  So (ignoring the 4-term rule for
submissions) we could submit a sequence with just 105 and 110, and add
comments saying that there might be earlier terms and that 2145 is also
in the sequence, but we shouldn't submit the sequence as starting with
105, 110, 2145.

Dean Hickerson
dean at math.ucdavis.edu





From tbaruchel at free.fr  Mon Nov  6 17:31:32 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Mon, 6 Nov 2006 17:31:32 +0100
Subject: A fast (written in assembly) library for continued fractions
In-Reply-To: <20061103221729.GB2467@>
References: <20061103221729.GB2467@>
Message-ID: <20061106163132.GA30587@>

On Ven, nov 03, 2006 at 11:17:29 +0100, Thomas Baruchel wrote:
> I wrote for my own purposes the following X86 assembly code, intended

The final version of my code can be found at
  http://baruchel.free.fr/~thomas/wp/?p=10
I give also (in the comments of the file) advices for using the functions
from gcc, gfortran, g95. The fastest continued fractions in the west...

Next step will be functions for computing the Pell equation, but
pure FPU stack will not be possible this time (only eight registers), thus
some memory access will be required :-(

Regards,

-- 
Thomas Baruchel
  http://baruchel.free.fr/~thomas/wp/





From Eric.Angelini at kntv.be  Mon Nov  6 18:57:00 2006
From: Eric.Angelini at kntv.be (Eric Angelini)
Date: Mon, 6 Nov 2006 18:57:00 +0100
Subject: Langford/Skolem-like sequence : odd + even terms
Message-ID: <F05775148D000C4B9321A5113D53CB7D191001@KNTV-SERVER01.kntv.local>


Hello SeqFans,
Another Langford/Skolem-like sequence, with a mixture of odd and
even terms -- plus two questions:
http://www.cetteadressecomportecinquantesignes.com/OddEvenSkol.htm
Best,
?.






From noe at sspectra.com  Mon Nov  6 19:04:49 2006
From: noe at sspectra.com (T. D. Noe)
Date: Mon, 6 Nov 2006 10:04:49 -0800
Subject: Cyclotomic polynomial for n=0
Message-ID: <p06230985c175281c5eee@[192.168.1.43]>

For n=1,2,3,... the cyclotomic polynomials are

x-1, x+1, x^2+x+1,...

It seems that there are two conventions for n=0:

x or 1

Mathematica says it is 1.  I think Maple uses x.  Can someone explain this?

Thanks,

Tony





From franktaw at netscape.net  Mon Nov  6 22:27:14 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 06 Nov 2006 16:27:14 -0500
Subject: Cyclotomic polynomial for n=0
In-Reply-To: <p06230985c175281c5eee@[192.168.1.43]>
References: <p06230985c175281c5eee@[192.168.1.43]>
Message-ID: <8C8D001766F0D57-BD0-2F58@WEBMAIL-MA14.sysops.aol.com>

Strictly speaking, the cyclotomic polynomial for n=0 is undefned.  The 
two packages have chosen different ways to extend the concept to n=0; 
it would have been perfectly OK for them to have returned some kind of 
error.

There are reasonable arguments for either choice.  A lot comes down to 
the basic property that the degree is phi(n); the question then is, 
what is phi(0)?  Again, this is strictly speaking undefined, but there 
are two reasonable choices: 0 and 1.  phi(0) = 0 gives 1 as the value 
of the 0th cyclotomic polynomial.  If we take phi(0) = 1, then we need 
to choose a root for the polynomial.  Choosing 0 as the root gives the 
nice property that the cyclotomic polynomials for n >= 0 constitute all 
the primitive monic polynomials all of whose roots lie in the unit 
disk; hence the selection of x as the polynomial.

If you take the formula Phi_n(x) = Product_{0<=k<n, gcd(k,n) = 1} e^{2 
k Pi i / n}, you get Phi_0(x) = 1.  On the other hand, if we regard 
this as a product over distinct residues modulo n that are relatively 
prime to n, we get x - e^{2 Pi i / 0}, which is undefined.

You can also make a case for 0 as the "correct" value.  Phi_n(x) is 
x^n-1, divided by all x - u for primitive dth roots of unity u for d a 
proper divisor of n.  x^0-1 is 0; you can divide it by anything you 
want, and the result is still 0.  (Of course, this becomes an infinite 
product in the denominator - but this is only a plausibility argument.)

Franklin T. Adams-Watters


-----Original Message-----
From: noe at sspectra.com

For n=1,2,3,... the cyclotomic polynomials are

x-1, x+1, x^2+x+1,...

It seems that there are two conventions for n=0:

x or 1

Mathematica says it is 1.  I think Maple uses x.  Can someone explain 
this?

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From zbi74583 at boat.zero.ad.jp  Tue Nov  7 02:34:32 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Tue, 7 Nov 2006 10:34:32 +0900
Subject: terms lack
References: <20061106093909.GA16454@math.ucdavis.edu>
Message-ID: <018901c7020c$e2322a30$ef73d7dc@FMC57937B6BE00>

    To Dean Hicherson

    Thanks for editing my sequences.
    And thanks for a good explanation about "completeness".
    Now, I understand that  the  purpose of OEIS is "Identification of 
sequences", it is necessary.

    Should A038362 be deleted?
    Still, I think that it should exist , because a sequence is a mapping 
from N to N, so a subset of the images is also a good information for 
understanding the mapping.

    Yasutoshi


----- Original Message ----- 
From: "Dean Hickerson" <dean at math.ucdavis.edu>
To: <seqfan at ext.jussieu.fr>
Sent: Monday, November 06, 2006 6:39 PM
Subject: Re: terms lack


> Mostly to Yasutoshi Kohmoto <zbi74583 at boat.zero.ad.jp>:
>
>> Recently I updated more terms of A054572, which are calculated by
>> Richard Mathar.
>> And Neil claimed or got angry.
>> He said, "You are red card. Don't submit a wrong sequences"
>> I am exaggerating a little.
>> These words are not exactly what he wrote.
>>
>> To Neil
>> They are correct, not "wrong".
>> It is only not complete.
>
> The terms that are listed in a sequence should be the start of the 
> sequence.
> If there are missing terms in between the listed ones, then the sequence
> entry is not just incomplete, it's wrong.
>
> If you've computed some terms of a sequence, but you're not sure if there
> are smaller ones that you've missed, then you should only include the ones
> that you're sure are at the start of the sequence.  You can add a comment
> about the larger terms, stating that it's not known if there are smaller 
> ones.
> For example, the "EXTENSIONS" lines for the sequence A000043 (Mersenne
> exponents) mention some recently discovered terms.  But there might be
> some smaller ones, so these are not included in the sequence.
>
>> I defined many "divisor" functions and calculated many Amicable Numbers 
>> and
>> Perfect Numbers using those functions.
>> And I submitted them as sequences to  OEIS.
>>
>> http://www.research.att.com/~njas/sequences/?q=kohmoto+amicable&sort=0&fmt=0&language=english&go=Search
>>
>> http://www.research.att.com/~njas/sequences/?q=kohmoto+perfect&sort=0&fmt=0&language=english&go=Search
>>
>> I suppose that they are almost complete but some of them might not be
>> complete.
>
> I've found missing terms for several of your sequences:
>
>
> Sequences A045613 and A045614 ("Super Unitary Amicable Number") consist
> of pairs of distinct numbers a and b such that
>
>    usigma(usigma(a)) = usigma(usigma(b)) = a+b.
>
> (The definitions don't make this clear, but A045613 contains the smaller
> members of the pairs and A045614 contains the larger members.  Also 
> they're
> sorted by their smallest members, so 155 comes before 142 in A045614.) 
> The
> first 3 pairs that you've listed are (105, 155), (110, 142) and (2145, 
> 3055).
> But you've missed several smaller ones:
>
>    (33, 35), (208, 224), (268, 272), (455, 601), (695, 1033), (812, 956),
>    (1609, 1847), (1808, 2512), (1892, 3004), (1913, 2407), (2096, 2224)
>
>
> A051594 and A051595 are "(-1)sigma amicable numbers".  (Again, the 
> definition
> doesn't make it clear that A051595 has the smaller terms and A051594 has 
> the
> larger terms.)  The first pair that you've listed is (1969706592, 
> 2236072608).
> But there are many smaller ones, starting with:
>
>    (429552, 466200), (497808, 604656), (800496, 1103760).
>
>
> A036471-A036474 are defined as "Amicable quadruple: 4 different numbers
> which satisfy sigma(a)=sigma(b)=sigma(c)=sigma(d)=a+b+c+d".  The first
> quadruple that you've listed has smallest element
> 342151462276356306033089201934180, but here's a much smaller one:
>
>    (3270960, 3361680, 3461040, 3834000)
>
> (I think that's the smallest one, but I could be wrong.)  If someone can
> determine the first several terms of this sequence, then it should be
> edited; otherwise it should be deleted.
>
>
> In some of your sequences, I wasn't able to find any smaller terms, but 
> the
> ones that you've listed are so large that I'm not at all confident that 
> they
> are the smallest ones.  For example, A038362 and A038363 are defined as
> "A Rational Amicable Number: two different numbers a, b which satisfies 
> the
> following equation: sigma(a)=sigma(b)=(a+b)^3/(a^2+b^2)", and the smallest
> pair that you've included is:
>
>    (26403469440047700, 30193441130006700)
>
> Do you have any reason to think that that's the smallest such pair?  If 
> not,
> and if you can't determine what the first few terms really are, then the
> sequence should be deleted.
>
>
> I've only looked at a few of your sequences, but that's enough to convince
> me that they should all be checked.
>
> Neil, I'll submit corrections for A045613, A045614, A051594, and A051595,
> but I'm not going to work on the others.  Maybe they should be deleted or
> marked as "probation" until the author or someone else fixes them.
>
> Dean Hickerson
> dean at math.ucdavis.edu
> 






From maxale at gmail.com  Tue Nov  7 22:14:38 2006
From: maxale at gmail.com (Max A.)
Date: Tue, 7 Nov 2006 13:14:38 -0800
Subject: square arrangements with bit-wise constraints
Message-ID: <d3dac270611071314i6023ba68h4e7e15d63da8c8b3@mail.gmail.com>

There is a problem at MathPages about arrangements of numbers
0,1,...,15 into 4x4 array with certain bit-wise constraints:
http://www.mathpages.com/home/kmath352.htm

This problem (and its variants) can be easily generalized to
(2^n)x(2^n) arrays, e.g.:

In how many distinct ways can the integers 0 through 2^(2n)-1 be arranged
in a (2^n)x(2^n) array such that the bit-wise OR over each row, each column,
and each diagonal is 2^(2n)-1, and the bit-wise AND over each row, column,
and diagonal is 0?

Does anybody wish to compute corresponding sequences and submit them to OEIS?

Regards,
Max





From maxale at gmail.com  Tue Nov  7 23:49:04 2006
From: maxale at gmail.com (Max A.)
Date: Tue, 7 Nov 2006 14:49:04 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>

On 11/3/06, franktaw at netscape.net <franktaw at netscape.net> wrote:

> In the following, b(i), i = 1..k is the finite sequence.  I'm starting
> all these
> sequences with the empty sequence, but I searched for them without
> that first term.  All are shown here in A066099 order.

What is good about this order comparing, say, to the prime
factorization order suggested by Marc?

I would vote for having a standard way to represent such sequences in
OEIS (well documented there, and maybe even having a dedicated
keyword). That is important when one come up with a multidimensional
sequence and want to query it in OEIS. Then it will be enough for
him/her to encode the sequence in the standard way and make a single
query.

Of course, there are many representations possible but one of them
should be called "standard", and every multidimensional sequence
should have this standard representation (among maybe others) in OEIS.

Neil, does that sound reasonable?

Regards,
Max





From mlb at well.com  Wed Nov  8 00:02:52 2006
From: mlb at well.com (Marc LeBrun)
Date: Tue, 07 Nov 2006 15:02:52 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com
 >
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
 <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>
Message-ID: <200611072304.kA7N3xSr010827@smtp.well.com>

 > does that sound reasonable?

It does, but in practice there are already several orderings in wide 
use (as Frank indicated).

Of course it's always possible to submit more than one ordering.

In fact sometimes different ones would be preferred in different contexts.

Or fancier (perhaps special-purpose) lookups could be used.

"That's the great thing about standards--there's so many to choose from!"







From franktaw at netscape.net  Wed Nov  8 00:56:58 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Tue, 07 Nov 2006 18:56:58 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com> <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>
Message-ID: <8C8D0DF8BAA6EAF-124-5E95@FWM-M16.sysops.aol.com>

Exactly what factorization order?  For most numbers, the sequence of
prime power exponents is not a sequence of positive integers.  If 2
represents the sequence (1), what sequence does 3 represent?

If doesn't work any better to use prime power exponents to represent
sequences of non-negative integers.  If 2 is (1) and 3 is (0,1), what 
do
you use to represent (1,0)?

There are ways around these difficulties, but they are pretty 
artificial.

A066099 sorts, first of all, by the number being composed; I think any
reasonable standard candidate should do that.  Second, it is simple,
with a definition tied in to the binary numbers.  Third, it is similar 
to the
Mathematica ordering for partitions (subsort in reverse lexicographic
order), which is one of the two standards for partition ordering in the
OEIS.

(The equivalent of the other, the Abramowitz and Stegun ordering,
was not yet in the OEIS; I did submit it the other day with A-number
A124734.  It is, in any event, not as simple to define as A066099.)

In any event, based on Neil saying to go ahead and submit these
sequences, I  have submitted them with comments stating that
A066099 is the standard ordering.

Franklin T. Adams-Watters


-----Original Message-----
From: maxale at gmail.com

  On 11/3/06, franktaw at netscape.net <franktaw at netscape.net> wrote:

  > In the following, b(i), i = 1..k is the finite sequence. I'm 
starting
 > all these
 > sequences with the empty sequence, but I searched for them without
 > that first term. All are shown here in A066099 order.

 What is good about this order comparing, say, to the prime
 factorization order suggested by Marc?

 I would vote for having a standard way to represent such sequences in
 OEIS (well documented there, and maybe even having a dedicated
 keyword). ...
________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From maxale at gmail.com  Wed Nov  8 01:42:29 2006
From: maxale at gmail.com (Max A.)
Date: Tue, 7 Nov 2006 16:42:29 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8D0DF8BAA6EAF-124-5E95@FWM-M16.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
	 <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>
	 <8C8D0DF8BAA6EAF-124-5E95@FWM-M16.sysops.aol.com>
Message-ID: <d3dac270611071642x66119c57o13f7bf6df6d2436a@mail.gmail.com>

On 11/7/06, franktaw at netscape.net <franktaw at netscape.net> wrote:

> Exactly what factorization order?  For most numbers, the sequence of
> prime power exponents is not a sequence of positive integers.  If 2
> represents the sequence (1), what sequence does 3 represent?
>
> If doesn't work any better to use prime power exponents to represent
> sequences of non-negative integers.  If 2 is (1) and 3 is (0,1), what
> do you use to represent (1,0)?

Actually, the sequences that I mentioned recently does not depend on
trailing zeros, so index (1,0) for them is the same as (1). That's why
the factorization order works well for them.

> A066099 sorts, first of all, by the number being composed; I think any
> reasonable standard candidate should do that.  Second, it is simple,
> with a definition tied in to the binary numbers.  Third, it is similar
> to the
> Mathematica ordering for partitions (subsort in reverse lexicographic
> order), which is one of the two standards for partition ordering in the
> OEIS.

I see now the advantages of the A066099 ordering. And I agree that it
is a good candidate for the "standard".

Max





From arndt at jjj.de  Wed Nov  8 03:54:45 2006
From: arndt at jjj.de (Joerg Arndt)
Date: Wed, 8 Nov 2006 13:54:45 +1100
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com> <20061102011002.GC6041@cochise.anu.edu.au> <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
Message-ID: <20061108025445.GC6252@cochise.anu.edu.au>

* Jonathan Post <jvospost3 at gmail.com> [Nov 08. 2006 13:29]:
> Thank you, Joerg.  That was interesting.  Does your generating function
> hypothesis hold when I extend the sequence by another 5 values, to a(n) = 0, 0,
> 1, 1, 3, 1, 7, 4,
> 14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again, assuming that
> I got the elementary algebra and arithmetic correct.

Alas, the function I gave was spurious, with the
additional terms no OGF is found.

The gp/script can be found here
http://www.jjj.de/pari/ggf.inc.gp
(cannot find Ralf original at the moment).


> If so, I should submit this sequence as by the two us!

Definitely your invention!
But thanks for the kind offer.


> [...]

best regards,    jj





From zbi74583 at boat.zero.ad.jp  Wed Nov  8 06:33:16 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Wed, 8 Nov 2006 14:33:16 +0900
Subject: N(2)=1+1/2^2+2/3^2+5/4^2+4/5^2+2/6^2+6/7^2+....
Message-ID: <02ba01c702f7$69237ea0$ef73d7dc@FMC57937B6BE00>

    Hi, Seqfans

    I asked Mathar to calculate the following
 number.
    N(s)=   Sum_{m=1 to infinity} (-1)Sigma(m)/m^s

     See the end of this mail

     I tried to find an identity like the following

         Zeta(s)Zeta(s-1)=Sum_{m=1 to infinity} Sigma(m)/m^s
         Zeta(s)/Zeta(s+1)=Sum_{m=1 to infinity} Phi(m)/m^s

    But it is too difficult.
    I transformed it a little as follows.

         N(s)=   Sum_{m=1 to infinity} (-1)Sigma(m)/m^s
         =   Product_{p_i} ( Sum_{k=0 to infinity}
(-1)Sigma(p_i^k)/p_i^(k*s) )
         =   Product_{p_i} ( Sum_{k=0 to infinity}
(Sigma(p_i^k)-2)/p_i^(k*s) )
         =   Product_{p_i} ( Sum_{k=0 to infinity}
 Sigma(p_i^k)/p_i^(k*s) -2/(p_i^s-1) )

        Some members of Seqfan might solve it easily.

    Yasutsoshi

----- Original Message ----- 
From: "Richard Mathar" <mathar at strw.leidenuniv.nl>
To: <zbi74583 at boat.zero.ad.jp>
Cc: <mathar at mail.strw.leidenuniv.nl>
Sent: Sunday, November 05, 2006 10:35 PM
Subject: Re: N(2)=1+1/2^2+2/3^2+5/4^2+4/5^2+2/6^2+6/7^2+....


>
> These are 5 tables of partial sums of N(s) up to some maximum
> m of sum_{m=1...max(m)) (-1)Sigma(m)/m^s. The first number
> is s, the second the maximum m up to which the sum has been accumulated,
> and finally the floating point number the partial sum.
> The case s=1 seems not to be a converging series; the partial sums are
> almost
> linear to the upper limit of m. (If your formula with the zeta-function is
> correct, this is expected since zeta(1)=infinty.) I am including some
> values
> for small maximum m so you can check the results for the basic cases.
> N(2) seems to be a converging series. For N(3) you could already get some
> pretty accurate estimate of the series limit.
>
> zeta(2)=1.644934067
> zeta(3)=1.202056903
> zeta(4)=1.082323234
> zeta(3)*zeta(2)=1.977304350
> zeta(4)*zeta(3)=1.301014115
> zeta(5)*zeta(4)=1.122291001
>
>
> PARI:
>
> \\ computes (-1)sigma(n)
> A049060(n)={
>        local(i,resul,rmax,p) ;
>        if(n==1,
>                return(1)
>        ) ;
>        i=factor(n) ;
>        rmax=matsize(i)[1] ;
>        resul=1 ;
>        for(r=1,rmax,
>                p=0 ;
>                for(j=1,i[r,2],
>                        p += i[r,1]^j ;
>                ) ;
>                resul *= p-1 ;
>        ) ;
>        return(resul) ;
> }
>
>
> N(s,mmax)={
> local(resul) ;
> resul=0.0 ;
> for(m=1,mmax,
> resul += A049060(m)/m^s ;
> if(m<10 || (m % 10000 == 0),
> print(s," ",m," ",resul) ;
> ) ;
> ) ;
> }
>
> {
> mmax=1000000 ;
> for(s=1,5,
> N(s,mmax) ;
> ) ;
> }
>
> Tables:
>
> 1 1 1.000000000000000000000000000
> 1 2 1.500000000000000000000000000
> 1 3 2.166666666666666666666666667
> 1 4 3.416666666666666666666666667
> 1 5 4.216666666666666666666666667
> 1 6 4.550000000000000000000000000
> 1 7 5.407142857142857142857142857
> 1 8 7.032142857142857142857142857
> 1 9 8.254365079365079365079365079
> 1 10000 8956.254813409229652481598757
> 1 20000 17913.54053798274101959639246
> 1 30000 26870.58900461199889420226700
> 1 40000 35827.55225952432056619824099
> 1 50000 44784.92216728336777991954546
> 1 60000 53741.30998083683291788901077
> 1 70000 62698.82877172292127401143322
> 1 80000 71656.42458494024747176547617
> 1 90000 80613.31691563743683765060924
> 1 100000 89570.20047616274019702521482
> 1 110000 98527.44018335441754406085900
> 1 120000 107484.2058258699552949703883
> 1 130000 116441.3450587298928159003627
> 1 140000 125399.1600101618061663172293
> 1 150000 134355.0716257765731578804073
> 1 160000 143312.6869188725594888173896
> 1 170000 152270.1691945778107893068833
> 1 180000 161227.3673088893702787242209
> 1 190000 170184.2283671443801174189235
> 1 200000 179141.9177161676803505348327
> 1 210000 188098.1905736334290746346704
> 1 220000 197055.5778178213299226211954
> 1 230000 206012.3243951074382665848232
> 1 240000 214969.0180191522602672455303
> 1 250000 223927.0626654355275622417046
> 1 260000 232884.0479228396547382227555
> 1 270000 241841.6897312605543987962985
> 1 280000 250798.1260711861562596104630
> 1 290000 259755.4115589713808307871029
> 1 300000 268712.2473108133199765189322
> 1 310000 277669.3610844430796369732363
> 1 320000 286626.7360084187087594659096
> 1 330000 295583.0702328420388811375076
> 1 340000 304540.9324662583651406115180
> 1 350000 313497.9488508493337302810705
> 1 360000 322455.2848872162239393519675
> 1 370000 331411.9952274305516780284586
> 1 380000 340369.3878688847801453424847
> 1 390000 349326.5442385613025662494469
> 1 400000 358283.7198151948112652311070
> 1 410000 367240.8596635115166082534173
> 1 420000 376197.2486760902447184900822
> 1 430000 385154.6524560274677028589761
> 1 440000 394112.0480841471143112935114
> 1 450000 403069.2697966051366843904756
> 1 460000 412026.1487890321657369929892
> 1 470000 420983.4928713199267860102478
> 1 480000 429940.0597652682271439943634
> 1 490000 438897.5981659322519223492273
> 1 500000 447855.4516433798944040122164
> 1 510000 456811.4127981064830642426529
> 1 520000 465769.1531758862533175249594
> 1 530000 474726.4930671377953578153300
> 1 540000 483683.7925672078781534825684
> 1 550000 492640.2315981062235193108053
> 1 560000 501597.6949987133986923353009
> 1 570000 510554.3958474732897001847922
> 1 580000 519511.5431567859764988621906
> 1 590000 528468.5920184442509092919859
> 1 600000 537425.2057958757909274862160
> 1 610000 546383.3300255139505983850944
> 1 620000 555340.2697592667964924572269
> 1 630000 564297.7073054137624068690112
> 1 640000 573254.2920689474814450061694
> 1 650000 582211.4873730487914773223047
> 1 660000 591168.5211525597522567544916
> 1 670000 600125.8589826114248021714650
> 1 680000 609083.0330009002048275663287
> 1 690000 618039.4919349372313479054900
> 1 700000 626997.2222632460664089943429
> 1 710000 635954.0917690439691078847726
> 1 720000 644911.6290093784488180074886
> 1 730000 653868.6304875870104467000556
> 1 740000 662825.3447013751015479514900
> 1 750000 671782.5558801465303249818702
> 1 760000 680739.9011662731435592584669
> 1 770000 689696.9475041453715157412648
> 1 780000 698653.1437839886472918561286
> 1 790000 707611.0985532863373721850039
> 1 800000 716568.7987693866776749447598
> 1 810000 725525.6575122445531974385646
> 1 820000 734482.4569919820693082893669
> 1 830000 743439.4779916042359124217226
> 1 840000 752396.3611940500092646796766
> 1 850000 761353.5394438445571430835057
> 1 860000 770310.9434931792736520368474
> 1 870000 779267.1927223886953325647496
> 1 880000 788225.1436674517253585125689
>
>
> 2 1 1.000000000000000000000000000
> 2 2 1.250000000000000000000000000
> 2 3 1.472222222222222222222222222
> 2 4 1.784722222222222222222222222
> 2 5 1.944722222222222222222222222
> 2 6 2.000277777777777777777777778
> 2 7 2.122726757369614512471655329
> 2 8 2.325851757369614512471655329
> 2 9 2.461654226505416981607457798
> 2 10000 8.619813177975507784214344698
> 2 20000 9.240635095871089404115753095
> 2 30000 9.603791897573761358292380061
> 2 40000 9.861458278261452185262848858
> 2 50000 10.06133043757656995196303496
> 2 60000 10.22462154552329778519365923
> 2 70000 10.36270038005636494621123847
> 2 80000 10.48231016697477680498962023
> 2 90000 10.58780522013752853231672862
> 2 100000 10.68217402235127296288785820
> 2 110000 10.76754463607533705360724580
> 2 120000 10.84547789729425765394094918
> 2 130000 10.91717268365027978460835188
> 2 140000 10.98355652885066079792761960
> 2 150000 11.04534545539722125239381859
> 2 160000 11.10315629233238669062836862
> 2 170000 11.15746025373895516747888760
> 2 180000 11.20865775307751148407150182
> 2 190000 11.25708463818456648077093060
> 2 200000 11.30303121795283056763425981
> 2 210000 11.34672871948779384291766276
> 2 220000 11.38839827063844308711114745
> 2 230000 11.42821240662906506074836109
> 2 240000 11.46633165190016188934526342
> 2 250000 11.50290004647923375585461222
> 2 260000 11.53802980091084626273394586
> 2 270000 11.57183602344116902681335304
> 2 280000 11.60440829561909536077171934
> 2 290000 11.63584045378869260629085740
> 2 300000 11.66620539683007648663354504
> 2 310000 11.69557551635064846235471407
> 2 320000 11.72401391041322342493145640
> 2 330000 11.75157395832622563415629569
> 2 340000 11.77831576251751436463182945
> 2 350000 11.80427986574616427809478607
> 2 360000 11.82951337960016320803463309
> 2 370000 11.85405377158818097821428014
> 2 380000 11.87794149361683265940241605
> 2 390000 11.90120806565142413868457075
> 2 400000 11.92388559358277288937420440
> 2 410000 11.94600303970164868968234173
> 2 420000 11.96758569454234998141373716
> 2 430000 11.98866287480559328733313634
> 2 440000 12.00925544388189350449936925
> 2 450000 12.02938482345842327394291528
> 2 460000 12.04907101416111155199518539
> 2 470000 12.06833481184766752219695603
> 2 480000 12.08719139777266186606623586
> 2 490000 12.10566116650605692754670219
> 2 500000 12.12375840131307130932177405
> 2 510000 12.14149351071909408890611230
> 2 520000 12.15888769016783248485026059
> 2 530000 12.17594975401341942285000014
> 2 540000 12.19269280126406301090981599
> 2 550000 12.20912705098196862278557687
> 2 560000 12.22526702472305560991043337
> 2 570000 12.24111996896626519121471994
> 2 580000 12.25669798166764128539680162
> 2 590000 12.27200952384435815319104573
> 2 600000 12.28706299338740013449483108
> 2 610000 12.30187012252190226181814868
> 2 620000 12.31643453756242668558457904
> 2 630000 12.33076670711890682146429120
> 2 640000 12.34487182648867321005497632
> 2 650000 12.35875920337989470739102627
> 2 660000 12.37243430644503889140869087
> 2 670000 12.38590421689421153982279985
> 2 680000 12.39917431914306915108546756
> 2 690000 12.41224965443445935379058168
> 2 700000 12.42513867918653815467126754
> 2 710000 12.43784365088579118428209464
> 2 720000 12.45037185509338005878377820
> 2 730000 12.46272650898502725121177200
> 2 740000 12.47491268376206272216237202
> 2 750000 12.48693594951832440956683550
> 2 760000 12.49880013638169511883229569
> 2 770000 12.51050884124405914256853750
> 2 780000 12.52206537108093533871333277
> 2 790000 12.53347692259745252540675575
> 2 800000 12.54474459716977128685280547
> 2 810000 12.55587125001645305770022096
> 2 820000 12.56686130746829341595469366
> 2 830000 12.57771842011223665163037795
> 2 840000 12.58844534265072464040923243
> 2 850000 12.59904566706465055360427300
> 2 860000 12.60952226922408394518254911
> 2 870000 12.61987642201166975414741555
> 2 880000 12.63011418303496833362065684
> 2 890000 12.64023571583693921823484395
> 2 900000 12.65024408926657285403820450
> 2 910000 12.66014049405210257196085418
> 2 920000 12.66993013960070545251999449
> 2 930000 12.67961329658082585267888159
> 2 940000 12.68919341076487592672709130
> 2 950000 12.69867217785123447166660908
> 2 960000 12.70805040456023104531710849
> 2 970000 12.71733330696500792660758365
> 2 980000 12.72652037909829933006161382
> 2 990000 12.73561385879130711822105048
> 2 1000000 12.74461639944508312563225494
>
>
> 3 1 1.000000000000000000000000000
> 3 2 1.125000000000000000000000000
> 3 3 1.199074074074074074074074074
> 3 4 1.277199074074074074074074074
> 3 5 1.309199074074074074074074074
> 3 6 1.318458333333333333333333333
> 3 7 1.335951044703595724003887269
> 3 8 1.361341669703595724003887269
> 3 9 1.376430832940907109463420877
> 3 10000 1.463913982110741753040591678
> 3 20000 1.463958761182127253891639226
> 3 30000 1.463973687899719174955107484
> 3 40000 1.463981151485566123839035784
> 3 50000 1.463985629911768534082474807
> 3 60000 1.463988615257603046255509048
> 3 70000 1.463990747941060935932260523
> 3 80000 1.463992347452072574407883095
> 3 90000 1.463993591418911296987275373
> 3 100000 1.463994586598910692322299337
> 3 110000 1.463995400874372718599240224
> 3 120000 1.463996079403452056486658683
> 3 130000 1.463996653569731083112186861
> 3 140000 1.463997145746936259101012298
> 3 150000 1.463997572212239449568643565
> 3 160000 1.463997945442508018944894819
> 3 170000 1.463998274756427172013405468
> 3 180000 1.463998567470471281896562070
> 3 190000 1.463998829362945706567178177
> 3 200000 1.463999065087923587578410681
> 3 210000 1.463999278329572278322382185
> 3 220000 1.463999472210290336272529678
> 3 230000 1.463999649219477985324795694
> 3 240000 1.463999811477588227637144335
> 3 250000 1.463999960777271571969049099
> 3 260000 1.464000098575628251036076097
> 3 270000 1.464000226175768714632044135
> 3 280000 1.464000344645861174695942201
> 3 290000 1.464000454956284414358238477
> 3 300000 1.464000557907605374141443674
> 3 310000 1.464000654220015163939180059
> 3 320000 1.464000744515464470055773720
> 3 330000 1.464000829328753859822237621
> 3 340000 1.464000909166687500211857215
> 3 350000 1.464000984435262908416401907
> 3 360000 1.464001055524739843653406744
> 3 370000 1.464001122766895437258623701
> 3 380000 1.464001186474841361410807209
> 3 390000 1.464001246914091200041356493
> 3 400000 1.464001304331492624240843891
> 3 410000 1.464001358947841911631088786
> 3 420000 1.464001410959152183436079651
> 3 430000 1.464001460557008749462202957
> 3 440000 1.464001507900298795165105625
> 3 450000 1.464001553138542176224706874
> 3 460000 1.464001596408268761316169011
> 3 470000 1.464001637838876141306278665
> 3 480000 1.464001677539798346550656570
> 3 490000 1.464001715624418180233792011
> 3 500000 1.464001752186865804286414230
> 3 510000 1.464001787308096983826068464
> 3 520000 1.464001821085252219783666269
> 3 530000 1.464001853586288170141300728
> 3 540000 1.464001884883421221311137792
> 3 550000 1.464001915039618326560450277
> 3 560000 1.464001944122159842425557477
> 3 570000 1.464001972181866862157721242
> 3 580000 1.464001999275380923272745020
> 3 590000 1.464002025450188768071703730
> 3 600000 1.464002050751299694154935270
> 3 610000 1.464002075226966770079630464
> 3 620000 1.464002098909928297454328876
> 3 630000 1.464002121842320025653578842
> 3 640000 1.464002144055977272720855659
> 3 650000 1.464002165587616796774276693
> 3 660000 1.464002186466412269783936185
> 3 670000 1.464002206722643544544226821
> 3 680000 1.464002226382732127695424983
> 3 690000 1.464002245471462412103956612
> 3 700000 1.464002264017435494080639785
> 3 710000 1.464002282039246813434225341
> 3 720000 1.464002299561748289079351576
> 3 730000 1.464002316603153605988980695
> 3 740000 1.464002333183468992396940641
> 3 750000 1.464002349322542550155865090
> 3 760000 1.464002365037130225625255935
> 3 770000 1.464002380343037616331619950
> 3 780000 1.464002395255086253184010138
> 3 790000 1.464002409792474848650673824
> 3 800000 1.464002423965997236326475471
> 3 810000 1.464002437788253358795504488
> 3 820000 1.464002451273302166423360831
> 3 830000 1.464002464433742258170070393
> 3 840000 1.464002477280648906020788669
> 3 850000 1.464002489825692224210435982
> 3 860000 1.464002502079291985535094261
> 3 870000 1.464002514049668308408128899
> 3 880000 1.464002525750211478039722139
> 3 890000 1.464002537187197706443946991
> 3 900000 1.464002548369947911638829284
> 3 910000 1.464002559305408714470469907
> 3 920000 1.464002570004676399632165939
> 3 930000 1.464002580473145840547517414
> 3 940000 1.464002590719441673840433620
> 3 950000 1.464002600750056430793286939
> 3 960000 1.464002610570359301053681776
> 3 970000 1.464002620190111512591642127
> 3 980000 1.464002629612902400536905838
> 3 990000 1.464002638845007329954067381
> 3 1000000 1.464002647892926094713436024
>
>
> 4 1 1.000000000000000000000000000
> 4 2 1.062500000000000000000000000
> 4 3 1.087191358024691358024691358
> 4 4 1.106722608024691358024691358
> 4 5 1.113122608024691358024691358
> 4 6 1.114665817901234567901234568
> 4 7 1.117164776668414909425599416
> 4 8 1.120338604793414909425599416
> 4 9 1.122015178486449507809992039
> 4 10000 1.126342859836933119635341727
> 4 20000 1.126342863195068461987333491
> 4 30000 1.126342863816977724856173635
> 4 40000 1.126342864034657367890755384
> 4 50000 1.126342864135419045299551969
> 4 60000 1.126342864190149270391012376
> 4 70000 1.126342864223154637804140480
> 4 80000 1.126342864244576264194082203
> 4 90000 1.126342864259261709833237965
> 4 100000 1.126342864269766231843710036
> 4 110000 1.126342864277538770311620766
> 4 120000 1.126342864283450137689488953
> 4 130000 1.126342864288050791700191891
> 4 140000 1.126342864291701518257416350
> 4 150000 1.126342864294646135946293287
> 4 160000 1.126342864297056569571605552
> 4 170000 1.126342864299054229967212718
> 4 180000 1.126342864300728234359584700
> 4 190000 1.126342864302144893188134103
> 4 200000 1.126342864303354525085913547
> 4 210000 1.126342864304395340180474300
> 4 220000 1.126342864305297594007409955
> 4 230000 1.126342864306084686590084820
> 4 240000 1.126342864306775457553279160
> 4 250000 1.126342864307385095784161975
> 4 260000 1.126342864307925686556878209
> 4 270000 1.126342864308407364997509003
> 4 280000 1.126342864308838305443671084
> 4 290000 1.126342864309225477016399704
> 4 300000 1.126342864309574563445429650
> 4 310000 1.126342864309890425673710622
> 4 320000 1.126342864310177149073405791
> 4 330000 1.126342864310438174074903581
> 4 340000 1.126342864310676548757732228
> 4 350000 1.126342864310894763694630232
> 4 360000 1.126342864311095054783294446
> 4 370000 1.126342864311279313873956099
> 4 380000 1.126342864311449231412772564
> 4 390000 1.126342864311606242436985681
> 4 400000 1.126342864311751625747966668
> 4 410000 1.126342864311886501047539562
> 4 420000 1.126342864312011847410167188
> 4 430000 1.126342864312128564203286214
> 4 440000 1.126342864312237413459871362
> 4 450000 1.126342864312339084988654298
> 4 460000 1.126342864312434194531858928
> 4 470000 1.126342864312523302682800842
> 4 480000 1.126342864312606892652019146
> 4 490000 1.126342864312685425831403550
> 4 500000 1.126342864312759296677644308
> 4 510000 1.126342864312828850304390296
> 4 520000 1.126342864312894443054768726
> 4 530000 1.126342864312956355217879326
> 4 540000 1.126342864313014859446213372
> 4 550000 1.126342864313070196425470463
> 4 560000 1.126342864313122601535952824
> 4 570000 1.126342864313172268506265819
> 4 580000 1.126342864313219391141266622
> 4 590000 1.126342864313264137597311253
> 4 600000 1.126342864313306663426890322
> 4 610000 1.126342864313347121770685419
> 4 620000 1.126342864313385633102777798
> 4 630000 1.126342864313422327185152818
> 4 640000 1.126342864313457311416964776
> 4 650000 1.126342864313490695748042037
> 4 660000 1.126342864313522573574899694
> 4 670000 1.126342864313553035740126664
> 4 680000 1.126342864313582163334291352
> 4 690000 1.126342864313610031536993753
> 4 700000 1.126342864313636717733889822
> 4 710000 1.126342864313662281830870319
> 4 720000 1.126342864313686789977325667
> 4 730000 1.126342864313710296430980716
> 4 740000 1.126342864313732855692110296
> 4 750000 1.126342864313754519823275656
> 4 760000 1.126342864313775334724375007
> 4 770000 1.126342864313795343268423855
> 4 780000 1.126342864313814585400999315
> 4 790000 1.126342864313833105102330919
> 4 800000 1.126342864313850934103761246
> 4 810000 1.126342864313868105235437633
> 4 820000 1.126342864313884651902605718
> 4 830000 1.126342864313900604514476466
> 4 840000 1.126342864313915990566003172
> 4 850000 1.126342864313930837276585851
> 4 860000 1.126342864313945169444268968
> 4 870000 1.126342864313959008481394737
> 4 880000 1.126342864313972380956143569
> 4 890000 1.126342864313985304497313295
> 4 900000 1.126342864313997799554982470
> 4 910000 1.126342864314009883288627525
> 4 920000 1.126342864314021576813991216
> 4 930000 1.126342864314032894395637362
> 4 940000 1.126342864314043853302500574
> 4 950000 1.126342864314054467992200910
> 4 960000 1.126342864314064751304010854
> 4 970000 1.126342864314074720218563997
> 4 980000 1.126342864314084384860927016
> 4 990000 1.126342864314093757784864329
> 4 1000000 1.126342864314102851387025112
>
>
> 5 1 1.000000000000000000000000000
> 5 2 1.031250000000000000000000000
> 5 3 1.039480452674897119341563786
> 5 4 1.044363265174897119341563786
> 5 5 1.045643265174897119341563786
> 5 6 1.045900466820987654320987654
> 5 7 1.046257460930584845967325490
> 5 8 1.046654189446209845967325490
> 5 9 1.046840475412102579121146892
> 5 10000 1.047127646962981639236350714
> 5 20000 1.047127646963242803683634380
> 5 30000 1.047127646963269060444667980
> 5 40000 1.047127646963275452376929668
> 5 50000 1.047127646963277728778610486
> 5 60000 1.047127646963278734909173763
> 5 70000 1.047127646963279246707166974
> 5 80000 1.047127646963279534023699805
> 5 90000 1.047127646963279707590251386
> 5 100000 1.047127646963279818572046343
> 5 110000 1.047127646963279892819669335
> 5 120000 1.047127646963279944352143863
> 5 130000 1.047127646963279981235720216
> 5 140000 1.047127646963280008327383488
> 5 150000 1.047127646963280028667159600
> 5 160000 1.047127646963280044239866946
> 5 170000 1.047127646963280056361640181
> 5 180000 1.047127646963280065937720739
> 5 190000 1.047127646963280073602739981
> 5 200000 1.047127646963280079811373591
> 5 210000 1.047127646963280084892514359
> 5 220000 1.047127646963280089092048700
> 5 230000 1.047127646963280092592526331
> 5 240000 1.047127646963280095533744603
> 5 250000 1.047127646963280098023437624
> 5 260000 1.047127646963280100144477685
> 5 270000 1.047127646963280101962983953
> 5 280000 1.047127646963280103530722491
> 5 290000 1.047127646963280104889770905
> 5 300000 1.047127646963280106073563423
> 5 310000 1.047127646963280107109545028
> 5 320000 1.047127646963280108020080532
> 5 330000 1.047127646963280108823485592
> 5 340000 1.047127646963280109535261396
> 5 350000 1.047127646963280110167943658
> 5 360000 1.047127646963280110732291221
> 5 370000 1.047127646963280111237235413
> 5 380000 1.047127646963280111690454876
> 5 390000 1.047127646963280112098366088
> 5 400000 1.047127646963280112466502499
> 5 410000 1.047127646963280112799594529
> 5 420000 1.047127646963280113101691720
> 5 430000 1.047127646963280113376369686
> 5 440000 1.047127646963280113626641318
> 5 450000 1.047127646963280113855154497
> 5 460000 1.047127646963280114064219587
> 5 470000 1.047127646963280114255879066
> 5 480000 1.047127646963280114431883567
> 5 490000 1.047127646963280114593830274
> 5 500000 1.047127646963280114743084164
> 5 510000 1.047127646963280114880831757
> 5 520000 1.047127646963280115008212055
> 5 530000 1.047127646963280115126153891
> 5 540000 1.047127646963280115235519959
> 5 550000 1.047127646963280115337066821
> 5 560000 1.047127646963280115431500449
> 5 570000 1.047127646963280115519415563
> 5 580000 1.047127646963280115601376090
> 5 590000 1.047127646963280115677873082
> 5 600000 1.047127646963280115749351712
> 5 610000 1.047127646963280115816230987
> 5 620000 1.047127646963280115878856405
> 5 630000 1.047127646963280115937571799
> 5 640000 1.047127646963280115992669514
> 5 650000 1.047127646963280116044432219
> 5 660000 1.047127646963280116093104355
> 5 670000 1.047127646963280116138915493
> 5 680000 1.047127646963280116182070550
> 5 690000 1.047127646963280116222756887
> 5 700000 1.047127646963280116261156889
> 5 710000 1.047127646963280116297420392
> 5 720000 1.047127646963280116331699689
> 5 730000 1.047127646963280116364124370
> 5 740000 1.047127646963280116394819088
> 5 750000 1.047127646963280116423900164
> 5 760000 1.047127646963280116451471135
> 5 770000 1.047127646963280116477627542
> 5 780000 1.047127646963280116502457450
> 5 790000 1.047127646963280116526050682
> 5 800000 1.047127646963280116548478239
> 5 810000 1.047127646963280116569809891
> 5 820000 1.047127646963280116590113535
> 5 830000 1.047127646963280116609450953
> 5 840000 1.047127646963280116627878223
> 5 850000 1.047127646963280116645449098
> 5 860000 1.047127646963280116662212610
> 5 870000 1.047127646963280116678212198
> 5 880000 1.047127646963280116693495678
> 5 890000 1.047127646963280116708099149
> 5 900000 1.047127646963280116722060660
> 5 910000 1.047127646963280116735413376
> 5 920000 1.047127646963280116748193681
> 5 930000 1.047127646963280116760429366
> 5 940000 1.047127646963280116772150557
> 5 950000 1.047127646963280116783383436
> 5 960000 1.047127646963280116794151686
> 5 970000 1.047127646963280116804482528
> 5 980000 1.047127646963280116814395317
> 5 990000 1.047127646963280116823911289
> 5 1000000 1.047127646963280116833050882
>
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From ralf at ark.in-berlin.de  Wed Nov  8 09:26:05 2006
From: ralf at ark.in-berlin.de (Ralf Stephan)
Date: Wed, 8 Nov 2006 09:26:05 +0100
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <20061108025445.GC6252@cochise.anu.edu.au>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com> <20061102011002.GC6041@cochise.anu.edu.au> <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com> <20061108025445.GC6252@cochise.anu.edu.au>
Message-ID: <20061108082605.GA21119@ark.in-berlin.de>

You wrote 
> * Jonathan Post <jvospost3 at gmail.com> [Nov 08. 2006 13:29]:
> > Thank you, Joerg.  That was interesting.  Does your generating function
> > hypothesis hold when I extend the sequence by another 5 values, to a(n) = 0, 0,
> > 1, 1, 3, 1, 7, 4,
> > 14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again, assuming that
> > I got the elementary algebra and arithmetic correct.
> 
> Alas, the function I gave was spurious, with the
> additional terms no OGF is found.

Quite possible. Always treat them as hypotheses, as Jonathan said.


ralf






From maxale at gmail.com  Wed Nov  8 11:27:40 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 8 Nov 2006 02:27:40 -0800
Subject: A038379 is all wrong
Message-ID: <d3dac270611080227lc6e0ad2qde71bd834c86d0f2@mail.gmail.com>

A038379 is defined as the "number of positive semi-definite real {0,1}
n X n matrices".
But the sequence and comments have nothing to do with the description.

In particular, all given values do not match the definition.
It lists 1, 3, 27, 729, 52649 while the correct values (that match the
definition) are 2, 7, 59, 1468, 102751 (I will be grateful if somebody
check these values).
For the simplest case n=1, it is clear that there two positive
semi-definite real {0,1}-matrices:
(0) and (1).

There is also a comment:
%C A038379 For n <= 4 a(n) = the upper bound 3^C(n,2).
Why the upper bound is 3^C(n,2) ? We have only two options for each
element of the matrix.
There would be an upper bound of this type if:
1) The main diagonal of a matrix is fixed, and the matrix is
symmetric. Then there are C(n,2) distinct non-diagonal elements whose
values should be assigned.
2) There are three (not two!) possible values for each non-diagonal
element of the matrix.

It seems that the definition of A038379 does not match the rest. Can
anybody find a definition of A038379 that is consistent with its
current content? If so, I will submit the real "number of positive
semi-definite real {0,1} n X n matrices" as a separate sequence.

Regards,
Max





From njas at research.att.com  Wed Nov  8 15:54:53 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 8 Nov 2006 09:54:53 -0500 (EST)
Subject: A038379 is all wrong
Message-ID: <200611081454.JAA83283@fry.research.att.com>

Max,   maybe I meant to say "Number of positive semi-definite real
matrices with entries {-1, 0, +1}." - can you check?

Neil





From maxale at gmail.com  Wed Nov  8 22:59:13 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 8 Nov 2006 13:59:13 -0800
Subject: A038379 is all wrong
In-Reply-To: <200611081454.JAA83283@fry.research.att.com>
References: <200611081454.JAA83283@fry.research.att.com>
Message-ID: <d3dac270611081359t117aab4ag1545e9c3130fa53a@mail.gmail.com>

On 11/8/06, N. J. A. Sloane <njas at research.att.com> wrote:
> Max,   maybe I meant to say "Number of positive semi-definite real
> matrices with entries {-1, 0, +1}." - can you check?

For the comment to make sense, the main diagonal should be fixed as well.
I will check most obvious variants.

Max





From jvospost3 at gmail.com  Thu Nov  9 02:51:54 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 8 Nov 2006 17:51:54 -0800
Subject: primes in arithmetic progression
In-Reply-To: <335cf8920611042302v7db1ca58xb8b67f872981b204@mail.gmail.com>
References: <200611041910.OAA62978@fry.research.att.com>
	 <5542af940611041355vd65aa5cr3a4a9df3cc348c15@mail.gmail.com>
	 <335cf8920611042302v7db1ca58xb8b67f872981b204@mail.gmail.com>
Message-ID: <5542af940611081751y69ee576as6473da25e91a2175@mail.gmail.com>

Sorry, Olivier.  I hope not to repeat the mistake.

Thank you again, Richard J. Mathar, for your corrections and extensions of 6
Nov 2006, which appear online today for A124570.  I'm surprised that nobody
looked in this direction, before you started it.  NJAS was right to see the
value, and 'kick things off" -- and I hope that it does not stop with my
clumsy shot at the semiprime case, which you so gallantly repaired.

I'm mostly satisfied with the results of myself and tens of millions of
others voting in the U.S. elections Tuesday 7 Nove 2006.  I feel as if the
whole country is being corrected and extended, to benefits both domestic and
global. But this is not the venue to discuss that.



On 11/4/06, Olivier Gerard <olivier.gerard at gmail.com> wrote:
>
> Hello Jonathan,
>
> Thanks for this message.
>
> Please remember NOT to send your posts to seqfan as HTML,
> even with a pure text version.
>
> regards,
>
> Olivier
>
>
> On 11/4/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> >
> > The analogue for 3-almost primes (or, as njas prefers, numbers which are
> > the product of exactly 3 primes) is left as an exercise for the reader...
> > I've just submitted the following.
> >
> > Subject: PRE-NUMBERED NEW SEQUENCE A124570 FROM Jonathan Vos Post
> >
> >
> >
>
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From jvospost3 at gmail.com  Thu Nov  9 04:11:11 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 8 Nov 2006 19:11:11 -0800
Subject: Scrabblomics; COMMENT FROM Jonathan Vos Post RE A113172
In-Reply-To: <20061108213820.28640.qmail@web55109.mail.re4.yahoo.com>
References: <20061108213820.28640.qmail@web55109.mail.re4.yahoo.com>
Message-ID: <5542af940611081911h2ff3a639v3a4b53b5c2927de2@mail.gmail.com>

Here's more fun, with a silly if plausibly motivated name.

Subject: NEW SEQUENCE FROM Jonathan Vos Post


%I A000001
%S A000001 1,
3, 3,
8, 10, 8,
9, 12, 12, 9,
4, 15, 19, 15, 4,
7, 15, 19, 19, 15, 7,
8, 18, 19, 21, 19, 18, 8,
9, 22, 20, 11, 11, 20, 22, 9,
4, 15, 17, 15, 18, 15, 17, 15, 4,
7, 15, 18, 18, 20, 20, 18, 18, 15, 7,
8, 18, 20, 22, 21, 11, 21, 22, 20, 18, 8,
9, 22, 21, 17, 19, 18, 18, 19, 17, 21, 22, 9
%N A000001 Pascrabble Triangle, by rows.
%C A000001 The apex of the triangle is 1.  Any other value is the
Scrabble value of English name for the number which is the sum of the
numbers above. This is generated the same way as A007318 Pascal's
triangle read by rows, except apply A113172 to each sum. The first
column of this triangle is 1, 3, 8, 9, 4, 7, 8, 9, 4, 7, 8, 9, 4, 7...
= iterations 1, A113172(1), A113172(A113172(1)),
A113172(A113172(A113172(1))).
The central pascrabble numbers T(2n+1,n) = 1, 10, 19, 21, 18, 11, 22, ...
%F A000001 T(1,1) = 1; for i > 1, T(i,j) = A113172(T(i-1, j-1)+T(i-1, j).
%e A000001 Triangle begins:
row.|.values in row
.1..|01
.2..|03.03
.3..|08.10.08
.4..|09.12.12.09
.5..|04.15.19.15.04
.6..|07.15.19.19.15.07
.7..|08.18.19.21.19.18.08
.8..|09.22.20.11.11.20.22.09
.9..|04.15.17.15.18.15.17.15.04
10..|07.15.18.18.20.20.18.18.15.07
11..|08.18.20.22.21.11.21.22.20.18.8
12..|09.22.21.17.19.18.18.19.17.21.22.09
%Y A000001 cf. A007318 Pascal's triangle read by rows, A113172.
%O A000001 1
%K A000001 ,easy,nonn,tabl,word,
%A A000001 Jonathan Vos Post (jvospost2 at yahoo.com), Nov 08 2006

I think that this cut & paste is pure text, and not HTML (as Olivier asked)
but I am not sure.

On 11/8/06, jonathan post <jvospost2 at yahoo.com> wrote:
>
> My backup copy (as no autoreply yet).
>
> Subject: COMMENT FROM Jonathan Vos Post RE A113172
>
>
> %I A113172
> %S A113172
> 13,3,6,8,7,10,10,8,9,4,3,9,12,11,11,13,14,12,12,8,12,15,18,20,19,22,22
> %N A113172 Scrabble value of English word for the
> number n.
> %C A113172 Is 12 the unique fixed point of Scrabble
> value of English word for the number n, where
> A113172(n) = n? 12 --> "TWELVE" --> 1+4+1+1+4+1 -->
> 12. For what n are A113172(n) prime? For what n are
> there cycles of what length for iterations n,
> A113172(n), A113172(A113172(n)),
> A113172(A113172(A113172(n))), ...? For what n does
> A113172(n) | n? These are key questions in general
> Scrabblomics.
> %F A113172 Scrabble value of English word for the
> number n.
> %O A113172 1
> %K A113172 ,easy,nonn,word,
> %A A113172 Jonathan Vos Post (jvospost2 at yahoo.com),
> Nov 08 2006
>
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From rkg at cpsc.ucalgary.ca  Thu Nov  9 20:26:26 2006
From: rkg at cpsc.ucalgary.ca (Richard Guy)
Date: Thu, 9 Nov 2006 12:26:26 -0700 (MST)
Subject: Small notation clean-up
In-Reply-To: <5542af940611081751y69ee576as6473da25e91a2175@mail.gmail.com>
References: <200611041910.OAA62978@fry.research.att.com> 
 <5542af940611041355vd65aa5cr3a4a9df3cc348c15@mail.gmail.com> 
 <335cf8920611042302v7db1ca58xb8b67f872981b204@mail.gmail.com>
 <5542af940611081751y69ee576as6473da25e91a2175@mail.gmail.com>
Message-ID: <Pine.LNX.4.63.0611091221220.23740@csl>

Would someone more competent & energetic than I
clean up the (usually unexplained)   S_n
notation that appears sporadically in OEIS ?

Evidently it means a ``star'' with  n  vertices,
and so is in fact the complete bipartite graph
K_(1,n-1), a notation which is more widely
known.   Best to all,   R.





From maxale at gmail.com  Thu Nov  9 20:35:39 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 9 Nov 2006 11:35:39 -0800
Subject: Small notation clean-up
In-Reply-To: <Pine.LNX.4.63.0611091221220.23740@csl>
References: <200611041910.OAA62978@fry.research.att.com>
	 <5542af940611041355vd65aa5cr3a4a9df3cc348c15@mail.gmail.com>
	 <335cf8920611042302v7db1ca58xb8b67f872981b204@mail.gmail.com>
	 <5542af940611081751y69ee576as6473da25e91a2175@mail.gmail.com>
	 <Pine.LNX.4.63.0611091221220.23740@csl>
Message-ID: <d3dac270611091135u58ec268au1f85498bf1ea978b@mail.gmail.com>

On 11/9/06, Richard Guy <rkg at cpsc.ucalgary.ca> wrote:
> Would someone more competent & energetic than I
> clean up the (usually unexplained)   S_n
> notation that appears sporadically in OEIS ?
>
> Evidently it means a ``star'' with  n  vertices,

It is also often used for the symmetric group of order n.

Max





From maxale at gmail.com  Thu Nov  9 23:11:00 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 9 Nov 2006 14:11:00 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <20061108025445.GC6252@cochise.anu.edu.au>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <20061102011002.GC6041@cochise.anu.edu.au>
	 <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
	 <20061108025445.GC6252@cochise.anu.edu.au>
Message-ID: <d3dac270611091411t71b55c5fl1087e070509b8fee@mail.gmail.com>

On 11/7/06, Joerg Arndt <arndt at jjj.de> wrote:
> * Jonathan Post <jvospost3 at gmail.com> [Nov 08. 2006 13:29]:
> > Thank you, Joerg.  That was interesting.  Does your generating function
> > hypothesis hold when I extend the sequence by another 5 values, to a(n) = 0, 0,
> > 1, 1, 3, 1, 7, 4,
> > 14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again, assuming that
> > I got the elementary algebra and arithmetic correct.
>
> Alas, the function I gave was spurious, with the
> additional terms no OGF is found.

Not surprising at all!
This sequence is not well-defined (see above in this discussion), so
one may view its terms as random numbers. How can they have an OGF? ;)

Max





From jvospost3 at gmail.com  Thu Nov  9 23:15:12 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 9 Nov 2006 14:15:12 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <d3dac270611091411t71b55c5fl1087e070509b8fee@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <20061102011002.GC6041@cochise.anu.edu.au>
	 <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
	 <20061108025445.GC6252@cochise.anu.edu.au>
	 <d3dac270611091411t71b55c5fl1087e070509b8fee@mail.gmail.com>
Message-ID: <5542af940611091415m4db107dei8659adb11bdcbffa@mail.gmail.com>

Max's position is well taken.  I think that "random" overstates it
somewhat.  I think that even careful definition leaves a residuum of chaos
in the proper sense.  I have not yet posted any seq based on this, out of
caution.  When I do, I'll probably post just one and see what happens.

On 11/9/06, Max A. <maxale at gmail.com> wrote:
>
> On 11/7/06, Joerg Arndt <arndt at jjj.de> wrote:
> > * Jonathan Post <jvospost3 at gmail.com> [Nov 08. 2006 13:29]:
> > > Thank you, Joerg.  That was interesting.  Does your generating
> function
> > > hypothesis hold when I extend the sequence by another 5 values, to
> a(n) = 0, 0,
> > > 1, 1, 3, 1, 7, 4,
> > > 14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again,
> assuming that
> > > I got the elementary algebra and arithmetic correct.
> >
> > Alas, the function I gave was spurious, with the
> > additional terms no OGF is found.
>
> Not surprising at all!
> This sequence is not well-defined (see above in this discussion), so
> one may view its terms as random numbers. How can they have an OGF? ;)
>
> Max
>
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From maxale at gmail.com  Thu Nov  9 23:41:15 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 9 Nov 2006 14:41:15 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <5542af940611091415m4db107dei8659adb11bdcbffa@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <20061102011002.GC6041@cochise.anu.edu.au>
	 <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
	 <20061108025445.GC6252@cochise.anu.edu.au>
	 <d3dac270611091411t71b55c5fl1087e070509b8fee@mail.gmail.com>
	 <5542af940611091415m4db107dei8659adb11bdcbffa@mail.gmail.com>
Message-ID: <d3dac270611091441la822b41v6a324c49ec2a1a76@mail.gmail.com>

On 11/9/06, Jonathan Post <jvospost3 at gmail.com> wrote:

> Max's position is well taken.  I think that "random" overstates it somewhat.

OK, it may not be random from your point of view since you followed
some rules (I believe) defining the terms of this sequence. The
problem is that you did not let us know these rules!
Remember the example I gave above? It was about different
representations of b(1) as 2-nd degree polynomials of w:
b(1) = w^2 = 2w^2+w+1 = 3w^2+2w+2 = ...
They all are equal (since w=(-1 + i*sqrt(3))/2) but at the same time
they all have different coefficients as polynomials of w.
For some (unknown to us) reason, you chose the first representation
that gave the 1st term of your sequence equal 0 (as the coefficient of
w^0 in b(1)=w^2). But you chose b(1)=2w^2+w+1, the first term would be
1; if you chose b(1)=3w^2+2w+2, it would be 2, etc.

There is yet another way to define your sequence precisely. This time
let's forget the exact value of w and keep in mind only that w^3 = 1.
Then every element of the sequence b(0) = w, b(1) = w^2, b(n) =
w*b(n-1) + b(n-2) have unique representation as a 2-nd degree
polynomial of w:
b(2) = -x + 1
b(3) = -2*x^2 + x
b(4) = x^2 + x - 3
b(5) = 3*x^2 - 4*x + 1
b(6) = -5*x^2 + 6
b(7) = -3*x^2 + 10*x - 6
b(8) = 15*x^2 - 6*x - 9
...
This gives us the following sequence of coefficients of w^0:
0, 0, 1, 0, -3, 1, 6, -6, -9
But it is different from your original sequence (again).

Max





From franktaw at netscape.net  Fri Nov 10 00:40:10 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 09 Nov 2006 18:40:10 -0500
Subject: Incorrect conjecture
Message-ID: <8C8D26F876C6EFF-76C-A121@WEBMAIL-MA17.sysops.aol.com>

The conjecture in http://www.research.att.com/~njas/sequences/A092903 
is wrong.  This conjecture is that A092903 is not the same as A005153.  
Translated, the opposite of this conjecture is that if every number up 
to m can be expressed as the sum of distinct divisors of m, then every 
number up to sigma(m) can be.  (The latter condition is known as m is 
practical.)

Following the link from A005153 to 
http://planetmath.org/encyclopedia/PracticalNumber.html, we find the 
lemma (paraphrased for non-graphical presentation): an integer m >= 2 
with factorization Product_{i=1}^k p_i^e_i with the p_i in ascending 
order is practical if and only if p_1 = 2 and, for 1 < i <= k, p_i <= 
sigma(Product_{j < i} p_j^e_j) + 1.

Proof: Let m =  Product_{i=1}^k p_i^e_i  be a number not of this form.  
If p_1 != 2, 2 cannot be represented as a sum.  If p_i > 
sigma(Product_{j<i} p_j^e_j) + 1, then sigma(Product_{j<i} p_j^e_j) + 1 
cannot be represented as the sum, and since this number is less than 
p_i, it is less than m.  Hence not every number up to m can be 
expressed as the sum of distinct divisors of m.   Thus if every number 
up to m is expressible as the sum of distinct divisors, so is every 
number up to sigma(m).  The converse is trivial, so the conditions are 
equivalent.

Hence A092903 is a duplicate of A005153.

Franklin T. Adams-Watters

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From davidwwilson at comcast.net  Fri Nov 10 03:06:32 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 9 Nov 2006 21:06:32 -0500
Subject: Explanation of A123937
Message-ID: <002f01c7046c$d7f7f880$6401a8c0@yourxhtr8hvc4p>

This is a hurried attempt to provide background on the genesis of A123937.  I am sure that a most of what I write here is a poor rehash of basic theory, and this whole post probably equates to a few lines of generating function theory.

For d and n in Z+, let A_d(n) be the set of integer points on the d-sphere of radius sqrt(n) centered at the origin, that is:

    [1] a_d(n) = #({P in Z^d : |P|^2 = n}).

A table of a_d(n) for a few small values of d and n:

Table of a_d(n)

 \d|  0    1    2    3    4    5     6     7      8      9      10
 n\|
---+--------------------------------------------------------------
 0 |  1    1    1    1    1    1     1     1      1      1       1
 1 |  0    2    4    6    8   10    12    14     16     18      20
 2 |  0    0    4   12   24   40    60    84    112    144     180
 3 |  0    0    0    8   32   80   160   280    448    672     960
 4 |  0    2    4    6   24   90   252   574   1136   2034    3380
 5 |  0    0    8   24   48  112   312   840   2016   4320    8424
 6 |  0    0    0   24   96  240   544  1288   3136   7392   16320
 7 |  0    0    0    0   64  320   960  2368   5504  12672   28800
 8 |  0    0    4   12   24  200  1020  3444   9328  22608   52020
 9 |  0    2    4   30  104  250   876  3542  12112  34802   88660
10 |  0    0    8   24  144  560  1560  4424  14112  44640  129064

Each sequence a_d is a column of this table. All of these column sequences already exist in the OEIS.

It turns out that a_d can be gotten by convolving the sequence j = (1,2,0,0,2,...) = A000122 d times with the convolutional identity sequence e = (1,0,0,0,0,...) = A000007. We could say that a_d = j^d as a "convolutional power".

The generating function of a_d is t(x)^d where t is the generating function of j. t turns out to be the Jacobi theta function theta_3, as noted on A000122. If we apply the finite difference transform to the sequence of generating functions {t(x)^d}, we obtain the sequence {(t(x)-1)^d}. Expanding these functions back to sequences gives the sequence of sequences {b_d} where b_d = (j-e)^d as a convolutional power. This allows us to constuct a table of b_d:

Table of b_d(n)

 \d|  0    1    2    3    4    5    6    7    8    9    10
 n\|
---+------------------------------------------------------
 0 |  1    0    0    0    0    0    0    0    0    0     0
 1 |  0    2    0    0    0    0    0    0    0    0     0
 2 |  0    0    4    0    0    0    0    0    0    0     0
 3 |  0    0    0    8    0    0    0    0    0    0     0
 4 |  0    2    0    0   16    0    0    0    0    0     0
 5 |  0    0    8    0    0   32    0    0    0    0     0
 6 |  0    0    0   24    0    0   64    0    0    0     0
 7 |  0    0    0    0   64    0    0  128    0    0     0
 8 |  0    0    4    0    0  160    0    0  256    0     0
 9 |  0    2    0   24    0    0  384    0    0  512     0
10 |  0    0    8    0   96    0    0  896    0    0  1024

It turns out that the process we used to get from the a_d table to the b_d table, namely, converting columns of a_d to generating functions, applying the finite difference transform to the gf's, and expanding the resulting functions back to columns of b_d, is the same as applying the finite difference transform to the rows of the a_d table to obtain rows of the b_d table.

An inductive proof starting with b_d = (j-e)^d where j-e = (0,2,0,0,2,...) shows that b_d(d) = 2^d and b_d(n) = 0 for n < d. This means that the nth element of the nth row of the b_d table is the last nonzero element in that row, which in turn implies that the nth row of the a_d table, that is, the number of integer points P with |P|^2 = n, is a polynomial of degree n in d.

If, instead of [1], we had started with

    [1] a_d'(n) = #({P in Z^d : |P|^2 <= n})

which counts points inside as well as on the d-sphere, the nth row of a_d' table would be gotten by adding termwise the rows 0 through n of the a_d table. The sequence a_d' has generating function t(x)^d/(1-x). Similarly, the nth row of the corresponding b_d' table is the sum of rows 0 through n of the b_d table, and sequence b_d has generating function (t(x)-1)^d)/(1-x). The table of b_d', omitting the superdiagonal zeroes, is sequence A123937.

The recurrence given in the title of A123937 follows from the relation

    b_d = (1,1,1,1,1,...) = A000012 if d = 0; (j-e) * b_{d-1}' if d >= 1.

I know this is a spotty exposition, but I'm sure the loose ends could be tied together by a real number theorist.

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From zakseidov at yahoo.com  Sat Nov 11 14:53:18 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sat, 11 Nov 2006 05:53:18 -0800 (PST)
Subject: A052530: to be edited
Message-ID: <20061111135318.71618.qmail@web38206.mail.mud.yahoo.com>

Both %F's and %t don't accord with entries %S 
(mainly with first term of %S),
and note this misprint(?):
"a(n)-4*a(n+1)+a(n+2)})"

%S A052530
1,2,8,30,112,418,1560,5822,21728,81090,302632,

%F A052530 a(0) = c, a(1) = p*c^3; a(n+2) =
p*c^2*a(n+1) - a(n), for p = 1, c = 2.

%F A052530 Recurrence: {a(0)=1, a(1)=2, a(2)=8, 
a(n)-4*a(n+1)+a(n+2)}

%t A052530 a[0] = c; a[1] = p*c^3; a[n_] := a[n] =
p*c^2*a[n - 1] - a[n - 2]; p = 1; c = 2; 
Table[ a[n], {n, 0, 20} ]

Anyone may wish to fix it? 
Zak


 
____________________________________________________________________________________
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From davidwwilson at comcast.net  Sat Nov 11 16:34:08 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sat, 11 Nov 2006 10:34:08 -0500
Subject: A124057 is subset of A045940
References: <20061105235118.90653.qmail@web38203.mail.mud.yahoo.com>
Message-ID: <004501c705a6$d4c0bea0$6501a8c0@yourxhtr8hvc4p>

Nah, there's lots of them:

Yah, lots:

8706123 24463374 32442848 32942943 36782289 48580623 55486248 57476573
59600365 59757774 62481222 62664810 62884590 63262374 63728124 64724373
65159575 65450824 69362487 70302087 70370223 70785924 71494773 72060272
72503682 73256910 73638422 74066874 74361858 75356070 75414702 78676623
83349123 84405123 88094895 92349423 93803982 96192782 97757955 97822374
98474661 99104550 100708374 100997575 101302623 101905622 101905623
102956103 105596334 110476143 111673374 112529493 112801974 113067422
114811332 114940375 115338894 115466874 116079774 118269150 119076174
120919623 123052110 123821970 123952542 124045207 124232373 124465624
124581534 126115623 126134874 126619623 127151127 128095749 129079248
129764789 131404623 131422575 132568225 132617814 133891623 133918134
134350623 134421174 135691143 138468069 142680174 142904430 142904431
145863150 148286886 151757655 152779624 153101583 153129248 153930510
154322523 155740023 155826423 158252994


----- Original Message ----- 
From: "zak seidov" <zakseidov at yahoo.com>
To: <seqfan at ext.jussieu.fr>
Sent: Sunday, November 05, 2006 6:51 PM
Subject: Re: A124057 is subset of A045940


> Just submitted:
> 
> My Q to seqfans:
> Really, there are no numbers n such that n, n+1, n+2
> and n+3 are products of 6 primes?
> No for n<7258290.
> Thanks, Zak






From njas at research.att.com  Sat Nov 11 22:52:30 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sat, 11 Nov 2006 16:52:30 -0500 (EST)
Subject: querying multidimensional sequences in OEIS
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
    <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
    <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <200611112152.QAA93259@fry.research.att.com>

Max, the other day you said:


Of course, there are many representations possible but one of them
should be called "standard", and every multidimensional sequence
should have this standard representation (among maybe others) in OEIS.

Neil, does that sound reasonable?


Me:  it is very hard to force a standard representation - there are
too many people who use the OEIS and don't spend much time
reading rules

even such a simple thing as the "offset" and "keywords" gives
a lot of people trouble.

in short, it would not work

Neil





From njas at research.att.com  Sun Nov 12 04:40:14 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sat, 11 Nov 2006 22:40:14 -0500 (EST)
Subject: Thanks to Tony Noe.
Message-ID: <200611120340.WAA18989@fry.research.att.com>


Seqfans:

in the past few weeks Tony Noe has corrected several hundred entries 
in the OEIS.  the kinds of errors he has found include numbers
broken by commas, numbers run together, dropped digits, duplicated 
numbers, and just plain wrong entries.

It is astonishing to me that there were so many errors.

And it suggests that there must be thousands of other errors
waiting to be found.

I would like to thank him for all his work.   Incidentally he
has also added hundreds of b-files.

Right now I am in the middle of a large batch of updates - there
will be a new version of the OEIS by dawn.

Tony:  thanks!


Neil





From neoneye at gmail.com  Sun Nov 12 08:27:28 2006
From: neoneye at gmail.com (Simon Strandgaard)
Date: Sun, 12 Nov 2006 08:27:28 +0100
Subject: growing binary trees
Message-ID: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>

I made up an interesting sequence:

1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 26, 33, 41,
50, 62, 77, 94, 115, 142, 174, 212, 260, 319,
389, 475, 582, 711, 867, 1060, 1296


Its a binary tree where leafs is getting
added when a nodes internal counter reaches 3.





below is my slow ruby program:

class Node
  def initialize
    @n = 1
    @childs = []
  end
  def count
    @childs.inject(@n){|n,child| n + child.count}
  end
  def grow
    return @n += 1 if @n < 3
    @childs.each{|child| child.grow }
    @childs << Node.new if @childs.size < 2
  end
end

result = []
node = Node.new
30.times do |i|
  result << node.count
  node.grow
end
p result







# btw: the grow method below outputs A000071: Fibonacci numbers - 1.
def grow
  return @n += 1 if @n < 1
  @childs.each{|child| child.grow }
  @childs << Node.new if @childs.size < 2
end


# btw: the grow method below outputs A054405: Row sums of array T as in A055215.
def grow
  return @n += 1 if @n < 2
  @childs.each{|child| child.grow }
  @childs << Node.new if @childs.size < 2
end




--
Simon Strandgaard





From ralf at ark.in-berlin.de  Sun Nov 12 10:09:02 2006
From: ralf at ark.in-berlin.de (Ralf Stephan)
Date: Sun, 12 Nov 2006 10:09:02 +0100
Subject: growing binary trees
In-Reply-To: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
Message-ID: <20061112090902.GA1205@ark.in-berlin.de>

> I made up an interesting sequence:
> 
> 1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 26, 33, 41,
> 50, 62, 77, 94, 115, 142, 174, 212, 260, 319,
> 389, 475, 582, 711, 867, 1060, 1296

According to superseeker:

                                     2      3      4
                        6 + 2 x + 2 x  - 3 x  - 4 x
                     [- ----------------------------, ogf]
                                5            3
                              -x  - 1 + x + x

Second cumulative sums of:

%I A017817
%S A017817 1,0,0,1,1,0,1,2,1,1,3,3,2,4,6,5,6,10,11,11,16,21,22,27,37,43,49,64,80,
%T A017817 92,113,144,172,205,257,316,377,462,573,693,839,1035,1266,1532,1874,
%U A017817 2301,2798,3406,4175,5099,6204,7581,9274,11303,13785,16855,20577,25088
%N A017817 a(0)=1, a(1)=a(2)=0, a(3)=1; a(n)=a(n-3)+a(n-4).
%F A017817 G.f.: 1/(1-x^3-x^4).
%F A017817 a(n)/a(n-1) tends to r = 1.2207440846..., a real root of x^4 - x - 1 = 0. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Oct 22 2006

First sums of:

Transformation T018 gave a match with:
%I A122511
%S A122511 1,1,1,2,2,2,3,4,4,5,7,8,9,12,15,17,21,27,32,38,48,59,70,86,107,129,156,
%T A122511 193,236,285,349,429,521,634,778,950,1155,1412,1728,2105,2567,3140,3833,
%U A122511 4672,5707,6973,8505,10379,12680,15478
%N A122511 Alternative method for A079398 using vector matrirx Markov.
%K A122511 nonn,uned
%A A122511 Roger Bagula (rlbagula(AT)sbcglobal.net), Sep 16 2006

which is IDENTICAL with

%I A079398
%S A079398 0,1,1,1,1,2,2,2,3,4,4,5,7,8,9,12,15,17,21,27,32,38,48,59,70,86,107,129,
%T A079398 156,193,236,285,349,429,521,634,778,950,1155,1412,1728,2105,2567,3140,
%U A079398 3833,4672,5707,6973,8505,10379,12680,15478,18884,23059,28158,34362
%N A079398 Conjectured values of d(n), the dimension of a Z-module in MZV(conv).
%C A079398 From the conjecture of Zagier, Drinfeld, Kontsevich and Goncharov (see link).
%C A079398 P(0)=P(1)=P(2)=P(3)=1, for m>3: P(m) = P(m-3) + P(m-4) is the 3rd sequence in the series: Fibonacci sequence, Padovan sequence, ... The Padovan sequence (whose ratio of successive terms approaches the plastic constant) is similar to the Perrin sequence. - Jonathan Vos Post (jvospost2(AT)yahoo.com), Jan 23 2005
%D A079398 Michel Waldschmidt, "Multiple Zeta values and Euler-Zagier numbers", in Number theory and discrete mathematics, International conference in honour of Srinivasa Ramanujan, Center for Advanced Study in Mathematics, Panjab University, Chandigarh, (Oct 02, 2000).
%H A079398 Michel Waldschmidt, <a href="http://www.institut.math.jussieu.fr/~miw/articles/pdf/Transparents.pdf">Multiple Zeta values and Euler-Zagier numbers</a>
%H A079398 Eric W. Weisstein, <a href="http://mathworld.wolfram.com/PadovanSequence.html">Padovan Sequence</a>.
%F A079398 a(1)=0 a(2)=a(3)=a(4)=1 for n>=4 a(n)=a(n-2)+a(n-3)
%F A079398 a(n)=sum{k=0..floor((n-1)/2), binomial(floor((n-k-1)/3), k)} (offset 0); a(n)=sum{k=0..floor(n/2), binomial(floor((n-k-1)/3), k)}-0^n. (offset 0). - Paul Barry (pbarry(AT)wit.ie), Jul 06 2004
%F A079398 For n>1, a(n) = P(n-2) where P(n) is defined by: P(0)=P(1)=P(2)=P(3)=1, for m>3: P(m) = P(m-3) + P(m-4). - Jonathan Vos Post (jvospost2(AT)yahoo.com), Jan 23 2005
%Y A079398 Cf. A000931.
%K A079398 nonn
%O A079398 1,6
%A A079398 Benoit Cloitre (abmt(AT)wanadoo.fr), Feb 16 2003


ralf





From neoneye at gmail.com  Sun Nov 12 16:54:56 2006
From: neoneye at gmail.com (Simon Strandgaard)
Date: Sun, 12 Nov 2006 16:54:56 +0100
Subject: growing binary trees
In-Reply-To: <20061112090902.GA1205@ark.in-berlin.de>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
	 <20061112090902.GA1205@ark.in-berlin.de>
Message-ID: <df1390cc0611120754n68a94a44k76bc07615c3bdb98@mail.gmail.com>

On 11/12/06, Ralf Stephan <ralf at ark.in-berlin.de> wrote:
> > I made up an interesting sequence:
> >
> > 1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 26, 33, 41,
> > 50, 62, 77, 94, 115, 142, 174, 212, 260, 319,
> > 389, 475, 582, 711, 867, 1060, 1296
>
> According to superseeker:
[snip]

I don't see how its already in oeis?

How is A017817, A122511, A079398 related to it?
There is no match between my numbers and their numbers.
There doesn't seem to be any constant I can add to
make them match. Im an amateur.


--
Simon Strandgaard





From ralf at ark.in-berlin.de  Sun Nov 12 17:18:14 2006
From: ralf at ark.in-berlin.de (Ralf Stephan)
Date: Sun, 12 Nov 2006 17:18:14 +0100
Subject: growing binary trees
In-Reply-To: <20061112090902.GA1205@ark.in-berlin.de>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com> <20061112090902.GA1205@ark.in-berlin.de>
Message-ID: <20061112161814.GA11423@ark.in-berlin.de>

Simon, sorry for the cryptic reply.

In other words:

Let n>=0, and your sequence a(n) = 1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 
26, 33, 41, 50, 62, 77, 94, 115, 142, 174, 212, 260, 319, ...

then 

A079398(n) = a(n-2) - a(n-1) 
A122511(n) = a(n+1) - a(n)      
A017817(n) = A122511(n+3) - A122511(n+2)

All of these sequences satisfy the recurrence
a(n) = a(n-2) + a(n-3) + c

and are thus related to the Padovan sequence A000931.
With your sequence, c appears to be 3 but this has
to be proved, and all of the above is only a conjecture,
based on solid evidence, though.


ralf







From neoneye at gmail.com  Sun Nov 12 17:30:14 2006
From: neoneye at gmail.com (Simon Strandgaard)
Date: Sun, 12 Nov 2006 17:30:14 +0100
Subject: growing binary trees
In-Reply-To: <20061112161814.GA11423@ark.in-berlin.de>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
	 <20061112090902.GA1205@ark.in-berlin.de>
	 <20061112161814.GA11423@ark.in-berlin.de>
Message-ID: <df1390cc0611120830u79249ff5ue1f9c588cde122d2@mail.gmail.com>

On 11/12/06, Ralf Stephan <ralf at ark.in-berlin.de> wrote:
[snip]
> All of these sequences satisfy the recurrence
> a(n) = a(n-2) + a(n-3) + c
>
> and are thus related to the Padovan sequence A000931.
> With your sequence, c appears to be 3 but this has
> to be proved, and all of the above is only a conjecture,
> based on solid evidence, though.
[snip]

I get it. Thanks for the explaination :-)

--
Simon Strandgaard





From njas at research.att.com  Sun Nov 12 17:38:06 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sun, 12 Nov 2006 11:38:06 -0500 (EST)
Subject: growing binary trees
Message-ID: <200611121638.LAA08650@fry.research.att.com>

I will add Simon S.'s sequence to the OEIS - it will be A123015 
in a few minutes

Can Ruby lines be combined onto one line?

NJAS





From njas at research.att.com  Sun Nov 12 18:22:45 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sun, 12 Nov 2006 12:22:45 -0500 (EST)
Subject: growing binary trees
Message-ID: <200611121722.MAA17909@fry.research.att.com>

About this sequence (it will be A123015):


1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 26, 33, 41,
50, 62, 77, 94, 115, 142, 174, 212, 260, 319,
389, 475, 582, 711, 867, 1060, 1296


Its a binary tree where leafs is getting
added when a nodes internal counter reaches 3.

- I really don't understand the definition.  What exactly is the
n-th term?

Neil Sloane





From neoneye at gmail.com  Sun Nov 12 19:10:23 2006
From: neoneye at gmail.com (Simon Strandgaard)
Date: Sun, 12 Nov 2006 19:10:23 +0100
Subject: growing binary trees
In-Reply-To: <200611121638.LAA08650@fry.research.att.com>
References: <200611121638.LAA08650@fry.research.att.com>
Message-ID: <df1390cc0611121010r2cf66135gfa52aff18c944540@mail.gmail.com>

On 11/12/06, N. J. A. Sloane <njas at research.att.com> wrote:
> I will add Simon S.'s sequence to the OEIS - it will be A123015
> in a few minutes
>
> Can Ruby lines be combined onto one line?


I cannot express this thing as math, and its difficult for me to
compact the code without loss of readability.


below is a 5 line version:

class Node; def initialize; @n = 1; @c = [] end
def count; @c.inject(@n){|n,c| n + c.count} end
def grow; return @n += 1 if @n < 3; @c.each{|c| c.grow }
@c << Node.new if @c.size < 2; end; end; r = []; node = Node.new
30.times { r << node.count; node.grow }; p r


here is the oneliner (less readable):

class C;def initialize;@n=1;@c=[] end;def n;@c.inject(@n){|n,c|n+c.n}
end;def g;return @n+=1 if @n<3; @c.each{|c|c.g};@c<<C.new if
@c.size<2;end;end;r=[];c=C.new;30.times{r<<c.n;c.g};p r



the n-th term is:
a(0)=1, a(1)=2, a(2)=3, a(3)=4, a(4)=6, a(5)=8, a(6)=10, a(7)=13, ...


code for n'th term is below:

class Node
 def initialize
   @n = 1
   @childs = []
 end
 def count
   @childs.inject(@n){|n,child| n + child.count}
 end
 def grow
   return @n += 1 if @n < 3
   @childs.each{|child| child.grow }
   @childs << Node.new if @childs.size < 2
 end
end

def a(n)
  node = Node.new
  n.times{ node.grow }
  node.count
end

0.upto(30) {|n| p a(n) }



visually it works like this:

step#0
  1


step#1
  2


step#2
  3


step#3
  3
 /
1


step#4
  3
 / \
2   1


step#5
  3
 / \
3   2


step#6
    3
   / \
  3   3
 /
1


step#7
      3
    /   \
  3       3
 / \     /
2   1   1


step#8
      3
    /   \
  3       3
 / \     / \
3   2   2   1



--
Simon Strandgaard





From maxale at gmail.com  Mon Nov 13 07:53:16 2006
From: maxale at gmail.com (Max A.)
Date: Sun, 12 Nov 2006 22:53:16 -0800
Subject: growing binary trees
In-Reply-To: <20061112090902.GA1205@ark.in-berlin.de>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
	 <20061112090902.GA1205@ark.in-berlin.de>
Message-ID: <d3dac270611122253m4930fb58qdf5d06ce4fbf8026@mail.gmail.com>

On 11/12/06, Ralf Stephan <ralf at ark.in-berlin.de> wrote:

> %I A122511
> %S A122511 1,1,1,2,2,2,3,4,4,5,7,8,9,12,15,17,21,27,32,38,48,59,70,86,107,129,156,
> %T A122511 193,236,285,349,429,521,634,778,950,1155,1412,1728,2105,2567,3140,3833,
> %U A122511 4672,5707,6973,8505,10379,12680,15478
> %N A122511 Alternative method for A079398 using vector matrirx Markov.
> %K A122511 nonn,uned
> %A A122511 Roger Bagula (rlbagula(AT)sbcglobal.net), Sep 16 2006
>
> which is IDENTICAL with
>
> %I A079398
> %S A079398 0,1,1,1,1,2,2,2,3,4,4,5,7,8,9,12,15,17,21,27,32,38,48,59,70,86,107,129,
> %T A079398 156,193,236,285,349,429,521,634,778,950,1155,1412,1728,2105,2567,3140,
> %U A079398 3833,4672,5707,6973,8505,10379,12680,15478,18884,23059,28158,34362
> %N A079398 Conjectured values of d(n), the dimension of a Z-module in MZV(conv).
> %C A079398 From the conjecture of Zagier, Drinfeld, Kontsevich and Goncharov (see link).
> %C A079398 P(0)=P(1)=P(2)=P(3)=1, for m>3: P(m) = P(m-3) + P(m-4) is the 3rd sequence in the series: Fibonacci sequence, Padovan sequence, ... The Padovan sequence (whose ratio of successive terms approaches the plastic constant) is similar to the Perrin sequence. - Jonathan Vos Post (jvospost2(AT)yahoo.com), Jan 23 2005
> %D A079398 Michel Waldschmidt, "Multiple Zeta values and Euler-Zagier numbers", in Number theory and discrete mathematics, International conference in honour of Srinivasa Ramanujan, Center for Advanced Study in Mathematics, Panjab University, Chandigarh, (Oct 02, 2000).
> %H A079398 Michel Waldschmidt, <a href="http://www.institut.math.jussieu.fr/~miw/articles/pdf/Transparents.pdf">Multiple Zeta values and Euler-Zagier numbers</a>
> %H A079398 Eric W. Weisstein, <a href="http://mathworld.wolfram.com/PadovanSequence.html">Padovan Sequence</a>.
> %F A079398 a(1)=0 a(2)=a(3)=a(4)=1 for n>=4 a(n)=a(n-2)+a(n-3)
> %F A079398 a(n)=sum{k=0..floor((n-1)/2), binomial(floor((n-k-1)/3), k)} (offset 0); a(n)=sum{k=0..floor(n/2), binomial(floor((n-k-1)/3), k)}-0^n. (offset 0). - Paul Barry (pbarry(AT)wit.ie), Jul 06 2004
> %F A079398 For n>1, a(n) = P(n-2) where P(n) is defined by: P(0)=P(1)=P(2)=P(3)=1, for m>3: P(m) = P(m-3) + P(m-4). - Jonathan Vos Post (jvospost2(AT)yahoo.com), Jan 23 2005
> %Y A079398 Cf. A000931.
> %K A079398 nonn
> %O A079398 1,6
> %A A079398 Benoit Cloitre (abmt(AT)wanadoo.fr), Feb 16 2003

Should not then A122511 be marked as a duplicate of A079398 with an
appropriate comment?
I do not see much sense in keeping them separate.

Max





From davidwwilson at comcast.net  Mon Nov 13 18:50:32 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Mon, 13 Nov 2006 12:50:32 -0500
Subject: A102567 = A106498
Message-ID: <000601c7074c$375f4d50$6501a8c0@yourxhtr8hvc4p>

A102567 and A106498 are duplicates and should be merged. The latter sequence has an associated b-file.
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From maxale at gmail.com  Mon Nov 13 19:24:29 2006
From: maxale at gmail.com (Max A.)
Date: Mon, 13 Nov 2006 10:24:29 -0800
Subject: A050259 and a new solution to 2^n = 3 (mod n)
Message-ID: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>

SeqFans,

In course of exploring all solutions of the congruence 2^n = 3 (mod n)
below 10^16, I have found a new solution:

n = 3468371109448915

This came as a big surprise to me since my original goal was to prove
that 8365386194032363 is the second term of A050259. But now it
appears to be at least the third term.

So far I've submitted the new solution as a comment to A050259, but
hopefully it will take its true place in this sequence soon.

Regards,
Max

P.S. See also http://mathworld.wolfram.com/2.html





From rkg at cpsc.ucalgary.ca  Mon Nov 13 20:15:20 2006
From: rkg at cpsc.ucalgary.ca (Richard Guy)
Date: Mon, 13 Nov 2006 12:15:20 -0700 (MST)
Subject: A050259 and a new solution to 2^n = 3 (mod n)
In-Reply-To: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>
References: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>
Message-ID: <Pine.LNX.4.63.0611131211270.24385@csl>

Congratulations!  May I know Max's full name,
so that he can earn undying fame in UPINT as
well as in OEIS ?   Thanks,   R.

On Mon, 13 Nov 2006, Max A. wrote:

> SeqFans,
>
> In course of exploring all solutions of the congruence 2^n = 3 
> (mod n)
> below 10^16, I have found a new solution:
>
> n = 3468371109448915
>
> This came as a big surprise to me since my original goal was 
> to prove
> that 8365386194032363 is the second term of A050259. But now 
> it
> appears to be at least the third term.
>
> So far I've submitted the new solution as a comment to 
> A050259, but
> hopefully it will take its true place in this sequence soon.
>
> Regards,
> Max
>
> P.S. See also http://mathworld.wolfram.com/2.html






From jvospost3 at gmail.com  Mon Nov 13 21:34:20 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Mon, 13 Nov 2006 12:34:20 -0800
Subject: A050259 and a new solution to 2^n = 3 (mod n)
In-Reply-To: <Pine.LNX.4.63.0611131211270.24385@csl>
References: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>
	 <Pine.LNX.4.63.0611131211270.24385@csl>
Message-ID: <5542af940611131234l54af8bcar76fcba8e87d0fb1b@mail.gmail.com>

Is this a hierarchy of fame, or a graph, or directed graph, or a lattice, or
what? What percentage of "nice" sequences are referenced in each successive
edition of UPINT (Unsolved problems in Number Theory)? How many OEIS seqs
per edition of UPINT, now that the two are more tightly interlinked? What
estimates does one make to extrapolate for future editions? Should there be
a "very nice" keyword for seqs that pose problems in UPINT, when they get
solved in later editions?

Only hypothetical for me, as I've had over 1400 seqs in OEIS and not one is
yet "nice", although some are comments on "nice" seqs. So if there is a
hierarchy of fame, I'm just above the lowest  level (or highest, depending
on indexing) in quality, which is most assuredly not correlated well with
quantity.  In Science Fiction, for instance, there are indeed a hierarchies
of fame, based on awards (Hugo, Nebula), coauthorships, pay, anthologies,
and the like.  I'm also just somewhat above the bottom, in part for my
coauthorships with Asimov, Bradbury, Clarke, and Heinlein.  I've done
research on Asimov Number as the Science Fiction equivalent of Erdos Number,
but that's another story.

By the way, I have the penultimate edition (so far) of UPINT as a gift from
the wonderful Tony Noe, when he bought the ultimate edition (so far). I echo
the applause by njas for all the hundreds of errors in OEIS found and
corrected by Tony, and the many many b-files.

Best,

Jonathan Vos Post

On 11/13/06, Richard Guy <rkg at cpsc.ucalgary.ca> wrote:
>
> Congratulations!  May I know Max's full name,
> so that he can earn undying fame in UPINT as
> well as in OEIS ?   Thanks,   R.
>
> On Mon, 13 Nov 2006, Max A. wrote:
>
> > SeqFans,
> >
> > In course of exploring all solutions of the congruence 2^n = 3
> > (mod n)
> > below 10^16, I have found a new solution:
> >
> > n = 3468371109448915
> >
> > This came as a big surprise to me since my original goal was
> > to prove
> > that 8365386194032363 is the second term of A050259. But now
> > it
> > appears to be at least the third term.
> >
> > So far I've submitted the new solution as a comment to
> > A050259, but
> > hopefully it will take its true place in this sequence soon.
> >
> > Regards,
> > Max
> >
> > P.S. See also http://mathworld.wolfram.com/2.html
>
>
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From r.rosenthal at web.de  Mon Nov 13 23:12:57 2006
From: r.rosenthal at web.de (Rainer Rosenthal)
Date: Mon, 13 Nov 2006 23:12:57 +0100
Subject: A050259 and a new solution to 2^n = 3 (mod n)
In-Reply-To: <Pine.LNX.4.63.0611131211270.24385@csl>
References: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com> <Pine.LNX.4.63.0611131211270.24385@csl>
Message-ID: <4558EDE9.2080404@web.de>

Richard Guy wrote:
> Congratulations!  May I know Max's full name,
> so that he can earn undying fame in UPINT as
> well as in OEIS ?   Thanks,   R.
> 
> On Mon, 13 Nov 2006, Max A. wrote:
>>In course of exploring all solutions of the congruence 2^n = 3 
>>(mod n) below 10^16, I have found a new solution:
>>
>>n = 3468371109448915
>>P.S. See also http://mathworld.wolfram.com/2.html

This is a lovely finding, which I will have to tell
immediately in the German Newsgroup de.sci.mathematik.

Congratulations from a friend of OEIS and UPINT and
of things that are not too difficult to comprehend :-)
Not only congratulations to Max for finding, but to all
of us for being witnesses.

Best regards,
Rainer Rosenthal
r.rosenthal at web.de






From franktaw at netscape.net  Tue Nov 14 00:01:52 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 13 Nov 2006 18:01:52 -0500
Subject: A102567 = A106498
In-Reply-To: <000601c7074c$375f4d50$6501a8c0@yourxhtr8hvc4p>
References: <000601c7074c$375f4d50$6501a8c0@yourxhtr8hvc4p>
Message-ID: <8C8D58ED7DCDA34-128-9CD7@FWM-R16.sysops.aol.com>

A116148 (n concatenated with n-1 gives the product of two numbers which 
differ by 2) is also a duplicate of these two.  The only possible 
exceptions would be numbers of the form 10^n-1; but then the 
concatenation is 10^{2n}+10^n-1, and 10^{2n}+10^n is never a square.

Franklin T. Adams-Watters


-----Original Message-----
From: davidwwilson at comcast.net

A102567 and A106498 are duplicates and should be merged.?The latter 
sequence has an associated b-file.




________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From maxale at gmail.com  Tue Nov 14 00:52:58 2006
From: maxale at gmail.com (Max A.)
Date: Mon, 13 Nov 2006 15:52:58 -0800
Subject: A050259 and a new solution to 2^n = 3 (mod n)
In-Reply-To: <Pine.LNX.4.63.0611131211270.24385@csl>
References: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>
	 <Pine.LNX.4.63.0611131211270.24385@csl>
Message-ID: <d3dac270611131552v64b33034n8ed43b34faf905a4@mail.gmail.com>

I answered Richard privately.
But if other people also want to refer to my full name, it is
Max Alekseyev

Regards,
Max

P.S. It is not a conspiracy but rather my way to beat spamming. So
far, spam emails never addressed me by my full name, and I can easily
recognize important emails that do contain my full name in the To:
field.

On 11/13/06, Richard Guy <rkg at cpsc.ucalgary.ca> wrote:
> Congratulations!  May I know Max's full name,
> so that he can earn undying fame in UPINT as
> well as in OEIS ?   Thanks,   R.
>
> On Mon, 13 Nov 2006, Max A. wrote:
>
> > SeqFans,
> >
> > In course of exploring all solutions of the congruence 2^n = 3
> > (mod n)
> > below 10^16, I have found a new solution:
> >
> > n = 3468371109448915
> >
> > This came as a big surprise to me since my original goal was
> > to prove
> > that 8365386194032363 is the second term of A050259. But now
> > it
> > appears to be at least the third term.
> >
> > So far I've submitted the new solution as a comment to
> > A050259, but
> > hopefully it will take its true place in this sequence soon.
> >
> > Regards,
> > Max
> >
> > P.S. See also http://mathworld.wolfram.com/2.html
>
>





From bowerc at usa.net  Tue Nov 14 02:10:04 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Mon, 13 Nov 2006 17:10:04 -0800
Subject: growing binary trees
Message-ID: <397kkNBJe5074S28.1163466604@cmsweb28.cms.usa.net>

 From Simon's diagrams, I can see this is analogous to Fibonacci's
rabbits. Basically, the behavior of the node is given by its age.
A node at age 0 or 1 grows and one at age 2 or 3 produces a new node.

 From this its easy to get a g.f. of

(1+x+x^2)/(1-x-x^3+x^5)

1 2 3 4 6 8 10 13 17 21 26 33 41 50 62 77 94 115 142 174 212 260 319 389
475 582 711 867 1060 1296 1581 1930 2359 2880 3514 4292 5242 6397 7809
9537

Also, the number of nodes has g.f.

1/(1-x-x^3+x^5)
1 1 1 2 3 3 4 6 7 8 11 14 16 20 26 31 37 47 58 69 85 106 128 155 192 235
284 348 428 520 633 777 949 1154 1411 1727 2104 2566 3139 3832

which isn't in the EIS either.

------ Original Message ------
From: "Simon Strandgaard" <neoneye at gmail.com>
To: njas at research.att.comCc: seqfan at ext.jussieu.fr
Subject: Re: growing binary trees

...

> visually it works like this:
> 
> step#0
>   1
> 
> 
> step#1
>   2
> 
> 
> step#2
>   3
> 
> 
> step#3
>   3
>  /
> 1
> 
> 
> step#4
>   3
>  / \
> 2   1
> 
> 
> step#5
>   3
>  / \
> 3   2
> 
> 
> step#6
>     3
>    / \
>   3   3
>  /
> 1
> 
> 
> step#7
>       3
>     /   \
>   3       3
>  / \     /
> 2   1   1
> 
> 
> step#8
>       3
>     /   \
>   3       3
>  / \     / \
> 3   2   2   1
> 
> 
> 
> --
> Simon Strandgaard
> 









From zbi74583 at boat.zero.ad.jp  Tue Nov 14 06:50:02 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Tue, 14 Nov 2006 14:50:02 +0900
Subject: A036471
Message-ID: <013d01c707b0$bedd9660$d967d7dc@FMC57937B6BE00>

    Hi, Seqfans

    An exhaustive search of A036471  with a computer up to 10^7 is almost 
impossible.

    So, using my algorithm, I searched small examples which are ?regular 
type?.



    See this site, there is an explanation of "type".



              http://amicable.homepage.dk/apstat.htm#typesys







    I found these examples.



    2^9*3^2*13*31*(5*11,71)*(7*23,191) 
11 digits

    2^15*3*5^2*11*31*257*(1439,19*71,23*59,29*47) 
16 digits

    2^3*3^4*5^2*(7*23,191)*(17*19,359) 
9 digits

    (2^3*19*41,2^5*199)*(3*5*7*13,3^2*5*7*139) 
8 digits

    (2^3*19*41,2^5*199)*(3^3*5*7*71,3^3*5*17*31) 
9 digits

    (2^3*19*41,2^5*199)*(3^2*7*13*5*17,3^2*7*13*107) 
9 digits

    2^7*3^2*13*(5*11,71)*(7*59,479) 
9 digits

    2^5*3^4*5^2*(17*19,359)*(23*29,719) 
11 digits







    Where, x*(y,z) means (x*z,x*y). (x,y)*(z,u)=(x*z,x*u,y*z,y*z)





    I think that if a smaller one than 3270960  exists, then it must be 
?irregular".



    Regurer type Amicable Quadruple are generated from ?seeds? which are 
two pairs or one quadruple of primes or almost primes, {x,y} and {z,u} or 
{x,y,z,u}.

    The numbers x,y,z,u of small Amicable Quadruple must satisfy the 
following conditions.

             .

    o Sigma(x)=Sigma(y) and Sigma(z)=Sigma(u), or 
Sigma(x)=Sigma(y)=Sigma(z)=Sigma(u)

    o all prime factors of x,y,z,u are small

    o all prime factors of x+y and z+u are small, or all  prime factors of 
x+y+z+u are small



    For example, if {x,y}={7*23,191} and {z,u}={17*19,359} then

    o Sigma(7*23)=Sigma(191), Sigma(17*19)=Sigma(359)

    o all prime factors of 7*23,191,17*19,359  are small

    o x+y=2^5*11, z+u=2*11*31

    So, (7*23,191)*(17*19,359) generates 2^3*3^4*5^2*(7*23,191)*(17*19,359)





    If we obtain all small seeds then we will calculate all small regular 
terms of A036471.



    Yasutoshi


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From g.resta at iit.cnr.it  Tue Nov 14 14:40:49 2006
From: g.resta at iit.cnr.it (Giovanni Resta)
Date: Tue, 14 Nov 2006 14:40:49 +0100
Subject: A036471
In-Reply-To: <013d01c707b0$bedd9660$d967d7dc@FMC57937B6BE00>
References: <013d01c707b0$bedd9660$d967d7dc@FMC57937B6BE00>
Message-ID: <4559C761.70305@iit.cnr.it>

kohmoto wrote:
>
> Hi, Seqfans
>
> An exhaustive search of A036471 with a computer up to 10^7 is almost
> impossible.
>
Here it is the result of an exhaustive search
with a computer up to 10^7, in the format
sigma -> {a,b,c,d} (with a,b,c,d<10^7)

13927680 -> {3270960, 3361680, 3461040, 3834000}
16248960 -> {3767400, 4090320, 4150440, 4240800}
19498752 -> {4651920, 4839120, 4918320, 5089392}
21228480 -> {4969440, 5116320, 5475960, 5666760}
24373440 -> {5682600, 5831280, 5920200, 6939360}
24998400 -> {5405400, 6029100, 6421800, 7142100}
27855360 -> {6514200, 6640200, 6768720, 7932240}
28304640 -> {6126120, 6541920, 7559640, 8076960}
28304640 -> {6126120, 6922080, 7280280, 7976160}
29030400 -> {6320160, 6902280, 7685496, 8122464}
29030400 -> {6977880, 7087080, 7227360, 7738080}
29877120 -> {7013160, 7436520, 7688520, 7738920}
29998080 -> {6819120, 7327320, 7715400, 8136240}
29998080 -> {6966960, 7054320, 7840560, 8136240}
32497920 -> {7706160, 7722000, 7948080, 9121680}
33022080 -> {7731360, 7852320, 8125920, 9312480}
33868800 -> {7469280, 8157240, 8873760, 9368520}
33868800 -> {7469280, 8157240, 9098460, 9143820}

Note that there are values which admit more than one quadruple,
and there are also quadruples which share 1 or 2 common values.

Here are some larger quadruples, but now there can be some
missing among these, because here I used an heuristic.

{8353800, 8920800, 10063200, 10159800}
{8288280, 9360540, 9830520, 10260180}
{8316000, 8759520, 10260180, 10403820}
{9258480, 9621360, 9661680, 10455984}
{9009000, 10174500, 10521000, 11179980}
{10048500, 10174500, 10188360, 10473120}
{9840600, 10098000, 10914120, 10930320}
{9923760, 10316880, 10747440, 10794960}
{9563400, 10085040, 11241720, 11964240}
{9646560, 10667160, 11502540, 11729340}
{9767520, 11202840, 11217960, 11357280}
{9828000, 10796940, 11362680, 12041820}
{10417680, 10538640, 10979280, 13061520}
{10450440, 11614680, 11802420, 13306860}
{10450440, 12179160, 12204360, 12340440}

etc.etc. (program is running).

Giovanni Resta






From davidwwilson at comcast.net  Tue Nov 14 18:00:27 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Tue, 14 Nov 2006 12:00:27 -0500
Subject: A023153 through A023161
References: <p06230918c17f1b479bcf@[192.168.1.43]>
Message-ID: <006d01c7080e$632ff480$6501a8c0@yourxhtr8hvc4p>

Tony Noe has found that A023153 through A023161 are not multiplicative as I 
had marked them.

At the time, I was convinced that in sequence {s_k(n) = k^(2^n)}, a cycle of 
period x (mod a) and a cycle of period y (mod b) implied a unique cycle of 
period xy (mod ab). However, if gcd(x, y) > 1, the cycles (mod a) and (mod 
b) can align in more than one way, producing more than one cycle (mod ab). 
For example, the cycle (2,4) (mod 7) and the cycle (4,7) (mod 9) produce two 
cycles (4,16) and (25,58) (mod 63). This means that sometimes f(ab) > 
f(a)f(b).

This problem affects A023153 through A023161, and Tony Noe has shown that 
indeed none of these are multiplicative.

NJAS: Please remove the "mult" keyword and any comments related to 
multiplicativity from A023153-A023161.

Tony: Since these sequences are almost multiplicative, but are not, and you 
seem to be investigating these sequences, could you please add lines such as

%C A023153 Not multiplicative: a(63) = 10, but a(7)*a(9) = 3*3 = 9.

Thanks for spotting my error, Tony. Seems I have been bitten by the dragon I 
released.

----- Original Message ----- 
From: "T. D. Noe" <noe at sspectra.com>
To: <davidwwilson at comcast.net>
Sent: Tuesday, November 14, 2006 2:06 AM
Subject: Question about A023161


> David, according to my calculations, this sequence fails to be
> multiplicative at a(667).
>
> Best regards,
>
> Tony
>
>
> -- 
> No virus found in this incoming message.
> Checked by AVG Free Edition.
> Version: 7.5.430 / Virus Database: 268.14.5/533 - Release Date: 11/13/2006 
> 8:56 PM
>
> 






From jvospost3 at gmail.com  Tue Nov 14 20:46:00 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Tue, 14 Nov 2006 11:46:00 -0800
Subject: A023153 through A023161
In-Reply-To: <006d01c7080e$632ff480$6501a8c0@yourxhtr8hvc4p>
References: <p06230918c17f1b479bcf@192.168.1.43>
	 <006d01c7080e$632ff480$6501a8c0@yourxhtr8hvc4p>
Message-ID: <5542af940611141146j1a3212b0u2d31815c6bee8387@mail.gmail.com>

"almost multiplicative" -- interesting notion.  What other "almost
multiplicative" are there in OEIS masquerading as multiplicative (i'm
guessing that the amazing Tony Noe is searching), and what are but with no
keyword on that? Can "almost multiplicative" be axiomatizeed, or is it just
a subjective interpretation of "almost"? What pattens are there in "almost
multiplicative" -- new sequences on where multiplicativity breaks down?

-- Jonathan Vos Post

On 11/14/06, David Wilson <davidwwilson at comcast.net> wrote:
>
> Tony Noe has found that A023153 through A023161 are not multiplicative as
> I
> had marked them.
>
> At the time, I was convinced that in sequence {s_k(n) = k^(2^n)}, a cycle
> of
> period x (mod a) and a cycle of period y (mod b) implied a unique cycle of
> period xy (mod ab). However, if gcd(x, y) > 1, the cycles (mod a) and (mod
> b) can align in more than one way, producing more than one cycle (mod ab).
> For example, the cycle (2,4) (mod 7) and the cycle (4,7) (mod 9) produce
> two
> cycles (4,16) and (25,58) (mod 63). This means that sometimes f(ab) >
> f(a)f(b).
>
> This problem affects A023153 through A023161, and Tony Noe has shown that
> indeed none of these are multiplicative.
>
> NJAS: Please remove the "mult" keyword and any comments related to
> multiplicativity from A023153-A023161.
>
> Tony: Since these sequences are almost multiplicative, but are not, and
> you
> seem to be investigating these sequences, could you please add lines such
> as
>
> %C A023153 Not multiplicative: a(63) = 10, but a(7)*a(9) = 3*3 = 9.
>
> Thanks for spotting my error, Tony. Seems I have been bitten by the dragon
> I
> released.
>
> ----- Original Message -----
> From: "T. D. Noe" <noe at sspectra.com>
> To: <davidwwilson at comcast.net>
> Sent: Tuesday, November 14, 2006 2:06 AM
> Subject: Question about A023161
>
>
> > David, according to my calculations, this sequence fails to be
> > multiplicative at a(667).
> >
> > Best regards,
> >
> > Tony
> >
> >
> > --
> > No virus found in this incoming message.
> > Checked by AVG Free Edition.
> > Version: 7.5.430 / Virus Database: 268.14.5/533 - Release Date:
> 11/13/2006
> > 8:56 PM
> >
> >
>
>
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From pauldhanna at juno.com  Wed Nov 15 05:04:51 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Tue, 14 Nov 2006 23:04:51 -0500
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
Message-ID: <20061114.230452.536.0.pauldhanna@juno.com>

Seqfans, 
    Could someone compute the initial terms of the following. 
I do not believe that it is in the OEIS (?). 
  
Number of labelled nodes in generation n of a rooted tree 
where a node with label k has k child nodes such that 
the label of each child node is the least unused label in 
the path from the root to that child node, 
and where root is labeled '1'. 
  
Thus every distinct path in the tree eventually forms 
a permutation of the positive integers. 
 
The wording needs to be made more precise and concise. 
 
Here are the initial nodes of the tree for generations 0..4: 
Gen 0: [1];
Gen 1: (1)->[2];
Gen 2: (2)->[3,4];
Gen 3: (3)->[4,5,6], (4)->[3,5,6,7];
Gen 4: (4)->[5,6,7,8], (5)->[4,6,7,8,9], (6)->[4,5,7,8,9,10], 
(3)->[5,6,7], (5)->[3,6,7,8,9], (6)->[3,5,7,8,9,10], 
(7)->[3,5,6,8,9,10,11]; ...
 
Thus, the number of nodes in generation n begins:
[1, 1, 2, 7, 36, 248, ...]
 
The maximum node in generation n begins:
[1, 2, 4, 7, 11, ...]
 
Would appreciate it if someone could compute more terms.
Thanks,
      Paul





From franktaw at netscape.net  Wed Nov 15 05:13:17 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Tue, 14 Nov 2006 23:13:17 -0500
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
In-Reply-To: <20061114.230452.536.0.pauldhanna@juno.com>
References: <20061114.230452.536.0.pauldhanna@juno.com>
Message-ID: <8C8D68382FC1E11-FEC-53E1@FWM-M20.sysops.aol.com>

You need to reword that a bit.  Your example makes clear that the
children of a given node must all have distinct labels, but you don't
say so.

Without that condition, the problem is pretty easy.  :-)

Franklin T. Adams-Watters


-----Original Message-----
From: pauldhanna at juno.com

...

Number of labelled nodes in generation n of a rooted tree
where a node with label k has k child nodes such that
the label of each child node is the least unused label in
the path from the root to that child node,
and where root is labeled '1'.

...
________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From njas at research.att.com  Wed Nov 15 06:13:02 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 15 Nov 2006 00:13:02 -0500 (EST)
Subject: king chickens
Message-ID: <200611150513.AAA96507@fry.research.att.com>

can anyone extend this?

%I A123553
%S A123553 1,2,12,104
%N A123553 A "king chicken" in a tournament graph (a directed labeled graph on n nodes) is a player A who for any other player B either beats B directly or beats someone who beats B.  Sequence gives total number of king chickens in all 2^(n(n-1)/2) tourna
%C A123553 H. G. Landau showed in 1951 that a player is a king chicken if and only if he has the highest score. There may be several king chickens in a tournament.
%D A123553 S. B. Maurer, The king chicken theorems, Math. Mag., 53 (1980), 67-80.
%e A123553 For n = 3 there are 8 tournaments: six of the form A beats B and C, and B beats C, with one king chicken (A), and two of the form A beats B beats C beats A, with three king chickens each (A or B or C), for a total of 6*1 + 2*3 = 12.
%K A123553 nonn,more,new
%O A123553 1,2
%A A123553 njas, Nov 14 2006

Neil





From bdm at cs.anu.edu.au  Wed Nov 15 06:37:18 2006
From: bdm at cs.anu.edu.au (Brendan McKay)
Date: Wed, 15 Nov 2006 16:37:18 +1100
Subject: king chickens
In-Reply-To: <200611150513.AAA96507@fry.research.att.com>
References: <200611150513.AAA96507@fry.research.att.com>
Message-ID: <20061115053717.GA15829@cs.anu.edu.au>


I don't have time right now, so if someone who knows how to use nauty
wants to jump into this, here is how to do it:
1. Fetch the unlabelled tournaments from
      http://cs.anu.edu.au/~bdm/data/digraphs.html
2. For each unlabelled tournament compute the automorphism group size A
   and the number of king chickens K.
3. The answer is the sum of K*(n!/A) over the unlabelled tournaments.

That will get the answer up to n=10 fairly easily.  The next case n=11
is plausible but more effort as there are nearly a billion tournaments.

Incidentally, the definition of "tournament" in the entry is incomplete.
It has to add that exactly one of the directed edges u->v, v->u is
present for each pair u,v of distinct vertices.

Brendan.


* N. J. A. Sloane <njas at research.att.com> [061115 16:13]:
> can anyone extend this?
> 
> %I A123553
> %S A123553 1,2,12,104
> %N A123553 A "king chicken" in a tournament graph (a directed labeled graph on n nodes) is a player A who for any other player B either beats B directly or beats someone who beats B.  Sequence gives total number of king chickens in all 2^(n(n-1)/2) tourna
> %C A123553 H. G. Landau showed in 1951 that a player is a king chicken if and only if he has the highest score. There may be several king chickens in a tournament.
> %D A123553 S. B. Maurer, The king chicken theorems, Math. Mag., 53 (1980), 67-80.
> %e A123553 For n = 3 there are 8 tournaments: six of the form A beats B and C, and B beats C, with one king chicken (A), and two of the form A beats B beats C beats A, with three king chickens each (A or B or C), for a total of 6*1 + 2*3 = 12.
> %K A123553 nonn,more,new
> %O A123553 1,2
> %A A123553 njas, Nov 14 2006
> 
> Neil





From g.resta at iit.cnr.it  Wed Nov 15 11:32:25 2006
From: g.resta at iit.cnr.it (Giovanni Resta)
Date: Wed, 15 Nov 2006 11:32:25 +0100
Subject: A question on A068480 
Message-ID: <455AECB9.5050308@iit.cnr.it>

A068480 is the sequence on numbers n such that gcd(n!-1,2^n+1)>1.

I noticed that for all the numbers listed gcd(n!-1,2^n+1) = 2n+1.

Is this a rule (proof ?) or a coincidence (counterexample ?) .

thanks,
giovanni.







From mathar at strw.leidenuniv.nl  Wed Nov 15 13:40:44 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Wed, 15 Nov 2006 13:40:44 +0100
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
Message-ID: <200611151240.kAFCeiKc009825@amer.strw.leidenuniv.nl>


pdh> From seqfan-owner at ext.jussieu.fr  Wed Nov 15 05:08:51 2006
pdh> To: seqfan at ext.jussieu.fr
pdh> Date: Tue, 14 Nov 2006 23:04:51 -0500
pdh> Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
pdh> From: "Paul D. Hanna" <pauldhanna at juno.com>
pdh> ...
pdh> Number of labelled nodes in generation n of a rooted tree 
pdh> where a node with label k has k child nodes such that 
pdh> the label of each child node is the least unused label in 
pdh> the path from the root to that child node, 
pdh> and where root is labeled '1'. 
pdh>   
pdh> Thus every distinct path in the tree eventually forms 
pdh> a permutation of the positive integers. 
pdh>  
pdh> The wording needs to be made more precise and concise. 
pdh>  
pdh> Here are the initial nodes of the tree for generations 0..4: 
pdh> Gen 0: [1];
pdh> Gen 1: (1)->[2];
pdh> Gen 2: (2)->[3,4];
pdh> Gen 3: (3)->[4,5,6], (4)->[3,5,6,7];
pdh> Gen 4: (4)->[5,6,7,8], (5)->[4,6,7,8,9], (6)->[4,5,7,8,9,10], 
pdh> (3)->[5,6,7], (5)->[3,6,7,8,9], (6)->[3,5,7,8,9,10], 
pdh> (7)->[3,5,6,8,9,10,11]; ...
pdh>  
pdh> Thus, the number of nodes in generation n begins:
pdh> [1, 1, 2, 7, 36, 248, ...]
pdh>  
pdh> The maximum node in generation n begins:
pdh> [1, 2, 4, 7, 11, ...]

My count of node numbers and maximum nodes is
Gen 0 1 1
Gen 1 1 2
Gen 2 2 4
Gen 3 7 7
Gen 4 36 11
Gen 5 248 16
Gen 6 2157 22
Gen 7 22761 29
Gen 8 283220 37
Gen 9 4068411 46

This list is limited by the simplicity of my program which keeps all the
trees in memory at the same time and runs into allocation problems if the
tenth generations are attempted.





From pauldhanna at juno.com  Wed Nov 15 14:34:15 2006
From: pauldhanna at juno.com (Paul D Hanna)
Date: Wed, 15 Nov 2006 13:34:15 GMT
Subject: Trees with Labeled Nodes that form Permutation of Positive Integer
	s
Message-ID: <20061115.053511.1712.764552@webmail42.nyc.untd.com>

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From franktaw at netscape.net  Wed Nov 15 18:34:02 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 15 Nov 2006 12:34:02 -0500
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
In-Reply-To: <20061114.230452.536.0.pauldhanna@juno.com>
References: <20061114.230452.536.0.pauldhanna@juno.com>
Message-ID: <8C8D6F3606C258E-3A4-7467@FWM-R09.sysops.aol.com>

From: pauldhanna at juno.com

>    Could someone compute the initial terms of the following.
>I do not believe that it is in the OEIS (?).
>
>Number of labelled nodes in generation n of a rooted tree
>where a node with label k has k child nodes such that
>the label of each child node is the least unused label in
>the path from the root to that child node,
excluding those used by earlier children of the same parent,
>and where root is labeled '1'.
>
>Thus every distinct path in the tree eventually forms
>a permutation of the positive integers.
>
>...

This isn't really true.  There are many paths in this tree
that do not form a permutation of the positive integers.
You get such a permutation iff, infinitely often, you choose
the first child from the current node.

It is true, of course, that any finite initial path can be
extended to a path that is a permuatation of the positive
integers.

It is an interesting problem to construct an interesting tree
where every path from the root is a permutation of the
positive integers.  I think the following construction works:
start with node 1, which arbitrarily has node 2 as a child
(by the construction that follows, 1 would have no children).
For each node with label m, add BigOmega(m) children, where
BigOmega(m) is the number of prime factors of m with
repetition; label these chilidren with the first positive integers
not occurring in the path from the root to the current node.

The tree starts:

......1
......|
......2
......|
......3
......|
......4
...../.\
....5...6
.../....|\
..6.....5.7
./.\....|.|
7...8...7.5
|../|\..|.|
8.7.9.A.8.8

Any time you hit a prime, there is only one child in the next
generation.  If you keep choosing non-prime children, the
primes start accumulating in the list of positive integers not
occurring in the path, until eventually all children of a node will
be prime.

One could of course use Omega(m), the number of distinct
prime divisors, instead of BigOmega(m).  I think tau(m)-1 would
also work.  A001511(m) (the power of 2 dividing 2m) works, too
- in this case, the single children are under the odd nodes.
________________________________________________________________________
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From qq-quet at mindspring.com  Wed Nov 15 19:23:30 2006
From: qq-quet at mindspring.com (Leroy Quet)
Date: Wed, 15 Nov 06 11:23:30 -0700
Subject: Max Score For A Game
Message-ID: <E1GkPUS-0005H1-00@pop04.mail.atl.earthlink.net>

Below is a game I invented (as posted to sci.math and rec.puzzles -- it 
is just one of many stupid silly simple games I have posted there).

I post it here to seq.fans because I wonder what the sequence is where 
the nth term is the maximum possible score for an n-by-n grid. (We could 
also have a table where the element in the mth column and nth row is the 
maximum score for an m-by-n grid.)
Of course, if there is a way to always get the score n^2 -1 for each 
n-by-n case, or m*n -1 for each m-by-n case (n or m equal 1 excluded), 
then there is be no need to calculate the sequence.

Thanks,
Leroy Quet

=====

Here is a solitaire game played on a square grid.
(I hope this game is more fun than my last game.)

Start by drawing an r-by-r grid on paper.
(I suggest an r of 5 or so for beginners.)
 
Put a "1" in each square of the left-most column
and in each square of the top-most row of the grid.

On each move the player chooses any one of the grid's
empty squares. This square is (m,n), which is the
square in the mth column from the left side of the
grid and in the nth row from the top.

The player then sums up all the integers already
written in the mth column and nth row.
So, if the grid looks like this, where the * is
the square the player has chosen to fill in next
with an integer, 
1 1 1 1
1 3
1 * 6
1 6

then the sum of the column 2's terms and row 3's
terms is 1+3+6 +1+6 = 17. Let this sum be s.

Next the player counts the number of integers in the
squares to the left of (m,n) and above (m,n)
(ie, the squares with coordinates (j,k), 1 <=j <=m,
1 <=k <=n) which are coprime to s.
If the count is c integers in the given rectangle
which are coprime to s, then the player writes c
in square (m,n).

Play continues until there is an integer in every
square of the grid.
The player's score is the number in the last square
he/she fills in.

Example game:
(r=4)
(View with fixed-width font.)

1 1 1 1   1 1 1 1   1 1 1 1   1 1 1 1 
1         1 3       1 3       1 3 5  
1         1         1   6     1   6
1         1         1         1

1 1 1 1   1 1 1 1   1 1 1 1
1 3 5 6   1 3 5 6   1 3 5 6
1   6     1   6     1 5 6
1         1 6       1 6

1 1 1 1   1 1 1 1   1 1 1 1
1 3 5 6   1 3 5 6   1 3 5 6
1 5 6     1 5 6     1 5 6 11
1 6   10  1 6 1110  1 6 1110

So the player gets 11 points.

(It seems that a good strategy is to
{unlike in my example} finish in the
lower right square, and to {as in my
example, 1+6+10+1+5+6=29} have a row
and column sum on the last move which
is a prime.)

Question: Is it always possible to get,
for an r-by-r grid (r >= 2), r^2 -1 points?

Thanks,
Leroy Quet





From davidwwilson at comcast.net  Wed Nov 15 19:48:18 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Wed, 15 Nov 2006 13:48:18 -0500
Subject: Domino questions
Message-ID: <001501c708e6$9e07a570$6501a8c0@yourxhtr8hvc4p>

1. On an n x n (m x n) checkerboard, what is the smallest number of dominoes that can be placed so that no further dominoes can be placed?

2. Suppose two players, A and B, alternately place dominoes on an n x n (m x n) checkerboard until no further play is possible. A plays to maximize the final number of dominoes on the board, B to minimize. What is the number of dominoes on the board at the end of the game if A moves first? If B moves first?
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From franktaw at netscape.net  Wed Nov 15 21:01:19 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 15 Nov 2006 15:01:19 -0500
Subject: Domino questions
In-Reply-To: <001501c708e6$9e07a570$6501a8c0@yourxhtr8hvc4p>
References: <001501c708e6$9e07a570$6501a8c0@yourxhtr8hvc4p>
Message-ID: <8C8D707F39EB04F-934-6F47@FWM-R01.sysops.aol.com>

From: davidwwilson at comcast.net
>1. On an n x n (m x n) checkerboard, what is the smallest number
>of dominoes that can be placed so that no further dominoes can
>be placed?

This might be ceiling(n^2/3), A008810.  Can someone please
check the reference there: "J. H. Conway and R. K. Guy, The Book
of Numbers, Copernicus Press, NY, 1996, number of red blocks in
Fig 2.5."
?
>2. Suppose two players, A and B, alternately place dominoes on an
>n x n (m x n) checkerboard until no further play is possible. A plays
>to maximize the final number of dominoes on the board, B to
>minimize. What is the number of dominoes on the board at the end
>of the game if A moves first? If B moves first?

You don't want much, do you?

(Note for anyone who may be confused - that comment means that
I think this is likely to be a hard problem.)

Franklin T. Adams-Watters
________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From njas at research.att.com  Wed Nov 15 23:40:20 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 15 Nov 2006 17:40:20 -0500 (EST)
Subject: request for a b-file for a core file
Message-ID: <200611152240.RAA60703@fry.research.att.com>


Could someone help produce a b-file for A000014?
(and A001678 along the way)

Here is the sequence:

%I A000014 M0320 N0118
%S A000014 0,1,1,0,1,1,2,2,4,5,10,14,26,42,78,132,249,445,842,1561,2988,5671,10981,
%T A000014 21209,41472,81181,160176,316749,629933,1256070,2515169,5049816,10172638,
%U A000014 20543579,41602425,84440886,171794492,350238175,715497037,1464407113
%N A000014 Number of series-reduced trees with n nodes.


I could use 200, or 500 or so terms

Notes on computing A14 and A1678:

looking at the H'book of Graph Theory, p. 525:

they define f(x) (essentially A1678 but with a different offset):
f(x) = 1*x+1*x^3+1*x^4+2*x^5+3*x^6+6*x^7+10*x^8+19*x^9+35*x^10+...

by

f = (x/(x+1))* exp( sum( f(x^i) / i, i=1..infinity ) )

then A14 is given by

(1+x)f(x) - ((1+x)/2) f(x)^2 + ((1-x)/2) f(x^2)

= x + x^2 + x^4 + x^5 + 2 x^6 + ...

= A14

I actually haven't checked these formulae - there are
slightly different versions in the entry for A14 itself.

Thanks!

Neil






From bowerc at usa.net  Thu Nov 16 01:35:58 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 15 Nov 2006 16:35:58 -0800
Subject: request for a b-file for a core file
Message-ID: <964kkPaI79512S23.1163637358@cmsweb23.cms.usa.net>

I just sent a b-file of terms 0-500 to Neil for A001678.
I'm still working on a version for A000014

------ Original Message ------
From: "N. J. A. Sloane" <njas at research.att.com>
To: seqfans at seqfan.net, seqfan at ext.jussieu.frCc: njas at research.att.com
Subject: request for a b-file for a core file

> 
> Could someone help produce a b-file for A000014?
> (and A001678 along the way)
> 
> Here is the sequence:
> 
> %I A000014 M0320 N0118
> %S A000014
0,1,1,0,1,1,2,2,4,5,10,14,26,42,78,132,249,445,842,1561,2988,5671,10981,
> %T A000014
21209,41472,81181,160176,316749,629933,1256070,2515169,5049816,10172638,
> %U A000014
20543579,41602425,84440886,171794492,350238175,715497037,1464407113
> %N A000014 Number of series-reduced trees with n nodes.
> 
> 
> I could use 200, or 500 or so terms
> 
> Notes on computing A14 and A1678:
> 
> looking at the H'book of Graph Theory, p. 525:
> 
> they define f(x) (essentially A1678 but with a different offset):
> f(x) = 1*x+1*x^3+1*x^4+2*x^5+3*x^6+6*x^7+10*x^8+19*x^9+35*x^10+...
> 
> by
> 
> f = (x/(x+1))* exp( sum( f(x^i) / i, i=1..infinity ) )
> 
> then A14 is given by
> 
> (1+x)f(x) - ((1+x)/2) f(x)^2 + ((1-x)/2) f(x^2)
> 
> = x + x^2 + x^4 + x^5 + 2 x^6 + ...
> 
> = A14
> 
> I actually haven't checked these formulae - there are
> slightly different versions in the entry for A14 itself.
> 
> Thanks!
> 
> Neil
> 
> 









From rkg at cpsc.ucalgary.ca  Thu Nov 16 01:50:35 2006
From: rkg at cpsc.ucalgary.ca (Richard Guy)
Date: Wed, 15 Nov 2006 17:50:35 -0700 (MST)
Subject: Domino questions
In-Reply-To: <8C8D707F39EB04F-934-6F47@FWM-R01.sysops.aol.com>
References: <001501c708e6$9e07a570$6501a8c0@yourxhtr8hvc4p>
 <8C8D707F39EB04F-934-6F47@FWM-R01.sysops.aol.com>
Message-ID: <Pine.LNX.4.63.0611151749420.25970@csl>

1.  Checked!  Unfortunately, in the printing,
red and orange are barely distinguishable.   R.

On Wed, 15 Nov 2006, franktaw at netscape.net wrote:

> From: davidwwilson at comcast.net
>> 1. On an n x n (m x n) checkerboard, what is the smallest number
>> of dominoes that can be placed so that no further dominoes can
>> be placed?
>
> This might be ceiling(n^2/3), A008810.  Can someone please
> check the reference there: "J. H. Conway and R. K. Guy, The Book
> of Numbers, Copernicus Press, NY, 1996, number of red blocks in
> Fig 2.5."
> ?
>> 2. Suppose two players, A and B, alternately place dominoes on an
>> n x n (m x n) checkerboard until no further play is possible. A plays
>> to maximize the final number of dominoes on the board, B to
>> minimize. What is the number of dominoes on the board at the end
>> of the game if A moves first? If B moves first?
>
> You don't want much, do you?
>
> (Note for anyone who may be confused - that comment means that
> I think this is likely to be a hard problem.)
>
> Franklin T. Adams-Watters
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading 
> spam and email virus protection.
>

From bowerc at usa.net  Thu Nov 16 02:01:10 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 15 Nov 2006 17:01:10 -0800
Subject: request for a b-file for a core file
Message-ID: <888kkPBak9480S24.1163638870@cmsweb24.cms.usa.net>

I just sent the b-file for A000014

Christian

------ Original Message ------
From: "Christian G. Bower" <bowerc at usa.net>
To: "seqfan" <seqfan at ext.jussieu.fr>
Subject: Re: request for a b-file for a core file

> I just sent a b-file of terms 0-500 to Neil for A001678.
> I'm still working on a version for A000014
> 








From zbi74583 at boat.zero.ad.jp  Thu Nov 16 04:45:46 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Thu, 16 Nov 2006 12:45:46 +0900
Subject: A036471
Message-ID: <011301c70931$b92d7500$ca73d7dc@FMC57937B6BE00>

    Thanks, Giovanni Resta.



    The smallest term!!

    It is great!

       .

    Now human being became to know much about A036471.       

    The first term is 3270960 and known last term is       

   

    



    

    

    n=32583657,    19.6million digits.



    Yasutoshi

    


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From maxale at gmail.com  Thu Nov 16 10:48:46 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 16 Nov 2006 01:48:46 -0800
Subject: A question on A068480
In-Reply-To: <455AECB9.5050308@iit.cnr.it>
References: <455AECB9.5050308@iit.cnr.it>
Message-ID: <d3dac270611160148q5f52517gbc511cc7b9fd042c@mail.gmail.com>

It is unlikely for n!-1 and 2^n+1 to have a common prime factor
(unless in a special case described below), especially since
1) all prime factors of n!-1 are greater than n;
2) all prime factors of 2^n+1 are of the form p=2kq+1 where q is a
divisor of n, and the multiplicative of 2 modulo p divides 2q.

The special case is when n is even and p=2n+1 is prime. In this case,
p divides both n!-1 and 2^n+1, and thus gcd(n!-1,2^n+1). And it is
very likely that gcd(n!-1,2^n+1) = p.

Overall, the statement "if gcd(n!-1,2^n+1)>1 then
gcd(n!-1,2^n+1)=2n+1" sounds plausible and I verified it for n up to
45,000. But it may be very hard to give a proof.

Max

On 11/15/06, Giovanni Resta <g.resta at iit.cnr.it> wrote:
> A068480 is the sequence on numbers n such that gcd(n!-1,2^n+1)>1.
>
> I noticed that for all the numbers listed gcd(n!-1,2^n+1) = 2n+1.
>
> Is this a rule (proof ?) or a coincidence (counterexample ?) .
>
> thanks,
> giovanni.
>
>
>





From maxale at gmail.com  Thu Nov 16 11:49:25 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 16 Nov 2006 02:49:25 -0800
Subject: A question on A068480
In-Reply-To: <d3dac270611160148q5f52517gbc511cc7b9fd042c@mail.gmail.com>
References: <455AECB9.5050308@iit.cnr.it>
	 <d3dac270611160148q5f52517gbc511cc7b9fd042c@mail.gmail.com>
Message-ID: <d3dac270611160249k774410b5p3b07705ed642b8ae@mail.gmail.com>

On 11/16/06, Max A. <maxale at gmail.com> wrote:

> The special case is when n is even and p=2n+1 is prime. In this case,
> p divides both n!-1 and 2^n+1, and thus gcd(n!-1,2^n+1). And it is
> very likely that gcd(n!-1,2^n+1) = p.

Oh, my bad! Correct statement:

The special case is when n is of the form 4k+1 and p=2n+1 is prime
(implying that 2 and -1 are non-squares modulo p), and p also belongs
to A058302.
In this case, p divides both n!-1 and 2^n+1, and thus gcd(n!-1,2^n+1).
And it is very likely that gcd(n!-1,2^n+1) = p.

The hypothesis about gcd(n!-1,2^n+1) implies that:
8k+3 belong to A058302 if and only if 4k+1 belong to A068480.

Max





From davidwwilson at comcast.net  Thu Nov 16 17:47:04 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 16 Nov 2006 11:47:04 -0500
Subject: b-files for power series expansions
Message-ID: <000601c7099e$d8c5e880$6501a8c0@yourxhtr8hvc4p>

There is a seqload of OEIS sequences that are expansions of real functions, which I guess means power series coefficients. An example is A000810.  Would Mma or some other math power tool be able to quickly generate b-files for these monsters?
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From mathar at strw.leidenuniv.nl  Thu Nov 16 18:44:15 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Thu, 16 Nov 2006 18:44:15 +0100
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
Message-ID: <200611161744.kAGHiFGX005987@amer.strw.leidenuniv.nl>


pdh> From seqfan-owner at ext.jussieu.fr  Wed Nov 15 05:08:51 2006
pdh> Return-Path: <seqfan-owner at ext.jussieu.fr>
pdh> To: seqfan at ext.jussieu.fr
pdh> Date: Tue, 14 Nov 2006 23:04:51 -0500
pdh> Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
pdh> From: "Paul D. Hanna" <pauldhanna at juno.com>
pdh> ...
pdh> Number of labelled nodes in generation n of a rooted tree 
pdh> where a node with label k has k child nodes such that 
pdh> the label of each child node is the least unused label in 
pdh> the path from the root to that child node, 
pdh> and where root is labeled '1'. 
pdh>   ...
pdh> Here are the initial nodes of the tree for generations 0..4: 
pdh> Gen 0: [1];
pdh> Gen 1: (1)->[2];
pdh> Gen 2: (2)->[3,4];
pdh> Gen 3: (3)->[4,5,6], (4)->[3,5,6,7];
pdh> Gen 4: (4)->[5,6,7,8], (5)->[4,6,7,8,9], (6)->[4,5,7,8,9,10], 
pdh> (3)->[5,6,7], (5)->[3,6,7,8,9], (6)->[3,5,7,8,9,10], 
pdh> (7)->[3,5,6,8,9,10,11]; ...
pdh>  
pdh> Thus, the number of nodes in generation n begins:
pdh> [1, 1, 2, 7, 36, 248, ...]
pdh>  
pdh> The maximum node in generation n begins:
pdh> [1, 2, 4, 7, 11, ...]
pdh>  
pdh> Would appreciate it if someone could compute more terms.
pdh> Thanks,
pdh>       Paul

rjm> My count of node numbers and maximum nodes is
rjm> Gen 0 1 1
rjm> Gen 1 1 2
rjm> Gen 2 2 4
rjm> Gen 3 7 7
rjm> Gen 4 36 11
rjm> Gen 5 248 16
rjm> Gen 6 2157 22
rjm> Gen 7 22761 29
rjm> Gen 8 283220 37
rjm> Gen 9 4068411 46

The maximum node label is trivially A000124 because the maximum node will be
generated from the child with itself the highest label, and this adds 'n' to
the maximum label for each new generation. This produces the arithmetic sum
sequence that is shown above as the last number in each row.

For those who have some spare computer cycles and Maple: below is a program
that runs recursively ("memory friendly") through the entire tree up to 
some maximum generation 'maxgen' (to be set/edited in the fourth but last line
below, maxgen=10 equivalent to Gen up to 9 above). It prints a film of
progresses made to count the nodes found so far at all levels, and should show
all numbers in the list monotonically increasing slowly towards
[1, 1, 2, 7, 36, 248, 2157, 22761, 283220, 4068411, 66367834, ...]
                                                    ^^^ 66367834 is Gen 10
I can also provide my older (but non-recursive, memory-limited) C++ version.

Maple V:

gen := proc(parents,maxgen,ocounts,lvl)
	local thislbl,lbl,childlbl,counts,npar ;
	counts := ocounts ;
	counts[lvl] := counts[lvl]+1 ;
	if nops(parents) < maxgen then
		thislbl := op(-1,parents) ;
		childlbl := 1 ;
		for lbl from 1 to thislbl do
			while ( childlbl in parents ) or ( childlbl = thislbl ) do
				childlbl := childlbl+1 ;
			od ;
			npar := [op(parents),childlbl] ;
			# this would generate a full list of all paths:
			# print("route",npar) ;
			if nops(counts) < lvl+1 then
				counts := [op(counts),0] ;
			fi ;
			counts := gen(npar,maxgen,counts,lvl+1) ;
			childlbl := childlbl+1 ;
		od ;
	fi ;
	# this produces a snapshot of the counts accumulated so far:
	# if the current node creation level is not too high, print it...
	if lvl <= maxgen -3 then
		print(counts) ;
	fi ;
	RETURN(counts) ;
end:

# Edit the maximum number below...total run time probably grows exponentially
# with the number chosen here.
maxgen := 8 ;
parents := [1,2] ;
n := [1,0] ;
gen(parents,maxgen,n,2) ;
print(%) ;





From franktaw at netscape.net  Thu Nov 16 18:59:28 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 16 Nov 2006 12:59:28 -0500
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
In-Reply-To: <200611161744.kAGHiFGX005987@amer.strw.leidenuniv.nl>
References: <200611161744.kAGHiFGX005987@amer.strw.leidenuniv.nl>
Message-ID: <8C8D7C018493BAF-87C-1422@FWM-M21.sysops.aol.com>

I don't know Maple well enough - or have enough time - to tell: does 
this keep each path separately, or does it keep a count of the number 
of paths with a given set of integers on the path?

The latter approach should be much faster and less memory intensive, 
once you get past the first few generations.  (Probably still 
exponential, though.)

Franklin T. Adams-Watters

-----Original Message-----
From: mathar at strw.leidenuniv.nl

...

Maple V:

gen := proc(parents,maxgen,ocounts,lvl)
    local thislbl,lbl,childlbl,counts,npar ;
    counts := ocounts ;
    counts[lvl] := counts[lvl]+1 ;
    if nops(parents) < maxgen then
        thislbl := op(-1,parents) ;
        childlbl := 1 ;
        for lbl from 1 to thislbl do
            while ( childlbl in parents ) or ( childlbl = thislbl ) do
                childlbl := childlbl+1 ;
            od ;
            npar := [op(parents),childlbl] ;
            # this would generate a full list of all paths:
            # print("route",npar) ;
            if nops(counts) < lvl+1 then
                counts := [op(counts),0] ;
            fi ;
            counts := gen(npar,maxgen,counts,lvl+1) ;
            childlbl := childlbl+1 ;
        od ;
    fi ;
    # this produces a snapshot of the counts accumulated so far:
    # if the current node creation level is not too high, print it...
    if lvl <= maxgen -3 then
        print(counts) ;
    fi ;
    RETURN(counts) ;
end:

# Edit the maximum number below...total run time probably grows 
exponentially
# with the number chosen here.
maxgen := 8 ;
parents := [1,2] ;
n := [1,0] ;
gen(parents,maxgen,n,2) ;
print(%) ;

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From zakseidov at yahoo.com  Thu Nov 16 19:33:20 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Thu, 16 Nov 2006 10:33:20 -0800 (PST)
Subject: b-files for power series expansions
In-Reply-To: <000601c7099e$d8c5e880$6501a8c0@yourxhtr8hvc4p>
Message-ID: <20061116183320.50052.qmail@web38201.mail.mud.yahoo.com>

Coefficients of the series
1, 1, 4, 13/3, 44/3, 242/15, 2408/45, 18581/315
differ from A000810 entries (?!),
Zak

A000810
1,1,8,26,352,1936,38528,297296,7869952,78098176
%N A000810 Expansion of (sin x + cos x)/cos 3x.


--- David Wilson <davidwwilson at comcast.net> wrote:

> There is a seqload of OEIS sequences that are
> expansions of real functions, which I guess means
> power series coefficients. An example is A000810. 
> Would Mma or some other math power tool be able to
> quickly generate b-files for these monsters?



 
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From franktaw at netscape.net  Thu Nov 16 20:25:13 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 16 Nov 2006 14:25:13 -0500
Subject: b-files for power series expansions
In-Reply-To: <20061116183320.50052.qmail@web38201.mail.mud.yahoo.com>
Message-ID: <8C8D7CC12CFC338-87C-19DE@FWM-M21.sysops.aol.com>

A000810 is the exponential generating function:

1/0!, 1/1!, 8/2!, 26/3!, 352/4!, ...

Franklin T. Adams-Watters


-----Original Message-----
From: zakseidov at yahoo.com

Coefficients of the series
1, 1, 4, 13/3, 44/3, 242/15, 2408/45, 18581/315
differ from A000810 entries (?!),
Zak

A000810
1,1,8,26,352,1936,38528,297296,7869952,78098176
%N A000810 Expansion of (sin x + cos x)/cos 3x.


--- David Wilson <davidwwilson at comcast.net> wrote:

> There is a seqload of OEIS sequences that are
> expansions of real functions, which I guess means
> power series coefficients. An example is A000810.
> Would Mma or some other math power tool be able to
> quickly generate b-files for these monsters?




_________________________________________________________________________
___________
Sponsored Link

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$420k for $1,399/mo. Calculate new payment!
www.LowerMyBills.com/lre

________________________________________________________________________
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From mathar at strw.leidenuniv.nl  Thu Nov 16 20:31:50 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Thu, 16 Nov 2006 20:31:50 +0100
Subject: b-files for power series expansions
Message-ID: <200611161931.kAGJVoHb015119@amer.strw.leidenuniv.nl>


dww> From seqfan-owner at ext.jussieu.fr  Thu Nov 16 17:49:20 2006
dww> From: "David Wilson" <davidwwilson at comcast.net>
dww> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
dww> Subject: b-files for power series expansions
dww> Date: Thu, 16 Nov 2006 11:47:04 -0500
dww> ...
dww> There is a seqload of OEIS sequences that are expansions of real =
dww> functions, which I guess means power series coefficients. An example is =
dww> A000810.  Would Mma or some other math power tool be able to quickly =
dww> generate b-files for these monsters?
dww>...

These are usually taylor series coefficients multiplied by n!
(the E.g.f. given in the definition). Where the functions are even,
the odd terms (coefficients equal to 0) are omitted.

I put some of them into
http://www.strw.leidenuniv.nl/~mathar/progs/b000810.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000813.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000816.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000819.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000822.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000825.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000828.txt

Unrelated are
http://www.strw.leidenuniv.nl/~mathar/progs/b007504.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b049060.txt





From mathar at strw.leidenuniv.nl  Thu Nov 16 22:05:41 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Thu, 16 Nov 2006 22:05:41 +0100
Subject: Re^2: Trees with Labeled Nodes that form Permutation of Positive Integers
Message-ID: <200611162105.kAGL5fkI022493@amer.strw.leidenuniv.nl>


ftaw> From seqfan-owner at ext.jussieu.fr  Thu Nov 16 19:01:33 2006
ftaw> Return-Path: <seqfan-owner at ext.jussieu.fr>
ftaw> To: seqfan at ext.jussieu.fr
ftaw> Subject: Re: Trees with Labeled Nodes that form Permutation of Positive Integers
ftaw> 
ftaw> I don't know Maple well enough - or have enough time - to tell: does 
ftaw> this keep each path separately, or does it keep a count of the number 
ftaw> of paths with a given set of integers on the path?
ftaw> 
ftaw> The latter approach should be much faster and less memory intensive, 
ftaw> once you get past the first few generations.  (Probably still 
ftaw> exponential, though.)
ftaw> 
ftaw> Franklin T. Adams-Watters
ftaw> 

There is no such optimization in the program shown below. For each
node, it descends into all branches/leafs independent of whether the
set of labels in the current node and higher up in the trunk is the same
or not the same as any set seen earlier. The state vector is
only the vector of counts (one number per generation) which is updated
at each level of recursion, and the single list of labels of the
ancestors of the current node.
So yes, there is room for optimization by introducing some memory
in here, certainly at the cost of increasing the machine memory requirement
to remember all the sub-counts of partial tree in some hash area.

The first action to become faster is to return to a compiled language.
Since this here does not involve any type of functionality which needs a number
theoretic package, one would try to exploit the fact that C would typically be a
factor of 5 to 10 faster than Maple. I don't want to appear selfish and
will therefore leave all this as an exercise to others :-)

RJM

rjm> -----Original Message-----
rjm> From: mathar at strw.leidenuniv.nl
rjm> 
rjm> ...
rjm> 
rjm> Maple V:
rjm> 
rjm> gen := proc(parents,maxgen,ocounts,lvl)
rjm>     local thislbl,lbl,childlbl,counts,npar ;
rjm>     counts := ocounts ;
rjm>     counts[lvl] := counts[lvl]+1 ;
rjm>     if nops(parents) < maxgen then
rjm>         thislbl := op(-1,parents) ;
rjm>         childlbl := 1 ;
rjm>         for lbl from 1 to thislbl do
rjm>             while ( childlbl in parents ) or ( childlbl = thislbl ) do
rjm>                 childlbl := childlbl+1 ;
rjm>             od ;
rjm>             npar := [op(parents),childlbl] ;
rjm>             # this would generate a full list of all paths:
rjm>             # print("route",npar) ;
rjm>             if nops(counts) < lvl+1 then
rjm>                 counts := [op(counts),0] ;
rjm>             fi ;
rjm>             counts := gen(npar,maxgen,counts,lvl+1) ;
rjm>             childlbl := childlbl+1 ;
rjm>         od ;
rjm>     fi ;
rjm>     # this produces a snapshot of the counts accumulated so far:
rjm>     # if the current node creation level is not too high, print it...
rjm>     if lvl <= maxgen -3 then
rjm>         print(counts) ;
rjm>     fi ;
rjm>     RETURN(counts) ;
rjm> end:
rjm> 
rjm> # Edit the maximum number below...total run time probably grows 
rjm> exponentially
rjm> # with the number chosen here.
rjm> maxgen := 8 ;
rjm> parents := [1,2] ;
rjm> n := [1,0] ;
rjm> gen(parents,maxgen,n,2) ;
rjm> print(%) ;





From franktaw at netscape.net  Fri Nov 17 00:46:06 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 16 Nov 2006 18:46:06 -0500
Subject: Re^2: Trees with Labeled Nodes that form Permutation of Positive
 Integers
In-Reply-To: <200611162105.kAGL5fkI022493@amer.strw.leidenuniv.nl>
References: <200611162105.kAGL5fkI022493@amer.strw.leidenuniv.nl>
Message-ID: <8C8D7F084CDAF74-554-265F@FWM-M18.sysops.aol.com>

No hashing is necessary.  There are (n-1)! possible sets of labels.  
Each
sorted sequence of labels has an increase of at most k between the kth
and k+1st label.  So you can take the first differences, subtract 1,
reverse, and treat as a base factorial number.  E.g., for 1,2,4,5,9, 
the
differences are 1,2,1,4; subtract one and reverse to get 3010,
3*3!+0*2!+1*1! = 19, so this will be at (zero-based) index 19.  (This 
is
easily reversible to get the label set from the index and level.)

For each label set, you'll need to keep a count of how many nodes
there are with each possible final label, to tell what to generate in
the next level.  Since 1 and 2 can't be labels after level 2, you only
need to keep n-2 such counters for each set.

We can tell that this algorithm is on the order of n! to compute level 
n.
The naive approach is proportional to the size of the nth level, which 
is
certainly greater; since the largest term at level n is C(n,2)+1 ~ 
n^2/2,
it is certainly no more than (n^2/2)!.  I'm guessing that is it more 
like
c^n n!, for some c between 1.6 and 2 (probably closer to 2, from the
data, though convergence is slow; k 2^n n!/n^2 gives a fairly good
fit to the available data).

Based on this, the eventual savings should be much larger than merely
switching to a compiled language.  I'll also leave this as an exercise,
though. :-)

Franklin T. Adams-Watters


-----Original Message-----
From: mathar at strw.leidenuniv.nl

ftaw> From seqfan-owner at ext.jussieu.fr  Thu Nov 16 19:01:33 2006
ftaw>
ftaw> I don't know Maple well enough - or have enough time - to tell: 
does
ftaw> this keep each path separately, or does it keep a count of the 
number
ftaw> of paths with a given set of integers on the path?
ftaw>
ftaw> The latter approach should be much faster and less memory 
intensive,
ftaw> once you get past the first few generations.  (Probably still
ftaw> exponential, though.)
ftaw>
ftaw> Franklin T. Adams-Watters
ftaw>

There is no such optimization in the program shown below. For each
node, it descends into all branches/leafs independent of whether the
set of labels in the current node and higher up in the trunk is the same
or not the same as any set seen earlier. The state vector is
only the vector of counts (one number per generation) which is updated
at each level of recursion, and the single list of labels of the
ancestors of the current node.
So yes, there is room for optimization by introducing some memory
in here, certainly at the cost of increasing the machine memory 
requirement
to remember all the sub-counts of partial tree in some hash area.

The first action to become faster is to return to a compiled language.
Since this here does not involve any type of functionality which needs 
a number
theoretic package, one would try to exploit the fact that C would 
typically be a
factor of 5 to 10 faster than Maple. I don't want to appear selfish and
will therefore leave all this as an exercise to others :-)

RJM

________________________________________________________________________
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From pauldhanna at juno.com  Fri Nov 17 07:13:46 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Fri, 17 Nov 2006 01:13:46 -0500
Subject: Sub-Fibonacci Tree
Message-ID: <20061117.011346.692.1.pauldhanna@juno.com>

Seqfans, 
     Consider the following "sub-Fibonacci tree" and in particular 
its associated sequence defined by: 
 
Sum of labels of nodes in generation n of a rooted tree in which a node 
with label k has child nodes assigned labels of successive integers 
beginning with k+1 such that the number of child nodes for each node 
is given by the label of the parent node, starting at generation n=0 
with a root node with label '1' followed by a child node with label '2'. 
 
(Perhaps someone could make this description more concise?!)
 
EXAMPLE.
The initial nodes of the tree for generations 0..5 are: 
gen 0: [1];
gen 1: [2];
gen 2: [3];
gen 3: [4,5];
gen 4: (4)->[5,6,7],(5)->[6,7,8];
gen 5: (5)->[6,7,8,9],(6)->[7,8,9,10],(7)->[8,9,10,11], 
(6)->[7,8,9,10,11],(7)->[8,9,10,11,12],(8)->[9,10,11,12,13];
  
By definition, there are 2 child nodes for node [3] since 
the parent of node [3] is [2]; likewise, 
there are 3 child nodes for nodes [4] and [5] (in gen.3) 
since the parent of both nodes has label [3] (in gen.2). 
 
Notice the minimum label in generation n is n+1, and 
the maximum label in generation n is Fibonacci(n+2).  
  
 From this tree I am interested in gleaning 2 sequences. 
  
SEQUENCE 1. 
I compute the number of nodes in generation n to begin: 
[1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368, ...]
 
which is equal (with offset) to A005270:
"Sub-Fibonacci sequences of length n."
1, 1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368 
 
SEQUENCE 2. 
But the following related sequence is NOT found in the OEIS: 
 
Sum of labels of nodes in generation n:
[1, 2, 3, 9, 39, 252, 2361, 32077, 631058, ...]
 
EXTENSION?
Could someone compute more terms of sequences 1 and 2? 
 
Thanks,
    Paul





From pauldhanna at juno.com  Fri Nov 17 08:55:47 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Fri, 17 Nov 2006 02:55:47 -0500
Subject: Sub-Fibonacci Tree
Message-ID: <20061117.025548.692.2.pauldhanna@juno.com>

Seqfans,
    Here I extend A005270 by one more term and also extend the sequence 
a(n) = sum of labels of nodes in generation n in the sub-Fibonacci tree.
I demonstrate a slightly faster method to compute these sequences. 
 
First, I am changing the description to be (hopefully) more clear: 
 
"Sum of labels of nodes in generation n of the sub-Fibonacci tree in
which 
a node with label k and parent node with label g, has g child nodes that 
are given labels beginning with k+1 through k+g; the tree starts at 
generation n=0 with a root node labeled '1' and a child node labeled
'2'."
 
Would appreciate any suggestions to change in wording.
  
 
EXTENDING A005270.
There is a faster way to compute b(n)=A005270(n+1): 
b(n) = "number of nodes in generation n of the sub-Fibonacci tree"
 
by employing the following statistic:
 
b(n+1) = sum of (parent label)*(label) over all nodes in generation n. 
 
For example, 
b(4) = 27 = 3*(4+5);
b(5) = 177 = 4*(5+6+7) + 5*(6+7+8);
b(6) = 1680 = 5*(6+7+8+9) + 6*(7+8+9+10) + 7*(8+9+10+11)+ 
 6*(7+8+9+10+11) + 7*(8+9+10+11+12) + 8*(9+10+11+12+13); ...

Using this method, I extend sequence A005270 by one more term:
1, 1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368, 13067353,
 

SUM OF LABELS SEQUENCE. 
In like manner, the computation of:
a(n) = "sum of labels of nodes in generation n of the sub-Fibonacci tree"

can be accomplished by the statistic:
 
a(n+1) = sum of (parent label)*(label) over all nodes in generation n 
+ sum of (parent label)*[label*(label+1)/2] over all nodes in generation
n-1. 
 
For example, 
a(2) = 3 = 1*2  + 1*( 1*2/2 ) ;
a(3) = 9 = 2*3  + 1*( 2*3/2 ) ;
a(4) = 39 = 3*(4+5)  + 2*( 3*4/2 ) ;
a(5) = 252 = 4*(5+6+7) + 5*(6+7+8)  + 3*( 4*5/2 + 5*6/2 ) ;
a(6) = 2361 = 5*(6+7+8+9) + 6*(7+8+9+10) + 7*(8+9+10+11) + 
 6*(7+8+9+10+11) + 7*(8+9+10+11+12) + 8*(9+10+11+12+13)  +
 4*( 5*6/2 + 6*7/2 + 7*8/2 ) + 5*( 6*7/2 + 7*8/2 + 8*9/2) ;
...
 
With this method, I extend the sum of labels in generation n to: 
1, 2, 3, 9, 39, 252, 2361, 32077, 631058, 18035534, 
 

Would someone please confirm my calculation of the terms: 
b(10) = A005270(11) = 13067353  and 
a(10) = 18035534
 
and maybe compute a few more terms? 
Thanks,
    Paul
----------------------------------------------------
> EXAMPLE.
> The initial nodes of the tree for generations 0..5 are: 
> gen 0: [1];
> gen 1: [2];
> gen 2: [3];
> gen 3: [4,5];
> gen 4: (4)->[5,6,7],(5)->[6,7,8];
> gen 5: (5)->[6,7,8,9],(6)->[7,8,9,10],(7)->[8,9,10,11], 
> (6)->[7,8,9,10,11],(7)->[8,9,10,11,12],(8)->[9,10,11,12,13];
>   
> By definition, there are 2 child nodes for node [3] since 
> the parent of node [3] is [2]; likewise, 
> there are 3 child nodes for nodes [4] and [5] (in gen.3) 
> since the parent of both nodes has label [3] (in gen.2). 
>  
> Notice the minimum label in generation n is n+1, and 
> the maximum label in generation n is Fibonacci(n+2).  
>   
> From this tree I am interested in gleaning 2 sequences. 
>   
> SEQUENCE 1. 
> I compute the number of nodes in generation n to begin: 
> [1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368, ...]
>  
> which is equal (with offset) to A005270:
> "Sub-Fibonacci sequences of length n."
> 1, 1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368 
>  
> SEQUENCE 2. 
> But the following related sequence is NOT found in the OEIS: 
>  
> Sum of labels of nodes in generation n:
> [1, 2, 3, 9, 39, 252, 2361, 32077, 631058, ...]
>  
> EXTENSION?
> Could someone compute more terms of sequences 1 and 2? 
>  
> Thanks,
>     Paul





From zbi74583 at boat.zero.ad.jp  Sat Nov 18 02:27:22 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Sat, 18 Nov 2006 10:27:22 +0900
Subject: A085058
Message-ID: <019201c70ab0$b44b3ce0$ca73d7dc@FMC57937B6BE00>

    Hi, Seqfans

    If 2^k(n) is the largest power of two dividing tangent number T(n) then 
numbers such that 2*n-k(n) are as follows.



    S : 2,3,2,4,2,3,2,5,2,....



    It looks like A085058.

    Are they the same seqences ?



    Yasutoshi


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From alec at mihailovs.com  Sun Nov 19 11:38:43 2006
From: alec at mihailovs.com (Alec Mihailovs)
Date: Sun, 19 Nov 2006 04:38:43 -0600
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
References: <200611161744.kAGHiFGX005987@amer.strw.leidenuniv.nl>
Message-ID: <000c01c70bc6$e36ead60$03fea8c0@Media>

From: "Richard Mathar" <mathar at strw.leidenuniv.nl>
Sent: Thursday, November 16, 2006 11:44 AM
>
> pdh> Date: Tue, 14 Nov 2006 23:04:51 -0500
> pdh> From: "Paul D. Hanna" <pauldhanna at juno.com>
> pdh> ...
> pdh> Number of labelled nodes in generation n of a rooted tree
> pdh> where a node with label k has k child nodes such that
> pdh> the label of each child node is the least unused label in
> pdh> the path from the root to that child node,
> pdh> and where root is labeled '1'.
> pdh>   ...
> pdh> Thus, the number of nodes in generation n begins:
> pdh> [1, 1, 2, 7, 36, 248, ...]
>
> [1, 1, 2, 7, 36, 248, 2157, 22761, 283220, 4068411, 66367834, ...]
>                                                    ^^^ 66367834 is Gen 10

Maple 10 can compile (some) procedures in C. I used the following 
back-tracking procedure,

f:=proc(n::integer[4],A::Array(datatype=integer[4]),
B::Array(datatype=integer[4]))::integer[4]; local c::integer[4],
i::integer[4],len::integer[4],m::integer[4]; c,len,m:=0,3,3;
while len>1 do if len=n then c:=c+1;m:=A[len];B[m]:=0;len:=len-1;
B[A[len]]:=B[A[len]]+1 elif B[A[len]]<=A[len] then for i from m+1
do if B[i]=0 then break fi od; len:=len+1;A[len]:=i;B[i]:=1;m:=2
else m:=A[len];B[m]:=0;len:=len-1;B[A[len]]:=B[A[len]]+1 fi od; c end:

cf:=Compiler:-Compile(f):

F:=proc(n::posint) local A,B;
if n<3 then 1 elif n=3 then 2 else
A:=Array([$1..3,0$(n-3)],datatype=integer[4]);
B:=Array([1$3,0$((n-2)*(n+1)/2)],datatype=integer[4]);
cf(n,A,B) fi end:

seq(F(n),n=1..12);

  1, 1, 2, 7, 36, 248, 2157, 22761, 283220, 4068411, 66367834,

        1213504295

That confirms Richard Mathar's and Paul D. Hanna's calculations and gives 
one more element of the sequence. Further elements could not be calculated 
on my 32-bit computer using this procedure, because they are greater than 
2^32. Changing the types of c and f from integer[4] to float[8] allows 
further calculations. I did that and after about an hour got the next 
element,

F(13);

                      24606397802


Alec Mihailovs
http://mihailovs.com/Alec/








From deutsch at duke.poly.edu  Mon Nov 20 17:33:59 2006
From: deutsch at duke.poly.edu (Emeric Deutsch)
Date: Mon, 20 Nov 2006 11:33:59 -0500 (EST)
Subject: help with a new sequence 
Message-ID: <Pine.GSO.4.62.0611201046020.12396@duke.poly.edu>

Seqfans,

I intend to submit to OEIS the following new sequence (triangle):

1,1,1,1,3,1,1,4,2,6,1,1,5,5,10,10,10,1,

Triangle read by rows: T(n,k) is the number of Dyck paths of
semilength n whose ascent lengths form the k-th partition of
the integer n; the partitions of n are ordered in the way
exemplified by [5], [4,1], [3,2], [3,1,1], [2,2,1], [2,1,1,1],
[1,1,1,1,1] (the "Mathematica" ordering).

Equivalently, T(n,k) is the number of ordered trees with n
edges whose node degrees form the k-th partition of the
integer n.

Example: T(5,3)=5 because the 3rd partition of 5 is [3,2] and
we have (UU)DD(UUU)DDD, (UUU)DDD(UU)DD, (UU)D(UUU)DDDD,
(UUU)D(UU)DDDD, and (UUU)DD(UU)DDD; here U=(1,1), D=(1,-1) and
the ascents are shown between parentheses.

Triangle starts:
1;
1,1;
1,3,1;
1,4,2,6,1;
1,5,5,10,10,10,1;

Row n has A000041(n) terms (=number of partitions of n).

------------

So far this is all I have; terms have been found by
straightforward counting. I am sure you can find more
terms, more facts, etc. I'd appreciate any collaboration.

Thanks, Emeric





From pauldhanna at juno.com  Mon Nov 20 19:39:48 2006
From: pauldhanna at juno.com (Paul D Hanna)
Date: Mon, 20 Nov 2006 18:39:48 GMT
Subject: help with a new sequence 
Message-ID: <20061120.104022.757.551642@webmail37.nyc.untd.com>

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From franktaw at netscape.net  Mon Nov 20 19:49:27 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 20 Nov 2006 13:49:27 -0500
Subject: help with a new sequence 
In-Reply-To: <20061120.104022.757.551642@webmail37.nyc.untd.com>
References: <20061120.104022.757.551642@webmail37.nyc.untd.com>
Message-ID: <8C8DAEBBD6AFC69-91C-B5F3@FWM-M20.sysops.aol.com>

This is A008284, a famous function.

Franklin T. Adams-Watters


-----Original Message-----
From: pauldhanna at juno.com

Seqfans,
??????Row n of a?related triangle?could?count?the number of terms in 
row n
of?Emerics triangle that?sum to?form the respective?terms of the 
Narayana triangle.
?
This triangle P?would begin:
1;
1, 1;
1, 1, 1;
1, 2, 1, 1;
1, 2, 2, 1, 1; ...

where row sums equal the partition numbers.
?
For example, row 3 of P?is [1, 2, 1, 1]
since Narayana row 3 is [1, 4+2, 6, 1]
formed from row 3 of?Emerics triangle: [1, 4, 2, 6, 1].
?
Is the resulting partition-related triangle P?already in the OEIS?

It?might have an interesting matrix inverse.
??? Paul
?
--?Emeric?Deutsch?<deutsch at duke.poly.edu>?wrote:
...
Triangle?starts:
1;
1,1;
1,3,1;
1,4,2,6,1;
1,5,5,10,10,10,1;

Row?n?has?A000041(n)?terms?(=number?of?partitions?of?n).
...
Thanks,?Emeric




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From franktaw at netscape.net  Mon Nov 20 20:55:23 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 20 Nov 2006 14:55:23 -0500
Subject: help with a new sequence 
In-Reply-To: <Pine.GSO.4.62.0611201046020.12396@duke.poly.edu>
References: <Pine.GSO.4.62.0611201046020.12396@duke.poly.edu>
Message-ID: <8C8DAF4F355FB9E-B7C-B33B@FWM-M19.sysops.aol.com>

How about if you look at the sequence (composition) of ascents, instead 
of the partition?

I think this starts (using the A066099 ordering):

1
1,1
1,2,1,1
1,3,2,3,1,2,1,1
1,4,3,6,2,5,3,1,1,3,2,2,1,3,4,1

Amazingly, this corresponds to Stern's diatomic array (version 3, 
A070879) for the first 18 terms.
Makes me wonder if A070879 can be interpreted as a function on 
compositions.

Franklin T. Adams-Watters


-----Original Message-----
From: deutsch at duke.poly.edu

Seqfans,?
?
I intend to submit to OEIS the following new sequence (triangle):?
?
1,1,1,1,3,1,1,4,2,6,1,1,5,5,10,10,10,1,?
?
Triangle read by rows: T(n,k) is the number of Dyck paths of?
semilength n whose ascent lengths form the k-th partition of?
the integer n; the partitions of n are ordered in the way?
exemplified by [5], [4,1], [3,2], [3,1,1], [2,2,1], [2,1,1,1],?
[1,1,1,1,1] (the "Mathematica" ordering).?
?
Equivalently, T(n,k) is the number of ordered trees with n?
edges whose node degrees form the k-th partition of the?
integer n.?
?
Example: T(5,3)=5 because the 3rd partition of 5 is [3,2] and?
we have (UU)DD(UUU)DDD, (UUU)DDD(UU)DD, (UU)D(UUU)DDDD,?
(UUU)D(UU)DDDD, and (UUU)DD(UU)DDD; here U=(1,1), D=(1,-1) and?
the ascents are shown between parentheses.?
?
Triangle starts:?
1;?
1,1;?
1,3,1;?
1,4,2,6,1;?
1,5,5,10,10,10,1;?
?
Row n has A000041(n) terms (=number of partitions of n).?
?
------------?
?
So far this is all I have; terms have been found by?
straightforward counting. I am sure you can find more?
terms, more facts, etc. I'd appreciate any collaboration.?
?
Thanks, Emeric?


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From franktaw at netscape.net  Mon Nov 20 22:54:07 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 20 Nov 2006 16:54:07 -0500
Subject: help with a new sequence 
In-Reply-To: <8C8DAF4F355FB9E-B7C-B33B@FWM-M19.sysops.aol.com>
References: <Pine.GSO.4.62.0611201046020.12396@duke.poly.edu> <8C8DAF4F355FB9E-B7C-B33B@FWM-M19.sysops.aol.com>
Message-ID: <8C8DB058980377A-B7C-BB3C@FWM-M19.sysops.aol.com>

Let f(a_0, a_1, ..., a_m) be the number of Dyck paths associated with 
the composition
[a_0, a_1, ..., a_m].   We have f(a_0) = 1; and for m > 0,

f(a_0, a_1, ..., a_m) = Sum_{i=0}^{a_0-1} f(a_1+i, a_2, ..., a_m).

Using this, I get the following table (including row 0, the empty 
composition):

1,
1,
1,1,
1,2,1,1,
1,3,2,3,1,2,1,1,
1,4,3,6,2,5,3,4,1,3,2,3,1,2,1,1,
1,5,4,10,3,9,6,10,2,7,5,9,3,7,4,5,1,4,3,6,2,5,3,4,1,3,2,3,1,2,1,1,
1,6,5,15,4,14,10,20,3,12,9,19,6,16,10,15,2,9,7,16,5,14,9,14,3,10,7,12,4,9
,5,6,1,5,4,
   10,3,9,6,10,2,7,5,9,3,7,4,5,1,4,3,6,2,5,3,4,1,3,2,3,1,2,1,1

And accumulating by partition, for Emeric's original table (but also 
including row 0):

1,
1,
1,1,
1,3,1,
1,4,2,6,1,
1,5,5,10,10,10,1,
1,6,6,15,3,30,20,5,30,15,1,
1,7,7,21,7,42,35,21,21,105,35,35,70,21,1,
1,8,8,28,8,56,56,4,56,28,168,70,28,84,168,280,56,14,140,140,28,1,
1,9,9,36,9,72,84,9,72,36,252,126,36,72,252,252,504,126,12,252,252,84,756,
630,84,
   126,420,252,36,1

Given a partition of n into k parts, [a_1^e_1,...,a_k^e_k], the number 
of
associated Dyck paths appears to be n!/(n-k+1)!/(Product (e_i)!).

Now, maybe somebody can prove that.

Franklin T. Adams-Watters


-----Original Message-----
From: franktaw at netscape.net

How about if you look at the sequence (composition) of ascents, instead 
of the partition??
?
I think this starts (using the A066099 ordering):?
?
1?
1,1?
1,2,1,1?
1,3,2,3,1,2,1,1?
1,4,3,6,2,5,3,1,1,3,2,2,1,3,4,1?
?
Amazingly, this corresponds to Stern's diatomic array (version 3, 
A070879) for the first 18 terms.?
Makes me wonder if A070879 can be interpreted as a function on 
compositions.?
?
Franklin T. Adams-Watters?
?
-----Original Message-----?
From: deutsch at duke.poly.edu?
?
Seqfans,??
??
I intend to submit to OEIS the following new sequence (triangle):??
??
1,1,1,1,3,1,1,4,2,6,1,1,5,5,10,10,10,1,??
??
Triangle read by rows: T(n,k) is the number of Dyck paths of??
semilength n whose ascent lengths form the k-th partition of??
the integer n; the partitions of n are ordered in the way??
exemplified by [5], [4,1], [3,2], [3,1,1], [2,2,1], [2,1,1,1],??
[1,1,1,1,1] (the "Mathematica" ordering).??
??
Equivalently, T(n,k) is the number of ordered trees with n??
edges whose node degrees form the k-th partition of the??
integer n.??
??
Example: T(5,3)=5 because the 3rd partition of 5 is [3,2] and??
we have (UU)DD(UUU)DDD, (UUU)DDD(UU)DD, (UU)D(UUU)DDDD,??
(UUU)D(UU)DDDD, and (UUU)DD(UU)DDD; here U=(1,1), D=(1,-1) and??
the ascents are shown between parentheses.??
??
Triangle starts:??
1;??
1,1;??
1,3,1;??
1,4,2,6,1;??
1,5,5,10,10,10,1;??
??
Row n has A000041(n) terms (=number of partitions of n).??
??
------------??
??
So far this is all I have; terms have been found by??
straightforward counting. I am sure you can find more??
terms, more facts, etc. I'd appreciate any collaboration.??
??
Thanks, Emeric??
?
________________________________________________________________________?

Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.?


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From njas at research.att.com  Mon Nov 20 23:25:04 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Mon, 20 Nov 2006 17:25:04 -0500 (EST)
Subject: two "find the next term" puzzles from Knuth Vol 4
Message-ID: <200611202225.RAA93655@fry.research.att.com>


the two puzzles (and their solutions) are

%I A123896
%S A123896 0,1,1,1,12,12,12,12,12,12,100,121,122,123,123
%N A123896 A123895(n^2).
...

%I A123902
%S A123902 0,1,12,100,112,121,122,123
%N A123902 A123896 sorted and uniqued.
...

which are based on this:

%I A123895
%S A123895 0,1,1,1,1,1,1,1,1,1,10,11,12,12,12,12,12,12,12,12,10,12,11,12,12,
%T A123895 12,12,12,12,12,10,12,12,11,12,12,12,12,12,12,10,12,12,12,11,12,12,
%U A123895 12,12,12,10,12,12,12,12,11,12,12,12,12,10,12,12,12,12,12,11,12,12
%N A123895 Restricted growth string for the (decimal expansion of the) number n.
%C A123895 Write n in base 10 prefixed with a 0. Read this string from left to r
ight. Write a 0 each time you see the first distinct digit (which is 0), write a
1 each time you see the second distinct digit, write a 2 each time you see the t
hird distinct digit, and so on. Finally, delete the leading zeros.
%D A123895 D. E. Knuth, TAOCP, Vol. 4, Section 7.2.1.5, Problems 4 and 5.
%e A123895 To find a(66041171): 066041171 -> 011023343 -> 11023343.
%Y A123895 Cf. A123896, A123902.
%K A123895 nonn,base,new
%O A123895 0,11
%A A123895 njas, Nov 20 2006

It would be nice if someone could extend the first two, and provide
maple or mma or other code for the third one

Neil






From franktaw at netscape.net  Mon Nov 20 23:59:41 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 20 Nov 2006 17:59:41 -0500
Subject: two "find the next term" puzzles from Knuth Vol 4
In-Reply-To: <200611202225.RAA93655@fry.research.att.com>
References: <200611202225.RAA93655@fry.research.att.com>
Message-ID: <8C8DB0EB222DED3-B7C-BF60@FWM-M19.sysops.aol.com>

A123896:

0 1 1 1 12 12 12 12 12 12 100 121 122 123 123 112 123 123 123 123 100 
112 121 123 123 123 121 123 123 123 100 123 1023 1023 1123 1223 1234 
1234 1222 1231 1200 1231 1234 1234 1234 1012 1223 1102 1203 1203 1200 
1203 1203 1203 1234 1023 1213 1234 1123 1234 1200 1234 1233 1232 1023 
1223 1234 1123 1231 1234 1200 1023 1234 1234 1234 1231 1223 1232 1023 
1234 1200 1213 1234 1223 1023 1223 1234 1234 1122 1234 1200

A123902:

0 1 12 100 112 121 122 123 1012 1023 1102 1122 1123 1200 1201 1203 1213 
1222 1223 1231 1232 1233 1234 10000 10012 10023 10102 10123 10201 10202 
10203 10213 10223 10231 10232 10234 11023 11102 11200 11203 11211 11213 
11221 11223 11232 11234 12003 12013 12023 12032 12033 12034 12100 12113 
12121 12131 12133 12134 12200 12203 12212 12213 12231 12232 12233 12234 
12300 12301 12304 12312 12313 12314 12321 12323 12324 12331 12332 12334 
12341 12342

Sorry, I didn't use Mathematica, Maple, or even PARI; I used Excel and 
VBA.

Here's a VBA program:

Public Function RestrictedGrowthString(ByVal x As String) As String
    Dim i As Long
    Dim dig As Integer
    Dim pos As Long

    For i = 1 To Len(x)
        If Mid(x, i, 1) = "0" Then
            RestrictedGrowthString = RestrictedGrowthString & "0"
        Else
            pos = InStr(x, Mid(x, i, 1))
            If pos = i Then
                dig = dig + 1
                 RestrictedGrowthString = RestrictedGrowthString & 
Format(dig)
            Else
                 RestrictedGrowthString = RestrictedGrowthString & 
Mid(RestrictedGrowthString, pos, 1)
            End If
        End If
    Next i
End Function

Franklin T. Adams-Watters


-----Original Message-----
From: njas at research.att.com

the two puzzles (and their solutions) are

%I A123896
%S A123896 0,1,1,1,12,12,12,12,12,12,100,121,122,123,123
%N A123896 A123895(n^2).
...

%I A123902
%S A123902 0,1,12,100,112,121,122,123
%N A123902 A123896 sorted and uniqued.
...

which are based on this:

%I A123895
%S A123895 
0,1,1,1,1,1,1,1,1,1,10,11,12,12,12,12,12,12,12,12,10,12,11,12,12,
%T A123895 
12,12,12,12,12,10,12,12,11,12,12,12,12,12,12,10,12,12,12,11,12,12,
%U A123895 
12,12,12,10,12,12,12,12,11,12,12,12,12,10,12,12,12,12,12,11,12,12
%N A123895 Restricted growth string for the (decimal expansion of the) 
number n.
%C A123895 Write n in base 10 prefixed with a 0. Read this string from 
left to r
ight. Write a 0 each time you see the first distinct digit (which is 
0), write a
1 each time you see the second distinct digit, write a 2 each time you 
see the t
hird distinct digit, and so on. Finally, delete the leading zeros.
%D A123895 D. E. Knuth, TAOCP, Vol. 4, Section 7.2.1.5, Problems 4 and 
5.
%e A123895 To find a(66041171): 066041171 -> 011023343 -> 11023343.
%Y A123895 Cf. A123896, A123902.
%K A123895 nonn,base,new
%O A123895 0,11
%A A123895 njas, Nov 20 2006

It would be nice if someone could extend the first two, and provide
maple or mma or other code for the third one

Neil


________________________________________________________________________
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From njas at research.att.com  Tue Nov 21 00:43:03 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Mon, 20 Nov 2006 18:43:03 -0500 (EST)
Subject: King chickens - I was wrong!
Message-ID: <200611202343.SAA88240@fry.research.att.com>


Martin Fuller pointed out that I was wrong.
A highest scorer in a tournament is a king chicken,
but not conversely.

So there are really two sequences, A123553 (K.C.'s),
now corrected by Martin, and A125031, the total number
of highest scorers, computed by Martin Fuller to n=11
(this is what Gordon Royle computed out to n=10).

Maurer also defines "Emperors", chickens
who peck everybody else, which is n*2^((n-1)(n-2)/2)
(it will be A123903).

Neil





From zakseidov at yahoo.com  Tue Nov 21 06:19:11 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Mon, 20 Nov 2006 21:19:11 -0800 (PST)
Subject: A125097.gif: (Very) nice patterns
Message-ID: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>

Neil, 
seqfans,

1. Have a look at this full graph -
aint these patterns (very) nice?!
It may be known, but still it's nice...

2. Is it possible to 
put "the extended graph" into OEIS?
In the standard graph no patterns are seen at all...

3. Neil removed  keyword "nice" from A125097
(as he did regularly with ny sequences)-
as if I insist that it's me who is "nice"
not it is the seq which is nice.

Thanks, Zak


 
____________________________________________________________________________________
Sponsored Link

$200,000 mortgage for $660/ mo - 
30/15 yr fixed, reduce debt - 
http://yahoo.ratemarketplace.com
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From maxale at gmail.com  Tue Nov 21 07:03:55 2006
From: maxale at gmail.com (Max A.)
Date: Mon, 20 Nov 2006 22:03:55 -0800
Subject: A125097.gif: (Very) nice patterns
In-Reply-To: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>
References: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>
Message-ID: <d3dac270611202203p2b9fff6s2bc522bc6662e735@mail.gmail.com>

Zak,

I do not see anything special about this graph.
Yes, it includes some points from the line y=2x+1, and, yes, it is
symmetric the region [0,500]x[0,1000] (i.e., a(n)=k iff
a((k-1)/2)=2n+1). But that does not make it any nicer.

Regards,
Max

On 11/20/06, zak seidov <zakseidov at yahoo.com> wrote:
> Neil,
> seqfans,
>
> 1. Have a look at this full graph -
> aint these patterns (very) nice?!
> It may be known, but still it's nice...
>
> 2. Is it possible to
> put "the extended graph" into OEIS?
> In the standard graph no patterns are seen at all...
>
> 3. Neil removed  keyword "nice" from A125097
> (as he did regularly with ny sequences)-
> as if I insist that it's me who is "nice"
> not it is the seq which is nice.
>
> Thanks, Zak
>
>
>
> ____________________________________________________________________________________
> Sponsored Link
>
> $200,000 mortgage for $660/ mo -
> 30/15 yr fixed, reduce debt -
> http://yahoo.ratemarketplace.com
>
>





From pauldhanna at juno.com  Tue Nov 21 11:07:54 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Tue, 21 Nov 2006 05:07:54 -0500
Subject: Formula for New Sequence A125054 ?
Message-ID: <20061121.050754.1448.0.pauldhanna@juno.com>

Seqfans,
     Could someone find a formula for this new sequence? 
Sequence: A125054  begins: 
 
1,3,21,327,9129,396363,24615741,2068052367,225742096209,
31048132997523,5252064083753061,1071525520294178007,
259439870666594250489,73542221109962636293083,
24125551094579137082039181,9068240688454120376775401247,
3871645204706420218816959159969,
(I can supply many more terms if needed).
 
The sequence forms the Central terms of a new triangle A125053 
(a variant of triangle A008301 - enumeration of binary trees). 
 
Triangle A125053 is nice since the first column (and row sums) 
form the Secant numbers A000364 (an unexpected result!). 
  
Below I define the triangle.
 
I would be very interested in a formula, perhaps an E.g.f.? 
(I know I am asking a lot, but Seqfans surprise me all the time 
with amazing formulas). 
  
Thanks, 
      Paul 
---------------------------------------------------------
  
If we write triangle A125053 like this: 
.......................... ...1; 
.................... ...1, ...3, ...1; 
.............. ...5, ..15, ..21, ..15, ...5; 
........ ..61, .183, .285, .327, .285, .183, ..61; 
.. 1385, 4155, 6681, 8475, 9129, 8475, 6681, 4155, 1385; 

then the first nonzero term is the sum of the previous row: 
1385 = 61 + 183 + 285 + 327 + 285 + 183 + 61,

the next term is 3 times the first:
4155 = 3*1385,

and the remaining terms in each row are obtained by the rule 
illustrated by: 
6681 = 2*4155 - 1385 - 4*61 ;
8475 = 2*6681 - 4155 - 4*183 ;
9129 = 2*8475 - 6681 - 4*285 ;
8475 = 2*9129 - 8475 - 4*327 ;
6681 = 2*8475 - 9129 - 4*285 ;
4155 = 2*6681 - 8475 - 4*183 ;
1385 = 2*4155 - 6681 - 4*61 .
 
An alternate recurrence is illustrated by:
4155 = 1385 + 2*(61 + 183 + 285 + 327 + 285 + 183 + 61);
6681 = 4155 + 2*(183 + 285 + 327 + 285 + 183);
8475 = 6681 + 2*(285 + 327 + 285);
9129 = 8475 + 2*(327); 
 
and then for k>n, k<=2n, T(n,k) = T(n,2*n-k).
 
END.





From pauldhanna at juno.com  Tue Nov 21 14:58:26 2006
From: pauldhanna at juno.com (Paul D Hanna)
Date: Tue, 21 Nov 2006 13:58:26 GMT
Subject: Formula for New Sequence A125054 ?
Message-ID: <20061121.055901.20413.609605@webmail43.nyc.untd.com>

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From antti.karttunen at gmail.com  Tue Nov 21 15:27:22 2006
From: antti.karttunen at gmail.com (Antti Karttunen)
Date: Tue, 21 Nov 2006 16:27:22 +0200
Subject: Binary tree terminology, again. Re: Formula for New Sequence A125054
 ?
In-Reply-To: <20061121.055901.20413.609605@webmail43.nyc.untd.com>
References: <20061121.055901.20413.609605@webmail43.nyc.untd.com>
Message-ID: <45630CCA.8000103@gmail.com>

Paul D Hanna wrote:

> Seqfans,
>
>      Could someone find a formula for this new sequence?
> Sequence: A125054  begins:
>  
> 1,3,21,327,9129,396363,24615741,2068052367,225742096209,
> 31048132997523,5252064083753061,1071525520294178007,
> 259439870666594250489,73542221109962636293083,
> 24125551094579137082039181,9068240688454120376775401247,
> 3871645204706420218816959159969,
> (I can supply many more terms if needed).
>  
> The sequence forms the Central terms of a new triangle A125053
> (a variant of triangle A008301 - enumeration of binary trees).
>
Is there any more helpful description for A008301 and A125053 ?
E.g. do they relate to non-planar binary trees
(as http://www.research.att.com/~njas/sequences/A001190 )
or to (rooted, unlabeled) planar binary trees, as Catalan numbers, 
A000108, do?
Labeled or unlabeled? Rooted or non-rooted?


Cumprimentos,

Antti Karttunen
writing from Porto, Portugal.

>  
> Triangle A125053 is nice since the first column (and row sums)
> form the Euler numbers A000364 (an unexpected result!).
>  
>
> Seqfans,
>     One should always try Superseeker first!
> Superseeker says:
>
> A125054 = Binomial transform of A000182 (e.g.f. tan(x))
>
> and it holds true.
>
> I did not expect such a simple answer! 
>     Paul
>

>  
>






From tbaruchel at free.fr  Tue Nov 21 15:47:28 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Tue, 21 Nov 2006 15:47:28 +0100 (CET)
Subject: Almost the number of divisors of n+1
Message-ID: <20061121154018.X1354@localhost.my.domain>

Hi, I have some number-theory related empirically collected datas to study.
As long as I can see, I have a link between each integer and the number
of divisors of its successor. But though the rule seems to work very
well in some cases: predecessors of prime numbers seem to appear 0 or 1 times,
predecessors of numbers like 12 or 24 seem to appear much more, etc. I am not
sure this is the very exact rule. Of course, the trouble may come from
the fact that my datas is quite hard to "extract", but maybe you know some
variants of
   a(n) is the number of divisors of (n+1)
Have you ever seen closely-related sequences like that one?

Regards,

-- 
Thomas Baruchel





From davidwwilson at comcast.net  Tue Nov 21 18:18:36 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Tue, 21 Nov 2006 12:18:36 -0500
Subject: Confirmation
Message-ID: <004601c70d91$14d8f3d0$6501a8c0@yourxhtr8hvc4p>

Define

    f(1) = 1
    f(n+1) = f(n) + (f(n) mod n+3364).

I wrote a computer program to find the smallest x with

    x+3365 divides f(x)

The value of x found is just under 8,350,000,000 and the corresponding f(x) just under 2^64. Although my program included an overflow check, I am still not comfortable because this value of f(x) is close to my 64-bit unsigned integer size.

Could someone with a fast computer, preferably a UNIX box, please confirm or refute my values?
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From maxale at gmail.com  Tue Nov 21 20:47:20 2006
From: maxale at gmail.com (Max A.)
Date: Tue, 21 Nov 2006 11:47:20 -0800
Subject: Confirmation
In-Reply-To: <004601c70d91$14d8f3d0$6501a8c0@yourxhtr8hvc4p>
References: <004601c70d91$14d8f3d0$6501a8c0@yourxhtr8hvc4p>
Message-ID: <d3dac270611211147i7b7ef94ev729014f094d24dc4@mail.gmail.com>

I've got

x = 8334201245
f(x) = 17364786429187398690

I've also checked that the value f(n) was non-decreasing during
computations. Since f(n+1) < f(n)+n+3364 and f(x)+x+3364 < 2^64, that
means no overflow in 64-bit unsigned integer happened.

Max

On 11/21/06, David Wilson <davidwwilson at comcast.net> wrote:
>
>
> Define
>
>     f(1) = 1
>     f(n+1) = f(n) + (f(n) mod n+3364).
>
> I wrote a computer program to find the smallest x with
>
>     x+3365 divides f(x)
>
> The value of x found is just under 8,350,000,000 and the corresponding f(x)
> just under 2^64. Although my program included an overflow check, I am still
> not comfortable because this value of f(x) is close to my 64-bit unsigned
> integer size.
>
> Could someone with a fast computer, preferably a UNIX box, please confirm or
> refute my values?





From all at abouthugo.de  Tue Nov 21 22:22:29 2006
From: all at abouthugo.de (Hugo Pfoertner)
Date: Tue, 21 Nov 2006 22:22:29 +0100
Subject: Wrong new terms in A085000
Message-ID: <45636E15.FD6E8D3F@abouthugo.de>

Seqfans, Neil,

the sequence A085000 "Maximal determinant of an n X n matrix using the
integers 1 to n^2."

that had previously the terms a(1)...a(6)
1,10,412,40800,6839492,1865999570,

has recently been extendend as follows:

762074188050,440477621828418, 345964972957674150

The comment says:
a(7) - a(9) were found by Richard Mueller
(richard.mueller(AT)informatik.hu-berlin.de) and Johannes Dewender
(johannes.dewender(AT)informatik.hu-berlin.de) in cooperation, Oct 22
2006 
Entry updated by Richard Mueller  Oct 22 2006

Unfortunately bigger counterexamples exist for all 3 terms:

d7:=det mat
((28, 9,34,41, 1,40,22),
 (47,25, 7,36,35,16, 8),
 (3, 27, 6,39,31,26,43),
 (38,21,32,14,18, 4,48),
 (19, 5,37,13,49,33,20),
 (29,42,15, 2,17,45,24),
 (11,46,44,30,23,12,10));
 
 Reduce: d7 := 762139309293

d8:= det mat
((13, 53, 40, 25,  2, 45, 58, 23),
 (21, 22,  8, 54, 24, 56, 18, 57),
 (60, 11, 36, 44, 12,  7, 49, 41),
 (15, 19, 48,  4, 52, 26, 38, 59),
 (20, 62, 34, 51, 42,  1, 17, 33),
 (37, 29,  3, 31, 61, 35, 55, 10),
 (30, 14, 63, 46, 39, 47, 16,  5),
 (64, 50, 28,  6, 27, 43,  9, 32));

Reduce: d8 := 440857916120379

d9:= det mat
((40, 48, 72, 26,  6, 50, 39, 78, 10),
 (23, 77, 42,  7, 37, 41, 74, 12, 56),
 (52, 47,  8, 54,  5, 64, 19, 44, 75),
 (59, 25, 13, 43, 58, 76, 63, 30,  3),
 (79, 36, 71, 60, 31, 18, 34,  2, 38),
 (32, 81, 22, 61, 67, 15, 17, 53, 21),
 (20, 24, 66, 27, 73, 68,  4, 33, 55),
 ( 1, 14, 45, 80, 35, 29, 69, 46, 49),
 (65, 16, 28, 11, 57,  9, 51, 70, 62));

Reduce: d9 := 346202123011922653

(Thanks to Rainer Rosenthal for X-checking the determinants with Maple).

None of the counterexamples is known to be the biggest possible
solution. The best value I've found in 2 CPU years for
a(7)=762140212575, but currently I don't have access to the computer
where I've found this solution and the corresponding arrangement of
matrix elements.

My suggestion is to remove the 3 new terms and the new Mathematica
program and to restore the previous state of the sequence. The
determination of a(8) and a(9) is currently beyond our available
computational capabilities. I will try to confirm a(7) and to find
improved lower bounds for a(8) and a(9) by running the slightly improved
program given at the link for some months on a cluster of Intel Itanium
2 CPUs.

In the meantime I suggest to add a comment:

Known lower bounds for a(7)...a(9) are
762140212575, 440857916120379, 346202123011922653,

Hugo Pfoertner





From mlb at well.com  Wed Nov 22 00:25:34 2006
From: mlb at well.com (Marc LeBrun)
Date: Tue, 21 Nov 2006 15:25:34 -0800
Subject: Almost the number of divisors of n+1
In-Reply-To: <20061121154018.X1354@localhost.my.domain>
References: <20061121154018.X1354@localhost.my.domain>
Message-ID: <200611212325.kALNPQh3019729@smtp.well.com>

 >  a(n) is the number of divisors of (n+1)
 > Have you ever seen closely-related sequences like that one?

Some random and rather vague ideas:

1. I faintly recall seeing something about algorithms for
factoring n involving properties of n-1 (I mean besides
the obvious connection to "Fermat's little theorem").
Perhaps in Knuth's TAOCP?

2. It sounds like you might have a case where the relationship
you're seeing, a(n) ~ d(n), is actually a first approximation
to some identity like

   a(n) = d(n) + F1(n) + F2(n) + ...

where the Fk depend on the factor structure of n or something.
So perhaps you might try to "take out" the d(n) and then look
further into the "error" and see what you can guess about how
the Fk's specifically vary with primes, highly composites, etc.
For example you might try "fitting" a(n) with Moebius-style sums.

3. There's some rather surprising known relationships involving
divisors--my favorite is one of Legendre's that is basically an
identity found by expanding elliptic functions in power series.
This suggests that you might want to express your relationships
as generating functions and see what emerges.

4. Sometimes there is simple underlying "combinatorial machinery".
For example I was once puzzled by some coefficients that turned
out to be formed by taking each divisor of n, *adding 1*, and then
multiplying them all together!  It took me a while to see that's
algebraically identical to summing the products of every possible
subset of the divisors.  But the "add 1" way of expressing this
was totally opaque, whereas the "every subset" explained them.

Happy hunting!







From zbi74583 at boat.zero.ad.jp  Wed Nov 22 02:15:26 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Wed, 22 Nov 2006 10:15:26 +0900
Subject: N(2)
Message-ID: <00a201c70dd3$b538af40$977fd7dc@FMC57937B6BE00>

    I wonder if N(2) goes infinity.

    Where N(s)=   Sum_{m=1 to infinity} (-1)Sigma(m)/m^s

    Mathar computed up to m=1000000



  s m N(s)

> 2 1 1.000000000000000000000000000
> 2 2 1.250000000000000000000000000
> 2 3 1.472222222222222222222222222
> 2 4 1.784722222222222222222222222
> 2 5 1.944722222222222222222222222
> 2 6 2.000277777777777777777777778
> 2 7 2.122726757369614512471655329
> 2 8 2.325851757369614512471655329
> 2 9 2.461654226505416981607457798
> 2 10000 8.619813177975507784214344698
> 2 20000 9.240635095871089404115753095
> 2 30000 9.603791897573761358292380061
> 2 40000 9.861458278261452185262848858
> 2 50000 10.06133043757656995196303496
> 2 60000 10.22462154552329778519365923
> 2 70000 10.36270038005636494621123847
> 2 80000 10.48231016697477680498962023
> 2 90000 10.58780522013752853231672862
> 2 100000 10.68217402235127296288785820
> 2 110000 10.76754463607533705360724580
> 2 120000 10.84547789729425765394094918
> 2 130000 10.91717268365027978460835188
> 2 140000 10.98355652885066079792761960
> 2 150000 11.04534545539722125239381859
> 2 160000 11.10315629233238669062836862
> 2 170000 11.15746025373895516747888760
> 2 180000 11.20865775307751148407150182
> 2 190000 11.25708463818456648077093060
> 2 200000 11.30303121795283056763425981
> 2 210000 11.34672871948779384291766276
> 2 220000 11.38839827063844308711114745
> 2 230000 11.42821240662906506074836109
> 2 240000 11.46633165190016188934526342
> 2 250000 11.50290004647923375585461222
> 2 260000 11.53802980091084626273394586
> 2 270000 11.57183602344116902681335304
> 2 280000 11.60440829561909536077171934
> 2 290000 11.63584045378869260629085740
> 2 300000 11.66620539683007648663354504
> 2 310000 11.69557551635064846235471407
> 2 320000 11.72401391041322342493145640
> 2 330000 11.75157395832622563415629569
> 2 340000 11.77831576251751436463182945
> 2 350000 11.80427986574616427809478607
> 2 360000 11.82951337960016320803463309
> 2 370000 11.85405377158818097821428014
> 2 380000 11.87794149361683265940241605
> 2 390000 11.90120806565142413868457075
> 2 400000 11.92388559358277288937420440
> 2 410000 11.94600303970164868968234173
> 2 420000 11.96758569454234998141373716
> 2 430000 11.98866287480559328733313634
> 2 440000 12.00925544388189350449936925
> 2 450000 12.02938482345842327394291528
> 2 460000 12.04907101416111155199518539
> 2 470000 12.06833481184766752219695603
> 2 480000 12.08719139777266186606623586
> 2 490000 12.10566116650605692754670219
> 2 500000 12.12375840131307130932177405
> 2 510000 12.14149351071909408890611230
> 2 520000 12.15888769016783248485026059
> 2 530000 12.17594975401341942285000014
> 2 540000 12.19269280126406301090981599
> 2 550000 12.20912705098196862278557687
> 2 560000 12.22526702472305560991043337
> 2 570000 12.24111996896626519121471994
> 2 580000 12.25669798166764128539680162
> 2 590000 12.27200952384435815319104573
> 2 600000 12.28706299338740013449483108
> 2 610000 12.30187012252190226181814868
> 2 620000 12.31643453756242668558457904
> 2 630000 12.33076670711890682146429120
> 2 640000 12.34487182648867321005497632
> 2 650000 12.35875920337989470739102627
> 2 660000 12.37243430644503889140869087
> 2 670000 12.38590421689421153982279985
> 2 680000 12.39917431914306915108546756
> 2 690000 12.41224965443445935379058168
> 2 700000 12.42513867918653815467126754
> 2 710000 12.43784365088579118428209464
> 2 720000 12.45037185509338005878377820
> 2 730000 12.46272650898502725121177200
> 2 740000 12.47491268376206272216237202
> 2 750000 12.48693594951832440956683550
> 2 760000 12.49880013638169511883229569
> 2 770000 12.51050884124405914256853750
> 2 780000 12.52206537108093533871333277
> 2 790000 12.53347692259745252540675575
> 2 800000 12.54474459716977128685280547
> 2 810000 12.55587125001645305770022096
> 2 820000 12.56686130746829341595469366
> 2 830000 12.57771842011223665163037795
> 2 840000 12.58844534265072464040923243
> 2 850000 12.59904566706465055360427300
> 2 860000 12.60952226922408394518254911
> 2 870000 12.61987642201166975414741555
> 2 880000 12.63011418303496833362065684
> 2 890000 12.64023571583693921823484395
> 2 900000 12.65024408926657285403820450
> 2 910000 12.66014049405210257196085418
> 2 920000 12.66993013960070545251999449
> 2 930000 12.67961329658082585267888159
> 2 940000 12.68919341076487592672709130
> 2 950000 12.69867217785123447166660908
> 2 960000 12.70805040456023104531710849
> 2 970000 12.71733330696500792660758365
> 2 980000 12.72652037909829933006161382
> 2 990000 12.73561385879130711822105048
> 2 1000000 12.74461639944508312563225494



         UnitaryPhi(m) < (-1)Sigma(m) < Sigma(m)

         So, U(2) < N(2) < S(2)=Zeta(1):*Zeta(2)

         Where,U(s) = Sum_{m=1 to infinity} UnitaryPhi(m)/m^s , S(s) = 
Sum_{m=1 to infinity} Sigma(m)/m^s



    I suppose that U(2) is already known.

    Does anyone know it?



    Yasutoshi
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From zbi74583 at boat.zero.ad.jp  Wed Nov 22 02:15:24 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Wed, 22 Nov 2006 10:15:24 +0900
Subject: A085058
References: <019201c70ab0$b44b3ce0$ca73d7dc@FMC57937B6BE00> <007501c70ac2$a6b28900$6303f0d5@speedy>
Message-ID: <00a101c70dd3$b38cd400$977fd7dc@FMC57937B6BE00>

   Is it an easy well known fact?

   I would like to know the reason why they are the same.



   Yasutoshi



  ----- Original Message ----- 
  From: Vladeta Jovovic
  To: kohmoto
  Sent: Saturday, November 18, 2006 12:35 PM
  Subject: Re: A085058



  Yes, they are the same sequence: see A101921.

  Best wishes,
  Vladeta Jovovic

    ----- Original Message ----- 
    From: kohmoto
    To: seqfan at ext.jussieu.fr
    Sent: Saturday, November 18, 2006 2:27 AM
    Subject: A085058


        Hi, Seqfans

        If 2^k(n) is the largest power of two dividing tangent number T(n) 
then numbers such that 2*n-k(n) are as follows.



        S : 2,3,2,4,2,3,2,5,2,....



        It looks like A085058.

        Are they the same seqences ?



        Yasutoshi




    __________ NOD32 1.1454 (20060321) Information __________

    This message was checked by NOD32 antivirus system.
    http://www.nod32.com
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From zbi74583 at boat.zero.ad.jp  Wed Nov 22 02:15:22 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Wed, 22 Nov 2006 10:15:22 +0900
Subject: A020955
Message-ID: <009801c70dd3$b0eb0000$977fd7dc@FMC57937B6BE00>

    Hi, Seafans.

    I suppose that no one understand what A020955 means.



    So, I explain it a little.

    First of all, see my home page and click ?Free Class?, recently I 
rewrote it.



         http://boat.zero.ad.jp/~zbi74583/another02.htm



    The idea ?ma??.an abbreviation of ?make??.is represented by a 
mapping from subsets of a set to members of the set..

    Because the idea ?ma? has the same property as a mapping.



    Number of subsets of a set which has N members is 2^N, and the 
possibility of images of each subsets is N, so number of the mapping 
subsets ->members is N^(2^N).



    But if the set is a model of Free Class then the mapping must satisfy 
some conditions.         .

    They are as follows.



    If the set is {0,1,2,3,?.n} then { } ma 0, {0} ma1, {0,1} ma 2, {0,1,2} 
ma 3,?.,{0,1,2,3,?.n} ma n .



    So, images of n+1 subsets are already decided.

    Hence the number of the mapping or the number of models which has N 
members is N^(2^N-n-1) .



    Yasutoshi


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From aplewe at sbcglobal.net  Wed Nov 22 02:59:20 2006
From: aplewe at sbcglobal.net (Andrew Plewe)
Date: Tue, 21 Nov 2006 17:59:20 -0800
Subject: Almost the number of divisors of n+1
In-Reply-To: <200611212325.kALNPQh3019729@smtp.well.com>
Message-ID: <200611220159.kAM1xMZG084485@shiva.jussieu.fr>

Marc said:

Some random and rather vague ideas:

1. I faintly recall seeing something about algorithms for
factoring n involving properties of n-1 (I mean besides
the obvious connection to "Fermat's little theorem").
Perhaps in Knuth's TAOCP? 



I have that book in front of me right now; he demonstrates how to use the
properties of n-1 to verify primality (pg. 375, in vol. 2, 2nd edition)
after applying Fermat's theorm. 

One a similar note, one could also try to solve:

n - 1 = a mod (a + 1) (or, more generally, n - x = a mod (a + x))

in which case a + 1 (or x) will be a factor of n, but I don't think that's a
particularly practical method -- finding a "mod difference" for n - x seems
to be just as hard as factoring n, even if you know the factors of n - x.

	-Andrew Plewe-

-----Original Message-----
From: Marc LeBrun [mailto:mlb at well.com] 
Sent: Tuesday, November 21, 2006 3:26 PM
To: seqfan at ext.jussieu.fr
Subject: Re: Almost the number of divisors of n+1

 >  a(n) is the number of divisors of (n+1)
 > Have you ever seen closely-related sequences like that one?

Some random and rather vague ideas:

1. I faintly recall seeing something about algorithms for
factoring n involving properties of n-1 (I mean besides
the obvious connection to "Fermat's little theorem").
Perhaps in Knuth's TAOCP?

2. It sounds like you might have a case where the relationship
you're seeing, a(n) ~ d(n), is actually a first approximation
to some identity like

   a(n) = d(n) + F1(n) + F2(n) + ...

where the Fk depend on the factor structure of n or something.
So perhaps you might try to "take out" the d(n) and then look
further into the "error" and see what you can guess about how
the Fk's specifically vary with primes, highly composites, etc.
For example you might try "fitting" a(n) with Moebius-style sums.

3. There's some rather surprising known relationships involving
divisors--my favorite is one of Legendre's that is basically an
identity found by expanding elliptic functions in power series.
This suggests that you might want to express your relationships
as generating functions and see what emerges.

4. Sometimes there is simple underlying "combinatorial machinery".
For example I was once puzzled by some coefficients that turned
out to be formed by taking each divisor of n, *adding 1*, and then
multiplying them all together!  It took me a while to see that's
algebraically identical to summing the products of every possible
subset of the divisors.  But the "add 1" way of expressing this
was totally opaque, whereas the "every subset" explained them.

Happy hunting!







From bowerc at usa.net  Wed Nov 22 03:31:46 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Tue, 21 Nov 2006 18:31:46 -0800
Subject: Assembly of nothing
Message-ID: <129kkVcEU2922S01.1164162706@cmsweb01.cms.usa.net>

This is about a problem with sequence transforms I've been thinking
about for a couple of years now. I call it the "assembly of nothing"
problem.

Assembly refers to creating a set of combinatorial structures
classified by number of nodes of the form A(B) where B is a
preexisting set of structures and A is a rule for how to place
the B-structures into the new structure.

One type of assembly is combinatorial substitution (or composition).
In this instance the "A" rule is itself a collection of structures.
The assembly is to take copies of "B" structures and use them as
the points of the "A" structure.

A variant of this assembly is one where certain pairs of points in
the "A" structure are set aside to be filled with the same "B"
structure (i.e. 2 copies of the same structure) or different
"B" structures.

What's particularly notable about these assemblies is that they
can be described by index series which are formal polynomial
series of the indexed variables x_1, x_2, x_3, ....

The "nothing" part of the title refers to "B" structures that have
a size (or number of points) of zero. The formulas detailing the
counting of these structures frequently forbid the "B" structure
from containing any zero sized members. In many cases this seems
like a necessary restriction, but in others, it can get in the
way of some useful combinatorial questions.

Let's consider 2 types of assemblies.

Sets of structures, allowing the same structure multiple times
(as implemented by the Euler transform for unlabeled structures
and the Exponential transform for labeled structures.)

Sets of structures, requiring all structures to be different
(as implemented by the Weigh transform for unlabeled structures.)

http://www.research.att.com/~njas/sequences/A000041
The partition sequence A000041 counts unlabeled sets of nonempty sets
hence it has g.f. EULER(x/(1-x)) if we consider EULER as a g.f.
transformation.

http://www.research.att.com/~njas/sequences/A000009
A000009 counts sets of nonempty sets all different from one another
with g.f. WEIGH(x/(1-x)).

in the first case, it makes sense that the "B" structure (the
nonempty sets) are not allowed to have 0 sized elements (i.e.
be empty). If we tried to count partitions of n and allowed n
to be partitioned into 0s, we would have infinitely many
possibilities. 1, 1+0, 1+0+0, ....

In the second case, we can allow empty "B" structures, since
each structure can only be used once giving us 4 partitions
of 3 into distinct parts: 3, 3+0, 1+2, 1+2+0, or in general
2*A000009(n) partitions of n into distinct parts allowing 0
as a part. Hence we would like to say that
WEIGH(1/(1-x)) = 2*WEIGH(x/(1-x)).

The commonly used formulas appear to support these observations.

The Euler transform as applied to a sequence is often rendered:

(1-x)^(-a1)*(1-x^2)^(-a2)*(1-x^3)^(-a3)*(1-x^4)^(-a4)*...

with the Weigh transform as

(1+x)^a1*(1+x^2)^a2*(1+x^3)^a3*(1+x^4)^a4*...

both of these formulas ignore the a0 term, but if we were to extend
these one term backward we would have Euler start

(1-1)^(-a0)*...

and Weigh

(1+1)^a0*...

and we can see that in Euler's case given any positive a0 we begin
with 0 raised to a negative power giving an infinite value and for
Weigh's case we start the product with 2^a0.

A similar effect is seen from the index series.

An index series is applied to a g.f. as follows:

If s(x_1, x_2, x_3, ...) is the index series, then it maps the
g.f. A(x) to s(A(x), A(x^2), A(x^3), ...) thus x_2^2 + x_3*x_4
maps as A(x^2)^2+A(x^3)*A(x^4).

Euler's i.s. is exp(x_1/1 + x_2/2 + x_3/3 + ...)

Weigh's i.s. is exp(x_1/1 - x_2/2 + x_3/3 - ...)

We can see that if A(x) has a positive constant term, then
that term will be multiplied by the divergent harmonic series
1+1/2+1/3+... and blow up. With Weigh we have the alternating
harmonic series 1-1/2+1/3-...=ln 2 giving it the 2^a0 character.

So in these 2 examples, the problem seems manageable, but if one
is looking for a general theory to cover all such assemblies or
all transforms defined by i.s., things are not so easy.

Consider 2 other assembly rules:

Ordered lists of structures, allowing the same structure multiple times
(as implemented by the Invert transform for unlabeled structures.)

Ordered lists of structures, requiring a structure not have a copy
of itself as a neighbor
(I call the unlabeled version of this the Carlitz transform. It's
not in the EIS canon yet.)

Invert's i.s. is 1/(1-x_1)
Carlitz's i.s. is 1/(1-x_1+x_2-x_3+...)

Invert creates compositions (or ordered partitions) the way Euler
creates partitions.

Hence we have INVERT(x/(1-x)) = (1-x)/(1-2x) = 1,1,2,4,8,16,...

Just as in the unordered partitions, we would not allow one to
compose n into 0s, there would be infinitely many choices.
Note that unlike the Euler transform which blows up for a0>0,
The Invert transform can be calculated for any a0 != 1, although
it's difficult to give a combinatorial interpretation to any
nonzero value.

http://www.research.att.com/~njas/sequences/A003242
Carlitz(x/(1-x)) gives us A003242 Carlitz compositions, those
where no two adjacent parts are equal. We can allow Carlitz
compositions to contain 0s since the separation of the 0s by other
numbers keeps it finite: Carlitz(1/(1-x)) = A114900.
http://www.research.att.com/~njas/sequences/A114900
However, the i.s. does not tell us how to do it. If A(x)=1 what
does one make of Carlitz(A(x)) = 1/(1-1+1-1+...)

I recently had an idea about how to deal with issues like this:
use an analytic interpretation of the g.f. of a sequence.

If a sequence represents structures contains only finitely many
structures, then the number of structures is A(1) where A(x) is
the g.f. My idea is to take the case where A(x) has a radius of
convergence (ROC) >0 and <=1 so that it cannot be evaluated by
adding the terms, and associating its size with A(1) of the
analytic extension of the g.f. Here in the Carlitz case we have
1-x+x^2-x^3+... = 1/(1+x) hence A(1) = 1/2 and 1/(1/2)=2, the
expected value for Carlitz compositions of 0 allowing 0 as a part.
(The 2 are the empty composition and zero.)

In this way I can evaluate a transform of a sequence whose only
nonzero term is a_0 as T(a_0*x)(1) and for more general sequences
I can apply the transform to a 2 dimensional sequence as follows.

If sequence has g.f. A(x) then let C(x,y) = y*A(x).

Apply T to C(x,y) getting D(x,y) as output:

Let d_n = [x^n] D(x,1)

I suspect this will be an adequate solution to many problems of
this sort, but I'm not sure how far this method can be trusted.
Part of the problem is I don't know my analytic function theory
as well as I should. (If anyone out there does, please chime in.)
More importantly, I've run across some examples that challenge
the method.

My first example involves the Euler transform. It's basically only
a problem because I would expect that if f(1)=g(1) then
(Tf)(1) = (Tg)(1) for any transform with both f and g in its
domain. It seems reasonable as 1^n=1 for all n so that i.s.'s
behave like g.f.'s when evaluated at 1. More importantly, the
choice to use the linear term in "C(x,y) = y*A(x)" as opposed
to "C(x,y) = y^2*A(x)" was arnitrary and it seemingly should
not matter which term I chose, so I would like some assurance
that I'll get the same answer following any such path.

Consider

EULER(0) = 1
EULER(x-x^2) = 1+x
EULER(x-x^3) = 1+x+x^2

For all of these, the inputs have f(1)=0, but the outputs have
g(x)=1,2 or 3

I think the reason for this anamoly may be in the instability
Euler has at 0. Just as Weigh behaves like 2^x, Euler behaves
like 0^(-x) giving 0 for negative values of x, infinity for
positive values and doing what we see here at x=0. 0^0 is
generally defined as 1, but analytically it can be anything and
that may be the limitation I'm running into.

My other example concerns me, because I'm interested in formulas
that can be applied to arbitrary sequences, not just integer
sequences. Most of our familiar transform formulas work fine for
arbitrary sequences, and it would be nice if all of them did,
however...


Ordered lists of structures, requiring all structures to be different
(as implemented by the AGK transform for unlabeled structures
found on
http://www.research.att.com/~njas/sequences/transforms2.html )

Obviously, this assembly makes sense for empty structures since
each structure can appear at most once. If I allow there to be
n types of empty structures in the input then I have
A003149(n) empty structures in the output.
http://www.research.att.com/~njas/sequences/A003149

While it seems quite plausible that this sequence can be
interpolated into a nice analytic function, the method I illustrated
above offers no help as for n anything other that a
nonnegative integer gives me a sequence with ROC of zero.

So the idea in posting this is to present some new ideas to the
sequence community, get some opinions from people who know some
things that I don't know and get some ideas about how to handle
difficult cases.

Christian








From davidwwilson at comcast.net  Wed Nov 22 04:22:55 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Tue, 21 Nov 2006 22:22:55 -0500
Subject: Confirmation
References: <200611211857.kALIvmai029436@amer.strw.leidenuniv.nl>
Message-ID: <001001c70de5$812f2190$6501a8c0@yourxhtr8hvc4p>

Thanks for your help, your results agree with my values.





From jawbrey at att.net  Wed Nov 22 04:28:31 2006
From: jawbrey at att.net (Jon Awbrey)
Date: Tue, 21 Nov 2006 22:28:31 -0500
Subject: Assembly of nothing
References: <129kkVcEU2922S01.1164162706@cmsweb01.cms.usa.net>
Message-ID: <4563C3DE.FB5134DC@att.net>

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o

This is one of the impulses that led to the construction of riffs and rotes.
It was probably in my freshman chemistry class, or later when I worked as a
hospital orderly and used to run errands for the nuclear medicine unit that
I was thinking about the techniques of molecular tagging, which is like the
business of labelling graphs, only you have to label molecules with bits of
physical structure, an extra neutron or the like, that alters the resulting
structure.

So the question arises whether you can have a class of structures that is
closed under self-labeling.  Somewhat roughly trying to use your notation,
if A(B) is the set of all structures where those of type A are labeled by
those of type B, can you have the doamin equation A = A(A) for any type A?

Or something like that ...

Jon Awbrey

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o
inquiry e-lab: http://stderr.org/pipermail/inquiry/
wikinfo: http://wikinfo.org/wiki.php?title=User:Jon_Awbrey
wp review: http://wikipediareview.com/index.php?showuser=398
o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o






From Annette.Warlich at t-online.de  Wed Nov 22 13:37:30 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Wed, 22 Nov 2006 13:37:30 +0100
Subject: Eigensequences
Message-ID: <4564448A.5010200@t-online.de>

In  a recent treatise of the pascal matrix and its
connections to the bernoulli-numbers I restated
their connection as an eigenvector/eigensequence-
relation.

Say P is the lower triangular matrix of binomial-
coefficients, Pj the matrix, where the column-signs
alternate and B the vector of bernoulli-numbers
(b_1 = +1/2)
then
   Pj * B = B

Now in connection with the subject of divergent
summmation I investigated powerseries, say
  V(x) = [1,x,x^2,x^3,... ]
in connection with Pj, like

  ??? = Pj * V(x)      // binomial-transform
  ??? = V(x)~ * Pj

(using ~ for the transpose)

Now since

   Pj * B = B

also

   V(x) * Pj * B = V(x) * B

and the effect of the coefficients of Pj
disappear, if I consider powerseries with
the bernoulli-numbers as coefficients:

 f(x) = b0 + b1*x + b2*x^2 + b3*x^3 + ...
      =    b0
         + (1*b0 - 1*b1)*x
         + (1*b0 - 2*b1 + 1*b2)*x^2
         + ...

---------
Now incidently I also found, that, using J as a diagonal-
unit-matrix with units with alternating signs
 J = diag(1,-1,1,-1,...)

such that also

 Pj  = P * J

then a matrix G, containing the bernoulli-numbers
in the first column (and shifted and scaled
in a certain manner in the follwing columns)

 Pj = G * J * G^-1

and thus, for instance, the above eigensequence-
property of the bernoulli-numbers can be extended
to an infinite set of coefficients arising from

   V(x) * Pj * G = V(x) * G * J

in the infinite number of columns of the result.

The second column, for instance adds to the above
f(x) the function g(x)

 g(x) =    1*b0*x + 2*b1*x^2 + 3*b2*x^3 + 4*b3*x^4 + ...
      = -  (    1* 0
             + (1* 0 - 1*1*b0)*x
             + (1* 0 - 2*1*b0 + 1*2*b1)*x^2
             + (1* 0 - 3*1*b0 + 3*2*b1 -  1*3*b2)*x^3
             + ...
      = -  (
               ( - 1*b0)*x
             + ( - 2*b0 + 2*b1)*x^2
             + ( - 3*b0 + 6*b1 -  3*b2)*x^3
             + ...

or, using x=1, the said eigensequence-operation, here
with the associated eigenvalue -1.

-------------------------------------
Even more interesting seems to me, that the said matrix
G can be decomposed into an eigensystem, where

  G = St2 * Z(1) * St2^-1

with St2 = matrix of stirling-numbers of 2'nd kind
     Z(s)= diag(1,  1/2^s,  1/3^s, 1/4^s,... )
the diagonal-matrix of the zeta-summands of power s.

That means, that even the bernoulli-numbers have
eigensequences related to them (even infinite many),
namely the columns of St2 , associated with the
eigenvalues (1,1/2,1/3,...) such that

  V(x) * G * St2*Z(1) = V(x)*St2

and for instance,

  f(x) = 1 + 1*x + 1*x^2 + 1*x^3 + ....  // taken from V(x)*St2

       =       1*b0                     // taken from V(x) * G * St2 * Z(1)
          + (1/2*b0 + b1)*m
          + (1/3*b0 + b1 +     b2 )*m^2
          + (1/4*b0 + b1 + 3/2*b2 +      b3 )*m^3
          + (1/5*b0 + b1 +   2*b2 +    2*b3 +     b4)*m^4
          + (1/6*b0 + b1 + 5/2*b2 + 10/3*b3 + 5/2*b4 +      b5)*m^5
           ...

which may look kind trivial, since the involved Stirling-
coefficients are all 1
Other coefficients occur in the next columns 1,2,3,..., for instance
in column 1:

   g(x)  =.  x + S2_1*x^2 + S3_1*x^3 + S4_1*x^4 + S5_1*x^5 + ....
         =              b0 * x
             ( 2/3*S2_1*b0 +      2*b1 ) * x^2
             ( 1/2*S3_1*b0 + 2*S2_1*b1 + 3     *b2 )*m^3
             ( 2/5*S4_1*b0 + 2*S3_1*b1 + 4*S2_1*b2 +    4     *b3 ) *m^4
             ( 1/3*S5_1*b0 + 2*S4_1*b1 + 5*S3_1*b2 + 20/3*S2_1*b3  +  5*b4) *m^5

and so on for the next columns.
With the involved Stirling-numbers evaluated we have then for this 2'nd
column:

   g(x)  =   .    2*x + 6*x^2 + 14*x^3 + 30*x^4  +  62*x^5
         =          2*b0 * x
            +   (   4*b0 +  4*b1)*x^2
            +   (   7*b0 + 12*b1 +  6*b2) *x^3
            +   (  12*b0 + 28*b1 + 24*b2 +  8*b3) *m^4
            +   (62/3*b0 + 60*b1 + 70*b2 + 40*b3 + 10*b4))))*m^5
              ...
which, if the bernoulli-numbers are inserted is obvious.
Using x=1 and written as sequences, we have another
eigensequence-expression and implicitely infinitely many.

This description was animated by Neil's article in the
integer-sequences-journal, which I came across recently -
though I do not know, if the above is actually of interest
related to the article (my current context is originally
the study of divergent summation, where these identities
popped up)

Regards-

Gottfried Helms







From Annette.Warlich at t-online.de  Wed Nov 22 13:46:11 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Wed, 22 Nov 2006 13:46:11 +0100
Subject: Eigensequences /editing corrections
Message-ID: <45644693.1000503@t-online.de>

// upps. In the previous article I've overlooked some
// needed editings, replacing the m,m^2,... by x,x^2 ...
// sorry. Repost here:


In  a recent treatise of the pascal matrix and its
connections to the bernoulli-numbers I restated
their connection as an eigenvector/eigensequence-
relation.

Say P is the lower triangular matrix of binomial-
coefficients, Pj the matrix, where the column-signs
alternate and B the vector of bernoulli-numbers
(b_1 = +1/2)
then
   Pj * B = B

Now in connection with the subject of divergent
summmation I investigated powerseries, say
  V(x) = [1,x,x^2,x^3,... ]
in connection with Pj, like

  ??? = Pj * V(x)      // binomial-transform
  ??? = V(x)~ * Pj

(using ~ for the transpose)

Now since

   Pj * B = B

also

   V(x) * Pj * B = V(x) * B

and the effect of the coefficients of Pj
disappear, if I consider powerseries with
the bernoulli-numbers as coefficients:

 f(x) = b0 + b1*x + b2*x^2 + b3*x^3 + ...
      =    b0
         + (1*b0 - 1*b1)*x
         + (1*b0 - 2*b1 + 1*b2)*x^2
         + ...

---------
Now incidently I also found, that, using J as a diagonal-
unit-matrix with units with alternating signs
 J = diag(1,-1,1,-1,...)

such that also

 Pj  = P * J

then a matrix G, containing the bernoulli-numbers
in the first column (and shifted and scaled
in a certain manner in the follwing columns)

 Pj = G * J * G^-1

and thus, for instance, the above eigensequence-
property of the bernoulli-numbers can be extended
to an infinite set of coefficients arising from

   V(x) * Pj * G = V(x) * G * J

in the infinite number of columns of the result.

The second column, for instance adds to the above
f(x) the function g(x)

 g(x) =    1*b0*x + 2*b1*x^2 + 3*b2*x^3 + 4*b3*x^4 + ...
      = -  (    1* 0
             + (1* 0 - 1*1*b0)*x
             + (1* 0 - 2*1*b0 + 1*2*b1)*x^2
             + (1* 0 - 3*1*b0 + 3*2*b1 -  1*3*b2)*x^3
             + ...
      = -  (
               ( - 1*b0)*x
             + ( - 2*b0 + 2*b1)*x^2
             + ( - 3*b0 + 6*b1 -  3*b2)*x^3
             + ...

or, using x=1, the said eigensequence-operation, here
with the associated eigenvalue -1.

-------------------------------------
Even more interesting seems to me, that the said matrix
G can be decomposed into an eigensystem, where

  G = St2 * Z(1) * St2^-1

with St2 = matrix of stirling-numbers of 2'nd kind
     Z(s)= diag(1,  1/2^s,  1/3^s, 1/4^s,... )
the diagonal-matrix of the zeta-summands of power s.

That means, that even the bernoulli-numbers have
eigensequences related to them (even infinite many),
namely the columns of St2 , associated with the
eigenvalues (1,1/2,1/3,...) such that

  V(x) * G * St2*Z(1) = V(x)*St2

and for instance,

  f(x) = 1 + 1*x + 1*x^2 + 1*x^3 + ....  // taken from V(x)*St2

       =       1*b0                     // taken from V(x) * G * St2 * Z(1)
          + (1/2*b0 + b1)*x
          + (1/3*b0 + b1 +     b2 )*x^2
          + (1/4*b0 + b1 + 3/2*b2 +      b3 )*x^3
          + (1/5*b0 + b1 +   2*b2 +    2*b3 +     b4)*x^4
          + (1/6*b0 + b1 + 5/2*b2 + 10/3*b3 + 5/2*b4 +      b5)*x^5
           ...

which may look kind trivial, since the involved Stirling-
coefficients are all 1
Other coefficients occur in the next columns 1,2,3,..., for instance
in column 1:

   g(x)  =.  x + S2_1*x^2 + S3_1*x^3 + S4_1*x^4 + S5_1*x^5 + ....
         =              b0 * x
             ( 2/3*S2_1*b0 +      2*b1 ) * x^2
             ( 1/2*S3_1*b0 + 2*S2_1*b1 + 3     *b2 )*x^3
             ( 2/5*S4_1*b0 + 2*S3_1*b1 + 4*S2_1*b2 +    4     *b3 ) *x^4
             ( 1/3*S5_1*b0 + 2*S4_1*b1 + 5*S3_1*b2 + 20/3*S2_1*b3  +  5*b4) *x^5

and so on for the next columns.
With the involved Stirling-numbers evaluated we have then for this 2'nd
column:

   g(x)  =   .    2*x + 6*x^2 + 14*x^3 + 30*x^4  +  62*x^5
         =          2*b0 * x
            +   (   4*b0 +  4*b1)*x^2
            +   (   7*b0 + 12*b1 +  6*b2) *x^3
            +   (  12*b0 + 28*b1 + 24*b2 +  8*b3) *x^4
            +   (62/3*b0 + 60*b1 + 70*b2 + 40*b3 + 10*b4) *x^5
              ...
which, if the bernoulli-numbers are inserted is obvious.
Using x=1 and written as sequences, we have another
eigensequence-expression and implicitely infinitely many.

This description was animated by Neil's article in the
integer-sequences-journal, which I came across recently -
though I do not know, if the above is actually of interest
related to the article (my current context is originally
the study of divergent summation, where these identities
popped up)

Regards-

Gottfried Helms








From njas at research.att.com  Wed Nov 22 17:58:32 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 22 Nov 2006 11:58:32 -0500 (EST)
Subject: A125097.gif: (Very) nice patterns
References: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>
Message-ID: <200611221658.LAA41364@fry.research.att.com>

The other day Zak said:
> Neil,
> seqfans,
>
> 1. Have a look at this full graph -
> aint these patterns (very) nice?!
> It may be known, but still it's nice...
>
> 2. Is it possible to
> put "the extended graph" into OEIS?
> In the standard graph no patterns are seen at all...
>
> 3. Neil removed  keyword "nice" from A125097
> (as he did regularly with ny sequences)-
> as if I insist that it's me who is "nice"
> not it is the seq which is nice.
>

Here is my reply:
1. i will use the b-file
2. we don't need the gif since the graph button will use the b-file
3. the definition of the sequence is
     a(n) is k such that k*(2n+1) mod 1000 = 123, or 2n+1 if no such k exists.
Now there is rule that a sequence should not depend on an arbitrary 
parameter. Here we have two of them. Why not:
a(n) is k such that k*(2n+1) mod 44445555 = 321345, or 2n+1 if no such k exists. ?
4. the sequence is not finite (a(1000) = 123)
5. although i know you are a nice guy, i do not think this sequence is nice.
Best regards
Neil





From grafix at csl.pl  Wed Nov 22 18:58:30 2006
From: grafix at csl.pl (Artur)
Date: Wed, 22 Nov 2006 18:58:30 +0100
Subject: All primes
In-Reply-To: <200611221658.LAA41364@fry.research.att.com>
References: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com> <200611221658.LAA41364@fry.research.att.com>
Message-ID: <op.tjfivsouat030q@tata>

Dear Members
I'm looking for members which have Mathematica Programme and can help in
finding the first number such that
(PrimeQ[2(n) + 1] == True) && (PrimeQ[4(
       n) + 1] == True) && (PrimeQ[6(n) + 1] == True) && (PrimeQ[8(
       n) + 1] == True) && (PrimeQ[10(n) + 1] == True) && (PrimeQ[12(
       n) + 1] == True) && (PrimeQ[14(n) + 1] == True) && (PrimeQ[16(
       n) + 1] == True) && (PrimeQ[18(n) + 1] == True) && (PrimeQ[20(
       n) + 1] == True) && (PrimeQ[22(n) + 1] == True)
these n is bigger than 4850000000000
I have Mathematica procedure which is 4000 much more quickest as above but  
for as big numbers PrimeQ working slow and will be good divided range of  
searching on few subranges and run each other on different computer.
If somebody can help let me know. I need information how long time  
computer can run without reset.
BEST REGARDS
ADRTUR JASINSKI






From rgwv at rgwv.com  Wed Nov 22 19:22:17 2006
From: rgwv at rgwv.com (Robert G. Wilson v)
Date: Wed, 22 Nov 2006 13:22:17 -0500
Subject: Whereabouts of Joe K. Crump & www.spacefire.com
Message-ID: <45649559.9030304@rgwv.com>

,

Does anyone on this list know the whereabouts of Joe K. Crump or the
status of the website www.spacefire.com ? Thanks in advance for any
information.

Bob.





From noe at sspectra.com  Wed Nov 22 19:27:50 2006
From: noe at sspectra.com (T. D. Noe)
Date: Wed, 22 Nov 2006 10:27:50 -0800
Subject: All primes
In-Reply-To: <op.tjfivsouat030q@tata>
References: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>
 <200611221658.LAA41364@fry.research.att.com> <op.tjfivsouat030q@tata>
Message-ID: <p06230905c18a46fef974@[192.168.1.43]>

At 7:04 PM +0100 11/22/06, Artur wrote:
>Dear Members
>I'm looking for members which have Mathematica Programme and can help in
>finding the first number such that
>(PrimeQ[2(n) + 1] == True) && (PrimeQ[4(
>       n) + 1] == True) && (PrimeQ[6(n) + 1] == True) && (PrimeQ[8(
>       n) + 1] == True) && (PrimeQ[10(n) + 1] == True) && (PrimeQ[12(
>       n) + 1] == True) && (PrimeQ[14(n) + 1] == True) && (PrimeQ[16(
>       n) + 1] == True) && (PrimeQ[18(n) + 1] == True) && (PrimeQ[20(
>       n) + 1] == True) && (PrimeQ[22(n) + 1] == True)
>these n is bigger than 4850000000000

See A071576.

Tony






From mathar at strw.leidenuniv.nl  Wed Nov 22 19:39:07 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Wed, 22 Nov 2006 19:39:07 +0100
Subject: Whereabouts of Joe K. Crump & www.spacefire.com
Message-ID: <200611221839.kAMId7db031776@amer.strw.leidenuniv.nl>


> From seqfan-owner at ext.jussieu.fr  Wed Nov 22 19:09:21 2006
> Date: Wed, 22 Nov 2006 13:22:17 -0500
> From: "Robert G. Wilson v" <rgwv at rgwv.com>
> To: "seqfan at ext.jussieu.fr" <SeqFan at ext.jussieu.fr>
> Subject: Whereabouts of Joe K. Crump & www.spacefire.com
> 
> Does anyone on this list know the whereabouts of Joe K. Crump or the
> status of the website www.spacefire.com ? Thanks in advance for any
> information.
..
According to Max (whom we now know as Alekseyev), there is a 
archive.org backup of his web page:

http://groups-beta.google.com/group/sci.math/browse_thread/thread/8d061b676bd7e32b/071ae01f53f9a1f5?hl=en#071ae01f53f9a1f5
(sci.math of Nov 14)
http://web.archive.org/web/20040609202646/www.spacefire.com/numbertheory/

An e-mail address of 2002 is shown in
http://www.primepuzzles.net/thepuzzlers/Crump.htm
and becomes
ssaywer1 at rr.carolina.com
in the archived web page of 2000.





From franktaw at netscape.net  Wed Nov 22 22:20:32 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 22 Nov 2006 16:20:32 -0500
Subject: Categories
Message-ID: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>

How many categories are there?

First, how many categories are there with n morphisms and k objects?  
This table starts:

 1
 2  1
 7  3 1
35 16 3 1

The first column is A058129, the number of monoids; the main diagonal 
is all 1's.  I am not
100% certain of the 16 in the final row.

Taking the row sums, we get:

1,3,11,55

the number of categories with n morphisms.  This is probably not in the 
OEIS (only
A001776 is possible - other matches become less than A058129).  The 
inverse Euler
transform,

1,2,8,41

is the number of connected categories with n morphisms; this is 
likewise probably not
in the OEIS (only A052447 is possible).

Can somebody generate more data?

Franklin T. Adams-Watters

A category is a collection of objects and morphisms; each morphism is 
from one object
to another (not necessarily different) object.  Where the destination 
of one morphism
is the source of a second, their composition is defined; composition is 
associative where
it is defined.  Each object has an identity morphism, which connects it 
to itself; this
is an identity when composed with morphisms coming in and with 
morphisms going
out.
________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From jvospost3 at gmail.com  Wed Nov 22 23:55:44 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 22 Nov 2006 14:55:44 -0800
Subject: Categories
In-Reply-To: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
Message-ID: <5542af940611221455r6e693b45kae8a9af38c6ee9eb@mail.gmail.com>

This actually relates to the enumeration results I've made on "prime"
endofunctions, namely endofunctions which are not unions of other prime
endofunctions nor categorical products of endofunctions. My partial results
here are in A125024, A124023, and a comment on A048888 that I submitted a
day or two ago about the enumeration of prime endofunctions in terms of
numbrals.

I have been working on enumerating not just endofunctions, but
(categorically) endomorphisms of endofunctions, (2-categorically)
endomorphisms of endomorphisms of endofunctions.  I've been posting partial
results on John Baez's "n-category cafe" and apparently just annoyed John
Baez, who didn't yet see that I was finding a new combinatorial result at
the base level (endofunctions) and working on the n-category hierarchy above
that, so those comments may already have been deleted from his blog.

That is, in counting all categories on n objects, most are trivially the
union of smaller disjoint categories (that is, the sagittal digraph is not
connected), or categorical products of smaller categories.  They key to
enumeration involves thus both a tree-enumeration, and the numbral part
which, per a recent paper in the Journal on Integer Sequences, relate to
balanced ordered trees.

Hard to summarize less cryptically, here, but I stand by in support of what
you're doing, and think my approach is genuinely related.

-- Jonathan Vos Post

On 11/22/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
> How many categories are there?
>
> First, how many categories are there with n morphisms and k objects?
> This table starts:
>
> 1
> 2  1
> 7  3 1
> 35 16 3 1
>
> The first column is A058129, the number of monoids; the main diagonal
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
>
> Taking the row sums, we get:
>
> 1,3,11,55
>
> the number of categories with n morphisms.  This is probably not in the
> OEIS (only
> A001776 is possible - other matches become less than A058129).  The
> inverse Euler
> transform,
>
> 1,2,8,41
>
> is the number of connected categories with n morphisms; this is
> likewise probably not
> in the OEIS (only A052447 is possible).
>
> Can somebody generate more data?
>
> Franklin T. Adams-Watters
>
> A category is a collection of objects and morphisms; each morphism is
> from one object
> to another (not necessarily different) object.  Where the destination
> of one morphism
> is the source of a second, their composition is defined; composition is
> associative where
> it is defined.  Each object has an identity morphism, which connects it
> to itself; this
> is an identity when composed with morphisms coming in and with
> morphisms going
> out.
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and
> industry-leading spam and email virus protection.
>
>
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From jvospost3 at gmail.com  Thu Nov 23 01:39:16 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 22 Nov 2006 16:39:16 -0800
Subject: Categories
In-Reply-To: <5542af940611221455r6e693b45kae8a9af38c6ee9eb@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <5542af940611221455r6e693b45kae8a9af38c6ee9eb@mail.gmail.com>
Message-ID: <5542af940611221639n2fde8e43j93dc69594ae9392b@mail.gmail.com>

Next question is thus: "How many 2-categories (bicategories) are there whose
objects are the categories on n objects?"

See:
Basic Bicategories
Tom Leinster (1998-10-04)  arXiv.org:math/9810017 A concise guide to very
basic bicategory theory, from the definition of a bicategory to the
coherence theorem. Comment: 11 pages; LaTeX 2e with Paul Taylor's diagram
macros

This is even harder to answer, because there seem to be at least 7 different
definitions of n-categories, of which categories = 1-categories, and I've
just raised the enumeration probem for n=2.

www.ima.umn.edu/categories/abstracts.html
Abstracts for the IMA 2004 Summer Program: n-Categories.
Ross Street has proposed two related definitions of n-category, one in 1987
and one in 2002.

See also:
Tom Leinster (2003-05-02) arXiv.org:math/0305049 Higher-dimensional category
theory is the study of n-categories, operads, braided monoidal categories,
and other such exotic structures. It draws its inspiration from areas as
diverse as topology, quantum algebra, mathematical physics, logic, and
theoretical computer science. This is the first book on ... Comment: Book,
410 pages

http://www.lepp.cornell.edu/spr/2002-04/msg0041086.html

In the beginning, there was nothing.
But with nothing came nothingness, the vacuity, which was something.
So now there was nothingness and somethingness,
vacuity and triviality, falsehood and truth,
which were 2 things.
So now there was a set of nothingness and somethingness,
and the whole realm of sets of elements
sprang up out of the set of nothingness and somethingness,
and sprouted functions between them
to relate them back to the set of nothingness and somethingness.
So now there was a category of sets and functions,
and the whole realm of categories of objects and morphisms
sprang up out of the category of sets and functions,
and sprouted functors between them,
to relate them back to the category of sets and functions,
which sprouted natural transformations between *them*,
to relate the relationships.
So now there was ...


-- Toby
   toby at math.ucr.edu

But your question is fundamental, and I am eager to see an answer.  I hope
it is not rude for me to be anticipating follow-ups.

Happy Thanksgiving,

Jonathan Vos Post

On 11/22/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> This actually relates to the enumeration results I've made on "prime"
> endofunctions, namely endofunctions which are not unions of other prime
> endofunctions nor categorical products of endofunctions. My partial results
> here are in A125024, A124023, and a comment on A048888 that I submitted a
> day or two ago about the enumeration of prime endofunctions in terms of
> numbrals.
>
> I have been working on enumerating not just endofunctions, but
> (categorically) endomorphisms of endofunctions, (2-categorically)
> endomorphisms of endomorphisms of endofunctions.  I've been posting partial
> results on John Baez's "n-category cafe" and apparently just annoyed John
> Baez, who didn't yet see that I was finding a new combinatorial result at
> the base level (endofunctions) and working on the n-category hierarchy above
> that, so those comments may already have been deleted from his blog.
>
> That is, in counting all categories on n objects, most are trivially the
> union of smaller disjoint categories (that is, the sagittal digraph is not
> connected), or categorical products of smaller categories.  They key to
> enumeration involves thus both a tree-enumeration, and the numbral part
> which, per a recent paper in the Journal on Integer Sequences, relate to
> balanced ordered trees.
>
> Hard to summarize less cryptically, here, but I stand by in support of
> what you're doing, and think my approach is genuinely related.
>
> -- Jonathan Vos Post
>
> On 11/22/06, franktaw at netscape.net <franktaw at netscape.net > wrote:
> >
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> > 1
> > 2  1
> > 7  3 1
> > 35 16 3 1
> >
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
> > Taking the row sums, we get:
> >
> > 1,3,11,55
> >
> > the number of categories with n morphisms.  This is probably not in the
> > OEIS (only
> > A001776 is possible - other matches become less than A058129).  The
> > inverse Euler
> > transform,
> >
> > 1,2,8,41
> >
> > is the number of connected categories with n morphisms; this is
> > likewise probably not
> > in the OEIS (only A052447 is possible).
> >
> > Can somebody generate more data?
> >
> > Franklin T. Adams-Watters
> >
> > A category is a collection of objects and morphisms; each morphism is
> > from one object
> > to another (not necessarily different) object.  Where the destination
> > of one morphism
> > is the source of a second, their composition is defined; composition is
> > associative where
> > it is defined.  Each object has an identity morphism, which connects it
> > to itself; this
> > is an identity when composed with morphisms coming in and with
> > morphisms going
> > out.
> > ________________________________________________________________________
> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > industry-leading spam and email virus protection.
> >
> >
>
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From maxale at gmail.com  Thu Nov 23 04:52:55 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 22 Nov 2006 19:52:55 -0800
Subject: Categories
In-Reply-To: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
Message-ID: <d3dac270611221952h2853749ev8fcb5ced9e2f80a7@mail.gmail.com>

On 11/22/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
> How many categories are there?
>
> First, how many categories are there with n morphisms and k objects?

Does that correspond to transitively closed digraphs with k (labeled?)
vertices and n edges?

Max





From maxale at gmail.com  Thu Nov 23 05:10:51 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 22 Nov 2006 20:10:51 -0800
Subject: Whereabouts of Joe K. Crump & www.spacefire.com
In-Reply-To: <200611221839.kAMId7db031776@amer.strw.leidenuniv.nl>
References: <200611221839.kAMId7db031776@amer.strw.leidenuniv.nl>
Message-ID: <d3dac270611222010m17a14a53x75b9de0cf2ffbe3a@mail.gmail.com>

On 11/22/06, Richard Mathar <mathar at strw.leidenuniv.nl> wrote:

> http://web.archive.org/web/20040609202646/www.spacefire.com/numbertheory/

I've put an updated version of Joe's Database of 2^x-3 factorizations at
http://www-cse.ucsd.edu/users/maxal/2nm3_db.txt

Please contribute if you like.

Max





From pauldhanna at juno.com  Thu Nov 23 07:19:41 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Thu, 23 Nov 2006 01:19:41 -0500
Subject: Formula for A112319? - Diagonals in Self-Compositions of (x+x^2)
Message-ID: <20061123.011941.272.1.pauldhanna@juno.com>

Seqfans, 
     Would someone like to try to find a formula for A112319:  
"Coefficients of x^n in the (n-1)-th self-composition of (x + x^2) for
n>=1."
1, 1, 2, 9, 64, 630, 7916, 121023, 2179556, 45179508, 1059312264, 
 
The related triangles A122888 and A122890 seem to indicate that 
A112319 may have a formula involving the Catalan numbers: 
http://www.research.att.com/~njas/sequences/A122890
 
Below I copy A112319 and triangle A122888 for your convenience. 
Other diagonals in self-compositions of (x+x^2) are A112317, A112320.
 
Be it a recurrence, series reversion, etc., any formula for these
diagonals 
could lead to other discoveries involving self-compositions of functions.

     Paul 
----------------------------------------------------------------

A122888
Triangle, read by rows, where row n lists the coefficients of x^k, 
k=1..2^n, in the n-th self-composition of (x + x^2) for n>=0. 

EXAMPLE  
Triangle begins: 
1; 
1, 1; 
1, 2, 2, 1; 
1, 3, 6, 9, 10, 8, 4, 1; 
1, 4, 12, 30, 64, 118, 188, 258, 302, 298, 244, 162, 84, 32, 8, 1; 

Multiplying the g.f. of column k by (1-x)^k, k>=1, with leading zeros, 
yields the g.f. of row k in the triangle A122890: 
1; 
0, 1; 
0, 0, 2; 
0, 0, 1, 5; 
0, 0, 0, 10, 14; 
0, 0, 0, 8, 70, 42; 
0, 0, 0, 4, 160, 424, 132; 
0, 0, 0, 1, 250, 1978, 2382, 429; 
0, 0, 0, 0, 302, 6276, 19508, 12804, 1430; ... 

in which the main diagonal is the Catalan numbers 
and the row sums form the factorials. 

----------------------------------------------------------------

A112319  
Coefficients of x^n in the (n-1)-th self-composition of (x + x^2) for
n>=1. 

1, 1, 2, 9, 64, 630, 7916, 121023, 2179556, 45179508, 1059312264, 
27715541568, 800423573676, 25289923553700, 867723362137464, 
32128443862364255, 1276818947065793736, 54208515369076658640, 
2448636361058495090816
 
FORMULA  
a(n) = [x^n] F_{n-1}(x) where F_n(x) = F_{n-1}(x+x^2) 
with F_1(x) = x+x^2 and F_0(x)=x for n>=1. 
 
EXAMPLE  
Initial terms in self-compositions of (x+x^2) are: 
F(x) = x + (1)*x^2 
F(F(x)) = x + 2*x^2 + (2)*x^3 + x^4 
F(F(F(x))) = x + 3*x^2 + 6*x^3+ (9)*x^4 +... 
F(F(F(F(x)))) = x + 4*x^2 + 12*x^3 + 30*x^4 + (64)*x^5 +... 
F(F(F(F(F(x))))) = x + 5*x^2 + 20*x^3 + 70*x^4 + 220*x^5 + (630)*x^6 +...
 
----------------------------------------------------------------
END.
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From jvospost3 at gmail.com  Thu Nov 23 07:34:45 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 22 Nov 2006 22:34:45 -0800
Subject: Categories
In-Reply-To: <d3dac270611221952h2853749ev8fcb5ced9e2f80a7@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <d3dac270611221952h2853749ev8fcb5ced9e2f80a7@mail.gmail.com>
Message-ID: <5542af940611222234r3ae28d48h1c3e6541143896a7@mail.gmail.com>

First, John Baez agrees that an earlier post of his partially answered my
follow-up question about 2-categories (some assembly required):

http://golem.ph.utexas.edu/category/2006/10/klein_2geometry_vi.html#c005157

 Re: Klein 2-Geometry VI

Tim writes:

So I'll wrap up and say what I think I've discovered. Just as
finite-dimensional vector spaces and their subspaces are intimately related
to finite sets and their subsets, so finite-dimensional 2-vector spaces and
their 2-subspaces are intimately related to finite graphs and their
subgraphs. These are obviously much more complicated and I'm not at all
confident my analyses here are correct, particularly the identification of
the various 2-Grassmannians associated to various graph maps. The basic idea
seems sound though.

Yeah, that's a great idea!

The way I see it, your fundamental idea is this. There's a 2-functor from
the 2-category of directed graphs to the 2-category of vector 2-spaces
(?2-term chain complexes). And, it goes like this:

   - Any directed graph gives a 2-term chain complex.
   - Any map between directed graphs gives a chain map between 2-term
   chain complexes.
   - And, there's also a kind of "homotopy between maps between directed
   graphs", which gives a chain homotopy between chain maps between 2-term
   chain complexes.

(Do graph theorists ever think about those homotopies? They should!)

All this generalizes further, to an (n+1)-functor from n-dimensional cell
complexes to (n+1)-term chain complexes.

I'm not sure how to get some amazing new insights into projective n-geometry
from this way of thinking - but that's not surprising, since I just read
your post 2 minutes ago. We should mull on it.

Thanks!
Posted by: John Baez <http://math.ucr.edu/home/baez/> on October 8, 2006
7:25 PM |

Second, I took Category theory in grad school 1976 or 1977, and forgot what
little I knew, and have relearned only some. So anyone in seqfans who
learned more and remebered more should jump in here, and don't worry about
making me look foolish, as I take the risk of exposing my ignorance every
time I post here.

If one asks: "how many categories are there with n morphisms and k objects?"
one is asking up to isomorphism. I'm making a first cut here off the top of
my head, which may be no more than an ill-informed guess.

The sagittal graph of a general morphism a.k.a. homomorphism (not restricted
to a monomorphism, epimorphism, isomorphism, endomorphism, or automorphism)
leaves us wondering what the structure of the objects are.  I think, by the
way that the question is asked, that they have no structure, i.e. that they
are just points. If they have structure, that affects everything.
unstructrured objects (points), then we still get different counts if the
vertices and arcs are labeled or unlabeled, and whether one allows loops or
not. if the edges are unlabeled, and loops are allowed, then any of the k
objects can be mapped to any of the k objects including itself, giving k^k,
if n = k.  If n>k then we can 't use more than k of the n because the
mapping must be functional, that is, only one arrow out of each point (but
multiple arrows to a point is allowed).

 If n<k then the first point can be mapped to any of the k points, and we
run out of arrows when we reach k^n.  Now we see why labeling matters, as if
they are unlabeled, we count only once in equivalence classes, but if
labeled each such mapping is distinct.  Each loop and arc is de facto
labeled by its source point and target point, as an ordered pair.  If you
want to allow other labels, you have to multiply by the proper combinational
multiplier.

So we're taking a summation of n=0 (just the k points and none map to any),
n=1 (one of k points maps to one of k), n=2 and so forth.

That starts the answer.  For k=n and any point can map to any, we have the
count of endomorphisms being the same as the count of endofunctions already
in OEIS.

Category theory has differing approaches depending on whether we have
digraphs, multgraphs,  reversible digraphs, reflexive digraphs, irreflexive
digraphs, and various tricks to represent undirected graphs with directed
graphs.

So exact definitions are necessary.

Sorry, just a first cut.  Not enough sleep.  Some time tomorrow before gests
and Thanksgiving dinner kicks in, but no promises.

Thanks for the dialog.

Best,

Jonathan Vos Post
<http://golem.ph.utexas.edu/cgi-bin/MT-3.0/sxp-comments.pl?entry_id=955;parent_id=5157>


On 11/22/06, Max A. <maxale at gmail.com> wrote:
>
> On 11/22/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
>
> Does that correspond to transitively closed digraphs with k (labeled?)
> vertices and n edges?
>
> Max
>
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From Roland.Bacher at ujf-grenoble.fr  Thu Nov 23 07:54:39 2006
From: Roland.Bacher at ujf-grenoble.fr (Roland Bacher)
Date: Thu, 23 Nov 2006 07:54:39 +0100
Subject: Categories
In-Reply-To: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
Message-ID: <20061123065439.GA28473@fourier.ujf-grenoble.fr>


This numbers can also be obtained as the numbers
of the following quotients of quivers:

Associate to a category with n morphisms and k objects 
the quiver with k vertices corresponding to the objects 
and a(X,Y) arrows directed from the object X to the object
Y if there are a(X,Y) morphisms from X into Y.

Given two morphism g:X-->Y, f:Y-->Z with composition
h=f o g:X-->Z, put the relation gf=h on the quiver algebra

The resulting quotient algebra has dimension n and a basis
given by the (simple) arrows.

This leads to the following "algorithm" for enumerating
all categories with n morphisms and k objects:

(a) enumerate all quivers (directed graphs) with k vertices and
n oriented edges.

(b) Associate the following "weight" to such a quiver as follows :

(b1)given a triplet of vertices X,Y,Z,
set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
(this counts the number of ways which associate a morphism
X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the 
number of oriented arrows starting at U and ending at W.

(b2) associate to a given quiver the weight 
\prod_{X,Y,Z} w(X,Y,Z)
where the product is over all triplets of vertices.

The total sum of such weighted quivers yields then the solution.

Roland Bacher



   


On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.net wrote:
> How many categories are there?
> 
> First, how many categories are there with n morphisms and k objects?  
> This table starts:
> 
> 1
> 2  1
> 7  3 1
> 35 16 3 1
> 
> The first column is A058129, the number of monoids; the main diagonal 
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
> 
> Taking the row sums, we get:
> 
> 1,3,11,55
> 
> the number of categories with n morphisms.  This is probably not in the 
> OEIS (only
> A001776 is possible - other matches become less than A058129).  The 
> inverse Euler
> transform,
> 
> 1,2,8,41
> 
> is the number of connected categories with n morphisms; this is 
> likewise probably not
> in the OEIS (only A052447 is possible).
> 
> Can somebody generate more data?
> 
> Franklin T. Adams-Watters
> 
> A category is a collection of objects and morphisms; each morphism is 
> from one object
> to another (not necessarily different) object.  Where the destination 
> of one morphism
> is the source of a second, their composition is defined; composition is 
> associative where
> it is defined.  Each object has an identity morphism, which connects it 
> to itself; this
> is an identity when composed with morphisms coming in and with 
> morphisms going
> out.
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and 
> industry-leading spam and email virus protection.
> 





From jvospost3 at gmail.com  Thu Nov 23 18:14:47 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 23 Nov 2006 09:14:47 -0800
Subject: Categories
In-Reply-To: <20061123065439.GA28473@fourier.ujf-grenoble.fr>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
Message-ID: <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>

Roland Bacher correctly refers to quivers as summarized in wikipedia at

http://en.wikipedia.org/wiki/Quiver_%28mathematics%29

On 11/22/06, Roland Bacher <Roland.Bacher at ujf-grenoble.fr> wrote:
>
>
> This numbers can also be obtained as the numbers
> of the following quotients of quivers:
>
> Associate to a category with n morphisms and k objects
> the quiver with k vertices corresponding to the objects
> and a(X,Y) arrows directed from the object X to the object
> Y if there are a(X,Y) morphisms from X into Y.
>
> Given two morphism g:X-->Y, f:Y-->Z with composition
> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
>
> The resulting quotient algebra has dimension n and a basis
> given by the (simple) arrows.
>
> This leads to the following "algorithm" for enumerating
> all categories with n morphisms and k objects:
>
> (a) enumerate all quivers (directed graphs) with k vertices and
> n oriented edges.
>
> (b) Associate the following "weight" to such a quiver as follows :
>
> (b1)given a triplet of vertices X,Y,Z,
> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> (this counts the number of ways which associate a morphism
> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> number of oriented arrows starting at U and ending at W.
>
> (b2) associate to a given quiver the weight
> \prod_{X,Y,Z} w(X,Y,Z)
> where the product is over all triplets of vertices.
>
> The total sum of such weighted quivers yields then the solution.
>
> Roland Bacher
>
>
>
>
>
>
> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.net wrote:
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> > 1
> > 2  1
> > 7  3 1
> > 35 16 3 1
> >
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
> > Taking the row sums, we get:
> >
> > 1,3,11,55
> >
> > the number of categories with n morphisms.  This is probably not in the
> > OEIS (only
> > A001776 is possible - other matches become less than A058129).  The
> > inverse Euler
> > transform,
> >
> > 1,2,8,41
> >
> > is the number of connected categories with n morphisms; this is
> > likewise probably not
> > in the OEIS (only A052447 is possible).
> >
> > Can somebody generate more data?
> >
> > Franklin T. Adams-Watters
> >
> > A category is a collection of objects and morphisms; each morphism is
> > from one object
> > to another (not necessarily different) object.  Where the destination
> > of one morphism
> > is the source of a second, their composition is defined; composition is
> > associative where
> > it is defined.  Each object has an identity morphism, which connects it
> > to itself; this
> > is an identity when composed with morphisms coming in and with
> > morphisms going
> > out.
> > ________________________________________________________________________
> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > industry-leading spam and email virus protection.
> >
>
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From jvospost3 at gmail.com  Thu Nov 23 21:10:51 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 23 Nov 2006 12:10:51 -0800
Subject: Categories
In-Reply-To: <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
Message-ID: <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>

*"Formalized Proof, Computation, and the Construction Problem in Algebraic
geometry", by Carlos Simpson.

[PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
2004<http://arxiv.org/pdf/math/0410224> File
Format: PDF/Adobe Acrobat - View as
HTML<http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5>
finite integer N, *how many categories* are there with N morphisms? What *
...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
(1965), 450-464.



On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> Roland Bacher correctly refers to quivers as summarized in wikipedia at
>
> http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
>
> On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> >
> >
> > This numbers can also be obtained as the numbers
> > of the following quotients of quivers:
> >
> > Associate to a category with n morphisms and k objects
> > the quiver with k vertices corresponding to the objects
> > and a(X,Y) arrows directed from the object X to the object
> > Y if there are a(X,Y) morphisms from X into Y.
> >
> > Given two morphism g:X-->Y, f:Y-->Z with composition
> > h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> >
> > The resulting quotient algebra has dimension n and a basis
> > given by the (simple) arrows.
> >
> > This leads to the following "algorithm" for enumerating
> > all categories with n morphisms and k objects:
> >
> > (a) enumerate all quivers (directed graphs) with k vertices and
> > n oriented edges.
> >
> > (b) Associate the following "weight" to such a quiver as follows :
> >
> > (b1)given a triplet of vertices X,Y,Z,
> > set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> > (this counts the number of ways which associate a morphism
> > X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> > number of oriented arrows starting at U and ending at W.
> >
> > (b2) associate to a given quiver the weight
> > \prod_{X,Y,Z} w(X,Y,Z)
> > where the product is over all triplets of vertices.
> >
> > The total sum of such weighted quivers yields then the solution.
> >
> > Roland Bacher
> >
> >
> >
> >
> >
> >
> > On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.net wrote:
> > > How many categories are there?
> > >
> > > First, how many categories are there with n morphisms and k objects?
> > > This table starts:
> > >
> > > 1
> > > 2  1
> > > 7  3 1
> > > 35 16 3 1
> > >
> > > The first column is A058129, the number of monoids; the main diagonal
> > > is all 1's.  I am not
> > > 100% certain of the 16 in the final row.
> > >
> > > Taking the row sums, we get:
> > >
> > > 1,3,11,55
> > >
> > > the number of categories with n morphisms.  This is probably not in
> > the
> > > OEIS (only
> > > A001776 is possible - other matches become less than A058129).  The
> > > inverse Euler
> > > transform,
> > >
> > > 1,2,8,41
> > >
> > > is the number of connected categories with n morphisms; this is
> > > likewise probably not
> > > in the OEIS (only A052447 is possible).
> > >
> > > Can somebody generate more data?
> > >
> > > Franklin T. Adams-Watters
> > >
> > > A category is a collection of objects and morphisms; each morphism is
> > > from one object
> > > to another (not necessarily different) object.  Where the destination
> > > of one morphism
> > > is the source of a second, their composition is defined; composition
> > is
> > > associative where
> > > it is defined.  Each object has an identity morphism, which connects
> > it
> > > to itself; this
> > > is an identity when composed with morphisms coming in and with
> > > morphisms going
> > > out.
> > >
> > ________________________________________________________________________
> > > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > > industry-leading spam and email virus protection.
> > >
> >
>
>
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From jvospost3 at gmail.com  Thu Nov 23 21:20:58 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 23 Nov 2006 12:20:58 -0800
Subject: Categories
In-Reply-To: <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
	 <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
Message-ID: <5542af940611231220oe45dce2rbee869c1dea94b9e@mail.gmail.com>

FOM: Set theory vs category theory: Representability of categories *Vaughan
Pratt* pratt at cs.Stanford.EDU
<fom%40cs.nyu.edu?Subject=FOM%3A%20Set%20theory%20vs%20category%20theory%3A%20Representability%20of%20categories&In-Reply-To=199801271058.LAA14876%40brics.dk>
*Tue Jan 27 13:37:42 EST 1998

http://cs.nyu.edu/pipermail/fom/1998-January/001021.html
*



A number of representation theorems were found by Czech algebraists and
category theorists during the latter half of the 1960's, representing
arbitrary small categories as categories of familiar concrete structures
and their homomorphisms.  Two notable representations are in terms
of semigroups, and of directed graphs.  In each case the theorem is
that every small category is representable as a category of semigroups
(directed graphs, etc.) and their homomorphisms (of the standard kind
in each case).

The criteria used here for representability of objects  a  and morphisms
f:a-&gt;b of a category C by respectively structures F(a) and homomorphisms
F(f): F(a)->F(b) of the concrete representing category D are as follows.

(i) F:C->D is a *functor*, that is, a graph homomorphism (viewing the
morphisms of C and D as edges of their respective underlying graphs)
that preserves composition and identities (a condition exactly analogous
to those of monoid homomorphism and group homomorphism).

(ii) F is *faithful*, i.e. for any pair a,b of objects of C, distinct
morphisms f:a-&gt;b of C are represented by distinct homomorphisms
F(f):F(a)->F(b) of D.

(iii) F is *full*, i.e. every homomorphism from F(a) to F(b) in D is
the representation under F of some morphism from a to b in C.

The intuitive meaning of full and faithful is that the representing
object is just as stiff as the object it represents when you wiggle both.
Fullness means it is at least as stiff (D offers no new transformations),
while faithfulness means it is at most as stiff (no transformations are
lost by identification).

Unfortunately the project of embedding arbitrary categories fully and
faithfully in familiar concrete categories has turned out not to have
the impact of its counterparts in group theory (permutations), Boolean
algebras (fields of sets), distributive lattices (rings of sets), etc.
These results have in consequence not been widely publicized in the
introductory categorical literature.

The most objectionable feature of these embeddings in my view (and I
am aware of no other serious objection) is their lack of respect for
concreteness itself.  If one takes for C a small *concrete* category,
such as all quotient groups of the additive group of natural numbers
and their group homomorphisms, the above-mentioned embedding represents
the finite group Z_2 as an infinite semigroup!  And if &quot;natural
numbers&quot;

is replaced by "reals", the corresponding embedding represents Z_2 as an
uncountable semigroup.  Such a lack of respect for concreteness would
seem to undermine the very point of representing abstract objects as
concrete structures.

In contrast a finite group of order n is representable by permutations of
n things, a finite Boolean algebra as a field of subsets of a finite set,
and so on.  In these better-known representation theorems, cardinality
is respected, typically to within at most two exponentials (which in
the infinite case means to within two beth numbers) and often much better.

This unsatisfactory situation with the concrete representation of
objects of arbitrary categories prompts the speculation that the notion
of category is too broad to admit of any improvement in the situation.
After all, any random graph G is made a category by adding to its
edges all paths in G, with composition defined as path concatenation
(the free category generated by G).  How could there possibly be a
cardinality-respecting representation of the vertices of a random graph
in terms of the objects of a familiar concrete category?  When one can
do any violence one wishes to the structure by adding an edge at random
between any two vertices, it would seem that any concrete representation
of each object would inevitably have a cardinality on the order of the
whole category.

But in fact there *is* a representation that does the job.  Let us make
two easily understood modifications to the usual notion of topological
space.

First, drop the requirement that the open sets be closed under arbitrary
union and finite intersection.  (The role of this traditional restriction
is in effect to limit the &quot;signature&quot; of a topological space
to just that
needed to express the notion of limit, or even less with the more discrete
spaces; dropping it greatly broadens the range of possible signatures.)

Second, in place of the customary two degrees of membership in the open
sets (ordinary sets), permit a set K of possible degrees of membership,
call such open sets fuzzy (Zadeh's fuzzy sets are the case K the reals).

Define continuity as usual for functions between topological spaces,
wording the definition in such a way however that its extension to
fuzzy sets is made clear.  Specifically, f:X-&gt;Y is continuous when for
every open set g:Y-&gt;K (here expressed as a characteristic function),
the composite

		   f      g
		X ---> Y ---> K

is an open set of X.  When K = {0,1} it is easily verified that this is
the standard notion of continuity.

Call such generalized topological spaces Chu spaces over K.  An even more
general notion was first studied in detail by Peter Chu in the 1970's,
a master's student of Michael Barr.  The more elementary version above
was first studied by Lafont and Streicher in 1991 (LICS) under the rubric
of games but the term &quot;Chu construction&quot; was already in the computer
science air in the late 1980's making it too late to change the name.

Definitions.

1.  A concrete category is a pair (C,U) where C is a category and U:C-&gt;Set
is a faithful functor, the "forgetful" functor giving the *underlying
set* of each object along with the realization of each morphism of C as
a function (a morphism of Set).

2.  A concrete functor F:(C,U)->(D,V) between two concrete categories
is one satisfying VF = U.  That is, the underlying set VF(a) of the
representation F(a) of a is the same as the underlying set U(a) of
a itself.

Remark.  The elements of all the representing sets of a small concrete
category (C,U) themselves form a single set, namely the disjoint or
marked union of all the sets in the category.  Call this set Elt(C,U),
the set of elements of (C,U).

A very minor technicality requires the notion of an *honest* concrete
category, namely one having, for all objects a and b, a morphism from
a to b if U(a) is empty (necessarily at most one, by concreteness).
Dishonest concrete categories cannot be represented as below because
the empty topological space has nowhere to store the information as to
which other spaces it does or does not have functions to.  (Thanks to
Peter Freyd for pointing out the lacuna in my proof that necessitated
this condition.)

Theorem.  Every small honest concrete category (C,U) embeds fully,
faithfully, and concretely in the category of Chu spaces over Elt(C,U).

(This theorem enjoys all the advantages of the previous embeddings, but
in addition satisfies the strongest possible concreteness requirement:
the representing object has the *same* underlying set as the object it
represents, and the representing morphisms viewed concretely are exactly
the same functions.)

Proof.  Let b be an object of C with underlying set X = U(b).  Represent
b by a generalized topological space X, having one open set Y_h for
each morphism h:b-&gt;c in C for some c.  Form Y_h by taking the degree of
membership of each point x in Y_h to be U(h)(x), i.e. the element of U(c)
which is the image of x under the underlying function U(h) of h.

It is then straightforward to show that this representation is faithful,
full, and concrete.

Obviously the burden of the obstacle that we have overcome has merely
been shifted to the open sets, the number of which must now be on the
order of the cardinality of the category.  But since when has the number
of open sets been an obstacle to topology?  The points of a space are
clearly visible, but our embedding has left these untouched.  Who has
ever seen an open set or cares about how many there are?

For more information on Chu spaces consult

	<A HREF="http://boole.stanford.edu/chuguide.html">http://boole.stanford.edu/chuguide.html</A>

My original interest in Chu spaces was as a model of concurrent
behavior.  More recently I have become interested in them as a primarily
set-theoretic foundation for mathematics that more directly emulates
what category theory has to offer than the extant litany of single-sorted
and and multisorted relational structures and algebras with and without
(ordinary) topology and their standard homomorphisms, all of which are
fully, faithfully, and concretely representable by Chu spaces.

---
Name: Vaughan Pratt
Position: Professor of Computer Science
Institution: Stanford University
Research interests: Foundations of computation and mathematics
For more information: <A
HREF="http://boole.stanford.edu">http://boole.stanford.edu</A>
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From maxale at gmail.com  Thu Nov 23 23:13:45 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 23 Nov 2006 14:13:45 -0800
Subject: connected components in subgraphs of the complete graph: A125205 - A125207
Message-ID: <d3dac270611231413v4ea7e950ifc9405a7ed04e057@mail.gmail.com>

SeqFans,

I've just sent to Neil the following three sequences that you may find
interesting.

It would be also interesting to compute unlabeled variants of these
sequences. If you like to contribute, please do.

Regards,
Max


%I A125205
%S A125205 1,2,1,3,6,3,1,4,18,30,24,15,6,1,5,40,135,250,295,282,215,120,45,10,1,6,
%T A125205 75,420,1385,3015,4800,6365,7170,6705,5065,3009,1365,455,105,15,1,7,126,
%U A125205 1050,5355,18690,47880,96796,166890,251370,329945,373947,362292,297115
%N A125205 Triangular array T(n,k) (n>=1, 0<=k<=n(n-1)/2) giving the
total number
           of connected components in all subgraphs (V,E') with |E'|=k of
           the complete labeled graph K_n=(V,E).
%F A125205 g.f.: Sum_{n,k}
T(n,k)*x^n/n!*y^k=(F(x,y)-1)*exp(F(x,y)-1)=G(x,y)*ln(G(x,y))
           where G(x,y)=Sum_{n=0..oo} (1+y)^(n(n-1)/2)*x^n/n!, and
F(x,y)=1+ln(G(x,y))
           is g.f. of A062734.
%e A125205 The array starts with
           1
           2, 1
           3, 6, 3, 1
           4, 18, 30, 24, 15, 6, 1
           5, 40, 135, 250, 295, 282, 215, 120, 45, 10, 1
           ...
%e A125205 T(3,1)=6 since there are three different subgraphs of K_3
with one edge,
           and each subgraph has two connected components.
%o A125205 (PARI) { reverse(v)=vector(length(v),i,v[length(v)+1-i]) }
           G=sum(n=0,6,(1+y)^(n*(n-1)/2)*x^n/n!); K=G*log(G);
           for(n=1,6,print(reverse(Vec(n!*polcoeff(K,n,x)))))
%Y A125205 Cf. A062734
%Y A125205 Cf. A125206 (row-reversed version), A125207 (row sums)
%K A125205 nonn,tabf
%O A125205 1,2
%A A125205 Max Alekseyev (maxal(AT)cs.ucsd.edu), Nov 23 2006



%I A125206
%S A125206 1,1,2,1,3,6,3,1,6,15,24,30,18,4,1,10,45,120,215,282,295,250,135,40,5,1,
%T A125206 15,105,455,1365,3009,5065,6705,7170,6365,4800,3015,1385,420,75,6,1,21,210,
%U A125206 1330,5985,20349,54271,116385,204225,297115,362292,373947,329945,251370
%N A125206 Triangular array T(n,k) (n>=1, 0<=k<=n(n-1)/2) giving the
total number
           of connected components in all subgraphs obtained from the complete
           labeled graph K_n by removing k edges.
%C A125206 Row-reversed version of A125205, see A125205 for further details
%e A125206 The array starts with
           1
           1, 2
           1, 3, 6, 3
           1, 6, 15, 24, 30, 18, 4
           1, 10, 45, 120, 215, 282, 295, 250, 135, 40, 5
           ...
%Y A125206 Cf. A125205 (row-reversed version), A125207 (row sums)
%K A125206 nonn,tabf
%O A125206 1,3
%A A125206 Max Alekseyev (maxal(AT)cs.ucsd.edu), Nov 23 2006




%I A125207
%S A125207 1,3,13,98,1398,39956,2354240,286394544,71225744048,35884971729760,36419817759267072,
%T A125207 74221711070826087424,303193538300703211111936,2480118087478081928075065344,
%U A125207 40601989279034990139321984265216,1329877330680067685563700135615633408
%N A125207 The total number of connected components in all subgraphs
obtained from
           the complete labeled graph K_n by removing zero or more edges.
%F A125207 E.g.f.: (F(x)-1)*exp(F(x)-1) = G(x)*ln(G(x)) where
           G(x)=Sum_{n=0..oo} 2^(n(n-1)/2)*x^n/n! and F(x)=1+ln(G(x))
is e.g.f. of A001187
%e A125207 For n=2, we have two graph on two vertices: complete and
empty, the former has
           one connected component while the latter has two connected
components. The total
           number of connected components is 3, which is a(2).
%o A125207 (PARI) G=sum(n=0,30,2^(n*(n-1)/2)*x^n/n!) + O(x31); v=Vec(G*log(G));
           for(i=1,length(v),v[i]*=i!); print(v)
%Y A125207 Cf. A001187, A125205, A125206
%K A125207 nonn
%O A125207 1,2
%A A125207 Max Alekseyev (maxal(AT)cs.ucsd.edu), Nov 23 2006





From davidwwilson at comcast.net  Fri Nov 24 01:39:43 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 23 Nov 2006 19:39:43 -0500
Subject: b-files for power series expansions
Message-ID: <01ad01c70f61$09624cf0$6501a8c0@yourxhtr8hvc4p>

I guess my point was, if someone has access to Mma or some other symbolic math package, it would be nice if they could write a program to generate b-files for sequences that are e.g.f. expansions of real functions. We could then generate a whole slew of b-files fairly easily.
 
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From maxale at gmail.com  Fri Nov 24 02:04:09 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 23 Nov 2006 17:04:09 -0800
Subject: Chestnut
In-Reply-To: <000601c6f538$1ca30fe0$6401a8c0@yourxhtr8hvc4p>
References: <000601c6f538$1ca30fe0$6401a8c0@yourxhtr8hvc4p>
Message-ID: <d3dac270611231704ua5487a7s547879c8e77174b7@mail.gmail.com>

On 10/21/06, David Wilson <davidwwilson at comcast.net> wrote:
>
>
>
>
> This is a chestnut that I am convinced it is true, but I cannot prove it. I
> have posed it on math-fun occasionally, and never got an answer. I would
> like to see it posed to some better minds that some of you might know, e.g,
> JHC or other number theorist.
>
> Let S and T be sets of real numbers. Call T a divider of S if some element
> of T lies strictly between any two elements of S.
>
> For integer n >= 1, let Fence(n) be the set of all rationals with
> denominator n, that is, { k/n : k in Z }.
>
> For real set S, let f(S) be the least n such that Fence(n) is a divider of
> S, if such an n exists.
>
> Let Recip(n) be the set of all integer reciprocals on [0,1] with denominator
> <= n, that is, { 1/b : 1 <= b <= n }
>
> Let Farey(n) be the set of all rationals on [0,1] with denominator <= n,
> that is, { a/b : 0 <= a <= b, 1 <= b <= n }
>
> Is f(Farey(n)) = f(Recip(n)) for every n?
>
> The apparently common sequences a(n) = f(Recip(n)) =? f(Farey(n)) is in the
> OEIS, I cannot find it in the short time I have.





From Roland.Bacher at ujf-grenoble.fr  Fri Nov 24 08:53:08 2006
From: Roland.Bacher at ujf-grenoble.fr (Roland Bacher)
Date: Fri, 24 Nov 2006 08:53:08 +0100
Subject: Categories
In-Reply-To: <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com> <20061123065439.GA28473@fourier.ujf-grenoble.fr> <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com> <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
Message-ID: <20061124075307.GA17864@fourier.ujf-grenoble.fr>


I realised that the formula below for enumerating categories
using quivers is completely wrong. It gives only an upper bound
since one has to require moreover associativity for the composition of 
morphisms. The quiver approach works however in principle but
is much more complex (and I guess this approach has been 
used in some of the previous approaches).   Roland Bacher

On Thu, Nov 23, 2006 at 12:10:51PM -0800, Jonathan Post wrote:
> *"Formalized Proof, Computation, and the Construction Problem in Algebraic
> geometry", by Carlos Simpson.
> 
> [PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
> 2004<http://arxiv.org/pdf/math/0410224> File
> Format: PDF/Adobe Acrobat - View as
> HTML<http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5>
> finite integer N, *how many categories* are there with N morphisms? What *
> ...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
> (1965), 450-464.
> 
> 
> 
> On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> >
> >Roland Bacher correctly refers to quivers as summarized in wikipedia at
> >
> >http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
> >
> >On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> >>
> >>
> >> This numbers can also be obtained as the numbers
> >> of the following quotients of quivers:
> >>
> >> Associate to a category with n morphisms and k objects
> >> the quiver with k vertices corresponding to the objects
> >> and a(X,Y) arrows directed from the object X to the object
> >> Y if there are a(X,Y) morphisms from X into Y.
> >>
> >> Given two morphism g:X-->Y, f:Y-->Z with composition
> >> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> >>
> >> The resulting quotient algebra has dimension n and a basis
> >> given by the (simple) arrows.
> >>
> >> This leads to the following "algorithm" for enumerating
> >> all categories with n morphisms and k objects:
> >>
> >> (a) enumerate all quivers (directed graphs) with k vertices and
> >> n oriented edges.
> >>
> >> (b) Associate the following "weight" to such a quiver as follows :
> >>
> >> (b1)given a triplet of vertices X,Y,Z,
> >> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> >> (this counts the number of ways which associate a morphism
> >> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> >> number of oriented arrows starting at U and ending at W.
> >>
> >> (b2) associate to a given quiver the weight
> >> \prod_{X,Y,Z} w(X,Y,Z)
> >> where the product is over all triplets of vertices.
> >>
> >> The total sum of such weighted quivers yields then the solution.
> >>
> >> Roland Bacher
> >>
> >>
> >>
> >>
> >>
> >>
> >> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.net wrote:
> >> > How many categories are there?
> >> >
> >> > First, how many categories are there with n morphisms and k objects?
> >> > This table starts:
> >> >
> >> > 1
> >> > 2  1
> >> > 7  3 1
> >> > 35 16 3 1
> >> >
> >> > The first column is A058129, the number of monoids; the main diagonal
> >> > is all 1's.  I am not
> >> > 100% certain of the 16 in the final row.
> >> >
> >> > Taking the row sums, we get:
> >> >
> >> > 1,3,11,55
> >> >
> >> > the number of categories with n morphisms.  This is probably not in
> >> the
> >> > OEIS (only
> >> > A001776 is possible - other matches become less than A058129).  The
> >> > inverse Euler
> >> > transform,
> >> >
> >> > 1,2,8,41
> >> >
> >> > is the number of connected categories with n morphisms; this is
> >> > likewise probably not
> >> > in the OEIS (only A052447 is possible).
> >> >
> >> > Can somebody generate more data?
> >> >
> >> > Franklin T. Adams-Watters
> >> >
> >> > A category is a collection of objects and morphisms; each morphism is
> >> > from one object
> >> > to another (not necessarily different) object.  Where the destination
> >> > of one morphism
> >> > is the source of a second, their composition is defined; composition
> >> is
> >> > associative where
> >> > it is defined.  Each object has an identity morphism, which connects
> >> it
> >> > to itself; this
> >> > is an identity when composed with morphisms coming in and with
> >> > morphisms going
> >> > out.
> >> >
> >> ________________________________________________________________________
> >> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> >> > industry-leading spam and email virus protection.
> >> >
> >>
> >
> >





From tbaruchel at free.fr  Fri Nov 24 15:12:13 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Fri, 24 Nov 2006 15:12:13 +0100 (CET)
Subject: Quadratic residues
Message-ID: <20061124150520.F1510@localhost.my.domain>


I asked recently about
>  a(n) is the number of divisors of (n+1)
> Have you ever seen closely-related sequences like that one?

After some hints and a
> Happy hunting!
I finally noticed this :
   a) prime numbers 4n+3 lead to the set of their quadratic residues
   b) prime numbers 4n+1 lead to the set of their quadratic non-residues
   c) odd composite numbers :
      15 --> quadratic residues
      21 --> difficult to understand at first sight
Since my datas are empirical and need much work it is difficult to get
much information. Even numbers are difficult to study also.
Have you ever heard about such a behaviour? In that case, what would
you predict for 21?

WARNING. When I speak of "quadratic residues", I mean the residue has
a GCD = 1 with the number (convention may vary).

Thank you for everything,

-- 
Thomas Baruchel





From tbaruchel at free.fr  Fri Nov 24 15:59:29 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Fri, 24 Nov 2006 15:59:29 +0100 (CET)
Subject: Quadratic residues
In-Reply-To: <20061124150520.F1510@localhost.my.domain>
References: <20061124150520.F1510@localhost.my.domain>
Message-ID: <20061124155641.B1631@localhost.my.domain>

On Fri, 24 Nov 2006, Thomas Baruchel wrote:
> Have you ever heard about such a behaviour? In that case, what would
> you predict for 21?

Forget it, I just found : Jacobi symbol + quadratic reciprocity theorem

Thanks.





From eclark at math.usf.edu  Fri Nov 24 16:28:29 2006
From: eclark at math.usf.edu (Edwin Clark)
Date: Fri, 24 Nov 2006 10:28:29 -0500 (EST)
Subject: Categories
In-Reply-To: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
Message-ID: <Pine.LNX.4.58.0611240958550.32273@shell.cas.usf.edu>

On Wed, 22 Nov 2006 franktaw at netscape.net wrote:

> How many categories are there?
> 
> First, how many categories are there with n morphisms and k objects?  
> This table starts:
> 
>  1
>  2  1
>  7  3 1
> 35 16 3 1
> 
> The first column is A058129, the number of monoids; the main diagonal 
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
> 

As you say, the number of one object categories on n objects is M(n)=the 
number of monoids with n elements. But this is known only up to n = 7
according to the OEIS. This seems to make the number of TWO object
categories with n morphisms a very difficult problem...Let alone THREE
object categories...

MathSciNet gives only one hit on "two object categor*":
------------------------------------------------------------------
Rosick\'y, Ji\v r?
Codensity and binding categories.
Comment. Math. Univ. Carolinae 16 (1975), no. 3, 515--529.

If there is no measurable cardinal, the author constructs a two-object 
category all of whose well-powered complete extensions are binding. 
-------------------------------------------------------------------
which is clearly not related to the enumeration problem. 

A search on "categor* with two objects" give a few more hits
but as far as I can see still no papers related to enumeration..







From jvospost3 at gmail.com  Fri Nov 24 17:51:12 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 24 Nov 2006 08:51:12 -0800
Subject: Categories
In-Reply-To: <Pine.LNX.4.58.0611240958550.32273@shell.cas.usf.edu>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <Pine.LNX.4.58.0611240958550.32273@shell.cas.usf.edu>
Message-ID: <5542af940611240851l23b0d23di56bf7bf28fe71f82@mail.gmail.com>

One should also count loops, i.e. an arrow from an object to itself.

Then there are 2 categories on one object, with and without the loop:
Notating V(C) = the vertices in category C (which we label in this
illustrartion but enumerate up to isomorphism, i.e. on forgetting labels;
and L(C) is the set of loops of C, and A(C) is the set of (proper) arcs.
That is, I've separated the loops out from their being a mere subset of
arcs, to aid in enumeration. Then each category is, for this enumeration, a
triple: C= V(C) U L(C) U A(C).

C_1,1 = (1,null,null);
C_1,2 = (1, 1->1,null).

There are 6 categories on 2 objects:
C_2,1 = ({1,2},null,null);
C_2,2 = ({1,2},null,(1->1))
note that the above is isomorphic to C_2,2 = ({1,2},null,(2->2))
C_2,3 = ({1,2},{(1->1),(2->2)},null);
C_2,4 = ({1,2},null,(1->2)) which is isomorphic to C_2,4 =
({1,2},null,(2->1));
C_2,5 = ({1,2},(1>1),(1->2)) which is isomorphic to ({1,2},(2>2),(2->1));
C_2,6 = ({1,2},null,(1>1),(2->1)).

There do seem to be 35 categories on 3 objects, which are essentially the
same as the 7 endomorphisms on 3 objects, with each loop either left in, or
thrown away.

The combinatorics are now straightforward.  We have the endofunction
enumeration, and binary choices on each loop.

I have many, many pages of drawings of these, but not wanting to either scan
them (not having a scanner) nor write out the explicit sets as above.

Is this clear enough, though, to proceed?

-- Jonathan
C_2,2 = ({1,2},(1->1),null)


On 11/24/06, Edwin Clark <eclark at math.usf.edu> wrote:
>
> On Wed, 22 Nov 2006 franktaw at netscape.net wrote:
>
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> >  1
> >  2  1
> >  7  3 1
> > 35 16 3 1
> >
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
>
> As you say, the number of one object categories on n objects is M(n)=the
> number of monoids with n elements. But this is known only up to n = 7
> according to the OEIS. This seems to make the number of TWO object
> categories with n morphisms a very difficult problem...Let alone THREE
> object categories...
>
> MathSciNet gives only one hit on "two object categor*":
> ------------------------------------------------------------------
> Rosick\'y, Ji\v r?
> Codensity and binding categories.
> Comment. Math. Univ. Carolinae 16 (1975), no. 3, 515--529.
>
> If there is no measurable cardinal, the author constructs a two-object
> category all of whose well-powered complete extensions are binding.
> -------------------------------------------------------------------
> which is clearly not related to the enumeration problem.
>
> A search on "categor* with two objects" give a few more hits
> but as far as I can see still no papers related to enumeration..
>
>
>
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From jvospost3 at gmail.com  Fri Nov 24 20:39:20 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 24 Nov 2006 11:39:20 -0800
Subject: Categories
In-Reply-To: <20061124075307.GA17864@fourier.ujf-grenoble.fr>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
	 <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
	 <20061124075307.GA17864@fourier.ujf-grenoble.fr>
Message-ID: <5542af940611241139sa6472eoc31f5980db42a918@mail.gmail.com>

Offline, I'm coordinating with Edwin Clark on definitions.

But I'm counting as lower bounds  (many pages of paper with drawings, too
lengthy for me to turn into ascii, don't have scanner):

2 categories on 1 point;
6 categories on 2 points;
16 categories on 3 points;
46 categories on 4 points;
116 categories on 2 points;

in each case where the underlying digraph is of an endofunction. The full
count on categories on n points is a weighted partial sum on these, related
to my enumeration of "prime" endofunctions being the sum of Fobonacci(n),
tribonacci(n), teranacci(n), ...

-- I can give examples on request.  Of course, I can always be off by 1 or
more on each count, through sheer blunder.  But I am being self-consistent
in methodology, and it can be done by software using the same algorithm,
carefully translated.

-- Jonathan


On 11/23/06, Roland Bacher <Roland.Bacher at ujf-grenoble.fr> wrote:
>
>
> I realised that the formula below for enumerating categories
> using quivers is completely wrong. It gives only an upper bound
> since one has to require moreover associativity for the composition of
> morphisms. The quiver approach works however in principle but
> is much more complex (and I guess this approach has been
> used in some of the previous approaches).   Roland Bacher
>
> On Thu, Nov 23, 2006 at 12:10:51PM -0800, Jonathan Post wrote:
> > *"Formalized Proof, Computation, and the Construction Problem in
> Algebraic
> > geometry", by Carlos Simpson.
> >
> > [PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
> > 2004<http://arxiv.org/pdf/math/0410224> File
> > Format: PDF/Adobe Acrobat - View as
> > HTML<
> http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5
> >
> > finite integer N, *how many categories* are there with N morphisms? What
> *
> > ...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
> > (1965), 450-464.
> >
> >
> >
> > On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> > >
> > >Roland Bacher correctly refers to quivers as summarized in wikipedia at
> > >
> > >http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
> > >
> > >On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> > >>
> > >>
> > >> This numbers can also be obtained as the numbers
> > >> of the following quotients of quivers:
> > >>
> > >> Associate to a category with n morphisms and k objects
> > >> the quiver with k vertices corresponding to the objects
> > >> and a(X,Y) arrows directed from the object X to the object
> > >> Y if there are a(X,Y) morphisms from X into Y.
> > >>
> > >> Given two morphism g:X-->Y, f:Y-->Z with composition
> > >> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> > >>
> > >> The resulting quotient algebra has dimension n and a basis
> > >> given by the (simple) arrows.
> > >>
> > >> This leads to the following "algorithm" for enumerating
> > >> all categories with n morphisms and k objects:
> > >>
> > >> (a) enumerate all quivers (directed graphs) with k vertices and
> > >> n oriented edges.
> > >>
> > >> (b) Associate the following "weight" to such a quiver as follows :
> > >>
> > >> (b1)given a triplet of vertices X,Y,Z,
> > >> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> > >> (this counts the number of ways which associate a morphism
> > >> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> > >> number of oriented arrows starting at U and ending at W.
> > >>
> > >> (b2) associate to a given quiver the weight
> > >> \prod_{X,Y,Z} w(X,Y,Z)
> > >> where the product is over all triplets of vertices.
> > >>
> > >> The total sum of such weighted quivers yields then the solution.
> > >>
> > >> Roland Bacher
> > >>
> > >>
> > >>
> > >>
> > >>
> > >>
> > >> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.netwrote:
> > >> > How many categories are there?
> > >> >
> > >> > First, how many categories are there with n morphisms and k
> objects?
> > >> > This table starts:
> > >> >
> > >> > 1
> > >> > 2  1
> > >> > 7  3 1
> > >> > 35 16 3 1
> > >> >
> > >> > The first column is A058129, the number of monoids; the main
> diagonal
> > >> > is all 1's.  I am not
> > >> > 100% certain of the 16 in the final row.
> > >> >
> > >> > Taking the row sums, we get:
> > >> >
> > >> > 1,3,11,55
> > >> >
> > >> > the number of categories with n morphisms.  This is probably not in
> > >> the
> > >> > OEIS (only
> > >> > A001776 is possible - other matches become less than A058129).  The
> > >> > inverse Euler
> > >> > transform,
> > >> >
> > >> > 1,2,8,41
> > >> >
> > >> > is the number of connected categories with n morphisms; this is
> > >> > likewise probably not
> > >> > in the OEIS (only A052447 is possible).
> > >> >
> > >> > Can somebody generate more data?
> > >> >
> > >> > Franklin T. Adams-Watters
> > >> >
> > >> > A category is a collection of objects and morphisms; each morphism
> is
> > >> > from one object
> > >> > to another (not necessarily different) object.  Where the
> destination
> > >> > of one morphism
> > >> > is the source of a second, their composition is defined;
> composition
> > >> is
> > >> > associative where
> > >> > it is defined.  Each object has an identity morphism, which
> connects
> > >> it
> > >> > to itself; this
> > >> > is an identity when composed with morphisms coming in and with
> > >> > morphisms going
> > >> > out.
> > >> >
> > >>
> ________________________________________________________________________
> > >> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > >> > industry-leading spam and email virus protection.
> > >> >
> > >>
> > >
> > >
>
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From jvospost3 at gmail.com  Fri Nov 24 23:41:12 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 24 Nov 2006 14:41:12 -0800
Subject: Categories
In-Reply-To: <5542af940611241139sa6472eoc31f5980db42a918@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
	 <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
	 <20061124075307.GA17864@fourier.ujf-grenoble.fr>
	 <5542af940611241139sa6472eoc31f5980db42a918@mail.gmail.com>
Message-ID: <5542af940611241441w68d51ab5pc6ba0832f740d4e8@mail.gmail.com>

Somehow the reply to Frank went astray.

The wikipedia entry on directed graph makes some of the definitional
distinctions I've made, and then some.  Further, it makes the useful
statement:

In Category theory,
a category can be considered a directed multigraph
with the objects as vertices and the morphisms as directed edges.
The functors between categories induce then some,
but not necessarily all, of the digraph morphisms.


-- Jonathan


On 11/24/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> Offline, I'm coordinating with Edwin Clark on definitions.
>
> But I'm counting as lower bounds  (many pages of paper with drawings, too
> lengthy for me to turn into ascii, don't have scanner):
>
> 2 categories on 1 point;
> 6 categories on 2 points;
> 16 categories on 3 points;
> 46 categories on 4 points;
> 116 categories on 2 points;
>
> in each case where the underlying digraph is of an endofunction. The full
> count on categories on n points is a weighted partial sum on these, related
> to my enumeration of "prime" endofunctions being the sum of Fobonacci(n),
> tribonacci(n), teranacci(n), ...
>
> -- I can give examples on request.  Of course, I can always be off by 1 or
> more on each count, through sheer blunder.  But I am being self-consistent
> in methodology, and it can be done by software using the same algorithm,
> carefully translated.
>
> -- Jonathan
>
>
> On 11/23/06, Roland Bacher <Roland.Bacher at ujf-grenoble.fr> wrote:
> >
> >
> > I realised that the formula below for enumerating categories
> > using quivers is completely wrong. It gives only an upper bound
> > since one has to require moreover associativity for the composition of
> > morphisms. The quiver approach works however in principle but
> > is much more complex (and I guess this approach has been
> > used in some of the previous approaches).   Roland Bacher
> >
> > On Thu, Nov 23, 2006 at 12:10:51PM -0800, Jonathan Post wrote:
> > > *"Formalized Proof, Computation, and the Construction Problem in
> > Algebraic
> > > geometry", by Carlos Simpson.
> > >
> > > [PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
> > > 2004<http://arxiv.org/pdf/math/0410224> File
> > > Format: PDF/Adobe Acrobat - View as
> > > HTML<http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5
> > >
> > > finite integer N, *how many categories* are there with N morphisms?
> > What *
> > > ...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
> > > (1965), 450-464.
> > >
> > >
> > >
> > > On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> > > >
> > > >Roland Bacher correctly refers to quivers as summarized in wikipedia
> > at
> > > >
> > > >http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
> > > >
> > > >On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> > > >>
> > > >>
> > > >> This numbers can also be obtained as the numbers
> > > >> of the following quotients of quivers:
> > > >>
> > > >> Associate to a category with n morphisms and k objects
> > > >> the quiver with k vertices corresponding to the objects
> > > >> and a(X,Y) arrows directed from the object X to the object
> > > >> Y if there are a(X,Y) morphisms from X into Y.
> > > >>
> > > >> Given two morphism g:X-->Y, f:Y-->Z with composition
> > > >> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> > > >>
> > > >> The resulting quotient algebra has dimension n and a basis
> > > >> given by the (simple) arrows.
> > > >>
> > > >> This leads to the following "algorithm" for enumerating
> > > >> all categories with n morphisms and k objects:
> > > >>
> > > >> (a) enumerate all quivers (directed graphs) with k vertices and
> > > >> n oriented edges.
> > > >>
> > > >> (b) Associate the following "weight" to such a quiver as follows :
> > > >>
> > > >> (b1)given a triplet of vertices X,Y,Z,
> > > >> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> > > >> (this counts the number of ways which associate a morphism
> > > >> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> > > >> number of oriented arrows starting at U and ending at W.
> > > >>
> > > >> (b2) associate to a given quiver the weight
> > > >> \prod_{X,Y,Z} w(X,Y,Z)
> > > >> where the product is over all triplets of vertices.
> > > >>
> > > >> The total sum of such weighted quivers yields then the solution.
> > > >>
> > > >> Roland Bacher
> > > >>
> > > >>
> > > >>
> > > >>
> > > >>
> > > >>
> > > >> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.netwrote:
> > > >> > How many categories are there?
> > > >> >
> > > >> > First, how many categories are there with n morphisms and k
> > objects?
> > > >> > This table starts:
> > > >> >
> > > >> > 1
> > > >> > 2  1
> > > >> > 7  3 1
> > > >> > 35 16 3 1
> > > >> >
> > > >> > The first column is A058129, the number of monoids; the main
> > diagonal
> > > >> > is all 1's.  I am not
> > > >> > 100% certain of the 16 in the final row.
> > > >> >
> > > >> > Taking the row sums, we get:
> > > >> >
> > > >> > 1,3,11,55
> > > >> >
> > > >> > the number of categories with n morphisms.  This is probably not
> > in
> > > >> the
> > > >> > OEIS (only
> > > >> > A001776 is possible - other matches become less than
> > A058129).  The
> > > >> > inverse Euler
> > > >> > transform,
> > > >> >
> > > >> > 1,2,8,41
> > > >> >
> > > >> > is the number of connected categories with n morphisms; this is
> > > >> > likewise probably not
> > > >> > in the OEIS (only A052447 is possible).
> > > >> >
> > > >> > Can somebody generate more data?
> > > >> >
> > > >> > Franklin T. Adams-Watters
> > > >> >
> > > >> > A category is a collection of objects and morphisms; each
> > morphism is
> > > >> > from one object
> > > >> > to another (not necessarily different) object.  Where the
> > destination
> > > >> > of one morphism
> > > >> > is the source of a second, their composition is defined;
> > composition
> > > >> is
> > > >> > associative where
> > > >> > it is defined.  Each object has an identity morphism, which
> > connects
> > > >> it
> > > >> > to itself; this
> > > >> > is an identity when composed with morphisms coming in and with
> > > >> > morphisms going
> > > >> > out.
> > > >> >
> > > >>
> > ________________________________________________________________________
> > > >> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > > >> > industry-leading spam and email virus protection.
> > > >> >
> > > >>
> > > >
> > > >
> >
>
>
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From njas at research.att.com  Sat Nov 25 11:51:53 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sat, 25 Nov 2006 05:51:53 -0500 (EST)
Subject: request for new member
Message-ID: <200611251051.FAA87094@fry.research.att.com>


Olivier:

Richard Guy sent me this:

>From rkg at cpsc.ucalgary.ca  Fri Nov 24 17:07:00 2006
Delivered-To: njas at research.att.com
Date: Fri, 24 Nov 2006 15:02:50 -0700 (MST)
From: Richard Guy <rkg at cpsc.ucalgary.ca>
X-X-Sender: rkg at csl
To: "Neil J. A. Sloane" <njas at research.att.com>
cc: "Saff, Kevin" <kevin.saff at gmail.com>
Subject: New Subscriber
X-Virus-Scanned: by amavisd-new at cpsc.ucalgary.ca
X-Spam-Checker-Version: SpamAssassin 3.0.6 (2005-12-07) on 
        mail-purple.research.att.com
X-Spam-Level: 
X-Spam-Status: No, score=-2.9 required=4.0 tests=AWL,BAYES_00 autolearn=ham 
        version=3.0.6

Neil,
      May Kevin Saff join the seq-fan list?

      His email address is as above.

      Thanks & best wishes,    R.

Neil







From Annette.Warlich at t-online.de  Sat Nov 25 20:26:49 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Sat, 25 Nov 2006 20:26:49 +0100
Subject: Curious binomial-identity /A002720
Message-ID: <456898F9.8020608@t-online.de>


By chance I came across this curious identity
involving the Pascal-triangle.

Assume a row n, say n=4 and the column n, combined
each weighted with the running factorial as in the example:


           1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
 ratio =   -----------------------------------------  -------------------
           1/0! + 4/1! + 6/2! + 4/3! + 1/4!           weighted row-sum


then

  ratio = e  (=exp(1))

The actual sums are the entries of A002720
 http://www.research.att.com/~njas/sequences/A002720

Regards -

Gottfried Helms






From Annette.Warlich at t-online.de  Sat Nov 25 20:41:53 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Sat, 25 Nov 2006 20:41:53 +0100
Subject: Curious binomial-identity /A002720 (small correction of prev.
 post)
In-Reply-To: <456898F9.8020608@t-online.de>
References: <456898F9.8020608@t-online.de>
Message-ID: <45689C81.2020302@t-online.de>

Am 25.11.2006 20:26 schrieb Gottfried Helms:
> By chance I came across this curious identity
> involving the Pascal-triangle.
> 
> Assume a row n, say n=4 and the column n, combined
> each weighted with the running factorial as in the example:
> 
> 
>            1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
>  ratio =   -----------------------------------------  -------------------
>            1/0! + 4/1! + 6/2! + 4/3! + 1/4!           weighted row-sum
> 
> 
> then
> 
>   ratio = e  (=exp(1))
> 

------------------------------------

> The actual sums are the entries of A002720
>  http://www.research.att.com/~njas/sequences/A002720
> 
that should be corrected; the entries in A002720 are

 A(n)  =  weighted-rowsum(n) * n!
       =  weighted-colsum(n) * n! / exp(1)

I forgot to mention the additional n!, since in numerator and
denominator of the above fraction they cancel out, sorry.

Gottfried Helms






From jvospost3 at gmail.com  Sat Nov 25 21:10:27 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 25 Nov 2006 12:10:27 -0800
Subject: Curious binomial-identity /A002720 (small correction of prev. post)
In-Reply-To: <45689C81.2020302@t-online.de>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de>
Message-ID: <5542af940611251210g5e4916bcvee22981abd609ca1@mail.gmail.com>

There are several formulae for A002720, including a(n) = Sum k!C(n, k)^2,
k=0..n.

What you write might be true in the asymptotic limit, but not for any term,
as each term is rational and dividing by e would make each term
transcendental.  I'm sure that you meant the limit, right?

On 11/25/06, Gottfried Helms <Annette.Warlich at t-online.de> wrote:
>
> Am 25.11.2006 20:26 schrieb Gottfried Helms:
> > By chance I came across this curious identity
> > involving the Pascal-triangle.
> >
> > Assume a row n, say n=4 and the column n, combined
> > each weighted with the running factorial as in the example:
> >
> >
> >            1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
> >  ratio =
> -----------------------------------------  -------------------
> >            1/0! + 4/1! + 6/2! + 4/3! + 1/4!           weighted row-sum
> >
> >
> > then
> >
> >   ratio = e  (=exp(1))
> >
>
> ------------------------------------
>
> > The actual sums are the entries of A002720
> >  http://www.research.att.com/~njas/sequences/A002720
> >
> that should be corrected; the entries in A002720 are
>
> A(n)  =  weighted-rowsum(n) * n!
>        =  weighted-colsum(n) * n! / exp(1)
>
> I forgot to mention the additional n!, since in numerator and
> denominator of the above fraction they cancel out, sorry.
>
> Gottfried Helms
>
>
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From Annette.Warlich at t-online.de  Sat Nov 25 22:41:35 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Sat, 25 Nov 2006 22:41:35 +0100
Subject: Curious binomial-identity /A002720 (small correction of prev.
 post)
In-Reply-To: <5542af940611251210g5e4916bcvee22981abd609ca1@mail.gmail.com>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de> <5542af940611251210g5e4916bcvee22981abd609ca1@mail.gmail.com>
Message-ID: <4568B88F.1020800@t-online.de>

Am 25.11.2006 21:10 schrieb Jonathan Post:
> There are several formulae for A002720, including  a(n) = Sum k!C(n,
> k)^2, k=0..n.
> 
> What you write might be true in the asymptotic limit, but not for any
> term, as each term is rational and dividing by e would make each term
> transcendental.  I'm sure that you meant the limit, right?

Yes; as the +... in the numerator should indicate. The sum of columns
have infinitely many terms.

I've seen the generation-formula involving the exp()-function; maybe
this translates in an obvious way (... only with gf's I don't have
*any* experience... sigh...) and the asymptotic formula, which looks
a bit more complicated, but additionally relates it to a pi-expression.

But that we have

   1/3! +  3/2! + 3/1! + 1/0!  = A(3) /3!
                       + 3/1!
                       + 6/2!
                       +10/3!
                       + ...
                        -----
    limit k->oo         A(3) /3! * exp(1)

 or

   1/4! +  4/3! + 6/2! + 4/1! + 1/0!  = A(4)/4!
                              + 4/1!
                              +10/2!
                              +20/3!
                              + ...
                              -----
    limit k->oo               A(4)/4! *exp(1)

has something of a certain beauty... :-)


Gottfried Helms







From jrbibers at indiana.edu  Sat Nov 25 22:53:40 2006
From: jrbibers at indiana.edu (Joseph Biberstine)
Date: Sat, 25 Nov 2006 16:53:40 -0500
Subject: Curious binomial-identity /A002720 (small correction of prev.
 post)
In-Reply-To: <45689C81.2020302@t-online.de>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de>
Message-ID: <4568BB64.9050703@indiana.edu>

The example you gave has an off-by-one row index as follows:

Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n+1)

Note that the row index is incremented.  Mathematica simplifies this
fraction to LaguerreL[-(n+1), 1]/LaguerreL[n+1, -1].  This is not equal
to Exp[1], but interestingly seems to approach Exp[1] monotonically from
below as n approaches infinity.

I'm sure you meant:

Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n)

Mathematica simplifies this to LaguerreL[-(n+1), 1]/LaguerreL[n, -1].
Although I can't get it to explicitly resolve that this is exactly
Exp[1], the values do seem to bear this truth out.

So, I believe your example should read: e.g. for n=3,

    1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
e = -----------------------------------------  -------------------
           1/0! + 3/1! + 3/2! + 1/3!           weighted row-sum

This does seem to hold for every col/row pair in the triangle.

Nice find :)
-JRB

Gottfried Helms wrote:
> Am 25.11.2006 20:26 schrieb Gottfried Helms:
>> By chance I came across this curious identity
>> involving the Pascal-triangle.
>>
>> Assume a row n, say n=4 and the column n, combined
>> each weighted with the running factorial as in the example:
>>
>>
>>            1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
>>  ratio =   -----------------------------------------  -------------------
>>            1/0! + 4/1! + 6/2! + 4/3! + 1/4!           weighted row-sum
>>
>>
>> then
>>
>>   ratio = e  (=exp(1))
>>
> 
> ------------------------------------
> 
>> The actual sums are the entries of A002720
>>  http://www.research.att.com/~njas/sequences/A002720
>>
> that should be corrected; the entries in A002720 are
> 
>  A(n)  =  weighted-rowsum(n) * n!
>        =  weighted-colsum(n) * n! / exp(1)
> 
> I forgot to mention the additional n!, since in numerator and
> denominator of the above fraction they cancel out, sorry.
> 
> Gottfried Helms
> 





From Annette.Warlich at t-online.de  Sun Nov 26 00:43:25 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Sun, 26 Nov 2006 00:43:25 +0100
Subject: Curious binomial-identity /A002720 (small correction of prev.
 post)
In-Reply-To: <4568BB64.9050703@indiana.edu>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de> <4568BB64.9050703@indiana.edu>
Message-ID: <4568D51D.8000305@uni-kassel.de>

Am 25.11.2006 22:53 schrieb Joseph Biberstine:
> The example you gave has an off-by-one row index as follows:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n+1)
> 
> Note that the row index is incremented.  Mathematica simplifies this
> fraction to LaguerreL[-(n+1), 1]/LaguerreL[n+1, -1].  This is not equal
> to Exp[1], but interestingly seems to approach Exp[1] monotonically from
> below as n approaches infinity.
> 
> I'm sure you meant:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n)
> 
> Mathematica simplifies this to LaguerreL[-(n+1), 1]/LaguerreL[n, -1].
> Although I can't get it to explicitly resolve that this is exactly
> Exp[1], the values do seem to bear this truth out.
> 
> So, I believe your example should read: e.g. for n=3,
> 
>     1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
> e = -----------------------------------------  -------------------
>            1/0! + 3/1! + 3/2! + 1/3!           weighted row-sum
> 
> This does seem to hold for every col/row pair in the triangle.

Yes, your correction is right - I didn't look right at the
matrix.


The connection to the Laguerre-polynomials is not too
much surprising; I'm dealing with a ceratin classes
of lower-triangular matrices, which are derived from the
logic of the computing of the Pascal-matrix by matrix-
exponentiation.

While

  Pascalmatrix     = Pk(1)=  matrixexp( subdiagonal(1; [1,2,3,4,5,...]))

the Laguerre-matrix occurs from

  Laguerrematrix(*) = Pk(2) = matrixexp( subdiagonal(1; [1,2^2,3^2,4^2,5^2,...]))
           ((*) means: in a signed and simply scaled form)


I also have
                      Pk(0) = matrixexp( subdiagonal(1; [1,2^0,3^0,4^0,5^0,...]))

(where the vectors are filled in the first principal subdiagonal of an
infinite square-matrix)




Now the current matrix was created by a somehow routinely
"completion of my toolbox" of pascal-like-matrices, as a
hadamard-product of

      Pf =    Pk(1) (x) Pk(0)

         =    1/0!
              1/1!  1/0!
              1/2!  2/1!  1/0!
              1/3!  3/2!  3/1!  1/0!
              1/4!  4/3!  6/2!  4/1!   1/0!
                ....


Then with a powerseries-column-vector

      V(x) = [1,x,x^2,x^3,....] ~

and a diagonal factorial-matrix (used for scaling)

     dF(m) = diag(0! , 1!^m, 2!^m .... )

I got

   dF(1)  *   Pf  * V(1) = A002720       // A002720 read as column-vector
                                         // Due to the entry of Paul Berry

and
   dF(1) *    Pf~ * V(1) = A002720  *exp(1)  // added today



As a table it looks like:

                                                 Row-Sum
         =    1/0!                              =  1/0!    =  A(0)/0!
              1/1!  1/0!                        =  2/1!    =  A(1)/1!
              1/2!  2/1!  1/0!                  =  7/2!    =  A(2)/2!
              1/3!  3/2!  3/1!  1/0!            = 34/3!    =  A(3)/3!
              1/4!  4/3!  6/2!  4/1!   1/0!     =209/4!    =  A(4)/4!
                ....
             --------------------------------------------------------
Col-sum   e*[  1    2/1!  7/2! 34/3! 209/4! ]

Note, that using

   dF(1)  *   Pf  * V(2) =          ???
   dF(1) *    Pf~ * V(2) = exp(2) * ???

  in the second an exponentiation of 2 occurs, and generally for an s

   dF(1) *    Pf~ * V(s) = exp(s) * ???

 with unknown, but -heuristically- integer sequences of coefficients in ???
 for integer s>=0 .


------------------------------------------------------------------------


The Pascal- and (unsigned,scaled) Laguerre-matrix are also
related by

  Pk(0)                              // "base"-matrix
  Pk(1) =   dF(1)* Pk(0) * dF(-1)    // Pascal-matrix
  Pk(2) =   dF(1)* Pk(1) * dF(-1)    // unsigned,scaled Laguerrematrix
 ...
(which is due to the construction via matrix-exponentiation
of the related subdiagonal-matrices)
so finally the occurence of the relation to the Laguerre-matrix
using Mathematica is not too surprising...

Regards -

Gottfried Helms







From zakseidov at yahoo.com  Sun Nov 26 20:58:31 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 26 Nov 2006 11:58:31 -0800 (PST)
Subject: A124080 10 times triangular numbers
Message-ID: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>

Neil, Zerinvary, seqfans,
Why this sequence?!
Thanks, Zak 

%I A124080
%S A124080
0,10,30,60,100,150,210,280,360,450,550,660,780,910,1050,1200,1360,1530,
%T A124080
1710,1900,2100,2310,2530,2760,3000,3250,3510,3780,4060,4350,4650,4960,
%U A124080
5280,5610,5950,6300,6660,7030,7410,7800,8200,8610,9030,9460,9900,10350
%N A124080 10 times triangular numbers.
%F A124080 a(n)=10*C(n,2), n>=1
%p A124080 [seq(10*binomial(n,2) ,n=1..51)];
%Y A124080 Cf. A028895, A046092, A045943, A002378,
A028896, A024966, A033996, A027468.
%K A124080 easy,nonn,new
%O A124080 0,2
%A A124080 Zerinvary Lajos
(zerinvarylajos(AT)yahoo.com), Nov 24 2006



 
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From jvospost3 at gmail.com  Sun Nov 26 22:23:25 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 26 Nov 2006 13:23:25 -0800
Subject: Categories
In-Reply-To: <20061124075307.GA17864@fourier.ujf-grenoble.fr>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
	 <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
	 <20061124075307.GA17864@fourier.ujf-grenoble.fr>
Message-ID: <5542af940611261323y13ada5f4p9e8f6e5d9051f93d@mail.gmail.com>

I can, with Edwin Clark's permission, forward to Frank my complete
correspondance with Edwin Clark about category enumeration via careful
definition.  I am asked not to cut & paste out of context.  But my emails to
Franklin T. Adams-Watters alone went astray, for whatever reason. so I am
not sure what to do.  I've also asked some Haskell programming experts about
Frank's enumeration question (properly citing Frank), as Haskell has monads
and functors and other categorical stuff built in.  I'll share any useful
answers that I get.

On 11/23/06, Roland Bacher <Roland.Bacher at ujf-grenoble.fr> wrote:
>
>
> I realised that the formula below for enumerating categories
> using quivers is completely wrong. It gives only an upper bound
> since one has to require moreover associativity for the composition of
> morphisms. The quiver approach works however in principle but
> is much more complex (and I guess this approach has been
> used in some of the previous approaches).   Roland Bacher
>
> On Thu, Nov 23, 2006 at 12:10:51PM -0800, Jonathan Post wrote:
> > *"Formalized Proof, Computation, and the Construction Problem in
> Algebraic
> > geometry", by Carlos Simpson.
> >
> > [PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
> > 2004<http://arxiv.org/pdf/math/0410224> File
> > Format: PDF/Adobe Acrobat - View as
> > HTML<
> http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5
> >
> > finite integer N, *how many categories* are there with N morphisms? What
> *
> > ...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
> > (1965), 450-464.
> >
> >
> >
> > On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> > >
> > >Roland Bacher correctly refers to quivers as summarized in wikipedia at
> > >
> > >http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
> > >
> > >On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> > >>
> > >>
> > >> This numbers can also be obtained as the numbers
> > >> of the following quotients of quivers:
> > >>
> > >> Associate to a category with n morphisms and k objects
> > >> the quiver with k vertices corresponding to the objects
> > >> and a(X,Y) arrows directed from the object X to the object
> > >> Y if there are a(X,Y) morphisms from X into Y.
> > >>
> > >> Given two morphism g:X-->Y, f:Y-->Z with composition
> > >> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> > >>
> > >> The resulting quotient algebra has dimension n and a basis
> > >> given by the (simple) arrows.
> > >>
> > >> This leads to the following "algorithm" for enumerating
> > >> all categories with n morphisms and k objects:
> > >>
> > >> (a) enumerate all quivers (directed graphs) with k vertices and
> > >> n oriented edges.
> > >>
> > >> (b) Associate the following "weight" to such a quiver as follows :
> > >>
> > >> (b1)given a triplet of vertices X,Y,Z,
> > >> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> > >> (this counts the number of ways which associate a morphism
> > >> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> > >> number of oriented arrows starting at U and ending at W.
> > >>
> > >> (b2) associate to a given quiver the weight
> > >> \prod_{X,Y,Z} w(X,Y,Z)
> > >> where the product is over all triplets of vertices.
> > >>
> > >> The total sum of such weighted quivers yields then the solution.
> > >>
> > >> Roland Bacher
> > >>
> > >>
> > >>
> > >>
> > >>
> > >>
> > >> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.netwrote:
> > >> > How many categories are there?
> > >> >
> > >> > First, how many categories are there with n morphisms and k
> objects?
> > >> > This table starts:
> > >> >
> > >> > 1
> > >> > 2  1
> > >> > 7  3 1
> > >> > 35 16 3 1
> > >> >
> > >> > The first column is A058129, the number of monoids; the main
> diagonal
> > >> > is all 1's.  I am not
> > >> > 100% certain of the 16 in the final row.
> > >> >
> > >> > Taking the row sums, we get:
> > >> >
> > >> > 1,3,11,55
> > >> >
> > >> > the number of categories with n morphisms.  This is probably not in
> > >> the
> > >> > OEIS (only
> > >> > A001776 is possible - other matches become less than A058129).  The
> > >> > inverse Euler
> > >> > transform,
> > >> >
> > >> > 1,2,8,41
> > >> >
> > >> > is the number of connected categories with n morphisms; this is
> > >> > likewise probably not
> > >> > in the OEIS (only A052447 is possible).
> > >> >
> > >> > Can somebody generate more data?
> > >> >
> > >> > Franklin T. Adams-Watters
> > >> >
> > >> > A category is a collection of objects and morphisms; each morphism
> is
> > >> > from one object
> > >> > to another (not necessarily different) object.  Where the
> destination
> > >> > of one morphism
> > >> > is the source of a second, their composition is defined;
> composition
> > >> is
> > >> > associative where
> > >> > it is defined.  Each object has an identity morphism, which
> connects
> > >> it
> > >> > to itself; this
> > >> > is an identity when composed with morphisms coming in and with
> > >> > morphisms going
> > >> > out.
> > >> >
> > >>
> ________________________________________________________________________
> > >> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > >> > industry-leading spam and email virus protection.
> > >> >
> > >>
> > >
> > >
>
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From jvospost3 at gmail.com  Sun Nov 26 22:27:11 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 26 Nov 2006 13:27:11 -0800
Subject: A124080 10 times triangular numbers
In-Reply-To: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
References: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
Message-ID: <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>

I wondered too.  It intersects the much more interesting
 A000537 <http://www.research.att.com/%7Enjas/sequences/A000537>
 Sum of first n cubes; or n-th triangular number squared.  at 100.

Not that all of my roughly 1 kilophi sequences and comments are interesting,
either.

-- Jonathan

On 11/26/06, zak seidov <zakseidov at yahoo.com> wrote:
>
> Neil, Zerinvary, seqfans,
> Why this sequence?!
> Thanks, Zak
>
> %I A124080
> %S A124080
> 0,10,30,60,100,150,210,280,360,450,550,660,780,910,1050,1200,1360,1530,
> %T A124080
> 1710,1900,2100,2310,2530,2760,3000,3250,3510,3780,4060,4350,4650,4960,
> %U A124080
> 5280,5610,5950,6300,6660,7030,7410,7800,8200,8610,9030,9460,9900,10350
> %N A124080 10 times triangular numbers.
> %F A124080 a(n)=10*C(n,2), n>=1
> %p A124080 [seq(10*binomial(n,2) ,n=1..51)];
> %Y A124080 Cf. A028895, A046092, A045943, A002378,
> A028896, A024966, A033996, A027468.
> %K A124080 easy,nonn,new
> %O A124080 0,2
> %A A124080 Zerinvary Lajos
> (zerinvarylajos(AT)yahoo.com), Nov 24 2006
>
>
>
>
>
> ____________________________________________________________________________________
> Do you Yahoo!?
> Everyone is raving about the all-new Yahoo! Mail beta.
> http://new.mail.yahoo.com
>
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From jvospost3 at gmail.com  Sun Nov 26 23:43:48 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 26 Nov 2006 14:43:48 -0800
Subject: A124080 10 times triangular numbers
In-Reply-To: <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>
References: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
	 <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>
Message-ID: <5542af940611261443g32cd3bf3lf2174cfc2d42e716@mail.gmail.com>

On the other hand, every sequence is related to some interesting sequence...
I just submitted by form the following:

 NEW SEQUENCE FROM Jonathan Vos Post


%I A000001
%S A000001 11, 29, 31, 59, 61, 101, 149, 151, 211, 281, 359, 449, 659,
661, 911, 1049, 1051, 1201, 1361, 1531, 1709, 1901, 2099, 2309, 2311,
2531, 2999, 3001
%N A000001 Primes of the form A124080 (10 times triangular numbers), +1
or -1.
%C A000001 The j, such that A124080(j)-1 is prime or A124080(j)+1 is
prime, where repetition means a twin prime, are 1, 2, 2, 3, 3, 4, 5, 5,
6, 7, 8, 9, 11, 11, 13, 14, 14, 15, 16, 17, 18, 19, 20, 21, 22, 24, 24.
%F A000001 {A124080(j)-1 when prime} U {A124080(j)+1 when prime} =
{i = 10*T(j)-1 such that i is prime} U {i = 10*T(j)+1 such that i is
prime} where T(j) = A000217(j) = j*(j+1)/2.
%e A000001 a(1) = A124080(1)+1 = (10*T(1)) - 1 = 10*(1*(1+1)/2) + 1 =
10+1 = 11 is prime.
a(2) = A124080(2)-1 = (10*T(2))-1 = 10*(2*(2+1)/2) - 1 = 30-1 = 29 is
prime.
a(3) = A124080(2)+1 = (10*T(2))+1 = 10*(2*(2+1)/2) + 1 = 30+1 = 31 is
prime.
%Y A000001 Cf. A000040, A000217, A028895, A046092, A045943, A002378,
A028896, A024966, A033996, A027468.
%O A000001 1,1
%K A000001 ,easy,nonn,
%A A000001 Jonathan Vos Post (jvospost2 at yahoo.com
<http://us.f551.mail.yahoo.com/ym/Compose?To=jvospost2 at yahoo.com&YY=86457&y5beta=yes&y5beta=yes&order=down&sort=date&pos=0&view=a&head=b>),
Nov 26 2006
RH
RA 192.20.225.32
RU
RI



On 11/26/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> I wondered too.  It intersects the much more interesting
>  A000537 <http://www.research.att.com/%7Enjas/sequences/A000537>
>  Sum of first n cubes; or n-th triangular number squared.  at 100.
>
> Not that all of my roughly 1 kilophi sequences and comments are
> interesting, either.
>
> -- Jonathan
>
> On 11/26/06, zak seidov <zakseidov at yahoo.com> wrote:
> >
> > Neil, Zerinvary, seqfans,
> > Why this sequence?!
> > Thanks, Zak
> >
> > %I A124080
> > %S A124080
> > 0,10,30,60,100,150,210,280,360,450,550,660,780,910,1050,1200,1360,1530,
> > %T A124080
> > 1710,1900,2100,2310,2530,2760,3000,3250,3510,3780,4060,4350,4650,4960,
> > %U A124080
> > 5280,5610,5950,6300,6660,7030,7410,7800,8200,8610,9030,9460,9900,10350
> > %N A124080 10 times triangular numbers.
> > %F A124080 a(n)=10*C(n,2), n>=1
> > %p A124080 [seq(10*binomial(n,2) ,n=1..51)];
> > %Y A124080 Cf. A028895, A046092, A045943, A002378,
> > A028896, A024966, A033996, A027468.
> > %K A124080 easy,nonn,new
> > %O A124080 0,2
> > %A A124080 Zerinvary Lajos
> > (zerinvarylajos(AT)yahoo.com), Nov 24 2006
> >
> >
> >
> >
> >
> > ____________________________________________________________________________________
> > Do you Yahoo!?
> > Everyone is raving about the all-new Yahoo! Mail beta.
> > http://new.mail.yahoo.com
> >
>
>
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From njas at research.att.com  Mon Nov 27 03:23:49 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sun, 26 Nov 2006 21:23:49 -0500 (EST)
Subject: A124080 10 times triangular numbers
References: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
    <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>
Message-ID: <200611270223.VAA41276@fry.research.att.com>

JVP, you submitted this sequence:
%N A000001 Primes of the form A124080 (10 times triangular numbers), +1
or -1.
I will use it, but I do not find it interesting.  Why?  Because it 
is derivative.  It is based on a sequence of little interest - acceptable
only because it was sent in by someone who perhaps does not know
what is worth being included in the database - and is a step further 
down the ladder towards the basement.

I seem to have to play the role of the singing teacher
at times:  so be it - this song was off-key!

Neil





From jvospost3 at gmail.com  Mon Nov 27 04:10:47 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 26 Nov 2006 19:10:47 -0800
Subject: A124080 10 times triangular numbers
In-Reply-To: <200611270223.VAA41276@fry.research.att.com>
References: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
	 <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>
	 <200611270223.VAA41276@fry.research.att.com>
Message-ID: <5542af940611261910x10195634teeac87bdf4bfd1e7@mail.gmail.com>

As usual, you're right, Neil.

I was trying to be nice to the newcomer, by showing him the collaborative
nature of the endeavor. I'm sure that his enthusiasm can be made fruitful.
I've certainly erred in judgment, many times, and you and your associate
editors have been splendid in guiding me towards better work.

Sometimes, as they sing: "you've got to be cruel to be kind."

Hence the "probation" keyword and, one of my useful suggestions to you and
your board, the "less" keyword.

Sorry for where that road leads, which I help pave with good intentions.

-- Jonathan

On 11/26/06, N. J. A. Sloane <njas at research.att.com> wrote:
>
> JVP, you submitted this sequence:
> %N A000001 Primes of the form A124080 (10 times triangular numbers), +1
> or -1.
> I will use it, but I do not find it interesting.  Why?  Because it
> is derivative.  It is based on a sequence of little interest - acceptable
> only because it was sent in by someone who perhaps does not know
> what is worth being included in the database - and is a step further
> down the ladder towards the basement.
>
> I seem to have to play the role of the singing teacher
> at times:  so be it - this song was off-key!
>
> Neil
>
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From pauldhanna at juno.com  Mon Nov 27 04:58:25 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Sun, 26 Nov 2006 22:58:25 -0500
Subject: Missing Table?
Message-ID: <20061126.225826.332.1.pauldhanna@juno.com>

Seqfans,
     Where is the rectangular table referred to by these sequences?
 
A040027  Main diagonal of an array of binomial recurrence coefficients.  
 0, 1, 1, 3, 9, 31, 121, 523, 2469, 12611, 69161, 404663,  
A045501  Second diagonal in table of binomial recurrence coefficients.  
 1, 1, 4, 14, 54, 233, 1101, 5625, 30846, 
A045499  Third diagonal in table of binomial recurrence coefficients. 
 1, 1, 5, 20, 85, 400, 2046, 11226, 65676, 407787, 2675410, 
A045500  Fourth diagonal in table of binomial recurrence coefficients. 
 1, 1, 6, 27, 125, 635, 3488, 20425, 126817, 831915, 5744784, 41618459 
(AUTHOR of these sequences:  H. W. Gould) 
 
The closest thing I can find is Emeric Deutsch's recent triangle A124496,

in which these sequences appear as columns. 
 
The lower half of the table (I guess) must look like this:
1, ...
1, 1, ...
1, 1, 3, ...
1, 1, 4, 9, ...
1, 1, 5, 14, 31, ...
1, 1, 6, 20, 54, 121, ...
1, 1, 7, 27, 85, 233, 523, ...
1, 1, 8, 35, 125, 400, 1101, 2469, ...
1, 1, 9, 44, 175, 635, 2046, 5625, 12611, ...
 
Is the complete table in the OEIS?  
If not, it certainly should be. 
     Paul
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From mathar at strw.leidenuniv.nl  Mon Nov 27 10:47:35 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Mon, 27 Nov 2006 10:47:35 +0100
Subject: Missing Table?
Message-ID: <200611270947.kAR9lZBJ001619@amer.strw.leidenuniv.nl>


The master table of A040027 could also be A046936..but this does not fit the
2nd, 3rd and 4th diagonal as well as A124496 does.

RJM

> From seqfan-owner at ext.jussieu.fr  Mon Nov 27 05:03:23 2006
> Return-Path: <seqfan-owner at ext.jussieu.fr>
> To: Seqfan at ext.jussieu.fr
> Cc: gould at math.wvu.edu
> Date: Sun, 26 Nov 2006 22:58:25 -0500
> Subject: Missing Table?
> From: "Paul D. Hanna" <pauldhanna at juno.com>
> 
> Seqfans,
>      Where is the rectangular table referred to by these sequences?
>  
> A040027  Main diagonal of an array of binomial recurrence coefficients.  
>  0, 1, 1, 3, 9, 31, 121, 523, 2469, 12611, 69161, 404663,  
> A045501  Second diagonal in table of binomial recurrence coefficients.  
>  1, 1, 4, 14, 54, 233, 1101, 5625, 30846, 
> A045499  Third diagonal in table of binomial recurrence coefficients. 
>  1, 1, 5, 20, 85, 400, 2046, 11226, 65676, 407787, 2675410, 
> A045500  Fourth diagonal in table of binomial recurrence coefficients. 
>  1, 1, 6, 27, 125, 635, 3488, 20425, 126817, 831915, 5744784, 41618459 
> (AUTHOR of these sequences:  H. W. Gould) 
>  
> The closest thing I can find is Emeric Deutsch's recent triangle A124496,
> 
> in which these sequences appear as columns. 
>  
> The lower half of the table (I guess) must look like this:
> 1, ...
> 1, 1, ...
> 1, 1, 3, ...
> 1, 1, 4, 9, ...
> 1, 1, 5, 14, 31, ...
> 1, 1, 6, 20, 54, 121, ...
> 1, 1, 7, 27, 85, 233, 523, ...
> 1, 1, 8, 35, 125, 400, 1101, 2469, ...
> 1, 1, 9, 44, 175, 635, 2046, 5625, 12611, ...
>  
> Is the complete table in the OEIS?  
> If not, it certainly should be. 
>      Paul





From tanyakh at TanyaKhovanova.com  Mon Nov 27 17:05:13 2006
From: tanyakh at TanyaKhovanova.com (Tanya Khovanova)
Date: Mon, 27 Nov 2006 08:05:13 -0800
Subject: practical numbers
Message-ID: <200611270805.AA74449328@TanyaKhovanova.com>


Hello seqfans,

There is some confusion with practical numbers. Wikipedia article on practical numbers: http://en.wikipedia.org/wiki/Practical_number
gives the following definition of a practical number:
A practical number or panarithmic number is a positive integer n such that all preceding positive integers are a sum of distinct divisors of n.

At the same time it refers to the sequence A005153 which gives a slightly different definition:
Practical numbers (first definition): all k <= sigma(n) are sums of distinct divisors of n. Also called panarithmic numbers.

Browsing other websites (including my own :-) ) I've found that the wiki definition is used more often. Moreover, my calculation with the my definition exactly matches the sequence A005153, though they are supposed to be different.

Please, clarify my confusion.

Tanya Khovanova






From mathar at strw.leidenuniv.nl  Mon Nov 27 17:54:41 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Mon, 27 Nov 2006 17:54:41 +0100
Subject: practical numbers
Message-ID: <200611271654.kARGsfOX022242@amer.strw.leidenuniv.nl>

> From seqfan-owner at ext.jussieu.fr  Mon Nov 27 17:07:28 2006
> Date: Mon, 27 Nov 2006 08:05:13 -0800
> From: "Tanya Khovanova" <tanyakh at tanyakhovanova.com>
> Reply-To: <tanyakh at tanyakhovanova.com>
> To: <seqfan at ext.jussieu.fr>
> Subject: practical numbers
> ...
> There is some confusion with practical numbers. Wikipedia article on practical numbers: http://en.wikipedia.org/wiki/Practical_numbe
> r
> gives the following definition of a practical number:
> A practical number or panarithmic number is a positive integer n such that all preceding positive integers are a sum of distinct div
> isors of n.
> 
> At the same time it refers to the sequence A005153 which gives a slightly different definition:
> Practical numbers (first definition): all k <= sigma(n) are sums of distinct divisors of n. Also called panarithmic numbers.
> 
> Browsing other websites (including my own :-) ) I've found that the wiki definition is used more often. Moreover, my calculation wit
> h the my definition exactly matches the sequence A005153, though they are supposed to be different.
> 
> Please, clarify my confusion.
> ...

The A005153 definition [with reference to the sigma(n)] is also used
in
http://www.dm.unipi.it/gauss-pages/melfi/public_html/articoli/jnt.ps
but the same author uses in his web page
http://members.unine.ch/giuseppe.melfi/pratica.html
just the other definition.
see also http://citeseer.ist.psu.edu/285.html
http://arxiv.org/abs/math.NT/0404555

B. M Stewart in Am J. Math 76 (4) (1954) p 779, available from
http://www.jstor.org, quotes Srinivasan [Current Science (1948), pp 179] with
the definition without the sigma.

E J Scourfield in J. Number Theory 62 (1) (1997) p 163 uses the definition
including the sigma:
http://www.emis.de/projects/EULER/detail?ide=1997saiaentidividens&matchno=7&matchtotal=18&q=cr%3AE*+Saias+

I have to stop here. The recent IAU resolution on what a useful definition of a
"planet" might be, in particular the status of Pluto, be has already caused
enough confusion, and we're still recovering from this.

Richard





From franktaw at netscape.net  Mon Nov 27 18:04:03 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 27 Nov 2006 12:04:03 -0500
Subject: practical numbers
In-Reply-To: <200611270805.AA74449328@TanyaKhovanova.com>
References: <200611270805.AA74449328@TanyaKhovanova.com>
Message-ID: <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com>

The definitions are equivalent.  The condition cited in the Wikipedia 
article, to make everything up to n a sum of the divisors of n, is 
sufficient to make everything up to sigma(n) be such a sum.

Franklin T. Adams-Watters


-----Original Message-----
From: tanyakh at TanyaKhovanova.com

Hello seqfans,

There is some confusion with practical numbers. Wikipedia article on 
practical
numbers: http://en.wikipedia.org/wiki/Practical_number
gives the following definition of a practical number:
A practical number or panarithmic number is a positive integer n such 
that all
preceding positive integers are a sum of distinct divisors of n.

At the same time it refers to the sequence A005153 which gives a 
slightly
different definition:
Practical numbers (first definition): all k <= sigma(n) are sums of 
distinct
divisors of n. Also called panarithmic numbers.

Browsing other websites (including my own :-) ) I've found that the 
wiki
definition is used more often. Moreover, my calculation with the my 
definition
exactly matches the sequence A005153, though they are supposed to be 
different.

Please, clarify my confusion.

Tanya Khovanova


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From davidwwilson at comcast.net  Mon Nov 27 18:30:31 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Mon, 27 Nov 2006 12:30:31 -0500
Subject: practical numbers
References: <200611270805.AA74449328@TanyaKhovanova.com> <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com>
Message-ID: <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p>

Assuming Franklin's observation to be correct, why not

%C A005153 Also, n such that all k <= n are sums of distinct divisors of n.

Some poorly-thought-out questions before I head off to work:

Are there not-quite-practical numbers, that is, numbers such that n-1 is the 
only number < n which is not the sum of distinct divisors of n?

If not, is there some interesting bound f(n) < n such that if all k <= f(n) 
are sums of distinct divisors of n, then n is also a sum of distinct 
divisors of n?

----- Original Message ----- 
From: <franktaw at netscape.net>
To: <tanyakh at TanyaKhovanova.com>; <seqfan at ext.jussieu.fr>
Sent: Monday, November 27, 2006 12:04 PM
Subject: Re: practical numbers


> The definitions are equivalent.  The condition cited in the Wikipedia 
> article, to make everything up to n a sum of the divisors of n, is 
> sufficient to make everything up to sigma(n) be such a sum.
>
> Franklin T. Adams-Watters






From davidwwilson at comcast.net  Mon Nov 27 18:42:56 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Mon, 27 Nov 2006 12:42:56 -0500
Subject: practical numbers
References: <200611270805.AA74449328@TanyaKhovanova.com> <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com> <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p>
Message-ID: <000601c7124b$7c8b0040$6501a8c0@yourxhtr8hvc4p>

Simce any number n is a repeated sum of any of its divisors, "sum of 
distinct divisors" seems like overkill.

Anyway, another question: What is the smallest number a(n) that is a sum of 
its distinct divisors in n ways? 






From franktaw at netscape.net  Mon Nov 27 20:12:37 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 27 Nov 2006 14:12:37 -0500
Subject: practical numbers
In-Reply-To: <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p>
References: <200611270805.AA74449328@TanyaKhovanova.com> <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com> <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p>
Message-ID: <8C8E06F230775CD-7BC-A6B4@FWM-M18.sysops.aol.com>

To answer the second question first, it suffices to show that every 
number up to and including ceiling(n/2) is a sum of the divisors of n.  
Except for n=3, that can be floor(n/2) instead.  (If n is odd, 2 is not 
a sum; if n is even, every m with n/2 <= m < n is a sum n/2 plus a 
number less than n/2.)  This answers the first question, essentially in 
the negative: 3 is the only such number.

More interesting, perhaps: what about numbers such that every number up 
to n, with only 1 exception, are the sum of divisors of n?  (Note that 
this has to be n in this case, not sigma(n), since the possible sums 
are symmetric with respect to sigma(n) - if a set S of divisors sums to 
m, then the complement of S, with respect to the set of divisors of n, 
sums to sigma(n) - m.)

70 is one such number, skipping only 4; I think 70p for any prime 5 <= 
p < 70 does the same.  945 is another, skipping 2.  (945 is the 
smallest odd abundant number.)  I think that these values (70, 350, 
490, 770, 910, 945) are the only ones up to 1000.

The number skipped is always even; in fact, it always one less than a 
prime divisor p of n (and 1 more than the sum of the divisors of a 
practical divisor of n - which implies that that practical divisor is a 
square or twice a square).  Except for n=3, there must be another prime 
divisor q of n with p < q < 2p.

Franklin T. Adams-Watters


-----Original Message-----
From: davidwwilson at comcast.net

Assuming Franklin's observation to be correct, why not?
?
%C A005153 Also, n such that all k <= n are sums of distinct divisors 
of n.?
?
Some poorly-thought-out questions before I head off to work:?
?
Are there not-quite-practical numbers, that is, numbers such that n-1 
is the only number < n which is not the sum of distinct divisors of n??
?
If not, is there some interesting bound f(n) < n such that if all k <= 
f(n) are sums of distinct divisors of n, then n is also a sum of 
distinct divisors of n??
?
----- Original Message ----- From: <franktaw at netscape.net>?
To: <tanyakh at TanyaKhovanova.com>; <seqfan at ext.jussieu.fr>?
Sent: Monday, November 27, 2006 12:04 PM?
Subject: Re: practical numbers?
?
> The definitions are equivalent. The condition cited in the Wikipedia 
 > article, to make everything up to n a sum of the divisors of n, is > 
sufficient to make everything up to sigma(n) be such a sum.?
>?
> Franklin T. Adams-Watters?
?


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From franktaw at netscape.net  Mon Nov 27 20:25:04 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 27 Nov 2006 14:25:04 -0500
Subject: practical numbers
In-Reply-To: <000601c7124b$7c8b0040$6501a8c0@yourxhtr8hvc4p>
References: <200611270805.AA74449328@TanyaKhovanova.com> <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com> <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p> <000601c7124b$7c8b0040$6501a8c0@yourxhtr8hvc4p>
Message-ID: <8C8E070DFC523C7-7BC-A7AB@FWM-M18.sysops.aol.com>

I don't think "sum of distinct divisors" is really overkill.  If you 
leave out the "distinct", you'll get people complaining "but every 
number is a sum of its divisors" - or worse, misunderstanding.

A096356.  (Which should really be edited to say "proper divisors" 
instead of just "divisors".  Adding an initial 1 at offset 0 would also 
make sense to me.)

Franklin T. Adams-Watters


-----Original Message-----
From: davidwwilson at comcast.net

Simce any number n is a repeated sum of any of its divisors, "sum of 
distinct divisors" seems like overkill.?
?
Anyway, another question: What is the smallest number a(n) that is a 
sum of its distinct divisors in n ways? ?


________________________________________________________________________
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industry-leading spam and email virus protection.






From mathar at strw.leidenuniv.nl  Mon Nov 27 21:46:49 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Mon, 27 Nov 2006 21:46:49 +0100
Subject: Re^2: practical numbers
Message-ID: <200611272046.kARKknOV015522@amer.strw.leidenuniv.nl>


> From seqfan-owner at ext.jussieu.fr  Mon Nov 27 18:44:49 2006
> From: "David Wilson" <davidwwilson at comcast.net>
> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
> Subject: Re: practical numbers
> ...
> Anyway, another question: What is the smallest number a(n) that is a sum of 
> its distinct divisors in n ways? 

For each n we get the "trivial" solution where the sum consists of n itself
as the only term. Further solutions, where the sum contains more than one term
(all of which are distinct divisors), are listed below in the format
n [list of terms that sum to n]:

6 [1, 2, 3]
12 [1, 2, 3, 6]
12 [2, 4, 6]
18 [1, 2, 6, 9]
18 [3, 6, 9]
20 [1, 4, 5, 10]
24 [1, 2, 3, 4, 6, 8]
24 [1, 2, 3, 6, 12]
24 [2, 4, 6, 12]
24 [1, 3, 8, 12]
24 [4, 8, 12]
28 [1, 2, 4, 7, 14]
30 [1, 3, 5, 6, 15]
30 [2, 3, 10, 15]
30 [5, 10, 15]
36 [2, 3, 4, 6, 9, 12]
36 [2, 3, 4, 9, 18]
36 [1, 2, 6, 9, 18]
36 [3, 6, 9, 18]
36 [1, 2, 3, 12, 18]
36 [2, 4, 12, 18]
36 [6, 12, 18]
40 [1, 2, 4, 5, 8, 20]
40 [1, 4, 5, 10, 20]
40 [2, 8, 10, 20]
42 [1, 6, 14, 21]
42 [7, 14, 21]


n=6 has 1 extra solution, n=12 or 18 have 2 extra solutions, n=20 has one,
n=24 has five etc.

Maple program to create that one (not checked!):

a := proc(n)
	local findx,d,f,found,i,count ;
	count :=0 :
	# decompose n into all sums that are candidates for the decomposition
	d := combinat[partition](n) ;
	# take each sum indivdually
	for findx from 1 to nops(d) do
		f := op(findx,d) ;
		# start with the assumption that it works
		found := true ;
		for i from 2 to nops(f) do
			# if two terms are equal in the Maple list, 
			# we flag this case as "not admitted"
			if op(i,f) = op(i-1,f) then
				found := false ;
				break ;
			fi ;
		od ;
		for i from 1 to nops(f) do
			# if any term is not a divisor,
			# we flag this case as "not admitted"
			if n mod op(i,f) <> 0 then
				found := false ;
				break ;
			fi ;
		od ;
		# If this sum seems to work and has more than one term, we list it
		if found and nops(f) > 1 then
			printf("%a %a\n",n,f) ;
			count := count+1 ;
		fi ;
	od ;
	RETURN(count) ;
end:

# loop over the most fundamental cases..
for n from 1 to 47 do
	a(n) ;
od ;

# Richard Mathar





From zbi74583 at boat.zero.ad.jp  Tue Nov 28 07:18:47 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Tue, 28 Nov 2006 15:18:47 +0900
Subject: A038260
Message-ID: <025d01c712b5$16134290$977fd7dc@FMC57937B6BE00>

    Neil wrote on comment line of A038260.

    >Surely there are smaller members! njas, Nov 07 2006



    Why do you think that?

    I searched the ?seeds? for ?regular? Rational Amicable Number again.

       .

    And {11*19, 239 } is the smallest one.           .

    So, I think that at least it is the first terms of regular Rational 
Amicable Pair.

            .

    Such a Diophantine equation which is high degree and is involved with 
Sigma function has many conditions  , so the solution must have many prime 
factors for having good property, hence it becomes to have many digits.



    I guess that 2^2*7*3*5^2*13*31*139*277*3877*11*19 is the smallest and if 
an irregular example exists then it has around 16 digits.



    Yasutoshi


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From Annette.Warlich at t-online.de  Wed Nov 29 08:38:17 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Wed, 29 Nov 2006 08:38:17 +0100
Subject: Curious binomial-identity /A002720
In-Reply-To: <4568BB64.9050703@indiana.edu>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de> <4568BB64.9050703@indiana.edu>
Message-ID: <456D38E9.8030900@t-online.de>

Am 25.11.2006 22:53 schrieb Joseph Biberstine:
> The example you gave has an off-by-one row index as follows:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n+1)
> 
> Note that the row index is incremented.  Mathematica simplifies this
> fraction to LaguerreL[-(n+1), 1]/LaguerreL[n+1, -1].  This is not equal
> to Exp[1], but interestingly seems to approach Exp[1] monotonically from
> below as n approaches infinity.
> 
> I'm sure you meant:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n)
> 
> Mathematica simplifies this to LaguerreL[-(n+1), 1]/LaguerreL[n, -1].
> Although I can't get it to explicitly resolve that this is exactly
> Exp[1], the values do seem to bear this truth out.


I think I got it. As usual I like to write that problem in
matrix-mode.
So with

     V(x)~ * Pf = B

defined as the (infinite) matrix product as set of formal powerseries
in x:

                     1/0!   .     .     .
                     1/1!  1/0!
                     1/2!  2/1!  1/0!
                     1/3!  3/2!  3/1!  1/0!
                 .....
                   -------------------------------
[1 x x^2 x^3...]  = [ b0   b1    b2     b3 .... ]

we have, using the apostroph ' for derivative wrt x:

   b0  =            exp(x)      /0!
   b1  =  x*  [ x  *exp(x) ]'   /1!
   b2 =   x^2*[ x^2*exp(x) ]''  /2!
   b3 =   x^3*[ x^3*exp(x) ]''' /3!
 ...
   bk =   x^k*[ x^k*exp(x) ]'(k)' /k!

Expanding the derivatives gives

(1):

   b0 *0! / exp(x) =   1
   b1 *1! / exp(x) =   1 + x

   b2 *2! / exp(x) =   (2 + 4x + x^2)
                   =   2!/0! + 2*2!/1!  + 1*2!/2! x^2

   b3 *3! / exp(x) =   (6 + 18x + 9x^2 + x^3)
                        3! + 3*3!/1! x + 3*3!/2! x^2 + 1 *3!/3! x^3

   b4 *4! / exp(x) =   (24   + 96x  + 72x^2   + 16x^3 + 1x^4)
                       (1*4! + 4*4!/1! x + 6*4!/2!x^2 + 4*4!/3!x^3 + 1*4!/4!*x^4)



and the rhs written as matrix, where also the k! of the k'th row is cancelled , gives

(2):
   b0 / exp(x)      =   1                              *  [1,  x,  x^2,  x^3, ....]~
   b1 / exp(x)      =   1/0!  1/1!
   b2 / exp(x)      =   1/0!  2/1!  1/2!
   b3 / exp(x)      =   1/0!  3/1!  3/2!   1/3!
     ...            =     ....

If we now consider the result when letting x->1 we can further use
the horizontal symmetry of the binomial-coefficients to reorder
the rows and just get the matrix Pf again as coefficients

(3):

 [b0 b1 b2 b3...]~ /exp(1) =    1/0!                            * [1, 1, 1, 1, 1, ...]~
                                1/1!   1/0!
                                1/2!   2/1!   1/0!
                                1/3!   3/2!   3/1!   1/0!


 so

   V(1)~ * Pf = B~
 or
           Pf~ * V(1) = B
 and
           Pf  * V(1) = B / exp(x)

The systematic of the derivation of matrix Pf as resultant
of the derivative-proceeding in (1) and (2) needs now a better
systematic proof, but I think, that could be done...


--------------------------------------------------

One could proceed one further step and state

 for a current row and column m of Pf we have

(4):
    V(x)~ * Pf[0..inf,m] = exp(x) * Pf[m,0..m] * x^(2m) * V(1/x)

or (Pari:)

   suminf( k=0, x^k*Pf[k+1,m+1]) = exp(x) * x^(2*m) *  sum(k=0,m,Pf[m+1,k+1]/x^k)

where for a row r and col c the entry Pf[r+1,c+1] may be replaced by

      Pf[r+1,c+1] = ch(r,c) / (r-c)!          // for r>=c
                  = 0                         // for r<c

where ch(a,b) is the binomial-coefficient.

----------------

The common notation for the above observation may be given as:

    oo            x^k               m    m              x^(m-k)
   sum   ch(k,m)*------   = exp(x) x  * sum   ch(m,k)* ---------
    k=m          (k-m)!                 k=0              (m-k)!

                where ch(a,b) is the binomial-coefficient.


Gottfried Helms






From bowerc at usa.net  Wed Nov 29 10:49:58 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 29 Nov 2006 01:49:58 -0800
Subject: Categories
Message-ID: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>


I have one more row to add

------ Original Message ------
From: franktaw at netscape.net
To: seqfan at ext.jussieu.fr
Subject: Categories

> How many categories are there?
> 
> First, how many categories are there with n morphisms and k objects?  
> This table starts:
> 
>  1
>  2  1
>  7  3 1
> 35 16 3 1
> 
228 77 20 3 1

> The first column is A058129, the number of monoids; the main diagonal 
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
> 
> Taking the row sums, we get:
> 
> 1,3,11,55
329
> 
> the number of categories with n morphisms.  This is probably not in the 
> OEIS (only
> A001776 is possible - other matches become less than A058129).

While the 329 is eerily close to A001776's 330, EULER(A058129) is a
lower limit for this sequence and we have 2982 vs 2345 for #6 there.
>  The 
> inverse Euler
> transform,
> 
> 1,2,8,41
258
> 
> is the number of connected categories with n morphisms; this is 
> likewise probably not
> in the OEIS (only A052447 is possible).
No longer possible
> 
> Can somebody generate more data?
> 
> Franklin T. Adams-Watters
> 

Christian

PS

I don't know if it will be feasible to collect enough data to do
this, but the columns of the triangle converge and that convergence
would make an interesting sequence in and of itself.








From zakseidov at yahoo.com  Wed Nov 29 13:05:28 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 29 Nov 2006 04:05:28 -0800 (PST)
Subject: Difference between neighbor square and cube
Message-ID: <566064.14049.qm@web38215.mail.mud.yahoo.com>

Just submitted (with no autoreply).
My Q is:
Is the Conjectured list (see %C).. 
old hat or what?
Thanks, Zak

%I A125643
%S A125643
0,0,1,1,4,8,9,16,25,27,36,49,64,64,81,100,121,125,144,169,196,216,225,256,289,324,343,361,400,441,484,512,529,576,625,676,729,729,784,841,900,961,1000,1024,1089,1156,1225,1296,1331,1369,1444,1521,1600,1681,1728,1764,1849,1936,2025,2116,2197
%N A125643 Squares and cubes (with repetitions)
%C A125643 Cf. A002760 Squares and cubes (without
repetitions).
Conjectured list of numbers not appeared as difference
between neighbor terms:
5,6,10,14,16,22,23,29,32,34,42,44,46,50,52,54,58,62,64,66,70,72,78,82,84,86,88,90,92,93,94,96,98,102,105,110,112,114,117,118,120,122,124,126,130,132,134,136,140,141,142,144,153,156,158,160,162,164,165,166,172,176,178,179,182,188,190,192,194
%Y A125643 A002760
%O A125643 0
%K A125643 ,nonn,
%A A125643 Zak Seidov  (zakseidov at gmail.com), Nov 29 2006


 
____________________________________________________________________________________
Do you Yahoo!?
Everyone is raving about the all-new Yahoo! Mail beta.
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From maxale at gmail.com  Wed Nov 29 13:37:32 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 29 Nov 2006 04:37:32 -0800
Subject: Difference between neighbor square and cube
In-Reply-To: <566064.14049.qm@web38215.mail.mud.yahoo.com>
References: <566064.14049.qm@web38215.mail.mud.yahoo.com>
Message-ID: <d3dac270611290437x23c81f2fq6fb37e5d662b801d@mail.gmail.com>

Zak,

Your conjecture is related to Mordell curves of the form y^2 = x^3 + n, see
http://mathworld.wolfram.com/MordellCurve.html
http://tnt.math.metro-u.ac.jp/simath/MORDELL/

The latter link can help to resolve the most difficult case of whether
a given number is the difference of a square and a cube. In
particular, for n=5 and n=-5 the corresponding equation y^2 = x^3 + n
has no positive solutions, meaning that 5 is not a difference of
positive square and cube (not necessary neighbors, so this is even
stronger than needed by conjecture).
It is also easy to see that 5 is not the difference of two cubes. And
5 can be represented in an unique way as the difference of two
squares: 5 = 3^2 - 2^2 but 2^2 and 3^2 are not neighbors in A125643
since they are separated by 2^3.
Therefore, 5 indeed never appears as the difference between neighbor
terms of A125643.

Max


On 11/29/06, zak seidov <zakseidov at yahoo.com> wrote:
> Just submitted (with no autoreply).
> My Q is:
> Is the Conjectured list (see %C)..
> old hat or what?
> Thanks, Zak
>
> %I A125643
> %S A125643
> 0,0,1,1,4,8,9,16,25,27,36,49,64,64,81,100,121,125,144,169,196,216,225,256,289,324,343,361,400,441,484,512,529,576,625,676,729,729,784,841,900,961,1000,1024,1089,1156,1225,1296,1331,1369,1444,1521,1600,1681,1728,1764,1849,1936,2025,2116,2197
> %N A125643 Squares and cubes (with repetitions)
> %C A125643 Cf. A002760 Squares and cubes (without
> repetitions).
> Conjectured list of numbers not appeared as difference
> between neighbor terms:
> 5,6,10,14,16,22,23,29,32,34,42,44,46,50,52,54,58,62,64,66,70,72,78,82,84,86,88,90,92,93,94,96,98,102,105,110,112,114,117,118,120,122,124,126,130,132,134,136,140,141,142,144,153,156,158,160,162,164,165,166,172,176,178,179,182,188,190,192,194
> %Y A125643 A002760
> %O A125643 0
> %K A125643 ,nonn,
> %A A125643 Zak Seidov  (zakseidov at gmail.com), Nov 29 2006
>
>
>
> ____________________________________________________________________________________
> Do you Yahoo!?
> Everyone is raving about the all-new Yahoo! Mail beta.
> http://new.mail.yahoo.com
>





From zakseidov at yahoo.com  Wed Nov 29 14:44:52 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 29 Nov 2006 05:44:52 -0800 (PST)
Subject: Difference between neighbor square and cube
In-Reply-To: <d3dac270611290437x23c81f2fq6fb37e5d662b801d@mail.gmail.com>
Message-ID: <948080.43313.qm@web38204.mail.mud.yahoo.com>

Max,
thanks a lot for your (excellent as usual) reply!!

Using it I found:
A054504(*
Numbers n such that Mordell's equation y^2 = x^3 + n
has no integral solutions. *)=
{6, 7, 11, 13, 14, 20, 21, 23, 29, 32, 34, 39, 42, 45,
46, 47, 51, 53, 58, 59, 60, 61, 62, 66, 67, 69, 70,
74, 75, 77, 78, 83, 84, 85, 86, 87, 88, 90, 93, 95,
96, 102, 103, 104, 109, 110, 111, 114, 115, 116, 118,
123, 124, 130, 133, 135, 137, 139, 140, 146, 147, 149,
153, 155};

A081121=(*
Numbers n such that Mordell's equation y^2 = x^3 - n
has no integral solutions. *){
3, 5, 6, 9, 10, 12, 14, 16, 17, 21, 22, 24, 29, 30,
31, 32, 33, 34, 36, 37, 38, 41, 42, 43, 46, 50, 51,
52, 57, 58, 59, 62, 65, 66, 68, 69, 70, 73, 75, 77,
78, 80, 82, 84, 85, 86, 88, 90, 91, 92, 93, 94, 96,
97, 98, 99};

Intersection[A081121,A054504]
ND={6,14,21,29,32,34,42,46,51,58,59,62,66,69,70,75,77,78,84,85,86,88,90,93,96}

And ND list gives 
(Absolute values of) Non-difference between square and
cube (not only neighbor as in %C A125643).

Hence any number in ID should be in %C A125643.
But 21 is in ND and absent in %C A125643!

Indeed in A125643,
we have:
..,64,81,100,121,125,...,
and 121-100=11^2-10^2=21!!

21 is not difference between neighbor square and cube,
but it is difference between neighbor terms in
A125643.

As a result: my subject 'Difference between neighbor
square and cube' is not correct,
and in %C A125643 we have 
'Difference between neighbor square/cube and
square/cube'.

Sorry and thanks to Max & all seqfans,
Zak  


--- "Max A." <maxale at gmail.com> wrote:

> Zak,
> 
> Your conjecture is related to Mordell curves of the
> form y^2 = x^3 + n, see
> http://mathworld.wolfram.com/MordellCurve.html
> http://tnt.math.metro-u.ac.jp/simath/MORDELL/
> 
> The latter link can help to resolve the most
> difficult case of whether
> a given number is the difference of a square and a
> cube. In
> particular, for n=5 and n=-5 the corresponding
> equation y^2 = x^3 + n
> has no positive solutions, meaning that 5 is not a
> difference of
> positive square and cube (not necessary neighbors,
> so this is even
> stronger than needed by conjecture).
> It is also easy to see that 5 is not the difference
> of two cubes. And
> 5 can be represented in an unique way as the
> difference of two
> squares: 5 = 3^2 - 2^2 but 2^2 and 3^2 are not
> neighbors in A125643
> since they are separated by 2^3.
> Therefore, 5 indeed never appears as the difference
> between neighbor
> terms of A125643.
> 
> Max
> 
> 
> On 11/29/06, zak seidov <zakseidov at yahoo.com> wrote:
> > Just submitted (with no autoreply).
> > My Q is:
> > Is the Conjectured list (see %C)..
> > old hat or what?
> > Thanks, Zak
> >
> > %I A125643
> > %S A125643
> >
>
0,0,1,1,4,8,9,16,25,27,36,49,64,64,81,100,121,125,144,169,196,216,225,256,289,324,343,361,400,441,484,512,529,576,625,676,729,729,784,841,900,961,1000,1024,1089,1156,1225,1296,1331,1369,1444,1521,1600,1681,1728,1764,1849,1936,2025,2116,2197
> > %N A125643 Squares and cubes (with repetitions)
> > %C A125643 Cf. A002760 Squares and cubes (without
> > repetitions).
> > Conjectured list of numbers not appeared as
> difference
> > between neighbor terms:
> >
>
5,6,10,14,16,22,23,29,32,34,42,44,46,50,52,54,58,62,64,66,70,72,78,82,84,86,88,90,92,93,94,96,98,102,105,110,112,114,117,118,120,122,124,126,130,132,134,136,140,141,142,144,153,156,158,160,162,164,165,166,172,176,178,179,182,188,190,192,194
> > %Y A125643 A002760
> > %O A125643 0
> > %K A125643 ,nonn,
> > %A A125643 Zak Seidov  (zakseidov at gmail.com), Nov
> 29 2006
> >
> >
> >
> >
>
____________________________________________________________________________________
> > Do you Yahoo!?
> > Everyone is raving about the all-new Yahoo! Mail
> beta.
> > http://new.mail.yahoo.com
> >
> 



 
____________________________________________________________________________________
Do you Yahoo!?
Everyone is raving about the all-new Yahoo! Mail beta.
http://new.mail.yahoo.com





From jvospost3 at gmail.com  Wed Nov 29 19:52:44 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 29 Nov 2006 10:52:44 -0800
Subject: Categories
In-Reply-To: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
References: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
Message-ID: <5542af940611291052l1a3b2654k7467715f3073dfde@mail.gmail.com>

Thank you, Christian.  That looks right.

Note that the partial sums of the first column are A118601 Number of monoids
(semigroups with identity) of order <=n.  This first column also equals the
row sums of A058137, which partitions the monoids of order n by the number k
of idempotents.






On 11/29/06, Christian G. Bower <bowerc at usa.net> wrote:
>
>
> I have one more row to add
>
> ------ Original Message ------
> Received: Wed, 22 Nov 2006 01:21:01 PM PST
> From: franktaw at netscape.net
> To: seqfan at ext.jussieu.fr
> Subject: Categories
>
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> >  1
> >  2  1
> >  7  3 1
> > 35 16 3 1
> >
> 228 77 20 3 1
>
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
> > Taking the row sums, we get:
> >
> > 1,3,11,55
> 329
> >
> > the number of categories with n morphisms.  This is probably not in the
> > OEIS (only
> > A001776 is possible - other matches become less than A058129).
>
> While the 329 is eerily close to A001776's 330, EULER(A058129) is a
> lower limit for this sequence and we have 2982 vs 2345 for #6 there.
> >  The
> > inverse Euler
> > transform,
> >
> > 1,2,8,41
> 258
> >
> > is the number of connected categories with n morphisms; this is
> > likewise probably not
> > in the OEIS (only A052447 is possible).
> No longer possible
> >
> > Can somebody generate more data?
> >
> > Franklin T. Adams-Watters
> >
>
> Christian
>
> PS
>
> I don't know if it will be feasible to collect enough data to do
> this, but the columns of the triangle converge and that convergence
> would make an interesting sequence in and of itself.
>
>
>
>
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From franktaw at netscape.net  Wed Nov 29 20:09:00 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 29 Nov 2006 14:09:00 -0500
Subject: Categories
In-Reply-To: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
References: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
Message-ID: <8C8E200F6031B3A-794-8706@FWM-M14.sysops.aol.com>

Thanks, Christian.

I think we need to look at one more table:  the number of connected
categories with n morphisms and k objects.  This starts:

  1
  2
  7 1
 35 6
228 28 2

(Row n has length ceiling(n/2).)

The table of the number of categories that I started with is a two-
dimensional Euler transform of this table.

The downward-sloping diagonal sums of this table are 1,3,15,???.
Shifting this left and taking the Euler transform gives the limiting
values Christian is referring to; starting with b(0):

1,3,21,???

For the table above, a(2n-1,n) has a purely algebraic or graph-
theoretic interpretation.  It is the number of connected
anti-transitive relations on n objects (meaning that if a R b and
b R c, then NOT (a R c)); equivalently, the number of bipartite
oriented trees, where each edge origin is in the same part.

If I haven't made any mistakes, this sequence starts:

1,1,2,3,6,10

This is not enough data to determine whether it is in the OEIS.

Franklin T. Adams-Watters


-----Original Message-----
From: bowerc at usa.net

I have one more row to add

------ Original Message ------

> How many categories are there?
>
> First, how many categories are there with n morphisms and k objects?
> This table starts:
>
>  1
>  2  1
>  7  3 1
> 35 16 3 1
>
228 77 20 3 1

> The first column is A058129, the number of monoids; the main diagonal
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
>
> Taking the row sums, we get:
>
> 1,3,11,55
329
>
> the number of categories with n morphisms.  This is probably not in 
the
> OEIS (only
> A001776 is possible - other matches become less than A058129).

While the 329 is eerily close to A001776's 330, EULER(A058129) is a
lower limit for this sequence and we have 2982 vs 2345 for #6 there.
>  The
> inverse Euler
> transform,
>
> 1,2,8,41
258
>
> is the number of connected categories with n morphisms; this is
> likewise probably not
> in the OEIS (only A052447 is possible).
No longer possible
>
> Can somebody generate more data?
>
> Franklin T. Adams-Watters
>

Christian

PS

I don't know if it will be feasible to collect enough data to do
this, but the columns of the triangle converge and that convergence
would make an interesting sequence in and of itself.

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From jvospost3 at gmail.com  Wed Nov 29 20:23:47 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 29 Nov 2006 11:23:47 -0800
Subject: Categories
In-Reply-To: <8C8E200F6031B3A-794-8706@FWM-M14.sysops.aol.com>
References: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
	 <8C8E200F6031B3A-794-8706@FWM-M14.sysops.aol.com>
Message-ID: <5542af940611291123l4f9c3385l283f3b2243dab1c2@mail.gmail.com>

The issue of "connected" in the underlying functional graph, as a subset of
the set of all digraphs, is further revealed by the distinction (definitions
at MathWorld) of Weakly Conncted versus Strongly Conncted Digraphs.

Every functional digraph is the disjoint union of connected digraphs.
That's why there is a recursion in the enumeration of F(n) = functional
graphs on n points. F(n) is equal or greater to the number of functional
graphs on n-1 points, unioned with an isolated point looping to itself, plus
the number of functional graphs on n-2 points unioned with any of the 3
functional graphs on 3 points, ..., throwing out any double or multiple
counts by equivalence class on isomorphisms of the unlabelled digraphs.

Note that MathWorld's correct definition of functional graph, and its
Mathematica, is identical to what I'd called endofunctions in earlier
posting of this thread.  Hence the enumeration problem that I solved for
"prime" endofunctions is identical to a subset of your category enumeration
problem, relating to the number of categories that are not unions nor
categorical products of smaller categories.

However, when I referred to 2-categories before, I should more correctly
have said Bicategories, the difference being based on whether associativity
is strict or up to 2-isomorphism.  Your enumeration of categories is more
formally explained in terms of Cat, the category of categories and functors,
but this is not the right venue for me to go into excruciating details.

On 11/29/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
> Thanks, Christian.
>
> I think we need to look at one more table:  the number of connected
> categories with n morphisms and k objects.  This starts:
>
>   1
>   2
>   7 1
> 35 6
> 228 28 2
>
> (Row n has length ceiling(n/2).)
>
> The table of the number of categories that I started with is a two-
> dimensional Euler transform of this table.
>
> The downward-sloping diagonal sums of this table are 1,3,15,???.
> Shifting this left and taking the Euler transform gives the limiting
> values Christian is referring to; starting with b(0):
>
> 1,3,21,???
>
> For the table above, a(2n-1,n) has a purely algebraic or graph-
> theoretic interpretation.  It is the number of connected
> anti-transitive relations on n objects (meaning that if a R b and
> b R c, then NOT (a R c)); equivalently, the number of bipartite
> oriented trees, where each edge origin is in the same part.
>
> If I haven't made any mistakes, this sequence starts:
>
> 1,1,2,3,6,10
>
> This is not enough data to determine whether it is in the OEIS.
>
> Franklin T. Adams-Watters
>
>
> -----Original Message-----
> From: bowerc at usa.net
>
> I have one more row to add
>
> ------ Original Message ------
> Received: Wed, 22 Nov 2006 01:21:01 PM PST
>
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> >  1
> >  2  1
> >  7  3 1
> > 35 16 3 1
> >
> 228 77 20 3 1
>
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
> > Taking the row sums, we get:
> >
> > 1,3,11,55
> 329
> >
> > the number of categories with n morphisms.  This is probably not in
> the
> > OEIS (only
> > A001776 is possible - other matches become less than A058129).
>
> While the 329 is eerily close to A001776's 330, EULER(A058129) is a
> lower limit for this sequence and we have 2982 vs 2345 for #6 there.
> >  The
> > inverse Euler
> > transform,
> >
> > 1,2,8,41
> 258
> >
> > is the number of connected categories with n morphisms; this is
> > likewise probably not
> > in the OEIS (only A052447 is possible).
> No longer possible
> >
> > Can somebody generate more data?
> >
> > Franklin T. Adams-Watters
> >
>
> Christian
>
> PS
>
> I don't know if it will be feasible to collect enough data to do
> this, but the columns of the triangle converge and that convergence
> would make an interesting sequence in and of itself.
>
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and
> industry-leading spam and email virus protection.
>
>
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From njas at research.att.com  Wed Nov 29 22:34:56 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 29 Nov 2006 16:34:56 -0500 (EST)
Subject: OEIS on vacation in December
Message-ID: <200611292134.QAA53261@fry.research.att.com>


This means:
- do send in sequences in published works or 
   that are important for your work
- do send important corrections, extensions, updates
- do send EDITED versions

- don't send in sequences that you made up (unless
   they are really beautiful - and if they
   are base dependent or involve primes 
   they are probably not)
- don't send very minor corrections

I will be doing occasional updates, but I won't be
working on the OEIS 24/7 !

Thanks!

Neil





From maxale at gmail.com  Wed Nov 29 22:45:38 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 29 Nov 2006 13:45:38 -0800
Subject: Difference between neighbor square and cube
In-Reply-To: <948080.43313.qm@web38204.mail.mud.yahoo.com>
References: <d3dac270611290437x23c81f2fq6fb37e5d662b801d@mail.gmail.com>
	 <948080.43313.qm@web38204.mail.mud.yahoo.com>
Message-ID: <d3dac270611291345n2ad5dae4id319f3bce8fca71c@mail.gmail.com>

On 11/29/06, zak seidov <zakseidov at yahoo.com> wrote:

> Intersection[A081121,A054504]
> ND={6,14,21,29,32,34,42,46,51,58,59,62,66,69,70,75,77,78,84,85,86,88,90,93,96}

That's A110223. But Robert G. Wilson posed as "conjectured".
I believe this sequence can be confirmed/verified using the
information about Mordell curves from
http://tnt.math.metro-u.ac.jp/simath/MORDELL/

Max





From bowerc at usa.net  Thu Nov 30 00:42:31 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 29 Nov 2006 15:42:31 -0800
Subject: Categories
Message-ID: <623kkCXpf3086S27.1164843751@cmsweb27.cms.usa.net>

ftaw
> I think we need to look at one more table:  the number of connected
> categories with n morphisms and k objects.  This starts:

>   1
>   2
>   7 1
>  35 6
> 228 28 2

> (Row n has length ceiling(n/2).)
...
> For the table above, a(2n-1,n) has a purely algebraic or graph-
> theoretic interpretation.  It is the number of connected
> anti-transitive relations on n objects (meaning that if a R b and
> b R c, then NOT (a R c)); equivalently, the number of bipartite
> oriented trees, where each edge origin is in the same part.
> 
> If I haven't made any mistakes, this sequence starts:
> 
> 1,1,2,3,6,10
> 
> This is not enough data to determine whether it is in the OEIS.

For n>1 it's A122086
http://www.research.att.com/~njas/sequences/A122086
A sequence which could use a nicer description and formula such as
a(n) = A000055(n) - A000081(n/2) where a sequence evaluated at a
noninteger is 0.
http://www.research.att.com/~njas/sequences/A000055
http://www.research.att.com/~njas/sequences/A000081

Christian








From bowerc at usa.net  Thu Nov 30 01:36:23 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 29 Nov 2006 16:36:23 -0800
Subject: Categories
Message-ID: <209kkdajX8954S10.1164846983@cmsweb10.cms.usa.net>

I have row 6 now:

> How many categories are there?
>
> First, how many categories are there with n morphisms and k objects?
> This table starts:
>
>  1
>  2  1
>  7  3 1
> 35 16 3 1
>
228 77 20 3 1
2237 485 111 21 3 1

> Taking the row sums, we get:
>
> 1,3,11,55
329
2858

>  The
> inverse Euler
> transform,
>
> 1,2,8,41
258
2407

> I think we need to look at one more table:  the number of connected
> categories with n morphisms and k objects.  This starts:

>   1
>   2
>   7 1
>  35 6
> 228 28 2
2237 159 11
31559 ?   ? 3

> (Row n has length ceiling(n/2).)
> 
> The table of the number of categories that I started with is a two-
> dimensional Euler transform of this table.
> 
> The downward-sloping diagonal sums of this table are 1,3,15,???.
77
> Shifting this left and taking the Euler transform gives the limiting
> values Christian is referring to; starting with b(0):
> 
> 1,3,21,???
132








From franktaw at netscape.net  Thu Nov 30 14:53:38 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 30 Nov 2006 08:53:38 -0500
Subject: Categories
In-Reply-To: <623kkCXpf3086S27.1164843751@cmsweb27.cms.usa.net>
References: <623kkCXpf3086S27.1164843751@cmsweb27.cms.usa.net>
Message-ID: <8C8E29E12708888-5D8-4EB@WEBMAIL-RA17.sysops.aol.com>

A typo there; that should be a(n) = 2*A000055(n) - A000081(n/2).

Franklin T. Adams-Watters


-----Original Message-----
From: bowerc at usa.net

...
For n>1 it's A122086
http://www.research.att.com/~njas/sequences/A122086
A sequence which could use a nicer description and formula such as
a(n) = A000055(n) - A000081(n/2) where a sequence evaluated at a
noninteger is 0.
http://www.research.att.com/~njas/sequences/A000055
http://www.research.att.com/~njas/sequences/A000081

Christian





________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From jvospost3 at gmail.com  Thu Nov 30 19:34:51 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 30 Nov 2006 10:34:51 -0800
Subject: Categories
In-Reply-To: <8C8E29E12708888-5D8-4EB@WEBMAIL-RA17.sysops.aol.com>
References: <623kkCXpf3086S27.1164843751@cmsweb27.cms.usa.net>
	 <8C8E29E12708888-5D8-4EB@WEBMAIL-RA17.sysops.aol.com>
Message-ID: <5542af940611301034t3bba13fdnadc86411facc535d@mail.gmail.com>

a(n) = 2*A000055(n) - A000081(n/2) = (2* Number of trees with n unlabeled
nodes) - Number of rooted trees with n/2 nodes (or connected functions with
a fixed point)

where a(n) is downwards sloping diagonal sums of main category table, or
what?

It isn't until we look at bicategories that we are forced to look at the
symmetries and structure of the individual categories being enumerated here,
as objects in the higher categories, and have to enumerate the morphisms of
the functions. A (finite, concrete) category whose underlying directed graph
has no symmetry (i.e. automorphism group of order 1) can only map to itself
or an identical copy of itself, point to point, loop to loop, arc to arc.
But any cycle of length >1 in that directed graph allows morphisms to other
directed graphs.  The directed graph C_n, a cycle of length n, can thus map
to the same graph but mapping each point to the next, and each arc to the
next, and so forth. That requires then analysis of cycles in directed
graphs, as by de Bruijn in 1952 and more deeply understood since then. This
is another enumeration problem, unnecessary to completing the answer to
Frank's original question, and more tied to my endofunctions analysis. Sorry
if my interruptions distracted you from the straightforward question.  Good
work, guys!

Funny how Category Theory folks tend to be addicted to theory for its own
sake, and less willing to "get their hands dirty" with anything as old
fashioned as enumeration.  OEIS helps us focus here, by demanding the
integer sequences!

On 11/30/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
> A typo there; that should be a(n) = 2*A000055(n) - A000081(n/2).
>
> Franklin T. Adams-Watters
>
>
> -----Original Message-----
> From: bowerc at usa.net
>
> ...
> For n>1 it's A122086
> http://www.research.att.com/~njas/sequences/A122086
> A sequence which could use a nicer description and formula such as
> a(n) = A000055(n) - A000081(n/2) where a sequence evaluated at a
> noninteger is 0.
> http://www.research.att.com/~njas/sequences/A000055
> http://www.research.att.com/~njas/sequences/A000081
>
> Christian
>
>
>
>
>
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and
> industry-leading spam and email virus protection.
>
>
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From all at abouthugo.de  Thu Nov 30 21:28:06 2006
From: all at abouthugo.de (Hugo Pfoertner)
Date: Thu, 30 Nov 2006 21:28:06 +0100
Subject: Difference between neighbor square and cube
References: <948080.43313.qm@web38204.mail.mud.yahoo.com>
Message-ID: <456F3ED6.D0619322@abouthugo.de>

zak seidov wrote:
> 
> Max,
> thanks a lot for your (excellent as usual) reply!!
> 
> Using it I found:
> A054504(*
> Numbers n such that Mordell's equation y^2 = x^3 + n
> has no integral solutions. *)=
> {6, 7, 11, 13, 14, 20, 21, 23, 29, 32, 34, 39, 42, 45,
> 46, 47, 51, 53, 58, 59, 60, 61, 62, 66, 67, 69, 70,
> 74, 75, 77, 78, 83, 84, 85, 86, 87, 88, 90, 93, 95,
> 96, 102, 103, 104, 109, 110, 111, 114, 115, 116, 118,
> 123, 124, 130, 133, 135, 137, 139, 140, 146, 147, 149,
> 153, 155};
> 
> A081121=(*
> Numbers n such that Mordell's equation y^2 = x^3 - n
> has no integral solutions. *){
> 3, 5, 6, 9, 10, 12, 14, 16, 17, 21, 22, 24, 29, 30,
> 31, 32, 33, 34, 36, 37, 38, 41, 42, 43, 46, 50, 51,
> 52, 57, 58, 59, 62, 65, 66, 68, 69, 70, 73, 75, 77,
> 78, 80, 82, 84, 85, 86, 88, 90, 91, 92, 93, 94, 96,
> 97, 98, 99};
> 
> Intersection[A081121,A054504]
> ND={6,14,21,29,32,34,42,46,51,58,59,62,66,69,70,75,77,78,84,85,86,88,90,93,96}
> 
> And ND list gives
> (Absolute values of) Non-difference between square and
> cube (not only neighbor as in %C A125643).
> 
> Hence any number in ID should be in %C A125643.
> But 21 is in ND and absent in %C A125643!
> 
> Indeed in A125643,
> we have:
> ..,64,81,100,121,125,...,
> and 121-100=11^2-10^2=21!!
> 
> 21 is not difference between neighbor square and cube,
> but it is difference between neighbor terms in
> A125643.
> 
> As a result: my subject 'Difference between neighbor
> square and cube' is not correct,
> and in %C A125643 we have
> 'Difference between neighbor square/cube and
> square/cube'.
> 
> Sorry and thanks to Max & all seqfans,
> Zak

Also closely related:

http://www.research.att.com/~njas/sequences/A087285
http://www.research.att.com/~njas/sequences/A087286
http://www.research.att.com/~njas/sequences/A088017

Hugo





From bowerc at usa.net  Thu Nov 30 22:29:20 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Thu, 30 Nov 2006 13:29:20 -0800
Subject: Categories
Message-ID: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>



------ Original Message ------
From: "Jonathan Post" <jvospost3 at gmail.com>
To: "franktaw at netscape.net" <franktaw at netscape.net>Cc: bowerc at usa.net,
seqfan at ext.jussieu.fr, jvospost2 at yahoo.com
Subject: Re: Categories

> a(n) = 2*A000055(n) - A000081(n/2) = (2* Number of trees with n unlabeled
> nodes) - Number of rooted trees with n/2 nodes (or connected functions with
> a fixed point)
> 
> where a(n) is downwards sloping diagonal sums of main category table, or
> what?

Actually it's the "corner" values from this table

>   *1*
>   2
>   7 *1*
>  35 6
> 228 28 *2*
> 2237 159 11
> 31559 ?   ? *3*

The downward sloping sums were an aide in calculating the convergent
sequence of the columns

The interpretation of these corners comes from the fact we have a
saturated category. It has 2n-1 elements, n of which are object
identities. It's connected, so it has a minimum of n-1 "crossover"
morphisms, it also has a maximum of n-1 cm's since there are only
n-1 elements left. Any connected graph with n points and n-1 edges is
a tree. The associativity of categories states that if we have an
a-b morphism (an edge on the graph) and a b-c we must have an a-c.
But if we have all 3, it's no longer a tree, hence we never have the
a-b and the b-c, hence each point has all edges either pointing in or
pointing out, hence the biparite graph analogy, hence a little
combinatorial magic.

...
> Funny how Category Theory folks tend to be addicted to theory for its > own
> sake, and less willing to "get their hands dirty" with anything as old
> fashioned as enumeration.  OEIS helps us focus here, by demanding the
> integer sequences!
> 

I try to stay out of philosophical stuff here, but sometimes it's too
hard to resist. I figure I'm guilty of spending too much time counting
stuff and not taking the theory in. Most of the world would not give a
hoot about either, so I'm glad there are folk spending time on what I
would not give any time to.

Still, I find it jarring any time I read a math paper and they write
about several things that can be represented as sequences, yet all they
give is a mathematical description, rarely do they cite the OEIS
reference or even give me the first n terms leaving me to calculate them
myself and hope I didn't make a mistake. If I wrote a paper, I'd
include that stuff, not just because I'm an OEIS nerd, but also because
I would want people to be able to find my paper who didn't know they
were looking for it.

Just a little more self indulgence here with my list of semigroup pipe
dreams. After doing sequences on semigroups
http://www.research.att.com/~njas/sequences/A027851
and monoids
http://www.research.att.com/~njas/sequences/A058129
I knew categories was one of the next logical steps, but I held off
because I knew it would be difficult. Perhaps I can spur someone else's
interest in related problems that may prove even more difficult.

There's transformation semigroups, the semigroup version of permutation
groups
http://www.research.att.com/~njas/sequences/A000638
I've seen TSs written about, but no attempts to enumerate them.
(Note, a TS is any set of endofunctions closed under composition)
There TS's close cousins which could be called:
transformation monoids
relation semigroups
relation monoids
(Composition of relations is such that (a,c) is in R1(R2) iff there
is b such that (a,b) in R1 and (b,c) in R2)
Then there are the generalizations of categories to semigroups and
groupoids. Note I mean the groupoid that is a set and closed operation
http://www.research.att.com/~njas/sequences/A001329
Not the group like category groupoid which might also be interesting
to count and not too difficult; oh terminology.
Anyway, with the semigroup and groupoid categories, there comes the
issue of treating objects and morphims. The ordinary (monoid like)
categories avoid this problem by having object identities among the
morphisms that can be associatied with the objects themselves. Without
that restriction, it's possible to have objects untouched by any
morphism or two structures with identical (or isomorphic) multiplication
tables, but not isomorphic when you take the objects into account.
Basically many of the same counting problems one has with multigraphs
(and probably the same solutions too.)








From jvospost3 at gmail.com  Thu Nov 30 22:53:42 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 30 Nov 2006 13:53:42 -0800
Subject: Categories
In-Reply-To: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>
References: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>
Message-ID: <5542af940611301353m486a4788wff2e4eb539b76943@mail.gmail.com>

I strongly agree with Christian.  My doctoral dissertation involved taking
the Krohn-Rhodes decomposition of the semigroup of differential operators of
a system of nonlinear differential equations that had been in the
mathematical Biology literature for decades, and routinely dismissed as not
capable of closed-form solution. Marvin Minsky (who had been the PhD advisor
of the eponymous John Rhodes) thought my work was the best use yet of his
student's theorem.  The great Stan Ulam (coinventor of cellular automata,
the H-bomb, and the nuclear pulse rocket) also thought it was cool. But only
half a dozen biologists in theb world knew that much about semigroups, and
some of them wrote to me from Edinburgh, Scotland, and places in the USSR.
Ulam died before we could coauthor; Semigroup Forum accepted my paper with
delight for a special issue, but wanted it reformatted, and I was laid off
from the company where I worked, and they erased on only machine-readable
version of the paper.  Equations were harder to typeset those decades
ago...  But yes, lots of so-called professionals go out of their way to
avoid citing the OEIS, or give any useful examples.  Also yes: semigroup
enumerations are worthwhile.  Also, you nailed the differences between those
and categories.

See for instance:

  A118581 <http://www.research.att.com/~njas/sequences/A118581> Number of
nonisomorphic *semigroups* of order <= n.

  A118099 <http://www.research.att.com/~njas/sequences/A118099> Number of
inverse *semigroups* of order <= n.

  A118100 <http://www.research.att.com/~njas/sequences/A118100>

Number of commutative *semigroups* of order <= n.

  A118601 <http://www.research.att.com/~njas/sequences/A118601> Number of
monoids (*semigroups* with identity) of order <= n.
  A113534 <http://www.research.att.com/~njas/sequences/A113534> Ascending
descending base exponent transform of the flipped tribonacci substitution (
A092782 <http://www.research.att.com/~njas/sequences/A092782>).
with its semigroup reference, and some others.

-- Jonathan


On 11/30/06, Christian G. Bower <bowerc at usa.net> wrote:
>
>
>
> ------ Original Message ------
> Received: Thu, 30 Nov 2006 10:34:55 AM PST
> From: "Jonathan Post" <jvospost3 at gmail.com>
> To: "franktaw at netscape.net" <franktaw at netscape.net>Cc: bowerc at usa.net,
> seqfan at ext.jussieu.fr, jvospost2 at yahoo.com
> Subject: Re: Categories
>
> > a(n) = 2*A000055(n) - A000081(n/2) = (2* Number of trees with n
> unlabeled
> > nodes) - Number of rooted trees with n/2 nodes (or connected functions
> with
> > a fixed point)
> >
> > where a(n) is downwards sloping diagonal sums of main category table, or
> > what?
>
> Actually it's the "corner" values from this table
>
> >   *1*
> >   2
> >   7 *1*
> >  35 6
> > 228 28 *2*
> > 2237 159 11
> > 31559 ?   ? *3*
>
> The downward sloping sums were an aide in calculating the convergent
> sequence of the columns
>
> The interpretation of these corners comes from the fact we have a
> saturated category. It has 2n-1 elements, n of which are object
> identities. It's connected, so it has a minimum of n-1 "crossover"
> morphisms, it also has a maximum of n-1 cm's since there are only
> n-1 elements left. Any connected graph with n points and n-1 edges is
> a tree. The associativity of categories states that if we have an
> a-b morphism (an edge on the graph) and a b-c we must have an a-c.
> But if we have all 3, it's no longer a tree, hence we never have the
> a-b and the b-c, hence each point has all edges either pointing in or
> pointing out, hence the biparite graph analogy, hence a little
> combinatorial magic.
>
> ...
> > Funny how Category Theory folks tend to be addicted to theory for its >
> own
> > sake, and less willing to "get their hands dirty" with anything as old
> > fashioned as enumeration.  OEIS helps us focus here, by demanding the
> > integer sequences!
> >
>
> I try to stay out of philosophical stuff here, but sometimes it's too
> hard to resist. I figure I'm guilty of spending too much time counting
> stuff and not taking the theory in. Most of the world would not give a
> hoot about either, so I'm glad there are folk spending time on what I
> would not give any time to.
>
> Still, I find it jarring any time I read a math paper and they write
> about several things that can be represented as sequences, yet all they
> give is a mathematical description, rarely do they cite the OEIS
> reference or even give me the first n terms leaving me to calculate them
> myself and hope I didn't make a mistake. If I wrote a paper, I'd
> include that stuff, not just because I'm an OEIS nerd, but also because
> I would want people to be able to find my paper who didn't know they
> were looking for it.
>
> Just a little more self indulgence here with my list of semigroup pipe
> dreams. After doing sequences on semigroups
> http://www.research.att.com/~njas/sequences/A027851
> and monoids
> http://www.research.att.com/~njas/sequences/A058129
> I knew categories was one of the next logical steps, but I held off
> because I knew it would be difficult. Perhaps I can spur someone else's
> interest in related problems that may prove even more difficult.
>
> There's transformation semigroups, the semigroup version of permutation
> groups
> http://www.research.att.com/~njas/sequences/A000638
> I've seen TSs written about, but no attempts to enumerate them.
> (Note, a TS is any set of endofunctions closed under composition)
> There TS's close cousins which could be called:
> transformation monoids
> relation semigroups
> relation monoids
> (Composition of relations is such that (a,c) is in R1(R2) iff there
> is b such that (a,b) in R1 and (b,c) in R2)
> Then there are the generalizations of categories to semigroups and
> groupoids. Note I mean the groupoid that is a set and closed operation
> http://www.research.att.com/~njas/sequences/A001329
> Not the group like category groupoid which might also be interesting
> to count and not too difficult; oh terminology.
> Anyway, with the semigroup and groupoid categories, there comes the
> issue of treating objects and morphims. The ordinary (monoid like)
> categories avoid this problem by having object identities among the
> morphisms that can be associatied with the objects themselves. Without
> that restriction, it's possible to have objects untouched by any
> morphism or two structures with identical (or isomorphic) multiplication
> tables, but not isomorphic when you take the objects into account.
> Basically many of the same counting problems one has with multigraphs
> (and probably the same solutions too.)
>
>
>
>
-------------- next part --------------
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From franktaw at netscape.net  Thu Nov 30 22:54:39 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 30 Nov 2006 16:54:39 -0500
Subject: Categories
In-Reply-To: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>
References: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>
Message-ID: <8C8E2E14509A6B2-934-A97D@FWM-M10.sysops.aol.com>

I'd like to put my plug in for more application in math.

I heard of a case some years ago where a paper was published about some 
kind of mathematical object (I think it was a kind of topological 
space, but it really doesn't matter).  Three or four more papers were 
published, establishing more and more properties for this type of 
object - until finally it was proved that they don't exist!  This 
wouldn't have happened if somebody had asked for an example at an early 
stage.

There is a strong tendency in mathematics to start at the end.  The 
researcher pursues a line of thought, which eventually leads to a 
spiffy proof.  The proof is then published, with no hint of the process 
by which it was reached.  This is a disservice to anybody who might use 
a similar approach to solve some other problem.  It is especially a 
disservice when presented to students.

On a more personal level, I find when looking a math paper, I want to 
know how this relates to problems that I am already interested in or at 
least familiar with.  If I can't get an answer to that, I have a hard 
time maintaining any interest.

Franklin T. Adams-Watters

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From zakseidov at yahoo.com  Wed Nov  1 06:43:12 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Tue, 31 Oct 2006 21:43:12 -0800 (PST)
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
Message-ID: <20061101054313.25333.qmail@web38210.mail.mud.yahoo.com>

Dear seqfans,
Bob, Ivars,

In A065577, 
a(1)=2 because 10=3+7=5+5?
Also a(2)=6 because
100=3+97=11+89=17+83=29+71=41+59=47+53?

The point is that in Ivars Peterson's
"Goldbach's Prime Pairs"
http://www.maa.org/mathland/mathtrek_8_21_00.html
we read:

" 
Note that Goldbach partitions take into account the
order of the primes that are summed. For example, 10
has two Goldbach partitions: 3 + 7 and 7 + 3.

Integer No. of Goldbach partitions 
10 2 
100 6 
1,000 28 
10,000 127 
100,000 810 
1,000,000 5,402 
10,000,000 38,807 
100,000,000 291,400
" 

Me: But in A065577 a(1)=2 and a(2)=6 
(and also other terms as I checked them)
without taking into account 
the order of the primes 
that are summed(?!)

Can anyoone (Bob? Ivars?) clarify this?
Thanks, Zak

BTW Two next terms are calculated by me
(for n=9, 10) as 2274205, 18200487.
Can anyone check these?
Thanks, Zak


%I A065577
%S A065577 2,6,28,127,810,5402,38807,291400
%N A065577 Number of Goldbach partitions of 10^n.
%H A065577 Ivars Peterson's MathTrek, <a
href="http://www.maa.org/mathland/mathtrek_8_21_00.html">Goldbach's
Prime Pairs</a>
%H A065577 Science News Online, week of Aug. 19, 2000;
Vol. 158, No. 8 <a href="http://www.sciencenews.org/
20000819/mathtrek.asp">Goldbach's Prime Pairs</a>
<skip>
%A A065577 Robert G. Wilson v (rgwv(AT)rgwv.com), Dec
01 2001




 
____________________________________________________________________________________
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From ralf at ark.in-berlin.de  Wed Nov  1 10:35:28 2006
From: ralf at ark.in-berlin.de (Ralf Stephan)
Date: Wed, 1 Nov 2006 10:35:28 +0100
Subject: Re^2: Error-finding program
In-Reply-To: <200610312150.k9VLoKfn018613@amer.strw.leidenuniv.nl>
References: <200610312150.k9VLoKfn018613@amer.strw.leidenuniv.nl>
Message-ID: <20061101093528.GA19017@ark.in-berlin.de>

> The definitions of A026468 and A026469 are
>  a(n) = least positive integer > a(n-1) and not a(i)^2 + a(j)^2 for 1<=i<=j<=n.
> and
>  a(n) = least positive integer > a(n-1) and not a(i)^2 + a(j)^2 for 1<=i<=j<=n.  	
> which looks quite identical. So why can A026468/9 be two different
> sequences? Is there a "not" too much or too little, or rather one of the 
> "less-equal" signs to be corrected to "less" or a starting condition on a(2)
> to be added?

My guess is that Clark started the same algorithm with 1, 2 and then
tried it with starting values 1, 3, so the definition of A026469 should 
be amended with ", starting with 1, 3".


ralf






From njas at research.att.com  Wed Nov  1 14:16:47 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 1 Nov 2006 08:16:47 -0500 (EST)
Subject: Re^2: Error-finding program
References: <200610312150.k9VLoKfn018613@amer.strw.leidenuniv.nl>
Message-ID: <200611011316.IAA53531@fry.research.att.com>

Ralf,  I think you've solved it!   Thanks!   Neil


My guess is that Clark started the same algorithm with 1, 2 and then
tried it with starting values 1, 3, so the definition of A026469 should 
be amended with ", starting with 1, 3".





From mathar at strw.leidenuniv.nl  Wed Nov  1 14:17:58 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Wed, 1 Nov 2006 14:17:58 +0100
Subject: A065577 Number of Goldbach partitions of 10^n?
Message-ID: <200611011317.kA1DHwmp018846@amer.strw.leidenuniv.nl>


zs> From seqfan-owner at ext.jussieu.fr  Wed Nov  1 06:44:53 2006
zs> Return-Path: <seqfan-owner at ext.jussieu.fr>
zs> Date: Tue, 31 Oct 2006 21:43:12 -0800 (PST)
zs> From: zak seidov <zakseidov at yahoo.com>
zs> Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
zs> To: seqfan at ext.jussieu.fr, ip at sciserv.org, rgwv at rgwv.com
zs> 
zs> Dear seqfans,
zs> Bob, Ivars,
zs> 
zs> In A065577, 
zs> a(1)=2 because 10=3+7=5+5?
zs> Also a(2)=6 because
zs> 100=3+97=11+89=17+83=29+71=41+59=47+53?
zs> ....

One might call A065577 explicitly the partitions without regard to order. The
conversion between the two counts is simple, see
http://mathworld.wolfram.com/GoldbachPartition.html

The table of n, ordered, and non-ordered partitions is

                                    1, 3, 2
                                   2, 12, 6
                                   3, 56, 28
                                  4, 254, 127
                                 5, 1620, 810
                                6, 10804, 5402
                                7, 77614, 38807

with the ordered partitions in the OEIS as A073610, the non-ordered in A061358 (?)
(Cf from A065577 to these two might also help to show the difference,
A065577(n)=A061358(10^n).)
With the exception of the n=1 leading term, the count of the
ordered partitions is twice that of the non-ordered partitions,
because 10^n/2 is not a prime then.

The Maple program to print the small table above:

A065577 := proc(n,orderd)
	local N,a,i,ip;
	N := 10^n ;
	a := 0 ;
	i := 1 ;
	ip := 2 ;
	while 2*ip <= N do
		if isprime(N-ip) then
			if orderd and ip <> N-ip then
				a := a+ 2;
			else
				a := a+ 1;
			fi ;
		fi ;
		i := i+1 ;
		ip := ithprime(i) ;
	od ;
	RETURN(a) ;
end:

for n from 1 to 20 do
	print(n,A065577(n,true),A065577(n,false)) ;
od ;

In PARI

A065577(n)={
	local(N,a,i,ip);
	N = 10^n ;
	a = 0 ;
	i = 1 ;
	ip = 2 ;
	while(2*ip <= N,
		if(isprime(N-ip),
			a++ ;
		) ;
		i++ ;
		\\ip = prime(i) ; \\ slower and needs -p switch
		ip = nextprime(ip+1) ; \\ pseudoprimes only
	) ;
	return(a) ;
}

{
	for(n=1,20,
		print(n," ",A065577(n)) ;
	)
}

The PARI output confirms the n=9 case quoted by Zak (I've not gone to n=10):

1 2
2 6
3 28
4 127
5 810
6 5402
7 38807
8 291400
9 2274205

-- Richard





From zakseidov at yahoo.com  Wed Nov  1 15:29:37 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 1 Nov 2006 06:29:37 -0800 (PST)
Subject: A065577 Number of Goldbach partitions of 10^n?
In-Reply-To: <200611011317.kA1DHwmp018846@amer.strw.leidenuniv.nl>
Message-ID: <20061101142937.98860.qmail@web38214.mail.mud.yahoo.com>

Richard,
thanks for your reply.
It seems this'd concern mainly Ivars!
Also could you please calculate cases n>=10) -
I guess this'd not take for you hours(days!)
as with my Mathematica.
thanks, Zak


--- Richard Mathar <mathar at strw.leidenuniv.nl> wrote:

> 
> zs> From seqfan-owner at ext.jussieu.fr  Wed Nov  1
> 06:44:53 2006
> zs> Return-Path: <seqfan-owner at ext.jussieu.fr>
> zs> Date: Tue, 31 Oct 2006 21:43:12 -0800 (PST)
> zs> From: zak seidov <zakseidov at yahoo.com>
> zs> Subject: A065577 Number of Goldbach partitions
> of 10^n? More terms?
> zs> To: seqfan at ext.jussieu.fr, ip at sciserv.org,
> rgwv at rgwv.com
> zs> 
> zs> Dear seqfans,
> zs> Bob, Ivars,
> zs> 
> zs> In A065577, 
> zs> a(1)=2 because 10=3+7=5+5?
> zs> Also a(2)=6 because
> zs> 100=3+97=11+89=17+83=29+71=41+59=47+53?
> zs> ....
> 
> One might call A065577 explicitly the partitions
> without regard to order. The
> conversion between the two counts is simple, see
> http://mathworld.wolfram.com/GoldbachPartition.html
> 
> The table of n, ordered, and non-ordered partitions
> is
> 
>                                     1, 3, 2
>                                    2, 12, 6
>                                    3, 56, 28
>                                   4, 254, 127
>                                  5, 1620, 810
>                                 6, 10804, 5402
>                                 7, 77614, 38807
> 
> with the ordered partitions in the OEIS as A073610,
> the non-ordered in A061358 (?)
> (Cf from A065577 to these two might also help to
> show the difference,
> A065577(n)=A061358(10^n).)
> With the exception of the n=1 leading term, the
> count of the
> ordered partitions is twice that of the non-ordered
> partitions,
> because 10^n/2 is not a prime then.
> 
> The Maple program to print the small table above:
> 
> A065577 := proc(n,orderd)
> 	local N,a,i,ip;
> 	N := 10^n ;
> 	a := 0 ;
> 	i := 1 ;
> 	ip := 2 ;
> 	while 2*ip <= N do
> 		if isprime(N-ip) then
> 			if orderd and ip <> N-ip then
> 				a := a+ 2;
> 			else
> 				a := a+ 1;
> 			fi ;
> 		fi ;
> 		i := i+1 ;
> 		ip := ithprime(i) ;
> 	od ;
> 	RETURN(a) ;
> end:
> 
> for n from 1 to 20 do
> 	print(n,A065577(n,true),A065577(n,false)) ;
> od ;
> 
> In PARI
> 
> A065577(n)={
> 	local(N,a,i,ip);
> 	N = 10^n ;
> 	a = 0 ;
> 	i = 1 ;
> 	ip = 2 ;
> 	while(2*ip <= N,
> 		if(isprime(N-ip),
> 			a++ ;
> 		) ;
> 		i++ ;
> 		\\ip = prime(i) ; \\ slower and needs -p switch
> 		ip = nextprime(ip+1) ; \\ pseudoprimes only
> 	) ;
> 	return(a) ;
> }
> 
> {
> 	for(n=1,20,
> 		print(n," ",A065577(n)) ;
> 	)
> }
> 
> The PARI output confirms the n=9 case quoted by Zak
> (I've not gone to n=10):
> 
> 1 2
> 2 6
> 3 28
> 4 127
> 5 810
> 6 5402
> 7 38807
> 8 291400
> 9 2274205
> 
> -- Richard
> 



 
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From tanyakh at TanyaKhovanova.com  Wed Nov  1 16:46:50 2006
From: tanyakh at TanyaKhovanova.com (Tanya Khovanova)
Date: Wed,  1 Nov 2006 07:46:50 -0800
Subject: What is the most famous sequence?
Message-ID: <200611010746.AA12650352@TanyaKhovanova.com>

What is the most famous sequence? The most basic (natural numbers?)? The most simple (0,0,0, ... or 1,1,1, ...?)? The most versatile (Catalan numbers?)?

I would be curious to know your opinion.

I tried to check the number of Google hits for some sequence numbers. Here is what I've got:
A000040 (prime numbers) 68,800
A000045 (Fibonacci)     66,000
A000108 (Catalan)       62,500
A000027 (natural)       47,500

Can someone write a program to find top ten? What is the most referenced sequence inside Encyclopedia? Can we find the most hits outside Encyclopedia? Is it easy to do?

Tanya


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From jvospost3 at gmail.com  Wed Nov  1 18:58:55 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 1 Nov 2006 09:58:55 -0800
Subject: New & old LEGO seqs; production cuts
Message-ID: <5542af940611010958me17f347w3bf2d2f2d6c9dec0@mail.gmail.com>

I like the new seqs about Lego, such as
 A123762 <http://www.research.att.com/%7Enjas/sequences/A123762>
 Number of ways, counted up to symmetry, to build a contiguous building with
n *LEGO* blocks of size 1x2.

Ironically timed, as *"Recent restructuring and production cuts have left
Lego unable to fill
orders<http://www.cbc.ca/money/story/2006/10/31/legoproduction.html>for
the upcoming holiday season. Affected products include Duplo bricks,
Lego City sets, and (horror of horrors!) Star Wars and Lego Technik
sets."*According to the article Lego stands to lose $127 million in
holiday sales.

http://www.cbc.ca/money/story/2006/10/31/legoproduction.html

It might be useful to add crossrefs to earlier seqs such as:

 A007576 <http://www.research.att.com/%7Enjas/sequences/A007576>
 Number of maximally stable towers of 2 X 2 *LEGO* blocks.

 A082679 <http://www.research.att.com/%7Enjas/sequences/A082679>
 Number of *Lego* towers, one piece per floor, where every floor is
perpendicular to the one below it (so we have a kind of 3-dimensional zigzag
pattern).
 -- Jonathan Vos Post
[wondering if my son's earlier intense exposure to Duplo and Lego led him
towards his double B.S. in Math and Computer Science]
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From dean at math.ucdavis.edu  Wed Nov  1 21:12:15 2006
From: dean at math.ucdavis.edu (Dean Hickerson)
Date: Wed, 1 Nov 2006 12:12:15 -0800
Subject: Re^2: Error-finding program
Message-ID: <20061101201215.GA2388@math.ucdavis.edu>

Ralf Stephan wrote:

> My guess is that Clark started the same algorithm with 1, 2 and then
> tried it with starting values 1, 3, so the definition of A026469 should 
> be amended with ", starting with 1, 3".

Actually, A026469 should just say "starting with 1", since the 3 follows
from the rest of the definition.  A026468 should say "starting with 1,2",
since the 2 is artificially added to the sequence.

Dean Hickerson
dean at math.ucdavis.edu





From franktaw at netscape.net  Thu Nov  2 02:04:26 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 01 Nov 2006 20:04:26 -0500
Subject: What is the most famous sequence?
In-Reply-To: <200611010746.AA12650352@TanyaKhovanova.com>
References: <200611010746.AA12650352@TanyaKhovanova.com>
Message-ID: <8C8CC31F9FF2FAD-370-91EC@WEBMAIL-MA05.sysops.aol.com>

I would have to say that the most famous sequence is the primes.

To get the most referenced inside the OEIS, just do a search that 
matches
everything (or almost everything) - I tend to use "-keyword:dead".  Be 
sure
you're sorting by references (for many searches, such as this one, the 
default
sort by relevance works, too).  You'll see the sequences sorted by how 
many
references there are.

The top ten are (with references as of 11/1/06 6 PM CST):

1666 A000040 Primes
1305 A000045 Fibonacci
1177 A000108 Catalan
1009 A000010 Totient function
 883 A000217 Triangular
 870 A000005 Number of Divisors
 808 A000203 Sum of Divisors
 674 A002275 Repunits
 616 A001358 Semiprimes
 594 A000142 Factorials

This tends to under-estimate the importance of sequences like the 
squares
and the powers of 2.  If somebody enters a formula for a sequence that
uses the totient function, they are likely to include a reference to 
A000010.
Adding a formula which squares an integer will not cause people to add 
a
reference to the squares, however.

Incidently, here are some searches, intended to identify how popular 
certain
topics are in the OEIS, with the number of sequences matched:

25267 Prime
17853 keyword:base
10904 Digit -maple:digit
 7087 Square (but this covers a lot more than just the squares)
 6384 Binomial
 5548 Power
 4904 Divisor
 4051 Polynomial
 3901 Keyword:cons
 3722 Transform
 3698 Partition
 3638 Factor
 3526 Keyword:cofr
 3448 Permutation
 3326 Pi
 3325 Fibonacci
 3308 Matrix OR Matrices (search each separately, then subtract number
        where both occur)
 3138 Group
 2601 Phi (includes both golden ratio and totient function)
 2238 Sigma
 2189 Graph
 2116 Composite
 1971 Tree
 1553 Keyword:frac
 1540 Vector
 1549 Catalan
 1511 Reverse
 1197 Factorial (but I can't search for "!" operator)
 1165 Modulo OR Modulus
 1159 Palindrome
 1051 Perfect
 1034 Semiprime

Obviously, many of these searches do not perfectly identify the 
intended
subject.

Franklin T. Adams-Watters


-----Original Message-----
From: tanyakh at TanyaKhovanova.com

What is the most famous sequence? The most basic (natural numbers?)? 
The most
simple (0,0,0, ... or 1,1,1, ...?)? The most versatile (Catalan 
numbers?)?

I would be curious to know your opinion.

I tried to check the number of Google hits for some sequence numbers. 
Here is
what I've got:
A000040 (prime numbers) 68,800
A000045 (Fibonacci)     66,000
A000108 (Catalan)       62,500
A000027 (natural)       47,500

Can someone write a program to find top ten? What is the most 
referenced
sequence inside Encyclopedia? Can we find the most hits outside 
Encyclopedia? Is
it easy to do?

Tanya

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From arndt at jjj.de  Thu Nov  2 02:10:02 2006
From: arndt at jjj.de (Joerg Arndt)
Date: Thu, 2 Nov 2006 12:10:02 +1100
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
Message-ID: <20061102011002.GC6041@cochise.anu.edu.au>

* Jonathan Post <jvospost3 at gmail.com> [Nov 01. 2006 10:14]:
> Eisenstein-Fibonacci sequences.  Cube root of unity
> analogues of square-root of unity A014291  Imaginary
> Rabbits.
> 
> Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
> where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
> = (-1 - i*sqrt(3))/2.
> 
> [...]
> The coefficients of 1 = a(n) = 0, 0, 1, 1, 3, 1, 7, 4,
> 14, 17, 28, 54, 66, 143, 182, 350, ...
> 
> which does not seem to be in OEIS.

Ralf Stephan's ggf() says that the OGF may be:

? g=ggf([1, 1, 3, 1, 7, 4,14, 17, 28, 54, 66, 143, 182, 350])
(x^5 - 3*x^3 + x + 1)/(-x^6 + 3*x^4 - x^3 - 3*x^2 + 1)
? factor(g)

[x - 1 1]

[x^4 + x^3 - 2*x^2 - 2*x - 1 1]

[x^2 + x - 1 -1]

[x^4 - x^3 - x^2 + x + 1 -1]


> [...]







From eclark at math.usf.edu  Thu Nov  2 03:28:27 2006
From: eclark at math.usf.edu (Edwin Clark)
Date: Wed, 1 Nov 2006 21:28:27 -0500 (EST)
Subject: What is the most famous sequence?
In-Reply-To: <200611010746.AA12650352@TanyaKhovanova.com>
References: <200611010746.AA12650352@TanyaKhovanova.com>
Message-ID: <Pine.LNX.4.58.0611012122360.14271@shell.cas.usf.edu>


> What is the most famous sequence?

Since the novel and movie "The Da Vince Code"-- not to mention 
Knuth's "Art of Computer Programming" it would have to be the 
Fibonacci numbers.







From zakseidov at yahoo.com  Thu Nov  2 07:12:28 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 1 Nov 2006 22:12:28 -0800 (PST)
Subject: A124450: no autoreply from OEIS
Message-ID: <20061102061228.7046.qmail@web38206.mail.mud.yahoo.com>

Just submitted:
A124450 (no autoreply from OEIS).
Hope someone may wish to add more terms:
Mathematica isn't the best in such calcs.
Thanks, Zak

%I A124450
%S A124450
5,47,491,4919,49877,499943,4999913,49999757,499999931,4999999937,49999999811,499999999769,4999999998431,49999999999619,499999999999769,4999999999998557,49999999999998887,499999999999999679,4999999999999999661,49999999999999998647
%N A124450 Lesser of pair of closest primes summed to
10^n.
%C A124450 Cf. A065577 Number of Goldbach partitions
of 10^n.
%F A124450 10^n - a(n) is prime.
%e A124450 10^1=5+5; 10^2=47+53; 10^3=491+509;
10^4=4919+5081; 10^5=49877=50123; 10^6=499943+500057;
10^7=4999913+5000087; 10^8=49999757+50000243;
10^9=499999931+500000069;
10^10=4999999937+5000000063}, etc.
%Y A124450 A065577
%O A124450 1
%K A124450 ,nonn,
%A A124450 Zak Seidov  (zakseidov at yahoo.com), Nov 02 2006


 
____________________________________________________________________________________
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From zakseidov at yahoo.com  Thu Nov  2 08:12:42 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 1 Nov 2006 23:12:42 -0800 (PST)
Subject: SEQ FROM Zak Seidov 
In-Reply-To: <200611020706.kA276utj016855@exspider.research.att.com>
Message-ID: <20061102071242.66437.qmail@web38208.mail.mud.yahoo.com>

Neil,
in contrast,
in this case I got immediate autoreply from OEIS.
Does submitting with ascribed or non-ascribed 
sequence number matter?
Zak 
--- The On-Line Encyclopedia of Integer Sequences
<oeis at research.att.com> wrote:

> The following is a copy of the email message that
> was sent to njas
> containing the sequence you submitted.
> 
> All greater than and less than signs have been
> replaced by their html
> equivalents.  They will be changed back when the
> message is processed.
> 
> This copy is just for your records.  No reply is
> expected.
>  Subject: NEW SEQUENCE FROM Zak Seidov 
> 
> 
> %I A000001
> %S A000001
>
5,,3,59,11,17,29,11,71,71,23,11,29,29,11,83,3,11,281,11,101,71,23,257,401,293,107,293,53,11,113,251,47,587,23,179,389,59,173,17,1427,83,431,53,563,593,41,47,239,383,431,1181,701,971,149,593,569,149,191,1973
> %N A000001 Lesser of pair of farsest primes summed
> to 10^n.
> %C A000001 Cf. A065577 Number of Goldbach partitions
> of 10^n, A124450 Lesser of pair of closest primes
> summed to 10^n.
> %F A000001 10^n - a(n) is prime.
> %e A000001 10^1 = 5 + 5, 10^2 = 3 + 97, 10^3 = 3 +
> 997, 10^4 = 59 + 9941, 10^5 = 11 + 99989, 10^6 = 17
> + 999983, 10^7 = 29 + 9999971, 10^8 = 11 + 99999989,
> 10^9 = 
> 71 + 999999929, 10^10 = 23 + 9999999977, etc.
> %Y A000001 A065577, A124450.
> %O A000001 1
> %K A000001 ,nonn,
> %A A000001 Zak Seidov  (zakseidov at yahoo.com), Nov 02
> 2006
> RH 
> RA 192.20.225.32
> RU 
> RI 
> 



 
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From jvospost3 at gmail.com  Thu Nov  2 08:44:39 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 1 Nov 2006 23:44:39 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <20061102011002.GC6041@cochise.anu.edu.au>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <20061102011002.GC6041@cochise.anu.edu.au>
Message-ID: <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>

Thank you, Joerg.  That was interesting.  Does your generating function
hypothesis hold when I extend the sequence by another 5 values, to a(n) = 0,
0, 1, 1, 3, 1, 7, 4,
14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again, assuming
that I got the elementary algebra and arithmetic correct.

If so, I should submit this sequence as by the two us!

Eisenstein-Fibonacci sequences.  Cube root of unity analogues of square-root
of unity A014291  Imaginary Rabbits.

[update of 1 Nov 06]

Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
= (-1 - i*sqrt(3))/2.

There are various integer sequences derived from this
complex sequence (such as the real part, the imaginary
part, the polynomial coefficient is of 1, w, and
w^2)Unless my arithmetic by hand is in error again,
is:

n   b(n)
0   w
1   w^2
2   w(w^2) + w = w^3 + w = 1 + w.
3   w(1 + w) + w^2 = w + w^2 + w^2 = 1 + 2w^2.
4   w(1 + 2w^2) + 1 + w = w + 2w^3 + 1 + w = 3 + 2w.
5   w(3 + 2w) + 1 + 2w^2 = 3w + 2w^2 + 1 + 2w^2.
   = 1 + 3w + 4w^2.
6  w(1 + 3w + 4w^2) + 3 + 2w = w + 3w^2 + 4w^3 + 3 +
2w
   = 7 + 3w + 3w^2.
7  w(7 + 3w + 3w^2) + 1 + 3w + 4w^2 = 7w + 3w^2 + 3w^3

  + 1 + 3w + 4w^2 = 4 + 10w + 7w^2.
8  w(4 + 10w + 7w^2) + 7 + 3w + 3w^2 = 4w + 10w^2 +
  7w^3 + 7 + 3w + 3w^2 = 14 + 7w + 13w^2.
9  w(14 + 7w + 13w^2) + 4 + 10w + 7w^2 = 14w + 7w^2 +
  13w^3 + 4 + 10w + 7w^2 = 17 + 24w + 14w^2.
10 w(17 + 24w + 14w^2) + 14 + 7w + 13w^2 = 17w + 24w^2
  + 14w^3 + 14 + 7w + 13w^2 = 28 + 24w + 37w^2.
11 w(28 + 24w + 37w^2) + 17 + 24w + 14w^2 = 28w +
  24w^2 + 37w^3 + 17 + 24w + 14w^2 = 54 + 52w +
38w^2.
12 w(54 + 52w + 38w^2) + 28 + 24w + 37w^2 = 54w +
  52w^2 + 38w^3 + 28 + 24w + 37w^2 = 66 + 78w +
89w^2.
13 w(66 + 78w + 89w^2) + 54 + 52w + 38w^2 =
  66w + 78w^2 + 89w^3 + 54 + 52w + 38w^2 =
  143 + 118w + 116w^2.
14 w(143 + 118w + 116w^2) + 66 + 78w + 89w^2 =
  143w + 118w^2 + 116w^3 + 66 + 78w + 89w^2 =
  182 + 221w + 207w^2.
15 w(182 + 221w + 207w^2) + 143 + 118w + 116w^2 =
  182w + 221w^2 + 207w^3 + 143 + 118w + 116w^2 =
  350 + 300w + 337w^2.

16 w(350 + 300w + 337w^2) + 350 + 300w + 337w^2 =
    350w + 300w^2 + 337w^3 + 350 + 300w + 337w^2 =
    687 +  650w + 637w^2.

17 w(687 +  650w + 637w^2) + 350 + 300w + 337w^2 =
    687w +  650w^2 + 637w^3 + 350 + 300w + 337w^2 =
    987 + 987w + 987w^2.

18  w(987 + 987w + 987w^2) + 687 +  650w + 637w^2 =
    987w + 987w^2 + 987w^3 + 687 +  650w + 637w^2 =
    1674 + 1637w + 1624w^2.

19  w(1674 + 1637w + 1624w^2) + 987 + 987w + 987w^2 =
     1674w + 1637w^2 + 1624w^3 + 987 + 987w + 987w^2
=
     2611 + 2661w + 2624w^2.

20  w(2611 + 2661w + 2624w^2) + 1674 + 1637w + 1624w^2
=
     2611w + 2661w^2 + 2624w^3 + 1674 + 1637w +
1624w^2 =
     4298 + 4248w + 4285w^2.


The coefficients of 1 = a(n) = 0, 0, 1, 1, 3, 1, 7, 4,
14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611,
4298

which does not seem to be in OEIS.

To excerpt my comment in A107890 (and some other
seqs):

"Eisenstein integers are of the form a + b*omega,
where a and b are ordinary integers, and omega = (-1 +
i*sqrt(3))/2 is a cube root of 1, the other cube roots
of 1 being 1 and omega^2 = (-1 - i*sqrt(3))/2.
Eisenstein integers are complex numbers that are also
members of the imaginary quadratic field Q(sqrt -3) =
Z[omega]. The sums, differences, and products of
Eisenstein integers are other Eisenstein integers."

Other sequences come from:
b(0) = w, b(1) = w^2, b(n) = b(n-1) + w*b(n-2),
and from the O, J notation of D?rrie (1965) in
Weisstein, Eric W. "Eisenstein Integer." From
MathWorld--A Wolfram Web Resource.
http://mathworld.wolfram.com/EisensteinInteger.html


Eisenstein-Fib.doc

=== end 2nd email ===

On 11/1/06, Joerg Arndt <arndt at jjj.de> wrote:
>
> * Jonathan Post <jvospost3 at gmail.com> [Nov 01. 2006 10:14]:
> > Eisenstein-Fibonacci sequences.  Cube root of unity
> > analogues of square-root of unity A014291  Imaginary
> > Rabbits.
> >
> > Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
> > where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
> > = (-1 - i*sqrt(3))/2.
> >
> > [...]
> > The coefficients of 1 = a(n) = 0, 0, 1, 1, 3, 1, 7, 4,
> > 14, 17, 28, 54, 66, 143, 182, 350, ...
> >
> > which does not seem to be in OEIS.
>
> Ralf Stephan's ggf() says that the OGF may be:
>
> ? g=ggf([1, 1, 3, 1, 7, 4,14, 17, 28, 54, 66, 143, 182, 350])
> (x^5 - 3*x^3 + x + 1)/(-x^6 + 3*x^4 - x^3 - 3*x^2 + 1)
> ? factor(g)
>
> [x - 1 1]
>
> [x^4 + x^3 - 2*x^2 - 2*x - 1 1]
>
> [x^2 + x - 1 -1]
>
> [x^4 - x^3 - x^2 + x + 1 -1]
>
>
> > [...]
>
>
>
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From maxale at gmail.com  Thu Nov  2 09:23:51 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 2 Nov 2006 00:23:51 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
Message-ID: <d3dac270611020023t17f583aep351f8e1e187148cc@mail.gmail.com>

On 10/22/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> Eisenstein-Fibonacci sequences.  Cube root of unity
>  analogues of square-root of unity A014291  Imaginary
>  Rabbits.
>
>  Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
>  where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
>  = (-1 - i*sqrt(3))/2.

First off, using the identity w^2 = -w-1, we can uniquely represent
b(n) as u(n)+v(n)*w where u(n) and v(n) are integers.
In particular, u(0)=0, v(0)=1, u(1)=-1, v(1)=-1.

Now the recurrence relation b(n) = w*b(n-1) + b(n-2) implies that
u(n) = u(n-2) - v(n-1)
v(n) = v(n-2) + u(n-1) - v(n-1)
that in turn implies that
[u(n-1),v(n-1),u(n),v(n)] = M*[u(n-2),v(n-2),u(n-1),v(n-1)]
where vectors are column-vectors and the matrix M is defined as
[ 0, 0, 1, 0 ]
[ 0, 0, 0, 1 ]
[ 1, 0, 0, -1 ]
[ 0, 1, 1, -1 ]

The characteristic polynomial of the matrix M is
x^4 + x^3 - x^2 - x + 1
meaning that u(n) and v(n) satisfies the recurrence
u(n) = - u(n-1) + u(n-2) + u(n-3) - u(n-4)
v(n) = - v(n-1) + v(n-2) + v(n-3) - v(n-4)
with u(0)=0, u(1)=-1, u(2)=1, u(3)=-2
and v(0)=1, v(1)=-1, v(2)=1, v(3)=-1.

>  There are various integer sequences derived from this
>  complex sequence (such as the real part, the imaginary
>  part, the polynomial coefficient is of 1, w, and
>  w^2)Unless my arithmetic by hand is in error again,
>  is:
>
>  n   b(n)
>  0   w
>  1   w^2
>  2   w(w^2) + w = w^3 + w = 1 + w.
>  3   w(1 + w) + w^2 = w + w^2 + w^2 = 1 + 2w^2.

It is not clear why you don't reduce the "polynomials" of w to the first degree.
Higher degree "polynomials" are not uniquely defined.
E.g., b(1)=w^2=2w^2+w+1=3w^2+2w+2=...
but if you reduce to the first degree "polynomial", there is an unique
representation:
b(1) = -w-1.

[...]

>  The coefficients of 1 = a(n) = 0, 0, 1, 1, 3, 1, 7, 4,
>  14, 17, 28, 54, 66, 143, 182, 350, ...

This sequence is not well-defined (see above).
Probably, what you are interested in is u(n):
0, -1, 1, -2, 2, -2, 1, 1, -4, 8, -12, 15, -15, 10, 2, -22, 49, -79, 104, -112

Max





From maxale at gmail.com  Thu Nov  2 13:06:51 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 2 Nov 2006 04:06:51 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <d3dac270611020023t17f583aep351f8e1e187148cc@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <d3dac270611020023t17f583aep351f8e1e187148cc@mail.gmail.com>
Message-ID: <d3dac270611020406l10f3deb8jb69c41cb2b267651@mail.gmail.com>

On 11/2/06, Max A. <maxale at gmail.com> wrote:

> >  Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
> >  where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
> >  = (-1 - i*sqrt(3))/2.
>
> First off, using the identity w^2 = -w-1, we can uniquely represent
> b(n) as u(n)+v(n)*w where u(n) and v(n) are integers.

[...]

> u(n) and v(n) satisfies the recurrence
> u(n) = - u(n-1) + u(n-2) + u(n-3) - u(n-4)
> v(n) = - v(n-1) + v(n-2) + v(n-3) - v(n-4)
> with u(0)=0, u(1)=-1, u(2)=1, u(3)=-2
> and v(0)=1, v(1)=-1, v(2)=1, v(3)=-1.

o.g.f. for u(n) is
-x/(1 + x - x^2 - x^3 + x^4)
and
o.g.f. for v(n) is
(1-x^2)/(1 + x - x^2 - x^3 + x^4)

Hence, o.g.f. for b(n) is
(-x + w*(1-x^2)) / (1 + x - x^2 - x^3 + x^4)

Max





From njas at research.att.com  Thu Nov  2 14:24:15 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Thu, 2 Nov 2006 08:24:15 -0500 (EST)
Subject: SEQ FROM Zak Seidov
Message-ID: <200611021324.IAA74292@fry.research.att.com>


Does submitting with ascribed or non-ascribed 
sequence number matter?
Zak 


Me:  I don't know.  But I see you are still using yahoo!
Why not switch to gmail?  Then you won't have these problems.

Best

Neil





From mathar at strw.leidenuniv.nl  Thu Nov  2 14:40:51 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Thu, 2 Nov 2006 14:40:51 +0100
Subject: Re^2:  Eisenstein-Fibonacci sequences
Message-ID: <200611021340.kA2DepPv014946@amer.strw.leidenuniv.nl>


ma> From seqfan-owner at ext.jussieu.fr  Thu Nov  2 13:12:31 2006
ma> Return-Path: <seqfan-owner at ext.jussieu.fr>
ma> Date: Thu, 2 Nov 2006 04:06:51 -0800
ma> From: "Max A." <maxale at gmail.com>
ma> To: "Jonathan Post" <jvospost3 at gmail.com>
ma> Subject: Re: Eisenstein-Fibonacci sequences
ma> Cc: "Sequence Fans Mailing List" <seqfan at ext.jussieu.fr>
ma> 
ma> On 11/2/06, Max A. <maxale at gmail.com> wrote:
ma> 
ma> > >  Let b(0) = w, b(1) = w^2, b(n) = w*b(n-1) + b(n-2),
ma> > >  where w = omega = (-1 + i*sqrt(3))/2 and w^2 = omega^2
ma> > >  = (-1 - i*sqrt(3))/2.
ma> >
ma> > First off, using the identity w^2 = -w-1, we can uniquely represent
ma> > b(n) as u(n)+v(n)*w where u(n) and v(n) are integers.
ma> 
ma> [...]
ma> 
ma> > u(n) and v(n) satisfies the recurrence
ma> > u(n) = - u(n-1) + u(n-2) + u(n-3) - u(n-4)
ma> > v(n) = - v(n-1) + v(n-2) + v(n-3) - v(n-4)
ma> > with u(0)=0, u(1)=-1, u(2)=1, u(3)=-2
ma> > and v(0)=1, v(1)=-1, v(2)=1, v(3)=-1.
ma> 
ma> o.g.f. for u(n) is
ma> -x/(1 + x - x^2 - x^3 + x^4)
ma> and
ma> o.g.f. for v(n) is
ma> (1-x^2)/(1 + x - x^2 - x^3 + x^4)
ma> 
ma> Hence, o.g.f. for b(n) is
ma> (-x + w*(1-x^2)) / (1 + x - x^2 - x^3 + x^4)
ma> 
ma> Max

The g.f. above [-x+w(1-x^2)]/[1+x-x^2-x^3+x^4] generates
the reduced  polynomial
                          2             3      4            5               6
w + (-1 - w) x + (1 + w) x  + (-2 - w) x  + 2 x  + (w - 2) x  + (-3 w + 1) x

                  7               8              9      10
     + (1 + 5 w) x  + (-4 - 7 w) x  + (8 + 8 w) x  + O(x  )

If one doesn't reduce the polynomial, one gets the simpler
g(x)=w/[1-w*x-x^2] which generates the non-reduced and more generic
b(0)+b(1) x+ b(2) x^2 + b(3) x^3+ ...
=

     2       3       2     4      2   3     5      3       4
w + w  x + (w  + w) x  + (w  + 2 w ) x  + (w  + 3 w  + w) x  +

      6      4      2   5     7      5      3       6
    (w  + 4 w  + 3 w ) x  + (w  + 5 w  + 6 w  + w) x  +

      8      6       4      2   7     9      7       5       3       8
    (w  + 6 w  + 10 w  + 4 w ) x  + (w  + 7 w  + 15 w  + 10 w  + w) x  +

      10      8       6       4      2   9      10
    (w   + 8 w  + 21 w  + 20 w  + 5 w ) x  + O(x  )

and which becomes (apart from an offset) A000045 for w=1, A052542 for w=2,
A052906 for w=3, A000007 for w=0, and signed variations of these for w=-1,-2 or -3..
The coefficients table of the non-reduced g.f. is in A102426, which is
essentially the same as A098925 and A092865.

R. Mathar





From tanyakh at TanyaKhovanova.com  Thu Nov  2 16:15:08 2006
From: tanyakh at TanyaKhovanova.com (Tanya Khovanova)
Date: Thu,  2 Nov 2006 07:15:08 -0800
Subject: What is the most famous sequence?
Message-ID: <200611020715.AA2085028026@TanyaKhovanova.com>

Would this mean that Fibonacci sequence is the most popularized?

---------- Original Message ----------------------------------
From: Edwin Clark <eclark at math.usf.edu>
Date:  Wed, 1 Nov 2006 21:28:27 -0500 (EST)

>
>> What is the most famous sequence?
>
>Since the novel and movie "The Da Vince Code"-- not to mention 
>Knuth's "Art of Computer Programming" it would have to be the 
>Fibonacci numbers.
>
>
>


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From davidwwilson at comcast.net  Thu Nov  2 16:46:38 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 2 Nov 2006 10:46:38 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
References: <20061101054313.25333.qmail@web38210.mail.mud.yahoo.com>
Message-ID: <008e01c6fe96$19f1bbe0$6401a8c0@yourxhtr8hvc4p>

I wrote a C++ program using sieve method and calculated

a(7) = 38807
a(8) = 291400
a(9) = 2274205
a(10) = 18200488

a(7), a(8) and a(9) agree with previous calculations.

However, Zak Seidov gave

However, my a(10) = 18200487.

So now I have to go root out the discrepancy. I am going to have my computer 
generate a file of all the partitions of 10^10 and do an independent check 
of the results, and hopefully uncover a problem in my program. 






From rsc at swtch.com  Thu Nov  2 17:26:57 2006
From: rsc at swtch.com (Russ Cox)
Date: Thu, 2 Nov 2006 11:26:57 -0500
Subject: SEQ FROM Zak Seidov
In-Reply-To: <20061102071242.66437.qmail@web38208.mail.mud.yahoo.com>
References: <200611020706.kA276utj016855@exspider.research.att.com>
	 <20061102071242.66437.qmail@web38208.mail.mud.yahoo.com>
Message-ID: <ee9e417a0611020826h1daaac8xa85b9f6cc4b0ebdf@mail.gmail.com>

As I said in my earlier mail, the semantics of the
Yahoo "spam protection" are such that if you
submit two sequences within some time period
(maybe 12 hours to a day), the first might get
delayed many hours while the second will go through
immediately.

Out of curiosity, do you have oeis at research.att.com
in your address book?

Russ





From davidwwilson at comcast.net  Thu Nov  2 17:29:52 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 2 Nov 2006 11:29:52 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
Message-ID: <001101c6fe9c$205d0c40$6401a8c0@yourxhtr8hvc4p>

Ok, that last message was incoherent.

For A065577 Zak Seidov got a(10) = 18200488, I got a(10) = 18200487.  I am 
checking my program. 






From pxp at rogers.com  Thu Nov  2 18:36:24 2006
From: pxp at rogers.com (Hans Havermann)
Date: Thu, 2 Nov 2006 12:36:24 -0500
Subject: A124450
In-Reply-To: <20061102061228.7046.qmail@web38206.mail.mud.yahoo.com>
References: <20061102061228.7046.qmail@web38206.mail.mud.yahoo.com>
Message-ID: <D7CB7013-474C-4B78-B336-E5DBA449F167@rogers.com>

zak seidov:

> Hope someone may wish to add more terms:
> Mathematica isn't the best in such calcs.

> %S A124450
> 5,47,491,4919,49877,499943,4999913,49999757,499999931,4999999937,49999 
> 999811,499999999769,4999999998431,49999999999619,499999999999769,49999 
> 99999998557,49999999999998887,499999999999999679,4999999999999999661,4 
> 9999999999999998647
> %N A124450 Lesser of pair of closest primes summed to 10^n.

I think you overstate the Mathematica difficulty. With no effort at  
any sort of efficiency in the programming, it still only takes me 8  
seconds to generate 50 terms.

Table[h =10^n/2; c=0; While[PrimeQ[h-c]==False || PrimeQ[h+c]==False,  
c++]; h-c, {n, 1, 50}]

{5, 47, 491, 4919, 49877, 499943, 4999913, 49999757, 499999931,  
4999999937, 49999999811, 499999999769, 4999999998431, 49999999999619,  
499999999999769, 4999999999998557, 49999999999998887,  
499999999999999679, 4999999999999999661, 49999999999999998647,  
499999999999999999637, 4999999999999999999481,  
49999999999999999998677, 499999999999999999998497,  
4999999999999999999999259, 49999999999999999999998779,  
499999999999999999999999043, 4999999999999999999999998947,  
49999999999999999999999999661, 499999999999999999999999994069,  
4999999999999999999999999997879, 49999999999999999999999999997699,  
499999999999999999999999999997969,  
4999999999999999999999999999995227,  
49999999999999999999999999999995263,  
499999999999999999999999999999989719,  
4999999999999999999999999999999998683,  
49999999999999999999999999999999999871,  
499999999999999999999999999999999996127,  
4999999999999999999999999999999999998557,  
49999999999999999999999999999999999999613,  
499999999999999999999999999999999999988231,  
4999999999999999999999999999999999999991729,  
49999999999999999999999999999999999999994663,  
499999999999999999999999999999999999999997117,  
4999999999999999999999999999999999999999992863,  
49999999999999999999999999999999999999999991807,  
499999999999999999999999999999999999999999991507,  
4999999999999999999999999999999999999999999993199,  
49999999999999999999999999999999999999999999997081}

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From davidwwilson at comcast.net  Thu Nov  2 18:45:17 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 2 Nov 2006 12:45:17 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
References: <20061102171720.72610.qmail@web38203.mail.mud.yahoo.com>
Message-ID: <002b01c6fea6$a96fcae0$6401a8c0@yourxhtr8hvc4p>

Well let's get it right this time

I got a(10) = 18200488, in agreement with Dick Mathar's value.





From jvospost3 at gmail.com  Thu Nov  2 19:01:55 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 2 Nov 2006 10:01:55 -0800
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
In-Reply-To: <002b01c6fea6$a96fcae0$6401a8c0@yourxhtr8hvc4p>
References: <20061102171720.72610.qmail@web38203.mail.mud.yahoo.com>
	 <002b01c6fea6$a96fcae0$6401a8c0@yourxhtr8hvc4p>
Message-ID: <5542af940611021001x27e8951akc9b931928d84333e@mail.gmail.com>

Amusing coincidence that the digits of a(10) = 18200488 consist entirely of
0 and powers of 2.  [base].

-- Jonathan Vos Post

On 11/2/06, David Wilson <davidwwilson at comcast.net> wrote:
>
> Well let's get it right this time
>
> I got a(10) = 18200488, in agreement with Dick Mathar's value.
>
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From jvospost3 at gmail.com  Thu Nov  2 21:54:44 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 2 Nov 2006 12:54:44 -0800
Subject: New & old LEGO seqs; production cuts
In-Reply-To: <C16FD8A8.20890%eilers@math.ku.dk>
References: <5542af940611010958me17f347w3bf2d2f2d6c9dec0@mail.gmail.com>
	 <C16FD8A8.20890%eilers@math.ku.dk>
Message-ID: <5542af940611021254w14b12a38y14021854e772de77@mail.gmail.com>

Dear Soren,

Thank you for a clear and thoughful answer.

My son (in the room with me now) agrees that baby/childhood play with Duplo
and Lego guided him towards his Math/Computer Science expertise today.

He and I agree, however, that it's a shame that the manufacturer doesn't
make 4-D Lego blocks, or, more properly, the ones that they do make are of
extension along the t-axis only in the trivial way, and cannot be freely
rotated in Minkowski space.

So, I wonder, how well does your approach to counting under symmetry work if
extended to an additional spacial dimension?

I suppose that we can consider an AxBxC 4-D Lego to be actually an AxBxCx1
Lego in Z^4. I think that adjacency is well-defined, all rotations are about
planes, and there is a relationship to 4-D analogues of polyominoes, i.e. to
polyhypercubes.

Best,

Jonathan Vos Post

On 11/2/06, S?ren Eilers <eilers at math.ku.dk> wrote:
>
>  Dear Jonathan
>
> Thank you, and yes, I think that LEGO inspires kids to learn about
> fundamental mathematical concepts such as symmetry.
>
> I've also considered adding crossrefs to the sequences you mentioned but
> decided against it.
>
> Regarding A007576: Watson definitely counts a subset of what we count in
> A123818, but as you can see in his paper he only looks at "planar" buildings
> where all blocks could be placed in, say, [0,2]xRxR. This means that one of
> the dimensions of the blocks used is irrelevant, and Watson could just have
> well have counted what we call FLAT structures with 1x2 blocks. Thus, IF one
> was to crossref I would suggest to do it to A123764, but perhaps this is a
> little hard to explain in the limited space of an OEIS entry.
>
> But Watson has even more restrictions; he looks only at MAX HEIGHT
> buildings with one block at each level. This is exactly what LEGO did wrong
> when they claimed that A123828(6)=102981500 [the number they wanted was
> (46^5+2^5)/2, but they got the last digit wrong too].
>
> Of course counting FLAT and MAX HEIGHT structures with 2x2 or 1x2 blocks
> is easy, the formula is 3^(n-1) or (3^(n-1)+1)/2 depending on whether or not
> you want to identify up to symmetry. Watson knows this and tries to count
> how many of these structures would tip over if placed on a flat surface,
> essentially computing the x-coordinate of the center of gravity. So I think
> the kind of LEGO counting done by Watson is quite different from the kind of
> LEGO counting we do.
>
>     {Btw, A007576 is indeed the same as A086821}
>     {If somebody was interested I could probably quite easily compute how
> many of "our" LEGO buildings are stable in the sense of Watson}
>
> Regarding Zeilberger's LEGO counts, they all relate, as far as I recall
> from a relatively hard look some time ago, to buildings with blocks of
> varying size, and hence are quite different in nature from what we do. In
> that case, n is not a number of blocks, but the total volume/weight of the
> structure, so that's a very different ball game.
>
> Best regards,
> Soren
>
>
>
>
>
> On 01/11/06 18:58, "Jonathan Post" <jvospost3 at gmail.com> wrote:
>
> I like the new seqs about Lego, such as
> A123762 <http://www.research.att.com/%7Enjas/sequences/A123762><http://www.research.att.com/%7Enjas/sequences/A123762>
>   Number of ways, counted up to symmetry, to build a contiguous building
> with n *LEGO* blocks of size 1x2.
>
> Ironically timed, as *"Recent restructuring and production cuts have left
> Lego unable to fill orders
> <http://www.cbc.ca/money/story/2006/10/31/legoproduction.html><http://www.cbc.ca/money/story/2006/10/31/legoproduction.html>for the upcoming holiday season. Affected products include Duplo bricks,
> Lego City sets, and (horror of horrors!) Star Wars and Lego Technik sets."
> * According to the article Lego stands to lose $127 million in holiday
> sales.
>
> http://www.cbc.ca/money/story/2006/10/31/legoproduction.html
>
> It might be useful to add crossrefs to earlier seqs such as:
>
> A007576 <http://www.research.att.com/%7Enjas/sequences/A007576><http://www.research.att.com/%7Enjas/sequences/A007576>
>   Number of maximally stable towers of 2 X 2 *LEGO* blocks.
>
>
> A082679 <http://www.research.att.com/%7Enjas/sequences/A082679><http://www.research.att.com/%7Enjas/sequences/A082679>
>   Number of *Lego* towers, one piece per floor, where every floor is
> perpendicular to the one below it (so we have a kind of 3-dimensional zigzag
> pattern).
>  -- Jonathan Vos Post
> [wondering if my son's earlier intense exposure to Duplo and Lego led him
> towards his double B.S. in Math and Computer Science]
>
>
>
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From davidwwilson at comcast.net  Thu Nov  2 22:49:42 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 2 Nov 2006 16:49:42 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
References: <20061102171720.72610.qmail@web38203.mail.mud.yahoo.com> <002b01c6fea6$a96fcae0$6401a8c0@yourxhtr8hvc4p>
Message-ID: <004e01c6fec8$ce34d100$6401a8c0@yourxhtr8hvc4p>

Regarding A065577:

I altered my counting program to generating a file listing, in increasing 
order, all primes p such that there exists prime q with p < q, p+q = 10^10. 
I wrote a verification program to read this file, check that the p are in 
increasing order, count the p, compute q = 10^10-p, check p < q and that p 
and q are both prime (using trial division).

The verification program has finished, and confirms that a(10) >= 18200488. 
This contradicts Zak Seidov's count a(10) = 18200487, but agrees with Dick 
Mathar's count a(10) = 18200488.

If anyone is interested, I am keeping the text file with the p values for a 
little while. It is 195MB, so e-mailing it is probably not a good option.






From zakseidov at yahoo.com  Fri Nov  3 04:09:56 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Thu, 2 Nov 2006 19:09:56 -0800 (PST)
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
In-Reply-To: <004e01c6fec8$ce34d100$6401a8c0@yourxhtr8hvc4p>
Message-ID: <20061103030956.55909.qmail@web38210.mail.mud.yahoo.com>

David, Dick, Bob, Ivars,
seqfans.

After recalculating (at two different PC's)
and rechecking I could finally 
find source of my error
(silly as usual) in one of codes,
and my final value is 
a(10) = 18200488 (NOT 18200487!)
which coincides now with the Dick-David value.

So, Bob, 
you may wish to give in A065577  the correct value 
a(10) = 18200488 with due refer to David Wilson  and
Richard Mathar (not me).
Also Ivars can extend his table.
Thanks, Zak

BTW1 Still, is it possible to find a(11)... with C++?

BIW2 Not all people even in this closed list
may be interested in Mmca specific problems,
so I''l be happy to have a list of you
particularly Mmca-involved.
Then we may circulate relevant messages
not to all seqfan community.
Please anyone interested write me personally
with subject "Me in Mmca", thanks.
Of course it's not intented to be closed list!
Hope this is not against Gerard' policy.

--- David Wilson <davidwwilson at comcast.net> wrote:

> Regarding A065577:
> 
> I altered my counting program to generating a file
> listing, in increasing 
> order, all primes p such that there exists prime q
> with p < q, p+q = 10^10. 
> I wrote a verification program to read this file,
> check that the p are in 
> increasing order, count the p, compute q = 10^10-p,
> check p < q and that p 
> and q are both prime (using trial division).
> 
> The verification program has finished, and confirms
> that a(10) >= 18200488. 
> This contradicts Zak Seidov's count a(10) =
> 18200487, but agrees with Dick 
> Mathar's count a(10) = 18200488.
> 
> If anyone is interested, I am keeping the text file
> with the p values for a 
> little while. It is 195MB, so e-mailing it is
> probably not a good option.
> 
> 



 
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From zakseidov at yahoo.com  Fri Nov  3 05:03:30 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Thu, 2 Nov 2006 20:03:30 -0800 (PST)
Subject: A123998: No primes except 3?
Message-ID: <20061103040330.2777.qmail@web38202.mail.mud.yahoo.com>

Why are there no prime numbers except of 3,
in this sequence?
Zak

A123998
={1,3,9,15,18,39,48,69,78,99,105,114,135,153,165,168,183,189,219,249,273,

288,300,303,309,330,345,363,405,414,438,468,483,498,504,534,585,618,
 639,648,699,714,729,765,804,813,828,879,933}; 
(* %N A123998 Numbers n for which 2n+1 and 4n+1 are
primes. *)



 
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From maxale at gmail.com  Fri Nov  3 06:41:20 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 2 Nov 2006 21:41:20 -0800
Subject: A123998: No primes except 3?
In-Reply-To: <20061103040330.2777.qmail@web38202.mail.mud.yahoo.com>
References: <20061103040330.2777.qmail@web38202.mail.mud.yahoo.com>
Message-ID: <d3dac270611022141x4d10f69evdd706982df6e3d28@mail.gmail.com>

Note that if n = 1 mod 3 then 2n+1 is not prime (except n=1);
and if n = 2 mod 3 then 4n+1 is not prime.
Therefore, n must be a multiple of 3, except n=1.

Max

If n>1 and both 2n+1, 4n+1 are primes

On 11/2/06, zak seidov <zakseidov at yahoo.com> wrote:
> Why are there no prime numbers except of 3,
> in this sequence?
> Zak
>
> A123998
> ={1,3,9,15,18,39,48,69,78,99,105,114,135,153,165,168,183,189,219,249,273,
>
> 288,300,303,309,330,345,363,405,414,438,468,483,498,504,534,585,618,
>  639,648,699,714,729,765,804,813,828,879,933};
> (* %N A123998 Numbers n for which 2n+1 and 4n+1 are
> primes. *)
>
>
>
>
> __________________________________________________________________________________________
> Check out the New Yahoo! Mail - Fire up a more powerful email and get things done faster.
> (http://advision.webevents.yahoo.com/mailbeta)
>
>





From zakseidov at yahoo.com  Fri Nov  3 07:01:51 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Thu, 2 Nov 2006 22:01:51 -0800 (PST)
Subject: A123998: No primes except 3?
In-Reply-To: <d3dac270611022141x4d10f69evdd706982df6e3d28@mail.gmail.com>
Message-ID: <20061103060151.3808.qmail@web38207.mail.mud.yahoo.com>

Great!
It'so easy if you are Max, not Zak!
Thanks, 
Zak

--- "Max A." <maxale at gmail.com> wrote:

> Note that if n = 1 mod 3 then 2n+1 is not prime
> (except n=1);
> and if n = 2 mod 3 then 4n+1 is not prime.
> Therefore, n must be a multiple of 3, except n=1.
> 
> Max
> 
> If n>1 and both 2n+1, 4n+1 are primes
> 
> On 11/2/06, zak seidov <zakseidov at yahoo.com> wrote:
> > Why are there no prime numbers except of 3,
> > in this sequence?
> > Zak
> >
> > A123998
> >
>
={1,3,9,15,18,39,48,69,78,99,105,114,135,153,165,168,183,189,219,249,273,
> >
> >
>
288,300,303,309,330,345,363,405,414,438,468,483,498,504,534,585,618,
> >  639,648,699,714,729,765,804,813,828,879,933};
> > (* %N A123998 Numbers n for which 2n+1 and 4n+1
> are
> > primes. *)
> >
> >
> >
> >
> >
>
__________________________________________________________________________________________
> > Check out the New Yahoo! Mail - Fire up a more
> powerful email and get things done faster.
> > (http://advision.webevents.yahoo.com/mailbeta)
> >
> >
> 



 
____________________________________________________________________________________
Get your email and see which of your friends are online - Right on the New Yahoo.com 
(http://www.yahoo.com/preview) 






From jeremy.gardiner at btinternet.com  Fri Nov  3 09:59:50 2006
From: jeremy.gardiner at btinternet.com (JEREMY GARDINER)
Date: Fri, 3 Nov 2006 08:59:50 +0000 (GMT)
Subject: What is the most famous sequence?
In-Reply-To: <200611020715.AA2085028026@TanyaKhovanova.com>
Message-ID: <20061103085950.5824.qmail@web86610.mail.ird.yahoo.com>

The natural numbers must be the most widely known sequence ...

Tanya Khovanova <tanyakh at TanyaKhovanova.com> wrote:  > What is the most famous sequence?
-------------- next part --------------
An HTML attachment was scrubbed...
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From maxale at gmail.com  Fri Nov  3 10:08:28 2006
From: maxale at gmail.com (Max A.)
Date: Fri, 3 Nov 2006 01:08:28 -0800
Subject: querying multidimensional sequences in OEIS
Message-ID: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>

I have a sequence indexed by vectors with non-negative integer components.
Is there a regular way to query such sequence in OEIS ?

As an example, let us consider a sequence of multinomial coefficients.
Any vector with non-negative integer components [k1, k2, ..., kn]
defines an element of this sequence:

a( [k1, k2, ..., kn] ) = (k1 + k2 +  ... + kn)! / (k1! * k2! * ... * kn!)

Assume that I know nothing about these numbers and have only a
black-box computing a(v) for any given vector v with non-negative
integer components. Can I anyhow find this sequence in OEIS and learn
that its elements called "multinomial coefficients" ?

Thanks,
Max





From wouter.meeussen at pandora.be  Fri Nov  3 12:39:40 2006
From: wouter.meeussen at pandora.be (wouter meeussen)
Date: Fri, 3 Nov 2006 12:39:40 +0100
Subject: download seqs resulting from search
Message-ID: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>

dear all,

if I search for all number triangles with row sums equal to the factorials,
I enter
"row sum" "factorial"
and get 9 pages of results in 'short' form.
(1) how can I improve on this query?
(2) how can I download them all in one go?

Wouter





From njas at research.att.com  Fri Nov  3 14:43:47 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Fri, 3 Nov 2006 08:43:47 -0500 (EST)
Subject: download seqs resulting from search
Message-ID: <200611031343.IAA51448@fry.research.att.com>

as i have said before, you can download up to 30 (or is it 50?) sequences
at once by using the old email lookup service

send email to  sequences at research.att.com  saying things like

lookup A000123
lookup A000125
lookup A012345

etc.

Neil





From franktaw at netscape.net  Fri Nov  3 16:18:26 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 10:18:26 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
Message-ID: <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>

I don't know of any standard for functions of an arbitrary finite 
sequence
of positive integers in the OEIS.

In your example, the function is commutative, which makes it 
essentially
a function of the multiset of values - or in other words, of a 
partition.  For
that there is a standard - two of them actually.  One is the 
enumeration
in Abramowitz and Stegun order, as specified in A036036, and the other
is the Mathematica ordering, as specified in A080577.

There are some relevant sequences in the database in both orders, some
only in the A&S order, and a few only in the Mma order.

Since the two orders are the same for partitions of up to 5 objects, 
one
often just search for an early part of the sequence.  If this does not 
suffice,
one should try both before giving up.

For example, the partitions of 5 objects are: [5], [4,1], [3,2], 
[3,1^2],
[2^2,1], [2,1^3], [1^5].  The multinomial coefficients for these are:

1,5,10,20,30,60,120

A search for this turns up A036038, the A&S version, and A078760, the
Mma version.

----
As for arbitrary finite sequences of positive integers, there are two 
main
candidates for an ordering: A066099 and A108244; these are the reverse
of each other.  I would probably go with A066099, as it is closest to 
the
ordering used for the partitions; but you can make an argument for 
either
one.  Possibly both should be used.

Off hand, I can't think of any interesting functions of finite 
sequences of
positive integers that depend on the order of the sequence.

Franklin T. Adams-Watters


-----Original Message-----
From: maxale at gmail.com

I have a sequence indexed by vectors with non-negative integer 
components.?
Is there a regular way to query such sequence in OEIS ??
?
As an example, let us consider a sequence of multinomial coefficients.?
Any vector with non-negative integer components [k1, k2, ..., kn]?
defines an element of this sequence:?
?
a( [k1, k2, ..., kn] ) = (k1 + k2 + ... + kn)! / (k1! * k2! * ... * 
kn!)?
?
Assume that I know nothing about these numbers and have only a?
black-box computing a(v) for any given vector v with non-negative?
integer components. Can I anyhow find this sequence in OEIS and learn?
that its elements called "multinomial coefficients" ??
?
Thanks,?
Max?


________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From rsc at swtch.com  Fri Nov  3 17:06:23 2006
From: rsc at swtch.com (Russ Cox)
Date: Fri, 3 Nov 2006 11:06:23 -0500
Subject: download seqs resulting from search
In-Reply-To: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>
References: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>
Message-ID: <ee9e417a0611030806q1490600ci75384340c6facd67@mail.gmail.com>

> if I search for all number triangles with row sums equal to the factorials,
> I enter
> "row sum" "factorial"
> and get 9 pages of results in 'short' form.
> (1) how can I improve on this query?

Not sure.

> (2) how can I download them all in one go?

Add &n=100 to the URL
and you will get 100 results per page.
Then click on "text" formatting and save the
resulting page.

http://www.research.att.com/~njas/sequences/?q=%22row+sums%22+factorial&n=100&fmt=3

Russ





From franktaw at netscape.net  Fri Nov  3 17:14:55 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 11:14:55 -0500
Subject: download seqs resulting from search
In-Reply-To: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>
References: <002001c6ff3c$c0040f50$45c7e0d5@pcwou>
Message-ID: <8C8CD7A55C060ED-FD0-B6C7@WEBMAIL-MA20.sysops.aol.com>

Try adding "keyword:tabl" to the search.  Then (probably) repeat with 
"keyword:tabf" instead - unless you aren't interested in irregular 
tables.  This won't be all in one go, but two isn't too bad.

Of course, there are undoubtably some tables which sum to factorials 
that don't have the words "row sum" in them.  And maybe some that just 
reference A000142 instead of using the word "factorial".

Franklin T. Adams-Watters


-----Original Message-----
From: wouter.meeussen at pandora.be

dear all,

if I search for all number triangles with row sums equal to the 
factorials,
I enter
"row sum" "factorial"
and get 9 pages of results in 'short' form.
(1) how can I improve on this query?
(2) how can I download them all in one go?

Wouter

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From mlb at well.com  Fri Nov  3 17:26:14 2006
From: mlb at well.com (Marc LeBrun)
Date: Fri, 03 Nov 2006 08:26:14 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <200611031625.kA3GP1mt017249@smtp.well.com>

 >=franktaw at netscape.net

Good advice...maybe it could be made part of the tips-n-tricks section?

 > just search for an early part of the sequence

Or sometimes you can search for a small set of "unusual" later, 
larger, terms that will act as a kind of signature.

 > Off hand, I can't think of any interesting functions of finite sequences of
 > positive integers that depend on the order of the sequence.

Exponents in prime factorizations, say.







From franktaw at netscape.net  Fri Nov  3 19:31:34 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 13:31:34 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <200611031625.kA3GP1mt017249@smtp.well.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com> <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com> <200611031625.kA3GP1mt017249@smtp.well.com>
Message-ID: <8C8CD8D6CD483AD-FD0-C3EF@WEBMAIL-MA20.sysops.aol.com>

Nice suggestion.  That would be A057335.

Franklin T. Adams-Watters


-----Original Message-----
From: mlb at well.com

>=franktaw at netscape.net?
> Off hand, I can't think of any interesting functions of finite 
sequences of?
> positive integers that depend on the order of the sequence.?
?
Exponents in prime factorizations, say.?
?


________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From maxale at gmail.com  Fri Nov  3 21:26:20 2006
From: maxale at gmail.com (Max A.)
Date: Fri, 3 Nov 2006 12:26:20 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <200611031625.kA3GP1mt017249@smtp.well.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <200611031625.kA3GP1mt017249@smtp.well.com>
Message-ID: <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com>

On 11/3/06, Marc LeBrun <mlb at well.com> wrote:

>  > Off hand, I can't think of any interesting functions of finite sequences of
>  > positive integers that depend on the order of the sequence.
>
> Exponents in prime factorizations, say.

Oh, that is a very nice suggestion.
Are there any multidimensional sequences represented this way in OEIS?

I have a number of multidimensional sequences in mind that I'd like to
see in OEIS.
E.g.: the coefficients of incomplete Bell polynomials:

b( [k1, k2, ..., kn] ) = (k1 + 2*k2 + ... + n*kn)! / (k1! * k2! * ...
* kn!) / (1!^k1 * 2!^k2 * ... * n!^kn)

and looking similar the number of permutations with given cycle structure:

c( [k1, k2, ..., kn] ) = (k1 + 2*k2 + ... + n*kn)! / (k1! * k2! * ...
* kn!) / (1^k1 * 2^k2 * ... * n^kn)

Max





From mlb at well.com  Fri Nov  3 22:41:00 2006
From: mlb at well.com (Marc LeBrun)
Date: Fri, 03 Nov 2006 13:41:00 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com
 >
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
 <200611031625.kA3GP1mt017249@smtp.well.com>
 <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com>
Message-ID: <200611032139.kA3LdZSV024106@smtp.well.com>

 > I have a number of multidimensional sequences in mind that I'd like to
 > see in OEIS.  E.g.: the coefficients of incomplete Bell polynomials:

Two possible ways to submit sequences of tuples T(1),T(2),...
where T(i)=[T(i,1),T(i,2),...] are:

A. Define a traversal order by a map (i,j) <--> n and set a(n) = T(i,j).
This gives the tabl/tabf style of entry.

B. Encode each tuple as a single number, say by using it as factor
exponents:  T(i) --> 2^T(i,1) * 3^T(i,2) * 5^T(i,3) *...

For example, for Pascal's triangle

A: 1; 1, 1; 1, 2, 1; 1, 3, 3, 1;...

B: 2, 6,    90,      47250,...

However there are a variety of possible ways to do these things--for example
you might subtract 1 from all the Pascal exponents, B': 1, 1, 3, 225,...

Or traverse in the order (1,1), (2,1), (1,2), (3,1), (1,3), (2,2), (1,4),...
which recursively interleaves sequences instead of the usual diagonalizing.

But this sort of thing can also be abused to breed swarms of silly sequences,
so it should be well-motivated.

It's nicest to use familiar schemes (eg as Frank noted for partitions).







From franktaw at netscape.net  Fri Nov  3 22:41:18 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 16:41:18 -0500
Subject: Possibly more errors in OEIS?
Message-ID: <8C8CDA7EE6FDEF2-12C-B6F@WEBMAIL-MA20.sysops.aol.com>

Looking at sequences with keyword:cons, there is a marked shortage of 
sequences with a given duplicate digit (excluding 0), compared with 
those with a given other pair of digits.  For example, restricting our 
attention to sequences not of the form "Decimal expansion of 1/n", and 
ignoring all pairs involving 0 or 1, there are an average of about 1710 
sequences  with any given digit pair, but an average of only about 1627 
with duplicate pairs.  10 of this difference is explained by the 10 
"selvage" sequences, which have no duplicate digits, but the rest is 
unexplained.

I'm guessing that there may be an input error here; that duplicate 
digits sometimes get combined at some point in the process, whether 
before or after submission.  I'm not sure how to test for this, short 
of examining the "cons" sequences and determining which are correct.

Mostly for Neil,

Sequences A081780 to A081796, "continued cotangent" sequences, all have 
the "cons" keyword, but they should not.

Also, sequence A114721 has the "cons" keyword; it should instead have 
the "frac" keyword.

Franklin T. Adams-Watters

________________________________________________________________________
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From jawbrey at att.net  Fri Nov  3 22:51:00 2006
From: jawbrey at att.net (Jon Awbrey)
Date: Fri, 03 Nov 2006 16:51:00 -0500
Subject: querying multidimensional sequences in OEIS
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <200611031625.kA3GP1mt017249@smtp.well.com>
	 <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com> <200611032139.kA3LdZSV024106@smtp.well.com>
Message-ID: <454BB9C4.CCCA8C55@att.net>

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o

you might get some ideas from looking at the
tables of "primal codes" or "primal functions"
that I compiled, for instance, here:

http://stderr.org/pipermail/inquiry/2005-July/002846.html

ja

Marc LeBrun wrote:
> 
>  > I have a number of multidimensional sequences in mind that I'd like to
>  > see in OEIS.  E.g.: the coefficients of incomplete Bell polynomials:
> 
> Two possible ways to submit sequences of tuples T(1),T(2),...
> where T(i)=[T(i,1),T(i,2),...] are:
> 
> A. Define a traversal order by a map (i,j) <--> n and set a(n) = T(i,j).
> This gives the tabl/tabf style of entry.
> 
> B. Encode each tuple as a single number, say by using it as factor
> exponents:  T(i) --> 2^T(i,1) * 3^T(i,2) * 5^T(i,3) *...
> 
> For example, for Pascal's triangle
> 
> A: 1; 1, 1; 1, 2, 1; 1, 3, 3, 1;...
> 
> B: 2, 6,    90,      47250,...
> 
> However there are a variety of possible ways to do these things--for example
> you might subtract 1 from all the Pascal exponents, B': 1, 1, 3, 225,...
> 
> Or traverse in the order (1,1), (2,1), (1,2), (3,1), (1,3), (2,2), (1,4),...
> which recursively interleaves sequences instead of the usual diagonalizing.
> 
> But this sort of thing can also be abused to breed swarms of silly sequences,
> so it should be well-motivated.
> 
> It's nicest to use familiar schemes (eg as Frank noted for partitions).

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o
inquiry e-lab: http://stderr.org/pipermail/inquiry/
citizendium development forums: http://smf.citizendium.org
http://www.textop.org/wiki/index.php?title=User:Jon_Awbrey
wikinfo: http://wikinfo.org/wiki.php?title=User:Jon_Awbrey
wp review: http://wikipediareview.com/index.php?showuser=398
o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o






From maxale at gmail.com  Fri Nov  3 23:02:04 2006
From: maxale at gmail.com (Max A.)
Date: Fri, 3 Nov 2006 14:02:04 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <200611032139.kA3LdZSV024106@smtp.well.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <200611031625.kA3GP1mt017249@smtp.well.com>
	 <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com>
	 <200611032139.kA3LdZSV024106@smtp.well.com>
Message-ID: <d3dac270611031402s62ff2746p3a5c749865678cec@mail.gmail.com>

On 11/3/06, Marc LeBrun <mlb at well.com> wrote:
>  > I have a number of multidimensional sequences in mind that I'd like to
>  > see in OEIS.  E.g.: the coefficients of incomplete Bell polynomials:
>
> Two possible ways to submit sequences of tuples T(1),T(2),...
> where T(i)=[T(i,1),T(i,2),...] are:
>
> A. Define a traversal order by a map (i,j) <--> n and set a(n) = T(i,j).
> This gives the tabl/tabf style of entry.
>
> B. Encode each tuple as a single number, say by using it as factor
> exponents:  T(i) --> 2^T(i,1) * 3^T(i,2) * 5^T(i,3) *...

As I understand, A is applicable only to the cases where the dimension
of tuples is fixed while B is perfectly fine for tuples with variable
dimensions (what I'm currently interested in).

>
> For example, for Pascal's triangle
>
> A: 1; 1, 1; 1, 2, 1; 1, 3, 3, 1;...
>
> B: 2, 6,    90,      47250,...

Does B apply to Pascal triangle?
The 5-th term corresponds to the tuple [0,0,5], what it is in Pascal triangle?

Max





From franktaw at netscape.net  Fri Nov  3 23:06:02 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 17:06:02 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com> <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>

<Me>
>Off hand, I can't think of any interesting functions of finite 
sequences of
>positive integers that depend on the order of the sequence.

Of course, I no sooner wrote that than I started thinking of such 
functions.
I'm not sure any of these are tremendously interesting, but they're at 
least
somewhat so.

In the following, b(i), i = 1..k is the finite sequence.  I'm starting 
all these
sequences with the empty sequence, but I searched for them without
that first term.  All are shown here in A066099 order.

Alternating sum
Sum_i (-1)^{i+1) b(i)
0,1,2,0,3,1,-1,1,4,2,0,2,-2,0,2,0

Binomial sum
Sum_i C(k-1,i-1) b(i)
0,1,2,2,3,3,3,4,4,4,4,5,4,6,5,8

Inverse binomial sum
Sum_i (-1)^{i+1) C(k-1,i-1) b(i)
0,1,2,0,3,1,-1,0,4,2,0,1,-2,-2,1,0

Weighted sum
sum_i i b(i)
0,1,2,3,3,4,5,6,4,5,6,7,7,8,9,10
This is A029931

Zero-based weighted sum
sum_i (i-1) b(i)
0,0,0,1,0,1,2,3,0,1,2,3,3,4,5,6

Sum of products of consecutive elements
sum_{i=1}^{k-1} b(i) b(i+1)
0,0,0,1,0,2,2,2,0,3,4,3,3,4,3,3

Number of rises
sum_{b(i)>b(i-1)} 1
0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,0

Number of falls
sum_{b(i)<b(i-1)} 1
0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0

Number of unchanged
sum_{b(i)=b(i-1)} 1
0,0,0,1,0,0,0,2,0,0,1,1,0,0,1,3

Number of non-rises
sum_{b(i)<=b(i-1)} 1
0,0,0,1,0,0,1,2,0,0,1,1,1,1,2,3

Number of non-falls
sum_{b(i)>=b(i-1)} 1
0,0,0,1,0,1,0,2,0,1,1,2,0,1,1,3

Number of monotonically increasing runs
1 + number of falls, but 0 for empty sequence
0,1,1,1,1,2,1,1,1,2,1,2,1,2,1,1

Number of monotonically decreasing runs
1 + number of rises, but 0 for empty sequence
0,1,1,1,1,1,2,1,1,1,1,1,2,2,2,1

Number of runs of equal terms
1 + number of unchanged, but 0 for empty sequence
0,1,1,2,1,1,1,3,1,1,2,2,1,1,2,4

Number of distinct non-empty subsequences
0,1,1,2,1,3,3,3,1,3,2,5,3,5,5,4
E.g., f(1,1) = 2, for the sequences [1], and [1,1].

Number of distinct subsequences
1 + number of distinct non-empty subsequences
1,2,2,3,2,4,4,4,2,4,3,6,4,6,6,5

Number of set partitions
List parts of set partition by their smallest element, and count the 
part sizes.
E.g., 1|2,4|3 (also known as {1,2,3,2}) would count for the sequence 
[1,2,1].
1,1,1,1,1,2,1,1,1,3,3,3,1,2,1,1

Number of permutations
List permutations in cycle form, sorted by the smallest element in the 
cycle,
and count the cycle lengths.
Number of set partitions * Product_i (b(i)-1)!
1,1,1,1,2,2,1,1,6,6,3,3,2,2,1,1

Number of partially ordered sets (unlabelled) by rank
The rank of an element in a poset is the length of the longest chain of 
which it
is the largest element.
1,1,1,1,1,2,1,1,1,3,4,3,1,2,1,1

Number of partially ordered sets (labelled) by rank
1,1,1,2,1,9,3,6,1,28,54,60,4,36,12,24

Number of partially ordered sets (naturally labelled) by rank
Naturally labelled means labels are consistent with the ordering.
1,1,1,1,1,4,1,1,1,11,13,8,1,4,1,1

Number of forests (unlabelled, unordered) with b(i) nodes at height i.
1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,2,2,1,1,1,1,2,1,1,1,1,1

... and, of course, various other forest options.

If there is agreement that this is the right order for these sequences, 
I'll try
to find time to submit them.
________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From tbaruchel at free.fr  Fri Nov  3 23:17:29 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Fri, 3 Nov 2006 23:17:29 +0100
Subject: [math-fun] A fast (written in assembly) library for continued
	fractions
Message-ID: <20061103221729.GB2467@>

I wrote for my own purposes the following X86 assembly code, intended
to be called from GFORTRAN. Such a thing should only work if you have
a similar configuration to mine, but anyway, it may interest you.
I wanted a fast library for computing the length in the continued
fraction expansion of sqrt(n) for any integer n that can fit in
a REAL*8 gfortran type for statistical purposes. I also included
a faster function that tells if this length is odd or even. Though
I am not an assembly expert, there should be a significant speed
improvement if you use this code rather than your own fortran
functions, since the whole process is performed on the FPU stack
with no load/store useless travels.

It may help you, it may not. You know about these kinds of software...

Instructions are at the beginning of the file, as comments.
Read them carefully. Code is below.

Regards.

------------------ FILE quad.s BEGINS BELOW --------------------

/*

  Compile your piece of code with something like
    gfortran -fno-underscoring mycode.f90 quad.s
  and add freely your own flags. Enjoy !
  Of course, if you don't want to use the -fno-underscoring
  option for some reason, you can add an underscore at the end
  of the global functions below, like that:
    .globl initfpu_
    initfpu_:
  it should work.

  Remember this piece of code has the following status:
    Seems to work for me, maybe it will also seem for you,
    maybe your computer will explode, maybe it will make the
    earth explode, maybe God will choose not to have created
    the world, think about it before using it.
  (Anyway, if you see some kind of bug, please let me know ;-)
  Remember that
    a) this is x86 assembly code;
    b) it uses the GNU AS (gas) syntax;
    c) it uses the gfortran calling conventions;
  (now you understand why you can't expect it will works ;-)

  Usage. In your program (or subroutine or whatever), declare
  the function you need like:
    
    logical :: oddperiod
    integer*8 :: period

  (the two subroutines don't need to be declared).

  Then use
    
    integer*4 :: oldfpustatus
    call initfpu(oldfpustatus)

  before any computation; if you don't do it, it won't work.
  When you don't need any more the library,
    
    call restorefpu(oldfpustatus)

  will restore the control register of the FPU.

  During all that time, you shouldn't be disturbed too much, 
  for your own mathematical operations, but
  remember the rounding mode of the FPU has been set to
  "round down" (rather than the usual "nearest").

  How does it work? The whole computation is performed on the
  stack of the FPU; only integer values are needed for these
  tasks, but they are stored as REAL*10, 80 bits reals. They
  are not converted to anything during the whole process, thus
  it should work for great integers stored in your own REAL*8,
  64 bits reals. Of course, it may happen (I think it will never,
  but didn't check) that an intermediate result can not fit in
  such a REAL*10 register; in that case, read above what is
  written about the end of the world...
  
*/



/*
  call initfpu(integer*4)
  =====================
    This subroutine should be called before any other call to the
    library; its purpose is to set the rounding control in the FPU
    control register to the "round down" mode.
    The subroutine puts in the integer variable the initial value
    of the FPU control register.

    If you want to use often this subroutine (which you shouldn't:
    speed being an issue, this subroutine is intended to be used
    once, then the functions billions of times, not the contrary ;-),
    maybe removing the "fnclex" can improve the speed, since I don't
    think it plays a great role here.
*/
.globl initfpu
initfpu:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        fnstcw  (%eax)
	fwait
        movw    (%eax), %dx
	andb	$0xf3, %dh
	orb	$0x4, %dh
        movw    %dx, 2(%eax)
	fnclex
	fldcw   2(%eax)
        fwait
        movw    $0, 2(%eax)
	leave
	ret

/*
  call restorefpu(integer*4)
  ========================
    This subroutine should be used for restoring the initial state of
    the FPU control register; the integer variable must contain the
    right value (which should have been previously set by the
    subroutine initfpu).
*/
.globl restorefpu
restorefpu:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
	fclex
	fldcw   (%eax)
	leave
	ret

/*
  sqrtint( real double precision : REAL*8 )
  =========================================
    This function returns the integer part of the square root
    of the given real. The return value has itself the real type.
*/
.global sqrtint
sqrtint:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
	fldl	(%eax)
	fsqrt
        frndint
	leave
	ret

/*
  oddperiod( real double precision : REAL*8 )
  ===========================================
    Despite the type of the real argument, the variable has to
    be some integer value stored as a double precision.
    The function has the logical type. It returns FALSE if
    the argument is a perfect square or if the continued
    fraction expansion of its square root has an even period
    length; it returns TRUE in the remaining case.
      oddperiod(dble(25)) --> FALSE
      oddperiod(dble(13)) --> TRUE
      oddperiod(dble(12)) --> FALSE
    If it may help, you can declare the function as an integer*4
    type rather than a logical one (output will be 0 or 1).

    Though it seems to work with integer*8, I am not absolutely
    sure it *always* will (unless you add the following instruction
      movl $0, %edx
    at the beginning of the code (for instance between the current
    second and third lines of code), but really why would you
    do it? Maybe something in the calling conventions is such that
    the register EDX already contains 0, but, again, who cares?

    If you read the instructions for the period() function, you
    will learn about some 2^32 limitations, but you don't have to
    care about that here: the length is not kept in any register
    during the computation and the code should work as long as the
    values in the FPU stack are the right one.
    
*/
.global oddperiod
oddperiod:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        fldln2
	fldl	(%eax)
        fld     %st
        fsqrt
        frndint
        fld     %st
        fld     %st
        fmul    %st, %st
        fld     %st(3)
        fsubp   %st, %st(1)
        fcom    %st(4)
        fstsw   %ax
        fwait
        sahf
        ja      oddperiod0
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
oddperiod2:
        movl    $0, %eax
        leave
        ret
oddperiod0:
        fld     %st(1)
        fadd    %st(3), %st
        fdiv    %st(1), %st
        frndint
        fld     %st(2)
        fld     %st(1)
        fmul    %st(3), %st
        fsubp   %st, %st(1)
        fxch    %st(3)
        fsub    %st(3), %st
        fabs
        fcomp   %st(6)
        fstsw   %ax
        fwait
        sahf
        ja      oddperiod1
        fld     %st(3)
        fadd    %st, %st
        fsubp   %st, %st(1)
        fcomp   %st(5)
        fstsw   %ax
        fwait
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        sahf
        ja      oddperiod2
        movl    $1, %eax
        leave
        ret
oddperiod1:
        ffree   %st
        fincstp
        fld     %st(1)
        fmul    %st, %st
        fld     %st(4)
        fsubp   %st, %st(1)
        fdivp   %st, %st(1)
        frndint
        jmp oddperiod0

/*
  period( real double precision : REAL*8 )
  ========================================
    Despite the type of the real argument, the variable has to
    be some integer value stored as a double precision.
    The function has integer*8 type, though the output
    can't be greater than 2^32-1 (this choice has been made
    for avoiding sign problem when the result is greater
    (or equal) then 2^31 and smaller then 2^32.
    The result is the length of period in the continued fraction
    expansion of the square root of the given integer.

    If speed is really for you an obsession, remember that it
    is highly unprobable that you use that library for periods
    between the two fatidic values. Maybe it will increase a little
    the speed to use integer*4 which allows to delete twice the line
        movl    $0,%edx
    (not the first time it appears, but the second and third one).
    If you try such a thing and see some speed improvement, please
    let me know.
*/
.global period
period:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        movl    $0, %edx
        fldln2
	fldl	(%eax)
        fld     %st
        fsqrt
        frndint
        fld     %st
        fld     %st
        fmul    %st, %st
        fld     %st(3)
        fsubp   %st, %st(1)
        fcom    %st(4)
        fstsw   %ax
        fwait
        sahf
        ja      period0
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        movl    %edx, %eax
        movl     $0, %edx
        leave
        ret
period0:
        inc     %edx
        fld     %st(1)
        fadd    %st(3), %st
        fdiv    %st(1), %st
        frndint
        fld     %st(2)
        fld     %st(1)
        fmul    %st(3), %st
        fsubp   %st, %st(1)
        fxch    %st(3)
        fsub    %st(3), %st
        fabs
        fcomp   %st(6)
        fstsw   %ax
        fwait
        sahf
        ja      period1
        fld     %st(3)
        fadd    %st, %st
        fsubp   %st, %st(1)
        fcomp   %st(5)
        fstsw   %ax
        fwait
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        sahf
        jb      period2
        add     %edx, %edx
period2:
        movl    %edx, %eax
        movl    $0,%edx
        leave
        ret
period1:
        ffree   %st
        fincstp
        fld     %st(1)
        fmul    %st, %st
        fld     %st(4)
        fsubp   %st, %st(1)
        fdivp   %st, %st(1)
        frndint
        jmp period0

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From tbaruchel at free.fr  Fri Nov  3 23:17:29 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Fri, 3 Nov 2006 23:17:29 +0100
Subject: A fast (written in assembly) library for continued fractions
Message-ID: <20061103221729.GB2467@>

I wrote for my own purposes the following X86 assembly code, intended
to be called from GFORTRAN. Such a thing should only work if you have
a similar configuration to mine, but anyway, it may interest you.
I wanted a fast library for computing the length in the continued
fraction expansion of sqrt(n) for any integer n that can fit in
a REAL*8 gfortran type for statistical purposes. I also included
a faster function that tells if this length is odd or even. Though
I am not an assembly expert, there should be a significant speed
improvement if you use this code rather than your own fortran
functions, since the whole process is performed on the FPU stack
with no load/store useless travels.

It may help you, it may not. You know about these kinds of software...

Instructions are at the beginning of the file, as comments.
Read them carefully. Code is below.

Regards.

------------------ FILE quad.s BEGINS BELOW --------------------

/*

  Compile your piece of code with something like
    gfortran -fno-underscoring mycode.f90 quad.s
  and add freely your own flags. Enjoy !
  Of course, if you don't want to use the -fno-underscoring
  option for some reason, you can add an underscore at the end
  of the global functions below, like that:
    .globl initfpu_
    initfpu_:
  it should work.

  Remember this piece of code has the following status:
    Seems to work for me, maybe it will also seem for you,
    maybe your computer will explode, maybe it will make the
    earth explode, maybe God will choose not to have created
    the world, think about it before using it.
  (Anyway, if you see some kind of bug, please let me know ;-)
  Remember that
    a) this is x86 assembly code;
    b) it uses the GNU AS (gas) syntax;
    c) it uses the gfortran calling conventions;
  (now you understand why you can't expect it will works ;-)

  Usage. In your program (or subroutine or whatever), declare
  the function you need like:
    
    logical :: oddperiod
    integer*8 :: period

  (the two subroutines don't need to be declared).

  Then use
    
    integer*4 :: oldfpustatus
    call initfpu(oldfpustatus)

  before any computation; if you don't do it, it won't work.
  When you don't need any more the library,
    
    call restorefpu(oldfpustatus)

  will restore the control register of the FPU.

  During all that time, you shouldn't be disturbed too much, 
  for your own mathematical operations, but
  remember the rounding mode of the FPU has been set to
  "round down" (rather than the usual "nearest").

  How does it work? The whole computation is performed on the
  stack of the FPU; only integer values are needed for these
  tasks, but they are stored as REAL*10, 80 bits reals. They
  are not converted to anything during the whole process, thus
  it should work for great integers stored in your own REAL*8,
  64 bits reals. Of course, it may happen (I think it will never,
  but didn't check) that an intermediate result can not fit in
  such a REAL*10 register; in that case, read above what is
  written about the end of the world...
  
*/



/*
  call initfpu(integer*4)
  =====================
    This subroutine should be called before any other call to the
    library; its purpose is to set the rounding control in the FPU
    control register to the "round down" mode.
    The subroutine puts in the integer variable the initial value
    of the FPU control register.

    If you want to use often this subroutine (which you shouldn't:
    speed being an issue, this subroutine is intended to be used
    once, then the functions billions of times, not the contrary ;-),
    maybe removing the "fnclex" can improve the speed, since I don't
    think it plays a great role here.
*/
.globl initfpu
initfpu:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        fnstcw  (%eax)
	fwait
        movw    (%eax), %dx
	andb	$0xf3, %dh
	orb	$0x4, %dh
        movw    %dx, 2(%eax)
	fnclex
	fldcw   2(%eax)
        fwait
        movw    $0, 2(%eax)
	leave
	ret

/*
  call restorefpu(integer*4)
  ========================
    This subroutine should be used for restoring the initial state of
    the FPU control register; the integer variable must contain the
    right value (which should have been previously set by the
    subroutine initfpu).
*/
.globl restorefpu
restorefpu:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
	fclex
	fldcw   (%eax)
	leave
	ret

/*
  sqrtint( real double precision : REAL*8 )
  =========================================
    This function returns the integer part of the square root
    of the given real. The return value has itself the real type.
*/
.global sqrtint
sqrtint:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
	fldl	(%eax)
	fsqrt
        frndint
	leave
	ret

/*
  oddperiod( real double precision : REAL*8 )
  ===========================================
    Despite the type of the real argument, the variable has to
    be some integer value stored as a double precision.
    The function has the logical type. It returns FALSE if
    the argument is a perfect square or if the continued
    fraction expansion of its square root has an even period
    length; it returns TRUE in the remaining case.
      oddperiod(dble(25)) --> FALSE
      oddperiod(dble(13)) --> TRUE
      oddperiod(dble(12)) --> FALSE
    If it may help, you can declare the function as an integer*4
    type rather than a logical one (output will be 0 or 1).

    Though it seems to work with integer*8, I am not absolutely
    sure it *always* will (unless you add the following instruction
      movl $0, %edx
    at the beginning of the code (for instance between the current
    second and third lines of code), but really why would you
    do it? Maybe something in the calling conventions is such that
    the register EDX already contains 0, but, again, who cares?

    If you read the instructions for the period() function, you
    will learn about some 2^32 limitations, but you don't have to
    care about that here: the length is not kept in any register
    during the computation and the code should work as long as the
    values in the FPU stack are the right one.
    
*/
.global oddperiod
oddperiod:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        fldln2
	fldl	(%eax)
        fld     %st
        fsqrt
        frndint
        fld     %st
        fld     %st
        fmul    %st, %st
        fld     %st(3)
        fsubp   %st, %st(1)
        fcom    %st(4)
        fstsw   %ax
        fwait
        sahf
        ja      oddperiod0
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
oddperiod2:
        movl    $0, %eax
        leave
        ret
oddperiod0:
        fld     %st(1)
        fadd    %st(3), %st
        fdiv    %st(1), %st
        frndint
        fld     %st(2)
        fld     %st(1)
        fmul    %st(3), %st
        fsubp   %st, %st(1)
        fxch    %st(3)
        fsub    %st(3), %st
        fabs
        fcomp   %st(6)
        fstsw   %ax
        fwait
        sahf
        ja      oddperiod1
        fld     %st(3)
        fadd    %st, %st
        fsubp   %st, %st(1)
        fcomp   %st(5)
        fstsw   %ax
        fwait
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        sahf
        ja      oddperiod2
        movl    $1, %eax
        leave
        ret
oddperiod1:
        ffree   %st
        fincstp
        fld     %st(1)
        fmul    %st, %st
        fld     %st(4)
        fsubp   %st, %st(1)
        fdivp   %st, %st(1)
        frndint
        jmp oddperiod0

/*
  period( real double precision : REAL*8 )
  ========================================
    Despite the type of the real argument, the variable has to
    be some integer value stored as a double precision.
    The function has integer*8 type, though the output
    can't be greater than 2^32-1 (this choice has been made
    for avoiding sign problem when the result is greater
    (or equal) then 2^31 and smaller then 2^32.
    The result is the length of period in the continued fraction
    expansion of the square root of the given integer.

    If speed is really for you an obsession, remember that it
    is highly unprobable that you use that library for periods
    between the two fatidic values. Maybe it will increase a little
    the speed to use integer*4 which allows to delete twice the line
        movl    $0,%edx
    (not the first time it appears, but the second and third one).
    If you try such a thing and see some speed improvement, please
    let me know.
*/
.global period
period:
	pushl	%ebp
	movl	%esp, %ebp
	movl	8(%ebp), %eax
        movl    $0, %edx
        fldln2
	fldl	(%eax)
        fld     %st
        fsqrt
        frndint
        fld     %st
        fld     %st
        fmul    %st, %st
        fld     %st(3)
        fsubp   %st, %st(1)
        fcom    %st(4)
        fstsw   %ax
        fwait
        sahf
        ja      period0
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        movl    %edx, %eax
        movl     $0, %edx
        leave
        ret
period0:
        inc     %edx
        fld     %st(1)
        fadd    %st(3), %st
        fdiv    %st(1), %st
        frndint
        fld     %st(2)
        fld     %st(1)
        fmul    %st(3), %st
        fsubp   %st, %st(1)
        fxch    %st(3)
        fsub    %st(3), %st
        fabs
        fcomp   %st(6)
        fstsw   %ax
        fwait
        sahf
        ja      period1
        fld     %st(3)
        fadd    %st, %st
        fsubp   %st, %st(1)
        fcomp   %st(5)
        fstsw   %ax
        fwait
        ffree   %st
        ffree   %st(1)
        ffree   %st(2)
        ffree   %st(3)
        ffree   %st(4)
        sahf
        jb      period2
        add     %edx, %edx
period2:
        movl    %edx, %eax
        movl    $0,%edx
        leave
        ret
period1:
        ffree   %st
        fincstp
        fld     %st(1)
        fmul    %st, %st
        fld     %st(4)
        fsubp   %st, %st(1)
        fdivp   %st, %st(1)
        frndint
        jmp period0





From jvospost3 at gmail.com  Fri Nov  3 23:36:07 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 3 Nov 2006 14:36:07 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com>

I think the example givem without A number, is:

 A007188 <http://www.research.att.com/%7Enjas/sequences/A007188>
 Multiplicative encoding of Pascal triangle: Product p(i+1)^C(n,i).
(Formerly M1722)
 +0
2
2, 6, 90, 47250, 66852843750, 2806877704512541816406250,
1216935896582703898519354781702537118597533386230468750
There might be other Godel-coded sequences, and the like, besides Conway's
brilliant "Fractran", above and beyond those with "decimal Godelization" --
but usually these quickly lead to numbers too large for OEIS.

-- Jonathan Vos Post

On 11/3/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
> <Me>
> >Off hand, I can't think of any interesting functions of finite
> sequences of
> >positive integers that depend on the order of the sequence.
>
> Of course, I no sooner wrote that than I started thinking of such
> functions.
> I'm not sure any of these are tremendously interesting, but they're at
> least
> somewhat so.
>
> In the following, b(i), i = 1..k is the finite sequence.  I'm starting
> all these
> sequences with the empty sequence, but I searched for them without
> that first term.  All are shown here in A066099 order.
>
> Alternating sum
> Sum_i (-1)^{i+1) b(i)
> 0,1,2,0,3,1,-1,1,4,2,0,2,-2,0,2,0
>
> Binomial sum
> Sum_i C(k-1,i-1) b(i)
> 0,1,2,2,3,3,3,4,4,4,4,5,4,6,5,8
>
> Inverse binomial sum
> Sum_i (-1)^{i+1) C(k-1,i-1) b(i)
> 0,1,2,0,3,1,-1,0,4,2,0,1,-2,-2,1,0
>
> Weighted sum
> sum_i i b(i)
> 0,1,2,3,3,4,5,6,4,5,6,7,7,8,9,10
> This is A029931
>
> Zero-based weighted sum
> sum_i (i-1) b(i)
> 0,0,0,1,0,1,2,3,0,1,2,3,3,4,5,6
>
> Sum of products of consecutive elements
> sum_{i=1}^{k-1} b(i) b(i+1)
> 0,0,0,1,0,2,2,2,0,3,4,3,3,4,3,3
>
> Number of rises
> sum_{b(i)>b(i-1)} 1
> 0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,0
>
> Number of falls
> sum_{b(i)<b(i-1)} 1
> 0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0
>
> Number of unchanged
> sum_{b(i)=b(i-1)} 1
> 0,0,0,1,0,0,0,2,0,0,1,1,0,0,1,3
>
> Number of non-rises
> sum_{b(i)<=b(i-1)} 1
> 0,0,0,1,0,0,1,2,0,0,1,1,1,1,2,3
>
> Number of non-falls
> sum_{b(i)>=b(i-1)} 1
> 0,0,0,1,0,1,0,2,0,1,1,2,0,1,1,3
>
> Number of monotonically increasing runs
> 1 + number of falls, but 0 for empty sequence
> 0,1,1,1,1,2,1,1,1,2,1,2,1,2,1,1
>
> Number of monotonically decreasing runs
> 1 + number of rises, but 0 for empty sequence
> 0,1,1,1,1,1,2,1,1,1,1,1,2,2,2,1
>
> Number of runs of equal terms
> 1 + number of unchanged, but 0 for empty sequence
> 0,1,1,2,1,1,1,3,1,1,2,2,1,1,2,4
>
> Number of distinct non-empty subsequences
> 0,1,1,2,1,3,3,3,1,3,2,5,3,5,5,4
> E.g., f(1,1) = 2, for the sequences [1], and [1,1].
>
> Number of distinct subsequences
> 1 + number of distinct non-empty subsequences
> 1,2,2,3,2,4,4,4,2,4,3,6,4,6,6,5
>
> Number of set partitions
> List parts of set partition by their smallest element, and count the
> part sizes.
> E.g., 1|2,4|3 (also known as {1,2,3,2}) would count for the sequence
> [1,2,1].
> 1,1,1,1,1,2,1,1,1,3,3,3,1,2,1,1
>
> Number of permutations
> List permutations in cycle form, sorted by the smallest element in the
> cycle,
> and count the cycle lengths.
> Number of set partitions * Product_i (b(i)-1)!
> 1,1,1,1,2,2,1,1,6,6,3,3,2,2,1,1
>
> Number of partially ordered sets (unlabelled) by rank
> The rank of an element in a poset is the length of the longest chain of
> which it
> is the largest element.
> 1,1,1,1,1,2,1,1,1,3,4,3,1,2,1,1
>
> Number of partially ordered sets (labelled) by rank
> 1,1,1,2,1,9,3,6,1,28,54,60,4,36,12,24
>
> Number of partially ordered sets (naturally labelled) by rank
> Naturally labelled means labels are consistent with the ordering.
> 1,1,1,1,1,4,1,1,1,11,13,8,1,4,1,1
>
> Number of forests (unlabelled, unordered) with b(i) nodes at height i.
> 1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,2,2,1,1,1,1,2,1,1,1,1,1
>
> ... and, of course, various other forest options.
>
> If there is agreement that this is the right order for these sequences,
> I'll try
> to find time to submit them.
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and
> industry-leading spam and email virus protection.
>
>
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From jvospost3 at gmail.com  Sat Nov  4 00:59:34 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 3 Nov 2006 15:59:34 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
	 <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com>
Message-ID: <5542af940611031559x7a45e57euad1d783fb83d82b@mail.gmail.com>

To give a specific example:

NEW SEQUENCE FROM Jonathan Vos Post

%I A000001
%S A000001 2, 6, 450, 2836181250, 81492043057751910481759423160156250,
4561157026363824997482074305569280581505536351717093893927260661169357729871499327113563125890139588096951624677718591308593750
%N A000001 Multiplicative encoding of Catalan's triangle: Product
p(i+1)^T(n,i).
%C A000001 This is to A009766 "Catalan's triangle T(n,k) (read by
rows)"  as A007188 "Multiplicative encoding of Pascal triangle: Product
p(i+1)^C(n,i)" is to A007318 "Pascal's triangle read by rows."
%F A000001 a(n) = PRODUCT[i=i..n] p(i+1)^T(n,i), where T(n,i), are
Cataln's triangle as in a009766.
%e A000001 a(1) = p(1)^T(1,1) = 2^1 = 2.
a(2) = p(1)^T(2,1) * p(2)^T(2,2) = 2^1 * 3^1 = 6.
a(3) = p(1)^T(3,1) * p(2)^T(3,2) * p(3)^T(3,3) = 2^1 * 3^2 * 5^2 = 450.
a(4) = 2^1 * 3^3 * 5^5 * 7^5 = 2836181250.
a(5) = 2^1 * 3^4 * 5^9 * 7^14 * 11^14 =
81492043057751910481759423160156250.
a(6) = 2^1 * 3^5 * 5^14 * 7^28 * 11^42 * 13^42.
%Y A000001 Cf. A007188, A007318, A009766.
%O A000001 1
%K A000001 ,easy,nonn,
%A A000001 Jonathan Vos Post (jvospost2 at yahoo.com
<http://us.f551.mail.yahoo.com/ym/Compose?To=jvospost2 at yahoo.com&YY=14244&y5beta=yes&y5beta=yes&order=down&sort=date&pos=0>),
Nov 03 2006
RH
RA 192.20.225.32



On 11/3/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> I think the example givem without A number, is:
>
>  A007188 <http://www.research.att.com/%7Enjas/sequences/A007188>
>  Multiplicative encoding of Pascal triangle: Product p(i+1)^C(n,i).
> (Formerly M1722)
>  +0
> 2
> 2, 6, 90, 47250, 66852843750, 2806877704512541816406250,
> 1216935896582703898519354781702537118597533386230468750
> There might be other Godel-coded sequences, and the like, besides Conway's
> brilliant "Fractran", above and beyond those with "decimal Godelization" --
> but usually these quickly lead to numbers too large for OEIS.
>
> -- Jonathan Vos Post
>
> On 11/3/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
> >
> > <Me>
> > >Off hand, I can't think of any interesting functions of finite
> > sequences of
> > >positive integers that depend on the order of the sequence.
> >
> > Of course, I no sooner wrote that than I started thinking of such
> > functions.
> > I'm not sure any of these are tremendously interesting, but they're at
> > least
> > somewhat so.
> >
> > In the following, b(i), i = 1..k is the finite sequence.  I'm starting
> > all these
> > sequences with the empty sequence, but I searched for them without
> > that first term.  All are shown here in A066099 order.
> >
> > Alternating sum
> > Sum_i (-1)^{i+1) b(i)
> > 0,1,2,0,3,1,-1,1,4,2,0,2,-2,0,2,0
> >
> > Binomial sum
> > Sum_i C(k-1,i-1) b(i)
> > 0,1,2,2,3,3,3,4,4,4,4,5,4,6,5,8
> >
> > Inverse binomial sum
> > Sum_i (-1)^{i+1) C(k-1,i-1) b(i)
> > 0,1,2,0,3,1,-1,0,4,2,0,1,-2,-2,1,0
> >
> > Weighted sum
> > sum_i i b(i)
> > 0,1,2,3,3,4,5,6,4,5,6,7,7,8,9,10
> > This is A029931
> >
> > Zero-based weighted sum
> > sum_i (i-1) b(i)
> > 0,0,0,1,0,1,2,3,0,1,2,3,3,4,5,6
> >
> > Sum of products of consecutive elements
> > sum_{i=1}^{k-1} b(i) b(i+1)
> > 0,0,0,1,0,2,2,2,0,3,4,3,3,4,3,3
> >
> > Number of rises
> > sum_{b(i)>b(i-1)} 1
> > 0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,0
> >
> > Number of falls
> > sum_{b(i)<b(i-1)} 1
> > 0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0
> >
> > Number of unchanged
> > sum_{b(i)=b(i-1)} 1
> > 0,0,0,1,0,0,0,2,0,0,1,1,0,0,1,3
> >
> > Number of non-rises
> > sum_{b(i)<=b(i-1)} 1
> > 0,0,0,1,0,0,1,2,0,0,1,1,1,1,2,3
> >
> > Number of non-falls
> > sum_{b(i)>=b(i-1)} 1
> > 0,0,0,1,0,1,0,2,0,1,1,2,0,1,1,3
> >
> > Number of monotonically increasing runs
> > 1 + number of falls, but 0 for empty sequence
> > 0,1,1,1,1,2,1,1,1,2,1,2,1,2,1,1
> >
> > Number of monotonically decreasing runs
> > 1 + number of rises, but 0 for empty sequence
> > 0,1,1,1,1,1,2,1,1,1,1,1,2,2,2,1
> >
> > Number of runs of equal terms
> > 1 + number of unchanged, but 0 for empty sequence
> > 0,1,1,2,1,1,1,3,1,1,2,2,1,1,2,4
> >
> > Number of distinct non-empty subsequences
> > 0,1,1,2,1,3,3,3,1,3,2,5,3,5,5,4
> > E.g., f(1,1) = 2, for the sequences [1], and [1,1].
> >
> > Number of distinct subsequences
> > 1 + number of distinct non-empty subsequences
> > 1,2,2,3,2,4,4,4,2,4,3,6,4,6,6,5
> >
> > Number of set partitions
> > List parts of set partition by their smallest element, and count the
> > part sizes.
> > E.g., 1|2,4|3 (also known as {1,2,3,2}) would count for the sequence
> > [1,2,1].
> > 1,1,1,1,1,2,1,1,1,3,3,3,1,2,1,1
> >
> > Number of permutations
> > List permutations in cycle form, sorted by the smallest element in the
> > cycle,
> > and count the cycle lengths.
> > Number of set partitions * Product_i (b(i)-1)!
> > 1,1,1,1,2,2,1,1,6,6,3,3,2,2,1,1
> >
> > Number of partially ordered sets (unlabelled) by rank
> > The rank of an element in a poset is the length of the longest chain of
> > which it
> > is the largest element.
> > 1,1,1,1,1,2,1,1,1,3,4,3,1,2,1,1
> >
> > Number of partially ordered sets (labelled) by rank
> > 1,1,1,2,1,9,3,6,1,28,54,60,4,36,12,24
> >
> > Number of partially ordered sets (naturally labelled) by rank
> > Naturally labelled means labels are consistent with the ordering.
> > 1,1,1,1,1,4,1,1,1,11,13,8,1,4,1,1
> >
> > Number of forests (unlabelled, unordered) with b(i) nodes at height i.
> > 1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,2,2,1,1,1,1,2,1,1,1,1,1
> >
> > ... and, of course, various other forest options.
> >
> > If there is agreement that this is the right order for these sequences,
> > I'll try
> > to find time to submit them.
> > ________________________________________________________________________
> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > industry-leading spam and email virus protection.
> >
> >
>
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From davidwwilson at comcast.net  Sat Nov  4 01:40:21 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Fri, 3 Nov 2006 19:40:21 -0500
Subject: A124015 thoughts
Message-ID: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p>

The description of A124015 is

%N A124015 Number of words with n letters in the National Scrabble Association Dictionary.

Some notes:

- It should be noted that the NSAD includes a restricted set of English words (words of 2 to 14 letters, no proper nouns or derivatives, no words with non-alphabetic characters (e.g, contractions, hyphenated words), and it is this restricted set that is being counted.

- The edition of the NSAD used to create A124015 should be specified, since the NSAD is continually edited.  While I do not think that A124015 should change with each new edition of the NSAD, because it may be referenced in other literature. On the other hand, I don't think that a new sequence should be created for each new edition of the NSAD unless the new edition exhibits some interesting statistical departure from the current NSAD (e.g, in some future NSAD, the number of 9-letter words exceeds the number of 8-letter words). This is because I believe that the value of A124015 is mainly to indicate a distribution of English word lengths.

- On the other hand, I have always understood that 4-letter words are most common in literature (where small words appear more often). It might be interesting to create a sequence counting words of length n in the King James Bible, War and Peace, or some other suitably large and stable piece of English literature (we don't have to go overboard on this, just one or two examples indicating word length distributions in literature).

- Another interesting idea: In various languages, what is the distribution of the number of letters in the names of numbers from one to (say) one million?

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From franktaw at netscape.net  Sat Nov  4 02:49:48 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Fri, 03 Nov 2006 20:49:48 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <5542af940611031559x7a45e57euad1d783fb83d82b@mail.gmail.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>	 <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com> <5542af940611031559x7a45e57euad1d783fb83d82b@mail.gmail.com>
Message-ID: <8C8CDCAA57554FD-ED8-5165@FWM-M11.sysops.aol.com>

This is OK, but it is the opposite of what we started out talking 
about.  This is a function from integers to sequences; but we are 
talking about functions from sequences to integers.  And not every 
integer encodes a sequence in this way; only those in A055932.

Franklin T. Adams-Watters


-----Original Message-----
From: jvospost3 at gmail.com

  To give a specific example:

 NEW SEQUENCE FROM Jonathan Vos Post

%I A000001
%S A000001 2, 6, 450, 2836181250, 81492043057751910481759423160156250,
4561157026363824997482074305569280581505536351717093893927260661169357729
871499327113563125890139588096951624677718591308593750

%N A000001 Multiplicative encoding of Catalan's triangle: Product
p(i+1)^T(n,i).
%C A000001 This is to A009766 "Catalan's triangle T(n,k) (read by
rows)"  as A007188 "Multiplicative encoding of Pascal triangle: Product

p(i+1)^C(n,i)" is to A007318 "Pascal's triangle read by rows."
%F A000001 a(n) = PRODUCT[i=i..n] p(i+1)^T(n,i), where T(n,i), are
Cataln's triangle as in a009766.
%e A000001 a(1) = p(1)^T(1,1) = 2^1 = 2.

a(2) = p(1)^T(2,1) * p(2)^T(2,2) = 2^1 * 3^1 = 6.
a(3) = p(1)^T(3,1) * p(2)^T(3,2) * p(3)^T(3,3) = 2^1 * 3^2 * 5^2 = 450.
a(4) = 2^1 * 3^3 * 5^5 * 7^5 = 2836181250.
a(5) = 2^1 * 3^4 * 5^9 * 7^14 * 11^14 =

81492043057751910481759423160156250.
a(6) = 2^1 * 3^5 * 5^14 * 7^28 * 11^42 * 13^42.
%Y A000001 Cf. A007188, A007318, A009766.
%O A000001 1
%K A000001 ,easy,nonn,
%A A000001 Jonathan Vos Post (
jvospost2 at yahoo.com), Nov 03 2006
RH
RA 192.20.225.32

On 11/3/06, Jonathan Post <jvospost3 at gmail.com> wrote: I think the 
example givem without A number, is:

   A007188
   Multiplicative encoding of Pascal triangle: Product p(i+1)^C(n,i).
(Formerly M1722)
   +0
2

  2, 6, 90, 47250, 66852843750, 2806877704512541816406250, 
1216935896582703898519354781702537118597533386230468750

  There might be other Godel-coded sequences, and the like, besides 
Conway's brilliant "Fractran", above and beyond those with "decimal 
Godelization" -- but usually these quickly lead to numbers too large 
for OEIS.

 -- Jonathan Vos Post

On 11/3/06, franktaw at netscape.net < franktaw at netscape.net> wrote: <Me>
>Off hand, I can't think of any interesting functions of finite
 sequences of
>positive integers that depend on the order of the sequence.

Of course, I no sooner wrote that than I started thinking of such
functions.
I'm not sure any of these are tremendously interesting, but they're at
least
somewhat so.

In the following, b(i), i = 1..k is the finite sequence. I'm starting
all these
sequences with the empty sequence, but I searched for them without
that first term. All are shown here in A066099 order.

Alternating sum
Sum_i (-1)^{i+1) b(i)
0,1,2,0,3,1,-1,1,4,2,0,2,-2,0,2,0

Binomial sum
Sum_i C(k-1,i-1) b(i)
0,1,2,2,3,3,3,4,4,4,4,5,4,6,5,8

Inverse binomial sum
Sum_i (-1)^{i+1) C(k-1,i-1) b(i)
0,1,2,0,3,1,-1,0,4,2,0,1,-2,-2,1,0

Weighted sum
sum_i i b(i)
0,1,2,3,3,4,5,6,4,5,6,7,7,8,9,10
This is A029931

Zero-based weighted sum
sum_i (i-1) b(i)
0,0,0,1,0,1,2,3,0,1,2,3,3,4,5,6

 Sum of products of consecutive elements
sum_{i=1}^{k-1} b(i) b(i+1)
0,0,0,1,0,2,2,2,0,3,4,3,3,4,3,3

Number of rises
sum_{b(i)>b(i-1)} 1
0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,0

Number of falls
sum_{b(i)<b(i-1)} 1
0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0

Number of unchanged
sum_{b(i)=b(i-1)} 1
0,0,0,1,0,0,0,2,0,0,1,1,0,0,1,3

Number of non-rises
sum_{b(i)<=b(i-1)} 1
0,0,0,1,0,0,1,2,0,0,1,1,1,1,2,3

Number of non-falls
sum_{b(i)>=b(i-1)} 1
0,0,0,1,0,1,0,2,0,1,1,2,0,1,1,3

Number of monotonically increasing runs
1 + number of falls, but 0 for empty sequence
0,1,1,1,1,2,1,1,1,2,1,2,1,2,1,1

Number of monotonically decreasing runs
1 + number of rises, but 0 for empty sequence
0,1,1,1,1,1,2,1,1,1,1,1,2,2,2,1

Number of runs of equal terms
1 + number of unchanged, but 0 for empty sequence
0,1,1,2,1,1,1,3,1,1,2,2,1,1,2,4

Number of distinct non-empty subsequences
0,1,1,2,1,3,3,3,1,3,2,5,3,5,5,4
E.g., f(1,1) = 2, for the sequences [1], and [1,1].

Number of distinct subsequences
1 + number of distinct non-empty subsequences
1,2,2,3,2,4,4,4,2,4,3,6,4,6,6,5

 Number of set partitions
List parts of set partition by their smallest element, and count the
part sizes.
E.g., 1|2,4|3 (also known as {1,2,3,2}) would count for the sequence
[1,2,1].
1,1,1,1,1,2,1,1,1,3,3,3,1,2,1,1

Number of permutations
List permutations in cycle form, sorted by the smallest element in the
cycle,
and count the cycle lengths.
Number of set partitions * Product_i (b(i)-1)!
1,1,1,1,2,2,1,1,6,6,3,3,2,2,1,1

Number of partially ordered sets (unlabelled) by rank
The rank of an element in a poset is the length of the longest chain of
which it
is the largest element.
1,1,1,1,1,2,1,1,1,3,4,3,1,2,1,1

Number of partially ordered sets (labelled) by rank
1,1,1,2,1,9,3,6,1,28,54,60,4,36,12,24

Number of partially ordered sets (naturally labelled) by rank
Naturally labelled means labels are consistent with the ordering.
1,1,1,1,1,4,1,1,1,11,13,8,1,4,1,1

 Number of forests (unlabelled, unordered) with b(i) nodes at height i.
1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,2,2,1,1,1,1,2,1,1,1,1,1

... and, of course, various other forest options.

If there is agreement that this is the right order for these sequences,
I'll try
to find time to submit them.
________________________________________________________________________
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industry-leading spam and email virus protection.








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From maxale at gmail.com  Sat Nov  4 03:09:57 2006
From: maxale at gmail.com (Max A.)
Date: Fri, 3 Nov 2006 18:09:57 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <200611032139.kA3LdZSV024106@smtp.well.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <200611031625.kA3GP1mt017249@smtp.well.com>
	 <d3dac270611031226x77574e3ele81e3cfce66edec3@mail.gmail.com>
	 <200611032139.kA3LdZSV024106@smtp.well.com>
Message-ID: <d3dac270611031809j6a1f850dj5a1846b40153dc0@mail.gmail.com>

On 11/3/06, Marc LeBrun <mlb at well.com> wrote:

> B. Encode each tuple as a single number, say by using it as factor
> exponents:  T(i) --> 2^T(i,1) * 3^T(i,2) * 5^T(i,3) *...

The sequences I've mentioned encoded this way:

I) Multinomial coefficients:

a( [k1, k2, ..., kn] ) = (k1 + k2 +  ... + kn)! / (k1! * k2! * ... * kn!)

1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 2, 1, 1, 3, 1, 3, 2, 2, 1,
4, 1, 2, 1, 3, 1, 6, 1, 1, 2, 2, 2, 6, 1, 2, 2, 4, 1, 6, 1, 3, 3, 2,
1, 5, 1, 3, 2, 3, 1, 4, 2, 4, 2, 2, 1, 12, 1, 2, 3, 1, 2, 6, 1, 3, 2,
6, 1, 10, 1, 2, 3, 3, 2, 6, 1, 5, 1, 2, 1, 12, 2, 2, 2, 4, 1, 12, 2,
3, 2, 2, 2, 6, 1, 3, 3, 6

II) Coefficients of incomplete Bell polynomials:

b( [k1, k2, ..., kn] ) = (k1 + 2*k2 + ... + n*kn)! / (k1! * k2! * ...
* kn!) / (1!^k1 * 2!^k2 * ... * n!^kn)

1, 1, 1, 1, 1, 3, 1, 1, 3, 4, 1, 6, 1, 5, 10, 1, 1, 15, 1, 10, 15, 6,
1, 10, 10, 7, 15, 15, 1, 60, 1, 1, 21, 8, 35, 45, 1, 9, 28, 20, 1,
105, 1, 21, 105, 10, 1, 15, 35, 70, 36, 28, 1, 105, 56, 35, 45, 11, 1,
210, 1, 12, 210, 1, 84, 168, 1, 36, 55, 280, 1, 105, 1, 13, 280, 45,
126, 252, 1, 35, 105, 14, 1, 420, 120, 15, 66, 56, 1, 840, 210, 55,
78, 16, 165, 21, 1, 315, 378, 280

III) The number of permutations with given cycle structure:

c( [k1, k2, ..., kn] ) = (k1 + 2*k2 + ... + n*kn)! / (k1! * k2! * ...
* kn!) / (1^k1 * 2^k2 * ... * n^kn)

1, 1, 1, 1, 2, 3, 6, 1, 3, 8, 24, 6, 120, 30, 20, 1, 720, 15, 5040,
20, 90, 144, 40320, 10, 40, 840, 15, 90, 362880, 120, 3628800, 1, 504,
5760, 420, 45, 39916800, 45360, 3360, 40, 479001600, 630, 6227020800,
504, 210, 403200, 87178291200, 15, 1260, 280, 25920, 3360,
1307674368000, 105, 2688, 210, 226800, 3991680, 20922789888000, 420,
355687428096000, 43545600, 1260, 1, 20160, 4032, 6402373705728000,
25920, 2217600, 3360, 121645100408832000, 105, 2432902008176640000,
518918400, 1120, 226800, 18144, 30240, 51090942171709440000, 70, 105,
6706022400, 1124000727777607680000, 2520, 172800, 93405312000,
23950080, 1344, 25852016738884976640000, 1680, 151200, 2217600,
283046400, 1394852659200, 1663200, 21, 620448401733239439360000,
11340, 9072, 1120

Max





From jawbrey at att.net  Sat Nov  4 03:26:36 2006
From: jawbrey at att.net (Jon Awbrey)
Date: Fri, 03 Nov 2006 21:26:36 -0500
Subject: querying multidimensional sequences in OEIS
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>	 <5542af940611031436w1d4a8081wc840fb7a2a02d41b@mail.gmail.com> <5542af940611031559x7a45e57euad1d783fb83d82b@mail.gmail.com> <8C8CDCAA57554FD-ED8-5165@FWM-M11.sysops.aol.com>
Message-ID: <454BFA5C.ACD4F79B@att.net>

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o

but every positive integer in N = {1, 2, 3, ...}
uniquely codes a finite partial function f : N -> N

http://stderr.org/pipermail/inquiry/2005-July/002846.html

ja

franktaw at netscape.net wrote:
> 
> This is OK, but it is the opposite of what we started out talking
> about.  This is a function from integers to sequences; but we are
> talking about functions from sequences to integers.  And not every
> integer encodes a sequence in this way; only those in A055932.
> 
> Franklin T. Adams-Watters

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o
inquiry e-lab: http://stderr.org/pipermail/inquiry/
citizendium development forums: http://smf.citizendium.org
http://www.textop.org/wiki/index.php?title=User:Jon_Awbrey
wikinfo: http://wikinfo.org/wiki.php?title=User:Jon_Awbrey
wp review: http://wikipediareview.com/index.php?showuser=398
o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o






From jvospost3 at gmail.com  Sat Nov  4 03:49:46 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 3 Nov 2006 18:49:46 -0800
Subject: A124015 thoughts
In-Reply-To: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p>
Message-ID: <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>

I mostly agree.  A large sample of text as actually used in a language, such
as a set of books, newspaper articles, transcripts of phone conversations,
or the like, is often called a "corpus" by those who undertake statistical
analyses.  Various government agencies have done so since well before World
War II, tapping the minds of as many of half the Mathematicians in the
world, with computational resources in advance of corporate and personal
users, and with the help of Mathematicians such as Turing.

There are statistical analyses of various editions of the Bible, of the
works of Shakespeare, and the like, with phalanxes of statistical analysts
on Literature.

Many of these are covered in specialized journals, of questionable value to
seqfans, and unnoticed here except when, for instance, a minor poem is
claimed by such methods to be by Shakespeare.

Hence I second the motion by David Wilson to start with an extremely
well-defined if artificial corpus, such as "In various languages, what is
the distribution of the number of letters in the names of numbers from one
to (say) one million?"

I have commented more extensively on Scrabble in other venues.  Do not ever
agree to use the OED (Oxford English Dictionary) as the official dictionary
when playing Scrabble.  Many misspellings by Tournament rumes are there.
Even though the OED is the definitive dictionary of the largest language on
Earth, and fascinating reading for various purposes, it is just too big and
too weird for game use.  Also, it makes it hard to distinguish a bluff from
a variant.

Now that I think of it, logarithmic trends can be interesting.  How about "what
is the distribution of the number of letters in the names of numbers from
one to one billion?"

-- Jonathan Vos Post

On 11/3/06, David Wilson <davidwwilson at comcast.net> wrote:
>
>  The description of A124015 is
>
> %N A124015 Number of words with n letters in the National Scrabble
> Association Dictionary.
>
> Some notes:
>
> - It should be noted that the NSAD includes a restricted set of English
> words (words of 2 to 14 letters, no proper nouns or derivatives, no words
> with non-alphabetic characters (e.g, contractions, hyphenated words), and
> it is this restricted set that is being counted.
>
> - The edition of the NSAD used to create A124015 should be specified,
> since the NSAD is continually edited.  While I do not think that A124015
> should change with each new edition of the NSAD, because it may be
> referenced in other literature. On the other hand, I don't think that a new
> sequence should be created for each new edition of the NSAD unless the new
> edition exhibits some interesting statistical departure from the current
> NSAD (e.g, in some future NSAD, the number of 9-letter words exceeds the
> number of 8-letter words). This is because I believe that the value of
> A124015 is mainly to indicate a distribution of English word lengths.
>
> - On the other hand, I have always understood that 4-letter words are most
> common in literature (where small words appear more often). It might be
> interesting to create a sequence counting words of length n in the King
> James Bible, War and Peace, or some other suitably large and stable piece of
> English literature (we don't have to go overboard on this, just one or two
> examples indicating word length distributions in literature).
>
> - Another interesting idea: In various languages, what is the distribution
> of the number of letters in the names of numbers from one to (say) one
> million?
>
>
>
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From davidwwilson at comcast.net  Sat Nov  4 17:58:32 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sat, 4 Nov 2006 11:58:32 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
References: <20061103030956.55909.qmail@web38210.mail.mud.yahoo.com>
Message-ID: <001701c70032$75ba7ff0$6401a8c0@yourxhtr8hvc4p>

----- Original Message ----- 
From: "zak seidov" <zakseidov at yahoo.com>
To: <seqfan at ext.jussieu.fr>
Sent: Thursday, November 02, 2006 10:09 PM
Subject: Re: A065577 Number of Goldbach partitions of 10^n? More terms?


> David, Dick, Bob, Ivars,
> seqfans.
>
> After recalculating (at two different PC's)
> and rechecking I could finally
> find source of my error
> (silly as usual) in one of codes,
> and my final value is
> a(10) = 18200488 (NOT 18200487!)
> which coincides now with the Dick-David value.
>
> So, Bob,
> you may wish to give in A065577  the correct value
> a(10) = 18200488 with due refer to David Wilson  and
> Richard Mathar (not me).

I don't remember if Richard Mathar or myself was first to find a(10) = 
18200488. It's not really important to me, since any value would be suspect 
until independently confirmed. The fact that your suspect value led you to a 
bug in your program simply serves as additional confirmation. I have no 
problem with all three of us getting attribution. But then, I have no 
problem with none of us getting attribution, as this computation strikes me 
as on the level of an undergraduate homework problem.

> Also Ivars can extend his table.
> Thanks, Zak
>
> BTW1 Still, is it possible to find a(11)... with C++?

Well, it is possible to find a(11) using a C++ sieving program on an aging 
PC in a few hours, and if you were to do so, you would find that a(11) = 
149091160. Methinks that finding a(12) using the same program on the same PC 
would take about a week, except that said PC is a shared resource.

> BIW2 Not all people even in this closed list
> may be interested in Mmca specific problems,
> so I''l be happy to have a list of you
> particularly Mmca-involved.
> Then we may circulate relevant messages
> not to all seqfan community.
> Please anyone interested write me personally
> with subject "Me in Mmca", thanks.
> Of course it's not intented to be closed list!
> Hope this is not against Gerard' policy.

I truly wish I could afford Mma (or better yet, a math-enabled PDF editor). 
If I'm on anybody's Christmas list...






From mathar at strw.leidenuniv.nl  Sat Nov  4 18:20:07 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Sat, 4 Nov 2006 18:20:07 +0100
Subject: primes in arithmetic progression
Message-ID: <200611041720.kA4HK70T027399@amer.strw.leidenuniv.nl>


Is there an OEIS table that shows the smallest prime p
of k (not necessarily consecutive) primes in arithmetic progression
with common difference d? The table would look similar to this one
below, and contain rather large primes where I am leaving blanks:

    k=1 k=2  k=3  k=4  k=5  k=6
d
0     2   2    2    2    2    2
2     2   3    3
4     2   3    3
6     2   5    5    5    5
8     2   3    3
10    2   3    3
12    2   5    5    5    5
14    2   3    3 
16    2   3
18    2   5    5    5
20    2   3
22    2   7
24    2   5    5   59
26    2   3
28    2   3    3
30    2   7    7    7     7    7
32    2   5
34    2   3    3
36    2   5    7   31

The row d=0 and the column k=1 are degenerate and filled with the
prime 2. All strides d are even. Example for row d=24 and column k=4:
The 4 numbers 59,59+24,59+2*24 and 59+3*24 are all primes.

http://www.research.att.com/~njas/sequences/Sindx_Pri.html#primes_AP

--Richard





From rsc at swtch.com  Sat Nov  4 18:24:53 2006
From: rsc at swtch.com (Russ Cox)
Date: Sat, 4 Nov 2006 12:24:53 -0500
Subject: A065577 Number of Goldbach partitions of 10^n? More terms?
In-Reply-To: <001701c70032$75ba7ff0$6401a8c0@yourxhtr8hvc4p>
References: <20061103030956.55909.qmail@web38210.mail.mud.yahoo.com>
	 <001701c70032$75ba7ff0$6401a8c0@yourxhtr8hvc4p>
Message-ID: <ee9e417a0611040924h7c62e2a6hc2ba0ec199c7a175@mail.gmail.com>

I have confirmed the values so far
and also computed that a(12) = 1243722370.
Submitted (with a(11)) to OEIS via form.

Russ





From jvospost3 at gmail.com  Sat Nov  4 19:33:14 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 4 Nov 2006 10:33:14 -0800
Subject: primes in arithmetic progression
In-Reply-To: <200611041720.kA4HK70T027399@amer.strw.leidenuniv.nl>
References: <200611041720.kA4HK70T027399@amer.strw.leidenuniv.nl>
Message-ID: <5542af940611041033s7a722e9fjf6ad6643f988477d@mail.gmail.com>

The semiprime analogue of this is a table that shows the smallest semiprime
S
of k (not necessarily consecutive) primes in arithmetic progression
with common difference d (once a 0 appears, the row has ended nonzero
values):

d   k+1  k=2  k=3  k=4  k=5  k=6  k=7 k=8
0   4      4       4      4       4       4       4 ...
1   4      9       33    0
2   4      4       91    213   1383 8129 3091 0
3   4      6       115
4   4      6       6      111
5   4      4       77
6   4      4
7   4      14    51
8   4      6      6       69

-- Fairly easy to extend, fairly easy to prove the first 0 in a row.

The same can be done for 3-almost primes, 4-almost primes, and so forth,
this making a 3-dimensional array.



On 11/4/06, Richard Mathar <mathar at strw.leidenuniv.nl> wrote:
>
>
> Is there an OEIS table that shows the smallest prime p
> of k (not necessarily consecutive) primes in arithmetic progression
> with common difference d? The table would look similar to this one
> below, and contain rather large primes where I am leaving blanks:
>
>     k=1 k=2  k=3  k=4  k=5  k=6
> d
> 0     2   2    2    2    2    2
> 2     2   3    3
> 4     2   3    3
> 6     2   5    5    5    5
> 8     2   3    3
> 10    2   3    3
> 12    2   5    5    5    5
> 14    2   3    3
> 16    2   3
> 18    2   5    5    5
> 20    2   3
> 22    2   7
> 24    2   5    5   59
> 26    2   3
> 28    2   3    3
> 30    2   7    7    7     7    7
> 32    2   5
> 34    2   3    3
> 36    2   5    7   31
>
> The row d=0 and the column k=1 are degenerate and filled with the
> prime 2. All strides d are even. Example for row d=24 and column k=4:
> The 4 numbers 59,59+24,59+2*24 and 59+3*24 are all primes.
>
> http://www.research.att.com/~njas/sequences/Sindx_Pri.html#primes_AP
>
> --Richard
>
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From jvospost3 at gmail.com  Sat Nov  4 19:37:19 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 4 Nov 2006 10:37:19 -0800
Subject: primes in arithmetic progression
In-Reply-To: <5542af940611041033s7a722e9fjf6ad6643f988477d@mail.gmail.com>
References: <200611041720.kA4HK70T027399@amer.strw.leidenuniv.nl>
	 <5542af940611041033s7a722e9fjf6ad6643f988477d@mail.gmail.com>
Message-ID: <5542af940611041037s636536abnf98a7eee7feb349e@mail.gmail.com>

Sorry, I mean: "The semiprime analogue of this is a table that shows the
smallest semiprime S of k (not necessarily consecutive) SEMIprimes in
arithmetic progression
 with common difference d."

On 11/4/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> The semiprime analogue of this is a table that shows the smallest
> semiprime S
> of k (not necessarily consecutive) primes in arithmetic progression
> with common difference d (once a 0 appears, the row has ended nonzero
> values):
>
> d   k+1  k=2  k=3  k=4  k=5  k=6  k=7 k=8
> 0   4      4       4      4       4       4       4 ...
> 1   4      9       33    0
> 2   4      4       91    213   1383 8129 3091 0
> 3   4      6       115
> 4   4      6       6      111
> 5   4      4       77
> 6   4      4
> 7   4      14    51
> 8   4      6      6       69
>
> -- Fairly easy to extend, fairly easy to prove the first 0 in a row.
>
> The same can be done for 3-almost primes, 4-almost primes, and so forth,
> this making a 3-dimensional array.
>
>
>
> On 11/4/06, Richard Mathar <mathar at strw.leidenuniv.nl> wrote:
> >
> >
> > Is there an OEIS table that shows the smallest prime p
> > of k (not necessarily consecutive) primes in arithmetic progression
> > with common difference d? The table would look similar to this one
> > below, and contain rather large primes where I am leaving blanks:
> >
> >     k=1 k=2  k=3  k=4  k=5  k=6
> > d
> > 0     2   2    2    2    2    2
> > 2     2   3    3
> > 4     2   3    3
> > 6     2   5    5    5    5
> > 8     2   3    3
> > 10    2   3    3
> > 12    2   5    5    5    5
> > 14    2   3    3
> > 16    2   3
> > 18    2   5    5    5
> > 20    2   3
> > 22    2   7
> > 24    2   5    5   59
> > 26    2   3
> > 28    2   3    3
> > 30    2   7    7    7     7    7
> > 32    2   5
> > 34    2   3    3
> > 36    2   5    7   31
> >
> > The row d=0 and the column k=1 are degenerate and filled with the
> > prime 2. All strides d are even. Example for row d=24 and column k=4:
> > The 4 numbers 59,59+24,59+2*24 and 59+3*24 are all primes.
> >
> > http://www.research.att.com/~njas/sequences/Sindx_Pri.html#primes_AP<http://www.research.att.com/%7Enjas/sequences/Sindx_Pri.html#primes_AP>
> >
> > --Richard
> >
>
>
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From njas at research.att.com  Sat Nov  4 20:10:10 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sat, 4 Nov 2006 14:10:10 -0500 (EST)
Subject: primes in arithmetic progression
Message-ID: <200611041910.OAA62978@fry.research.att.com>

Just to get the ball rolling, I am adding this entry:

%I A124064
%S A124064 2,2,2,2,3,2,2,3,3,2,2,5,3
%N A124064 Array read by antidiagonals: T(d,k) (k >= 1, d = 2,4,6,8,...) = smallest prime p of k (not necessarily consecutive) primes in arithmetic progression with common difference d.
%C A124064 The row d=0 and the column k=1 are degenerate and are filled with the prime 2.
%O A124064 1,1
%K A124064 nonn,tabl,more
%e A124064 Array begins:
%e A124064 d.\...k=1.k=2..k=3..k=4..k=5..k=6
%e A124064 0..|..2...2....2....2....2....2
%e A124064 2..|..2...3....3
%e A124064 4..|..2...3....3
%e A124064 6..|..2...5....5....5....5
%e A124064 8..|..2...3....3
%e A124064 10.|..2...3....3
%e A124064 12.|..2...5....5....5....5
%e A124064 14.|..2...3....3.
%e A124064 16.|..2...3
%e A124064 18.|..2...5....5....5
%e A124064 20.|..2...3
%e A124064 22.|..2...7
%e A124064 24.|..2...5....5...59
%e A124064 26.|..2...3
%e A124064 28.|..2...3....3
%e A124064 30.|..2...7....7....7.....7....7
%e A124064 32.|..2...5
%e A124064 34.|..2...3....3
%e A124064 36.|..2...5....7...31
%e A124064 Example for row d=24 and column k=4: the 4 numbers 59,59+24,59+2*24 and 59+3*24 are all primes.
%A A124064 Richard Mathar (mathar(AT)strw.leidenuniv.nl), Nov 04 2006




NJAS





From davidwwilson at comcast.net  Sat Nov  4 21:10:45 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sat, 4 Nov 2006 15:10:45 -0500
Subject: Observation on A053669
Message-ID: <000601c7004d$507dc510$6401a8c0@yourxhtr8hvc4p>

It seems to me that

%N A053669 Smallest prime co-prime to n.

could just as easily be described

%N A053669 Smallest number > 1 coprime to n.

and the fact that a(n) is prime falls out.  This definition would also simplify the program.
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From mathar at strw.leidenuniv.nl  Sat Nov  4 21:45:17 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Sat, 4 Nov 2006 21:45:17 +0100
Subject: Observation on A053669
Message-ID: <200611042045.kA4KjHeV011396@amer.strw.leidenuniv.nl>


dww> From seqfan-owner at ext.jussieu.fr  Sat Nov  4 21:12:32 2006
dww> Return-Path: <seqfan-owner at ext.jussieu.fr>
dww> From: "David Wilson" <davidwwilson at comcast.net>
dww> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
dww> Subject: Observation on A053669
dww> ...
dww> It seems to me that
dww> 
dww> %N A053669 Smallest prime co-prime to n.
dww> 
dww> could just as easily be described
dww> 
dww> %N A053669 Smallest number > 1 coprime to n.
dww> 
dww> and the fact that a(n) is prime falls out.  This definition would also =
dww> simplify the program.
dww>...

Yes. If the smallest prime co-prime to n would not be a prime,
it would be built by at least two prime factors, the smaller of which
would be the correct entry (proof by reductio ad absurdum).

It would still be useful to keep in a comment that the numbers are all prime
(this might speed up some searches for the numbers...)

I'd propose to list in a comment the first occurrences of the first primes:
first 5 at a(6)
first 7 is a(30)
first 11 is a(210)
first 13 is a(2310)
first 17 is a(30030)
first 19 is a(510510)
where the indices are 6=2*3, 30=2*3*5, 210=2*3*5*7, 2*3*5*7*11 etc,
(the primorials, I guess, as in A002110)

-- RJM





From jvospost3 at gmail.com  Sat Nov  4 22:55:31 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 4 Nov 2006 13:55:31 -0800
Subject: primes in arithmetic progression
In-Reply-To: <200611041910.OAA62978@fry.research.att.com>
References: <200611041910.OAA62978@fry.research.att.com>
Message-ID: <5542af940611041355vd65aa5cr3a4a9df3cc348c15@mail.gmail.com>

The analogue for 3-almost primes (or, as njas prefers, numbers which are the
product of exactly 3 primes) is left as an exercise for the reader...  I've
just submitted the following.

Subject: PRE-NUMBERED NEW SEQUENCE A124570 FROM Jonathan Vos Post


%I A124570
%S A124570 4, 4, 4, 4, 9, 4, 4, 4, 33, 4, 4, 6, 91, 0, 4, 4, 6, 115,
213, 0, 4, 4, 4, 6, 0, 1383, 0, 4, 4, 4, 77, 111, 0, 8129, 0, 4, 4, 14
%N A124570 Array read by antidiagonals: T(d,k) (k >= 1, d =
1,2,3,4,5,6,...) = smallest semiprime s of k (not necessarily consecutive)
semiprimes in arithmetic progression with common difference d.
%C A124570 Semiprime analogue of A124064 Array read by antidiagonals:
T(d,k) (k >= 1, d = 2,4,6,8,...) = smallest prime p of k (not
necessarily consecutive) primes in arithmetic progression with common
difference d. The row d=0 and the column k=1 are degenerate and are filled with
the semiprime 4. Once a 0 occurs in a row, there are no more nonzero
vales in the row. For example, There cannot be 4 semiprimes in arithmetic
progression with common difference 3, starting with k, because modulo 4
we have {k, k+3, k+6, k+9} == {k+0, k+3, k+2, k+1} and one of these
must be divisible by 4, hence a nonsemiprime (eliminating k = 4 by
inspection).
%e A124570 Array begins:
d.\...k=1.k=2..k=3..k=4..k=5..k=6..k=7..k=8
0..|..4...4....4....4....4....4....4....4
1..|..4...9....33...0....0....0....0....0
2..|..4...4....91...213..1383.8129.3091.0
3..|..4...6....115..0....0....0....0....0
4..|..4...6....6....6....111...
5..|..4...4....77...
6..|..4...4....
7..|..4...14...51...
8..|..4...6....6....69.
Example for row 3: 115 = 5 * 23 is semiprime, 115+3 = 118 = 2 * 59 is
semiprime, and 115+3+3 = 121 = 11^2 is semiprime, so T(3,3) = 115.
%Y A124570 Cf. A000040, A001358, A056809, A070552, A092125, A092126,
A092127, A092128, A092129, A124064.
%O A124570 1,1
%K A124570 ,easy,more,nonn,tabl,
%A A124570 Jonathan Vos Post (jvospost2 at yahoo.com
<http://us.f551.mail.yahoo.com/ym/Compose?To=jvospost2 at yahoo.com&YY=99063&y5beta=yes&y5beta=yes&order=down&sort=date&pos=0&view=a&head=b>),
Nov 04 2006
RH
RA 192.20.225.32
RU
RI



On 11/4/06, N. J. A. Sloane <njas at research.att.com> wrote:
>
> Just to get the ball rolling, I am adding this entry:
>
> %I A124064
> %S A124064 2,2,2,2,3,2,2,3,3,2,2,5,3
> %N A124064 Array read by antidiagonals: T(d,k) (k >= 1, d = 2,4,6,8,...) =
> smallest prime p of k (not necessarily consecutive) primes in arithmetic
> progression with common difference d.
> %C A124064 The row d=0 and the column k=1 are degenerate and are filled
> with the prime 2.
> %O A124064 1,1
> %K A124064 nonn,tabl,more
> %e A124064 Array begins:
> %e A124064 d.\...k=1.k=2..k=3..k=4..k=5..k=6
> %e A124064 0..|..2...2....2....2....2....2
> %e A124064 2..|..2...3....3
> %e A124064 4..|..2...3....3
> %e A124064 6..|..2...5....5....5....5
> %e A124064 8..|..2...3....3
> %e A124064 10.|..2...3....3
> %e A124064 12.|..2...5....5....5....5
> %e A124064 14.|..2...3....3.
> %e A124064 16.|..2...3
> %e A124064 18.|..2...5....5....5
> %e A124064 20.|..2...3
> %e A124064 22.|..2...7
> %e A124064 24.|..2...5....5...59
> %e A124064 26.|..2...3
> %e A124064 28.|..2...3....3
> %e A124064 30.|..2...7....7....7.....7....7
> %e A124064 32.|..2...5
> %e A124064 34.|..2...3....3
> %e A124064 36.|..2...5....7...31
> %e A124064 Example for row d=24 and column k=4: the 4 numbers
> 59,59+24,59+2*24 and 59+3*24 are all primes.
> %A A124064 Richard Mathar (mathar(AT)strw.leidenuniv.nl), Nov 04 2006
>
>
>
>
> NJAS
>
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From jrbibers at indiana.edu  Sat Nov  4 23:17:45 2006
From: jrbibers at indiana.edu (Joseph Biberstine)
Date: Sat, 04 Nov 2006 17:17:45 -0500
Subject: A124015 thoughts
In-Reply-To: <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
Message-ID: <454D1189.8090102@indiana.edu>

	For anyone interested, this Mathematica function, eng[], will (as
usual, barring bugs) give the American English name for integers on 0 to
10^21-1.  Extension of the upper bound should be plain to implement
(just one line of code -- observe the trend).

	All printing is lowercase and without hyphens.  In holding that "and"
should only ever be read for radix points, it is never printed here.
Note this implementation uses a text postprocessor and so is inherently
ugly.  Also note that the helper memoizes for better or worse.  Please,
of course, report all bugs.

	Attached are some diagrams of the predictable distribution of
StringLength[eng[n]] (note this includes spaces).  The third ranges up
to 10^9, sampling every 10^6.

	Here is sample output from random naturals on 0,999999:

 40573, forty thousand five hundred seventy three
108097, one hundred eight thousand ninety seven
539799, five hundred thirty nine thousand seven hundred ninety nine
136868, one hundred thirty six thousand eight hundred sixty eight
597841, five hundred ninety seven thousand eight hundred forty one
272460, two hundred seventy two thousand four hundred sixty
135550, one hundred thirty five thousand five hundred fifty
402424, four hundred two thousand four hundred twenty four
543617, five hundred forty three thousand six hundred seventeen
868945, eight hundred sixty eight thousand nine hundred forty five
951779, nine hundred fifty one thousand seven hundred seventy nine
742845, seven hundred forty two thousand eight hundred forty five
 84140, eighty four thousand one hundred forty
680145, six hundred eighty thousand one hundred forty five
 82734, eighty two thousand seven hundred thirty four
154758, one hundred fifty four thousand seven hundred fifty eight
961084, nine hundred sixty one thousand eighty four
360944, three hundred sixty thousand nine hundred forty four
574416, five hundred seventy four thousand four hundred sixteen
164099, one hundred sixty four thousand ninety nine

	Here is the code:

Clear[eng, engRaw, n];
eng[n_] := (
  pre = engRaw[n];
  If[pre == "", pre = "zero"]; (* fixes "zero" *)
  pre = FixedPoint[StringReplace[#, "  " -> " "] &, pre] ; (* no
consecutive spaces *)
  pre = FixedPoint[StringReplace[#, RegularExpression[" $"] :> ""] &,
pre] ; (* no trailing spaces *)
  pre);
(* This raw function relies heavily on the postprocessor, which fixes
"zero" and deals with whitespace *)
engRaw[n_] := engRaw[n] = (* memo *)
  Which[
    n == 0, "",(* fixed in postprocessing *)
    n == 1, "one",
    n == 2, "two",
    n == 3, "three",
    n == 4, "four",
    n == 5, "five",
    n == 6, "six",
    n == 7, "seven",
    n == 8, "eight",
    n == 9, "nine",
    n == 10, "ten",
    n == 11, "eleven",
    n == 12, "twelve",
    n == 13, "thirteen",
    n == 14, "fourteen",
    n == 15, "fifteen",
    n == 16, "sixteen",
    n == 17, "seventeen",
    n == 18, "eighteen",
    n == 19, "nineteen",
    20 <= n <= 29, "twenty " <> engRaw[Mod[n, 10]],
    30 <= n <= 39, "thirty " <> engRaw[Mod[n, 10]],
    40 <= n <= 49, "forty " <> engRaw[Mod[n, 10]],
    50 <= n <= 59, "fifty " <> engRaw[Mod[n, 10]],
    60 <= n <= 69, "sixty " <> engRaw[Mod[n, 10]],
    70 <= n <= 79, "seventy " <> engRaw[Mod[n, 10]],
    80 <= n <= 89, "eighty " <> engRaw[Mod[n, 10]],
    90 <= n <= 99, "ninety " <> engRaw[Mod[n, 10]],
    10^2 <= n <= 10^3 - 1, engRaw[Quotient[n, 10^2]] <> " hundred " <>
engRaw[Mod[n, 10^2]],
    10^3 <= n <= 10^6 - 1, engRaw[Quotient[n, 10^3]] <> " thousand " <>
engRaw[Mod[n, 10^3]],
    10^6 <= n <= 10^9 - 1, engRaw[Quotient[n, 10^6]] <> " million " <>
engRaw[Mod[n, 10^6]],
    10^9 <= n <= 10^12 - 1, engRaw[Quotient[n, 10^9]] <> " billion " <>
engRaw[Mod[n, 10^9]],
    10^12 <= n <= 10^15 - 1, engRaw[Quotient[n, 10^12]] <> " trillion "
<> engRaw[Mod[n, 10^12]],
    10^15 <= n <= 10^18 - 1, engRaw[Quotient[n, 10^15]] <> " quadrillion
" <> engRaw[Mod[n, 10^15]],
    10^18 <= n <= 10^21 - 1, engRaw[Quotient[n, 10^18]] <> " quintillion
" <> engRaw[Mod[n, 10^18]],
    True, Return["Naturals beyond 10^21-1 are not supported"];];

-JRB

Jonathan Post wrote:
> Now that I think of it, logarithmic trends can be interesting.  How
> about "what is the distribution of the number of letters in the names of
> numbers from one to one billion?"
> 
> -- Jonathan Vos Post
> 
> On 11/3/06, *David Wilson* <davidwwilson at comcast.net
> <mailto:davidwwilson at comcast.net>> wrote:
> 
>     The description of A124015 is
>      
>     %N A124015 Number of words with n letters in the National Scrabble
>     Association Dictionary.
>      
>     Some notes:
>      
>     - It should be noted that the NSAD includes a restricted set of
>     English words (words of 2 to 14 letters, no proper nouns or
>     derivatives, no words with non-alphabetic characters (e.g,
>     contractions, hyphenated words), and it is this restricted set that
>     is being counted.
>      
>     - The edition of the NSAD used to create A124015 should be
>     specified, since the NSAD is continually edited.  While I do not
>     think that A124015 should change with each new edition of the NSAD,
>     because it may be referenced in other literature. On the other hand,
>     I don't think that a new sequence should be created for each new
>     edition of the NSAD unless the new edition exhibits some interesting
>     statistical departure from the current NSAD (e.g, in some future
>     NSAD, the number of 9-letter words exceeds the number of 8-letter
>     words). This is because I believe that the value of A124015 is
>     mainly to indicate a distribution of English word lengths.
>      
>     - On the other hand, I have always understood that 4-letter words
>     are most common in literature (where small words appear more often).
>     It might be interesting to create a sequence counting words of
>     length n in the King James Bible, War and Peace, or some other
>     suitably large and stable piece of English literature (we don't have
>     to go overboard on this, just one or two examples indicating word
>     length distributions in literature).
>      
>     - Another interesting idea: In various languages, what is the
>     distribution of the number of letters in the names of numbers from
>     one to (say) one million?
>      
>      
> 
> 
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From jvospost3 at gmail.com  Sun Nov  5 02:05:28 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 4 Nov 2006 17:05:28 -0800
Subject: A124015 thoughts
In-Reply-To: <454D1189.8090102@indiana.edu>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p>
	 <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
	 <454D1189.8090102@indiana.edu>
Message-ID: <5542af940611041705u66643b27u66c836a95a2b9a1@mail.gmail.com>

Nice code.  And what fascinating graphs! They have both periodicities and a
fractal appearance.  Somehwere in there is the equivalent of Benford's Law,
and Zipf's Law, on lengths names of numbers (as opposed to digit
distribution and power law).

On 11/4/06, Joseph Biberstine <jrbibers at indiana.edu> wrote:
>
>         For anyone interested, this Mathematica function, eng[], will (as
> usual, barring bugs) give the American English name for integers on 0 to
> 10^21-1.  Extension of the upper bound should be plain to implement
> (just one line of code -- observe the trend).
>
>         All printing is lowercase and without hyphens.  In holding that
> "and"
> should only ever be read for radix points, it is never printed here.
> Note this implementation uses a text postprocessor and so is inherently
> ugly.  Also note that the helper memoizes for better or worse.  Please,
> of course, report all bugs.
>
>         Attached are some diagrams of the predictable distribution of
> StringLength[eng[n]] (note this includes spaces).  The third ranges up
> to 10^9, sampling every 10^6.
>
>         Here is sample output from random naturals on 0,999999:
>
> 40573, forty thousand five hundred seventy three
> 108097, one hundred eight thousand ninety seven
> 539799, five hundred thirty nine thousand seven hundred ninety nine
> 136868, one hundred thirty six thousand eight hundred sixty eight
> 597841, five hundred ninety seven thousand eight hundred forty one
> 272460, two hundred seventy two thousand four hundred sixty
> 135550, one hundred thirty five thousand five hundred fifty
> 402424, four hundred two thousand four hundred twenty four
> 543617, five hundred forty three thousand six hundred seventeen
> 868945, eight hundred sixty eight thousand nine hundred forty five
> 951779, nine hundred fifty one thousand seven hundred seventy nine
> 742845, seven hundred forty two thousand eight hundred forty five
> 84140, eighty four thousand one hundred forty
> 680145, six hundred eighty thousand one hundred forty five
> 82734, eighty two thousand seven hundred thirty four
> 154758, one hundred fifty four thousand seven hundred fifty eight
> 961084, nine hundred sixty one thousand eighty four
> 360944, three hundred sixty thousand nine hundred forty four
> 574416, five hundred seventy four thousand four hundred sixteen
> 164099, one hundred sixty four thousand ninety nine
>
>         Here is the code:
>
> Clear[eng, engRaw, n];
> eng[n_] := (
>   pre = engRaw[n];
>   If[pre == "", pre = "zero"]; (* fixes "zero" *)
>   pre = FixedPoint[StringReplace[#, "  " -> " "] &, pre] ; (* no
> consecutive spaces *)
>   pre = FixedPoint[StringReplace[#, RegularExpression[" $"] :> ""] &,
> pre] ; (* no trailing spaces *)
>   pre);
> (* This raw function relies heavily on the postprocessor, which fixes
> "zero" and deals with whitespace *)
> engRaw[n_] := engRaw[n] = (* memo *)
>   Which[
>     n == 0, "",(* fixed in postprocessing *)
>     n == 1, "one",
>     n == 2, "two",
>     n == 3, "three",
>     n == 4, "four",
>     n == 5, "five",
>     n == 6, "six",
>     n == 7, "seven",
>     n == 8, "eight",
>     n == 9, "nine",
>     n == 10, "ten",
>     n == 11, "eleven",
>     n == 12, "twelve",
>     n == 13, "thirteen",
>     n == 14, "fourteen",
>     n == 15, "fifteen",
>     n == 16, "sixteen",
>     n == 17, "seventeen",
>     n == 18, "eighteen",
>     n == 19, "nineteen",
>     20 <= n <= 29, "twenty " <> engRaw[Mod[n, 10]],
>     30 <= n <= 39, "thirty " <> engRaw[Mod[n, 10]],
>     40 <= n <= 49, "forty " <> engRaw[Mod[n, 10]],
>     50 <= n <= 59, "fifty " <> engRaw[Mod[n, 10]],
>     60 <= n <= 69, "sixty " <> engRaw[Mod[n, 10]],
>     70 <= n <= 79, "seventy " <> engRaw[Mod[n, 10]],
>     80 <= n <= 89, "eighty " <> engRaw[Mod[n, 10]],
>     90 <= n <= 99, "ninety " <> engRaw[Mod[n, 10]],
>     10^2 <= n <= 10^3 - 1, engRaw[Quotient[n, 10^2]] <> " hundred " <>
> engRaw[Mod[n, 10^2]],
>     10^3 <= n <= 10^6 - 1, engRaw[Quotient[n, 10^3]] <> " thousand " <>
> engRaw[Mod[n, 10^3]],
>     10^6 <= n <= 10^9 - 1, engRaw[Quotient[n, 10^6]] <> " million " <>
> engRaw[Mod[n, 10^6]],
>     10^9 <= n <= 10^12 - 1, engRaw[Quotient[n, 10^9]] <> " billion " <>
> engRaw[Mod[n, 10^9]],
>     10^12 <= n <= 10^15 - 1, engRaw[Quotient[n, 10^12]] <> " trillion "
> <> engRaw[Mod[n, 10^12]],
>     10^15 <= n <= 10^18 - 1, engRaw[Quotient[n, 10^15]] <> " quadrillion
> " <> engRaw[Mod[n, 10^15]],
>     10^18 <= n <= 10^21 - 1, engRaw[Quotient[n, 10^18]] <> " quintillion
> " <> engRaw[Mod[n, 10^18]],
>     True, Return["Naturals beyond 10^21-1 are not supported"];];
>
> -JRB
>
> Jonathan Post wrote:
> > Now that I think of it, logarithmic trends can be interesting.  How
> > about "what is the distribution of the number of letters in the names of
> > numbers from one to one billion?"
> >
> > -- Jonathan Vos Post
> >
> > On 11/3/06, *David Wilson* <davidwwilson at comcast.net
> > <mailto:davidwwilson at comcast.net>> wrote:
> >
> >     The description of A124015 is
> >
> >     %N A124015 Number of words with n letters in the National Scrabble
> >     Association Dictionary.
> >
> >     Some notes:
> >
> >     - It should be noted that the NSAD includes a restricted set of
> >     English words (words of 2 to 14 letters, no proper nouns or
> >     derivatives, no words with non-alphabetic characters (e.g,
> >     contractions, hyphenated words), and it is this restricted set that
> >     is being counted.
> >
> >     - The edition of the NSAD used to create A124015 should be
> >     specified, since the NSAD is continually edited.  While I do not
> >     think that A124015 should change with each new edition of the NSAD,
> >     because it may be referenced in other literature. On the other hand,
> >     I don't think that a new sequence should be created for each new
> >     edition of the NSAD unless the new edition exhibits some interesting
> >     statistical departure from the current NSAD (e.g, in some future
> >     NSAD, the number of 9-letter words exceeds the number of 8-letter
> >     words). This is because I believe that the value of A124015 is
> >     mainly to indicate a distribution of English word lengths.
> >
> >     - On the other hand, I have always understood that 4-letter words
> >     are most common in literature (where small words appear more often).
> >     It might be interesting to create a sequence counting words of
> >     length n in the King James Bible, War and Peace, or some other
> >     suitably large and stable piece of English literature (we don't have
> >     to go overboard on this, just one or two examples indicating word
> >     length distributions in literature).
> >
> >     - Another interesting idea: In various languages, what is the
> >     distribution of the number of letters in the names of numbers from
> >     one to (say) one million?
> >
> >
> >
> >
>
>
>
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From pxp at rogers.com  Sun Nov  5 02:29:38 2006
From: pxp at rogers.com (Hans Havermann)
Date: Sat, 4 Nov 2006 20:29:38 -0500
Subject: A124015 thoughts
In-Reply-To: <454D1189.8090102@indiana.edu>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com> <454D1189.8090102@indiana.edu>
Message-ID: <3E87B5F0-736C-466D-B4EC-C2D438AA1BA3@rogers.com>

Joseph Biberstine:

> 	For anyone interested, this Mathematica function, eng[], will (as  
> usual, barring bugs) give the American English name for integers on  
> 0 to 10^21-1.  Extension of the upper bound should be plain to  
> implement (just one line of code -- observe the trend).

I forget where I found the following bit of elegant code, but it  
essentially does the same thing as yours (except use a hyphen instead  
of a space between the tens and units for tens > 20):


Ones[n_, pad_:0] := If[Mod[pad, 10] > 0, "-", ""] <> {"", "one",  
"two", "three", "four", "five", "six", "seven", "eight", "nine",  
"ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen",  
"sixteen", "seventeen", "eighteen", "nineteen"}[[n + 1]]

Tens[n_] := {"", "twenty", "thirty", "forty", "fifty", "sixty",  
"seventy", "eighty", "ninety"}[[n]]

LastThree[n_] := LastThree[n] = Module[{h = Quotient[n, 100], m = Mod 
[n, 100]}, If[n > 99, Ones[h, 0] <> " hundred ", ""] <> If[m < 20,  
Ones[m], Tens[Quotient[m, 10]] <> Ones[Mod[m, 10], m]]]

Powers[n_] := Powers[n] = {"", " thousand", " million", " billion", "  
trillion", " quadrillion", " quintillion", " sextillion", "  
septillion", " octillion", "nonillion", " decillion", " undecillion",  
" duodecillion", " tredecillion", " quattuordecillion", "  
quindecillion", " sexdecillion", " septendecillion", "  
octodecillion", " novemdecillion", " vigintillion"}[[n]]

(*NumberName[0] := "zero"*)

NumberName[n_Integer] := Module[{db = Reverse[MapIndexed[If[#1 == 0,  
"", LastThree[#1] <> Powers[#2[[1]]] <> " "] &, FromDigits[#, 10] & / 
@ Reverse /@ Partition[Reverse[IntegerDigits[n, 10]], 3, 3, {1, 1},  
{}]]]}, StringDrop[StringJoin @@ db, -1]]

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From zbi74583 at boat.zero.ad.jp  Sun Nov  5 04:58:03 2006
From: zbi74583 at boat.zero.ad.jp (koh)
Date: Sun, 05 Nov 2006 12:58:03 +0900
Subject: terms lack
Message-ID: <20061105035802.zbi74583@boat.zero.ad.jp>

    Recently I updated more terms of A054572, which are calculated by Richard Mathar.
    And Neil claimed or got angry.
    He said, "You are red card. Don't submit a wrong sequences"
    I am exaggerating a little.
    These words are not exactly what he wrote.



    To Neil
    They are correct, not "wrong". 
    It is only not complete.

    In early time, I didn't do a computer search, so there is a possibility that my old submitted sequences are not complete.
    Now, I calculate both theoretically and with a computer.



    To Seqfans
    I defined many "divisor" functions and calculated many Amicable Numbers and Perfect Numbers using those functions.
    And I submitted them as sequences to  OEIS.

        http://www.research.att.com/~njas/sequences/?q=kohmoto+amicable&sort=0&fmt=0&language=english&go=Search        
        
        http://www.research.att.com/~njas/sequences/?q=kohmoto+perfect&sort=0&fmt=0&language=english&go=Search
   
    I suppose that they are almost complete but some of them might not be complete.

    I wish some one do a computer search about them.
    I think that A054572 is the worst case, because I gave only two terms up to 10^4.
    But sometimes my theory gives a complete set of examples.
    See A099105  , I was sure that these three terms are all, and a computer search also said it is right.

    Yasutoshi
    





From deutsch at duke.poly.edu  Sun Nov  5 05:30:30 2006
From: deutsch at duke.poly.edu (Emeric Deutsch)
Date: Sat, 4 Nov 2006 23:30:30 -0500 (EST)
Subject: No subject
Message-ID: <Pine.GSO.4.62.0611041740160.21855@duke.poly.edu>

Dear Seqfans,
We (Lou Shapiro and myself) would like to submit a few
sequences similar to A124497 (shown at the bottom of this
message). They will refer to 1-2-3-4 trees, 1-2-3-4-5
trees, etc. rather than to 1-2-3 trees. We can describe
the corresponding g.f.'s but we need help in finding the
terms themselves.

Here is the description of all the g.f.'s. Let M[k](z) be
the g.f. of the 1-2-...-k trees with thinning limbs. I will
describe these functions; consequently, we may disregard
the objects counted by them.

Denote by C[k](z) the g.f. of the k-ary trees. C[k]
satisfies C[k]=1+zC[k]^k and the coefficient of z^n
in its series expansion is
(*)      binom(kn,n)/[(k-1)n+1].

I will write Mk for M[k] and Ck for C[k].

Obviously, M1(z)=1/(1-z) (path-trees). The next one is 
M2(z)=M1(z)C2(z^2*M1), i.e. M1 times C2 evaluated at
z^2*M1. This gives no problem since
C2(z)=[1-sqrt(1-4z)]/(2z). We obtain A090344.

Next M3(z)=M2(z)C3(z^3*M2^2); since there is an
explicit expression for C3(z), namely
C3(z)=2/sqrt(3z)*sin[(1/3)*arcsin(sqrt(27z/4))],
this yields easily A124497.

 From here on we need help.
M4(z)=M3(z)C4(z^4*M3^3); 
Most probably sequence starts with
1,1,2,4,10,24,62,160,425,1140,3105 (I used some terms
of the series of C4(z); see (*)).

M5(z)=M4(z)C5(z^5*M4^4);
Most probably sequence starts with
1,1,2,4,10,25,67,180,495,1375,3871 (I used some terms
of the series of C5(z); see (*)).

And so on.

These sequences approach A124344:
1,1,2,4,10,25,68,187,530,1523,4447,13121,39107,...

Many thanks.
Emeric

--------------------------------
A124497  Number of 1-2-3 trees with n edges and with thinning limbs. A 
1-2-3 tree is an ordered tree with vertices of outdegree at most 3. A 
rooted tree with thinning limbs is such that if a node has k children, all 
its children have at most k children.

1, 1, 2, 4, 9, 20, 48, 116, 288, 724, 1849, 4768, 12423, 32628, 86342, 
229952, 616042, 1659012, 4489101, 12199521, 33284546, 91140797, 250396629, 
690043032, 1907022197, 5284167884, 14677681554, 40862469713, 114001697975 
(list; graph; listen)
OFFSET  0,3
FORMULA  G.f.=H*T(H^2*z^3), where 
T=2/sqrt(3*x)*sin((1/3)*arcsin(sqrt(27*x/4))) (solution of T=1+zT^3, 
T(0)=1), H=C(z^2/(1-z))/(1-z), and C(x)=[1-sqrt(1-4x)]/(2x) is the Catalan 
function. More generally, if M[k](z) is the g.f. of the 1-2-...-k trees 
with thinning limbs and C[k](z)=1+z*{C[k](z)}^k is the g.f. of the k-ary 
trees, then M[k](z)=M[k-1](z)C[k](M[k-1]^(k-1)*z^k).
MAPLE  C:=x->(1-sqrt(1-4*x))/2/x: 
T:=x->2/sqrt(3*x)*sin((1/3)*arcsin(sqrt(27*x/4))): M2:=C(z^2/(1-z))/(1-z): 
G:=M2*T(M2^2*z^3): Gser:=series(G, z=0, 40): seq(coeff(Gser, z, n), 
n=0..33);
CROSSREFS  Cf. A090344, A124344.
KEYWORD  nonn,new
AUTHOR  Emeric Deutsch and Louis Shapiro (deutsch(AT)duke.poly.edu, 
lshapiro(AT)Howard.edu), Nov 04 2006





From zakseidov at yahoo.com  Sun Nov  5 07:28:49 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sat, 4 Nov 2006 22:28:49 -0800 (PST)
Subject: Observation on A053669
In-Reply-To: <200611042045.kA4KjHeV011396@amer.strw.leidenuniv.nl>
Message-ID: <20061105062849.51615.qmail@web38208.mail.mud.yahoo.com>

Yes, 
and because a(n_odd)=2
keep only even n's.
Zak

--- Richard Mathar <mathar at strw.leidenuniv.nl> wrote:

> 
> dww> From seqfan-owner at ext.jussieu.fr  Sat Nov  4
> 21:12:32 2006
> dww> Return-Path: <seqfan-owner at ext.jussieu.fr>
> dww> From: "David Wilson" <davidwwilson at comcast.net>
> dww> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
> dww> Subject: Observation on A053669
> dww> ...
> dww> It seems to me that
> dww> 
> dww> %N A053669 Smallest prime co-prime to n.
> dww> 
> dww> could just as easily be described
> dww> 
> dww> %N A053669 Smallest number > 1 coprime to n.
> dww> 
> dww> and the fact that a(n) is prime falls out. 
> This definition would also =
> dww> simplify the program.
> dww>...
> 
> Yes. If the smallest prime co-prime to n would not
> be a prime,
> it would be built by at least two prime factors, the
> smaller of which
> would be the correct entry (proof by reductio ad
> absurdum).
> 
> It would still be useful to keep in a comment that
> the numbers are all prime
> (this might speed up some searches for the
> numbers...)
> 
> I'd propose to list in a comment the first
> occurrences of the first primes:
> first 5 at a(6)
> first 7 is a(30)
> first 11 is a(210)
> first 13 is a(2310)
> first 17 is a(30030)
> first 19 is a(510510)
> where the indices are 6=2*3, 30=2*3*5, 210=2*3*5*7,
> 2*3*5*7*11 etc,
> (the primorials, I guess, as in A002110)
> 
> -- RJM
> 



 
__________________________________________________________________________________________
Check out the New Yahoo! Mail - Fire up a more powerful email and get things done faster. 
(http://advision.webevents.yahoo.com/mailbeta) 






From davidwwilson at comcast.net  Sun Nov  5 07:38:34 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sun, 5 Nov 2006 01:38:34 -0500
Subject: Notes on A124064 (was Re: primes in arithmetic progression)
References: <200611041910.OAA62978@fry.research.att.com>
Message-ID: <003101c700a5$05184930$6401a8c0@yourxhtr8hvc4p>

This is one of those cases where I think we have jumped the gun, creating a 
sequence without first determining the best structure for that sequence.

Richard Mather proposed a table of T(d,k) = the smallest possible initial 
element of a nondecreasing AP of k primes with difference d. Clearly, T(d,k) 
is defined only for d >= 0 and k >= 1.

For d = 0, it is pretty easy to show T(0, k) = 2 for any k >= 1. For d >= 1, 
however, T(d,k) is defined only for a finite number of k.

Proof: Suppose there exists AP P of primes with difference d >= 1 and 2d+2 
elements. Let c = d+1. Because P is an arithmetic progression of 2c elements 
with difference d, and c is coprime to d, P will have two elements divisible 
by c, call them cx and cy. Whereas P is an increasing sequence of positive 
integers, we can assume that 0 < cx < cy. Hence 0 < x < y, and y >= 2. But 
this makes cy composite, contrary to fact that P is a sequence of primes. 
Hence P does not exist, that is to say, any AP of primes with difference d 
 >= 1 has at most 2d+1 elements, and T(d,k) is undefined for k > 2d+1. QED.

In fact, assuming the k-tuple conjecture is true, we can give the precise 
domain of T for d >= 1: If p is the smallest prime not dividing d, then 
T(d,k) is defined precisely for 1 <= k <= p if the AP of p elements with 
initial element p and difference d is a prime AP, and for 1 <= k <= p-1 
otherwise.

This means that the table of T(d,k) looks as follows:

  \k|  1  2  3  4  5  6  7  8  9 10
  d\|
----+------------------------------
  0 |  2  2  2  2  2  2  2  2  2  2 ...
  1 |  2  2
  2 |  2  3  3
  3 |  2  2
  4 |  2  3  3
  5 |  2  2
  6 |  2  5  5  5  5
  7 |  2
  8 |  2  3  3
  9 |  2  2
 10 |  2  3  3
 11 |  2  2
 12 |  2  5  5  5  5
 13 |  2
 14 |  2  3  3
 15 |  2  2
 16 |  2  3
 17 |  2  2
 18 |  2  5  5  5
 19 |  2
 20 |  2  3  3
 ...

Two important points:

- Sometimes in the OEIS, we are tempted to pare down sequences to remove 
uninteresting values in order to emphasize the interesting values. In such 
cases, I think that the OEIS should include a main sequence that is 
complete, and additional pared-down sequences if they seem to augment our 
understanding of the main sequence. In this case, Mather's sequence included 
T(d,k) only for even d, presumably because T(d,k) is uninteresting for odd 
d. My table above includes T(d,k) for both even and odd d. In fact, it turns 
out that for odd d, T(d,1) = 2 and T(d,2) = 2 if 2+d is prime, not 
scintillating behavior, but not trivial either. My feeling is that the main 
sequence for T(d,k) (presumably A124064) should include all the values of 
T(d,k), and if restricting T(d,k) to even d seems to augment our 
understanding, a second pared-down sequence should be made referring back to 
the complete main sequence.

- For d >= 1 and sufficient k, T(d,k) is undefined. The standard 
diagonalization technique won't work for this irregular table. That is why 
NJAS's proposed A124064 ends where it does, because the next value, T(2,4), 
does not exist (it is not a matter of doing more computation, T(2,4) does 
not exist because there is no AP of 4 primes with difference 2). I can see 
two possible ways to address this problem, the (icky) option of making 
T(d,k) = 0 if undefined, or the (less icky) option of requiring the APs in 
question to be strictly increasing. This would dispense with the infinite d 
= 0 line and leave us with a list of finite sequences which we could 
concatenate to build A124064, vis:

2,2,2,3,3,2,2,2,3,3,2,2,2,5,5,5,5,2,2,3,3,2,2,2,3,3,2,2,2,5,5,5,5,2,2,
3,3,2,2,2,3,2,2,2,5,5,5,2,2,3,3,2,2,2,7,2,2,5,5,59,2,2,3,2,2,2,3,3,2,2,
2,7,7,7,7,7,2,2,5,2,2,3,3,2,2,2,5,7,31,2,2,3,3,2,2,2,3,3,2,2,2,5,5,5,5

We could put the line lengths in another sequence.

2,3,2,3,2,5,1,3,2,3,2,5,1,3,2,2,2,4,1,3,2,2,1,4,1,2,2,3,2,6,1,2,1,3,2,4,
1,3,2,3,2,5,1,2,2,2,1,5,1,3,2,2,1,4,1,2,2,2,2,6,1,2,1,3,2,4,1,3,2,2,2,4,
1,2,1,2,2,4,1,3,2,2,1,4,1,2,2,2,1,6,1,2,1,3,2,5,1,3,2,2,2,4,1,3,2,2,2,4

----- Original Message ----- 
From: "N. J. A. Sloane" <njas at research.att.com>
To: <seqfan at ext.jussieu.fr>; "Richard Mathar" <mathar at strw.leidenuniv.nl>
Sent: Saturday, November 04, 2006 2:10 PM
Subject: Re: primes in arithmetic progression


> Just to get the ball rolling, I am adding this entry:
>
> %I A124064
> %S A124064 2,2,2,2,3,2,2,3,3,2,2,5,3
> %N A124064 Array read by antidiagonals: T(d,k) (k >= 1, d = 2,4,6,8,...) = 
> smallest prime p of k (not necessarily consecutive) primes in arithmetic 
> progression with common difference d.
> %C A124064 The row d=0 and the column k=1 are degenerate and are filled 
> with the prime 2.
> %O A124064 1,1
> %K A124064 nonn,tabl,more
> %e A124064 Array begins:
> %e A124064 d.\...k=1.k=2..k=3..k=4..k=5..k=6
> %e A124064 0..|..2...2....2....2....2....2
> %e A124064 2..|..2...3....3
> %e A124064 4..|..2...3....3
> %e A124064 6..|..2...5....5....5....5
> %e A124064 8..|..2...3....3
> %e A124064 10.|..2...3....3
> %e A124064 12.|..2...5....5....5....5
> %e A124064 14.|..2...3....3.
> %e A124064 16.|..2...3
> %e A124064 18.|..2...5....5....5
> %e A124064 20.|..2...3
> %e A124064 22.|..2...7
> %e A124064 24.|..2...5....5...59
> %e A124064 26.|..2...3
> %e A124064 28.|..2...3....3
> %e A124064 30.|..2...7....7....7.....7....7
> %e A124064 32.|..2...5
> %e A124064 34.|..2...3....3
> %e A124064 36.|..2...5....7...31
> %e A124064 Example for row d=24 and column k=4: the 4 numbers 
> 59,59+24,59+2*24 and 59+3*24 are all primes.
> %A A124064 Richard Mathar (mathar(AT)strw.leidenuniv.nl), Nov 04 2006
>
>
>
>
> NJAS
>
>
> -- 
> No virus found in this incoming message.
> Checked by AVG Free Edition.
> Version: 7.5.430 / Virus Database: 268.13.27/517 - Release Date: 11/3/2006 
> 4:30 PM
> 






From davidwwilson at comcast.net  Sun Nov  5 07:40:23 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sun, 5 Nov 2006 01:40:23 -0500
Subject: Observation on A053669
References: <200611042045.kA4KjHeV011396@amer.strw.leidenuniv.nl>
Message-ID: <003b01c700a5$45a9d3b0$6401a8c0@yourxhtr8hvc4p>

Another possible description of A053669:

Smallest prime non-divisor of n.

----- Original Message ----- 
From: "Richard Mathar" <mathar at strw.leidenuniv.nl>
To: <seqfan at ext.jussieu.fr>
Sent: Saturday, November 04, 2006 3:45 PM
Subject: Re: Observation on A053669


>
> dww> From seqfan-owner at ext.jussieu.fr  Sat Nov  4 21:12:32 2006
> dww> Return-Path: <seqfan-owner at ext.jussieu.fr>
> dww> From: "David Wilson" <davidwwilson at comcast.net>
> dww> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
> dww> Subject: Observation on A053669
> dww> ...
> dww> It seems to me that
> dww>
> dww> %N A053669 Smallest prime co-prime to n.
> dww>
> dww> could just as easily be described
> dww>
> dww> %N A053669 Smallest number > 1 coprime to n.
> dww>
> dww> and the fact that a(n) is prime falls out.  This definition would 
> also =
> dww> simplify the program.
> dww>...
>
> Yes. If the smallest prime co-prime to n would not be a prime,
> it would be built by at least two prime factors, the smaller of which
> would be the correct entry (proof by reductio ad absurdum).
>
> It would still be useful to keep in a comment that the numbers are all 
> prime
> (this might speed up some searches for the numbers...)
>
> I'd propose to list in a comment the first occurrences of the first 
> primes:
> first 5 at a(6)
> first 7 is a(30)
> first 11 is a(210)
> first 13 is a(2310)
> first 17 is a(30030)
> first 19 is a(510510)
> where the indices are 6=2*3, 30=2*3*5, 210=2*3*5*7, 2*3*5*7*11 etc,
> (the primorials, I guess, as in A002110)
>
> -- RJM
>
>
> -- 
> No virus found in this incoming message.
> Checked by AVG Free Edition.
> Version: 7.5.430 / Virus Database: 268.13.27/517 - Release Date: 11/3/2006 
> 4:30 PM
>
> 






From pauldhanna at juno.com  Sun Nov  5 07:49:40 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Sun, 5 Nov 2006 01:49:40 -0500
Subject: A124497: 1-2-3-...-k Trees
Message-ID: <20061105.014940.1648.0.pauldhanna@juno.com>

Emeric (and Seqfans),
     What a fascinating problem.
Although I will not be using this formula now, 
let's extend your formula for k-ary trees: 
(*)  [z^n] C[k](z)^m  =  binomial(kn+m-1,n)*m/[(k-1)n+m] 
to give the coefficient of z^n in the m-th power of C[k](z). 
Note that the logarithm is obtained similarly: 
(**) [z^n] log( C[k](z) ) =  binomial(kn-1,n)/[(k-1)n], for n>0.
No doubt you are already aware of this lovely facts; 
and they may simplify some series representations involved here. 
 
I will not attempt to give a simpler g.f. for your M[n], where:
   M[n+1](z) = M[n](z)*C[n](z^n*M[n-1]^(n-1)); 
 
but using PARI I will list the first 31 terms for n=2..10 (below). 
 
You have some impressive results so far. 
     Paul 
 
(PARI)
{M=1/(1-x+O(x^31));
 for(n=2,10,
   M=M*sum(k=0,30,binomial(n*k,k)/((n-1)*k+1)*(x^n*M^(n-1))^k);
   print("n="n":");print(Vec(M))
 )
}

n=2: 
[1, 1, 2, 3, 6, 11, 23, 47, 102, 221, 493, 1105, 2516, 5763, 13328, 
30995, 72556, 170655, 403351, 957135, 2279948, 5449013, 13063596, 
31406517, 75701508, 182902337, 442885683, 1074604289, 2612341856, 
6361782007, 15518343597]
n=3: 
[1, 1, 2, 4, 9, 20, 48, 116, 288, 724, 1849, 4768, 12423, 32628, 
86342, 229952, 616042, 1659012, 4489101, 12199521, 33284546, 
91140797, 250396629, 690043032, 1907022197, 5284167884, 14677681554, 
40862469713, 114001697975, 318681491431, 892495078849]
n=4: 
[1, 1, 2, 4, 10, 24, 62, 160, 425, 1140, 3105, 8528, 23643, 66008, 
185526, 524384, 1489810, 4251852, 12184745, 35048405, 101156752, 
292865417, 850314803, 2475327088, 7223400899, 21126670372, 
61920289652, 181838859665, 534980035009, 1576653291583, 
4654122552827]
n=5: 
[1, 1, 2, 4, 10, 25, 67, 180, 495, 1375, 3871, 10993, 31493, 90843, 
263686, 769466, 2256135, 6643082, 19634705, 58232350, 173242381, 
516860717, 1546035258, 4635543843, 13929569399, 41943013047, 
126532961332, 382396277940, 1157548532794, 3509416800398, 
10655157352027]
n=6: 
[1, 1, 2, 4, 10, 25, 68, 186, 522, 1479, 4246, 12289, 35872, 105411, 
311662, 926270, 2765778, 8292296, 24953437, 75338686, 228140842, 
692733127, 2108652750, 6433255041, 19668210742, 60247367313, 
184879648441, 568281131800, 1749490180723, 5393738767856, 
16651705792875]
n=7: 
[1, 1, 2, 4, 10, 25, 68, 187, 529, 1514, 4393, 12856, 37944, 112740, 
337017, 1012671, 3056978, 9265765, 28187304, 86028764, 263339173, 
808256040, 2486804090, 7668374150, 23694936466, 73354869098, 
227489892754, 706643494680, 2198340096395, 6848600059320, 
21363953916171]
n=8: 
[1, 1, 2, 4, 10, 25, 68, 187, 530, 1522, 4437, 13056, 38766, 115908, 
348733, 1054775, 3205251, 9780165, 29951916, 92029644, 283606809, 
876333336, 2714448570, 8426790446, 26213887250, 81699432762, 
255072076566, 797641572824, 2498068134817, 7834443329344, 
24602515926899]
n=9: 
[1, 1, 2, 4, 10, 25, 68, 187, 530, 1523, 4446, 13110, 39030, 117060, 
353404, 1072811, 3272526, 10024884, 30825490, 95103360, 294300321, 
913203240, 2840650881, 8856206258, 27667828196, 86601978030, 
271545341505, 852829532176, 2682483667081, 8449321174753, 
26648690288909]
n=10: 
[1, 1, 2, 4, 10, 25, 68, 187, 530, 1523, 4447, 13120, 39095, 117400, 
354974, 1079493, 3299426, 10128934, 31216245, 96538200, 299477771, 
931628670, 2905496841, 9082358938, 28450647731, 89294760932, 
280759352295, 884216374006, 2788990466531, 8809540396813, 
27863492984436]

END. 
----------------------------------------------------------
On Sat, 4 Nov 2006 23:30:30 -0500 (EST) Emeric Deutsch
<deutsch at duke.poly.edu> writes:
> Dear Seqfans,
> We (Lou Shapiro and myself) would like to submit a few
> sequences similar to A124497 (shown at the bottom of this
> message). They will refer to 1-2-3-4 trees, 1-2-3-4-5
> trees, etc. rather than to 1-2-3 trees. We can describe
> the corresponding g.f.'s but we need help in finding the
> terms themselves.
> 
> Here is the description of all the g.f.'s. Let M[k](z) be
> the g.f. of the 1-2-...-k trees with thinning limbs. I will
> describe these functions; consequently, we may disregard
> the objects counted by them.
> 
> Denote by C[k](z) the g.f. of the k-ary trees. C[k]
> satisfies C[k]=1+zC[k]^k and the coefficient of z^n
> in its series expansion is
> (*)      binom(kn,n)/[(k-1)n+1].
> 
> I will write Mk for M[k] and Ck for C[k].
> 
> Obviously, M1(z)=1/(1-z) (path-trees). The next one is 
> M2(z)=M1(z)C2(z^2*M1), i.e. M1 times C2 evaluated at
> z^2*M1. This gives no problem since
> C2(z)=[1-sqrt(1-4z)]/(2z). We obtain A090344.
> 
> Next M3(z)=M2(z)C3(z^3*M2^2); since there is an
> explicit expression for C3(z), namely
> C3(z)=2/sqrt(3z)*sin[(1/3)*arcsin(sqrt(27z/4))],
> this yields easily A124497.
> 
> From here on we need help.
> M4(z)=M3(z)C4(z^4*M3^3); 
> Most probably sequence starts with
> 1,1,2,4,10,24,62,160,425,1140,3105 (I used some terms
> of the series of C4(z); see (*)).
> 
> M5(z)=M4(z)C5(z^5*M4^4);
> Most probably sequence starts with
> 1,1,2,4,10,25,67,180,495,1375,3871 (I used some terms
> of the series of C5(z); see (*)).
> 
> And so on.
> 
> These sequences approach A124344:
> 1,1,2,4,10,25,68,187,530,1523,4447,13121,39107,...
> 
> Many thanks.
> Emeric





From pauldhanna at juno.com  Sun Nov  5 08:04:15 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Sun, 5 Nov 2006 02:04:15 -0500
Subject: A124497: 1-2-3-...-k Trees
Message-ID: <20061105.020415.1648.1.pauldhanna@juno.com>

Emeric (and Seqfans),
     Correction to my typo in prior e-mail:
> I will not attempt to give a simpler g.f. for your M[n], where:
>    M[n+1](z) = M[n](z)*C[n](z^n*M[n-1]^(n-1)); 
 
that should have been:
 
M[1](z) = 1/(1-z);
 
M[n+1](z) = M[n](z)*C[n+1](z^(n+1)*M[n]^n); 
 
This was more clearly given by Emeric, but I correct myself for the sake
of accuracy. 
    Paul





From zakseidov at yahoo.com  Sun Nov  5 09:49:35 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 00:49:35 -0800 (PST)
Subject: A124057 is subset of A045940
Message-ID: <20061105084935.7574.qmail@web38212.mail.mud.yahoo.com>

A124057 is subset of A045940
from which more terms for A124057 
can be retrieved:

602,603,1083,2012,2091,2522,2523,2524,2634,2763,3243,3355,4202,4203,4921,4922,4923,5034,5035,5132,5203,5282,5283,5785,5882,5954,5972,6092,6212,6476,6962,6985,7730,7731,7945,8393,8825,8956,8972


%N A045940 Numbers n such that factorizations of n
through n+3 have same number of primes (including 
               multiplicities).

%N A124057 Numbers n such that n, n+1, n+2 and n+3 are
products of exactly 3 primes.



 
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From jvospost3 at gmail.com  Sun Nov  5 17:53:45 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 5 Nov 2006 08:53:45 -0800
Subject: A124057 is subset of A045940
In-Reply-To: <20061105084935.7574.qmail@web38212.mail.mud.yahoo.com>
References: <20061105084935.7574.qmail@web38212.mail.mud.yahoo.com>
Message-ID: <5542af940611050853h50af8ae5ue5004d32f982bef0@mail.gmail.com>

Dear Zak.

You're right.  I knew that; it's in my handwritten notes.  I should have you
submit that as a comment, and also have A045940 added as a cf to A124057,
and also show your extension. Thank you!

Shalom,

Jonathan

On 11/5/06, zak seidov <zakseidov at yahoo.com> wrote:
>
> A124057 is subset of A045940
> from which more terms for A124057
> can be retrieved:
>
>
> 602,603,1083,2012,2091,2522,2523,2524,2634,2763,3243,3355,4202,4203,4921,4922,4923,5034,5035,5132,5203,5282,5283,5785,5882,5954,5972,6092,6212,6476,6962,6985,7730,7731,7945,8393,8825,8956,8972
>
>
> %N A045940 Numbers n such that factorizations of n
> through n+3 have same number of primes (including
>                multiplicities).
>
> %N A124057 Numbers n such that n, n+1, n+2 and n+3 are
> products of exactly 3 primes.
>
>
>
>
>
> ____________________________________________________________________________________
> Access over 1 million songs - Yahoo! Music Unlimited
> (http://music.yahoo.com/unlimited)
>
>
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From franktaw at netscape.net  Sun Nov  5 18:35:13 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Sun, 05 Nov 2006 12:35:13 -0500
Subject: A124015 thoughts
In-Reply-To: <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com>
Message-ID: <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com>

Some years ago, researchers at Brown University put together a 
collection of text from a large variety of source to serve as a 
cross-section of English usage, for study of this kind of question.  If 
you're going to add a sequence with statistics of this sort, that would 
be the place to start.

See, e.g., Wikipedia: http://en.wikipedia.org/wiki/Brown_Corpus.

Franklin T. Adams-Watters

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From bdm at cs.anu.edu.au  Sun Nov  5 23:40:24 2006
From: bdm at cs.anu.edu.au (Brendan McKay)
Date: Mon, 6 Nov 2006 09:40:24 +1100
Subject: A124015 thoughts
In-Reply-To: <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com> <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com>
Message-ID: <20061105224024.GC28253@cs.anu.edu.au>

* franktaw at netscape.net <franktaw at netscape.net> [061106 04:35]:
> Some years ago, researchers at Brown University put together a 
> collection of text from a large variety of source to serve as a 
> cross-section of English usage, for study of this kind of question.  If 
> you're going to add a sequence with statistics of this sort, that would 
> be the place to start.
> 
> See, e.g., Wikipedia: http://en.wikipedia.org/wiki/Brown_Corpus.
> 
> Franklin T. Adams-Watters

Yes, but I don't know if statistics based on sampling belong in OEIS.
They are not precise well-defined quantities.  Counts derived from
single texts are more concrete, but even then there are multiple
editions, even changes between printings of the same edition.

Brendan.





From zakseidov at yahoo.com  Sun Nov  5 23:53:40 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 14:53:40 -0800 (PST)
Subject: Don't restrict OEIS sequence domains!
In-Reply-To: <001501c70116$e2fabd20$6401a8c0@yourxhtr8hvc4p>
Message-ID: <20061105225340.19195.qmail@web38215.mail.mud.yahoo.com>

David, 

I simply remember Neil's recommendation
(somewhere in his tip's) to omit trivial cases
e.g. if each second term is zero.

And you may wish to 
put in %C A053699 
some of your observations, 
e.g., a(n_odd)=2.
Zak

--- David Wilson <davidwwilson at comcast.net> wrote:

> ----- Original Message ----- 
> From: "zak seidov" <zakseidov at yahoo.com>
> To: <seqfan at ext.jussieu.fr>
> Sent: Sunday, November 05, 2006 1:28 AM
> Subject: Re: Observation on A053669
> 
> 
> > Yes,
> > and because a(n_odd)=2
> > keep only even n's.
> > Zak
> 
> > [Earlier included message elided]
> ---------------------------
> 
> Let's take a look at Zak's recommendation to
> restrict the domain of A053699 
> to even n:
> 
> A053699(n) currently looks like:
> 
>
2,3,2,3,2,5,2,3,2,3,2,5,2,3,2,3,2,5,2,3,2,3,2,5,2,3,2,3,2,5,2,3,2,3,2,7,...
> 
> But a(k) = 2 for odd k, these values are
> uninteresting, so list only a(2n):
> 
>
3,3,5,3,3,5,3,3,5,3,3,5,3,3,7,3,3,5,3,3,5,3,3,5,3,3,5,3,3,7,3,3,5,3,3,5,...
> 
> But now a(k) = 3 unless 6 divides k, so we should
> list only a(6n):
> 
>
5,5,5,5,7,5,5,5,5,7,5,5,5,5,7,5,5,5,5,7,5,5,5,5,7,5,5,5,5,7,5,5,5,5,11,..
> 
> But again, a(k) = 5 except when 30 divides k, so
> let's eject these boring 
> values and keep only a(30n):
> 
>
7,7,7,7,7,7,11,7,7,7,7,7,7,11,7,7,7,7,7,7,11,7,7,7,7,7,7,11,7,7,7,7,7,...
> 
> But look at all the uninteresting 7's. Maybe we
> should list only a(210n)...
> 
> I think you get the picture. In the case of this
> sequence, we repeatedly 
> eliminate uninteresting domain values, but we never
> flush out the elusive 
> "interesting domain" of the sequence. Indeed, we can
> eliminate the entire 
> domain as uninteresting, whereas I think the
> original sequence is indeed 
> quite interesting.
> 
> Clearly, there are times when restricted domains are
> useful. For example, 
> since the areas of Pythagorean triangles are always
> divisible by 6, it would 
> seem reasonable, in the name of compactness, to
> prefer "Number of 
> Pythagorean triangles with area 6n" over "Number of
> Pythagorean triangles of 
> area n", since the latter would have a(n) = 0 for n
> != 6k. Nevertheless, I 
> would advocate having both sequences in the OEIS,
> because we never know if 
> the user will be searching for the compact or
> inflated version of the 
> sequence.
> 
> Similarly, if we adopt Zak's recommendation to
> restrict A065577 to a domain 
> of even values, and then someone looks up
> 
> 2,3,2,3,2,5,2,3,2,3,2,5
> 
> they will probably miss the most straightforward
> explanation of their 
> sequence.
> 
> In short: Don't restrict domains of OEIS sequences
> unless it makes clear 
> sense to do so, and if you do include a sequence on
> some restricted 
> "interesting" domain, have pity on the OEIS users
> and also include the same 
> sequence on its fullest possible domain.
> 
> ----- Original Message ----- 
> From: "zak seidov" <zakseidov at yahoo.com>
> To: <seqfan at ext.jussieu.fr>
> Sent: Sunday, November 05, 2006 1:28 AM
> Subject: Re: Observation on A053669
> 
> 
> > Yes,
> > and because a(n_odd)=2
> > keep only even n's.
> > Zak 
> 
> 



 
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From franktaw at netscape.net  Mon Nov  6 00:26:46 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Sun, 05 Nov 2006 18:26:46 -0500
Subject: A124015 thoughts
In-Reply-To: <20061105224024.GC28253@cs.anu.edu.au>
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com> <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com> <20061105224024.GC28253@cs.anu.edu.au>
Message-ID: <8C8CF48FECF25AB-958-F877@FWM-M05.sysops.aol.com>

I'm not saying one way or the other whether this sort of thing belongs
in the OEIS; just that, if it is going to be there, the Brown Corpus
would be the best source for it.

The Brown Corpus is about as concrete as you can get in this area.  It
is very well defined and a standard research reference.

Franklin T. Adams-Watters


-----Original Message-----
From: bdm at cs.anu.edu.au

  * franktaw at netscape.net <franktaw at netscape.net> [061106 04:35]:
> Some years ago, researchers at Brown University put together a
> collection of text from a large variety of source to serve as a
> cross-section of English usage, for study of this kind of question.  
If
> you're going to add a sequence with statistics of this sort, that 
would
> be the place to start.
>
> See, e.g., Wikipedia: http://en.wikipedia.org/wiki/Brown_Corpus.
>
> Franklin T. Adams-Watters

Yes, but I don't know if statistics based on sampling belong in OEIS.
They are not precise well-defined quantities.  Counts derived from
single texts are more concrete, but even then there are multiple
editions, even changes between printings of the same edition.

Brendan.


________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From zakseidov at yahoo.com  Mon Nov  6 00:51:18 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 15:51:18 -0800 (PST)
Subject: A124057 is subset of A045940
Message-ID: <20061105235118.90653.qmail@web38203.mail.mud.yahoo.com>

Just submitted:

My Q to seqfans:
Really, there are no numbers n such that n, n+1, n+2
and n+3 are products of 6 primes?
No for n<7258290.
Thanks, Zak


%I A124728
%S A124728
4023,7314,9162,12122,12123,16674,19434,19940,23874,24723,29094,33234,35124,35125,39234,42182,42183,44163,45175,46988,49147,51793,52854,52855,54584,54585,54663,58375,63594,64074,64075,64323,64491,64712,64713,65943,67507,67962,70262,71124,75123,75446,77283,80792,80793,81025,81963,81975,86823
%N A124728 Numbers n such that n, n+1, n+2 and n+3 are
products of 4 primes.
%C A124728 Subset of A045940  Numbers n such that
factorizations of n through n+3 have same number of
primes (including multiplicities).
Cf. A124057, A124729  Numbers n such that n, n+1, n+2
and n+3 are products of exactly 3,5 primes. 
There are no numbers n such that n, n+1, n+2 and n+3
are products of exactly 6 primes(?)
%e A124728 4023=3^3*149, 4024=2^3*503, 4025=5^2*7*23,
4026=2*3*11*61 (all products of 4 primes).
%Y A124728 A045940, A124057, A124729.
%O A124728 1
%K A124728 ,nonn,
%A A124728 Zak Seidov  (zakseidov at yahoo.com), Nov 05
2006


%I A124729
%S A124729
57967,491875,543303,584647,632148,632149,715374,824523,878875,914823,930123,931623,955448,964143,995874
%N A124729 Numbers n such that n, n+1, n+2 and n+3 are
products of 5 primes.
%C A124729 Subset of A045940  Numbers n such that
factorizations of n through n+3 have same number of
primes (including multiplicities).
Cf. A124057, A124728  Numbers n such that n, n+1, n+2
and n+3 are products of exactly 3,4 primes. 
There are no numbers n such that n, n+1, n+2 and n+3
are products of exactly 6 primes(?)
%e A124729 57967=7^3*13^2, 57968=2^4*3623,
57969=3^3*19*113, 57970=2*5*11*17*31 (all product of 5
primes (including multiplicities)};
632148 is the first number such that n through n+4 are
5-almost primes.
%Y A124729 A045940, A124057, A124728.
%O A124729 1
%K A124729 ,nonn,
%A A124729 Zak Seidov  (zakseidov at yahoo.com), Nov 05
2006



 
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From zakseidov at yahoo.com  Mon Nov  6 02:02:27 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 17:02:27 -0800 (PST)
Subject: n..n+6 are all N-almost primes
Message-ID: <20061106010227.58041.qmail@web38206.mail.mud.yahoo.com>

Just dubmitted.

My Q to gurus:
Are there longer set(s) of successive numbers
n..n+m (m>6)
which are products of the same number (N) of primes
(including multiplicities)?

It seems that for  larger N's, 
the larger m's may be(?).

Thanks, Zak
  
%I A000001
%S A000001
211673,298433,355923,381353,460801,506521,540292,568729,690593,705953,737633,741305,921529
%N A000001 Numbers n such that factorizations of
n..n+6 have same number of primes (including
multiplicities).
%C A000001 Subset of A045940  Numbers n such that
factorizations of n..n+3 have same number of primes
(including multiplicities).
%e A000001 211673=7*11*2749, 211674=2*3*35279,
211675=5^2*8467, 211676=2^2*52919, 211677=3*37*1907,
211677=2*109*971, 211679=13*19*857 all 3-almost
primes;
355923=3^2*71*557, 355924=2^2*101*881,
355925=5^22*23*619, 355926=2*3*137*433,
355927=11*13*19*131, 355928=2^3*44491,
355929=3*7*17*997 all 4-almost primes.
%Y A000001 A045940.
%O A000001 1
%K A000001 ,nonn,
%A A000001 Zak Seidov  (zakseidov at yahoo.com), Nov 05
2006



 
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From zakseidov at yahoo.com  Mon Nov  6 03:49:26 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 5 Nov 2006 18:49:26 -0800 (PST)
Subject: n..n+6 are all N-almost primes
In-Reply-To: <20061106010227.58041.qmail@web38206.mail.mud.yahoo.com>
Message-ID: <20061106024926.64870.qmail@web38203.mail.mud.yahoo.com>

--- zak seidov <zakseidov at yahoo.com> wrote:
> Are there longer set(s) of successive numbers
> n..n+m (m>6)
> which are products of the same number (N) of primes
> (including multiplicities)?

Me: Just submitted:
%I A000001
%S A000001
3405122,3405123,6612470,8360103,8520321,9306710
%N A000001 Numbers n such that n..n+7 are products of
the same number of primes.
%C A000001 Note that because 3405130=2*5*167*2039 is
also 4-almost prime,
3405122 is the first n such that numbers n..n+8 are
products of the same number N of primes (N=4).
%e A000001 3405122=2*7*29*8387, 3405123=3^2*19*19913,
3405124=2^2*127*6703, 3405125=5^3*27241,
3405126=2*3*59*9619, 3405127=11*23*43*313,
3405128=2^3*425641, 3405129=3*7*13*12473 all 4-almost
primes.
%O A000001 1
%K A000001 ,nonn,
%A A000001 Zak Seidov  (zakseidov at yahoo.com), Nov 05
2006

Now we need to find numbers n such that n..n+m (m>8)
are N-almost primes (n'd be >10^7).

Thanks, Zak

BTW The "annoying" ad at the bottom of message is 
from Yahoo!Mail - not me, sorry.

<original message>
> Just dubmitted.
> 
> My Q to gurus:
> Are there longer set(s) of successive numbers
> n..n+m (m>6)
> which are products of the same number (N) of primes
> (including multiplicities)?
> 
> It seems that for  larger N's, 
> the larger m's may be(?).
> 
> Thanks, Zak
>   
> %I A000001
> %S A000001
>
211673,298433,355923,381353,460801,506521,540292,568729,690593,705953,737633,741305,921529
> %N A000001 Numbers n such that factorizations of
> n..n+6 have same number of primes (including
> multiplicities).
> %C A000001 Subset of A045940  Numbers n such that
> factorizations of n..n+3 have same number of primes
> (including multiplicities).
> %e A000001 211673=7*11*2749, 211674=2*3*35279,
> 211675=5^2*8467, 211676=2^2*52919, 211677=3*37*1907,
> 211677=2*109*971, 211679=13*19*857 all 3-almost
> primes;
> 355923=3^2*71*557, 355924=2^2*101*881,
> 355925=5^22*23*619, 355926=2*3*137*433,
> 355927=11*13*19*131, 355928=2^3*44491,
> 355929=3*7*17*997 all 4-almost primes.
> %Y A000001 A045940.
> %O A000001 1
> %K A000001 ,nonn,
> %A A000001 Zak Seidov  (zakseidov at yahoo.com), Nov 05
> 2006
> 




 
____________________________________________________________________________________
Access over 1 million songs - Yahoo! Music Unlimited 
(http://music.yahoo.com/unlimited)






From davidwwilson at comcast.net  Mon Nov  6 04:10:17 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sun, 5 Nov 2006 22:10:17 -0500
Subject: A124015 thoughts
References: <000601c6ffa9$cfe95890$6401a8c0@yourxhtr8hvc4p> <5542af940611031849k32eb1687oc17e423c11cb03a9@mail.gmail.com> <8C8CF17E2474CE1-958-EFD9@FWM-M05.sysops.aol.com> <20061105224024.GC28253@cs.anu.edu.au> <8C8CF48FECF25AB-958-F877@FWM-M05.sysops.aol.com>
Message-ID: <001901c70151$16106360$6401a8c0@yourxhtr8hvc4p>

Sounds great. Where do I get the Brown Corpus in computer-readable form?

----- Original Message ----- 
From: <franktaw at netscape.net>
To: <seqfan at ext.jussieu.fr>
Sent: Sunday, November 05, 2006 6:26 PM
Subject: Re: A124015 thoughts


> I'm not saying one way or the other whether this sort of thing belongs
> in the OEIS; just that, if it is going to be there, the Brown Corpus
> would be the best source for it.
>
> The Brown Corpus is about as concrete as you can get in this area.  It
> is very well defined and a standard research reference.
>
> Franklin T. Adams-Watters
>
>
> -----Original Message-----
> From: bdm at cs.anu.edu.au
>
>  * franktaw at netscape.net <franktaw at netscape.net> [061106 04:35]:
>> Some years ago, researchers at Brown University put together a
>> collection of text from a large variety of source to serve as a
>> cross-section of English usage, for study of this kind of question.
> If
>> you're going to add a sequence with statistics of this sort, that
> would
>> be the place to start.
>>
>> See, e.g., Wikipedia: http://en.wikipedia.org/wiki/Brown_Corpus.
>>
>> Franklin T. Adams-Watters
>
> Yes, but I don't know if statistics based on sampling belong in OEIS.
> They are not precise well-defined quantities.  Counts derived from
> single texts are more concrete, but even then there are multiple
> editions, even changes between printings of the same edition.
>
> Brendan.
>
>
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading 
> spam and email virus protection.
>
>
>
> -- 
> No virus found in this incoming message.
> Checked by AVG Free Edition.
> Version: 7.5.430 / Virus Database: 268.13.28/518 - Release Date: 11/4/2006 
> 5:30 PM
>
> 






From dean at math.ucdavis.edu  Mon Nov  6 10:39:09 2006
From: dean at math.ucdavis.edu (Dean Hickerson)
Date: Mon, 6 Nov 2006 01:39:09 -0800
Subject: terms lack
Message-ID: <20061106093909.GA16454@math.ucdavis.edu>

Mostly to Yasutoshi Kohmoto <zbi74583 at boat.zero.ad.jp>:

> Recently I updated more terms of A054572, which are calculated by
> Richard Mathar.
> And Neil claimed or got angry.
> He said, "You are red card. Don't submit a wrong sequences"
> I am exaggerating a little.
> These words are not exactly what he wrote.
>
> To Neil
> They are correct, not "wrong".
> It is only not complete.

The terms that are listed in a sequence should be the start of the sequence.
If there are missing terms in between the listed ones, then the sequence
entry is not just incomplete, it's wrong.

If you've computed some terms of a sequence, but you're not sure if there
are smaller ones that you've missed, then you should only include the ones
that you're sure are at the start of the sequence.  You can add a comment
about the larger terms, stating that it's not known if there are smaller ones.
For example, the "EXTENSIONS" lines for the sequence A000043 (Mersenne
exponents) mention some recently discovered terms.  But there might be
some smaller ones, so these are not included in the sequence.

> I defined many "divisor" functions and calculated many Amicable Numbers and
> Perfect Numbers using those functions.
> And I submitted them as sequences to  OEIS.
>
> http://www.research.att.com/~njas/sequences/?q=kohmoto+amicable&sort=0&fmt=0&language=english&go=Search
>
> http://www.research.att.com/~njas/sequences/?q=kohmoto+perfect&sort=0&fmt=0&language=english&go=Search
>
> I suppose that they are almost complete but some of them might not be
> complete.

I've found missing terms for several of your sequences:


Sequences A045613 and A045614 ("Super Unitary Amicable Number") consist
of pairs of distinct numbers a and b such that

    usigma(usigma(a)) = usigma(usigma(b)) = a+b.

(The definitions don't make this clear, but A045613 contains the smaller
members of the pairs and A045614 contains the larger members.  Also they're
sorted by their smallest members, so 155 comes before 142 in A045614.)  The
first 3 pairs that you've listed are (105, 155), (110, 142) and (2145, 3055).
But you've missed several smaller ones:

    (33, 35), (208, 224), (268, 272), (455, 601), (695, 1033), (812, 956),
    (1609, 1847), (1808, 2512), (1892, 3004), (1913, 2407), (2096, 2224)


A051594 and A051595 are "(-1)sigma amicable numbers".  (Again, the definition
doesn't make it clear that A051595 has the smaller terms and A051594 has the
larger terms.)  The first pair that you've listed is (1969706592, 2236072608).
But there are many smaller ones, starting with:

    (429552, 466200), (497808, 604656), (800496, 1103760).


A036471-A036474 are defined as "Amicable quadruple: 4 different numbers
which satisfy sigma(a)=sigma(b)=sigma(c)=sigma(d)=a+b+c+d".  The first
quadruple that you've listed has smallest element
342151462276356306033089201934180, but here's a much smaller one:

    (3270960, 3361680, 3461040, 3834000)

(I think that's the smallest one, but I could be wrong.)  If someone can
determine the first several terms of this sequence, then it should be
edited; otherwise it should be deleted.


In some of your sequences, I wasn't able to find any smaller terms, but the
ones that you've listed are so large that I'm not at all confident that they
are the smallest ones.  For example, A038362 and A038363 are defined as
"A Rational Amicable Number: two different numbers a, b which satisfies the
following equation: sigma(a)=sigma(b)=(a+b)^3/(a^2+b^2)", and the smallest
pair that you've included is:

    (26403469440047700, 30193441130006700)

Do you have any reason to think that that's the smallest such pair?  If not,
and if you can't determine what the first few terms really are, then the
sequence should be deleted.


I've only looked at a few of your sequences, but that's enough to convince
me that they should all be checked.

Neil, I'll submit corrections for A045613, A045614, A051594, and A051595,
but I'm not going to work on the others.  Maybe they should be deleted or
marked as "probation" until the author or someone else fixes them.

Dean Hickerson
dean at math.ucdavis.edu





From bdm at cs.anu.edu.au  Mon Nov  6 12:45:30 2006
From: bdm at cs.anu.edu.au (Brendan McKay)
Date: Mon, 6 Nov 2006 22:45:30 +1100
Subject: terms lack
In-Reply-To: <20061106093909.GA16454@math.ucdavis.edu>
References: <20061106093909.GA16454@math.ucdavis.edu>
Message-ID: <20061106114530.GB4232@cs.anu.edu.au>

* Dean Hickerson <dean at math.ucdavis.edu> [061106 20:40]:
> The terms that are listed in a sequence should be the start of the sequence.
> If there are missing terms in between the listed ones, then the sequence
> entry is not just incomplete, it's wrong.
> 
> If you've computed some terms of a sequence, but you're not sure if there
> are smaller ones that you've missed, then you should only include the ones
> that you're sure are at the start of the sequence.  You can add a comment
> about the larger terms, stating that it's not known if there are smaller ones.
> For example, the "EXTENSIONS" lines for the sequence A000043 (Mersenne
> exponents) mention some recently discovered terms.  But there might be
> some smaller ones, so these are not included in the sequence.
 
Is there an official rule about this?  I'm not sure it is correct.
Of course we should not submit sequences with possible gaps, but
suppose we can compute a(10) thru a(20) but have trouble with
a(1) thru a(9).  I don't see why we can't submit a(10..20) with
an OFFSET of 10 and a comment explaining why a(1..9) are absent.

Brendan.





From Annette.Warlich at t-online.de  Mon Nov  6 13:05:20 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Mon, 06 Nov 2006 13:05:20 +0100
Subject: Fibbinary sequence
Message-ID: <454F2500.2020107@t-online.de>

A courious result by chance:

A003714	Fibbinary numbers:
       if n = F_i1+F_i2+...+F_ik is the Zeckendorf representation of n
       (i.e. write n in Fibonacci number system)
       then
         a(n) = 2^{i1-2}+2^{i2-2}+...+2^{ik-2}.

	0, 1, 2, 4, 5, 8, 9, 10, 16, 17, 18, 20, 21, 32, 33, 34, 36, 37,
      40, 41, 42, 64, 65, 66, 68, 69, 72, 73, 74, 80, 81, 82, 84, 85,
     128, 129, 130, 132, 133, 136, 137, 138, 144, 145, 146, 148, 149,
     160, 161, 162, 164, 165, 168, 169, 170, 256, 257, 258, 260

I was checking some variants of the Pascal-matrix and got the
following
Let Px =
      1       .    .     .     .    .  .  .
      1      -1    .     .     .    .  .  .
    3/2      -3    1     .     .    .  .  .
      4      -9    6    -1     .    .  .  .
     15     -35   30   -10     1    .  .  .
  137/2  -345/2  165   -75    15   -1  .  .
 1491/4 -2009/2 1050  -560 315/2  -21  1  .
   2374   -6720 7588 -4515  1540 -294 28 -1

then the index r of the rows, r=1..inf(??) , which contain only integers, is A003714
      1, 2, 4, 5, 8, 9, 10, 16, 17, 18, 20, 21, 32, 33, 34, 36, 37,
     40, 41, 42, 64,

(although I checked this only up to N=64)


Explanation: -----------------------------------------------------------

The matrix Px is constructed as a variant of Pj
Pj:
 1  .  .   .  .   . .  .
 1 -1  .   .  .   . .  .
 1 -2  1   .  .   . .  .
 1 -3  3  -1  .   . .  .
 1 -4  6  -4  1   . .  .
 1 -5 10 -10  5  -1 .  .
 1 -6 15 -20 15  -6 1  .
 1 -7 21 -35 35 -21 7 -1

which is the column-signed version of the Pascalmatrix P
   Pj = P*J
where J is diagonal unit with alternating signs (diag(1,-1,1,-1,...))

The matrix S2 of the stirling-numbers of 2'nd kind
 1   .   .    .    .   .  . .
 1   1   .    .    .   .  . .
 1   3   1    .    .   .  . .
 1   7   6    1    .   .  . .
 1  15  25   10    1   .  . .
 1  31  90   65   15   1  . .
 1  63 301  350  140  21  1 .
 1 127 966 1701 1050 266 28 1

and S1 of the stirling numbers of first kind (where also S1 = S2^-1)
     1     .      .    .     .   .   . .
    -1     1      .    .     .   .   . .
     2    -3      1    .     .   .   . .
    -6    11     -6    1     .   .   . .
    24   -50     35  -10     1   .   . .
  -120   274   -225   85   -15   1   . .
   720 -1764   1624 -735   175 -21   1 .
 -5040 13068 -13132 6769 -1960 322 -28 1

with the eigenvalues Z(1) = diag(1,1/2,1/3,1/4) such that

   G = S2 * Z(1) * S2^-1

G:
     1     .    .     .   .    .   .   .
   1/2   1/2    .     .   .    .   .   .
   1/6   1/2  1/3     .   .    .   .   .
     0   1/4  1/2   1/4   .    .   .   .
 -1/30     0  1/3   1/2 1/5    .   .   .
     0 -1/12    0  5/12 1/2  1/6   .   .
  1/42     0 -1/6     0 1/2  1/2 1/7   .
     0  1/12    0 -7/24   0 7/12 1/2 1/8

form an eigensystem of G.
G ist interesting as it contains the bernoulli-
numbers, and coefficients of the integrals of the
bernoulli-functions.



Also G is Eigenmatrix of the signed Pascal-matrix Pj,
with the eigenvalues J = diag(1,-1,1,-1,...)

  Pj = G * J * G^-1

and Pj is used in divergent Euler-summation, Hasse's
zeta-summation and elsewhere.




Now I tried a variant of P by varying G, where the eigenvalues
are not Z(1), but are V(1/2) = diag(1,1/2,1/4,1/8,....) for some
analyses about summing zeta series, since Z(1) and V(1/2) are
closely related in this subject and complementary/mutually supplementary
in certain manners.

Now with the G-variant

  Gx  = S2 * V(1/2) * S2^-1
Gx:
     1      .       .        .      .      .    .     .
   1/2    1/2       .        .      .      .    .     .
     0    3/4     1/4        .      .      .    .     .
  -1/4    3/8     3/4      1/8      .      .    .     .
     0   -5/8   15/16      5/8   1/16      .    .     .
   1/2 -15/16  -15/32    45/32  15/32   1/32    .     .
     0    7/4 -105/32    35/64 105/64  21/64 1/64     .
 -17/8 147/32    7/32 -735/128  35/16 105/64 7/32 1/128

I come to

  Px =         Gx               * J *       Gx^-1
     =  ( S2 * V(1/2) * S2^-1 ) * J * (S2 * V(2) * S2^-1)

(while Pj is
  Pj =         G                * J *       G ^-1
     =  ( S2 * Z(-1) * S2^-1 )  * J * (S2 * Z(1) * S2^-1)
  )

and as shown above:

Px:
      1       .    .     .     .    .  .  .
      1      -1    .     .     .    .  .  .
    3/2      -3    1     .     .    .  .  .
      4      -9    6    -1     .    .  .  .
     15     -35   30   -10     1    .  .  .
  137/2  -345/2  165   -75    15   -1  .  .
 1491/4 -2009/2 1050  -560 315/2  -21  1  .
   2374   -6720 7588 -4515  1540 -294 28 -1

and the rows containing only integers, are uniquely
indicated by A003714, up to r=64, where the rownumber r
starts with 1.

Don't have an idea, how this coincidence could be interesting -
I thought I'd just sentd the observation..

Regards -

Gottfried Helms







From Labos at ana.sote.hu  Mon Nov  6 13:38:39 2006
From: Labos at ana.sote.hu (=?ISO-8859-2?Q?L=DFbos_Elem=DAr?=)
Date: Mon, 6 Nov 2006 13:38:39 MET-1MEST
Subject: terms lack
In-Reply-To: <20061106114530.GB4232@cs.anu.edu.au>
References: <20061106093909.GA16454@math.ucdavis.edu>
Message-ID: <6578DDB2A59@ana.sote.hu>

On 6 Nov 06, at 22:45, Brendan McKay wrote:

> * Dean Hickerson <dean at math.ucdavis.edu> [061106 20:40]:
> > The terms that are listed in a sequence should be the start of the sequence.
> > If there are missing terms in between the listed ones, then the sequence
> > entry is not just incomplete, it's wrong.
> > 
> > If you've computed some terms of a sequence, but you're not sure if there
> > are smaller ones that you've missed, then you should only include the ones
> > that you're sure are at the start of the sequence.  You can add a comment
> > about the larger terms, stating that it's not known if there are smaller ones.
> > For example, the "EXTENSIONS" lines for the sequence A000043 (Mersenne
> > exponents) mention some recently discovered terms.  But there might be
> > some smaller ones, so these are not included in the sequence.
>  
> Is there an official rule about this?  I'm not sure it is correct.
> Of course we should not submit sequences with possible gaps, but
> suppose we can compute a(10) thru a(20) but have trouble with
> a(1) thru a(9).  I don't see why we can't submit a(10..20) with
> an OFFSET of 10 and a comment explaining why a(1..9) are absent.
> 
> Brendan.

There are 2 cases :
 1  -- You cannot even know how many terms are missing.
          This is essentally the case of eg Mersenne primes..
-------------------
 2. --  You can specify the missing terms by list-
          position. In this second case you can introduce 
          a marker signal to point to missing position.

Handling of the cases can follow different standard..
The general case can be worthwhile to publish if 
so many hard computation is behind. 
Ordering terms somehow it can be mentioned that
the sequence is lacunar.
With correct description the situation, the case is
neither wrong, nor good. Perhaps unnecessary...

The empty sequence represents the full lacunarity.
Quite correctly. However the empty sequence is
not unique. 

Labos 






From dean at math.ucdavis.edu  Mon Nov  6 14:22:58 2006
From: dean at math.ucdavis.edu (Dean Hickerson)
Date: Mon, 6 Nov 2006 05:22:58 -0800
Subject: terms lack
Message-ID: <20061106132258.GA19536@math.ucdavis.edu>

Mostly to Brendan McKay

> Is there an official rule about this?  I'm not sure it is correct.
> Of course we should not submit sequences with possible gaps, but
> suppose we can compute a(10) thru a(20) but have trouble with
> a(1) thru a(9).  I don't see why we can't submit a(10..20) with
> an OFFSET of 10 and a comment explaining why a(1..9) are absent.

Sure, that's fine.  In that case the sequence that's in the OEIS starts
with a(10), and it doesn't have any gaps.  It's part of a longer, partially
unknown, one that starts with a(1), but the sequence that it's part of
isn't in the OEIS.

But in cases like Kohmoto's sequences that wouldn't work.  For example,
suppose we know from hand calculation (as Kohmoto did for A045613), that
105, 110, and 2145 are in the sequence.  Presumably we can test that
there are no terms between 105 and 110, but we can't tell if there are
terms below 105 or between 110 and 2145.  So (ignoring the 4-term rule for
submissions) we could submit a sequence with just 105 and 110, and add
comments saying that there might be earlier terms and that 2145 is also
in the sequence, but we shouldn't submit the sequence as starting with
105, 110, 2145.

Dean Hickerson
dean at math.ucdavis.edu





From tbaruchel at free.fr  Mon Nov  6 17:31:32 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Mon, 6 Nov 2006 17:31:32 +0100
Subject: A fast (written in assembly) library for continued fractions
In-Reply-To: <20061103221729.GB2467@>
References: <20061103221729.GB2467@>
Message-ID: <20061106163132.GA30587@>

On Ven, nov 03, 2006 at 11:17:29 +0100, Thomas Baruchel wrote:
> I wrote for my own purposes the following X86 assembly code, intended

The final version of my code can be found at
  http://baruchel.free.fr/~thomas/wp/?p=10
I give also (in the comments of the file) advices for using the functions
from gcc, gfortran, g95. The fastest continued fractions in the west...

Next step will be functions for computing the Pell equation, but
pure FPU stack will not be possible this time (only eight registers), thus
some memory access will be required :-(

Regards,

-- 
Thomas Baruchel
  http://baruchel.free.fr/~thomas/wp/





From Eric.Angelini at kntv.be  Mon Nov  6 18:57:00 2006
From: Eric.Angelini at kntv.be (Eric Angelini)
Date: Mon, 6 Nov 2006 18:57:00 +0100
Subject: Langford/Skolem-like sequence : odd + even terms
Message-ID: <F05775148D000C4B9321A5113D53CB7D191001@KNTV-SERVER01.kntv.local>


Hello SeqFans,
Another Langford/Skolem-like sequence, with a mixture of odd and
even terms -- plus two questions:
http://www.cetteadressecomportecinquantesignes.com/OddEvenSkol.htm
Best,
?.






From noe at sspectra.com  Mon Nov  6 19:04:49 2006
From: noe at sspectra.com (T. D. Noe)
Date: Mon, 6 Nov 2006 10:04:49 -0800
Subject: Cyclotomic polynomial for n=0
Message-ID: <p06230985c175281c5eee@[192.168.1.43]>

For n=1,2,3,... the cyclotomic polynomials are

x-1, x+1, x^2+x+1,...

It seems that there are two conventions for n=0:

x or 1

Mathematica says it is 1.  I think Maple uses x.  Can someone explain this?

Thanks,

Tony





From franktaw at netscape.net  Mon Nov  6 22:27:14 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 06 Nov 2006 16:27:14 -0500
Subject: Cyclotomic polynomial for n=0
In-Reply-To: <p06230985c175281c5eee@[192.168.1.43]>
References: <p06230985c175281c5eee@[192.168.1.43]>
Message-ID: <8C8D001766F0D57-BD0-2F58@WEBMAIL-MA14.sysops.aol.com>

Strictly speaking, the cyclotomic polynomial for n=0 is undefned.  The 
two packages have chosen different ways to extend the concept to n=0; 
it would have been perfectly OK for them to have returned some kind of 
error.

There are reasonable arguments for either choice.  A lot comes down to 
the basic property that the degree is phi(n); the question then is, 
what is phi(0)?  Again, this is strictly speaking undefined, but there 
are two reasonable choices: 0 and 1.  phi(0) = 0 gives 1 as the value 
of the 0th cyclotomic polynomial.  If we take phi(0) = 1, then we need 
to choose a root for the polynomial.  Choosing 0 as the root gives the 
nice property that the cyclotomic polynomials for n >= 0 constitute all 
the primitive monic polynomials all of whose roots lie in the unit 
disk; hence the selection of x as the polynomial.

If you take the formula Phi_n(x) = Product_{0<=k<n, gcd(k,n) = 1} e^{2 
k Pi i / n}, you get Phi_0(x) = 1.  On the other hand, if we regard 
this as a product over distinct residues modulo n that are relatively 
prime to n, we get x - e^{2 Pi i / 0}, which is undefined.

You can also make a case for 0 as the "correct" value.  Phi_n(x) is 
x^n-1, divided by all x - u for primitive dth roots of unity u for d a 
proper divisor of n.  x^0-1 is 0; you can divide it by anything you 
want, and the result is still 0.  (Of course, this becomes an infinite 
product in the denominator - but this is only a plausibility argument.)

Franklin T. Adams-Watters


-----Original Message-----
From: noe at sspectra.com

For n=1,2,3,... the cyclotomic polynomials are

x-1, x+1, x^2+x+1,...

It seems that there are two conventions for n=0:

x or 1

Mathematica says it is 1.  I think Maple uses x.  Can someone explain 
this?

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From zbi74583 at boat.zero.ad.jp  Tue Nov  7 02:34:32 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Tue, 7 Nov 2006 10:34:32 +0900
Subject: terms lack
References: <20061106093909.GA16454@math.ucdavis.edu>
Message-ID: <018901c7020c$e2322a30$ef73d7dc@FMC57937B6BE00>

    To Dean Hicherson

    Thanks for editing my sequences.
    And thanks for a good explanation about "completeness".
    Now, I understand that  the  purpose of OEIS is "Identification of 
sequences", it is necessary.

    Should A038362 be deleted?
    Still, I think that it should exist , because a sequence is a mapping 
from N to N, so a subset of the images is also a good information for 
understanding the mapping.

    Yasutoshi


----- Original Message ----- 
From: "Dean Hickerson" <dean at math.ucdavis.edu>
To: <seqfan at ext.jussieu.fr>
Sent: Monday, November 06, 2006 6:39 PM
Subject: Re: terms lack


> Mostly to Yasutoshi Kohmoto <zbi74583 at boat.zero.ad.jp>:
>
>> Recently I updated more terms of A054572, which are calculated by
>> Richard Mathar.
>> And Neil claimed or got angry.
>> He said, "You are red card. Don't submit a wrong sequences"
>> I am exaggerating a little.
>> These words are not exactly what he wrote.
>>
>> To Neil
>> They are correct, not "wrong".
>> It is only not complete.
>
> The terms that are listed in a sequence should be the start of the 
> sequence.
> If there are missing terms in between the listed ones, then the sequence
> entry is not just incomplete, it's wrong.
>
> If you've computed some terms of a sequence, but you're not sure if there
> are smaller ones that you've missed, then you should only include the ones
> that you're sure are at the start of the sequence.  You can add a comment
> about the larger terms, stating that it's not known if there are smaller 
> ones.
> For example, the "EXTENSIONS" lines for the sequence A000043 (Mersenne
> exponents) mention some recently discovered terms.  But there might be
> some smaller ones, so these are not included in the sequence.
>
>> I defined many "divisor" functions and calculated many Amicable Numbers 
>> and
>> Perfect Numbers using those functions.
>> And I submitted them as sequences to  OEIS.
>>
>> http://www.research.att.com/~njas/sequences/?q=kohmoto+amicable&sort=0&fmt=0&language=english&go=Search
>>
>> http://www.research.att.com/~njas/sequences/?q=kohmoto+perfect&sort=0&fmt=0&language=english&go=Search
>>
>> I suppose that they are almost complete but some of them might not be
>> complete.
>
> I've found missing terms for several of your sequences:
>
>
> Sequences A045613 and A045614 ("Super Unitary Amicable Number") consist
> of pairs of distinct numbers a and b such that
>
>    usigma(usigma(a)) = usigma(usigma(b)) = a+b.
>
> (The definitions don't make this clear, but A045613 contains the smaller
> members of the pairs and A045614 contains the larger members.  Also 
> they're
> sorted by their smallest members, so 155 comes before 142 in A045614.) 
> The
> first 3 pairs that you've listed are (105, 155), (110, 142) and (2145, 
> 3055).
> But you've missed several smaller ones:
>
>    (33, 35), (208, 224), (268, 272), (455, 601), (695, 1033), (812, 956),
>    (1609, 1847), (1808, 2512), (1892, 3004), (1913, 2407), (2096, 2224)
>
>
> A051594 and A051595 are "(-1)sigma amicable numbers".  (Again, the 
> definition
> doesn't make it clear that A051595 has the smaller terms and A051594 has 
> the
> larger terms.)  The first pair that you've listed is (1969706592, 
> 2236072608).
> But there are many smaller ones, starting with:
>
>    (429552, 466200), (497808, 604656), (800496, 1103760).
>
>
> A036471-A036474 are defined as "Amicable quadruple: 4 different numbers
> which satisfy sigma(a)=sigma(b)=sigma(c)=sigma(d)=a+b+c+d".  The first
> quadruple that you've listed has smallest element
> 342151462276356306033089201934180, but here's a much smaller one:
>
>    (3270960, 3361680, 3461040, 3834000)
>
> (I think that's the smallest one, but I could be wrong.)  If someone can
> determine the first several terms of this sequence, then it should be
> edited; otherwise it should be deleted.
>
>
> In some of your sequences, I wasn't able to find any smaller terms, but 
> the
> ones that you've listed are so large that I'm not at all confident that 
> they
> are the smallest ones.  For example, A038362 and A038363 are defined as
> "A Rational Amicable Number: two different numbers a, b which satisfies 
> the
> following equation: sigma(a)=sigma(b)=(a+b)^3/(a^2+b^2)", and the smallest
> pair that you've included is:
>
>    (26403469440047700, 30193441130006700)
>
> Do you have any reason to think that that's the smallest such pair?  If 
> not,
> and if you can't determine what the first few terms really are, then the
> sequence should be deleted.
>
>
> I've only looked at a few of your sequences, but that's enough to convince
> me that they should all be checked.
>
> Neil, I'll submit corrections for A045613, A045614, A051594, and A051595,
> but I'm not going to work on the others.  Maybe they should be deleted or
> marked as "probation" until the author or someone else fixes them.
>
> Dean Hickerson
> dean at math.ucdavis.edu
> 






From maxale at gmail.com  Tue Nov  7 22:14:38 2006
From: maxale at gmail.com (Max A.)
Date: Tue, 7 Nov 2006 13:14:38 -0800
Subject: square arrangements with bit-wise constraints
Message-ID: <d3dac270611071314i6023ba68h4e7e15d63da8c8b3@mail.gmail.com>

There is a problem at MathPages about arrangements of numbers
0,1,...,15 into 4x4 array with certain bit-wise constraints:
http://www.mathpages.com/home/kmath352.htm

This problem (and its variants) can be easily generalized to
(2^n)x(2^n) arrays, e.g.:

In how many distinct ways can the integers 0 through 2^(2n)-1 be arranged
in a (2^n)x(2^n) array such that the bit-wise OR over each row, each column,
and each diagonal is 2^(2n)-1, and the bit-wise AND over each row, column,
and diagonal is 0?

Does anybody wish to compute corresponding sequences and submit them to OEIS?

Regards,
Max





From maxale at gmail.com  Tue Nov  7 23:49:04 2006
From: maxale at gmail.com (Max A.)
Date: Tue, 7 Nov 2006 14:49:04 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>

On 11/3/06, franktaw at netscape.net <franktaw at netscape.net> wrote:

> In the following, b(i), i = 1..k is the finite sequence.  I'm starting
> all these
> sequences with the empty sequence, but I searched for them without
> that first term.  All are shown here in A066099 order.

What is good about this order comparing, say, to the prime
factorization order suggested by Marc?

I would vote for having a standard way to represent such sequences in
OEIS (well documented there, and maybe even having a dedicated
keyword). That is important when one come up with a multidimensional
sequence and want to query it in OEIS. Then it will be enough for
him/her to encode the sequence in the standard way and make a single
query.

Of course, there are many representations possible but one of them
should be called "standard", and every multidimensional sequence
should have this standard representation (among maybe others) in OEIS.

Neil, does that sound reasonable?

Regards,
Max





From mlb at well.com  Wed Nov  8 00:02:52 2006
From: mlb at well.com (Marc LeBrun)
Date: Tue, 07 Nov 2006 15:02:52 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com
 >
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
 <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>
Message-ID: <200611072304.kA7N3xSr010827@smtp.well.com>

 > does that sound reasonable?

It does, but in practice there are already several orderings in wide 
use (as Frank indicated).

Of course it's always possible to submit more than one ordering.

In fact sometimes different ones would be preferred in different contexts.

Or fancier (perhaps special-purpose) lookups could be used.

"That's the great thing about standards--there's so many to choose from!"







From franktaw at netscape.net  Wed Nov  8 00:56:58 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Tue, 07 Nov 2006 18:56:58 -0500
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com> <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>
Message-ID: <8C8D0DF8BAA6EAF-124-5E95@FWM-M16.sysops.aol.com>

Exactly what factorization order?  For most numbers, the sequence of
prime power exponents is not a sequence of positive integers.  If 2
represents the sequence (1), what sequence does 3 represent?

If doesn't work any better to use prime power exponents to represent
sequences of non-negative integers.  If 2 is (1) and 3 is (0,1), what 
do
you use to represent (1,0)?

There are ways around these difficulties, but they are pretty 
artificial.

A066099 sorts, first of all, by the number being composed; I think any
reasonable standard candidate should do that.  Second, it is simple,
with a definition tied in to the binary numbers.  Third, it is similar 
to the
Mathematica ordering for partitions (subsort in reverse lexicographic
order), which is one of the two standards for partition ordering in the
OEIS.

(The equivalent of the other, the Abramowitz and Stegun ordering,
was not yet in the OEIS; I did submit it the other day with A-number
A124734.  It is, in any event, not as simple to define as A066099.)

In any event, based on Neil saying to go ahead and submit these
sequences, I  have submitted them with comments stating that
A066099 is the standard ordering.

Franklin T. Adams-Watters


-----Original Message-----
From: maxale at gmail.com

  On 11/3/06, franktaw at netscape.net <franktaw at netscape.net> wrote:

  > In the following, b(i), i = 1..k is the finite sequence. I'm 
starting
 > all these
 > sequences with the empty sequence, but I searched for them without
 > that first term. All are shown here in A066099 order.

 What is good about this order comparing, say, to the prime
 factorization order suggested by Marc?

 I would vote for having a standard way to represent such sequences in
 OEIS (well documented there, and maybe even having a dedicated
 keyword). ...
________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From maxale at gmail.com  Wed Nov  8 01:42:29 2006
From: maxale at gmail.com (Max A.)
Date: Tue, 7 Nov 2006 16:42:29 -0800
Subject: querying multidimensional sequences in OEIS
In-Reply-To: <8C8D0DF8BAA6EAF-124-5E95@FWM-M16.sysops.aol.com>
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
	 <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
	 <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
	 <d3dac270611071449u3b3b2464qe85550276fe249f1@mail.gmail.com>
	 <8C8D0DF8BAA6EAF-124-5E95@FWM-M16.sysops.aol.com>
Message-ID: <d3dac270611071642x66119c57o13f7bf6df6d2436a@mail.gmail.com>

On 11/7/06, franktaw at netscape.net <franktaw at netscape.net> wrote:

> Exactly what factorization order?  For most numbers, the sequence of
> prime power exponents is not a sequence of positive integers.  If 2
> represents the sequence (1), what sequence does 3 represent?
>
> If doesn't work any better to use prime power exponents to represent
> sequences of non-negative integers.  If 2 is (1) and 3 is (0,1), what
> do you use to represent (1,0)?

Actually, the sequences that I mentioned recently does not depend on
trailing zeros, so index (1,0) for them is the same as (1). That's why
the factorization order works well for them.

> A066099 sorts, first of all, by the number being composed; I think any
> reasonable standard candidate should do that.  Second, it is simple,
> with a definition tied in to the binary numbers.  Third, it is similar
> to the
> Mathematica ordering for partitions (subsort in reverse lexicographic
> order), which is one of the two standards for partition ordering in the
> OEIS.

I see now the advantages of the A066099 ordering. And I agree that it
is a good candidate for the "standard".

Max





From arndt at jjj.de  Wed Nov  8 03:54:45 2006
From: arndt at jjj.de (Joerg Arndt)
Date: Wed, 8 Nov 2006 13:54:45 +1100
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com> <20061102011002.GC6041@cochise.anu.edu.au> <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
Message-ID: <20061108025445.GC6252@cochise.anu.edu.au>

* Jonathan Post <jvospost3 at gmail.com> [Nov 08. 2006 13:29]:
> Thank you, Joerg.  That was interesting.  Does your generating function
> hypothesis hold when I extend the sequence by another 5 values, to a(n) = 0, 0,
> 1, 1, 3, 1, 7, 4,
> 14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again, assuming that
> I got the elementary algebra and arithmetic correct.

Alas, the function I gave was spurious, with the
additional terms no OGF is found.

The gp/script can be found here
http://www.jjj.de/pari/ggf.inc.gp
(cannot find Ralf original at the moment).


> If so, I should submit this sequence as by the two us!

Definitely your invention!
But thanks for the kind offer.


> [...]

best regards,    jj





From zbi74583 at boat.zero.ad.jp  Wed Nov  8 06:33:16 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Wed, 8 Nov 2006 14:33:16 +0900
Subject: N(2)=1+1/2^2+2/3^2+5/4^2+4/5^2+2/6^2+6/7^2+....
Message-ID: <02ba01c702f7$69237ea0$ef73d7dc@FMC57937B6BE00>

    Hi, Seqfans

    I asked Mathar to calculate the following
 number.
    N(s)=   Sum_{m=1 to infinity} (-1)Sigma(m)/m^s

     See the end of this mail

     I tried to find an identity like the following

         Zeta(s)Zeta(s-1)=Sum_{m=1 to infinity} Sigma(m)/m^s
         Zeta(s)/Zeta(s+1)=Sum_{m=1 to infinity} Phi(m)/m^s

    But it is too difficult.
    I transformed it a little as follows.

         N(s)=   Sum_{m=1 to infinity} (-1)Sigma(m)/m^s
         =   Product_{p_i} ( Sum_{k=0 to infinity}
(-1)Sigma(p_i^k)/p_i^(k*s) )
         =   Product_{p_i} ( Sum_{k=0 to infinity}
(Sigma(p_i^k)-2)/p_i^(k*s) )
         =   Product_{p_i} ( Sum_{k=0 to infinity}
 Sigma(p_i^k)/p_i^(k*s) -2/(p_i^s-1) )

        Some members of Seqfan might solve it easily.

    Yasutsoshi

----- Original Message ----- 
From: "Richard Mathar" <mathar at strw.leidenuniv.nl>
To: <zbi74583 at boat.zero.ad.jp>
Cc: <mathar at mail.strw.leidenuniv.nl>
Sent: Sunday, November 05, 2006 10:35 PM
Subject: Re: N(2)=1+1/2^2+2/3^2+5/4^2+4/5^2+2/6^2+6/7^2+....


>
> These are 5 tables of partial sums of N(s) up to some maximum
> m of sum_{m=1...max(m)) (-1)Sigma(m)/m^s. The first number
> is s, the second the maximum m up to which the sum has been accumulated,
> and finally the floating point number the partial sum.
> The case s=1 seems not to be a converging series; the partial sums are
> almost
> linear to the upper limit of m. (If your formula with the zeta-function is
> correct, this is expected since zeta(1)=infinty.) I am including some
> values
> for small maximum m so you can check the results for the basic cases.
> N(2) seems to be a converging series. For N(3) you could already get some
> pretty accurate estimate of the series limit.
>
> zeta(2)=1.644934067
> zeta(3)=1.202056903
> zeta(4)=1.082323234
> zeta(3)*zeta(2)=1.977304350
> zeta(4)*zeta(3)=1.301014115
> zeta(5)*zeta(4)=1.122291001
>
>
> PARI:
>
> \\ computes (-1)sigma(n)
> A049060(n)={
>        local(i,resul,rmax,p) ;
>        if(n==1,
>                return(1)
>        ) ;
>        i=factor(n) ;
>        rmax=matsize(i)[1] ;
>        resul=1 ;
>        for(r=1,rmax,
>                p=0 ;
>                for(j=1,i[r,2],
>                        p += i[r,1]^j ;
>                ) ;
>                resul *= p-1 ;
>        ) ;
>        return(resul) ;
> }
>
>
> N(s,mmax)={
> local(resul) ;
> resul=0.0 ;
> for(m=1,mmax,
> resul += A049060(m)/m^s ;
> if(m<10 || (m % 10000 == 0),
> print(s," ",m," ",resul) ;
> ) ;
> ) ;
> }
>
> {
> mmax=1000000 ;
> for(s=1,5,
> N(s,mmax) ;
> ) ;
> }
>
> Tables:
>
> 1 1 1.000000000000000000000000000
> 1 2 1.500000000000000000000000000
> 1 3 2.166666666666666666666666667
> 1 4 3.416666666666666666666666667
> 1 5 4.216666666666666666666666667
> 1 6 4.550000000000000000000000000
> 1 7 5.407142857142857142857142857
> 1 8 7.032142857142857142857142857
> 1 9 8.254365079365079365079365079
> 1 10000 8956.254813409229652481598757
> 1 20000 17913.54053798274101959639246
> 1 30000 26870.58900461199889420226700
> 1 40000 35827.55225952432056619824099
> 1 50000 44784.92216728336777991954546
> 1 60000 53741.30998083683291788901077
> 1 70000 62698.82877172292127401143322
> 1 80000 71656.42458494024747176547617
> 1 90000 80613.31691563743683765060924
> 1 100000 89570.20047616274019702521482
> 1 110000 98527.44018335441754406085900
> 1 120000 107484.2058258699552949703883
> 1 130000 116441.3450587298928159003627
> 1 140000 125399.1600101618061663172293
> 1 150000 134355.0716257765731578804073
> 1 160000 143312.6869188725594888173896
> 1 170000 152270.1691945778107893068833
> 1 180000 161227.3673088893702787242209
> 1 190000 170184.2283671443801174189235
> 1 200000 179141.9177161676803505348327
> 1 210000 188098.1905736334290746346704
> 1 220000 197055.5778178213299226211954
> 1 230000 206012.3243951074382665848232
> 1 240000 214969.0180191522602672455303
> 1 250000 223927.0626654355275622417046
> 1 260000 232884.0479228396547382227555
> 1 270000 241841.6897312605543987962985
> 1 280000 250798.1260711861562596104630
> 1 290000 259755.4115589713808307871029
> 1 300000 268712.2473108133199765189322
> 1 310000 277669.3610844430796369732363
> 1 320000 286626.7360084187087594659096
> 1 330000 295583.0702328420388811375076
> 1 340000 304540.9324662583651406115180
> 1 350000 313497.9488508493337302810705
> 1 360000 322455.2848872162239393519675
> 1 370000 331411.9952274305516780284586
> 1 380000 340369.3878688847801453424847
> 1 390000 349326.5442385613025662494469
> 1 400000 358283.7198151948112652311070
> 1 410000 367240.8596635115166082534173
> 1 420000 376197.2486760902447184900822
> 1 430000 385154.6524560274677028589761
> 1 440000 394112.0480841471143112935114
> 1 450000 403069.2697966051366843904756
> 1 460000 412026.1487890321657369929892
> 1 470000 420983.4928713199267860102478
> 1 480000 429940.0597652682271439943634
> 1 490000 438897.5981659322519223492273
> 1 500000 447855.4516433798944040122164
> 1 510000 456811.4127981064830642426529
> 1 520000 465769.1531758862533175249594
> 1 530000 474726.4930671377953578153300
> 1 540000 483683.7925672078781534825684
> 1 550000 492640.2315981062235193108053
> 1 560000 501597.6949987133986923353009
> 1 570000 510554.3958474732897001847922
> 1 580000 519511.5431567859764988621906
> 1 590000 528468.5920184442509092919859
> 1 600000 537425.2057958757909274862160
> 1 610000 546383.3300255139505983850944
> 1 620000 555340.2697592667964924572269
> 1 630000 564297.7073054137624068690112
> 1 640000 573254.2920689474814450061694
> 1 650000 582211.4873730487914773223047
> 1 660000 591168.5211525597522567544916
> 1 670000 600125.8589826114248021714650
> 1 680000 609083.0330009002048275663287
> 1 690000 618039.4919349372313479054900
> 1 700000 626997.2222632460664089943429
> 1 710000 635954.0917690439691078847726
> 1 720000 644911.6290093784488180074886
> 1 730000 653868.6304875870104467000556
> 1 740000 662825.3447013751015479514900
> 1 750000 671782.5558801465303249818702
> 1 760000 680739.9011662731435592584669
> 1 770000 689696.9475041453715157412648
> 1 780000 698653.1437839886472918561286
> 1 790000 707611.0985532863373721850039
> 1 800000 716568.7987693866776749447598
> 1 810000 725525.6575122445531974385646
> 1 820000 734482.4569919820693082893669
> 1 830000 743439.4779916042359124217226
> 1 840000 752396.3611940500092646796766
> 1 850000 761353.5394438445571430835057
> 1 860000 770310.9434931792736520368474
> 1 870000 779267.1927223886953325647496
> 1 880000 788225.1436674517253585125689
>
>
> 2 1 1.000000000000000000000000000
> 2 2 1.250000000000000000000000000
> 2 3 1.472222222222222222222222222
> 2 4 1.784722222222222222222222222
> 2 5 1.944722222222222222222222222
> 2 6 2.000277777777777777777777778
> 2 7 2.122726757369614512471655329
> 2 8 2.325851757369614512471655329
> 2 9 2.461654226505416981607457798
> 2 10000 8.619813177975507784214344698
> 2 20000 9.240635095871089404115753095
> 2 30000 9.603791897573761358292380061
> 2 40000 9.861458278261452185262848858
> 2 50000 10.06133043757656995196303496
> 2 60000 10.22462154552329778519365923
> 2 70000 10.36270038005636494621123847
> 2 80000 10.48231016697477680498962023
> 2 90000 10.58780522013752853231672862
> 2 100000 10.68217402235127296288785820
> 2 110000 10.76754463607533705360724580
> 2 120000 10.84547789729425765394094918
> 2 130000 10.91717268365027978460835188
> 2 140000 10.98355652885066079792761960
> 2 150000 11.04534545539722125239381859
> 2 160000 11.10315629233238669062836862
> 2 170000 11.15746025373895516747888760
> 2 180000 11.20865775307751148407150182
> 2 190000 11.25708463818456648077093060
> 2 200000 11.30303121795283056763425981
> 2 210000 11.34672871948779384291766276
> 2 220000 11.38839827063844308711114745
> 2 230000 11.42821240662906506074836109
> 2 240000 11.46633165190016188934526342
> 2 250000 11.50290004647923375585461222
> 2 260000 11.53802980091084626273394586
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>
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From ralf at ark.in-berlin.de  Wed Nov  8 09:26:05 2006
From: ralf at ark.in-berlin.de (Ralf Stephan)
Date: Wed, 8 Nov 2006 09:26:05 +0100
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <20061108025445.GC6252@cochise.anu.edu.au>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com> <20061102011002.GC6041@cochise.anu.edu.au> <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com> <20061108025445.GC6252@cochise.anu.edu.au>
Message-ID: <20061108082605.GA21119@ark.in-berlin.de>

You wrote 
> * Jonathan Post <jvospost3 at gmail.com> [Nov 08. 2006 13:29]:
> > Thank you, Joerg.  That was interesting.  Does your generating function
> > hypothesis hold when I extend the sequence by another 5 values, to a(n) = 0, 0,
> > 1, 1, 3, 1, 7, 4,
> > 14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again, assuming that
> > I got the elementary algebra and arithmetic correct.
> 
> Alas, the function I gave was spurious, with the
> additional terms no OGF is found.

Quite possible. Always treat them as hypotheses, as Jonathan said.


ralf






From maxale at gmail.com  Wed Nov  8 11:27:40 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 8 Nov 2006 02:27:40 -0800
Subject: A038379 is all wrong
Message-ID: <d3dac270611080227lc6e0ad2qde71bd834c86d0f2@mail.gmail.com>

A038379 is defined as the "number of positive semi-definite real {0,1}
n X n matrices".
But the sequence and comments have nothing to do with the description.

In particular, all given values do not match the definition.
It lists 1, 3, 27, 729, 52649 while the correct values (that match the
definition) are 2, 7, 59, 1468, 102751 (I will be grateful if somebody
check these values).
For the simplest case n=1, it is clear that there two positive
semi-definite real {0,1}-matrices:
(0) and (1).

There is also a comment:
%C A038379 For n <= 4 a(n) = the upper bound 3^C(n,2).
Why the upper bound is 3^C(n,2) ? We have only two options for each
element of the matrix.
There would be an upper bound of this type if:
1) The main diagonal of a matrix is fixed, and the matrix is
symmetric. Then there are C(n,2) distinct non-diagonal elements whose
values should be assigned.
2) There are three (not two!) possible values for each non-diagonal
element of the matrix.

It seems that the definition of A038379 does not match the rest. Can
anybody find a definition of A038379 that is consistent with its
current content? If so, I will submit the real "number of positive
semi-definite real {0,1} n X n matrices" as a separate sequence.

Regards,
Max





From njas at research.att.com  Wed Nov  8 15:54:53 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 8 Nov 2006 09:54:53 -0500 (EST)
Subject: A038379 is all wrong
Message-ID: <200611081454.JAA83283@fry.research.att.com>

Max,   maybe I meant to say "Number of positive semi-definite real
matrices with entries {-1, 0, +1}." - can you check?

Neil





From maxale at gmail.com  Wed Nov  8 22:59:13 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 8 Nov 2006 13:59:13 -0800
Subject: A038379 is all wrong
In-Reply-To: <200611081454.JAA83283@fry.research.att.com>
References: <200611081454.JAA83283@fry.research.att.com>
Message-ID: <d3dac270611081359t117aab4ag1545e9c3130fa53a@mail.gmail.com>

On 11/8/06, N. J. A. Sloane <njas at research.att.com> wrote:
> Max,   maybe I meant to say "Number of positive semi-definite real
> matrices with entries {-1, 0, +1}." - can you check?

For the comment to make sense, the main diagonal should be fixed as well.
I will check most obvious variants.

Max





From jvospost3 at gmail.com  Thu Nov  9 02:51:54 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 8 Nov 2006 17:51:54 -0800
Subject: primes in arithmetic progression
In-Reply-To: <335cf8920611042302v7db1ca58xb8b67f872981b204@mail.gmail.com>
References: <200611041910.OAA62978@fry.research.att.com>
	 <5542af940611041355vd65aa5cr3a4a9df3cc348c15@mail.gmail.com>
	 <335cf8920611042302v7db1ca58xb8b67f872981b204@mail.gmail.com>
Message-ID: <5542af940611081751y69ee576as6473da25e91a2175@mail.gmail.com>

Sorry, Olivier.  I hope not to repeat the mistake.

Thank you again, Richard J. Mathar, for your corrections and extensions of 6
Nov 2006, which appear online today for A124570.  I'm surprised that nobody
looked in this direction, before you started it.  NJAS was right to see the
value, and 'kick things off" -- and I hope that it does not stop with my
clumsy shot at the semiprime case, which you so gallantly repaired.

I'm mostly satisfied with the results of myself and tens of millions of
others voting in the U.S. elections Tuesday 7 Nove 2006.  I feel as if the
whole country is being corrected and extended, to benefits both domestic and
global. But this is not the venue to discuss that.



On 11/4/06, Olivier Gerard <olivier.gerard at gmail.com> wrote:
>
> Hello Jonathan,
>
> Thanks for this message.
>
> Please remember NOT to send your posts to seqfan as HTML,
> even with a pure text version.
>
> regards,
>
> Olivier
>
>
> On 11/4/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> >
> > The analogue for 3-almost primes (or, as njas prefers, numbers which are
> > the product of exactly 3 primes) is left as an exercise for the reader...
> > I've just submitted the following.
> >
> > Subject: PRE-NUMBERED NEW SEQUENCE A124570 FROM Jonathan Vos Post
> >
> >
> >
>
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From jvospost3 at gmail.com  Thu Nov  9 04:11:11 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 8 Nov 2006 19:11:11 -0800
Subject: Scrabblomics; COMMENT FROM Jonathan Vos Post RE A113172
In-Reply-To: <20061108213820.28640.qmail@web55109.mail.re4.yahoo.com>
References: <20061108213820.28640.qmail@web55109.mail.re4.yahoo.com>
Message-ID: <5542af940611081911h2ff3a639v3a4b53b5c2927de2@mail.gmail.com>

Here's more fun, with a silly if plausibly motivated name.

Subject: NEW SEQUENCE FROM Jonathan Vos Post


%I A000001
%S A000001 1,
3, 3,
8, 10, 8,
9, 12, 12, 9,
4, 15, 19, 15, 4,
7, 15, 19, 19, 15, 7,
8, 18, 19, 21, 19, 18, 8,
9, 22, 20, 11, 11, 20, 22, 9,
4, 15, 17, 15, 18, 15, 17, 15, 4,
7, 15, 18, 18, 20, 20, 18, 18, 15, 7,
8, 18, 20, 22, 21, 11, 21, 22, 20, 18, 8,
9, 22, 21, 17, 19, 18, 18, 19, 17, 21, 22, 9
%N A000001 Pascrabble Triangle, by rows.
%C A000001 The apex of the triangle is 1.  Any other value is the
Scrabble value of English name for the number which is the sum of the
numbers above. This is generated the same way as A007318 Pascal's
triangle read by rows, except apply A113172 to each sum. The first
column of this triangle is 1, 3, 8, 9, 4, 7, 8, 9, 4, 7, 8, 9, 4, 7...
= iterations 1, A113172(1), A113172(A113172(1)),
A113172(A113172(A113172(1))).
The central pascrabble numbers T(2n+1,n) = 1, 10, 19, 21, 18, 11, 22, ...
%F A000001 T(1,1) = 1; for i > 1, T(i,j) = A113172(T(i-1, j-1)+T(i-1, j).
%e A000001 Triangle begins:
row.|.values in row
.1..|01
.2..|03.03
.3..|08.10.08
.4..|09.12.12.09
.5..|04.15.19.15.04
.6..|07.15.19.19.15.07
.7..|08.18.19.21.19.18.08
.8..|09.22.20.11.11.20.22.09
.9..|04.15.17.15.18.15.17.15.04
10..|07.15.18.18.20.20.18.18.15.07
11..|08.18.20.22.21.11.21.22.20.18.8
12..|09.22.21.17.19.18.18.19.17.21.22.09
%Y A000001 cf. A007318 Pascal's triangle read by rows, A113172.
%O A000001 1
%K A000001 ,easy,nonn,tabl,word,
%A A000001 Jonathan Vos Post (jvospost2 at yahoo.com), Nov 08 2006

I think that this cut & paste is pure text, and not HTML (as Olivier asked)
but I am not sure.

On 11/8/06, jonathan post <jvospost2 at yahoo.com> wrote:
>
> My backup copy (as no autoreply yet).
>
> Subject: COMMENT FROM Jonathan Vos Post RE A113172
>
>
> %I A113172
> %S A113172
> 13,3,6,8,7,10,10,8,9,4,3,9,12,11,11,13,14,12,12,8,12,15,18,20,19,22,22
> %N A113172 Scrabble value of English word for the
> number n.
> %C A113172 Is 12 the unique fixed point of Scrabble
> value of English word for the number n, where
> A113172(n) = n? 12 --> "TWELVE" --> 1+4+1+1+4+1 -->
> 12. For what n are A113172(n) prime? For what n are
> there cycles of what length for iterations n,
> A113172(n), A113172(A113172(n)),
> A113172(A113172(A113172(n))), ...? For what n does
> A113172(n) | n? These are key questions in general
> Scrabblomics.
> %F A113172 Scrabble value of English word for the
> number n.
> %O A113172 1
> %K A113172 ,easy,nonn,word,
> %A A113172 Jonathan Vos Post (jvospost2 at yahoo.com),
> Nov 08 2006
>
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From rkg at cpsc.ucalgary.ca  Thu Nov  9 20:26:26 2006
From: rkg at cpsc.ucalgary.ca (Richard Guy)
Date: Thu, 9 Nov 2006 12:26:26 -0700 (MST)
Subject: Small notation clean-up
In-Reply-To: <5542af940611081751y69ee576as6473da25e91a2175@mail.gmail.com>
References: <200611041910.OAA62978@fry.research.att.com> 
 <5542af940611041355vd65aa5cr3a4a9df3cc348c15@mail.gmail.com> 
 <335cf8920611042302v7db1ca58xb8b67f872981b204@mail.gmail.com>
 <5542af940611081751y69ee576as6473da25e91a2175@mail.gmail.com>
Message-ID: <Pine.LNX.4.63.0611091221220.23740@csl>

Would someone more competent & energetic than I
clean up the (usually unexplained)   S_n
notation that appears sporadically in OEIS ?

Evidently it means a ``star'' with  n  vertices,
and so is in fact the complete bipartite graph
K_(1,n-1), a notation which is more widely
known.   Best to all,   R.





From maxale at gmail.com  Thu Nov  9 20:35:39 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 9 Nov 2006 11:35:39 -0800
Subject: Small notation clean-up
In-Reply-To: <Pine.LNX.4.63.0611091221220.23740@csl>
References: <200611041910.OAA62978@fry.research.att.com>
	 <5542af940611041355vd65aa5cr3a4a9df3cc348c15@mail.gmail.com>
	 <335cf8920611042302v7db1ca58xb8b67f872981b204@mail.gmail.com>
	 <5542af940611081751y69ee576as6473da25e91a2175@mail.gmail.com>
	 <Pine.LNX.4.63.0611091221220.23740@csl>
Message-ID: <d3dac270611091135u58ec268au1f85498bf1ea978b@mail.gmail.com>

On 11/9/06, Richard Guy <rkg at cpsc.ucalgary.ca> wrote:
> Would someone more competent & energetic than I
> clean up the (usually unexplained)   S_n
> notation that appears sporadically in OEIS ?
>
> Evidently it means a ``star'' with  n  vertices,

It is also often used for the symmetric group of order n.

Max





From maxale at gmail.com  Thu Nov  9 23:11:00 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 9 Nov 2006 14:11:00 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <20061108025445.GC6252@cochise.anu.edu.au>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <20061102011002.GC6041@cochise.anu.edu.au>
	 <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
	 <20061108025445.GC6252@cochise.anu.edu.au>
Message-ID: <d3dac270611091411t71b55c5fl1087e070509b8fee@mail.gmail.com>

On 11/7/06, Joerg Arndt <arndt at jjj.de> wrote:
> * Jonathan Post <jvospost3 at gmail.com> [Nov 08. 2006 13:29]:
> > Thank you, Joerg.  That was interesting.  Does your generating function
> > hypothesis hold when I extend the sequence by another 5 values, to a(n) = 0, 0,
> > 1, 1, 3, 1, 7, 4,
> > 14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again, assuming that
> > I got the elementary algebra and arithmetic correct.
>
> Alas, the function I gave was spurious, with the
> additional terms no OGF is found.

Not surprising at all!
This sequence is not well-defined (see above in this discussion), so
one may view its terms as random numbers. How can they have an OGF? ;)

Max





From jvospost3 at gmail.com  Thu Nov  9 23:15:12 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 9 Nov 2006 14:15:12 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <d3dac270611091411t71b55c5fl1087e070509b8fee@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <20061102011002.GC6041@cochise.anu.edu.au>
	 <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
	 <20061108025445.GC6252@cochise.anu.edu.au>
	 <d3dac270611091411t71b55c5fl1087e070509b8fee@mail.gmail.com>
Message-ID: <5542af940611091415m4db107dei8659adb11bdcbffa@mail.gmail.com>

Max's position is well taken.  I think that "random" overstates it
somewhat.  I think that even careful definition leaves a residuum of chaos
in the proper sense.  I have not yet posted any seq based on this, out of
caution.  When I do, I'll probably post just one and see what happens.

On 11/9/06, Max A. <maxale at gmail.com> wrote:
>
> On 11/7/06, Joerg Arndt <arndt at jjj.de> wrote:
> > * Jonathan Post <jvospost3 at gmail.com> [Nov 08. 2006 13:29]:
> > > Thank you, Joerg.  That was interesting.  Does your generating
> function
> > > hypothesis hold when I extend the sequence by another 5 values, to
> a(n) = 0, 0,
> > > 1, 1, 3, 1, 7, 4,
> > > 14, 17, 28, 54, 66, 143, 182, 350, 687, 987, 2611, 4298 ? Again,
> assuming that
> > > I got the elementary algebra and arithmetic correct.
> >
> > Alas, the function I gave was spurious, with the
> > additional terms no OGF is found.
>
> Not surprising at all!
> This sequence is not well-defined (see above in this discussion), so
> one may view its terms as random numbers. How can they have an OGF? ;)
>
> Max
>
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From maxale at gmail.com  Thu Nov  9 23:41:15 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 9 Nov 2006 14:41:15 -0800
Subject: Eisenstein-Fibonacci sequences
In-Reply-To: <5542af940611091415m4db107dei8659adb11bdcbffa@mail.gmail.com>
References: <5542af940610221358k58b10d14q25d0d9825ed04974@mail.gmail.com>
	 <20061102011002.GC6041@cochise.anu.edu.au>
	 <5542af940611012344s6b5c6218j6849a4e1d0493d3d@mail.gmail.com>
	 <20061108025445.GC6252@cochise.anu.edu.au>
	 <d3dac270611091411t71b55c5fl1087e070509b8fee@mail.gmail.com>
	 <5542af940611091415m4db107dei8659adb11bdcbffa@mail.gmail.com>
Message-ID: <d3dac270611091441la822b41v6a324c49ec2a1a76@mail.gmail.com>

On 11/9/06, Jonathan Post <jvospost3 at gmail.com> wrote:

> Max's position is well taken.  I think that "random" overstates it somewhat.

OK, it may not be random from your point of view since you followed
some rules (I believe) defining the terms of this sequence. The
problem is that you did not let us know these rules!
Remember the example I gave above? It was about different
representations of b(1) as 2-nd degree polynomials of w:
b(1) = w^2 = 2w^2+w+1 = 3w^2+2w+2 = ...
They all are equal (since w=(-1 + i*sqrt(3))/2) but at the same time
they all have different coefficients as polynomials of w.
For some (unknown to us) reason, you chose the first representation
that gave the 1st term of your sequence equal 0 (as the coefficient of
w^0 in b(1)=w^2). But you chose b(1)=2w^2+w+1, the first term would be
1; if you chose b(1)=3w^2+2w+2, it would be 2, etc.

There is yet another way to define your sequence precisely. This time
let's forget the exact value of w and keep in mind only that w^3 = 1.
Then every element of the sequence b(0) = w, b(1) = w^2, b(n) =
w*b(n-1) + b(n-2) have unique representation as a 2-nd degree
polynomial of w:
b(2) = -x + 1
b(3) = -2*x^2 + x
b(4) = x^2 + x - 3
b(5) = 3*x^2 - 4*x + 1
b(6) = -5*x^2 + 6
b(7) = -3*x^2 + 10*x - 6
b(8) = 15*x^2 - 6*x - 9
...
This gives us the following sequence of coefficients of w^0:
0, 0, 1, 0, -3, 1, 6, -6, -9
But it is different from your original sequence (again).

Max





From franktaw at netscape.net  Fri Nov 10 00:40:10 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 09 Nov 2006 18:40:10 -0500
Subject: Incorrect conjecture
Message-ID: <8C8D26F876C6EFF-76C-A121@WEBMAIL-MA17.sysops.aol.com>

The conjecture in http://www.research.att.com/~njas/sequences/A092903 
is wrong.  This conjecture is that A092903 is not the same as A005153.  
Translated, the opposite of this conjecture is that if every number up 
to m can be expressed as the sum of distinct divisors of m, then every 
number up to sigma(m) can be.  (The latter condition is known as m is 
practical.)

Following the link from A005153 to 
http://planetmath.org/encyclopedia/PracticalNumber.html, we find the 
lemma (paraphrased for non-graphical presentation): an integer m >= 2 
with factorization Product_{i=1}^k p_i^e_i with the p_i in ascending 
order is practical if and only if p_1 = 2 and, for 1 < i <= k, p_i <= 
sigma(Product_{j < i} p_j^e_j) + 1.

Proof: Let m =  Product_{i=1}^k p_i^e_i  be a number not of this form.  
If p_1 != 2, 2 cannot be represented as a sum.  If p_i > 
sigma(Product_{j<i} p_j^e_j) + 1, then sigma(Product_{j<i} p_j^e_j) + 1 
cannot be represented as the sum, and since this number is less than 
p_i, it is less than m.  Hence not every number up to m can be 
expressed as the sum of distinct divisors of m.   Thus if every number 
up to m is expressible as the sum of distinct divisors, so is every 
number up to sigma(m).  The converse is trivial, so the conditions are 
equivalent.

Hence A092903 is a duplicate of A005153.

Franklin T. Adams-Watters

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From davidwwilson at comcast.net  Fri Nov 10 03:06:32 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 9 Nov 2006 21:06:32 -0500
Subject: Explanation of A123937
Message-ID: <002f01c7046c$d7f7f880$6401a8c0@yourxhtr8hvc4p>

This is a hurried attempt to provide background on the genesis of A123937.  I am sure that a most of what I write here is a poor rehash of basic theory, and this whole post probably equates to a few lines of generating function theory.

For d and n in Z+, let A_d(n) be the set of integer points on the d-sphere of radius sqrt(n) centered at the origin, that is:

    [1] a_d(n) = #({P in Z^d : |P|^2 = n}).

A table of a_d(n) for a few small values of d and n:

Table of a_d(n)

 \d|  0    1    2    3    4    5     6     7      8      9      10
 n\|
---+--------------------------------------------------------------
 0 |  1    1    1    1    1    1     1     1      1      1       1
 1 |  0    2    4    6    8   10    12    14     16     18      20
 2 |  0    0    4   12   24   40    60    84    112    144     180
 3 |  0    0    0    8   32   80   160   280    448    672     960
 4 |  0    2    4    6   24   90   252   574   1136   2034    3380
 5 |  0    0    8   24   48  112   312   840   2016   4320    8424
 6 |  0    0    0   24   96  240   544  1288   3136   7392   16320
 7 |  0    0    0    0   64  320   960  2368   5504  12672   28800
 8 |  0    0    4   12   24  200  1020  3444   9328  22608   52020
 9 |  0    2    4   30  104  250   876  3542  12112  34802   88660
10 |  0    0    8   24  144  560  1560  4424  14112  44640  129064

Each sequence a_d is a column of this table. All of these column sequences already exist in the OEIS.

It turns out that a_d can be gotten by convolving the sequence j = (1,2,0,0,2,...) = A000122 d times with the convolutional identity sequence e = (1,0,0,0,0,...) = A000007. We could say that a_d = j^d as a "convolutional power".

The generating function of a_d is t(x)^d where t is the generating function of j. t turns out to be the Jacobi theta function theta_3, as noted on A000122. If we apply the finite difference transform to the sequence of generating functions {t(x)^d}, we obtain the sequence {(t(x)-1)^d}. Expanding these functions back to sequences gives the sequence of sequences {b_d} where b_d = (j-e)^d as a convolutional power. This allows us to constuct a table of b_d:

Table of b_d(n)

 \d|  0    1    2    3    4    5    6    7    8    9    10
 n\|
---+------------------------------------------------------
 0 |  1    0    0    0    0    0    0    0    0    0     0
 1 |  0    2    0    0    0    0    0    0    0    0     0
 2 |  0    0    4    0    0    0    0    0    0    0     0
 3 |  0    0    0    8    0    0    0    0    0    0     0
 4 |  0    2    0    0   16    0    0    0    0    0     0
 5 |  0    0    8    0    0   32    0    0    0    0     0
 6 |  0    0    0   24    0    0   64    0    0    0     0
 7 |  0    0    0    0   64    0    0  128    0    0     0
 8 |  0    0    4    0    0  160    0    0  256    0     0
 9 |  0    2    0   24    0    0  384    0    0  512     0
10 |  0    0    8    0   96    0    0  896    0    0  1024

It turns out that the process we used to get from the a_d table to the b_d table, namely, converting columns of a_d to generating functions, applying the finite difference transform to the gf's, and expanding the resulting functions back to columns of b_d, is the same as applying the finite difference transform to the rows of the a_d table to obtain rows of the b_d table.

An inductive proof starting with b_d = (j-e)^d where j-e = (0,2,0,0,2,...) shows that b_d(d) = 2^d and b_d(n) = 0 for n < d. This means that the nth element of the nth row of the b_d table is the last nonzero element in that row, which in turn implies that the nth row of the a_d table, that is, the number of integer points P with |P|^2 = n, is a polynomial of degree n in d.

If, instead of [1], we had started with

    [1] a_d'(n) = #({P in Z^d : |P|^2 <= n})

which counts points inside as well as on the d-sphere, the nth row of a_d' table would be gotten by adding termwise the rows 0 through n of the a_d table. The sequence a_d' has generating function t(x)^d/(1-x). Similarly, the nth row of the corresponding b_d' table is the sum of rows 0 through n of the b_d table, and sequence b_d has generating function (t(x)-1)^d)/(1-x). The table of b_d', omitting the superdiagonal zeroes, is sequence A123937.

The recurrence given in the title of A123937 follows from the relation

    b_d = (1,1,1,1,1,...) = A000012 if d = 0; (j-e) * b_{d-1}' if d >= 1.

I know this is a spotty exposition, but I'm sure the loose ends could be tied together by a real number theorist.

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From zakseidov at yahoo.com  Sat Nov 11 14:53:18 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sat, 11 Nov 2006 05:53:18 -0800 (PST)
Subject: A052530: to be edited
Message-ID: <20061111135318.71618.qmail@web38206.mail.mud.yahoo.com>

Both %F's and %t don't accord with entries %S 
(mainly with first term of %S),
and note this misprint(?):
"a(n)-4*a(n+1)+a(n+2)})"

%S A052530
1,2,8,30,112,418,1560,5822,21728,81090,302632,

%F A052530 a(0) = c, a(1) = p*c^3; a(n+2) =
p*c^2*a(n+1) - a(n), for p = 1, c = 2.

%F A052530 Recurrence: {a(0)=1, a(1)=2, a(2)=8, 
a(n)-4*a(n+1)+a(n+2)}

%t A052530 a[0] = c; a[1] = p*c^3; a[n_] := a[n] =
p*c^2*a[n - 1] - a[n - 2]; p = 1; c = 2; 
Table[ a[n], {n, 0, 20} ]

Anyone may wish to fix it? 
Zak


 
____________________________________________________________________________________
Want to start your own business?
Learn how on Yahoo! Small Business.
http://smallbusiness.yahoo.com/r-index





From davidwwilson at comcast.net  Sat Nov 11 16:34:08 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Sat, 11 Nov 2006 10:34:08 -0500
Subject: A124057 is subset of A045940
References: <20061105235118.90653.qmail@web38203.mail.mud.yahoo.com>
Message-ID: <004501c705a6$d4c0bea0$6501a8c0@yourxhtr8hvc4p>

Nah, there's lots of them:

Yah, lots:

8706123 24463374 32442848 32942943 36782289 48580623 55486248 57476573
59600365 59757774 62481222 62664810 62884590 63262374 63728124 64724373
65159575 65450824 69362487 70302087 70370223 70785924 71494773 72060272
72503682 73256910 73638422 74066874 74361858 75356070 75414702 78676623
83349123 84405123 88094895 92349423 93803982 96192782 97757955 97822374
98474661 99104550 100708374 100997575 101302623 101905622 101905623
102956103 105596334 110476143 111673374 112529493 112801974 113067422
114811332 114940375 115338894 115466874 116079774 118269150 119076174
120919623 123052110 123821970 123952542 124045207 124232373 124465624
124581534 126115623 126134874 126619623 127151127 128095749 129079248
129764789 131404623 131422575 132568225 132617814 133891623 133918134
134350623 134421174 135691143 138468069 142680174 142904430 142904431
145863150 148286886 151757655 152779624 153101583 153129248 153930510
154322523 155740023 155826423 158252994


----- Original Message ----- 
From: "zak seidov" <zakseidov at yahoo.com>
To: <seqfan at ext.jussieu.fr>
Sent: Sunday, November 05, 2006 6:51 PM
Subject: Re: A124057 is subset of A045940


> Just submitted:
> 
> My Q to seqfans:
> Really, there are no numbers n such that n, n+1, n+2
> and n+3 are products of 6 primes?
> No for n<7258290.
> Thanks, Zak






From njas at research.att.com  Sat Nov 11 22:52:30 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sat, 11 Nov 2006 16:52:30 -0500 (EST)
Subject: querying multidimensional sequences in OEIS
References: <d3dac270611030108q1072b47ex3c065a53e4c61348@mail.gmail.com>
    <8C8CD7271AC8D39-FD0-B08C@WEBMAIL-MA20.sysops.aol.com>
    <8C8CDAB62830D6B-12C-D5A@WEBMAIL-MA20.sysops.aol.com>
Message-ID: <200611112152.QAA93259@fry.research.att.com>

Max, the other day you said:


Of course, there are many representations possible but one of them
should be called "standard", and every multidimensional sequence
should have this standard representation (among maybe others) in OEIS.

Neil, does that sound reasonable?


Me:  it is very hard to force a standard representation - there are
too many people who use the OEIS and don't spend much time
reading rules

even such a simple thing as the "offset" and "keywords" gives
a lot of people trouble.

in short, it would not work

Neil





From njas at research.att.com  Sun Nov 12 04:40:14 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sat, 11 Nov 2006 22:40:14 -0500 (EST)
Subject: Thanks to Tony Noe.
Message-ID: <200611120340.WAA18989@fry.research.att.com>


Seqfans:

in the past few weeks Tony Noe has corrected several hundred entries 
in the OEIS.  the kinds of errors he has found include numbers
broken by commas, numbers run together, dropped digits, duplicated 
numbers, and just plain wrong entries.

It is astonishing to me that there were so many errors.

And it suggests that there must be thousands of other errors
waiting to be found.

I would like to thank him for all his work.   Incidentally he
has also added hundreds of b-files.

Right now I am in the middle of a large batch of updates - there
will be a new version of the OEIS by dawn.

Tony:  thanks!


Neil





From neoneye at gmail.com  Sun Nov 12 08:27:28 2006
From: neoneye at gmail.com (Simon Strandgaard)
Date: Sun, 12 Nov 2006 08:27:28 +0100
Subject: growing binary trees
Message-ID: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>

I made up an interesting sequence:

1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 26, 33, 41,
50, 62, 77, 94, 115, 142, 174, 212, 260, 319,
389, 475, 582, 711, 867, 1060, 1296


Its a binary tree where leafs is getting
added when a nodes internal counter reaches 3.





below is my slow ruby program:

class Node
  def initialize
    @n = 1
    @childs = []
  end
  def count
    @childs.inject(@n){|n,child| n + child.count}
  end
  def grow
    return @n += 1 if @n < 3
    @childs.each{|child| child.grow }
    @childs << Node.new if @childs.size < 2
  end
end

result = []
node = Node.new
30.times do |i|
  result << node.count
  node.grow
end
p result







# btw: the grow method below outputs A000071: Fibonacci numbers - 1.
def grow
  return @n += 1 if @n < 1
  @childs.each{|child| child.grow }
  @childs << Node.new if @childs.size < 2
end


# btw: the grow method below outputs A054405: Row sums of array T as in A055215.
def grow
  return @n += 1 if @n < 2
  @childs.each{|child| child.grow }
  @childs << Node.new if @childs.size < 2
end




--
Simon Strandgaard





From ralf at ark.in-berlin.de  Sun Nov 12 10:09:02 2006
From: ralf at ark.in-berlin.de (Ralf Stephan)
Date: Sun, 12 Nov 2006 10:09:02 +0100
Subject: growing binary trees
In-Reply-To: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
Message-ID: <20061112090902.GA1205@ark.in-berlin.de>

> I made up an interesting sequence:
> 
> 1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 26, 33, 41,
> 50, 62, 77, 94, 115, 142, 174, 212, 260, 319,
> 389, 475, 582, 711, 867, 1060, 1296

According to superseeker:

                                     2      3      4
                        6 + 2 x + 2 x  - 3 x  - 4 x
                     [- ----------------------------, ogf]
                                5            3
                              -x  - 1 + x + x

Second cumulative sums of:

%I A017817
%S A017817 1,0,0,1,1,0,1,2,1,1,3,3,2,4,6,5,6,10,11,11,16,21,22,27,37,43,49,64,80,
%T A017817 92,113,144,172,205,257,316,377,462,573,693,839,1035,1266,1532,1874,
%U A017817 2301,2798,3406,4175,5099,6204,7581,9274,11303,13785,16855,20577,25088
%N A017817 a(0)=1, a(1)=a(2)=0, a(3)=1; a(n)=a(n-3)+a(n-4).
%F A017817 G.f.: 1/(1-x^3-x^4).
%F A017817 a(n)/a(n-1) tends to r = 1.2207440846..., a real root of x^4 - x - 1 = 0. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Oct 22 2006

First sums of:

Transformation T018 gave a match with:
%I A122511
%S A122511 1,1,1,2,2,2,3,4,4,5,7,8,9,12,15,17,21,27,32,38,48,59,70,86,107,129,156,
%T A122511 193,236,285,349,429,521,634,778,950,1155,1412,1728,2105,2567,3140,3833,
%U A122511 4672,5707,6973,8505,10379,12680,15478
%N A122511 Alternative method for A079398 using vector matrirx Markov.
%K A122511 nonn,uned
%A A122511 Roger Bagula (rlbagula(AT)sbcglobal.net), Sep 16 2006

which is IDENTICAL with

%I A079398
%S A079398 0,1,1,1,1,2,2,2,3,4,4,5,7,8,9,12,15,17,21,27,32,38,48,59,70,86,107,129,
%T A079398 156,193,236,285,349,429,521,634,778,950,1155,1412,1728,2105,2567,3140,
%U A079398 3833,4672,5707,6973,8505,10379,12680,15478,18884,23059,28158,34362
%N A079398 Conjectured values of d(n), the dimension of a Z-module in MZV(conv).
%C A079398 From the conjecture of Zagier, Drinfeld, Kontsevich and Goncharov (see link).
%C A079398 P(0)=P(1)=P(2)=P(3)=1, for m>3: P(m) = P(m-3) + P(m-4) is the 3rd sequence in the series: Fibonacci sequence, Padovan sequence, ... The Padovan sequence (whose ratio of successive terms approaches the plastic constant) is similar to the Perrin sequence. - Jonathan Vos Post (jvospost2(AT)yahoo.com), Jan 23 2005
%D A079398 Michel Waldschmidt, "Multiple Zeta values and Euler-Zagier numbers", in Number theory and discrete mathematics, International conference in honour of Srinivasa Ramanujan, Center for Advanced Study in Mathematics, Panjab University, Chandigarh, (Oct 02, 2000).
%H A079398 Michel Waldschmidt, <a href="http://www.institut.math.jussieu.fr/~miw/articles/pdf/Transparents.pdf">Multiple Zeta values and Euler-Zagier numbers</a>
%H A079398 Eric W. Weisstein, <a href="http://mathworld.wolfram.com/PadovanSequence.html">Padovan Sequence</a>.
%F A079398 a(1)=0 a(2)=a(3)=a(4)=1 for n>=4 a(n)=a(n-2)+a(n-3)
%F A079398 a(n)=sum{k=0..floor((n-1)/2), binomial(floor((n-k-1)/3), k)} (offset 0); a(n)=sum{k=0..floor(n/2), binomial(floor((n-k-1)/3), k)}-0^n. (offset 0). - Paul Barry (pbarry(AT)wit.ie), Jul 06 2004
%F A079398 For n>1, a(n) = P(n-2) where P(n) is defined by: P(0)=P(1)=P(2)=P(3)=1, for m>3: P(m) = P(m-3) + P(m-4). - Jonathan Vos Post (jvospost2(AT)yahoo.com), Jan 23 2005
%Y A079398 Cf. A000931.
%K A079398 nonn
%O A079398 1,6
%A A079398 Benoit Cloitre (abmt(AT)wanadoo.fr), Feb 16 2003


ralf





From neoneye at gmail.com  Sun Nov 12 16:54:56 2006
From: neoneye at gmail.com (Simon Strandgaard)
Date: Sun, 12 Nov 2006 16:54:56 +0100
Subject: growing binary trees
In-Reply-To: <20061112090902.GA1205@ark.in-berlin.de>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
	 <20061112090902.GA1205@ark.in-berlin.de>
Message-ID: <df1390cc0611120754n68a94a44k76bc07615c3bdb98@mail.gmail.com>

On 11/12/06, Ralf Stephan <ralf at ark.in-berlin.de> wrote:
> > I made up an interesting sequence:
> >
> > 1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 26, 33, 41,
> > 50, 62, 77, 94, 115, 142, 174, 212, 260, 319,
> > 389, 475, 582, 711, 867, 1060, 1296
>
> According to superseeker:
[snip]

I don't see how its already in oeis?

How is A017817, A122511, A079398 related to it?
There is no match between my numbers and their numbers.
There doesn't seem to be any constant I can add to
make them match. Im an amateur.


--
Simon Strandgaard





From ralf at ark.in-berlin.de  Sun Nov 12 17:18:14 2006
From: ralf at ark.in-berlin.de (Ralf Stephan)
Date: Sun, 12 Nov 2006 17:18:14 +0100
Subject: growing binary trees
In-Reply-To: <20061112090902.GA1205@ark.in-berlin.de>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com> <20061112090902.GA1205@ark.in-berlin.de>
Message-ID: <20061112161814.GA11423@ark.in-berlin.de>

Simon, sorry for the cryptic reply.

In other words:

Let n>=0, and your sequence a(n) = 1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 
26, 33, 41, 50, 62, 77, 94, 115, 142, 174, 212, 260, 319, ...

then 

A079398(n) = a(n-2) - a(n-1) 
A122511(n) = a(n+1) - a(n)      
A017817(n) = A122511(n+3) - A122511(n+2)

All of these sequences satisfy the recurrence
a(n) = a(n-2) + a(n-3) + c

and are thus related to the Padovan sequence A000931.
With your sequence, c appears to be 3 but this has
to be proved, and all of the above is only a conjecture,
based on solid evidence, though.


ralf







From neoneye at gmail.com  Sun Nov 12 17:30:14 2006
From: neoneye at gmail.com (Simon Strandgaard)
Date: Sun, 12 Nov 2006 17:30:14 +0100
Subject: growing binary trees
In-Reply-To: <20061112161814.GA11423@ark.in-berlin.de>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
	 <20061112090902.GA1205@ark.in-berlin.de>
	 <20061112161814.GA11423@ark.in-berlin.de>
Message-ID: <df1390cc0611120830u79249ff5ue1f9c588cde122d2@mail.gmail.com>

On 11/12/06, Ralf Stephan <ralf at ark.in-berlin.de> wrote:
[snip]
> All of these sequences satisfy the recurrence
> a(n) = a(n-2) + a(n-3) + c
>
> and are thus related to the Padovan sequence A000931.
> With your sequence, c appears to be 3 but this has
> to be proved, and all of the above is only a conjecture,
> based on solid evidence, though.
[snip]

I get it. Thanks for the explaination :-)

--
Simon Strandgaard





From njas at research.att.com  Sun Nov 12 17:38:06 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sun, 12 Nov 2006 11:38:06 -0500 (EST)
Subject: growing binary trees
Message-ID: <200611121638.LAA08650@fry.research.att.com>

I will add Simon S.'s sequence to the OEIS - it will be A123015 
in a few minutes

Can Ruby lines be combined onto one line?

NJAS





From njas at research.att.com  Sun Nov 12 18:22:45 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sun, 12 Nov 2006 12:22:45 -0500 (EST)
Subject: growing binary trees
Message-ID: <200611121722.MAA17909@fry.research.att.com>

About this sequence (it will be A123015):


1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 26, 33, 41,
50, 62, 77, 94, 115, 142, 174, 212, 260, 319,
389, 475, 582, 711, 867, 1060, 1296


Its a binary tree where leafs is getting
added when a nodes internal counter reaches 3.

- I really don't understand the definition.  What exactly is the
n-th term?

Neil Sloane





From neoneye at gmail.com  Sun Nov 12 19:10:23 2006
From: neoneye at gmail.com (Simon Strandgaard)
Date: Sun, 12 Nov 2006 19:10:23 +0100
Subject: growing binary trees
In-Reply-To: <200611121638.LAA08650@fry.research.att.com>
References: <200611121638.LAA08650@fry.research.att.com>
Message-ID: <df1390cc0611121010r2cf66135gfa52aff18c944540@mail.gmail.com>

On 11/12/06, N. J. A. Sloane <njas at research.att.com> wrote:
> I will add Simon S.'s sequence to the OEIS - it will be A123015
> in a few minutes
>
> Can Ruby lines be combined onto one line?


I cannot express this thing as math, and its difficult for me to
compact the code without loss of readability.


below is a 5 line version:

class Node; def initialize; @n = 1; @c = [] end
def count; @c.inject(@n){|n,c| n + c.count} end
def grow; return @n += 1 if @n < 3; @c.each{|c| c.grow }
@c << Node.new if @c.size < 2; end; end; r = []; node = Node.new
30.times { r << node.count; node.grow }; p r


here is the oneliner (less readable):

class C;def initialize;@n=1;@c=[] end;def n;@c.inject(@n){|n,c|n+c.n}
end;def g;return @n+=1 if @n<3; @c.each{|c|c.g};@c<<C.new if
@c.size<2;end;end;r=[];c=C.new;30.times{r<<c.n;c.g};p r



the n-th term is:
a(0)=1, a(1)=2, a(2)=3, a(3)=4, a(4)=6, a(5)=8, a(6)=10, a(7)=13, ...


code for n'th term is below:

class Node
 def initialize
   @n = 1
   @childs = []
 end
 def count
   @childs.inject(@n){|n,child| n + child.count}
 end
 def grow
   return @n += 1 if @n < 3
   @childs.each{|child| child.grow }
   @childs << Node.new if @childs.size < 2
 end
end

def a(n)
  node = Node.new
  n.times{ node.grow }
  node.count
end

0.upto(30) {|n| p a(n) }



visually it works like this:

step#0
  1


step#1
  2


step#2
  3


step#3
  3
 /
1


step#4
  3
 / \
2   1


step#5
  3
 / \
3   2


step#6
    3
   / \
  3   3
 /
1


step#7
      3
    /   \
  3       3
 / \     /
2   1   1


step#8
      3
    /   \
  3       3
 / \     / \
3   2   2   1



--
Simon Strandgaard





From maxale at gmail.com  Mon Nov 13 07:53:16 2006
From: maxale at gmail.com (Max A.)
Date: Sun, 12 Nov 2006 22:53:16 -0800
Subject: growing binary trees
In-Reply-To: <20061112090902.GA1205@ark.in-berlin.de>
References: <df1390cc0611112327jdf924eej15ce5b969dcaa4b1@mail.gmail.com>
	 <20061112090902.GA1205@ark.in-berlin.de>
Message-ID: <d3dac270611122253m4930fb58qdf5d06ce4fbf8026@mail.gmail.com>

On 11/12/06, Ralf Stephan <ralf at ark.in-berlin.de> wrote:

> %I A122511
> %S A122511 1,1,1,2,2,2,3,4,4,5,7,8,9,12,15,17,21,27,32,38,48,59,70,86,107,129,156,
> %T A122511 193,236,285,349,429,521,634,778,950,1155,1412,1728,2105,2567,3140,3833,
> %U A122511 4672,5707,6973,8505,10379,12680,15478
> %N A122511 Alternative method for A079398 using vector matrirx Markov.
> %K A122511 nonn,uned
> %A A122511 Roger Bagula (rlbagula(AT)sbcglobal.net), Sep 16 2006
>
> which is IDENTICAL with
>
> %I A079398
> %S A079398 0,1,1,1,1,2,2,2,3,4,4,5,7,8,9,12,15,17,21,27,32,38,48,59,70,86,107,129,
> %T A079398 156,193,236,285,349,429,521,634,778,950,1155,1412,1728,2105,2567,3140,
> %U A079398 3833,4672,5707,6973,8505,10379,12680,15478,18884,23059,28158,34362
> %N A079398 Conjectured values of d(n), the dimension of a Z-module in MZV(conv).
> %C A079398 From the conjecture of Zagier, Drinfeld, Kontsevich and Goncharov (see link).
> %C A079398 P(0)=P(1)=P(2)=P(3)=1, for m>3: P(m) = P(m-3) + P(m-4) is the 3rd sequence in the series: Fibonacci sequence, Padovan sequence, ... The Padovan sequence (whose ratio of successive terms approaches the plastic constant) is similar to the Perrin sequence. - Jonathan Vos Post (jvospost2(AT)yahoo.com), Jan 23 2005
> %D A079398 Michel Waldschmidt, "Multiple Zeta values and Euler-Zagier numbers", in Number theory and discrete mathematics, International conference in honour of Srinivasa Ramanujan, Center for Advanced Study in Mathematics, Panjab University, Chandigarh, (Oct 02, 2000).
> %H A079398 Michel Waldschmidt, <a href="http://www.institut.math.jussieu.fr/~miw/articles/pdf/Transparents.pdf">Multiple Zeta values and Euler-Zagier numbers</a>
> %H A079398 Eric W. Weisstein, <a href="http://mathworld.wolfram.com/PadovanSequence.html">Padovan Sequence</a>.
> %F A079398 a(1)=0 a(2)=a(3)=a(4)=1 for n>=4 a(n)=a(n-2)+a(n-3)
> %F A079398 a(n)=sum{k=0..floor((n-1)/2), binomial(floor((n-k-1)/3), k)} (offset 0); a(n)=sum{k=0..floor(n/2), binomial(floor((n-k-1)/3), k)}-0^n. (offset 0). - Paul Barry (pbarry(AT)wit.ie), Jul 06 2004
> %F A079398 For n>1, a(n) = P(n-2) where P(n) is defined by: P(0)=P(1)=P(2)=P(3)=1, for m>3: P(m) = P(m-3) + P(m-4). - Jonathan Vos Post (jvospost2(AT)yahoo.com), Jan 23 2005
> %Y A079398 Cf. A000931.
> %K A079398 nonn
> %O A079398 1,6
> %A A079398 Benoit Cloitre (abmt(AT)wanadoo.fr), Feb 16 2003

Should not then A122511 be marked as a duplicate of A079398 with an
appropriate comment?
I do not see much sense in keeping them separate.

Max





From davidwwilson at comcast.net  Mon Nov 13 18:50:32 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Mon, 13 Nov 2006 12:50:32 -0500
Subject: A102567 = A106498
Message-ID: <000601c7074c$375f4d50$6501a8c0@yourxhtr8hvc4p>

A102567 and A106498 are duplicates and should be merged. The latter sequence has an associated b-file.
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From maxale at gmail.com  Mon Nov 13 19:24:29 2006
From: maxale at gmail.com (Max A.)
Date: Mon, 13 Nov 2006 10:24:29 -0800
Subject: A050259 and a new solution to 2^n = 3 (mod n)
Message-ID: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>

SeqFans,

In course of exploring all solutions of the congruence 2^n = 3 (mod n)
below 10^16, I have found a new solution:

n = 3468371109448915

This came as a big surprise to me since my original goal was to prove
that 8365386194032363 is the second term of A050259. But now it
appears to be at least the third term.

So far I've submitted the new solution as a comment to A050259, but
hopefully it will take its true place in this sequence soon.

Regards,
Max

P.S. See also http://mathworld.wolfram.com/2.html





From rkg at cpsc.ucalgary.ca  Mon Nov 13 20:15:20 2006
From: rkg at cpsc.ucalgary.ca (Richard Guy)
Date: Mon, 13 Nov 2006 12:15:20 -0700 (MST)
Subject: A050259 and a new solution to 2^n = 3 (mod n)
In-Reply-To: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>
References: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>
Message-ID: <Pine.LNX.4.63.0611131211270.24385@csl>

Congratulations!  May I know Max's full name,
so that he can earn undying fame in UPINT as
well as in OEIS ?   Thanks,   R.

On Mon, 13 Nov 2006, Max A. wrote:

> SeqFans,
>
> In course of exploring all solutions of the congruence 2^n = 3 
> (mod n)
> below 10^16, I have found a new solution:
>
> n = 3468371109448915
>
> This came as a big surprise to me since my original goal was 
> to prove
> that 8365386194032363 is the second term of A050259. But now 
> it
> appears to be at least the third term.
>
> So far I've submitted the new solution as a comment to 
> A050259, but
> hopefully it will take its true place in this sequence soon.
>
> Regards,
> Max
>
> P.S. See also http://mathworld.wolfram.com/2.html






From jvospost3 at gmail.com  Mon Nov 13 21:34:20 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Mon, 13 Nov 2006 12:34:20 -0800
Subject: A050259 and a new solution to 2^n = 3 (mod n)
In-Reply-To: <Pine.LNX.4.63.0611131211270.24385@csl>
References: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>
	 <Pine.LNX.4.63.0611131211270.24385@csl>
Message-ID: <5542af940611131234l54af8bcar76fcba8e87d0fb1b@mail.gmail.com>

Is this a hierarchy of fame, or a graph, or directed graph, or a lattice, or
what? What percentage of "nice" sequences are referenced in each successive
edition of UPINT (Unsolved problems in Number Theory)? How many OEIS seqs
per edition of UPINT, now that the two are more tightly interlinked? What
estimates does one make to extrapolate for future editions? Should there be
a "very nice" keyword for seqs that pose problems in UPINT, when they get
solved in later editions?

Only hypothetical for me, as I've had over 1400 seqs in OEIS and not one is
yet "nice", although some are comments on "nice" seqs. So if there is a
hierarchy of fame, I'm just above the lowest  level (or highest, depending
on indexing) in quality, which is most assuredly not correlated well with
quantity.  In Science Fiction, for instance, there are indeed a hierarchies
of fame, based on awards (Hugo, Nebula), coauthorships, pay, anthologies,
and the like.  I'm also just somewhat above the bottom, in part for my
coauthorships with Asimov, Bradbury, Clarke, and Heinlein.  I've done
research on Asimov Number as the Science Fiction equivalent of Erdos Number,
but that's another story.

By the way, I have the penultimate edition (so far) of UPINT as a gift from
the wonderful Tony Noe, when he bought the ultimate edition (so far). I echo
the applause by njas for all the hundreds of errors in OEIS found and
corrected by Tony, and the many many b-files.

Best,

Jonathan Vos Post

On 11/13/06, Richard Guy <rkg at cpsc.ucalgary.ca> wrote:
>
> Congratulations!  May I know Max's full name,
> so that he can earn undying fame in UPINT as
> well as in OEIS ?   Thanks,   R.
>
> On Mon, 13 Nov 2006, Max A. wrote:
>
> > SeqFans,
> >
> > In course of exploring all solutions of the congruence 2^n = 3
> > (mod n)
> > below 10^16, I have found a new solution:
> >
> > n = 3468371109448915
> >
> > This came as a big surprise to me since my original goal was
> > to prove
> > that 8365386194032363 is the second term of A050259. But now
> > it
> > appears to be at least the third term.
> >
> > So far I've submitted the new solution as a comment to
> > A050259, but
> > hopefully it will take its true place in this sequence soon.
> >
> > Regards,
> > Max
> >
> > P.S. See also http://mathworld.wolfram.com/2.html
>
>
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From r.rosenthal at web.de  Mon Nov 13 23:12:57 2006
From: r.rosenthal at web.de (Rainer Rosenthal)
Date: Mon, 13 Nov 2006 23:12:57 +0100
Subject: A050259 and a new solution to 2^n = 3 (mod n)
In-Reply-To: <Pine.LNX.4.63.0611131211270.24385@csl>
References: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com> <Pine.LNX.4.63.0611131211270.24385@csl>
Message-ID: <4558EDE9.2080404@web.de>

Richard Guy wrote:
> Congratulations!  May I know Max's full name,
> so that he can earn undying fame in UPINT as
> well as in OEIS ?   Thanks,   R.
> 
> On Mon, 13 Nov 2006, Max A. wrote:
>>In course of exploring all solutions of the congruence 2^n = 3 
>>(mod n) below 10^16, I have found a new solution:
>>
>>n = 3468371109448915
>>P.S. See also http://mathworld.wolfram.com/2.html

This is a lovely finding, which I will have to tell
immediately in the German Newsgroup de.sci.mathematik.

Congratulations from a friend of OEIS and UPINT and
of things that are not too difficult to comprehend :-)
Not only congratulations to Max for finding, but to all
of us for being witnesses.

Best regards,
Rainer Rosenthal
r.rosenthal at web.de






From franktaw at netscape.net  Tue Nov 14 00:01:52 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 13 Nov 2006 18:01:52 -0500
Subject: A102567 = A106498
In-Reply-To: <000601c7074c$375f4d50$6501a8c0@yourxhtr8hvc4p>
References: <000601c7074c$375f4d50$6501a8c0@yourxhtr8hvc4p>
Message-ID: <8C8D58ED7DCDA34-128-9CD7@FWM-R16.sysops.aol.com>

A116148 (n concatenated with n-1 gives the product of two numbers which 
differ by 2) is also a duplicate of these two.  The only possible 
exceptions would be numbers of the form 10^n-1; but then the 
concatenation is 10^{2n}+10^n-1, and 10^{2n}+10^n is never a square.

Franklin T. Adams-Watters


-----Original Message-----
From: davidwwilson at comcast.net

A102567 and A106498 are duplicates and should be merged.?The latter 
sequence has an associated b-file.




________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From maxale at gmail.com  Tue Nov 14 00:52:58 2006
From: maxale at gmail.com (Max A.)
Date: Mon, 13 Nov 2006 15:52:58 -0800
Subject: A050259 and a new solution to 2^n = 3 (mod n)
In-Reply-To: <Pine.LNX.4.63.0611131211270.24385@csl>
References: <d3dac270611131024n559b18e2hd20847652d21f1f5@mail.gmail.com>
	 <Pine.LNX.4.63.0611131211270.24385@csl>
Message-ID: <d3dac270611131552v64b33034n8ed43b34faf905a4@mail.gmail.com>

I answered Richard privately.
But if other people also want to refer to my full name, it is
Max Alekseyev

Regards,
Max

P.S. It is not a conspiracy but rather my way to beat spamming. So
far, spam emails never addressed me by my full name, and I can easily
recognize important emails that do contain my full name in the To:
field.

On 11/13/06, Richard Guy <rkg at cpsc.ucalgary.ca> wrote:
> Congratulations!  May I know Max's full name,
> so that he can earn undying fame in UPINT as
> well as in OEIS ?   Thanks,   R.
>
> On Mon, 13 Nov 2006, Max A. wrote:
>
> > SeqFans,
> >
> > In course of exploring all solutions of the congruence 2^n = 3
> > (mod n)
> > below 10^16, I have found a new solution:
> >
> > n = 3468371109448915
> >
> > This came as a big surprise to me since my original goal was
> > to prove
> > that 8365386194032363 is the second term of A050259. But now
> > it
> > appears to be at least the third term.
> >
> > So far I've submitted the new solution as a comment to
> > A050259, but
> > hopefully it will take its true place in this sequence soon.
> >
> > Regards,
> > Max
> >
> > P.S. See also http://mathworld.wolfram.com/2.html
>
>





From bowerc at usa.net  Tue Nov 14 02:10:04 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Mon, 13 Nov 2006 17:10:04 -0800
Subject: growing binary trees
Message-ID: <397kkNBJe5074S28.1163466604@cmsweb28.cms.usa.net>

 From Simon's diagrams, I can see this is analogous to Fibonacci's
rabbits. Basically, the behavior of the node is given by its age.
A node at age 0 or 1 grows and one at age 2 or 3 produces a new node.

 From this its easy to get a g.f. of

(1+x+x^2)/(1-x-x^3+x^5)

1 2 3 4 6 8 10 13 17 21 26 33 41 50 62 77 94 115 142 174 212 260 319 389
475 582 711 867 1060 1296 1581 1930 2359 2880 3514 4292 5242 6397 7809
9537

Also, the number of nodes has g.f.

1/(1-x-x^3+x^5)
1 1 1 2 3 3 4 6 7 8 11 14 16 20 26 31 37 47 58 69 85 106 128 155 192 235
284 348 428 520 633 777 949 1154 1411 1727 2104 2566 3139 3832

which isn't in the EIS either.

------ Original Message ------
From: "Simon Strandgaard" <neoneye at gmail.com>
To: njas at research.att.comCc: seqfan at ext.jussieu.fr
Subject: Re: growing binary trees

...

> visually it works like this:
> 
> step#0
>   1
> 
> 
> step#1
>   2
> 
> 
> step#2
>   3
> 
> 
> step#3
>   3
>  /
> 1
> 
> 
> step#4
>   3
>  / \
> 2   1
> 
> 
> step#5
>   3
>  / \
> 3   2
> 
> 
> step#6
>     3
>    / \
>   3   3
>  /
> 1
> 
> 
> step#7
>       3
>     /   \
>   3       3
>  / \     /
> 2   1   1
> 
> 
> step#8
>       3
>     /   \
>   3       3
>  / \     / \
> 3   2   2   1
> 
> 
> 
> --
> Simon Strandgaard
> 









From zbi74583 at boat.zero.ad.jp  Tue Nov 14 06:50:02 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Tue, 14 Nov 2006 14:50:02 +0900
Subject: A036471
Message-ID: <013d01c707b0$bedd9660$d967d7dc@FMC57937B6BE00>

    Hi, Seqfans

    An exhaustive search of A036471  with a computer up to 10^7 is almost 
impossible.

    So, using my algorithm, I searched small examples which are ?regular 
type?.



    See this site, there is an explanation of "type".



              http://amicable.homepage.dk/apstat.htm#typesys







    I found these examples.



    2^9*3^2*13*31*(5*11,71)*(7*23,191) 
11 digits

    2^15*3*5^2*11*31*257*(1439,19*71,23*59,29*47) 
16 digits

    2^3*3^4*5^2*(7*23,191)*(17*19,359) 
9 digits

    (2^3*19*41,2^5*199)*(3*5*7*13,3^2*5*7*139) 
8 digits

    (2^3*19*41,2^5*199)*(3^3*5*7*71,3^3*5*17*31) 
9 digits

    (2^3*19*41,2^5*199)*(3^2*7*13*5*17,3^2*7*13*107) 
9 digits

    2^7*3^2*13*(5*11,71)*(7*59,479) 
9 digits

    2^5*3^4*5^2*(17*19,359)*(23*29,719) 
11 digits







    Where, x*(y,z) means (x*z,x*y). (x,y)*(z,u)=(x*z,x*u,y*z,y*z)





    I think that if a smaller one than 3270960  exists, then it must be 
?irregular".



    Regurer type Amicable Quadruple are generated from ?seeds? which are 
two pairs or one quadruple of primes or almost primes, {x,y} and {z,u} or 
{x,y,z,u}.

    The numbers x,y,z,u of small Amicable Quadruple must satisfy the 
following conditions.

             .

    o Sigma(x)=Sigma(y) and Sigma(z)=Sigma(u), or 
Sigma(x)=Sigma(y)=Sigma(z)=Sigma(u)

    o all prime factors of x,y,z,u are small

    o all prime factors of x+y and z+u are small, or all  prime factors of 
x+y+z+u are small



    For example, if {x,y}={7*23,191} and {z,u}={17*19,359} then

    o Sigma(7*23)=Sigma(191), Sigma(17*19)=Sigma(359)

    o all prime factors of 7*23,191,17*19,359  are small

    o x+y=2^5*11, z+u=2*11*31

    So, (7*23,191)*(17*19,359) generates 2^3*3^4*5^2*(7*23,191)*(17*19,359)





    If we obtain all small seeds then we will calculate all small regular 
terms of A036471.



    Yasutoshi


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From g.resta at iit.cnr.it  Tue Nov 14 14:40:49 2006
From: g.resta at iit.cnr.it (Giovanni Resta)
Date: Tue, 14 Nov 2006 14:40:49 +0100
Subject: A036471
In-Reply-To: <013d01c707b0$bedd9660$d967d7dc@FMC57937B6BE00>
References: <013d01c707b0$bedd9660$d967d7dc@FMC57937B6BE00>
Message-ID: <4559C761.70305@iit.cnr.it>

kohmoto wrote:
>
> Hi, Seqfans
>
> An exhaustive search of A036471 with a computer up to 10^7 is almost
> impossible.
>
Here it is the result of an exhaustive search
with a computer up to 10^7, in the format
sigma -> {a,b,c,d} (with a,b,c,d<10^7)

13927680 -> {3270960, 3361680, 3461040, 3834000}
16248960 -> {3767400, 4090320, 4150440, 4240800}
19498752 -> {4651920, 4839120, 4918320, 5089392}
21228480 -> {4969440, 5116320, 5475960, 5666760}
24373440 -> {5682600, 5831280, 5920200, 6939360}
24998400 -> {5405400, 6029100, 6421800, 7142100}
27855360 -> {6514200, 6640200, 6768720, 7932240}
28304640 -> {6126120, 6541920, 7559640, 8076960}
28304640 -> {6126120, 6922080, 7280280, 7976160}
29030400 -> {6320160, 6902280, 7685496, 8122464}
29030400 -> {6977880, 7087080, 7227360, 7738080}
29877120 -> {7013160, 7436520, 7688520, 7738920}
29998080 -> {6819120, 7327320, 7715400, 8136240}
29998080 -> {6966960, 7054320, 7840560, 8136240}
32497920 -> {7706160, 7722000, 7948080, 9121680}
33022080 -> {7731360, 7852320, 8125920, 9312480}
33868800 -> {7469280, 8157240, 8873760, 9368520}
33868800 -> {7469280, 8157240, 9098460, 9143820}

Note that there are values which admit more than one quadruple,
and there are also quadruples which share 1 or 2 common values.

Here are some larger quadruples, but now there can be some
missing among these, because here I used an heuristic.

{8353800, 8920800, 10063200, 10159800}
{8288280, 9360540, 9830520, 10260180}
{8316000, 8759520, 10260180, 10403820}
{9258480, 9621360, 9661680, 10455984}
{9009000, 10174500, 10521000, 11179980}
{10048500, 10174500, 10188360, 10473120}
{9840600, 10098000, 10914120, 10930320}
{9923760, 10316880, 10747440, 10794960}
{9563400, 10085040, 11241720, 11964240}
{9646560, 10667160, 11502540, 11729340}
{9767520, 11202840, 11217960, 11357280}
{9828000, 10796940, 11362680, 12041820}
{10417680, 10538640, 10979280, 13061520}
{10450440, 11614680, 11802420, 13306860}
{10450440, 12179160, 12204360, 12340440}

etc.etc. (program is running).

Giovanni Resta






From davidwwilson at comcast.net  Tue Nov 14 18:00:27 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Tue, 14 Nov 2006 12:00:27 -0500
Subject: A023153 through A023161
References: <p06230918c17f1b479bcf@[192.168.1.43]>
Message-ID: <006d01c7080e$632ff480$6501a8c0@yourxhtr8hvc4p>

Tony Noe has found that A023153 through A023161 are not multiplicative as I 
had marked them.

At the time, I was convinced that in sequence {s_k(n) = k^(2^n)}, a cycle of 
period x (mod a) and a cycle of period y (mod b) implied a unique cycle of 
period xy (mod ab). However, if gcd(x, y) > 1, the cycles (mod a) and (mod 
b) can align in more than one way, producing more than one cycle (mod ab). 
For example, the cycle (2,4) (mod 7) and the cycle (4,7) (mod 9) produce two 
cycles (4,16) and (25,58) (mod 63). This means that sometimes f(ab) > 
f(a)f(b).

This problem affects A023153 through A023161, and Tony Noe has shown that 
indeed none of these are multiplicative.

NJAS: Please remove the "mult" keyword and any comments related to 
multiplicativity from A023153-A023161.

Tony: Since these sequences are almost multiplicative, but are not, and you 
seem to be investigating these sequences, could you please add lines such as

%C A023153 Not multiplicative: a(63) = 10, but a(7)*a(9) = 3*3 = 9.

Thanks for spotting my error, Tony. Seems I have been bitten by the dragon I 
released.

----- Original Message ----- 
From: "T. D. Noe" <noe at sspectra.com>
To: <davidwwilson at comcast.net>
Sent: Tuesday, November 14, 2006 2:06 AM
Subject: Question about A023161


> David, according to my calculations, this sequence fails to be
> multiplicative at a(667).
>
> Best regards,
>
> Tony
>
>
> -- 
> No virus found in this incoming message.
> Checked by AVG Free Edition.
> Version: 7.5.430 / Virus Database: 268.14.5/533 - Release Date: 11/13/2006 
> 8:56 PM
>
> 






From jvospost3 at gmail.com  Tue Nov 14 20:46:00 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Tue, 14 Nov 2006 11:46:00 -0800
Subject: A023153 through A023161
In-Reply-To: <006d01c7080e$632ff480$6501a8c0@yourxhtr8hvc4p>
References: <p06230918c17f1b479bcf@192.168.1.43>
	 <006d01c7080e$632ff480$6501a8c0@yourxhtr8hvc4p>
Message-ID: <5542af940611141146j1a3212b0u2d31815c6bee8387@mail.gmail.com>

"almost multiplicative" -- interesting notion.  What other "almost
multiplicative" are there in OEIS masquerading as multiplicative (i'm
guessing that the amazing Tony Noe is searching), and what are but with no
keyword on that? Can "almost multiplicative" be axiomatizeed, or is it just
a subjective interpretation of "almost"? What pattens are there in "almost
multiplicative" -- new sequences on where multiplicativity breaks down?

-- Jonathan Vos Post

On 11/14/06, David Wilson <davidwwilson at comcast.net> wrote:
>
> Tony Noe has found that A023153 through A023161 are not multiplicative as
> I
> had marked them.
>
> At the time, I was convinced that in sequence {s_k(n) = k^(2^n)}, a cycle
> of
> period x (mod a) and a cycle of period y (mod b) implied a unique cycle of
> period xy (mod ab). However, if gcd(x, y) > 1, the cycles (mod a) and (mod
> b) can align in more than one way, producing more than one cycle (mod ab).
> For example, the cycle (2,4) (mod 7) and the cycle (4,7) (mod 9) produce
> two
> cycles (4,16) and (25,58) (mod 63). This means that sometimes f(ab) >
> f(a)f(b).
>
> This problem affects A023153 through A023161, and Tony Noe has shown that
> indeed none of these are multiplicative.
>
> NJAS: Please remove the "mult" keyword and any comments related to
> multiplicativity from A023153-A023161.
>
> Tony: Since these sequences are almost multiplicative, but are not, and
> you
> seem to be investigating these sequences, could you please add lines such
> as
>
> %C A023153 Not multiplicative: a(63) = 10, but a(7)*a(9) = 3*3 = 9.
>
> Thanks for spotting my error, Tony. Seems I have been bitten by the dragon
> I
> released.
>
> ----- Original Message -----
> From: "T. D. Noe" <noe at sspectra.com>
> To: <davidwwilson at comcast.net>
> Sent: Tuesday, November 14, 2006 2:06 AM
> Subject: Question about A023161
>
>
> > David, according to my calculations, this sequence fails to be
> > multiplicative at a(667).
> >
> > Best regards,
> >
> > Tony
> >
> >
> > --
> > No virus found in this incoming message.
> > Checked by AVG Free Edition.
> > Version: 7.5.430 / Virus Database: 268.14.5/533 - Release Date:
> 11/13/2006
> > 8:56 PM
> >
> >
>
>
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From pauldhanna at juno.com  Wed Nov 15 05:04:51 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Tue, 14 Nov 2006 23:04:51 -0500
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
Message-ID: <20061114.230452.536.0.pauldhanna@juno.com>

Seqfans, 
    Could someone compute the initial terms of the following. 
I do not believe that it is in the OEIS (?). 
  
Number of labelled nodes in generation n of a rooted tree 
where a node with label k has k child nodes such that 
the label of each child node is the least unused label in 
the path from the root to that child node, 
and where root is labeled '1'. 
  
Thus every distinct path in the tree eventually forms 
a permutation of the positive integers. 
 
The wording needs to be made more precise and concise. 
 
Here are the initial nodes of the tree for generations 0..4: 
Gen 0: [1];
Gen 1: (1)->[2];
Gen 2: (2)->[3,4];
Gen 3: (3)->[4,5,6], (4)->[3,5,6,7];
Gen 4: (4)->[5,6,7,8], (5)->[4,6,7,8,9], (6)->[4,5,7,8,9,10], 
(3)->[5,6,7], (5)->[3,6,7,8,9], (6)->[3,5,7,8,9,10], 
(7)->[3,5,6,8,9,10,11]; ...
 
Thus, the number of nodes in generation n begins:
[1, 1, 2, 7, 36, 248, ...]
 
The maximum node in generation n begins:
[1, 2, 4, 7, 11, ...]
 
Would appreciate it if someone could compute more terms.
Thanks,
      Paul





From franktaw at netscape.net  Wed Nov 15 05:13:17 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Tue, 14 Nov 2006 23:13:17 -0500
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
In-Reply-To: <20061114.230452.536.0.pauldhanna@juno.com>
References: <20061114.230452.536.0.pauldhanna@juno.com>
Message-ID: <8C8D68382FC1E11-FEC-53E1@FWM-M20.sysops.aol.com>

You need to reword that a bit.  Your example makes clear that the
children of a given node must all have distinct labels, but you don't
say so.

Without that condition, the problem is pretty easy.  :-)

Franklin T. Adams-Watters


-----Original Message-----
From: pauldhanna at juno.com

...

Number of labelled nodes in generation n of a rooted tree
where a node with label k has k child nodes such that
the label of each child node is the least unused label in
the path from the root to that child node,
and where root is labeled '1'.

...
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From njas at research.att.com  Wed Nov 15 06:13:02 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 15 Nov 2006 00:13:02 -0500 (EST)
Subject: king chickens
Message-ID: <200611150513.AAA96507@fry.research.att.com>

can anyone extend this?

%I A123553
%S A123553 1,2,12,104
%N A123553 A "king chicken" in a tournament graph (a directed labeled graph on n nodes) is a player A who for any other player B either beats B directly or beats someone who beats B.  Sequence gives total number of king chickens in all 2^(n(n-1)/2) tourna
%C A123553 H. G. Landau showed in 1951 that a player is a king chicken if and only if he has the highest score. There may be several king chickens in a tournament.
%D A123553 S. B. Maurer, The king chicken theorems, Math. Mag., 53 (1980), 67-80.
%e A123553 For n = 3 there are 8 tournaments: six of the form A beats B and C, and B beats C, with one king chicken (A), and two of the form A beats B beats C beats A, with three king chickens each (A or B or C), for a total of 6*1 + 2*3 = 12.
%K A123553 nonn,more,new
%O A123553 1,2
%A A123553 njas, Nov 14 2006

Neil





From bdm at cs.anu.edu.au  Wed Nov 15 06:37:18 2006
From: bdm at cs.anu.edu.au (Brendan McKay)
Date: Wed, 15 Nov 2006 16:37:18 +1100
Subject: king chickens
In-Reply-To: <200611150513.AAA96507@fry.research.att.com>
References: <200611150513.AAA96507@fry.research.att.com>
Message-ID: <20061115053717.GA15829@cs.anu.edu.au>


I don't have time right now, so if someone who knows how to use nauty
wants to jump into this, here is how to do it:
1. Fetch the unlabelled tournaments from
      http://cs.anu.edu.au/~bdm/data/digraphs.html
2. For each unlabelled tournament compute the automorphism group size A
   and the number of king chickens K.
3. The answer is the sum of K*(n!/A) over the unlabelled tournaments.

That will get the answer up to n=10 fairly easily.  The next case n=11
is plausible but more effort as there are nearly a billion tournaments.

Incidentally, the definition of "tournament" in the entry is incomplete.
It has to add that exactly one of the directed edges u->v, v->u is
present for each pair u,v of distinct vertices.

Brendan.


* N. J. A. Sloane <njas at research.att.com> [061115 16:13]:
> can anyone extend this?
> 
> %I A123553
> %S A123553 1,2,12,104
> %N A123553 A "king chicken" in a tournament graph (a directed labeled graph on n nodes) is a player A who for any other player B either beats B directly or beats someone who beats B.  Sequence gives total number of king chickens in all 2^(n(n-1)/2) tourna
> %C A123553 H. G. Landau showed in 1951 that a player is a king chicken if and only if he has the highest score. There may be several king chickens in a tournament.
> %D A123553 S. B. Maurer, The king chicken theorems, Math. Mag., 53 (1980), 67-80.
> %e A123553 For n = 3 there are 8 tournaments: six of the form A beats B and C, and B beats C, with one king chicken (A), and two of the form A beats B beats C beats A, with three king chickens each (A or B or C), for a total of 6*1 + 2*3 = 12.
> %K A123553 nonn,more,new
> %O A123553 1,2
> %A A123553 njas, Nov 14 2006
> 
> Neil





From g.resta at iit.cnr.it  Wed Nov 15 11:32:25 2006
From: g.resta at iit.cnr.it (Giovanni Resta)
Date: Wed, 15 Nov 2006 11:32:25 +0100
Subject: A question on A068480 
Message-ID: <455AECB9.5050308@iit.cnr.it>

A068480 is the sequence on numbers n such that gcd(n!-1,2^n+1)>1.

I noticed that for all the numbers listed gcd(n!-1,2^n+1) = 2n+1.

Is this a rule (proof ?) or a coincidence (counterexample ?) .

thanks,
giovanni.







From mathar at strw.leidenuniv.nl  Wed Nov 15 13:40:44 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Wed, 15 Nov 2006 13:40:44 +0100
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
Message-ID: <200611151240.kAFCeiKc009825@amer.strw.leidenuniv.nl>


pdh> From seqfan-owner at ext.jussieu.fr  Wed Nov 15 05:08:51 2006
pdh> To: seqfan at ext.jussieu.fr
pdh> Date: Tue, 14 Nov 2006 23:04:51 -0500
pdh> Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
pdh> From: "Paul D. Hanna" <pauldhanna at juno.com>
pdh> ...
pdh> Number of labelled nodes in generation n of a rooted tree 
pdh> where a node with label k has k child nodes such that 
pdh> the label of each child node is the least unused label in 
pdh> the path from the root to that child node, 
pdh> and where root is labeled '1'. 
pdh>   
pdh> Thus every distinct path in the tree eventually forms 
pdh> a permutation of the positive integers. 
pdh>  
pdh> The wording needs to be made more precise and concise. 
pdh>  
pdh> Here are the initial nodes of the tree for generations 0..4: 
pdh> Gen 0: [1];
pdh> Gen 1: (1)->[2];
pdh> Gen 2: (2)->[3,4];
pdh> Gen 3: (3)->[4,5,6], (4)->[3,5,6,7];
pdh> Gen 4: (4)->[5,6,7,8], (5)->[4,6,7,8,9], (6)->[4,5,7,8,9,10], 
pdh> (3)->[5,6,7], (5)->[3,6,7,8,9], (6)->[3,5,7,8,9,10], 
pdh> (7)->[3,5,6,8,9,10,11]; ...
pdh>  
pdh> Thus, the number of nodes in generation n begins:
pdh> [1, 1, 2, 7, 36, 248, ...]
pdh>  
pdh> The maximum node in generation n begins:
pdh> [1, 2, 4, 7, 11, ...]

My count of node numbers and maximum nodes is
Gen 0 1 1
Gen 1 1 2
Gen 2 2 4
Gen 3 7 7
Gen 4 36 11
Gen 5 248 16
Gen 6 2157 22
Gen 7 22761 29
Gen 8 283220 37
Gen 9 4068411 46

This list is limited by the simplicity of my program which keeps all the
trees in memory at the same time and runs into allocation problems if the
tenth generations are attempted.





From pauldhanna at juno.com  Wed Nov 15 14:34:15 2006
From: pauldhanna at juno.com (Paul D Hanna)
Date: Wed, 15 Nov 2006 13:34:15 GMT
Subject: Trees with Labeled Nodes that form Permutation of Positive Integer
	s
Message-ID: <20061115.053511.1712.764552@webmail42.nyc.untd.com>

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From franktaw at netscape.net  Wed Nov 15 18:34:02 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 15 Nov 2006 12:34:02 -0500
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
In-Reply-To: <20061114.230452.536.0.pauldhanna@juno.com>
References: <20061114.230452.536.0.pauldhanna@juno.com>
Message-ID: <8C8D6F3606C258E-3A4-7467@FWM-R09.sysops.aol.com>

From: pauldhanna at juno.com

>    Could someone compute the initial terms of the following.
>I do not believe that it is in the OEIS (?).
>
>Number of labelled nodes in generation n of a rooted tree
>where a node with label k has k child nodes such that
>the label of each child node is the least unused label in
>the path from the root to that child node,
excluding those used by earlier children of the same parent,
>and where root is labeled '1'.
>
>Thus every distinct path in the tree eventually forms
>a permutation of the positive integers.
>
>...

This isn't really true.  There are many paths in this tree
that do not form a permutation of the positive integers.
You get such a permutation iff, infinitely often, you choose
the first child from the current node.

It is true, of course, that any finite initial path can be
extended to a path that is a permuatation of the positive
integers.

It is an interesting problem to construct an interesting tree
where every path from the root is a permutation of the
positive integers.  I think the following construction works:
start with node 1, which arbitrarily has node 2 as a child
(by the construction that follows, 1 would have no children).
For each node with label m, add BigOmega(m) children, where
BigOmega(m) is the number of prime factors of m with
repetition; label these chilidren with the first positive integers
not occurring in the path from the root to the current node.

The tree starts:

......1
......|
......2
......|
......3
......|
......4
...../.\
....5...6
.../....|\
..6.....5.7
./.\....|.|
7...8...7.5
|../|\..|.|
8.7.9.A.8.8

Any time you hit a prime, there is only one child in the next
generation.  If you keep choosing non-prime children, the
primes start accumulating in the list of positive integers not
occurring in the path, until eventually all children of a node will
be prime.

One could of course use Omega(m), the number of distinct
prime divisors, instead of BigOmega(m).  I think tau(m)-1 would
also work.  A001511(m) (the power of 2 dividing 2m) works, too
- in this case, the single children are under the odd nodes.
________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From qq-quet at mindspring.com  Wed Nov 15 19:23:30 2006
From: qq-quet at mindspring.com (Leroy Quet)
Date: Wed, 15 Nov 06 11:23:30 -0700
Subject: Max Score For A Game
Message-ID: <E1GkPUS-0005H1-00@pop04.mail.atl.earthlink.net>

Below is a game I invented (as posted to sci.math and rec.puzzles -- it 
is just one of many stupid silly simple games I have posted there).

I post it here to seq.fans because I wonder what the sequence is where 
the nth term is the maximum possible score for an n-by-n grid. (We could 
also have a table where the element in the mth column and nth row is the 
maximum score for an m-by-n grid.)
Of course, if there is a way to always get the score n^2 -1 for each 
n-by-n case, or m*n -1 for each m-by-n case (n or m equal 1 excluded), 
then there is be no need to calculate the sequence.

Thanks,
Leroy Quet

=====

Here is a solitaire game played on a square grid.
(I hope this game is more fun than my last game.)

Start by drawing an r-by-r grid on paper.
(I suggest an r of 5 or so for beginners.)
 
Put a "1" in each square of the left-most column
and in each square of the top-most row of the grid.

On each move the player chooses any one of the grid's
empty squares. This square is (m,n), which is the
square in the mth column from the left side of the
grid and in the nth row from the top.

The player then sums up all the integers already
written in the mth column and nth row.
So, if the grid looks like this, where the * is
the square the player has chosen to fill in next
with an integer, 
1 1 1 1
1 3
1 * 6
1 6

then the sum of the column 2's terms and row 3's
terms is 1+3+6 +1+6 = 17. Let this sum be s.

Next the player counts the number of integers in the
squares to the left of (m,n) and above (m,n)
(ie, the squares with coordinates (j,k), 1 <=j <=m,
1 <=k <=n) which are coprime to s.
If the count is c integers in the given rectangle
which are coprime to s, then the player writes c
in square (m,n).

Play continues until there is an integer in every
square of the grid.
The player's score is the number in the last square
he/she fills in.

Example game:
(r=4)
(View with fixed-width font.)

1 1 1 1   1 1 1 1   1 1 1 1   1 1 1 1 
1         1 3       1 3       1 3 5  
1         1         1   6     1   6
1         1         1         1

1 1 1 1   1 1 1 1   1 1 1 1
1 3 5 6   1 3 5 6   1 3 5 6
1   6     1   6     1 5 6
1         1 6       1 6

1 1 1 1   1 1 1 1   1 1 1 1
1 3 5 6   1 3 5 6   1 3 5 6
1 5 6     1 5 6     1 5 6 11
1 6   10  1 6 1110  1 6 1110

So the player gets 11 points.

(It seems that a good strategy is to
{unlike in my example} finish in the
lower right square, and to {as in my
example, 1+6+10+1+5+6=29} have a row
and column sum on the last move which
is a prime.)

Question: Is it always possible to get,
for an r-by-r grid (r >= 2), r^2 -1 points?

Thanks,
Leroy Quet





From davidwwilson at comcast.net  Wed Nov 15 19:48:18 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Wed, 15 Nov 2006 13:48:18 -0500
Subject: Domino questions
Message-ID: <001501c708e6$9e07a570$6501a8c0@yourxhtr8hvc4p>

1. On an n x n (m x n) checkerboard, what is the smallest number of dominoes that can be placed so that no further dominoes can be placed?

2. Suppose two players, A and B, alternately place dominoes on an n x n (m x n) checkerboard until no further play is possible. A plays to maximize the final number of dominoes on the board, B to minimize. What is the number of dominoes on the board at the end of the game if A moves first? If B moves first?
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From franktaw at netscape.net  Wed Nov 15 21:01:19 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 15 Nov 2006 15:01:19 -0500
Subject: Domino questions
In-Reply-To: <001501c708e6$9e07a570$6501a8c0@yourxhtr8hvc4p>
References: <001501c708e6$9e07a570$6501a8c0@yourxhtr8hvc4p>
Message-ID: <8C8D707F39EB04F-934-6F47@FWM-R01.sysops.aol.com>

From: davidwwilson at comcast.net
>1. On an n x n (m x n) checkerboard, what is the smallest number
>of dominoes that can be placed so that no further dominoes can
>be placed?

This might be ceiling(n^2/3), A008810.  Can someone please
check the reference there: "J. H. Conway and R. K. Guy, The Book
of Numbers, Copernicus Press, NY, 1996, number of red blocks in
Fig 2.5."
?
>2. Suppose two players, A and B, alternately place dominoes on an
>n x n (m x n) checkerboard until no further play is possible. A plays
>to maximize the final number of dominoes on the board, B to
>minimize. What is the number of dominoes on the board at the end
>of the game if A moves first? If B moves first?

You don't want much, do you?

(Note for anyone who may be confused - that comment means that
I think this is likely to be a hard problem.)

Franklin T. Adams-Watters
________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From njas at research.att.com  Wed Nov 15 23:40:20 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 15 Nov 2006 17:40:20 -0500 (EST)
Subject: request for a b-file for a core file
Message-ID: <200611152240.RAA60703@fry.research.att.com>


Could someone help produce a b-file for A000014?
(and A001678 along the way)

Here is the sequence:

%I A000014 M0320 N0118
%S A000014 0,1,1,0,1,1,2,2,4,5,10,14,26,42,78,132,249,445,842,1561,2988,5671,10981,
%T A000014 21209,41472,81181,160176,316749,629933,1256070,2515169,5049816,10172638,
%U A000014 20543579,41602425,84440886,171794492,350238175,715497037,1464407113
%N A000014 Number of series-reduced trees with n nodes.


I could use 200, or 500 or so terms

Notes on computing A14 and A1678:

looking at the H'book of Graph Theory, p. 525:

they define f(x) (essentially A1678 but with a different offset):
f(x) = 1*x+1*x^3+1*x^4+2*x^5+3*x^6+6*x^7+10*x^8+19*x^9+35*x^10+...

by

f = (x/(x+1))* exp( sum( f(x^i) / i, i=1..infinity ) )

then A14 is given by

(1+x)f(x) - ((1+x)/2) f(x)^2 + ((1-x)/2) f(x^2)

= x + x^2 + x^4 + x^5 + 2 x^6 + ...

= A14

I actually haven't checked these formulae - there are
slightly different versions in the entry for A14 itself.

Thanks!

Neil






From bowerc at usa.net  Thu Nov 16 01:35:58 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 15 Nov 2006 16:35:58 -0800
Subject: request for a b-file for a core file
Message-ID: <964kkPaI79512S23.1163637358@cmsweb23.cms.usa.net>

I just sent a b-file of terms 0-500 to Neil for A001678.
I'm still working on a version for A000014

------ Original Message ------
From: "N. J. A. Sloane" <njas at research.att.com>
To: seqfans at seqfan.net, seqfan at ext.jussieu.frCc: njas at research.att.com
Subject: request for a b-file for a core file

> 
> Could someone help produce a b-file for A000014?
> (and A001678 along the way)
> 
> Here is the sequence:
> 
> %I A000014 M0320 N0118
> %S A000014
0,1,1,0,1,1,2,2,4,5,10,14,26,42,78,132,249,445,842,1561,2988,5671,10981,
> %T A000014
21209,41472,81181,160176,316749,629933,1256070,2515169,5049816,10172638,
> %U A000014
20543579,41602425,84440886,171794492,350238175,715497037,1464407113
> %N A000014 Number of series-reduced trees with n nodes.
> 
> 
> I could use 200, or 500 or so terms
> 
> Notes on computing A14 and A1678:
> 
> looking at the H'book of Graph Theory, p. 525:
> 
> they define f(x) (essentially A1678 but with a different offset):
> f(x) = 1*x+1*x^3+1*x^4+2*x^5+3*x^6+6*x^7+10*x^8+19*x^9+35*x^10+...
> 
> by
> 
> f = (x/(x+1))* exp( sum( f(x^i) / i, i=1..infinity ) )
> 
> then A14 is given by
> 
> (1+x)f(x) - ((1+x)/2) f(x)^2 + ((1-x)/2) f(x^2)
> 
> = x + x^2 + x^4 + x^5 + 2 x^6 + ...
> 
> = A14
> 
> I actually haven't checked these formulae - there are
> slightly different versions in the entry for A14 itself.
> 
> Thanks!
> 
> Neil
> 
> 









From rkg at cpsc.ucalgary.ca  Thu Nov 16 01:50:35 2006
From: rkg at cpsc.ucalgary.ca (Richard Guy)
Date: Wed, 15 Nov 2006 17:50:35 -0700 (MST)
Subject: Domino questions
In-Reply-To: <8C8D707F39EB04F-934-6F47@FWM-R01.sysops.aol.com>
References: <001501c708e6$9e07a570$6501a8c0@yourxhtr8hvc4p>
 <8C8D707F39EB04F-934-6F47@FWM-R01.sysops.aol.com>
Message-ID: <Pine.LNX.4.63.0611151749420.25970@csl>

1.  Checked!  Unfortunately, in the printing,
red and orange are barely distinguishable.   R.

On Wed, 15 Nov 2006, franktaw at netscape.net wrote:

> From: davidwwilson at comcast.net
>> 1. On an n x n (m x n) checkerboard, what is the smallest number
>> of dominoes that can be placed so that no further dominoes can
>> be placed?
>
> This might be ceiling(n^2/3), A008810.  Can someone please
> check the reference there: "J. H. Conway and R. K. Guy, The Book
> of Numbers, Copernicus Press, NY, 1996, number of red blocks in
> Fig 2.5."
> ?
>> 2. Suppose two players, A and B, alternately place dominoes on an
>> n x n (m x n) checkerboard until no further play is possible. A plays
>> to maximize the final number of dominoes on the board, B to
>> minimize. What is the number of dominoes on the board at the end
>> of the game if A moves first? If B moves first?
>
> You don't want much, do you?
>
> (Note for anyone who may be confused - that comment means that
> I think this is likely to be a hard problem.)
>
> Franklin T. Adams-Watters
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading 
> spam and email virus protection.
>

From bowerc at usa.net  Thu Nov 16 02:01:10 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 15 Nov 2006 17:01:10 -0800
Subject: request for a b-file for a core file
Message-ID: <888kkPBak9480S24.1163638870@cmsweb24.cms.usa.net>

I just sent the b-file for A000014

Christian

------ Original Message ------
From: "Christian G. Bower" <bowerc at usa.net>
To: "seqfan" <seqfan at ext.jussieu.fr>
Subject: Re: request for a b-file for a core file

> I just sent a b-file of terms 0-500 to Neil for A001678.
> I'm still working on a version for A000014
> 








From zbi74583 at boat.zero.ad.jp  Thu Nov 16 04:45:46 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Thu, 16 Nov 2006 12:45:46 +0900
Subject: A036471
Message-ID: <011301c70931$b92d7500$ca73d7dc@FMC57937B6BE00>

    Thanks, Giovanni Resta.



    The smallest term!!

    It is great!

       .

    Now human being became to know much about A036471.       

    The first term is 3270960 and known last term is       

   

    



    

    

    n=32583657,    19.6million digits.



    Yasutoshi

    


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From maxale at gmail.com  Thu Nov 16 10:48:46 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 16 Nov 2006 01:48:46 -0800
Subject: A question on A068480
In-Reply-To: <455AECB9.5050308@iit.cnr.it>
References: <455AECB9.5050308@iit.cnr.it>
Message-ID: <d3dac270611160148q5f52517gbc511cc7b9fd042c@mail.gmail.com>

It is unlikely for n!-1 and 2^n+1 to have a common prime factor
(unless in a special case described below), especially since
1) all prime factors of n!-1 are greater than n;
2) all prime factors of 2^n+1 are of the form p=2kq+1 where q is a
divisor of n, and the multiplicative of 2 modulo p divides 2q.

The special case is when n is even and p=2n+1 is prime. In this case,
p divides both n!-1 and 2^n+1, and thus gcd(n!-1,2^n+1). And it is
very likely that gcd(n!-1,2^n+1) = p.

Overall, the statement "if gcd(n!-1,2^n+1)>1 then
gcd(n!-1,2^n+1)=2n+1" sounds plausible and I verified it for n up to
45,000. But it may be very hard to give a proof.

Max

On 11/15/06, Giovanni Resta <g.resta at iit.cnr.it> wrote:
> A068480 is the sequence on numbers n such that gcd(n!-1,2^n+1)>1.
>
> I noticed that for all the numbers listed gcd(n!-1,2^n+1) = 2n+1.
>
> Is this a rule (proof ?) or a coincidence (counterexample ?) .
>
> thanks,
> giovanni.
>
>
>





From maxale at gmail.com  Thu Nov 16 11:49:25 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 16 Nov 2006 02:49:25 -0800
Subject: A question on A068480
In-Reply-To: <d3dac270611160148q5f52517gbc511cc7b9fd042c@mail.gmail.com>
References: <455AECB9.5050308@iit.cnr.it>
	 <d3dac270611160148q5f52517gbc511cc7b9fd042c@mail.gmail.com>
Message-ID: <d3dac270611160249k774410b5p3b07705ed642b8ae@mail.gmail.com>

On 11/16/06, Max A. <maxale at gmail.com> wrote:

> The special case is when n is even and p=2n+1 is prime. In this case,
> p divides both n!-1 and 2^n+1, and thus gcd(n!-1,2^n+1). And it is
> very likely that gcd(n!-1,2^n+1) = p.

Oh, my bad! Correct statement:

The special case is when n is of the form 4k+1 and p=2n+1 is prime
(implying that 2 and -1 are non-squares modulo p), and p also belongs
to A058302.
In this case, p divides both n!-1 and 2^n+1, and thus gcd(n!-1,2^n+1).
And it is very likely that gcd(n!-1,2^n+1) = p.

The hypothesis about gcd(n!-1,2^n+1) implies that:
8k+3 belong to A058302 if and only if 4k+1 belong to A068480.

Max





From davidwwilson at comcast.net  Thu Nov 16 17:47:04 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 16 Nov 2006 11:47:04 -0500
Subject: b-files for power series expansions
Message-ID: <000601c7099e$d8c5e880$6501a8c0@yourxhtr8hvc4p>

There is a seqload of OEIS sequences that are expansions of real functions, which I guess means power series coefficients. An example is A000810.  Would Mma or some other math power tool be able to quickly generate b-files for these monsters?
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From mathar at strw.leidenuniv.nl  Thu Nov 16 18:44:15 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Thu, 16 Nov 2006 18:44:15 +0100
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
Message-ID: <200611161744.kAGHiFGX005987@amer.strw.leidenuniv.nl>


pdh> From seqfan-owner at ext.jussieu.fr  Wed Nov 15 05:08:51 2006
pdh> Return-Path: <seqfan-owner at ext.jussieu.fr>
pdh> To: seqfan at ext.jussieu.fr
pdh> Date: Tue, 14 Nov 2006 23:04:51 -0500
pdh> Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
pdh> From: "Paul D. Hanna" <pauldhanna at juno.com>
pdh> ...
pdh> Number of labelled nodes in generation n of a rooted tree 
pdh> where a node with label k has k child nodes such that 
pdh> the label of each child node is the least unused label in 
pdh> the path from the root to that child node, 
pdh> and where root is labeled '1'. 
pdh>   ...
pdh> Here are the initial nodes of the tree for generations 0..4: 
pdh> Gen 0: [1];
pdh> Gen 1: (1)->[2];
pdh> Gen 2: (2)->[3,4];
pdh> Gen 3: (3)->[4,5,6], (4)->[3,5,6,7];
pdh> Gen 4: (4)->[5,6,7,8], (5)->[4,6,7,8,9], (6)->[4,5,7,8,9,10], 
pdh> (3)->[5,6,7], (5)->[3,6,7,8,9], (6)->[3,5,7,8,9,10], 
pdh> (7)->[3,5,6,8,9,10,11]; ...
pdh>  
pdh> Thus, the number of nodes in generation n begins:
pdh> [1, 1, 2, 7, 36, 248, ...]
pdh>  
pdh> The maximum node in generation n begins:
pdh> [1, 2, 4, 7, 11, ...]
pdh>  
pdh> Would appreciate it if someone could compute more terms.
pdh> Thanks,
pdh>       Paul

rjm> My count of node numbers and maximum nodes is
rjm> Gen 0 1 1
rjm> Gen 1 1 2
rjm> Gen 2 2 4
rjm> Gen 3 7 7
rjm> Gen 4 36 11
rjm> Gen 5 248 16
rjm> Gen 6 2157 22
rjm> Gen 7 22761 29
rjm> Gen 8 283220 37
rjm> Gen 9 4068411 46

The maximum node label is trivially A000124 because the maximum node will be
generated from the child with itself the highest label, and this adds 'n' to
the maximum label for each new generation. This produces the arithmetic sum
sequence that is shown above as the last number in each row.

For those who have some spare computer cycles and Maple: below is a program
that runs recursively ("memory friendly") through the entire tree up to 
some maximum generation 'maxgen' (to be set/edited in the fourth but last line
below, maxgen=10 equivalent to Gen up to 9 above). It prints a film of
progresses made to count the nodes found so far at all levels, and should show
all numbers in the list monotonically increasing slowly towards
[1, 1, 2, 7, 36, 248, 2157, 22761, 283220, 4068411, 66367834, ...]
                                                    ^^^ 66367834 is Gen 10
I can also provide my older (but non-recursive, memory-limited) C++ version.

Maple V:

gen := proc(parents,maxgen,ocounts,lvl)
	local thislbl,lbl,childlbl,counts,npar ;
	counts := ocounts ;
	counts[lvl] := counts[lvl]+1 ;
	if nops(parents) < maxgen then
		thislbl := op(-1,parents) ;
		childlbl := 1 ;
		for lbl from 1 to thislbl do
			while ( childlbl in parents ) or ( childlbl = thislbl ) do
				childlbl := childlbl+1 ;
			od ;
			npar := [op(parents),childlbl] ;
			# this would generate a full list of all paths:
			# print("route",npar) ;
			if nops(counts) < lvl+1 then
				counts := [op(counts),0] ;
			fi ;
			counts := gen(npar,maxgen,counts,lvl+1) ;
			childlbl := childlbl+1 ;
		od ;
	fi ;
	# this produces a snapshot of the counts accumulated so far:
	# if the current node creation level is not too high, print it...
	if lvl <= maxgen -3 then
		print(counts) ;
	fi ;
	RETURN(counts) ;
end:

# Edit the maximum number below...total run time probably grows exponentially
# with the number chosen here.
maxgen := 8 ;
parents := [1,2] ;
n := [1,0] ;
gen(parents,maxgen,n,2) ;
print(%) ;





From franktaw at netscape.net  Thu Nov 16 18:59:28 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 16 Nov 2006 12:59:28 -0500
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
In-Reply-To: <200611161744.kAGHiFGX005987@amer.strw.leidenuniv.nl>
References: <200611161744.kAGHiFGX005987@amer.strw.leidenuniv.nl>
Message-ID: <8C8D7C018493BAF-87C-1422@FWM-M21.sysops.aol.com>

I don't know Maple well enough - or have enough time - to tell: does 
this keep each path separately, or does it keep a count of the number 
of paths with a given set of integers on the path?

The latter approach should be much faster and less memory intensive, 
once you get past the first few generations.  (Probably still 
exponential, though.)

Franklin T. Adams-Watters

-----Original Message-----
From: mathar at strw.leidenuniv.nl

...

Maple V:

gen := proc(parents,maxgen,ocounts,lvl)
    local thislbl,lbl,childlbl,counts,npar ;
    counts := ocounts ;
    counts[lvl] := counts[lvl]+1 ;
    if nops(parents) < maxgen then
        thislbl := op(-1,parents) ;
        childlbl := 1 ;
        for lbl from 1 to thislbl do
            while ( childlbl in parents ) or ( childlbl = thislbl ) do
                childlbl := childlbl+1 ;
            od ;
            npar := [op(parents),childlbl] ;
            # this would generate a full list of all paths:
            # print("route",npar) ;
            if nops(counts) < lvl+1 then
                counts := [op(counts),0] ;
            fi ;
            counts := gen(npar,maxgen,counts,lvl+1) ;
            childlbl := childlbl+1 ;
        od ;
    fi ;
    # this produces a snapshot of the counts accumulated so far:
    # if the current node creation level is not too high, print it...
    if lvl <= maxgen -3 then
        print(counts) ;
    fi ;
    RETURN(counts) ;
end:

# Edit the maximum number below...total run time probably grows 
exponentially
# with the number chosen here.
maxgen := 8 ;
parents := [1,2] ;
n := [1,0] ;
gen(parents,maxgen,n,2) ;
print(%) ;

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From zakseidov at yahoo.com  Thu Nov 16 19:33:20 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Thu, 16 Nov 2006 10:33:20 -0800 (PST)
Subject: b-files for power series expansions
In-Reply-To: <000601c7099e$d8c5e880$6501a8c0@yourxhtr8hvc4p>
Message-ID: <20061116183320.50052.qmail@web38201.mail.mud.yahoo.com>

Coefficients of the series
1, 1, 4, 13/3, 44/3, 242/15, 2408/45, 18581/315
differ from A000810 entries (?!),
Zak

A000810
1,1,8,26,352,1936,38528,297296,7869952,78098176
%N A000810 Expansion of (sin x + cos x)/cos 3x.


--- David Wilson <davidwwilson at comcast.net> wrote:

> There is a seqload of OEIS sequences that are
> expansions of real functions, which I guess means
> power series coefficients. An example is A000810. 
> Would Mma or some other math power tool be able to
> quickly generate b-files for these monsters?



 
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From franktaw at netscape.net  Thu Nov 16 20:25:13 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 16 Nov 2006 14:25:13 -0500
Subject: b-files for power series expansions
In-Reply-To: <20061116183320.50052.qmail@web38201.mail.mud.yahoo.com>
Message-ID: <8C8D7CC12CFC338-87C-19DE@FWM-M21.sysops.aol.com>

A000810 is the exponential generating function:

1/0!, 1/1!, 8/2!, 26/3!, 352/4!, ...

Franklin T. Adams-Watters


-----Original Message-----
From: zakseidov at yahoo.com

Coefficients of the series
1, 1, 4, 13/3, 44/3, 242/15, 2408/45, 18581/315
differ from A000810 entries (?!),
Zak

A000810
1,1,8,26,352,1936,38528,297296,7869952,78098176
%N A000810 Expansion of (sin x + cos x)/cos 3x.


--- David Wilson <davidwwilson at comcast.net> wrote:

> There is a seqload of OEIS sequences that are
> expansions of real functions, which I guess means
> power series coefficients. An example is A000810.
> Would Mma or some other math power tool be able to
> quickly generate b-files for these monsters?




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From mathar at strw.leidenuniv.nl  Thu Nov 16 20:31:50 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Thu, 16 Nov 2006 20:31:50 +0100
Subject: b-files for power series expansions
Message-ID: <200611161931.kAGJVoHb015119@amer.strw.leidenuniv.nl>


dww> From seqfan-owner at ext.jussieu.fr  Thu Nov 16 17:49:20 2006
dww> From: "David Wilson" <davidwwilson at comcast.net>
dww> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
dww> Subject: b-files for power series expansions
dww> Date: Thu, 16 Nov 2006 11:47:04 -0500
dww> ...
dww> There is a seqload of OEIS sequences that are expansions of real =
dww> functions, which I guess means power series coefficients. An example is =
dww> A000810.  Would Mma or some other math power tool be able to quickly =
dww> generate b-files for these monsters?
dww>...

These are usually taylor series coefficients multiplied by n!
(the E.g.f. given in the definition). Where the functions are even,
the odd terms (coefficients equal to 0) are omitted.

I put some of them into
http://www.strw.leidenuniv.nl/~mathar/progs/b000810.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000813.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000816.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000819.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000822.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000825.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b000828.txt

Unrelated are
http://www.strw.leidenuniv.nl/~mathar/progs/b007504.txt
http://www.strw.leidenuniv.nl/~mathar/progs/b049060.txt





From mathar at strw.leidenuniv.nl  Thu Nov 16 22:05:41 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Thu, 16 Nov 2006 22:05:41 +0100
Subject: Re^2: Trees with Labeled Nodes that form Permutation of Positive Integers
Message-ID: <200611162105.kAGL5fkI022493@amer.strw.leidenuniv.nl>


ftaw> From seqfan-owner at ext.jussieu.fr  Thu Nov 16 19:01:33 2006
ftaw> Return-Path: <seqfan-owner at ext.jussieu.fr>
ftaw> To: seqfan at ext.jussieu.fr
ftaw> Subject: Re: Trees with Labeled Nodes that form Permutation of Positive Integers
ftaw> 
ftaw> I don't know Maple well enough - or have enough time - to tell: does 
ftaw> this keep each path separately, or does it keep a count of the number 
ftaw> of paths with a given set of integers on the path?
ftaw> 
ftaw> The latter approach should be much faster and less memory intensive, 
ftaw> once you get past the first few generations.  (Probably still 
ftaw> exponential, though.)
ftaw> 
ftaw> Franklin T. Adams-Watters
ftaw> 

There is no such optimization in the program shown below. For each
node, it descends into all branches/leafs independent of whether the
set of labels in the current node and higher up in the trunk is the same
or not the same as any set seen earlier. The state vector is
only the vector of counts (one number per generation) which is updated
at each level of recursion, and the single list of labels of the
ancestors of the current node.
So yes, there is room for optimization by introducing some memory
in here, certainly at the cost of increasing the machine memory requirement
to remember all the sub-counts of partial tree in some hash area.

The first action to become faster is to return to a compiled language.
Since this here does not involve any type of functionality which needs a number
theoretic package, one would try to exploit the fact that C would typically be a
factor of 5 to 10 faster than Maple. I don't want to appear selfish and
will therefore leave all this as an exercise to others :-)

RJM

rjm> -----Original Message-----
rjm> From: mathar at strw.leidenuniv.nl
rjm> 
rjm> ...
rjm> 
rjm> Maple V:
rjm> 
rjm> gen := proc(parents,maxgen,ocounts,lvl)
rjm>     local thislbl,lbl,childlbl,counts,npar ;
rjm>     counts := ocounts ;
rjm>     counts[lvl] := counts[lvl]+1 ;
rjm>     if nops(parents) < maxgen then
rjm>         thislbl := op(-1,parents) ;
rjm>         childlbl := 1 ;
rjm>         for lbl from 1 to thislbl do
rjm>             while ( childlbl in parents ) or ( childlbl = thislbl ) do
rjm>                 childlbl := childlbl+1 ;
rjm>             od ;
rjm>             npar := [op(parents),childlbl] ;
rjm>             # this would generate a full list of all paths:
rjm>             # print("route",npar) ;
rjm>             if nops(counts) < lvl+1 then
rjm>                 counts := [op(counts),0] ;
rjm>             fi ;
rjm>             counts := gen(npar,maxgen,counts,lvl+1) ;
rjm>             childlbl := childlbl+1 ;
rjm>         od ;
rjm>     fi ;
rjm>     # this produces a snapshot of the counts accumulated so far:
rjm>     # if the current node creation level is not too high, print it...
rjm>     if lvl <= maxgen -3 then
rjm>         print(counts) ;
rjm>     fi ;
rjm>     RETURN(counts) ;
rjm> end:
rjm> 
rjm> # Edit the maximum number below...total run time probably grows 
rjm> exponentially
rjm> # with the number chosen here.
rjm> maxgen := 8 ;
rjm> parents := [1,2] ;
rjm> n := [1,0] ;
rjm> gen(parents,maxgen,n,2) ;
rjm> print(%) ;





From franktaw at netscape.net  Fri Nov 17 00:46:06 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 16 Nov 2006 18:46:06 -0500
Subject: Re^2: Trees with Labeled Nodes that form Permutation of Positive
 Integers
In-Reply-To: <200611162105.kAGL5fkI022493@amer.strw.leidenuniv.nl>
References: <200611162105.kAGL5fkI022493@amer.strw.leidenuniv.nl>
Message-ID: <8C8D7F084CDAF74-554-265F@FWM-M18.sysops.aol.com>

No hashing is necessary.  There are (n-1)! possible sets of labels.  
Each
sorted sequence of labels has an increase of at most k between the kth
and k+1st label.  So you can take the first differences, subtract 1,
reverse, and treat as a base factorial number.  E.g., for 1,2,4,5,9, 
the
differences are 1,2,1,4; subtract one and reverse to get 3010,
3*3!+0*2!+1*1! = 19, so this will be at (zero-based) index 19.  (This 
is
easily reversible to get the label set from the index and level.)

For each label set, you'll need to keep a count of how many nodes
there are with each possible final label, to tell what to generate in
the next level.  Since 1 and 2 can't be labels after level 2, you only
need to keep n-2 such counters for each set.

We can tell that this algorithm is on the order of n! to compute level 
n.
The naive approach is proportional to the size of the nth level, which 
is
certainly greater; since the largest term at level n is C(n,2)+1 ~ 
n^2/2,
it is certainly no more than (n^2/2)!.  I'm guessing that is it more 
like
c^n n!, for some c between 1.6 and 2 (probably closer to 2, from the
data, though convergence is slow; k 2^n n!/n^2 gives a fairly good
fit to the available data).

Based on this, the eventual savings should be much larger than merely
switching to a compiled language.  I'll also leave this as an exercise,
though. :-)

Franklin T. Adams-Watters


-----Original Message-----
From: mathar at strw.leidenuniv.nl

ftaw> From seqfan-owner at ext.jussieu.fr  Thu Nov 16 19:01:33 2006
ftaw>
ftaw> I don't know Maple well enough - or have enough time - to tell: 
does
ftaw> this keep each path separately, or does it keep a count of the 
number
ftaw> of paths with a given set of integers on the path?
ftaw>
ftaw> The latter approach should be much faster and less memory 
intensive,
ftaw> once you get past the first few generations.  (Probably still
ftaw> exponential, though.)
ftaw>
ftaw> Franklin T. Adams-Watters
ftaw>

There is no such optimization in the program shown below. For each
node, it descends into all branches/leafs independent of whether the
set of labels in the current node and higher up in the trunk is the same
or not the same as any set seen earlier. The state vector is
only the vector of counts (one number per generation) which is updated
at each level of recursion, and the single list of labels of the
ancestors of the current node.
So yes, there is room for optimization by introducing some memory
in here, certainly at the cost of increasing the machine memory 
requirement
to remember all the sub-counts of partial tree in some hash area.

The first action to become faster is to return to a compiled language.
Since this here does not involve any type of functionality which needs 
a number
theoretic package, one would try to exploit the fact that C would 
typically be a
factor of 5 to 10 faster than Maple. I don't want to appear selfish and
will therefore leave all this as an exercise to others :-)

RJM

________________________________________________________________________
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From pauldhanna at juno.com  Fri Nov 17 07:13:46 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Fri, 17 Nov 2006 01:13:46 -0500
Subject: Sub-Fibonacci Tree
Message-ID: <20061117.011346.692.1.pauldhanna@juno.com>

Seqfans, 
     Consider the following "sub-Fibonacci tree" and in particular 
its associated sequence defined by: 
 
Sum of labels of nodes in generation n of a rooted tree in which a node 
with label k has child nodes assigned labels of successive integers 
beginning with k+1 such that the number of child nodes for each node 
is given by the label of the parent node, starting at generation n=0 
with a root node with label '1' followed by a child node with label '2'. 
 
(Perhaps someone could make this description more concise?!)
 
EXAMPLE.
The initial nodes of the tree for generations 0..5 are: 
gen 0: [1];
gen 1: [2];
gen 2: [3];
gen 3: [4,5];
gen 4: (4)->[5,6,7],(5)->[6,7,8];
gen 5: (5)->[6,7,8,9],(6)->[7,8,9,10],(7)->[8,9,10,11], 
(6)->[7,8,9,10,11],(7)->[8,9,10,11,12],(8)->[9,10,11,12,13];
  
By definition, there are 2 child nodes for node [3] since 
the parent of node [3] is [2]; likewise, 
there are 3 child nodes for nodes [4] and [5] (in gen.3) 
since the parent of both nodes has label [3] (in gen.2). 
 
Notice the minimum label in generation n is n+1, and 
the maximum label in generation n is Fibonacci(n+2).  
  
 From this tree I am interested in gleaning 2 sequences. 
  
SEQUENCE 1. 
I compute the number of nodes in generation n to begin: 
[1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368, ...]
 
which is equal (with offset) to A005270:
"Sub-Fibonacci sequences of length n."
1, 1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368 
 
SEQUENCE 2. 
But the following related sequence is NOT found in the OEIS: 
 
Sum of labels of nodes in generation n:
[1, 2, 3, 9, 39, 252, 2361, 32077, 631058, ...]
 
EXTENSION?
Could someone compute more terms of sequences 1 and 2? 
 
Thanks,
    Paul





From pauldhanna at juno.com  Fri Nov 17 08:55:47 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Fri, 17 Nov 2006 02:55:47 -0500
Subject: Sub-Fibonacci Tree
Message-ID: <20061117.025548.692.2.pauldhanna@juno.com>

Seqfans,
    Here I extend A005270 by one more term and also extend the sequence 
a(n) = sum of labels of nodes in generation n in the sub-Fibonacci tree.
I demonstrate a slightly faster method to compute these sequences. 
 
First, I am changing the description to be (hopefully) more clear: 
 
"Sum of labels of nodes in generation n of the sub-Fibonacci tree in
which 
a node with label k and parent node with label g, has g child nodes that 
are given labels beginning with k+1 through k+g; the tree starts at 
generation n=0 with a root node labeled '1' and a child node labeled
'2'."
 
Would appreciate any suggestions to change in wording.
  
 
EXTENDING A005270.
There is a faster way to compute b(n)=A005270(n+1): 
b(n) = "number of nodes in generation n of the sub-Fibonacci tree"
 
by employing the following statistic:
 
b(n+1) = sum of (parent label)*(label) over all nodes in generation n. 
 
For example, 
b(4) = 27 = 3*(4+5);
b(5) = 177 = 4*(5+6+7) + 5*(6+7+8);
b(6) = 1680 = 5*(6+7+8+9) + 6*(7+8+9+10) + 7*(8+9+10+11)+ 
 6*(7+8+9+10+11) + 7*(8+9+10+11+12) + 8*(9+10+11+12+13); ...

Using this method, I extend sequence A005270 by one more term:
1, 1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368, 13067353,
 

SUM OF LABELS SEQUENCE. 
In like manner, the computation of:
a(n) = "sum of labels of nodes in generation n of the sub-Fibonacci tree"

can be accomplished by the statistic:
 
a(n+1) = sum of (parent label)*(label) over all nodes in generation n 
+ sum of (parent label)*[label*(label+1)/2] over all nodes in generation
n-1. 
 
For example, 
a(2) = 3 = 1*2  + 1*( 1*2/2 ) ;
a(3) = 9 = 2*3  + 1*( 2*3/2 ) ;
a(4) = 39 = 3*(4+5)  + 2*( 3*4/2 ) ;
a(5) = 252 = 4*(5+6+7) + 5*(6+7+8)  + 3*( 4*5/2 + 5*6/2 ) ;
a(6) = 2361 = 5*(6+7+8+9) + 6*(7+8+9+10) + 7*(8+9+10+11) + 
 6*(7+8+9+10+11) + 7*(8+9+10+11+12) + 8*(9+10+11+12+13)  +
 4*( 5*6/2 + 6*7/2 + 7*8/2 ) + 5*( 6*7/2 + 7*8/2 + 8*9/2) ;
...
 
With this method, I extend the sum of labels in generation n to: 
1, 2, 3, 9, 39, 252, 2361, 32077, 631058, 18035534, 
 

Would someone please confirm my calculation of the terms: 
b(10) = A005270(11) = 13067353  and 
a(10) = 18035534
 
and maybe compute a few more terms? 
Thanks,
    Paul
----------------------------------------------------
> EXAMPLE.
> The initial nodes of the tree for generations 0..5 are: 
> gen 0: [1];
> gen 1: [2];
> gen 2: [3];
> gen 3: [4,5];
> gen 4: (4)->[5,6,7],(5)->[6,7,8];
> gen 5: (5)->[6,7,8,9],(6)->[7,8,9,10],(7)->[8,9,10,11], 
> (6)->[7,8,9,10,11],(7)->[8,9,10,11,12],(8)->[9,10,11,12,13];
>   
> By definition, there are 2 child nodes for node [3] since 
> the parent of node [3] is [2]; likewise, 
> there are 3 child nodes for nodes [4] and [5] (in gen.3) 
> since the parent of both nodes has label [3] (in gen.2). 
>  
> Notice the minimum label in generation n is n+1, and 
> the maximum label in generation n is Fibonacci(n+2).  
>   
> From this tree I am interested in gleaning 2 sequences. 
>   
> SEQUENCE 1. 
> I compute the number of nodes in generation n to begin: 
> [1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368, ...]
>  
> which is equal (with offset) to A005270:
> "Sub-Fibonacci sequences of length n."
> 1, 1, 1, 1, 2, 6, 27, 177, 1680, 23009, 455368 
>  
> SEQUENCE 2. 
> But the following related sequence is NOT found in the OEIS: 
>  
> Sum of labels of nodes in generation n:
> [1, 2, 3, 9, 39, 252, 2361, 32077, 631058, ...]
>  
> EXTENSION?
> Could someone compute more terms of sequences 1 and 2? 
>  
> Thanks,
>     Paul





From zbi74583 at boat.zero.ad.jp  Sat Nov 18 02:27:22 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Sat, 18 Nov 2006 10:27:22 +0900
Subject: A085058
Message-ID: <019201c70ab0$b44b3ce0$ca73d7dc@FMC57937B6BE00>

    Hi, Seqfans

    If 2^k(n) is the largest power of two dividing tangent number T(n) then 
numbers such that 2*n-k(n) are as follows.



    S : 2,3,2,4,2,3,2,5,2,....



    It looks like A085058.

    Are they the same seqences ?



    Yasutoshi


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From alec at mihailovs.com  Sun Nov 19 11:38:43 2006
From: alec at mihailovs.com (Alec Mihailovs)
Date: Sun, 19 Nov 2006 04:38:43 -0600
Subject: Trees with Labeled Nodes that form Permutation of Positive Integers
References: <200611161744.kAGHiFGX005987@amer.strw.leidenuniv.nl>
Message-ID: <000c01c70bc6$e36ead60$03fea8c0@Media>

From: "Richard Mathar" <mathar at strw.leidenuniv.nl>
Sent: Thursday, November 16, 2006 11:44 AM
>
> pdh> Date: Tue, 14 Nov 2006 23:04:51 -0500
> pdh> From: "Paul D. Hanna" <pauldhanna at juno.com>
> pdh> ...
> pdh> Number of labelled nodes in generation n of a rooted tree
> pdh> where a node with label k has k child nodes such that
> pdh> the label of each child node is the least unused label in
> pdh> the path from the root to that child node,
> pdh> and where root is labeled '1'.
> pdh>   ...
> pdh> Thus, the number of nodes in generation n begins:
> pdh> [1, 1, 2, 7, 36, 248, ...]
>
> [1, 1, 2, 7, 36, 248, 2157, 22761, 283220, 4068411, 66367834, ...]
>                                                    ^^^ 66367834 is Gen 10

Maple 10 can compile (some) procedures in C. I used the following 
back-tracking procedure,

f:=proc(n::integer[4],A::Array(datatype=integer[4]),
B::Array(datatype=integer[4]))::integer[4]; local c::integer[4],
i::integer[4],len::integer[4],m::integer[4]; c,len,m:=0,3,3;
while len>1 do if len=n then c:=c+1;m:=A[len];B[m]:=0;len:=len-1;
B[A[len]]:=B[A[len]]+1 elif B[A[len]]<=A[len] then for i from m+1
do if B[i]=0 then break fi od; len:=len+1;A[len]:=i;B[i]:=1;m:=2
else m:=A[len];B[m]:=0;len:=len-1;B[A[len]]:=B[A[len]]+1 fi od; c end:

cf:=Compiler:-Compile(f):

F:=proc(n::posint) local A,B;
if n<3 then 1 elif n=3 then 2 else
A:=Array([$1..3,0$(n-3)],datatype=integer[4]);
B:=Array([1$3,0$((n-2)*(n+1)/2)],datatype=integer[4]);
cf(n,A,B) fi end:

seq(F(n),n=1..12);

  1, 1, 2, 7, 36, 248, 2157, 22761, 283220, 4068411, 66367834,

        1213504295

That confirms Richard Mathar's and Paul D. Hanna's calculations and gives 
one more element of the sequence. Further elements could not be calculated 
on my 32-bit computer using this procedure, because they are greater than 
2^32. Changing the types of c and f from integer[4] to float[8] allows 
further calculations. I did that and after about an hour got the next 
element,

F(13);

                      24606397802


Alec Mihailovs
http://mihailovs.com/Alec/








From deutsch at duke.poly.edu  Mon Nov 20 17:33:59 2006
From: deutsch at duke.poly.edu (Emeric Deutsch)
Date: Mon, 20 Nov 2006 11:33:59 -0500 (EST)
Subject: help with a new sequence 
Message-ID: <Pine.GSO.4.62.0611201046020.12396@duke.poly.edu>

Seqfans,

I intend to submit to OEIS the following new sequence (triangle):

1,1,1,1,3,1,1,4,2,6,1,1,5,5,10,10,10,1,

Triangle read by rows: T(n,k) is the number of Dyck paths of
semilength n whose ascent lengths form the k-th partition of
the integer n; the partitions of n are ordered in the way
exemplified by [5], [4,1], [3,2], [3,1,1], [2,2,1], [2,1,1,1],
[1,1,1,1,1] (the "Mathematica" ordering).

Equivalently, T(n,k) is the number of ordered trees with n
edges whose node degrees form the k-th partition of the
integer n.

Example: T(5,3)=5 because the 3rd partition of 5 is [3,2] and
we have (UU)DD(UUU)DDD, (UUU)DDD(UU)DD, (UU)D(UUU)DDDD,
(UUU)D(UU)DDDD, and (UUU)DD(UU)DDD; here U=(1,1), D=(1,-1) and
the ascents are shown between parentheses.

Triangle starts:
1;
1,1;
1,3,1;
1,4,2,6,1;
1,5,5,10,10,10,1;

Row n has A000041(n) terms (=number of partitions of n).

------------

So far this is all I have; terms have been found by
straightforward counting. I am sure you can find more
terms, more facts, etc. I'd appreciate any collaboration.

Thanks, Emeric





From pauldhanna at juno.com  Mon Nov 20 19:39:48 2006
From: pauldhanna at juno.com (Paul D Hanna)
Date: Mon, 20 Nov 2006 18:39:48 GMT
Subject: help with a new sequence 
Message-ID: <20061120.104022.757.551642@webmail37.nyc.untd.com>

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From franktaw at netscape.net  Mon Nov 20 19:49:27 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 20 Nov 2006 13:49:27 -0500
Subject: help with a new sequence 
In-Reply-To: <20061120.104022.757.551642@webmail37.nyc.untd.com>
References: <20061120.104022.757.551642@webmail37.nyc.untd.com>
Message-ID: <8C8DAEBBD6AFC69-91C-B5F3@FWM-M20.sysops.aol.com>

This is A008284, a famous function.

Franklin T. Adams-Watters


-----Original Message-----
From: pauldhanna at juno.com

Seqfans,
??????Row n of a?related triangle?could?count?the number of terms in 
row n
of?Emerics triangle that?sum to?form the respective?terms of the 
Narayana triangle.
?
This triangle P?would begin:
1;
1, 1;
1, 1, 1;
1, 2, 1, 1;
1, 2, 2, 1, 1; ...

where row sums equal the partition numbers.
?
For example, row 3 of P?is [1, 2, 1, 1]
since Narayana row 3 is [1, 4+2, 6, 1]
formed from row 3 of?Emerics triangle: [1, 4, 2, 6, 1].
?
Is the resulting partition-related triangle P?already in the OEIS?

It?might have an interesting matrix inverse.
??? Paul
?
--?Emeric?Deutsch?<deutsch at duke.poly.edu>?wrote:
...
Triangle?starts:
1;
1,1;
1,3,1;
1,4,2,6,1;
1,5,5,10,10,10,1;

Row?n?has?A000041(n)?terms?(=number?of?partitions?of?n).
...
Thanks,?Emeric




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From franktaw at netscape.net  Mon Nov 20 20:55:23 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 20 Nov 2006 14:55:23 -0500
Subject: help with a new sequence 
In-Reply-To: <Pine.GSO.4.62.0611201046020.12396@duke.poly.edu>
References: <Pine.GSO.4.62.0611201046020.12396@duke.poly.edu>
Message-ID: <8C8DAF4F355FB9E-B7C-B33B@FWM-M19.sysops.aol.com>

How about if you look at the sequence (composition) of ascents, instead 
of the partition?

I think this starts (using the A066099 ordering):

1
1,1
1,2,1,1
1,3,2,3,1,2,1,1
1,4,3,6,2,5,3,1,1,3,2,2,1,3,4,1

Amazingly, this corresponds to Stern's diatomic array (version 3, 
A070879) for the first 18 terms.
Makes me wonder if A070879 can be interpreted as a function on 
compositions.

Franklin T. Adams-Watters


-----Original Message-----
From: deutsch at duke.poly.edu

Seqfans,?
?
I intend to submit to OEIS the following new sequence (triangle):?
?
1,1,1,1,3,1,1,4,2,6,1,1,5,5,10,10,10,1,?
?
Triangle read by rows: T(n,k) is the number of Dyck paths of?
semilength n whose ascent lengths form the k-th partition of?
the integer n; the partitions of n are ordered in the way?
exemplified by [5], [4,1], [3,2], [3,1,1], [2,2,1], [2,1,1,1],?
[1,1,1,1,1] (the "Mathematica" ordering).?
?
Equivalently, T(n,k) is the number of ordered trees with n?
edges whose node degrees form the k-th partition of the?
integer n.?
?
Example: T(5,3)=5 because the 3rd partition of 5 is [3,2] and?
we have (UU)DD(UUU)DDD, (UUU)DDD(UU)DD, (UU)D(UUU)DDDD,?
(UUU)D(UU)DDDD, and (UUU)DD(UU)DDD; here U=(1,1), D=(1,-1) and?
the ascents are shown between parentheses.?
?
Triangle starts:?
1;?
1,1;?
1,3,1;?
1,4,2,6,1;?
1,5,5,10,10,10,1;?
?
Row n has A000041(n) terms (=number of partitions of n).?
?
------------?
?
So far this is all I have; terms have been found by?
straightforward counting. I am sure you can find more?
terms, more facts, etc. I'd appreciate any collaboration.?
?
Thanks, Emeric?


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From franktaw at netscape.net  Mon Nov 20 22:54:07 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 20 Nov 2006 16:54:07 -0500
Subject: help with a new sequence 
In-Reply-To: <8C8DAF4F355FB9E-B7C-B33B@FWM-M19.sysops.aol.com>
References: <Pine.GSO.4.62.0611201046020.12396@duke.poly.edu> <8C8DAF4F355FB9E-B7C-B33B@FWM-M19.sysops.aol.com>
Message-ID: <8C8DB058980377A-B7C-BB3C@FWM-M19.sysops.aol.com>

Let f(a_0, a_1, ..., a_m) be the number of Dyck paths associated with 
the composition
[a_0, a_1, ..., a_m].   We have f(a_0) = 1; and for m > 0,

f(a_0, a_1, ..., a_m) = Sum_{i=0}^{a_0-1} f(a_1+i, a_2, ..., a_m).

Using this, I get the following table (including row 0, the empty 
composition):

1,
1,
1,1,
1,2,1,1,
1,3,2,3,1,2,1,1,
1,4,3,6,2,5,3,4,1,3,2,3,1,2,1,1,
1,5,4,10,3,9,6,10,2,7,5,9,3,7,4,5,1,4,3,6,2,5,3,4,1,3,2,3,1,2,1,1,
1,6,5,15,4,14,10,20,3,12,9,19,6,16,10,15,2,9,7,16,5,14,9,14,3,10,7,12,4,9
,5,6,1,5,4,
   10,3,9,6,10,2,7,5,9,3,7,4,5,1,4,3,6,2,5,3,4,1,3,2,3,1,2,1,1

And accumulating by partition, for Emeric's original table (but also 
including row 0):

1,
1,
1,1,
1,3,1,
1,4,2,6,1,
1,5,5,10,10,10,1,
1,6,6,15,3,30,20,5,30,15,1,
1,7,7,21,7,42,35,21,21,105,35,35,70,21,1,
1,8,8,28,8,56,56,4,56,28,168,70,28,84,168,280,56,14,140,140,28,1,
1,9,9,36,9,72,84,9,72,36,252,126,36,72,252,252,504,126,12,252,252,84,756,
630,84,
   126,420,252,36,1

Given a partition of n into k parts, [a_1^e_1,...,a_k^e_k], the number 
of
associated Dyck paths appears to be n!/(n-k+1)!/(Product (e_i)!).

Now, maybe somebody can prove that.

Franklin T. Adams-Watters


-----Original Message-----
From: franktaw at netscape.net

How about if you look at the sequence (composition) of ascents, instead 
of the partition??
?
I think this starts (using the A066099 ordering):?
?
1?
1,1?
1,2,1,1?
1,3,2,3,1,2,1,1?
1,4,3,6,2,5,3,1,1,3,2,2,1,3,4,1?
?
Amazingly, this corresponds to Stern's diatomic array (version 3, 
A070879) for the first 18 terms.?
Makes me wonder if A070879 can be interpreted as a function on 
compositions.?
?
Franklin T. Adams-Watters?
?
-----Original Message-----?
From: deutsch at duke.poly.edu?
?
Seqfans,??
??
I intend to submit to OEIS the following new sequence (triangle):??
??
1,1,1,1,3,1,1,4,2,6,1,1,5,5,10,10,10,1,??
??
Triangle read by rows: T(n,k) is the number of Dyck paths of??
semilength n whose ascent lengths form the k-th partition of??
the integer n; the partitions of n are ordered in the way??
exemplified by [5], [4,1], [3,2], [3,1,1], [2,2,1], [2,1,1,1],??
[1,1,1,1,1] (the "Mathematica" ordering).??
??
Equivalently, T(n,k) is the number of ordered trees with n??
edges whose node degrees form the k-th partition of the??
integer n.??
??
Example: T(5,3)=5 because the 3rd partition of 5 is [3,2] and??
we have (UU)DD(UUU)DDD, (UUU)DDD(UU)DD, (UU)D(UUU)DDDD,??
(UUU)D(UU)DDDD, and (UUU)DD(UU)DDD; here U=(1,1), D=(1,-1) and??
the ascents are shown between parentheses.??
??
Triangle starts:??
1;??
1,1;??
1,3,1;??
1,4,2,6,1;??
1,5,5,10,10,10,1;??
??
Row n has A000041(n) terms (=number of partitions of n).??
??
------------??
??
So far this is all I have; terms have been found by??
straightforward counting. I am sure you can find more??
terms, more facts, etc. I'd appreciate any collaboration.??
??
Thanks, Emeric??
?
________________________________________________________________________?

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From njas at research.att.com  Mon Nov 20 23:25:04 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Mon, 20 Nov 2006 17:25:04 -0500 (EST)
Subject: two "find the next term" puzzles from Knuth Vol 4
Message-ID: <200611202225.RAA93655@fry.research.att.com>


the two puzzles (and their solutions) are

%I A123896
%S A123896 0,1,1,1,12,12,12,12,12,12,100,121,122,123,123
%N A123896 A123895(n^2).
...

%I A123902
%S A123902 0,1,12,100,112,121,122,123
%N A123902 A123896 sorted and uniqued.
...

which are based on this:

%I A123895
%S A123895 0,1,1,1,1,1,1,1,1,1,10,11,12,12,12,12,12,12,12,12,10,12,11,12,12,
%T A123895 12,12,12,12,12,10,12,12,11,12,12,12,12,12,12,10,12,12,12,11,12,12,
%U A123895 12,12,12,10,12,12,12,12,11,12,12,12,12,10,12,12,12,12,12,11,12,12
%N A123895 Restricted growth string for the (decimal expansion of the) number n.
%C A123895 Write n in base 10 prefixed with a 0. Read this string from left to r
ight. Write a 0 each time you see the first distinct digit (which is 0), write a
1 each time you see the second distinct digit, write a 2 each time you see the t
hird distinct digit, and so on. Finally, delete the leading zeros.
%D A123895 D. E. Knuth, TAOCP, Vol. 4, Section 7.2.1.5, Problems 4 and 5.
%e A123895 To find a(66041171): 066041171 -> 011023343 -> 11023343.
%Y A123895 Cf. A123896, A123902.
%K A123895 nonn,base,new
%O A123895 0,11
%A A123895 njas, Nov 20 2006

It would be nice if someone could extend the first two, and provide
maple or mma or other code for the third one

Neil






From franktaw at netscape.net  Mon Nov 20 23:59:41 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 20 Nov 2006 17:59:41 -0500
Subject: two "find the next term" puzzles from Knuth Vol 4
In-Reply-To: <200611202225.RAA93655@fry.research.att.com>
References: <200611202225.RAA93655@fry.research.att.com>
Message-ID: <8C8DB0EB222DED3-B7C-BF60@FWM-M19.sysops.aol.com>

A123896:

0 1 1 1 12 12 12 12 12 12 100 121 122 123 123 112 123 123 123 123 100 
112 121 123 123 123 121 123 123 123 100 123 1023 1023 1123 1223 1234 
1234 1222 1231 1200 1231 1234 1234 1234 1012 1223 1102 1203 1203 1200 
1203 1203 1203 1234 1023 1213 1234 1123 1234 1200 1234 1233 1232 1023 
1223 1234 1123 1231 1234 1200 1023 1234 1234 1234 1231 1223 1232 1023 
1234 1200 1213 1234 1223 1023 1223 1234 1234 1122 1234 1200

A123902:

0 1 12 100 112 121 122 123 1012 1023 1102 1122 1123 1200 1201 1203 1213 
1222 1223 1231 1232 1233 1234 10000 10012 10023 10102 10123 10201 10202 
10203 10213 10223 10231 10232 10234 11023 11102 11200 11203 11211 11213 
11221 11223 11232 11234 12003 12013 12023 12032 12033 12034 12100 12113 
12121 12131 12133 12134 12200 12203 12212 12213 12231 12232 12233 12234 
12300 12301 12304 12312 12313 12314 12321 12323 12324 12331 12332 12334 
12341 12342

Sorry, I didn't use Mathematica, Maple, or even PARI; I used Excel and 
VBA.

Here's a VBA program:

Public Function RestrictedGrowthString(ByVal x As String) As String
    Dim i As Long
    Dim dig As Integer
    Dim pos As Long

    For i = 1 To Len(x)
        If Mid(x, i, 1) = "0" Then
            RestrictedGrowthString = RestrictedGrowthString & "0"
        Else
            pos = InStr(x, Mid(x, i, 1))
            If pos = i Then
                dig = dig + 1
                 RestrictedGrowthString = RestrictedGrowthString & 
Format(dig)
            Else
                 RestrictedGrowthString = RestrictedGrowthString & 
Mid(RestrictedGrowthString, pos, 1)
            End If
        End If
    Next i
End Function

Franklin T. Adams-Watters


-----Original Message-----
From: njas at research.att.com

the two puzzles (and their solutions) are

%I A123896
%S A123896 0,1,1,1,12,12,12,12,12,12,100,121,122,123,123
%N A123896 A123895(n^2).
...

%I A123902
%S A123902 0,1,12,100,112,121,122,123
%N A123902 A123896 sorted and uniqued.
...

which are based on this:

%I A123895
%S A123895 
0,1,1,1,1,1,1,1,1,1,10,11,12,12,12,12,12,12,12,12,10,12,11,12,12,
%T A123895 
12,12,12,12,12,10,12,12,11,12,12,12,12,12,12,10,12,12,12,11,12,12,
%U A123895 
12,12,12,10,12,12,12,12,11,12,12,12,12,10,12,12,12,12,12,11,12,12
%N A123895 Restricted growth string for the (decimal expansion of the) 
number n.
%C A123895 Write n in base 10 prefixed with a 0. Read this string from 
left to r
ight. Write a 0 each time you see the first distinct digit (which is 
0), write a
1 each time you see the second distinct digit, write a 2 each time you 
see the t
hird distinct digit, and so on. Finally, delete the leading zeros.
%D A123895 D. E. Knuth, TAOCP, Vol. 4, Section 7.2.1.5, Problems 4 and 
5.
%e A123895 To find a(66041171): 066041171 -> 011023343 -> 11023343.
%Y A123895 Cf. A123896, A123902.
%K A123895 nonn,base,new
%O A123895 0,11
%A A123895 njas, Nov 20 2006

It would be nice if someone could extend the first two, and provide
maple or mma or other code for the third one

Neil


________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From njas at research.att.com  Tue Nov 21 00:43:03 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Mon, 20 Nov 2006 18:43:03 -0500 (EST)
Subject: King chickens - I was wrong!
Message-ID: <200611202343.SAA88240@fry.research.att.com>


Martin Fuller pointed out that I was wrong.
A highest scorer in a tournament is a king chicken,
but not conversely.

So there are really two sequences, A123553 (K.C.'s),
now corrected by Martin, and A125031, the total number
of highest scorers, computed by Martin Fuller to n=11
(this is what Gordon Royle computed out to n=10).

Maurer also defines "Emperors", chickens
who peck everybody else, which is n*2^((n-1)(n-2)/2)
(it will be A123903).

Neil





From zakseidov at yahoo.com  Tue Nov 21 06:19:11 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Mon, 20 Nov 2006 21:19:11 -0800 (PST)
Subject: A125097.gif: (Very) nice patterns
Message-ID: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>

Neil, 
seqfans,

1. Have a look at this full graph -
aint these patterns (very) nice?!
It may be known, but still it's nice...

2. Is it possible to 
put "the extended graph" into OEIS?
In the standard graph no patterns are seen at all...

3. Neil removed  keyword "nice" from A125097
(as he did regularly with ny sequences)-
as if I insist that it's me who is "nice"
not it is the seq which is nice.

Thanks, Zak


 
____________________________________________________________________________________
Sponsored Link

$200,000 mortgage for $660/ mo - 
30/15 yr fixed, reduce debt - 
http://yahoo.ratemarketplace.com
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From maxale at gmail.com  Tue Nov 21 07:03:55 2006
From: maxale at gmail.com (Max A.)
Date: Mon, 20 Nov 2006 22:03:55 -0800
Subject: A125097.gif: (Very) nice patterns
In-Reply-To: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>
References: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>
Message-ID: <d3dac270611202203p2b9fff6s2bc522bc6662e735@mail.gmail.com>

Zak,

I do not see anything special about this graph.
Yes, it includes some points from the line y=2x+1, and, yes, it is
symmetric the region [0,500]x[0,1000] (i.e., a(n)=k iff
a((k-1)/2)=2n+1). But that does not make it any nicer.

Regards,
Max

On 11/20/06, zak seidov <zakseidov at yahoo.com> wrote:
> Neil,
> seqfans,
>
> 1. Have a look at this full graph -
> aint these patterns (very) nice?!
> It may be known, but still it's nice...
>
> 2. Is it possible to
> put "the extended graph" into OEIS?
> In the standard graph no patterns are seen at all...
>
> 3. Neil removed  keyword "nice" from A125097
> (as he did regularly with ny sequences)-
> as if I insist that it's me who is "nice"
> not it is the seq which is nice.
>
> Thanks, Zak
>
>
>
> ____________________________________________________________________________________
> Sponsored Link
>
> $200,000 mortgage for $660/ mo -
> 30/15 yr fixed, reduce debt -
> http://yahoo.ratemarketplace.com
>
>





From pauldhanna at juno.com  Tue Nov 21 11:07:54 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Tue, 21 Nov 2006 05:07:54 -0500
Subject: Formula for New Sequence A125054 ?
Message-ID: <20061121.050754.1448.0.pauldhanna@juno.com>

Seqfans,
     Could someone find a formula for this new sequence? 
Sequence: A125054  begins: 
 
1,3,21,327,9129,396363,24615741,2068052367,225742096209,
31048132997523,5252064083753061,1071525520294178007,
259439870666594250489,73542221109962636293083,
24125551094579137082039181,9068240688454120376775401247,
3871645204706420218816959159969,
(I can supply many more terms if needed).
 
The sequence forms the Central terms of a new triangle A125053 
(a variant of triangle A008301 - enumeration of binary trees). 
 
Triangle A125053 is nice since the first column (and row sums) 
form the Secant numbers A000364 (an unexpected result!). 
  
Below I define the triangle.
 
I would be very interested in a formula, perhaps an E.g.f.? 
(I know I am asking a lot, but Seqfans surprise me all the time 
with amazing formulas). 
  
Thanks, 
      Paul 
---------------------------------------------------------
  
If we write triangle A125053 like this: 
.......................... ...1; 
.................... ...1, ...3, ...1; 
.............. ...5, ..15, ..21, ..15, ...5; 
........ ..61, .183, .285, .327, .285, .183, ..61; 
.. 1385, 4155, 6681, 8475, 9129, 8475, 6681, 4155, 1385; 

then the first nonzero term is the sum of the previous row: 
1385 = 61 + 183 + 285 + 327 + 285 + 183 + 61,

the next term is 3 times the first:
4155 = 3*1385,

and the remaining terms in each row are obtained by the rule 
illustrated by: 
6681 = 2*4155 - 1385 - 4*61 ;
8475 = 2*6681 - 4155 - 4*183 ;
9129 = 2*8475 - 6681 - 4*285 ;
8475 = 2*9129 - 8475 - 4*327 ;
6681 = 2*8475 - 9129 - 4*285 ;
4155 = 2*6681 - 8475 - 4*183 ;
1385 = 2*4155 - 6681 - 4*61 .
 
An alternate recurrence is illustrated by:
4155 = 1385 + 2*(61 + 183 + 285 + 327 + 285 + 183 + 61);
6681 = 4155 + 2*(183 + 285 + 327 + 285 + 183);
8475 = 6681 + 2*(285 + 327 + 285);
9129 = 8475 + 2*(327); 
 
and then for k>n, k<=2n, T(n,k) = T(n,2*n-k).
 
END.





From pauldhanna at juno.com  Tue Nov 21 14:58:26 2006
From: pauldhanna at juno.com (Paul D Hanna)
Date: Tue, 21 Nov 2006 13:58:26 GMT
Subject: Formula for New Sequence A125054 ?
Message-ID: <20061121.055901.20413.609605@webmail43.nyc.untd.com>

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From antti.karttunen at gmail.com  Tue Nov 21 15:27:22 2006
From: antti.karttunen at gmail.com (Antti Karttunen)
Date: Tue, 21 Nov 2006 16:27:22 +0200
Subject: Binary tree terminology, again. Re: Formula for New Sequence A125054
 ?
In-Reply-To: <20061121.055901.20413.609605@webmail43.nyc.untd.com>
References: <20061121.055901.20413.609605@webmail43.nyc.untd.com>
Message-ID: <45630CCA.8000103@gmail.com>

Paul D Hanna wrote:

> Seqfans,
>
>      Could someone find a formula for this new sequence?
> Sequence: A125054  begins:
>  
> 1,3,21,327,9129,396363,24615741,2068052367,225742096209,
> 31048132997523,5252064083753061,1071525520294178007,
> 259439870666594250489,73542221109962636293083,
> 24125551094579137082039181,9068240688454120376775401247,
> 3871645204706420218816959159969,
> (I can supply many more terms if needed).
>  
> The sequence forms the Central terms of a new triangle A125053
> (a variant of triangle A008301 - enumeration of binary trees).
>
Is there any more helpful description for A008301 and A125053 ?
E.g. do they relate to non-planar binary trees
(as http://www.research.att.com/~njas/sequences/A001190 )
or to (rooted, unlabeled) planar binary trees, as Catalan numbers, 
A000108, do?
Labeled or unlabeled? Rooted or non-rooted?


Cumprimentos,

Antti Karttunen
writing from Porto, Portugal.

>  
> Triangle A125053 is nice since the first column (and row sums)
> form the Euler numbers A000364 (an unexpected result!).
>  
>
> Seqfans,
>     One should always try Superseeker first!
> Superseeker says:
>
> A125054 = Binomial transform of A000182 (e.g.f. tan(x))
>
> and it holds true.
>
> I did not expect such a simple answer! 
>     Paul
>

>  
>






From tbaruchel at free.fr  Tue Nov 21 15:47:28 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Tue, 21 Nov 2006 15:47:28 +0100 (CET)
Subject: Almost the number of divisors of n+1
Message-ID: <20061121154018.X1354@localhost.my.domain>

Hi, I have some number-theory related empirically collected datas to study.
As long as I can see, I have a link between each integer and the number
of divisors of its successor. But though the rule seems to work very
well in some cases: predecessors of prime numbers seem to appear 0 or 1 times,
predecessors of numbers like 12 or 24 seem to appear much more, etc. I am not
sure this is the very exact rule. Of course, the trouble may come from
the fact that my datas is quite hard to "extract", but maybe you know some
variants of
   a(n) is the number of divisors of (n+1)
Have you ever seen closely-related sequences like that one?

Regards,

-- 
Thomas Baruchel





From davidwwilson at comcast.net  Tue Nov 21 18:18:36 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Tue, 21 Nov 2006 12:18:36 -0500
Subject: Confirmation
Message-ID: <004601c70d91$14d8f3d0$6501a8c0@yourxhtr8hvc4p>

Define

    f(1) = 1
    f(n+1) = f(n) + (f(n) mod n+3364).

I wrote a computer program to find the smallest x with

    x+3365 divides f(x)

The value of x found is just under 8,350,000,000 and the corresponding f(x) just under 2^64. Although my program included an overflow check, I am still not comfortable because this value of f(x) is close to my 64-bit unsigned integer size.

Could someone with a fast computer, preferably a UNIX box, please confirm or refute my values?
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From maxale at gmail.com  Tue Nov 21 20:47:20 2006
From: maxale at gmail.com (Max A.)
Date: Tue, 21 Nov 2006 11:47:20 -0800
Subject: Confirmation
In-Reply-To: <004601c70d91$14d8f3d0$6501a8c0@yourxhtr8hvc4p>
References: <004601c70d91$14d8f3d0$6501a8c0@yourxhtr8hvc4p>
Message-ID: <d3dac270611211147i7b7ef94ev729014f094d24dc4@mail.gmail.com>

I've got

x = 8334201245
f(x) = 17364786429187398690

I've also checked that the value f(n) was non-decreasing during
computations. Since f(n+1) < f(n)+n+3364 and f(x)+x+3364 < 2^64, that
means no overflow in 64-bit unsigned integer happened.

Max

On 11/21/06, David Wilson <davidwwilson at comcast.net> wrote:
>
>
> Define
>
>     f(1) = 1
>     f(n+1) = f(n) + (f(n) mod n+3364).
>
> I wrote a computer program to find the smallest x with
>
>     x+3365 divides f(x)
>
> The value of x found is just under 8,350,000,000 and the corresponding f(x)
> just under 2^64. Although my program included an overflow check, I am still
> not comfortable because this value of f(x) is close to my 64-bit unsigned
> integer size.
>
> Could someone with a fast computer, preferably a UNIX box, please confirm or
> refute my values?





From all at abouthugo.de  Tue Nov 21 22:22:29 2006
From: all at abouthugo.de (Hugo Pfoertner)
Date: Tue, 21 Nov 2006 22:22:29 +0100
Subject: Wrong new terms in A085000
Message-ID: <45636E15.FD6E8D3F@abouthugo.de>

Seqfans, Neil,

the sequence A085000 "Maximal determinant of an n X n matrix using the
integers 1 to n^2."

that had previously the terms a(1)...a(6)
1,10,412,40800,6839492,1865999570,

has recently been extendend as follows:

762074188050,440477621828418, 345964972957674150

The comment says:
a(7) - a(9) were found by Richard Mueller
(richard.mueller(AT)informatik.hu-berlin.de) and Johannes Dewender
(johannes.dewender(AT)informatik.hu-berlin.de) in cooperation, Oct 22
2006 
Entry updated by Richard Mueller  Oct 22 2006

Unfortunately bigger counterexamples exist for all 3 terms:

d7:=det mat
((28, 9,34,41, 1,40,22),
 (47,25, 7,36,35,16, 8),
 (3, 27, 6,39,31,26,43),
 (38,21,32,14,18, 4,48),
 (19, 5,37,13,49,33,20),
 (29,42,15, 2,17,45,24),
 (11,46,44,30,23,12,10));
 
 Reduce: d7 := 762139309293

d8:= det mat
((13, 53, 40, 25,  2, 45, 58, 23),
 (21, 22,  8, 54, 24, 56, 18, 57),
 (60, 11, 36, 44, 12,  7, 49, 41),
 (15, 19, 48,  4, 52, 26, 38, 59),
 (20, 62, 34, 51, 42,  1, 17, 33),
 (37, 29,  3, 31, 61, 35, 55, 10),
 (30, 14, 63, 46, 39, 47, 16,  5),
 (64, 50, 28,  6, 27, 43,  9, 32));

Reduce: d8 := 440857916120379

d9:= det mat
((40, 48, 72, 26,  6, 50, 39, 78, 10),
 (23, 77, 42,  7, 37, 41, 74, 12, 56),
 (52, 47,  8, 54,  5, 64, 19, 44, 75),
 (59, 25, 13, 43, 58, 76, 63, 30,  3),
 (79, 36, 71, 60, 31, 18, 34,  2, 38),
 (32, 81, 22, 61, 67, 15, 17, 53, 21),
 (20, 24, 66, 27, 73, 68,  4, 33, 55),
 ( 1, 14, 45, 80, 35, 29, 69, 46, 49),
 (65, 16, 28, 11, 57,  9, 51, 70, 62));

Reduce: d9 := 346202123011922653

(Thanks to Rainer Rosenthal for X-checking the determinants with Maple).

None of the counterexamples is known to be the biggest possible
solution. The best value I've found in 2 CPU years for
a(7)=762140212575, but currently I don't have access to the computer
where I've found this solution and the corresponding arrangement of
matrix elements.

My suggestion is to remove the 3 new terms and the new Mathematica
program and to restore the previous state of the sequence. The
determination of a(8) and a(9) is currently beyond our available
computational capabilities. I will try to confirm a(7) and to find
improved lower bounds for a(8) and a(9) by running the slightly improved
program given at the link for some months on a cluster of Intel Itanium
2 CPUs.

In the meantime I suggest to add a comment:

Known lower bounds for a(7)...a(9) are
762140212575, 440857916120379, 346202123011922653,

Hugo Pfoertner





From mlb at well.com  Wed Nov 22 00:25:34 2006
From: mlb at well.com (Marc LeBrun)
Date: Tue, 21 Nov 2006 15:25:34 -0800
Subject: Almost the number of divisors of n+1
In-Reply-To: <20061121154018.X1354@localhost.my.domain>
References: <20061121154018.X1354@localhost.my.domain>
Message-ID: <200611212325.kALNPQh3019729@smtp.well.com>

 >  a(n) is the number of divisors of (n+1)
 > Have you ever seen closely-related sequences like that one?

Some random and rather vague ideas:

1. I faintly recall seeing something about algorithms for
factoring n involving properties of n-1 (I mean besides
the obvious connection to "Fermat's little theorem").
Perhaps in Knuth's TAOCP?

2. It sounds like you might have a case where the relationship
you're seeing, a(n) ~ d(n), is actually a first approximation
to some identity like

   a(n) = d(n) + F1(n) + F2(n) + ...

where the Fk depend on the factor structure of n or something.
So perhaps you might try to "take out" the d(n) and then look
further into the "error" and see what you can guess about how
the Fk's specifically vary with primes, highly composites, etc.
For example you might try "fitting" a(n) with Moebius-style sums.

3. There's some rather surprising known relationships involving
divisors--my favorite is one of Legendre's that is basically an
identity found by expanding elliptic functions in power series.
This suggests that you might want to express your relationships
as generating functions and see what emerges.

4. Sometimes there is simple underlying "combinatorial machinery".
For example I was once puzzled by some coefficients that turned
out to be formed by taking each divisor of n, *adding 1*, and then
multiplying them all together!  It took me a while to see that's
algebraically identical to summing the products of every possible
subset of the divisors.  But the "add 1" way of expressing this
was totally opaque, whereas the "every subset" explained them.

Happy hunting!







From zbi74583 at boat.zero.ad.jp  Wed Nov 22 02:15:26 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Wed, 22 Nov 2006 10:15:26 +0900
Subject: N(2)
Message-ID: <00a201c70dd3$b538af40$977fd7dc@FMC57937B6BE00>

    I wonder if N(2) goes infinity.

    Where N(s)=   Sum_{m=1 to infinity} (-1)Sigma(m)/m^s

    Mathar computed up to m=1000000



  s m N(s)

> 2 1 1.000000000000000000000000000
> 2 2 1.250000000000000000000000000
> 2 3 1.472222222222222222222222222
> 2 4 1.784722222222222222222222222
> 2 5 1.944722222222222222222222222
> 2 6 2.000277777777777777777777778
> 2 7 2.122726757369614512471655329
> 2 8 2.325851757369614512471655329
> 2 9 2.461654226505416981607457798
> 2 10000 8.619813177975507784214344698
> 2 20000 9.240635095871089404115753095
> 2 30000 9.603791897573761358292380061
> 2 40000 9.861458278261452185262848858
> 2 50000 10.06133043757656995196303496
> 2 60000 10.22462154552329778519365923
> 2 70000 10.36270038005636494621123847
> 2 80000 10.48231016697477680498962023
> 2 90000 10.58780522013752853231672862
> 2 100000 10.68217402235127296288785820
> 2 110000 10.76754463607533705360724580
> 2 120000 10.84547789729425765394094918
> 2 130000 10.91717268365027978460835188
> 2 140000 10.98355652885066079792761960
> 2 150000 11.04534545539722125239381859
> 2 160000 11.10315629233238669062836862
> 2 170000 11.15746025373895516747888760
> 2 180000 11.20865775307751148407150182
> 2 190000 11.25708463818456648077093060
> 2 200000 11.30303121795283056763425981
> 2 210000 11.34672871948779384291766276
> 2 220000 11.38839827063844308711114745
> 2 230000 11.42821240662906506074836109
> 2 240000 11.46633165190016188934526342
> 2 250000 11.50290004647923375585461222
> 2 260000 11.53802980091084626273394586
> 2 270000 11.57183602344116902681335304
> 2 280000 11.60440829561909536077171934
> 2 290000 11.63584045378869260629085740
> 2 300000 11.66620539683007648663354504
> 2 310000 11.69557551635064846235471407
> 2 320000 11.72401391041322342493145640
> 2 330000 11.75157395832622563415629569
> 2 340000 11.77831576251751436463182945
> 2 350000 11.80427986574616427809478607
> 2 360000 11.82951337960016320803463309
> 2 370000 11.85405377158818097821428014
> 2 380000 11.87794149361683265940241605
> 2 390000 11.90120806565142413868457075
> 2 400000 11.92388559358277288937420440
> 2 410000 11.94600303970164868968234173
> 2 420000 11.96758569454234998141373716
> 2 430000 11.98866287480559328733313634
> 2 440000 12.00925544388189350449936925
> 2 450000 12.02938482345842327394291528
> 2 460000 12.04907101416111155199518539
> 2 470000 12.06833481184766752219695603
> 2 480000 12.08719139777266186606623586
> 2 490000 12.10566116650605692754670219
> 2 500000 12.12375840131307130932177405
> 2 510000 12.14149351071909408890611230
> 2 520000 12.15888769016783248485026059
> 2 530000 12.17594975401341942285000014
> 2 540000 12.19269280126406301090981599
> 2 550000 12.20912705098196862278557687
> 2 560000 12.22526702472305560991043337
> 2 570000 12.24111996896626519121471994
> 2 580000 12.25669798166764128539680162
> 2 590000 12.27200952384435815319104573
> 2 600000 12.28706299338740013449483108
> 2 610000 12.30187012252190226181814868
> 2 620000 12.31643453756242668558457904
> 2 630000 12.33076670711890682146429120
> 2 640000 12.34487182648867321005497632
> 2 650000 12.35875920337989470739102627
> 2 660000 12.37243430644503889140869087
> 2 670000 12.38590421689421153982279985
> 2 680000 12.39917431914306915108546756
> 2 690000 12.41224965443445935379058168
> 2 700000 12.42513867918653815467126754
> 2 710000 12.43784365088579118428209464
> 2 720000 12.45037185509338005878377820
> 2 730000 12.46272650898502725121177200
> 2 740000 12.47491268376206272216237202
> 2 750000 12.48693594951832440956683550
> 2 760000 12.49880013638169511883229569
> 2 770000 12.51050884124405914256853750
> 2 780000 12.52206537108093533871333277
> 2 790000 12.53347692259745252540675575
> 2 800000 12.54474459716977128685280547
> 2 810000 12.55587125001645305770022096
> 2 820000 12.56686130746829341595469366
> 2 830000 12.57771842011223665163037795
> 2 840000 12.58844534265072464040923243
> 2 850000 12.59904566706465055360427300
> 2 860000 12.60952226922408394518254911
> 2 870000 12.61987642201166975414741555
> 2 880000 12.63011418303496833362065684
> 2 890000 12.64023571583693921823484395
> 2 900000 12.65024408926657285403820450
> 2 910000 12.66014049405210257196085418
> 2 920000 12.66993013960070545251999449
> 2 930000 12.67961329658082585267888159
> 2 940000 12.68919341076487592672709130
> 2 950000 12.69867217785123447166660908
> 2 960000 12.70805040456023104531710849
> 2 970000 12.71733330696500792660758365
> 2 980000 12.72652037909829933006161382
> 2 990000 12.73561385879130711822105048
> 2 1000000 12.74461639944508312563225494



         UnitaryPhi(m) < (-1)Sigma(m) < Sigma(m)

         So, U(2) < N(2) < S(2)=Zeta(1):*Zeta(2)

         Where,U(s) = Sum_{m=1 to infinity} UnitaryPhi(m)/m^s , S(s) = 
Sum_{m=1 to infinity} Sigma(m)/m^s



    I suppose that U(2) is already known.

    Does anyone know it?



    Yasutoshi
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From zbi74583 at boat.zero.ad.jp  Wed Nov 22 02:15:24 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Wed, 22 Nov 2006 10:15:24 +0900
Subject: A085058
References: <019201c70ab0$b44b3ce0$ca73d7dc@FMC57937B6BE00> <007501c70ac2$a6b28900$6303f0d5@speedy>
Message-ID: <00a101c70dd3$b38cd400$977fd7dc@FMC57937B6BE00>

   Is it an easy well known fact?

   I would like to know the reason why they are the same.



   Yasutoshi



  ----- Original Message ----- 
  From: Vladeta Jovovic
  To: kohmoto
  Sent: Saturday, November 18, 2006 12:35 PM
  Subject: Re: A085058



  Yes, they are the same sequence: see A101921.

  Best wishes,
  Vladeta Jovovic

    ----- Original Message ----- 
    From: kohmoto
    To: seqfan at ext.jussieu.fr
    Sent: Saturday, November 18, 2006 2:27 AM
    Subject: A085058


        Hi, Seqfans

        If 2^k(n) is the largest power of two dividing tangent number T(n) 
then numbers such that 2*n-k(n) are as follows.



        S : 2,3,2,4,2,3,2,5,2,....



        It looks like A085058.

        Are they the same seqences ?



        Yasutoshi




    __________ NOD32 1.1454 (20060321) Information __________

    This message was checked by NOD32 antivirus system.
    http://www.nod32.com
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From zbi74583 at boat.zero.ad.jp  Wed Nov 22 02:15:22 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Wed, 22 Nov 2006 10:15:22 +0900
Subject: A020955
Message-ID: <009801c70dd3$b0eb0000$977fd7dc@FMC57937B6BE00>

    Hi, Seafans.

    I suppose that no one understand what A020955 means.



    So, I explain it a little.

    First of all, see my home page and click ?Free Class?, recently I 
rewrote it.



         http://boat.zero.ad.jp/~zbi74583/another02.htm



    The idea ?ma??.an abbreviation of ?make??.is represented by a 
mapping from subsets of a set to members of the set..

    Because the idea ?ma? has the same property as a mapping.



    Number of subsets of a set which has N members is 2^N, and the 
possibility of images of each subsets is N, so number of the mapping 
subsets ->members is N^(2^N).



    But if the set is a model of Free Class then the mapping must satisfy 
some conditions.         .

    They are as follows.



    If the set is {0,1,2,3,?.n} then { } ma 0, {0} ma1, {0,1} ma 2, {0,1,2} 
ma 3,?.,{0,1,2,3,?.n} ma n .



    So, images of n+1 subsets are already decided.

    Hence the number of the mapping or the number of models which has N 
members is N^(2^N-n-1) .



    Yasutoshi


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From aplewe at sbcglobal.net  Wed Nov 22 02:59:20 2006
From: aplewe at sbcglobal.net (Andrew Plewe)
Date: Tue, 21 Nov 2006 17:59:20 -0800
Subject: Almost the number of divisors of n+1
In-Reply-To: <200611212325.kALNPQh3019729@smtp.well.com>
Message-ID: <200611220159.kAM1xMZG084485@shiva.jussieu.fr>

Marc said:

Some random and rather vague ideas:

1. I faintly recall seeing something about algorithms for
factoring n involving properties of n-1 (I mean besides
the obvious connection to "Fermat's little theorem").
Perhaps in Knuth's TAOCP? 



I have that book in front of me right now; he demonstrates how to use the
properties of n-1 to verify primality (pg. 375, in vol. 2, 2nd edition)
after applying Fermat's theorm. 

One a similar note, one could also try to solve:

n - 1 = a mod (a + 1) (or, more generally, n - x = a mod (a + x))

in which case a + 1 (or x) will be a factor of n, but I don't think that's a
particularly practical method -- finding a "mod difference" for n - x seems
to be just as hard as factoring n, even if you know the factors of n - x.

	-Andrew Plewe-

-----Original Message-----
From: Marc LeBrun [mailto:mlb at well.com] 
Sent: Tuesday, November 21, 2006 3:26 PM
To: seqfan at ext.jussieu.fr
Subject: Re: Almost the number of divisors of n+1

 >  a(n) is the number of divisors of (n+1)
 > Have you ever seen closely-related sequences like that one?

Some random and rather vague ideas:

1. I faintly recall seeing something about algorithms for
factoring n involving properties of n-1 (I mean besides
the obvious connection to "Fermat's little theorem").
Perhaps in Knuth's TAOCP?

2. It sounds like you might have a case where the relationship
you're seeing, a(n) ~ d(n), is actually a first approximation
to some identity like

   a(n) = d(n) + F1(n) + F2(n) + ...

where the Fk depend on the factor structure of n or something.
So perhaps you might try to "take out" the d(n) and then look
further into the "error" and see what you can guess about how
the Fk's specifically vary with primes, highly composites, etc.
For example you might try "fitting" a(n) with Moebius-style sums.

3. There's some rather surprising known relationships involving
divisors--my favorite is one of Legendre's that is basically an
identity found by expanding elliptic functions in power series.
This suggests that you might want to express your relationships
as generating functions and see what emerges.

4. Sometimes there is simple underlying "combinatorial machinery".
For example I was once puzzled by some coefficients that turned
out to be formed by taking each divisor of n, *adding 1*, and then
multiplying them all together!  It took me a while to see that's
algebraically identical to summing the products of every possible
subset of the divisors.  But the "add 1" way of expressing this
was totally opaque, whereas the "every subset" explained them.

Happy hunting!







From bowerc at usa.net  Wed Nov 22 03:31:46 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Tue, 21 Nov 2006 18:31:46 -0800
Subject: Assembly of nothing
Message-ID: <129kkVcEU2922S01.1164162706@cmsweb01.cms.usa.net>

This is about a problem with sequence transforms I've been thinking
about for a couple of years now. I call it the "assembly of nothing"
problem.

Assembly refers to creating a set of combinatorial structures
classified by number of nodes of the form A(B) where B is a
preexisting set of structures and A is a rule for how to place
the B-structures into the new structure.

One type of assembly is combinatorial substitution (or composition).
In this instance the "A" rule is itself a collection of structures.
The assembly is to take copies of "B" structures and use them as
the points of the "A" structure.

A variant of this assembly is one where certain pairs of points in
the "A" structure are set aside to be filled with the same "B"
structure (i.e. 2 copies of the same structure) or different
"B" structures.

What's particularly notable about these assemblies is that they
can be described by index series which are formal polynomial
series of the indexed variables x_1, x_2, x_3, ....

The "nothing" part of the title refers to "B" structures that have
a size (or number of points) of zero. The formulas detailing the
counting of these structures frequently forbid the "B" structure
from containing any zero sized members. In many cases this seems
like a necessary restriction, but in others, it can get in the
way of some useful combinatorial questions.

Let's consider 2 types of assemblies.

Sets of structures, allowing the same structure multiple times
(as implemented by the Euler transform for unlabeled structures
and the Exponential transform for labeled structures.)

Sets of structures, requiring all structures to be different
(as implemented by the Weigh transform for unlabeled structures.)

http://www.research.att.com/~njas/sequences/A000041
The partition sequence A000041 counts unlabeled sets of nonempty sets
hence it has g.f. EULER(x/(1-x)) if we consider EULER as a g.f.
transformation.

http://www.research.att.com/~njas/sequences/A000009
A000009 counts sets of nonempty sets all different from one another
with g.f. WEIGH(x/(1-x)).

in the first case, it makes sense that the "B" structure (the
nonempty sets) are not allowed to have 0 sized elements (i.e.
be empty). If we tried to count partitions of n and allowed n
to be partitioned into 0s, we would have infinitely many
possibilities. 1, 1+0, 1+0+0, ....

In the second case, we can allow empty "B" structures, since
each structure can only be used once giving us 4 partitions
of 3 into distinct parts: 3, 3+0, 1+2, 1+2+0, or in general
2*A000009(n) partitions of n into distinct parts allowing 0
as a part. Hence we would like to say that
WEIGH(1/(1-x)) = 2*WEIGH(x/(1-x)).

The commonly used formulas appear to support these observations.

The Euler transform as applied to a sequence is often rendered:

(1-x)^(-a1)*(1-x^2)^(-a2)*(1-x^3)^(-a3)*(1-x^4)^(-a4)*...

with the Weigh transform as

(1+x)^a1*(1+x^2)^a2*(1+x^3)^a3*(1+x^4)^a4*...

both of these formulas ignore the a0 term, but if we were to extend
these one term backward we would have Euler start

(1-1)^(-a0)*...

and Weigh

(1+1)^a0*...

and we can see that in Euler's case given any positive a0 we begin
with 0 raised to a negative power giving an infinite value and for
Weigh's case we start the product with 2^a0.

A similar effect is seen from the index series.

An index series is applied to a g.f. as follows:

If s(x_1, x_2, x_3, ...) is the index series, then it maps the
g.f. A(x) to s(A(x), A(x^2), A(x^3), ...) thus x_2^2 + x_3*x_4
maps as A(x^2)^2+A(x^3)*A(x^4).

Euler's i.s. is exp(x_1/1 + x_2/2 + x_3/3 + ...)

Weigh's i.s. is exp(x_1/1 - x_2/2 + x_3/3 - ...)

We can see that if A(x) has a positive constant term, then
that term will be multiplied by the divergent harmonic series
1+1/2+1/3+... and blow up. With Weigh we have the alternating
harmonic series 1-1/2+1/3-...=ln 2 giving it the 2^a0 character.

So in these 2 examples, the problem seems manageable, but if one
is looking for a general theory to cover all such assemblies or
all transforms defined by i.s., things are not so easy.

Consider 2 other assembly rules:

Ordered lists of structures, allowing the same structure multiple times
(as implemented by the Invert transform for unlabeled structures.)

Ordered lists of structures, requiring a structure not have a copy
of itself as a neighbor
(I call the unlabeled version of this the Carlitz transform. It's
not in the EIS canon yet.)

Invert's i.s. is 1/(1-x_1)
Carlitz's i.s. is 1/(1-x_1+x_2-x_3+...)

Invert creates compositions (or ordered partitions) the way Euler
creates partitions.

Hence we have INVERT(x/(1-x)) = (1-x)/(1-2x) = 1,1,2,4,8,16,...

Just as in the unordered partitions, we would not allow one to
compose n into 0s, there would be infinitely many choices.
Note that unlike the Euler transform which blows up for a0>0,
The Invert transform can be calculated for any a0 != 1, although
it's difficult to give a combinatorial interpretation to any
nonzero value.

http://www.research.att.com/~njas/sequences/A003242
Carlitz(x/(1-x)) gives us A003242 Carlitz compositions, those
where no two adjacent parts are equal. We can allow Carlitz
compositions to contain 0s since the separation of the 0s by other
numbers keeps it finite: Carlitz(1/(1-x)) = A114900.
http://www.research.att.com/~njas/sequences/A114900
However, the i.s. does not tell us how to do it. If A(x)=1 what
does one make of Carlitz(A(x)) = 1/(1-1+1-1+...)

I recently had an idea about how to deal with issues like this:
use an analytic interpretation of the g.f. of a sequence.

If a sequence represents structures contains only finitely many
structures, then the number of structures is A(1) where A(x) is
the g.f. My idea is to take the case where A(x) has a radius of
convergence (ROC) >0 and <=1 so that it cannot be evaluated by
adding the terms, and associating its size with A(1) of the
analytic extension of the g.f. Here in the Carlitz case we have
1-x+x^2-x^3+... = 1/(1+x) hence A(1) = 1/2 and 1/(1/2)=2, the
expected value for Carlitz compositions of 0 allowing 0 as a part.
(The 2 are the empty composition and zero.)

In this way I can evaluate a transform of a sequence whose only
nonzero term is a_0 as T(a_0*x)(1) and for more general sequences
I can apply the transform to a 2 dimensional sequence as follows.

If sequence has g.f. A(x) then let C(x,y) = y*A(x).

Apply T to C(x,y) getting D(x,y) as output:

Let d_n = [x^n] D(x,1)

I suspect this will be an adequate solution to many problems of
this sort, but I'm not sure how far this method can be trusted.
Part of the problem is I don't know my analytic function theory
as well as I should. (If anyone out there does, please chime in.)
More importantly, I've run across some examples that challenge
the method.

My first example involves the Euler transform. It's basically only
a problem because I would expect that if f(1)=g(1) then
(Tf)(1) = (Tg)(1) for any transform with both f and g in its
domain. It seems reasonable as 1^n=1 for all n so that i.s.'s
behave like g.f.'s when evaluated at 1. More importantly, the
choice to use the linear term in "C(x,y) = y*A(x)" as opposed
to "C(x,y) = y^2*A(x)" was arnitrary and it seemingly should
not matter which term I chose, so I would like some assurance
that I'll get the same answer following any such path.

Consider

EULER(0) = 1
EULER(x-x^2) = 1+x
EULER(x-x^3) = 1+x+x^2

For all of these, the inputs have f(1)=0, but the outputs have
g(x)=1,2 or 3

I think the reason for this anamoly may be in the instability
Euler has at 0. Just as Weigh behaves like 2^x, Euler behaves
like 0^(-x) giving 0 for negative values of x, infinity for
positive values and doing what we see here at x=0. 0^0 is
generally defined as 1, but analytically it can be anything and
that may be the limitation I'm running into.

My other example concerns me, because I'm interested in formulas
that can be applied to arbitrary sequences, not just integer
sequences. Most of our familiar transform formulas work fine for
arbitrary sequences, and it would be nice if all of them did,
however...


Ordered lists of structures, requiring all structures to be different
(as implemented by the AGK transform for unlabeled structures
found on
http://www.research.att.com/~njas/sequences/transforms2.html )

Obviously, this assembly makes sense for empty structures since
each structure can appear at most once. If I allow there to be
n types of empty structures in the input then I have
A003149(n) empty structures in the output.
http://www.research.att.com/~njas/sequences/A003149

While it seems quite plausible that this sequence can be
interpolated into a nice analytic function, the method I illustrated
above offers no help as for n anything other that a
nonnegative integer gives me a sequence with ROC of zero.

So the idea in posting this is to present some new ideas to the
sequence community, get some opinions from people who know some
things that I don't know and get some ideas about how to handle
difficult cases.

Christian








From davidwwilson at comcast.net  Wed Nov 22 04:22:55 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Tue, 21 Nov 2006 22:22:55 -0500
Subject: Confirmation
References: <200611211857.kALIvmai029436@amer.strw.leidenuniv.nl>
Message-ID: <001001c70de5$812f2190$6501a8c0@yourxhtr8hvc4p>

Thanks for your help, your results agree with my values.





From jawbrey at att.net  Wed Nov 22 04:28:31 2006
From: jawbrey at att.net (Jon Awbrey)
Date: Tue, 21 Nov 2006 22:28:31 -0500
Subject: Assembly of nothing
References: <129kkVcEU2922S01.1164162706@cmsweb01.cms.usa.net>
Message-ID: <4563C3DE.FB5134DC@att.net>

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o

This is one of the impulses that led to the construction of riffs and rotes.
It was probably in my freshman chemistry class, or later when I worked as a
hospital orderly and used to run errands for the nuclear medicine unit that
I was thinking about the techniques of molecular tagging, which is like the
business of labelling graphs, only you have to label molecules with bits of
physical structure, an extra neutron or the like, that alters the resulting
structure.

So the question arises whether you can have a class of structures that is
closed under self-labeling.  Somewhat roughly trying to use your notation,
if A(B) is the set of all structures where those of type A are labeled by
those of type B, can you have the doamin equation A = A(A) for any type A?

Or something like that ...

Jon Awbrey

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o
inquiry e-lab: http://stderr.org/pipermail/inquiry/
wikinfo: http://wikinfo.org/wiki.php?title=User:Jon_Awbrey
wp review: http://wikipediareview.com/index.php?showuser=398
o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o






From Annette.Warlich at t-online.de  Wed Nov 22 13:37:30 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Wed, 22 Nov 2006 13:37:30 +0100
Subject: Eigensequences
Message-ID: <4564448A.5010200@t-online.de>

In  a recent treatise of the pascal matrix and its
connections to the bernoulli-numbers I restated
their connection as an eigenvector/eigensequence-
relation.

Say P is the lower triangular matrix of binomial-
coefficients, Pj the matrix, where the column-signs
alternate and B the vector of bernoulli-numbers
(b_1 = +1/2)
then
   Pj * B = B

Now in connection with the subject of divergent
summmation I investigated powerseries, say
  V(x) = [1,x,x^2,x^3,... ]
in connection with Pj, like

  ??? = Pj * V(x)      // binomial-transform
  ??? = V(x)~ * Pj

(using ~ for the transpose)

Now since

   Pj * B = B

also

   V(x) * Pj * B = V(x) * B

and the effect of the coefficients of Pj
disappear, if I consider powerseries with
the bernoulli-numbers as coefficients:

 f(x) = b0 + b1*x + b2*x^2 + b3*x^3 + ...
      =    b0
         + (1*b0 - 1*b1)*x
         + (1*b0 - 2*b1 + 1*b2)*x^2
         + ...

---------
Now incidently I also found, that, using J as a diagonal-
unit-matrix with units with alternating signs
 J = diag(1,-1,1,-1,...)

such that also

 Pj  = P * J

then a matrix G, containing the bernoulli-numbers
in the first column (and shifted and scaled
in a certain manner in the follwing columns)

 Pj = G * J * G^-1

and thus, for instance, the above eigensequence-
property of the bernoulli-numbers can be extended
to an infinite set of coefficients arising from

   V(x) * Pj * G = V(x) * G * J

in the infinite number of columns of the result.

The second column, for instance adds to the above
f(x) the function g(x)

 g(x) =    1*b0*x + 2*b1*x^2 + 3*b2*x^3 + 4*b3*x^4 + ...
      = -  (    1* 0
             + (1* 0 - 1*1*b0)*x
             + (1* 0 - 2*1*b0 + 1*2*b1)*x^2
             + (1* 0 - 3*1*b0 + 3*2*b1 -  1*3*b2)*x^3
             + ...
      = -  (
               ( - 1*b0)*x
             + ( - 2*b0 + 2*b1)*x^2
             + ( - 3*b0 + 6*b1 -  3*b2)*x^3
             + ...

or, using x=1, the said eigensequence-operation, here
with the associated eigenvalue -1.

-------------------------------------
Even more interesting seems to me, that the said matrix
G can be decomposed into an eigensystem, where

  G = St2 * Z(1) * St2^-1

with St2 = matrix of stirling-numbers of 2'nd kind
     Z(s)= diag(1,  1/2^s,  1/3^s, 1/4^s,... )
the diagonal-matrix of the zeta-summands of power s.

That means, that even the bernoulli-numbers have
eigensequences related to them (even infinite many),
namely the columns of St2 , associated with the
eigenvalues (1,1/2,1/3,...) such that

  V(x) * G * St2*Z(1) = V(x)*St2

and for instance,

  f(x) = 1 + 1*x + 1*x^2 + 1*x^3 + ....  // taken from V(x)*St2

       =       1*b0                     // taken from V(x) * G * St2 * Z(1)
          + (1/2*b0 + b1)*m
          + (1/3*b0 + b1 +     b2 )*m^2
          + (1/4*b0 + b1 + 3/2*b2 +      b3 )*m^3
          + (1/5*b0 + b1 +   2*b2 +    2*b3 +     b4)*m^4
          + (1/6*b0 + b1 + 5/2*b2 + 10/3*b3 + 5/2*b4 +      b5)*m^5
           ...

which may look kind trivial, since the involved Stirling-
coefficients are all 1
Other coefficients occur in the next columns 1,2,3,..., for instance
in column 1:

   g(x)  =.  x + S2_1*x^2 + S3_1*x^3 + S4_1*x^4 + S5_1*x^5 + ....
         =              b0 * x
             ( 2/3*S2_1*b0 +      2*b1 ) * x^2
             ( 1/2*S3_1*b0 + 2*S2_1*b1 + 3     *b2 )*m^3
             ( 2/5*S4_1*b0 + 2*S3_1*b1 + 4*S2_1*b2 +    4     *b3 ) *m^4
             ( 1/3*S5_1*b0 + 2*S4_1*b1 + 5*S3_1*b2 + 20/3*S2_1*b3  +  5*b4) *m^5

and so on for the next columns.
With the involved Stirling-numbers evaluated we have then for this 2'nd
column:

   g(x)  =   .    2*x + 6*x^2 + 14*x^3 + 30*x^4  +  62*x^5
         =          2*b0 * x
            +   (   4*b0 +  4*b1)*x^2
            +   (   7*b0 + 12*b1 +  6*b2) *x^3
            +   (  12*b0 + 28*b1 + 24*b2 +  8*b3) *m^4
            +   (62/3*b0 + 60*b1 + 70*b2 + 40*b3 + 10*b4))))*m^5
              ...
which, if the bernoulli-numbers are inserted is obvious.
Using x=1 and written as sequences, we have another
eigensequence-expression and implicitely infinitely many.

This description was animated by Neil's article in the
integer-sequences-journal, which I came across recently -
though I do not know, if the above is actually of interest
related to the article (my current context is originally
the study of divergent summation, where these identities
popped up)

Regards-

Gottfried Helms







From Annette.Warlich at t-online.de  Wed Nov 22 13:46:11 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Wed, 22 Nov 2006 13:46:11 +0100
Subject: Eigensequences /editing corrections
Message-ID: <45644693.1000503@t-online.de>

// upps. In the previous article I've overlooked some
// needed editings, replacing the m,m^2,... by x,x^2 ...
// sorry. Repost here:


In  a recent treatise of the pascal matrix and its
connections to the bernoulli-numbers I restated
their connection as an eigenvector/eigensequence-
relation.

Say P is the lower triangular matrix of binomial-
coefficients, Pj the matrix, where the column-signs
alternate and B the vector of bernoulli-numbers
(b_1 = +1/2)
then
   Pj * B = B

Now in connection with the subject of divergent
summmation I investigated powerseries, say
  V(x) = [1,x,x^2,x^3,... ]
in connection with Pj, like

  ??? = Pj * V(x)      // binomial-transform
  ??? = V(x)~ * Pj

(using ~ for the transpose)

Now since

   Pj * B = B

also

   V(x) * Pj * B = V(x) * B

and the effect of the coefficients of Pj
disappear, if I consider powerseries with
the bernoulli-numbers as coefficients:

 f(x) = b0 + b1*x + b2*x^2 + b3*x^3 + ...
      =    b0
         + (1*b0 - 1*b1)*x
         + (1*b0 - 2*b1 + 1*b2)*x^2
         + ...

---------
Now incidently I also found, that, using J as a diagonal-
unit-matrix with units with alternating signs
 J = diag(1,-1,1,-1,...)

such that also

 Pj  = P * J

then a matrix G, containing the bernoulli-numbers
in the first column (and shifted and scaled
in a certain manner in the follwing columns)

 Pj = G * J * G^-1

and thus, for instance, the above eigensequence-
property of the bernoulli-numbers can be extended
to an infinite set of coefficients arising from

   V(x) * Pj * G = V(x) * G * J

in the infinite number of columns of the result.

The second column, for instance adds to the above
f(x) the function g(x)

 g(x) =    1*b0*x + 2*b1*x^2 + 3*b2*x^3 + 4*b3*x^4 + ...
      = -  (    1* 0
             + (1* 0 - 1*1*b0)*x
             + (1* 0 - 2*1*b0 + 1*2*b1)*x^2
             + (1* 0 - 3*1*b0 + 3*2*b1 -  1*3*b2)*x^3
             + ...
      = -  (
               ( - 1*b0)*x
             + ( - 2*b0 + 2*b1)*x^2
             + ( - 3*b0 + 6*b1 -  3*b2)*x^3
             + ...

or, using x=1, the said eigensequence-operation, here
with the associated eigenvalue -1.

-------------------------------------
Even more interesting seems to me, that the said matrix
G can be decomposed into an eigensystem, where

  G = St2 * Z(1) * St2^-1

with St2 = matrix of stirling-numbers of 2'nd kind
     Z(s)= diag(1,  1/2^s,  1/3^s, 1/4^s,... )
the diagonal-matrix of the zeta-summands of power s.

That means, that even the bernoulli-numbers have
eigensequences related to them (even infinite many),
namely the columns of St2 , associated with the
eigenvalues (1,1/2,1/3,...) such that

  V(x) * G * St2*Z(1) = V(x)*St2

and for instance,

  f(x) = 1 + 1*x + 1*x^2 + 1*x^3 + ....  // taken from V(x)*St2

       =       1*b0                     // taken from V(x) * G * St2 * Z(1)
          + (1/2*b0 + b1)*x
          + (1/3*b0 + b1 +     b2 )*x^2
          + (1/4*b0 + b1 + 3/2*b2 +      b3 )*x^3
          + (1/5*b0 + b1 +   2*b2 +    2*b3 +     b4)*x^4
          + (1/6*b0 + b1 + 5/2*b2 + 10/3*b3 + 5/2*b4 +      b5)*x^5
           ...

which may look kind trivial, since the involved Stirling-
coefficients are all 1
Other coefficients occur in the next columns 1,2,3,..., for instance
in column 1:

   g(x)  =.  x + S2_1*x^2 + S3_1*x^3 + S4_1*x^4 + S5_1*x^5 + ....
         =              b0 * x
             ( 2/3*S2_1*b0 +      2*b1 ) * x^2
             ( 1/2*S3_1*b0 + 2*S2_1*b1 + 3     *b2 )*x^3
             ( 2/5*S4_1*b0 + 2*S3_1*b1 + 4*S2_1*b2 +    4     *b3 ) *x^4
             ( 1/3*S5_1*b0 + 2*S4_1*b1 + 5*S3_1*b2 + 20/3*S2_1*b3  +  5*b4) *x^5

and so on for the next columns.
With the involved Stirling-numbers evaluated we have then for this 2'nd
column:

   g(x)  =   .    2*x + 6*x^2 + 14*x^3 + 30*x^4  +  62*x^5
         =          2*b0 * x
            +   (   4*b0 +  4*b1)*x^2
            +   (   7*b0 + 12*b1 +  6*b2) *x^3
            +   (  12*b0 + 28*b1 + 24*b2 +  8*b3) *x^4
            +   (62/3*b0 + 60*b1 + 70*b2 + 40*b3 + 10*b4) *x^5
              ...
which, if the bernoulli-numbers are inserted is obvious.
Using x=1 and written as sequences, we have another
eigensequence-expression and implicitely infinitely many.

This description was animated by Neil's article in the
integer-sequences-journal, which I came across recently -
though I do not know, if the above is actually of interest
related to the article (my current context is originally
the study of divergent summation, where these identities
popped up)

Regards-

Gottfried Helms








From njas at research.att.com  Wed Nov 22 17:58:32 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 22 Nov 2006 11:58:32 -0500 (EST)
Subject: A125097.gif: (Very) nice patterns
References: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>
Message-ID: <200611221658.LAA41364@fry.research.att.com>

The other day Zak said:
> Neil,
> seqfans,
>
> 1. Have a look at this full graph -
> aint these patterns (very) nice?!
> It may be known, but still it's nice...
>
> 2. Is it possible to
> put "the extended graph" into OEIS?
> In the standard graph no patterns are seen at all...
>
> 3. Neil removed  keyword "nice" from A125097
> (as he did regularly with ny sequences)-
> as if I insist that it's me who is "nice"
> not it is the seq which is nice.
>

Here is my reply:
1. i will use the b-file
2. we don't need the gif since the graph button will use the b-file
3. the definition of the sequence is
     a(n) is k such that k*(2n+1) mod 1000 = 123, or 2n+1 if no such k exists.
Now there is rule that a sequence should not depend on an arbitrary 
parameter. Here we have two of them. Why not:
a(n) is k such that k*(2n+1) mod 44445555 = 321345, or 2n+1 if no such k exists. ?
4. the sequence is not finite (a(1000) = 123)
5. although i know you are a nice guy, i do not think this sequence is nice.
Best regards
Neil





From grafix at csl.pl  Wed Nov 22 18:58:30 2006
From: grafix at csl.pl (Artur)
Date: Wed, 22 Nov 2006 18:58:30 +0100
Subject: All primes
In-Reply-To: <200611221658.LAA41364@fry.research.att.com>
References: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com> <200611221658.LAA41364@fry.research.att.com>
Message-ID: <op.tjfivsouat030q@tata>

Dear Members
I'm looking for members which have Mathematica Programme and can help in
finding the first number such that
(PrimeQ[2(n) + 1] == True) && (PrimeQ[4(
       n) + 1] == True) && (PrimeQ[6(n) + 1] == True) && (PrimeQ[8(
       n) + 1] == True) && (PrimeQ[10(n) + 1] == True) && (PrimeQ[12(
       n) + 1] == True) && (PrimeQ[14(n) + 1] == True) && (PrimeQ[16(
       n) + 1] == True) && (PrimeQ[18(n) + 1] == True) && (PrimeQ[20(
       n) + 1] == True) && (PrimeQ[22(n) + 1] == True)
these n is bigger than 4850000000000
I have Mathematica procedure which is 4000 much more quickest as above but  
for as big numbers PrimeQ working slow and will be good divided range of  
searching on few subranges and run each other on different computer.
If somebody can help let me know. I need information how long time  
computer can run without reset.
BEST REGARDS
ADRTUR JASINSKI






From rgwv at rgwv.com  Wed Nov 22 19:22:17 2006
From: rgwv at rgwv.com (Robert G. Wilson v)
Date: Wed, 22 Nov 2006 13:22:17 -0500
Subject: Whereabouts of Joe K. Crump & www.spacefire.com
Message-ID: <45649559.9030304@rgwv.com>

,

Does anyone on this list know the whereabouts of Joe K. Crump or the
status of the website www.spacefire.com ? Thanks in advance for any
information.

Bob.





From noe at sspectra.com  Wed Nov 22 19:27:50 2006
From: noe at sspectra.com (T. D. Noe)
Date: Wed, 22 Nov 2006 10:27:50 -0800
Subject: All primes
In-Reply-To: <op.tjfivsouat030q@tata>
References: <20061121051911.12979.qmail@web38212.mail.mud.yahoo.com>
 <200611221658.LAA41364@fry.research.att.com> <op.tjfivsouat030q@tata>
Message-ID: <p06230905c18a46fef974@[192.168.1.43]>

At 7:04 PM +0100 11/22/06, Artur wrote:
>Dear Members
>I'm looking for members which have Mathematica Programme and can help in
>finding the first number such that
>(PrimeQ[2(n) + 1] == True) && (PrimeQ[4(
>       n) + 1] == True) && (PrimeQ[6(n) + 1] == True) && (PrimeQ[8(
>       n) + 1] == True) && (PrimeQ[10(n) + 1] == True) && (PrimeQ[12(
>       n) + 1] == True) && (PrimeQ[14(n) + 1] == True) && (PrimeQ[16(
>       n) + 1] == True) && (PrimeQ[18(n) + 1] == True) && (PrimeQ[20(
>       n) + 1] == True) && (PrimeQ[22(n) + 1] == True)
>these n is bigger than 4850000000000

See A071576.

Tony






From mathar at strw.leidenuniv.nl  Wed Nov 22 19:39:07 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Wed, 22 Nov 2006 19:39:07 +0100
Subject: Whereabouts of Joe K. Crump & www.spacefire.com
Message-ID: <200611221839.kAMId7db031776@amer.strw.leidenuniv.nl>


> From seqfan-owner at ext.jussieu.fr  Wed Nov 22 19:09:21 2006
> Date: Wed, 22 Nov 2006 13:22:17 -0500
> From: "Robert G. Wilson v" <rgwv at rgwv.com>
> To: "seqfan at ext.jussieu.fr" <SeqFan at ext.jussieu.fr>
> Subject: Whereabouts of Joe K. Crump & www.spacefire.com
> 
> Does anyone on this list know the whereabouts of Joe K. Crump or the
> status of the website www.spacefire.com ? Thanks in advance for any
> information.
..
According to Max (whom we now know as Alekseyev), there is a 
archive.org backup of his web page:

http://groups-beta.google.com/group/sci.math/browse_thread/thread/8d061b676bd7e32b/071ae01f53f9a1f5?hl=en#071ae01f53f9a1f5
(sci.math of Nov 14)
http://web.archive.org/web/20040609202646/www.spacefire.com/numbertheory/

An e-mail address of 2002 is shown in
http://www.primepuzzles.net/thepuzzlers/Crump.htm
and becomes
ssaywer1 at rr.carolina.com
in the archived web page of 2000.





From franktaw at netscape.net  Wed Nov 22 22:20:32 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 22 Nov 2006 16:20:32 -0500
Subject: Categories
Message-ID: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>

How many categories are there?

First, how many categories are there with n morphisms and k objects?  
This table starts:

 1
 2  1
 7  3 1
35 16 3 1

The first column is A058129, the number of monoids; the main diagonal 
is all 1's.  I am not
100% certain of the 16 in the final row.

Taking the row sums, we get:

1,3,11,55

the number of categories with n morphisms.  This is probably not in the 
OEIS (only
A001776 is possible - other matches become less than A058129).  The 
inverse Euler
transform,

1,2,8,41

is the number of connected categories with n morphisms; this is 
likewise probably not
in the OEIS (only A052447 is possible).

Can somebody generate more data?

Franklin T. Adams-Watters

A category is a collection of objects and morphisms; each morphism is 
from one object
to another (not necessarily different) object.  Where the destination 
of one morphism
is the source of a second, their composition is defined; composition is 
associative where
it is defined.  Each object has an identity morphism, which connects it 
to itself; this
is an identity when composed with morphisms coming in and with 
morphisms going
out.
________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From jvospost3 at gmail.com  Wed Nov 22 23:55:44 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 22 Nov 2006 14:55:44 -0800
Subject: Categories
In-Reply-To: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
Message-ID: <5542af940611221455r6e693b45kae8a9af38c6ee9eb@mail.gmail.com>

This actually relates to the enumeration results I've made on "prime"
endofunctions, namely endofunctions which are not unions of other prime
endofunctions nor categorical products of endofunctions. My partial results
here are in A125024, A124023, and a comment on A048888 that I submitted a
day or two ago about the enumeration of prime endofunctions in terms of
numbrals.

I have been working on enumerating not just endofunctions, but
(categorically) endomorphisms of endofunctions, (2-categorically)
endomorphisms of endomorphisms of endofunctions.  I've been posting partial
results on John Baez's "n-category cafe" and apparently just annoyed John
Baez, who didn't yet see that I was finding a new combinatorial result at
the base level (endofunctions) and working on the n-category hierarchy above
that, so those comments may already have been deleted from his blog.

That is, in counting all categories on n objects, most are trivially the
union of smaller disjoint categories (that is, the sagittal digraph is not
connected), or categorical products of smaller categories.  They key to
enumeration involves thus both a tree-enumeration, and the numbral part
which, per a recent paper in the Journal on Integer Sequences, relate to
balanced ordered trees.

Hard to summarize less cryptically, here, but I stand by in support of what
you're doing, and think my approach is genuinely related.

-- Jonathan Vos Post

On 11/22/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
> How many categories are there?
>
> First, how many categories are there with n morphisms and k objects?
> This table starts:
>
> 1
> 2  1
> 7  3 1
> 35 16 3 1
>
> The first column is A058129, the number of monoids; the main diagonal
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
>
> Taking the row sums, we get:
>
> 1,3,11,55
>
> the number of categories with n morphisms.  This is probably not in the
> OEIS (only
> A001776 is possible - other matches become less than A058129).  The
> inverse Euler
> transform,
>
> 1,2,8,41
>
> is the number of connected categories with n morphisms; this is
> likewise probably not
> in the OEIS (only A052447 is possible).
>
> Can somebody generate more data?
>
> Franklin T. Adams-Watters
>
> A category is a collection of objects and morphisms; each morphism is
> from one object
> to another (not necessarily different) object.  Where the destination
> of one morphism
> is the source of a second, their composition is defined; composition is
> associative where
> it is defined.  Each object has an identity morphism, which connects it
> to itself; this
> is an identity when composed with morphisms coming in and with
> morphisms going
> out.
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and
> industry-leading spam and email virus protection.
>
>
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From jvospost3 at gmail.com  Thu Nov 23 01:39:16 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 22 Nov 2006 16:39:16 -0800
Subject: Categories
In-Reply-To: <5542af940611221455r6e693b45kae8a9af38c6ee9eb@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <5542af940611221455r6e693b45kae8a9af38c6ee9eb@mail.gmail.com>
Message-ID: <5542af940611221639n2fde8e43j93dc69594ae9392b@mail.gmail.com>

Next question is thus: "How many 2-categories (bicategories) are there whose
objects are the categories on n objects?"

See:
Basic Bicategories
Tom Leinster (1998-10-04)  arXiv.org:math/9810017 A concise guide to very
basic bicategory theory, from the definition of a bicategory to the
coherence theorem. Comment: 11 pages; LaTeX 2e with Paul Taylor's diagram
macros

This is even harder to answer, because there seem to be at least 7 different
definitions of n-categories, of which categories = 1-categories, and I've
just raised the enumeration probem for n=2.

www.ima.umn.edu/categories/abstracts.html
Abstracts for the IMA 2004 Summer Program: n-Categories.
Ross Street has proposed two related definitions of n-category, one in 1987
and one in 2002.

See also:
Tom Leinster (2003-05-02) arXiv.org:math/0305049 Higher-dimensional category
theory is the study of n-categories, operads, braided monoidal categories,
and other such exotic structures. It draws its inspiration from areas as
diverse as topology, quantum algebra, mathematical physics, logic, and
theoretical computer science. This is the first book on ... Comment: Book,
410 pages

http://www.lepp.cornell.edu/spr/2002-04/msg0041086.html

In the beginning, there was nothing.
But with nothing came nothingness, the vacuity, which was something.
So now there was nothingness and somethingness,
vacuity and triviality, falsehood and truth,
which were 2 things.
So now there was a set of nothingness and somethingness,
and the whole realm of sets of elements
sprang up out of the set of nothingness and somethingness,
and sprouted functions between them
to relate them back to the set of nothingness and somethingness.
So now there was a category of sets and functions,
and the whole realm of categories of objects and morphisms
sprang up out of the category of sets and functions,
and sprouted functors between them,
to relate them back to the category of sets and functions,
which sprouted natural transformations between *them*,
to relate the relationships.
So now there was ...


-- Toby
   toby at math.ucr.edu

But your question is fundamental, and I am eager to see an answer.  I hope
it is not rude for me to be anticipating follow-ups.

Happy Thanksgiving,

Jonathan Vos Post

On 11/22/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> This actually relates to the enumeration results I've made on "prime"
> endofunctions, namely endofunctions which are not unions of other prime
> endofunctions nor categorical products of endofunctions. My partial results
> here are in A125024, A124023, and a comment on A048888 that I submitted a
> day or two ago about the enumeration of prime endofunctions in terms of
> numbrals.
>
> I have been working on enumerating not just endofunctions, but
> (categorically) endomorphisms of endofunctions, (2-categorically)
> endomorphisms of endomorphisms of endofunctions.  I've been posting partial
> results on John Baez's "n-category cafe" and apparently just annoyed John
> Baez, who didn't yet see that I was finding a new combinatorial result at
> the base level (endofunctions) and working on the n-category hierarchy above
> that, so those comments may already have been deleted from his blog.
>
> That is, in counting all categories on n objects, most are trivially the
> union of smaller disjoint categories (that is, the sagittal digraph is not
> connected), or categorical products of smaller categories.  They key to
> enumeration involves thus both a tree-enumeration, and the numbral part
> which, per a recent paper in the Journal on Integer Sequences, relate to
> balanced ordered trees.
>
> Hard to summarize less cryptically, here, but I stand by in support of
> what you're doing, and think my approach is genuinely related.
>
> -- Jonathan Vos Post
>
> On 11/22/06, franktaw at netscape.net <franktaw at netscape.net > wrote:
> >
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> > 1
> > 2  1
> > 7  3 1
> > 35 16 3 1
> >
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
> > Taking the row sums, we get:
> >
> > 1,3,11,55
> >
> > the number of categories with n morphisms.  This is probably not in the
> > OEIS (only
> > A001776 is possible - other matches become less than A058129).  The
> > inverse Euler
> > transform,
> >
> > 1,2,8,41
> >
> > is the number of connected categories with n morphisms; this is
> > likewise probably not
> > in the OEIS (only A052447 is possible).
> >
> > Can somebody generate more data?
> >
> > Franklin T. Adams-Watters
> >
> > A category is a collection of objects and morphisms; each morphism is
> > from one object
> > to another (not necessarily different) object.  Where the destination
> > of one morphism
> > is the source of a second, their composition is defined; composition is
> > associative where
> > it is defined.  Each object has an identity morphism, which connects it
> > to itself; this
> > is an identity when composed with morphisms coming in and with
> > morphisms going
> > out.
> > ________________________________________________________________________
> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > industry-leading spam and email virus protection.
> >
> >
>
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From maxale at gmail.com  Thu Nov 23 04:52:55 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 22 Nov 2006 19:52:55 -0800
Subject: Categories
In-Reply-To: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
Message-ID: <d3dac270611221952h2853749ev8fcb5ced9e2f80a7@mail.gmail.com>

On 11/22/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
> How many categories are there?
>
> First, how many categories are there with n morphisms and k objects?

Does that correspond to transitively closed digraphs with k (labeled?)
vertices and n edges?

Max





From maxale at gmail.com  Thu Nov 23 05:10:51 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 22 Nov 2006 20:10:51 -0800
Subject: Whereabouts of Joe K. Crump & www.spacefire.com
In-Reply-To: <200611221839.kAMId7db031776@amer.strw.leidenuniv.nl>
References: <200611221839.kAMId7db031776@amer.strw.leidenuniv.nl>
Message-ID: <d3dac270611222010m17a14a53x75b9de0cf2ffbe3a@mail.gmail.com>

On 11/22/06, Richard Mathar <mathar at strw.leidenuniv.nl> wrote:

> http://web.archive.org/web/20040609202646/www.spacefire.com/numbertheory/

I've put an updated version of Joe's Database of 2^x-3 factorizations at
http://www-cse.ucsd.edu/users/maxal/2nm3_db.txt

Please contribute if you like.

Max





From pauldhanna at juno.com  Thu Nov 23 07:19:41 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Thu, 23 Nov 2006 01:19:41 -0500
Subject: Formula for A112319? - Diagonals in Self-Compositions of (x+x^2)
Message-ID: <20061123.011941.272.1.pauldhanna@juno.com>

Seqfans, 
     Would someone like to try to find a formula for A112319:  
"Coefficients of x^n in the (n-1)-th self-composition of (x + x^2) for
n>=1."
1, 1, 2, 9, 64, 630, 7916, 121023, 2179556, 45179508, 1059312264, 
 
The related triangles A122888 and A122890 seem to indicate that 
A112319 may have a formula involving the Catalan numbers: 
http://www.research.att.com/~njas/sequences/A122890
 
Below I copy A112319 and triangle A122888 for your convenience. 
Other diagonals in self-compositions of (x+x^2) are A112317, A112320.
 
Be it a recurrence, series reversion, etc., any formula for these
diagonals 
could lead to other discoveries involving self-compositions of functions.

     Paul 
----------------------------------------------------------------

A122888
Triangle, read by rows, where row n lists the coefficients of x^k, 
k=1..2^n, in the n-th self-composition of (x + x^2) for n>=0. 

EXAMPLE  
Triangle begins: 
1; 
1, 1; 
1, 2, 2, 1; 
1, 3, 6, 9, 10, 8, 4, 1; 
1, 4, 12, 30, 64, 118, 188, 258, 302, 298, 244, 162, 84, 32, 8, 1; 

Multiplying the g.f. of column k by (1-x)^k, k>=1, with leading zeros, 
yields the g.f. of row k in the triangle A122890: 
1; 
0, 1; 
0, 0, 2; 
0, 0, 1, 5; 
0, 0, 0, 10, 14; 
0, 0, 0, 8, 70, 42; 
0, 0, 0, 4, 160, 424, 132; 
0, 0, 0, 1, 250, 1978, 2382, 429; 
0, 0, 0, 0, 302, 6276, 19508, 12804, 1430; ... 

in which the main diagonal is the Catalan numbers 
and the row sums form the factorials. 

----------------------------------------------------------------

A112319  
Coefficients of x^n in the (n-1)-th self-composition of (x + x^2) for
n>=1. 

1, 1, 2, 9, 64, 630, 7916, 121023, 2179556, 45179508, 1059312264, 
27715541568, 800423573676, 25289923553700, 867723362137464, 
32128443862364255, 1276818947065793736, 54208515369076658640, 
2448636361058495090816
 
FORMULA  
a(n) = [x^n] F_{n-1}(x) where F_n(x) = F_{n-1}(x+x^2) 
with F_1(x) = x+x^2 and F_0(x)=x for n>=1. 
 
EXAMPLE  
Initial terms in self-compositions of (x+x^2) are: 
F(x) = x + (1)*x^2 
F(F(x)) = x + 2*x^2 + (2)*x^3 + x^4 
F(F(F(x))) = x + 3*x^2 + 6*x^3+ (9)*x^4 +... 
F(F(F(F(x)))) = x + 4*x^2 + 12*x^3 + 30*x^4 + (64)*x^5 +... 
F(F(F(F(F(x))))) = x + 5*x^2 + 20*x^3 + 70*x^4 + 220*x^5 + (630)*x^6 +...
 
----------------------------------------------------------------
END.
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From jvospost3 at gmail.com  Thu Nov 23 07:34:45 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 22 Nov 2006 22:34:45 -0800
Subject: Categories
In-Reply-To: <d3dac270611221952h2853749ev8fcb5ced9e2f80a7@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <d3dac270611221952h2853749ev8fcb5ced9e2f80a7@mail.gmail.com>
Message-ID: <5542af940611222234r3ae28d48h1c3e6541143896a7@mail.gmail.com>

First, John Baez agrees that an earlier post of his partially answered my
follow-up question about 2-categories (some assembly required):

http://golem.ph.utexas.edu/category/2006/10/klein_2geometry_vi.html#c005157

 Re: Klein 2-Geometry VI

Tim writes:

So I'll wrap up and say what I think I've discovered. Just as
finite-dimensional vector spaces and their subspaces are intimately related
to finite sets and their subsets, so finite-dimensional 2-vector spaces and
their 2-subspaces are intimately related to finite graphs and their
subgraphs. These are obviously much more complicated and I'm not at all
confident my analyses here are correct, particularly the identification of
the various 2-Grassmannians associated to various graph maps. The basic idea
seems sound though.

Yeah, that's a great idea!

The way I see it, your fundamental idea is this. There's a 2-functor from
the 2-category of directed graphs to the 2-category of vector 2-spaces
(?2-term chain complexes). And, it goes like this:

   - Any directed graph gives a 2-term chain complex.
   - Any map between directed graphs gives a chain map between 2-term
   chain complexes.
   - And, there's also a kind of "homotopy between maps between directed
   graphs", which gives a chain homotopy between chain maps between 2-term
   chain complexes.

(Do graph theorists ever think about those homotopies? They should!)

All this generalizes further, to an (n+1)-functor from n-dimensional cell
complexes to (n+1)-term chain complexes.

I'm not sure how to get some amazing new insights into projective n-geometry
from this way of thinking - but that's not surprising, since I just read
your post 2 minutes ago. We should mull on it.

Thanks!
Posted by: John Baez <http://math.ucr.edu/home/baez/> on October 8, 2006
7:25 PM |

Second, I took Category theory in grad school 1976 or 1977, and forgot what
little I knew, and have relearned only some. So anyone in seqfans who
learned more and remebered more should jump in here, and don't worry about
making me look foolish, as I take the risk of exposing my ignorance every
time I post here.

If one asks: "how many categories are there with n morphisms and k objects?"
one is asking up to isomorphism. I'm making a first cut here off the top of
my head, which may be no more than an ill-informed guess.

The sagittal graph of a general morphism a.k.a. homomorphism (not restricted
to a monomorphism, epimorphism, isomorphism, endomorphism, or automorphism)
leaves us wondering what the structure of the objects are.  I think, by the
way that the question is asked, that they have no structure, i.e. that they
are just points. If they have structure, that affects everything.
unstructrured objects (points), then we still get different counts if the
vertices and arcs are labeled or unlabeled, and whether one allows loops or
not. if the edges are unlabeled, and loops are allowed, then any of the k
objects can be mapped to any of the k objects including itself, giving k^k,
if n = k.  If n>k then we can 't use more than k of the n because the
mapping must be functional, that is, only one arrow out of each point (but
multiple arrows to a point is allowed).

 If n<k then the first point can be mapped to any of the k points, and we
run out of arrows when we reach k^n.  Now we see why labeling matters, as if
they are unlabeled, we count only once in equivalence classes, but if
labeled each such mapping is distinct.  Each loop and arc is de facto
labeled by its source point and target point, as an ordered pair.  If you
want to allow other labels, you have to multiply by the proper combinational
multiplier.

So we're taking a summation of n=0 (just the k points and none map to any),
n=1 (one of k points maps to one of k), n=2 and so forth.

That starts the answer.  For k=n and any point can map to any, we have the
count of endomorphisms being the same as the count of endofunctions already
in OEIS.

Category theory has differing approaches depending on whether we have
digraphs, multgraphs,  reversible digraphs, reflexive digraphs, irreflexive
digraphs, and various tricks to represent undirected graphs with directed
graphs.

So exact definitions are necessary.

Sorry, just a first cut.  Not enough sleep.  Some time tomorrow before gests
and Thanksgiving dinner kicks in, but no promises.

Thanks for the dialog.

Best,

Jonathan Vos Post
<http://golem.ph.utexas.edu/cgi-bin/MT-3.0/sxp-comments.pl?entry_id=955;parent_id=5157>


On 11/22/06, Max A. <maxale at gmail.com> wrote:
>
> On 11/22/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
>
> Does that correspond to transitively closed digraphs with k (labeled?)
> vertices and n edges?
>
> Max
>
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From Roland.Bacher at ujf-grenoble.fr  Thu Nov 23 07:54:39 2006
From: Roland.Bacher at ujf-grenoble.fr (Roland Bacher)
Date: Thu, 23 Nov 2006 07:54:39 +0100
Subject: Categories
In-Reply-To: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
Message-ID: <20061123065439.GA28473@fourier.ujf-grenoble.fr>


This numbers can also be obtained as the numbers
of the following quotients of quivers:

Associate to a category with n morphisms and k objects 
the quiver with k vertices corresponding to the objects 
and a(X,Y) arrows directed from the object X to the object
Y if there are a(X,Y) morphisms from X into Y.

Given two morphism g:X-->Y, f:Y-->Z with composition
h=f o g:X-->Z, put the relation gf=h on the quiver algebra

The resulting quotient algebra has dimension n and a basis
given by the (simple) arrows.

This leads to the following "algorithm" for enumerating
all categories with n morphisms and k objects:

(a) enumerate all quivers (directed graphs) with k vertices and
n oriented edges.

(b) Associate the following "weight" to such a quiver as follows :

(b1)given a triplet of vertices X,Y,Z,
set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
(this counts the number of ways which associate a morphism
X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the 
number of oriented arrows starting at U and ending at W.

(b2) associate to a given quiver the weight 
\prod_{X,Y,Z} w(X,Y,Z)
where the product is over all triplets of vertices.

The total sum of such weighted quivers yields then the solution.

Roland Bacher



   


On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.net wrote:
> How many categories are there?
> 
> First, how many categories are there with n morphisms and k objects?  
> This table starts:
> 
> 1
> 2  1
> 7  3 1
> 35 16 3 1
> 
> The first column is A058129, the number of monoids; the main diagonal 
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
> 
> Taking the row sums, we get:
> 
> 1,3,11,55
> 
> the number of categories with n morphisms.  This is probably not in the 
> OEIS (only
> A001776 is possible - other matches become less than A058129).  The 
> inverse Euler
> transform,
> 
> 1,2,8,41
> 
> is the number of connected categories with n morphisms; this is 
> likewise probably not
> in the OEIS (only A052447 is possible).
> 
> Can somebody generate more data?
> 
> Franklin T. Adams-Watters
> 
> A category is a collection of objects and morphisms; each morphism is 
> from one object
> to another (not necessarily different) object.  Where the destination 
> of one morphism
> is the source of a second, their composition is defined; composition is 
> associative where
> it is defined.  Each object has an identity morphism, which connects it 
> to itself; this
> is an identity when composed with morphisms coming in and with 
> morphisms going
> out.
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and 
> industry-leading spam and email virus protection.
> 





From jvospost3 at gmail.com  Thu Nov 23 18:14:47 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 23 Nov 2006 09:14:47 -0800
Subject: Categories
In-Reply-To: <20061123065439.GA28473@fourier.ujf-grenoble.fr>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
Message-ID: <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>

Roland Bacher correctly refers to quivers as summarized in wikipedia at

http://en.wikipedia.org/wiki/Quiver_%28mathematics%29

On 11/22/06, Roland Bacher <Roland.Bacher at ujf-grenoble.fr> wrote:
>
>
> This numbers can also be obtained as the numbers
> of the following quotients of quivers:
>
> Associate to a category with n morphisms and k objects
> the quiver with k vertices corresponding to the objects
> and a(X,Y) arrows directed from the object X to the object
> Y if there are a(X,Y) morphisms from X into Y.
>
> Given two morphism g:X-->Y, f:Y-->Z with composition
> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
>
> The resulting quotient algebra has dimension n and a basis
> given by the (simple) arrows.
>
> This leads to the following "algorithm" for enumerating
> all categories with n morphisms and k objects:
>
> (a) enumerate all quivers (directed graphs) with k vertices and
> n oriented edges.
>
> (b) Associate the following "weight" to such a quiver as follows :
>
> (b1)given a triplet of vertices X,Y,Z,
> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> (this counts the number of ways which associate a morphism
> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> number of oriented arrows starting at U and ending at W.
>
> (b2) associate to a given quiver the weight
> \prod_{X,Y,Z} w(X,Y,Z)
> where the product is over all triplets of vertices.
>
> The total sum of such weighted quivers yields then the solution.
>
> Roland Bacher
>
>
>
>
>
>
> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.net wrote:
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> > 1
> > 2  1
> > 7  3 1
> > 35 16 3 1
> >
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
> > Taking the row sums, we get:
> >
> > 1,3,11,55
> >
> > the number of categories with n morphisms.  This is probably not in the
> > OEIS (only
> > A001776 is possible - other matches become less than A058129).  The
> > inverse Euler
> > transform,
> >
> > 1,2,8,41
> >
> > is the number of connected categories with n morphisms; this is
> > likewise probably not
> > in the OEIS (only A052447 is possible).
> >
> > Can somebody generate more data?
> >
> > Franklin T. Adams-Watters
> >
> > A category is a collection of objects and morphisms; each morphism is
> > from one object
> > to another (not necessarily different) object.  Where the destination
> > of one morphism
> > is the source of a second, their composition is defined; composition is
> > associative where
> > it is defined.  Each object has an identity morphism, which connects it
> > to itself; this
> > is an identity when composed with morphisms coming in and with
> > morphisms going
> > out.
> > ________________________________________________________________________
> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > industry-leading spam and email virus protection.
> >
>
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From jvospost3 at gmail.com  Thu Nov 23 21:10:51 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 23 Nov 2006 12:10:51 -0800
Subject: Categories
In-Reply-To: <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
Message-ID: <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>

*"Formalized Proof, Computation, and the Construction Problem in Algebraic
geometry", by Carlos Simpson.

[PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
2004<http://arxiv.org/pdf/math/0410224> File
Format: PDF/Adobe Acrobat - View as
HTML<http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5>
finite integer N, *how many categories* are there with N morphisms? What *
...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
(1965), 450-464.



On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> Roland Bacher correctly refers to quivers as summarized in wikipedia at
>
> http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
>
> On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> >
> >
> > This numbers can also be obtained as the numbers
> > of the following quotients of quivers:
> >
> > Associate to a category with n morphisms and k objects
> > the quiver with k vertices corresponding to the objects
> > and a(X,Y) arrows directed from the object X to the object
> > Y if there are a(X,Y) morphisms from X into Y.
> >
> > Given two morphism g:X-->Y, f:Y-->Z with composition
> > h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> >
> > The resulting quotient algebra has dimension n and a basis
> > given by the (simple) arrows.
> >
> > This leads to the following "algorithm" for enumerating
> > all categories with n morphisms and k objects:
> >
> > (a) enumerate all quivers (directed graphs) with k vertices and
> > n oriented edges.
> >
> > (b) Associate the following "weight" to such a quiver as follows :
> >
> > (b1)given a triplet of vertices X,Y,Z,
> > set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> > (this counts the number of ways which associate a morphism
> > X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> > number of oriented arrows starting at U and ending at W.
> >
> > (b2) associate to a given quiver the weight
> > \prod_{X,Y,Z} w(X,Y,Z)
> > where the product is over all triplets of vertices.
> >
> > The total sum of such weighted quivers yields then the solution.
> >
> > Roland Bacher
> >
> >
> >
> >
> >
> >
> > On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.net wrote:
> > > How many categories are there?
> > >
> > > First, how many categories are there with n morphisms and k objects?
> > > This table starts:
> > >
> > > 1
> > > 2  1
> > > 7  3 1
> > > 35 16 3 1
> > >
> > > The first column is A058129, the number of monoids; the main diagonal
> > > is all 1's.  I am not
> > > 100% certain of the 16 in the final row.
> > >
> > > Taking the row sums, we get:
> > >
> > > 1,3,11,55
> > >
> > > the number of categories with n morphisms.  This is probably not in
> > the
> > > OEIS (only
> > > A001776 is possible - other matches become less than A058129).  The
> > > inverse Euler
> > > transform,
> > >
> > > 1,2,8,41
> > >
> > > is the number of connected categories with n morphisms; this is
> > > likewise probably not
> > > in the OEIS (only A052447 is possible).
> > >
> > > Can somebody generate more data?
> > >
> > > Franklin T. Adams-Watters
> > >
> > > A category is a collection of objects and morphisms; each morphism is
> > > from one object
> > > to another (not necessarily different) object.  Where the destination
> > > of one morphism
> > > is the source of a second, their composition is defined; composition
> > is
> > > associative where
> > > it is defined.  Each object has an identity morphism, which connects
> > it
> > > to itself; this
> > > is an identity when composed with morphisms coming in and with
> > > morphisms going
> > > out.
> > >
> > ________________________________________________________________________
> > > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > > industry-leading spam and email virus protection.
> > >
> >
>
>
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From jvospost3 at gmail.com  Thu Nov 23 21:20:58 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 23 Nov 2006 12:20:58 -0800
Subject: Categories
In-Reply-To: <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
	 <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
Message-ID: <5542af940611231220oe45dce2rbee869c1dea94b9e@mail.gmail.com>

FOM: Set theory vs category theory: Representability of categories *Vaughan
Pratt* pratt at cs.Stanford.EDU
<fom%40cs.nyu.edu?Subject=FOM%3A%20Set%20theory%20vs%20category%20theory%3A%20Representability%20of%20categories&In-Reply-To=199801271058.LAA14876%40brics.dk>
*Tue Jan 27 13:37:42 EST 1998

http://cs.nyu.edu/pipermail/fom/1998-January/001021.html
*



A number of representation theorems were found by Czech algebraists and
category theorists during the latter half of the 1960's, representing
arbitrary small categories as categories of familiar concrete structures
and their homomorphisms.  Two notable representations are in terms
of semigroups, and of directed graphs.  In each case the theorem is
that every small category is representable as a category of semigroups
(directed graphs, etc.) and their homomorphisms (of the standard kind
in each case).

The criteria used here for representability of objects  a  and morphisms
f:a-&gt;b of a category C by respectively structures F(a) and homomorphisms
F(f): F(a)->F(b) of the concrete representing category D are as follows.

(i) F:C->D is a *functor*, that is, a graph homomorphism (viewing the
morphisms of C and D as edges of their respective underlying graphs)
that preserves composition and identities (a condition exactly analogous
to those of monoid homomorphism and group homomorphism).

(ii) F is *faithful*, i.e. for any pair a,b of objects of C, distinct
morphisms f:a-&gt;b of C are represented by distinct homomorphisms
F(f):F(a)->F(b) of D.

(iii) F is *full*, i.e. every homomorphism from F(a) to F(b) in D is
the representation under F of some morphism from a to b in C.

The intuitive meaning of full and faithful is that the representing
object is just as stiff as the object it represents when you wiggle both.
Fullness means it is at least as stiff (D offers no new transformations),
while faithfulness means it is at most as stiff (no transformations are
lost by identification).

Unfortunately the project of embedding arbitrary categories fully and
faithfully in familiar concrete categories has turned out not to have
the impact of its counterparts in group theory (permutations), Boolean
algebras (fields of sets), distributive lattices (rings of sets), etc.
These results have in consequence not been widely publicized in the
introductory categorical literature.

The most objectionable feature of these embeddings in my view (and I
am aware of no other serious objection) is their lack of respect for
concreteness itself.  If one takes for C a small *concrete* category,
such as all quotient groups of the additive group of natural numbers
and their group homomorphisms, the above-mentioned embedding represents
the finite group Z_2 as an infinite semigroup!  And if &quot;natural
numbers&quot;

is replaced by "reals", the corresponding embedding represents Z_2 as an
uncountable semigroup.  Such a lack of respect for concreteness would
seem to undermine the very point of representing abstract objects as
concrete structures.

In contrast a finite group of order n is representable by permutations of
n things, a finite Boolean algebra as a field of subsets of a finite set,
and so on.  In these better-known representation theorems, cardinality
is respected, typically to within at most two exponentials (which in
the infinite case means to within two beth numbers) and often much better.

This unsatisfactory situation with the concrete representation of
objects of arbitrary categories prompts the speculation that the notion
of category is too broad to admit of any improvement in the situation.
After all, any random graph G is made a category by adding to its
edges all paths in G, with composition defined as path concatenation
(the free category generated by G).  How could there possibly be a
cardinality-respecting representation of the vertices of a random graph
in terms of the objects of a familiar concrete category?  When one can
do any violence one wishes to the structure by adding an edge at random
between any two vertices, it would seem that any concrete representation
of each object would inevitably have a cardinality on the order of the
whole category.

But in fact there *is* a representation that does the job.  Let us make
two easily understood modifications to the usual notion of topological
space.

First, drop the requirement that the open sets be closed under arbitrary
union and finite intersection.  (The role of this traditional restriction
is in effect to limit the &quot;signature&quot; of a topological space
to just that
needed to express the notion of limit, or even less with the more discrete
spaces; dropping it greatly broadens the range of possible signatures.)

Second, in place of the customary two degrees of membership in the open
sets (ordinary sets), permit a set K of possible degrees of membership,
call such open sets fuzzy (Zadeh's fuzzy sets are the case K the reals).

Define continuity as usual for functions between topological spaces,
wording the definition in such a way however that its extension to
fuzzy sets is made clear.  Specifically, f:X-&gt;Y is continuous when for
every open set g:Y-&gt;K (here expressed as a characteristic function),
the composite

		   f      g
		X ---> Y ---> K

is an open set of X.  When K = {0,1} it is easily verified that this is
the standard notion of continuity.

Call such generalized topological spaces Chu spaces over K.  An even more
general notion was first studied in detail by Peter Chu in the 1970's,
a master's student of Michael Barr.  The more elementary version above
was first studied by Lafont and Streicher in 1991 (LICS) under the rubric
of games but the term &quot;Chu construction&quot; was already in the computer
science air in the late 1980's making it too late to change the name.

Definitions.

1.  A concrete category is a pair (C,U) where C is a category and U:C-&gt;Set
is a faithful functor, the "forgetful" functor giving the *underlying
set* of each object along with the realization of each morphism of C as
a function (a morphism of Set).

2.  A concrete functor F:(C,U)->(D,V) between two concrete categories
is one satisfying VF = U.  That is, the underlying set VF(a) of the
representation F(a) of a is the same as the underlying set U(a) of
a itself.

Remark.  The elements of all the representing sets of a small concrete
category (C,U) themselves form a single set, namely the disjoint or
marked union of all the sets in the category.  Call this set Elt(C,U),
the set of elements of (C,U).

A very minor technicality requires the notion of an *honest* concrete
category, namely one having, for all objects a and b, a morphism from
a to b if U(a) is empty (necessarily at most one, by concreteness).
Dishonest concrete categories cannot be represented as below because
the empty topological space has nowhere to store the information as to
which other spaces it does or does not have functions to.  (Thanks to
Peter Freyd for pointing out the lacuna in my proof that necessitated
this condition.)

Theorem.  Every small honest concrete category (C,U) embeds fully,
faithfully, and concretely in the category of Chu spaces over Elt(C,U).

(This theorem enjoys all the advantages of the previous embeddings, but
in addition satisfies the strongest possible concreteness requirement:
the representing object has the *same* underlying set as the object it
represents, and the representing morphisms viewed concretely are exactly
the same functions.)

Proof.  Let b be an object of C with underlying set X = U(b).  Represent
b by a generalized topological space X, having one open set Y_h for
each morphism h:b-&gt;c in C for some c.  Form Y_h by taking the degree of
membership of each point x in Y_h to be U(h)(x), i.e. the element of U(c)
which is the image of x under the underlying function U(h) of h.

It is then straightforward to show that this representation is faithful,
full, and concrete.

Obviously the burden of the obstacle that we have overcome has merely
been shifted to the open sets, the number of which must now be on the
order of the cardinality of the category.  But since when has the number
of open sets been an obstacle to topology?  The points of a space are
clearly visible, but our embedding has left these untouched.  Who has
ever seen an open set or cares about how many there are?

For more information on Chu spaces consult

	<A HREF="http://boole.stanford.edu/chuguide.html">http://boole.stanford.edu/chuguide.html</A>

My original interest in Chu spaces was as a model of concurrent
behavior.  More recently I have become interested in them as a primarily
set-theoretic foundation for mathematics that more directly emulates
what category theory has to offer than the extant litany of single-sorted
and and multisorted relational structures and algebras with and without
(ordinary) topology and their standard homomorphisms, all of which are
fully, faithfully, and concretely representable by Chu spaces.

---
Name: Vaughan Pratt
Position: Professor of Computer Science
Institution: Stanford University
Research interests: Foundations of computation and mathematics
For more information: <A
HREF="http://boole.stanford.edu">http://boole.stanford.edu</A>
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From maxale at gmail.com  Thu Nov 23 23:13:45 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 23 Nov 2006 14:13:45 -0800
Subject: connected components in subgraphs of the complete graph: A125205 - A125207
Message-ID: <d3dac270611231413v4ea7e950ifc9405a7ed04e057@mail.gmail.com>

SeqFans,

I've just sent to Neil the following three sequences that you may find
interesting.

It would be also interesting to compute unlabeled variants of these
sequences. If you like to contribute, please do.

Regards,
Max


%I A125205
%S A125205 1,2,1,3,6,3,1,4,18,30,24,15,6,1,5,40,135,250,295,282,215,120,45,10,1,6,
%T A125205 75,420,1385,3015,4800,6365,7170,6705,5065,3009,1365,455,105,15,1,7,126,
%U A125205 1050,5355,18690,47880,96796,166890,251370,329945,373947,362292,297115
%N A125205 Triangular array T(n,k) (n>=1, 0<=k<=n(n-1)/2) giving the
total number
           of connected components in all subgraphs (V,E') with |E'|=k of
           the complete labeled graph K_n=(V,E).
%F A125205 g.f.: Sum_{n,k}
T(n,k)*x^n/n!*y^k=(F(x,y)-1)*exp(F(x,y)-1)=G(x,y)*ln(G(x,y))
           where G(x,y)=Sum_{n=0..oo} (1+y)^(n(n-1)/2)*x^n/n!, and
F(x,y)=1+ln(G(x,y))
           is g.f. of A062734.
%e A125205 The array starts with
           1
           2, 1
           3, 6, 3, 1
           4, 18, 30, 24, 15, 6, 1
           5, 40, 135, 250, 295, 282, 215, 120, 45, 10, 1
           ...
%e A125205 T(3,1)=6 since there are three different subgraphs of K_3
with one edge,
           and each subgraph has two connected components.
%o A125205 (PARI) { reverse(v)=vector(length(v),i,v[length(v)+1-i]) }
           G=sum(n=0,6,(1+y)^(n*(n-1)/2)*x^n/n!); K=G*log(G);
           for(n=1,6,print(reverse(Vec(n!*polcoeff(K,n,x)))))
%Y A125205 Cf. A062734
%Y A125205 Cf. A125206 (row-reversed version), A125207 (row sums)
%K A125205 nonn,tabf
%O A125205 1,2
%A A125205 Max Alekseyev (maxal(AT)cs.ucsd.edu), Nov 23 2006



%I A125206
%S A125206 1,1,2,1,3,6,3,1,6,15,24,30,18,4,1,10,45,120,215,282,295,250,135,40,5,1,
%T A125206 15,105,455,1365,3009,5065,6705,7170,6365,4800,3015,1385,420,75,6,1,21,210,
%U A125206 1330,5985,20349,54271,116385,204225,297115,362292,373947,329945,251370
%N A125206 Triangular array T(n,k) (n>=1, 0<=k<=n(n-1)/2) giving the
total number
           of connected components in all subgraphs obtained from the complete
           labeled graph K_n by removing k edges.
%C A125206 Row-reversed version of A125205, see A125205 for further details
%e A125206 The array starts with
           1
           1, 2
           1, 3, 6, 3
           1, 6, 15, 24, 30, 18, 4
           1, 10, 45, 120, 215, 282, 295, 250, 135, 40, 5
           ...
%Y A125206 Cf. A125205 (row-reversed version), A125207 (row sums)
%K A125206 nonn,tabf
%O A125206 1,3
%A A125206 Max Alekseyev (maxal(AT)cs.ucsd.edu), Nov 23 2006




%I A125207
%S A125207 1,3,13,98,1398,39956,2354240,286394544,71225744048,35884971729760,36419817759267072,
%T A125207 74221711070826087424,303193538300703211111936,2480118087478081928075065344,
%U A125207 40601989279034990139321984265216,1329877330680067685563700135615633408
%N A125207 The total number of connected components in all subgraphs
obtained from
           the complete labeled graph K_n by removing zero or more edges.
%F A125207 E.g.f.: (F(x)-1)*exp(F(x)-1) = G(x)*ln(G(x)) where
           G(x)=Sum_{n=0..oo} 2^(n(n-1)/2)*x^n/n! and F(x)=1+ln(G(x))
is e.g.f. of A001187
%e A125207 For n=2, we have two graph on two vertices: complete and
empty, the former has
           one connected component while the latter has two connected
components. The total
           number of connected components is 3, which is a(2).
%o A125207 (PARI) G=sum(n=0,30,2^(n*(n-1)/2)*x^n/n!) + O(x31); v=Vec(G*log(G));
           for(i=1,length(v),v[i]*=i!); print(v)
%Y A125207 Cf. A001187, A125205, A125206
%K A125207 nonn
%O A125207 1,2
%A A125207 Max Alekseyev (maxal(AT)cs.ucsd.edu), Nov 23 2006





From davidwwilson at comcast.net  Fri Nov 24 01:39:43 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Thu, 23 Nov 2006 19:39:43 -0500
Subject: b-files for power series expansions
Message-ID: <01ad01c70f61$09624cf0$6501a8c0@yourxhtr8hvc4p>

I guess my point was, if someone has access to Mma or some other symbolic math package, it would be nice if they could write a program to generate b-files for sequences that are e.g.f. expansions of real functions. We could then generate a whole slew of b-files fairly easily.
 
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From maxale at gmail.com  Fri Nov 24 02:04:09 2006
From: maxale at gmail.com (Max A.)
Date: Thu, 23 Nov 2006 17:04:09 -0800
Subject: Chestnut
In-Reply-To: <000601c6f538$1ca30fe0$6401a8c0@yourxhtr8hvc4p>
References: <000601c6f538$1ca30fe0$6401a8c0@yourxhtr8hvc4p>
Message-ID: <d3dac270611231704ua5487a7s547879c8e77174b7@mail.gmail.com>

On 10/21/06, David Wilson <davidwwilson at comcast.net> wrote:
>
>
>
>
> This is a chestnut that I am convinced it is true, but I cannot prove it. I
> have posed it on math-fun occasionally, and never got an answer. I would
> like to see it posed to some better minds that some of you might know, e.g,
> JHC or other number theorist.
>
> Let S and T be sets of real numbers. Call T a divider of S if some element
> of T lies strictly between any two elements of S.
>
> For integer n >= 1, let Fence(n) be the set of all rationals with
> denominator n, that is, { k/n : k in Z }.
>
> For real set S, let f(S) be the least n such that Fence(n) is a divider of
> S, if such an n exists.
>
> Let Recip(n) be the set of all integer reciprocals on [0,1] with denominator
> <= n, that is, { 1/b : 1 <= b <= n }
>
> Let Farey(n) be the set of all rationals on [0,1] with denominator <= n,
> that is, { a/b : 0 <= a <= b, 1 <= b <= n }
>
> Is f(Farey(n)) = f(Recip(n)) for every n?
>
> The apparently common sequences a(n) = f(Recip(n)) =? f(Farey(n)) is in the
> OEIS, I cannot find it in the short time I have.





From Roland.Bacher at ujf-grenoble.fr  Fri Nov 24 08:53:08 2006
From: Roland.Bacher at ujf-grenoble.fr (Roland Bacher)
Date: Fri, 24 Nov 2006 08:53:08 +0100
Subject: Categories
In-Reply-To: <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com> <20061123065439.GA28473@fourier.ujf-grenoble.fr> <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com> <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
Message-ID: <20061124075307.GA17864@fourier.ujf-grenoble.fr>


I realised that the formula below for enumerating categories
using quivers is completely wrong. It gives only an upper bound
since one has to require moreover associativity for the composition of 
morphisms. The quiver approach works however in principle but
is much more complex (and I guess this approach has been 
used in some of the previous approaches).   Roland Bacher

On Thu, Nov 23, 2006 at 12:10:51PM -0800, Jonathan Post wrote:
> *"Formalized Proof, Computation, and the Construction Problem in Algebraic
> geometry", by Carlos Simpson.
> 
> [PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
> 2004<http://arxiv.org/pdf/math/0410224> File
> Format: PDF/Adobe Acrobat - View as
> HTML<http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5>
> finite integer N, *how many categories* are there with N morphisms? What *
> ...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
> (1965), 450-464.
> 
> 
> 
> On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> >
> >Roland Bacher correctly refers to quivers as summarized in wikipedia at
> >
> >http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
> >
> >On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> >>
> >>
> >> This numbers can also be obtained as the numbers
> >> of the following quotients of quivers:
> >>
> >> Associate to a category with n morphisms and k objects
> >> the quiver with k vertices corresponding to the objects
> >> and a(X,Y) arrows directed from the object X to the object
> >> Y if there are a(X,Y) morphisms from X into Y.
> >>
> >> Given two morphism g:X-->Y, f:Y-->Z with composition
> >> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> >>
> >> The resulting quotient algebra has dimension n and a basis
> >> given by the (simple) arrows.
> >>
> >> This leads to the following "algorithm" for enumerating
> >> all categories with n morphisms and k objects:
> >>
> >> (a) enumerate all quivers (directed graphs) with k vertices and
> >> n oriented edges.
> >>
> >> (b) Associate the following "weight" to such a quiver as follows :
> >>
> >> (b1)given a triplet of vertices X,Y,Z,
> >> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> >> (this counts the number of ways which associate a morphism
> >> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> >> number of oriented arrows starting at U and ending at W.
> >>
> >> (b2) associate to a given quiver the weight
> >> \prod_{X,Y,Z} w(X,Y,Z)
> >> where the product is over all triplets of vertices.
> >>
> >> The total sum of such weighted quivers yields then the solution.
> >>
> >> Roland Bacher
> >>
> >>
> >>
> >>
> >>
> >>
> >> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.net wrote:
> >> > How many categories are there?
> >> >
> >> > First, how many categories are there with n morphisms and k objects?
> >> > This table starts:
> >> >
> >> > 1
> >> > 2  1
> >> > 7  3 1
> >> > 35 16 3 1
> >> >
> >> > The first column is A058129, the number of monoids; the main diagonal
> >> > is all 1's.  I am not
> >> > 100% certain of the 16 in the final row.
> >> >
> >> > Taking the row sums, we get:
> >> >
> >> > 1,3,11,55
> >> >
> >> > the number of categories with n morphisms.  This is probably not in
> >> the
> >> > OEIS (only
> >> > A001776 is possible - other matches become less than A058129).  The
> >> > inverse Euler
> >> > transform,
> >> >
> >> > 1,2,8,41
> >> >
> >> > is the number of connected categories with n morphisms; this is
> >> > likewise probably not
> >> > in the OEIS (only A052447 is possible).
> >> >
> >> > Can somebody generate more data?
> >> >
> >> > Franklin T. Adams-Watters
> >> >
> >> > A category is a collection of objects and morphisms; each morphism is
> >> > from one object
> >> > to another (not necessarily different) object.  Where the destination
> >> > of one morphism
> >> > is the source of a second, their composition is defined; composition
> >> is
> >> > associative where
> >> > it is defined.  Each object has an identity morphism, which connects
> >> it
> >> > to itself; this
> >> > is an identity when composed with morphisms coming in and with
> >> > morphisms going
> >> > out.
> >> >
> >> ________________________________________________________________________
> >> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> >> > industry-leading spam and email virus protection.
> >> >
> >>
> >
> >





From tbaruchel at free.fr  Fri Nov 24 15:12:13 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Fri, 24 Nov 2006 15:12:13 +0100 (CET)
Subject: Quadratic residues
Message-ID: <20061124150520.F1510@localhost.my.domain>


I asked recently about
>  a(n) is the number of divisors of (n+1)
> Have you ever seen closely-related sequences like that one?

After some hints and a
> Happy hunting!
I finally noticed this :
   a) prime numbers 4n+3 lead to the set of their quadratic residues
   b) prime numbers 4n+1 lead to the set of their quadratic non-residues
   c) odd composite numbers :
      15 --> quadratic residues
      21 --> difficult to understand at first sight
Since my datas are empirical and need much work it is difficult to get
much information. Even numbers are difficult to study also.
Have you ever heard about such a behaviour? In that case, what would
you predict for 21?

WARNING. When I speak of "quadratic residues", I mean the residue has
a GCD = 1 with the number (convention may vary).

Thank you for everything,

-- 
Thomas Baruchel





From tbaruchel at free.fr  Fri Nov 24 15:59:29 2006
From: tbaruchel at free.fr (Thomas Baruchel)
Date: Fri, 24 Nov 2006 15:59:29 +0100 (CET)
Subject: Quadratic residues
In-Reply-To: <20061124150520.F1510@localhost.my.domain>
References: <20061124150520.F1510@localhost.my.domain>
Message-ID: <20061124155641.B1631@localhost.my.domain>

On Fri, 24 Nov 2006, Thomas Baruchel wrote:
> Have you ever heard about such a behaviour? In that case, what would
> you predict for 21?

Forget it, I just found : Jacobi symbol + quadratic reciprocity theorem

Thanks.





From eclark at math.usf.edu  Fri Nov 24 16:28:29 2006
From: eclark at math.usf.edu (Edwin Clark)
Date: Fri, 24 Nov 2006 10:28:29 -0500 (EST)
Subject: Categories
In-Reply-To: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
Message-ID: <Pine.LNX.4.58.0611240958550.32273@shell.cas.usf.edu>

On Wed, 22 Nov 2006 franktaw at netscape.net wrote:

> How many categories are there?
> 
> First, how many categories are there with n morphisms and k objects?  
> This table starts:
> 
>  1
>  2  1
>  7  3 1
> 35 16 3 1
> 
> The first column is A058129, the number of monoids; the main diagonal 
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
> 

As you say, the number of one object categories on n objects is M(n)=the 
number of monoids with n elements. But this is known only up to n = 7
according to the OEIS. This seems to make the number of TWO object
categories with n morphisms a very difficult problem...Let alone THREE
object categories...

MathSciNet gives only one hit on "two object categor*":
------------------------------------------------------------------
Rosick\'y, Ji\v r?
Codensity and binding categories.
Comment. Math. Univ. Carolinae 16 (1975), no. 3, 515--529.

If there is no measurable cardinal, the author constructs a two-object 
category all of whose well-powered complete extensions are binding. 
-------------------------------------------------------------------
which is clearly not related to the enumeration problem. 

A search on "categor* with two objects" give a few more hits
but as far as I can see still no papers related to enumeration..







From jvospost3 at gmail.com  Fri Nov 24 17:51:12 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 24 Nov 2006 08:51:12 -0800
Subject: Categories
In-Reply-To: <Pine.LNX.4.58.0611240958550.32273@shell.cas.usf.edu>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <Pine.LNX.4.58.0611240958550.32273@shell.cas.usf.edu>
Message-ID: <5542af940611240851l23b0d23di56bf7bf28fe71f82@mail.gmail.com>

One should also count loops, i.e. an arrow from an object to itself.

Then there are 2 categories on one object, with and without the loop:
Notating V(C) = the vertices in category C (which we label in this
illustrartion but enumerate up to isomorphism, i.e. on forgetting labels;
and L(C) is the set of loops of C, and A(C) is the set of (proper) arcs.
That is, I've separated the loops out from their being a mere subset of
arcs, to aid in enumeration. Then each category is, for this enumeration, a
triple: C= V(C) U L(C) U A(C).

C_1,1 = (1,null,null);
C_1,2 = (1, 1->1,null).

There are 6 categories on 2 objects:
C_2,1 = ({1,2},null,null);
C_2,2 = ({1,2},null,(1->1))
note that the above is isomorphic to C_2,2 = ({1,2},null,(2->2))
C_2,3 = ({1,2},{(1->1),(2->2)},null);
C_2,4 = ({1,2},null,(1->2)) which is isomorphic to C_2,4 =
({1,2},null,(2->1));
C_2,5 = ({1,2},(1>1),(1->2)) which is isomorphic to ({1,2},(2>2),(2->1));
C_2,6 = ({1,2},null,(1>1),(2->1)).

There do seem to be 35 categories on 3 objects, which are essentially the
same as the 7 endomorphisms on 3 objects, with each loop either left in, or
thrown away.

The combinatorics are now straightforward.  We have the endofunction
enumeration, and binary choices on each loop.

I have many, many pages of drawings of these, but not wanting to either scan
them (not having a scanner) nor write out the explicit sets as above.

Is this clear enough, though, to proceed?

-- Jonathan
C_2,2 = ({1,2},(1->1),null)


On 11/24/06, Edwin Clark <eclark at math.usf.edu> wrote:
>
> On Wed, 22 Nov 2006 franktaw at netscape.net wrote:
>
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> >  1
> >  2  1
> >  7  3 1
> > 35 16 3 1
> >
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
>
> As you say, the number of one object categories on n objects is M(n)=the
> number of monoids with n elements. But this is known only up to n = 7
> according to the OEIS. This seems to make the number of TWO object
> categories with n morphisms a very difficult problem...Let alone THREE
> object categories...
>
> MathSciNet gives only one hit on "two object categor*":
> ------------------------------------------------------------------
> Rosick\'y, Ji\v r?
> Codensity and binding categories.
> Comment. Math. Univ. Carolinae 16 (1975), no. 3, 515--529.
>
> If there is no measurable cardinal, the author constructs a two-object
> category all of whose well-powered complete extensions are binding.
> -------------------------------------------------------------------
> which is clearly not related to the enumeration problem.
>
> A search on "categor* with two objects" give a few more hits
> but as far as I can see still no papers related to enumeration..
>
>
>
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From jvospost3 at gmail.com  Fri Nov 24 20:39:20 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 24 Nov 2006 11:39:20 -0800
Subject: Categories
In-Reply-To: <20061124075307.GA17864@fourier.ujf-grenoble.fr>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
	 <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
	 <20061124075307.GA17864@fourier.ujf-grenoble.fr>
Message-ID: <5542af940611241139sa6472eoc31f5980db42a918@mail.gmail.com>

Offline, I'm coordinating with Edwin Clark on definitions.

But I'm counting as lower bounds  (many pages of paper with drawings, too
lengthy for me to turn into ascii, don't have scanner):

2 categories on 1 point;
6 categories on 2 points;
16 categories on 3 points;
46 categories on 4 points;
116 categories on 2 points;

in each case where the underlying digraph is of an endofunction. The full
count on categories on n points is a weighted partial sum on these, related
to my enumeration of "prime" endofunctions being the sum of Fobonacci(n),
tribonacci(n), teranacci(n), ...

-- I can give examples on request.  Of course, I can always be off by 1 or
more on each count, through sheer blunder.  But I am being self-consistent
in methodology, and it can be done by software using the same algorithm,
carefully translated.

-- Jonathan


On 11/23/06, Roland Bacher <Roland.Bacher at ujf-grenoble.fr> wrote:
>
>
> I realised that the formula below for enumerating categories
> using quivers is completely wrong. It gives only an upper bound
> since one has to require moreover associativity for the composition of
> morphisms. The quiver approach works however in principle but
> is much more complex (and I guess this approach has been
> used in some of the previous approaches).   Roland Bacher
>
> On Thu, Nov 23, 2006 at 12:10:51PM -0800, Jonathan Post wrote:
> > *"Formalized Proof, Computation, and the Construction Problem in
> Algebraic
> > geometry", by Carlos Simpson.
> >
> > [PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
> > 2004<http://arxiv.org/pdf/math/0410224> File
> > Format: PDF/Adobe Acrobat - View as
> > HTML<
> http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5
> >
> > finite integer N, *how many categories* are there with N morphisms? What
> *
> > ...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
> > (1965), 450-464.
> >
> >
> >
> > On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> > >
> > >Roland Bacher correctly refers to quivers as summarized in wikipedia at
> > >
> > >http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
> > >
> > >On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> > >>
> > >>
> > >> This numbers can also be obtained as the numbers
> > >> of the following quotients of quivers:
> > >>
> > >> Associate to a category with n morphisms and k objects
> > >> the quiver with k vertices corresponding to the objects
> > >> and a(X,Y) arrows directed from the object X to the object
> > >> Y if there are a(X,Y) morphisms from X into Y.
> > >>
> > >> Given two morphism g:X-->Y, f:Y-->Z with composition
> > >> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> > >>
> > >> The resulting quotient algebra has dimension n and a basis
> > >> given by the (simple) arrows.
> > >>
> > >> This leads to the following "algorithm" for enumerating
> > >> all categories with n morphisms and k objects:
> > >>
> > >> (a) enumerate all quivers (directed graphs) with k vertices and
> > >> n oriented edges.
> > >>
> > >> (b) Associate the following "weight" to such a quiver as follows :
> > >>
> > >> (b1)given a triplet of vertices X,Y,Z,
> > >> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> > >> (this counts the number of ways which associate a morphism
> > >> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> > >> number of oriented arrows starting at U and ending at W.
> > >>
> > >> (b2) associate to a given quiver the weight
> > >> \prod_{X,Y,Z} w(X,Y,Z)
> > >> where the product is over all triplets of vertices.
> > >>
> > >> The total sum of such weighted quivers yields then the solution.
> > >>
> > >> Roland Bacher
> > >>
> > >>
> > >>
> > >>
> > >>
> > >>
> > >> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.netwrote:
> > >> > How many categories are there?
> > >> >
> > >> > First, how many categories are there with n morphisms and k
> objects?
> > >> > This table starts:
> > >> >
> > >> > 1
> > >> > 2  1
> > >> > 7  3 1
> > >> > 35 16 3 1
> > >> >
> > >> > The first column is A058129, the number of monoids; the main
> diagonal
> > >> > is all 1's.  I am not
> > >> > 100% certain of the 16 in the final row.
> > >> >
> > >> > Taking the row sums, we get:
> > >> >
> > >> > 1,3,11,55
> > >> >
> > >> > the number of categories with n morphisms.  This is probably not in
> > >> the
> > >> > OEIS (only
> > >> > A001776 is possible - other matches become less than A058129).  The
> > >> > inverse Euler
> > >> > transform,
> > >> >
> > >> > 1,2,8,41
> > >> >
> > >> > is the number of connected categories with n morphisms; this is
> > >> > likewise probably not
> > >> > in the OEIS (only A052447 is possible).
> > >> >
> > >> > Can somebody generate more data?
> > >> >
> > >> > Franklin T. Adams-Watters
> > >> >
> > >> > A category is a collection of objects and morphisms; each morphism
> is
> > >> > from one object
> > >> > to another (not necessarily different) object.  Where the
> destination
> > >> > of one morphism
> > >> > is the source of a second, their composition is defined;
> composition
> > >> is
> > >> > associative where
> > >> > it is defined.  Each object has an identity morphism, which
> connects
> > >> it
> > >> > to itself; this
> > >> > is an identity when composed with morphisms coming in and with
> > >> > morphisms going
> > >> > out.
> > >> >
> > >>
> ________________________________________________________________________
> > >> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > >> > industry-leading spam and email virus protection.
> > >> >
> > >>
> > >
> > >
>
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From jvospost3 at gmail.com  Fri Nov 24 23:41:12 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Fri, 24 Nov 2006 14:41:12 -0800
Subject: Categories
In-Reply-To: <5542af940611241139sa6472eoc31f5980db42a918@mail.gmail.com>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
	 <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
	 <20061124075307.GA17864@fourier.ujf-grenoble.fr>
	 <5542af940611241139sa6472eoc31f5980db42a918@mail.gmail.com>
Message-ID: <5542af940611241441w68d51ab5pc6ba0832f740d4e8@mail.gmail.com>

Somehow the reply to Frank went astray.

The wikipedia entry on directed graph makes some of the definitional
distinctions I've made, and then some.  Further, it makes the useful
statement:

In Category theory,
a category can be considered a directed multigraph
with the objects as vertices and the morphisms as directed edges.
The functors between categories induce then some,
but not necessarily all, of the digraph morphisms.


-- Jonathan


On 11/24/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> Offline, I'm coordinating with Edwin Clark on definitions.
>
> But I'm counting as lower bounds  (many pages of paper with drawings, too
> lengthy for me to turn into ascii, don't have scanner):
>
> 2 categories on 1 point;
> 6 categories on 2 points;
> 16 categories on 3 points;
> 46 categories on 4 points;
> 116 categories on 2 points;
>
> in each case where the underlying digraph is of an endofunction. The full
> count on categories on n points is a weighted partial sum on these, related
> to my enumeration of "prime" endofunctions being the sum of Fobonacci(n),
> tribonacci(n), teranacci(n), ...
>
> -- I can give examples on request.  Of course, I can always be off by 1 or
> more on each count, through sheer blunder.  But I am being self-consistent
> in methodology, and it can be done by software using the same algorithm,
> carefully translated.
>
> -- Jonathan
>
>
> On 11/23/06, Roland Bacher <Roland.Bacher at ujf-grenoble.fr> wrote:
> >
> >
> > I realised that the formula below for enumerating categories
> > using quivers is completely wrong. It gives only an upper bound
> > since one has to require moreover associativity for the composition of
> > morphisms. The quiver approach works however in principle but
> > is much more complex (and I guess this approach has been
> > used in some of the previous approaches).   Roland Bacher
> >
> > On Thu, Nov 23, 2006 at 12:10:51PM -0800, Jonathan Post wrote:
> > > *"Formalized Proof, Computation, and the Construction Problem in
> > Algebraic
> > > geometry", by Carlos Simpson.
> > >
> > > [PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
> > > 2004<http://arxiv.org/pdf/math/0410224> File
> > > Format: PDF/Adobe Acrobat - View as
> > > HTML<http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5
> > >
> > > finite integer N, *how many categories* are there with N morphisms?
> > What *
> > > ...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
> > > (1965), 450-464.
> > >
> > >
> > >
> > > On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> > > >
> > > >Roland Bacher correctly refers to quivers as summarized in wikipedia
> > at
> > > >
> > > >http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
> > > >
> > > >On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> > > >>
> > > >>
> > > >> This numbers can also be obtained as the numbers
> > > >> of the following quotients of quivers:
> > > >>
> > > >> Associate to a category with n morphisms and k objects
> > > >> the quiver with k vertices corresponding to the objects
> > > >> and a(X,Y) arrows directed from the object X to the object
> > > >> Y if there are a(X,Y) morphisms from X into Y.
> > > >>
> > > >> Given two morphism g:X-->Y, f:Y-->Z with composition
> > > >> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> > > >>
> > > >> The resulting quotient algebra has dimension n and a basis
> > > >> given by the (simple) arrows.
> > > >>
> > > >> This leads to the following "algorithm" for enumerating
> > > >> all categories with n morphisms and k objects:
> > > >>
> > > >> (a) enumerate all quivers (directed graphs) with k vertices and
> > > >> n oriented edges.
> > > >>
> > > >> (b) Associate the following "weight" to such a quiver as follows :
> > > >>
> > > >> (b1)given a triplet of vertices X,Y,Z,
> > > >> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> > > >> (this counts the number of ways which associate a morphism
> > > >> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> > > >> number of oriented arrows starting at U and ending at W.
> > > >>
> > > >> (b2) associate to a given quiver the weight
> > > >> \prod_{X,Y,Z} w(X,Y,Z)
> > > >> where the product is over all triplets of vertices.
> > > >>
> > > >> The total sum of such weighted quivers yields then the solution.
> > > >>
> > > >> Roland Bacher
> > > >>
> > > >>
> > > >>
> > > >>
> > > >>
> > > >>
> > > >> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.netwrote:
> > > >> > How many categories are there?
> > > >> >
> > > >> > First, how many categories are there with n morphisms and k
> > objects?
> > > >> > This table starts:
> > > >> >
> > > >> > 1
> > > >> > 2  1
> > > >> > 7  3 1
> > > >> > 35 16 3 1
> > > >> >
> > > >> > The first column is A058129, the number of monoids; the main
> > diagonal
> > > >> > is all 1's.  I am not
> > > >> > 100% certain of the 16 in the final row.
> > > >> >
> > > >> > Taking the row sums, we get:
> > > >> >
> > > >> > 1,3,11,55
> > > >> >
> > > >> > the number of categories with n morphisms.  This is probably not
> > in
> > > >> the
> > > >> > OEIS (only
> > > >> > A001776 is possible - other matches become less than
> > A058129).  The
> > > >> > inverse Euler
> > > >> > transform,
> > > >> >
> > > >> > 1,2,8,41
> > > >> >
> > > >> > is the number of connected categories with n morphisms; this is
> > > >> > likewise probably not
> > > >> > in the OEIS (only A052447 is possible).
> > > >> >
> > > >> > Can somebody generate more data?
> > > >> >
> > > >> > Franklin T. Adams-Watters
> > > >> >
> > > >> > A category is a collection of objects and morphisms; each
> > morphism is
> > > >> > from one object
> > > >> > to another (not necessarily different) object.  Where the
> > destination
> > > >> > of one morphism
> > > >> > is the source of a second, their composition is defined;
> > composition
> > > >> is
> > > >> > associative where
> > > >> > it is defined.  Each object has an identity morphism, which
> > connects
> > > >> it
> > > >> > to itself; this
> > > >> > is an identity when composed with morphisms coming in and with
> > > >> > morphisms going
> > > >> > out.
> > > >> >
> > > >>
> > ________________________________________________________________________
> > > >> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > > >> > industry-leading spam and email virus protection.
> > > >> >
> > > >>
> > > >
> > > >
> >
>
>
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From njas at research.att.com  Sat Nov 25 11:51:53 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sat, 25 Nov 2006 05:51:53 -0500 (EST)
Subject: request for new member
Message-ID: <200611251051.FAA87094@fry.research.att.com>


Olivier:

Richard Guy sent me this:

>From rkg at cpsc.ucalgary.ca  Fri Nov 24 17:07:00 2006
Delivered-To: njas at research.att.com
Date: Fri, 24 Nov 2006 15:02:50 -0700 (MST)
From: Richard Guy <rkg at cpsc.ucalgary.ca>
X-X-Sender: rkg at csl
To: "Neil J. A. Sloane" <njas at research.att.com>
cc: "Saff, Kevin" <kevin.saff at gmail.com>
Subject: New Subscriber
X-Virus-Scanned: by amavisd-new at cpsc.ucalgary.ca
X-Spam-Checker-Version: SpamAssassin 3.0.6 (2005-12-07) on 
        mail-purple.research.att.com
X-Spam-Level: 
X-Spam-Status: No, score=-2.9 required=4.0 tests=AWL,BAYES_00 autolearn=ham 
        version=3.0.6

Neil,
      May Kevin Saff join the seq-fan list?

      His email address is as above.

      Thanks & best wishes,    R.

Neil







From Annette.Warlich at t-online.de  Sat Nov 25 20:26:49 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Sat, 25 Nov 2006 20:26:49 +0100
Subject: Curious binomial-identity /A002720
Message-ID: <456898F9.8020608@t-online.de>


By chance I came across this curious identity
involving the Pascal-triangle.

Assume a row n, say n=4 and the column n, combined
each weighted with the running factorial as in the example:


           1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
 ratio =   -----------------------------------------  -------------------
           1/0! + 4/1! + 6/2! + 4/3! + 1/4!           weighted row-sum


then

  ratio = e  (=exp(1))

The actual sums are the entries of A002720
 http://www.research.att.com/~njas/sequences/A002720

Regards -

Gottfried Helms






From Annette.Warlich at t-online.de  Sat Nov 25 20:41:53 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Sat, 25 Nov 2006 20:41:53 +0100
Subject: Curious binomial-identity /A002720 (small correction of prev.
 post)
In-Reply-To: <456898F9.8020608@t-online.de>
References: <456898F9.8020608@t-online.de>
Message-ID: <45689C81.2020302@t-online.de>

Am 25.11.2006 20:26 schrieb Gottfried Helms:
> By chance I came across this curious identity
> involving the Pascal-triangle.
> 
> Assume a row n, say n=4 and the column n, combined
> each weighted with the running factorial as in the example:
> 
> 
>            1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
>  ratio =   -----------------------------------------  -------------------
>            1/0! + 4/1! + 6/2! + 4/3! + 1/4!           weighted row-sum
> 
> 
> then
> 
>   ratio = e  (=exp(1))
> 

------------------------------------

> The actual sums are the entries of A002720
>  http://www.research.att.com/~njas/sequences/A002720
> 
that should be corrected; the entries in A002720 are

 A(n)  =  weighted-rowsum(n) * n!
       =  weighted-colsum(n) * n! / exp(1)

I forgot to mention the additional n!, since in numerator and
denominator of the above fraction they cancel out, sorry.

Gottfried Helms






From jvospost3 at gmail.com  Sat Nov 25 21:10:27 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sat, 25 Nov 2006 12:10:27 -0800
Subject: Curious binomial-identity /A002720 (small correction of prev. post)
In-Reply-To: <45689C81.2020302@t-online.de>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de>
Message-ID: <5542af940611251210g5e4916bcvee22981abd609ca1@mail.gmail.com>

There are several formulae for A002720, including a(n) = Sum k!C(n, k)^2,
k=0..n.

What you write might be true in the asymptotic limit, but not for any term,
as each term is rational and dividing by e would make each term
transcendental.  I'm sure that you meant the limit, right?

On 11/25/06, Gottfried Helms <Annette.Warlich at t-online.de> wrote:
>
> Am 25.11.2006 20:26 schrieb Gottfried Helms:
> > By chance I came across this curious identity
> > involving the Pascal-triangle.
> >
> > Assume a row n, say n=4 and the column n, combined
> > each weighted with the running factorial as in the example:
> >
> >
> >            1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
> >  ratio =
> -----------------------------------------  -------------------
> >            1/0! + 4/1! + 6/2! + 4/3! + 1/4!           weighted row-sum
> >
> >
> > then
> >
> >   ratio = e  (=exp(1))
> >
>
> ------------------------------------
>
> > The actual sums are the entries of A002720
> >  http://www.research.att.com/~njas/sequences/A002720
> >
> that should be corrected; the entries in A002720 are
>
> A(n)  =  weighted-rowsum(n) * n!
>        =  weighted-colsum(n) * n! / exp(1)
>
> I forgot to mention the additional n!, since in numerator and
> denominator of the above fraction they cancel out, sorry.
>
> Gottfried Helms
>
>
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From Annette.Warlich at t-online.de  Sat Nov 25 22:41:35 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Sat, 25 Nov 2006 22:41:35 +0100
Subject: Curious binomial-identity /A002720 (small correction of prev.
 post)
In-Reply-To: <5542af940611251210g5e4916bcvee22981abd609ca1@mail.gmail.com>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de> <5542af940611251210g5e4916bcvee22981abd609ca1@mail.gmail.com>
Message-ID: <4568B88F.1020800@t-online.de>

Am 25.11.2006 21:10 schrieb Jonathan Post:
> There are several formulae for A002720, including  a(n) = Sum k!C(n,
> k)^2, k=0..n.
> 
> What you write might be true in the asymptotic limit, but not for any
> term, as each term is rational and dividing by e would make each term
> transcendental.  I'm sure that you meant the limit, right?

Yes; as the +... in the numerator should indicate. The sum of columns
have infinitely many terms.

I've seen the generation-formula involving the exp()-function; maybe
this translates in an obvious way (... only with gf's I don't have
*any* experience... sigh...) and the asymptotic formula, which looks
a bit more complicated, but additionally relates it to a pi-expression.

But that we have

   1/3! +  3/2! + 3/1! + 1/0!  = A(3) /3!
                       + 3/1!
                       + 6/2!
                       +10/3!
                       + ...
                        -----
    limit k->oo         A(3) /3! * exp(1)

 or

   1/4! +  4/3! + 6/2! + 4/1! + 1/0!  = A(4)/4!
                              + 4/1!
                              +10/2!
                              +20/3!
                              + ...
                              -----
    limit k->oo               A(4)/4! *exp(1)

has something of a certain beauty... :-)


Gottfried Helms







From jrbibers at indiana.edu  Sat Nov 25 22:53:40 2006
From: jrbibers at indiana.edu (Joseph Biberstine)
Date: Sat, 25 Nov 2006 16:53:40 -0500
Subject: Curious binomial-identity /A002720 (small correction of prev.
 post)
In-Reply-To: <45689C81.2020302@t-online.de>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de>
Message-ID: <4568BB64.9050703@indiana.edu>

The example you gave has an off-by-one row index as follows:

Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n+1)

Note that the row index is incremented.  Mathematica simplifies this
fraction to LaguerreL[-(n+1), 1]/LaguerreL[n+1, -1].  This is not equal
to Exp[1], but interestingly seems to approach Exp[1] monotonically from
below as n approaches infinity.

I'm sure you meant:

Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n)

Mathematica simplifies this to LaguerreL[-(n+1), 1]/LaguerreL[n, -1].
Although I can't get it to explicitly resolve that this is exactly
Exp[1], the values do seem to bear this truth out.

So, I believe your example should read: e.g. for n=3,

    1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
e = -----------------------------------------  -------------------
           1/0! + 3/1! + 3/2! + 1/3!           weighted row-sum

This does seem to hold for every col/row pair in the triangle.

Nice find :)
-JRB

Gottfried Helms wrote:
> Am 25.11.2006 20:26 schrieb Gottfried Helms:
>> By chance I came across this curious identity
>> involving the Pascal-triangle.
>>
>> Assume a row n, say n=4 and the column n, combined
>> each weighted with the running factorial as in the example:
>>
>>
>>            1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
>>  ratio =   -----------------------------------------  -------------------
>>            1/0! + 4/1! + 6/2! + 4/3! + 1/4!           weighted row-sum
>>
>>
>> then
>>
>>   ratio = e  (=exp(1))
>>
> 
> ------------------------------------
> 
>> The actual sums are the entries of A002720
>>  http://www.research.att.com/~njas/sequences/A002720
>>
> that should be corrected; the entries in A002720 are
> 
>  A(n)  =  weighted-rowsum(n) * n!
>        =  weighted-colsum(n) * n! / exp(1)
> 
> I forgot to mention the additional n!, since in numerator and
> denominator of the above fraction they cancel out, sorry.
> 
> Gottfried Helms
> 





From Annette.Warlich at t-online.de  Sun Nov 26 00:43:25 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Sun, 26 Nov 2006 00:43:25 +0100
Subject: Curious binomial-identity /A002720 (small correction of prev.
 post)
In-Reply-To: <4568BB64.9050703@indiana.edu>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de> <4568BB64.9050703@indiana.edu>
Message-ID: <4568D51D.8000305@uni-kassel.de>

Am 25.11.2006 22:53 schrieb Joseph Biberstine:
> The example you gave has an off-by-one row index as follows:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n+1)
> 
> Note that the row index is incremented.  Mathematica simplifies this
> fraction to LaguerreL[-(n+1), 1]/LaguerreL[n+1, -1].  This is not equal
> to Exp[1], but interestingly seems to approach Exp[1] monotonically from
> below as n approaches infinity.
> 
> I'm sure you meant:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n)
> 
> Mathematica simplifies this to LaguerreL[-(n+1), 1]/LaguerreL[n, -1].
> Although I can't get it to explicitly resolve that this is exactly
> Exp[1], the values do seem to bear this truth out.
> 
> So, I believe your example should read: e.g. for n=3,
> 
>     1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
> e = -----------------------------------------  -------------------
>            1/0! + 3/1! + 3/2! + 1/3!           weighted row-sum
> 
> This does seem to hold for every col/row pair in the triangle.

Yes, your correction is right - I didn't look right at the
matrix.


The connection to the Laguerre-polynomials is not too
much surprising; I'm dealing with a ceratin classes
of lower-triangular matrices, which are derived from the
logic of the computing of the Pascal-matrix by matrix-
exponentiation.

While

  Pascalmatrix     = Pk(1)=  matrixexp( subdiagonal(1; [1,2,3,4,5,...]))

the Laguerre-matrix occurs from

  Laguerrematrix(*) = Pk(2) = matrixexp( subdiagonal(1; [1,2^2,3^2,4^2,5^2,...]))
           ((*) means: in a signed and simply scaled form)


I also have
                      Pk(0) = matrixexp( subdiagonal(1; [1,2^0,3^0,4^0,5^0,...]))

(where the vectors are filled in the first principal subdiagonal of an
infinite square-matrix)




Now the current matrix was created by a somehow routinely
"completion of my toolbox" of pascal-like-matrices, as a
hadamard-product of

      Pf =    Pk(1) (x) Pk(0)

         =    1/0!
              1/1!  1/0!
              1/2!  2/1!  1/0!
              1/3!  3/2!  3/1!  1/0!
              1/4!  4/3!  6/2!  4/1!   1/0!
                ....


Then with a powerseries-column-vector

      V(x) = [1,x,x^2,x^3,....] ~

and a diagonal factorial-matrix (used for scaling)

     dF(m) = diag(0! , 1!^m, 2!^m .... )

I got

   dF(1)  *   Pf  * V(1) = A002720       // A002720 read as column-vector
                                         // Due to the entry of Paul Berry

and
   dF(1) *    Pf~ * V(1) = A002720  *exp(1)  // added today



As a table it looks like:

                                                 Row-Sum
         =    1/0!                              =  1/0!    =  A(0)/0!
              1/1!  1/0!                        =  2/1!    =  A(1)/1!
              1/2!  2/1!  1/0!                  =  7/2!    =  A(2)/2!
              1/3!  3/2!  3/1!  1/0!            = 34/3!    =  A(3)/3!
              1/4!  4/3!  6/2!  4/1!   1/0!     =209/4!    =  A(4)/4!
                ....
             --------------------------------------------------------
Col-sum   e*[  1    2/1!  7/2! 34/3! 209/4! ]

Note, that using

   dF(1)  *   Pf  * V(2) =          ???
   dF(1) *    Pf~ * V(2) = exp(2) * ???

  in the second an exponentiation of 2 occurs, and generally for an s

   dF(1) *    Pf~ * V(s) = exp(s) * ???

 with unknown, but -heuristically- integer sequences of coefficients in ???
 for integer s>=0 .


------------------------------------------------------------------------


The Pascal- and (unsigned,scaled) Laguerre-matrix are also
related by

  Pk(0)                              // "base"-matrix
  Pk(1) =   dF(1)* Pk(0) * dF(-1)    // Pascal-matrix
  Pk(2) =   dF(1)* Pk(1) * dF(-1)    // unsigned,scaled Laguerrematrix
 ...
(which is due to the construction via matrix-exponentiation
of the related subdiagonal-matrices)
so finally the occurence of the relation to the Laguerre-matrix
using Mathematica is not too surprising...

Regards -

Gottfried Helms







From zakseidov at yahoo.com  Sun Nov 26 20:58:31 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Sun, 26 Nov 2006 11:58:31 -0800 (PST)
Subject: A124080 10 times triangular numbers
Message-ID: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>

Neil, Zerinvary, seqfans,
Why this sequence?!
Thanks, Zak 

%I A124080
%S A124080
0,10,30,60,100,150,210,280,360,450,550,660,780,910,1050,1200,1360,1530,
%T A124080
1710,1900,2100,2310,2530,2760,3000,3250,3510,3780,4060,4350,4650,4960,
%U A124080
5280,5610,5950,6300,6660,7030,7410,7800,8200,8610,9030,9460,9900,10350
%N A124080 10 times triangular numbers.
%F A124080 a(n)=10*C(n,2), n>=1
%p A124080 [seq(10*binomial(n,2) ,n=1..51)];
%Y A124080 Cf. A028895, A046092, A045943, A002378,
A028896, A024966, A033996, A027468.
%K A124080 easy,nonn,new
%O A124080 0,2
%A A124080 Zerinvary Lajos
(zerinvarylajos(AT)yahoo.com), Nov 24 2006



 
____________________________________________________________________________________
Do you Yahoo!?
Everyone is raving about the all-new Yahoo! Mail beta.
http://new.mail.yahoo.com





From jvospost3 at gmail.com  Sun Nov 26 22:23:25 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 26 Nov 2006 13:23:25 -0800
Subject: Categories
In-Reply-To: <20061124075307.GA17864@fourier.ujf-grenoble.fr>
References: <8C8DC932D93028F-B90-4484@FWM-M05.sysops.aol.com>
	 <20061123065439.GA28473@fourier.ujf-grenoble.fr>
	 <5542af940611230914g2c7b773i32a265bdc5b33c18@mail.gmail.com>
	 <5542af940611231210p5426fee8i9542123458aee474@mail.gmail.com>
	 <20061124075307.GA17864@fourier.ujf-grenoble.fr>
Message-ID: <5542af940611261323y13ada5f4p9e8f6e5d9051f93d@mail.gmail.com>

I can, with Edwin Clark's permission, forward to Frank my complete
correspondance with Edwin Clark about category enumeration via careful
definition.  I am asked not to cut & paste out of context.  But my emails to
Franklin T. Adams-Watters alone went astray, for whatever reason. so I am
not sure what to do.  I've also asked some Haskell programming experts about
Frank's enumeration question (properly citing Frank), as Haskell has monads
and functors and other categorical stuff built in.  I'll share any useful
answers that I get.

On 11/23/06, Roland Bacher <Roland.Bacher at ujf-grenoble.fr> wrote:
>
>
> I realised that the formula below for enumerating categories
> using quivers is completely wrong. It gives only an upper bound
> since one has to require moreover associativity for the composition of
> morphisms. The quiver approach works however in principle but
> is much more complex (and I guess this approach has been
> used in some of the previous approaches).   Roland Bacher
>
> On Thu, Nov 23, 2006 at 12:10:51PM -0800, Jonathan Post wrote:
> > *"Formalized Proof, Computation, and the Construction Problem in
> Algebraic
> > geometry", by Carlos Simpson.
> >
> > [PDF]* arXiv:*math*.AG/0410224 v1 8 Oct
> > 2004<http://arxiv.org/pdf/math/0410224> File
> > Format: PDF/Adobe Acrobat - View as
> > HTML<
> http://72.14.253.104/search?q=cache:8FWyl8lGqRwJ:arxiv.org/pdf/math/0410224+%22how+many+categories%22+math+theory&hl=en&gl=us&ct=clnk&cd=5
> >
> > finite integer N, *how many categories* are there with N morphisms? What
> *
> > ...* Algebraic *theory* of machines, I. Trans. Amer. *Math*. Soc. 116
> > (1965), 450-464.
> >
> >
> >
> > On 11/23/06, Jonathan Post <jvospost3 at gmail.com> wrote:
> > >
> > >Roland Bacher correctly refers to quivers as summarized in wikipedia at
> > >
> > >http://en.wikipedia.org/wiki/Quiver_%28mathematics%29
> > >
> > >On 11/22/06, Roland Bacher < Roland.Bacher at ujf-grenoble.fr> wrote:
> > >>
> > >>
> > >> This numbers can also be obtained as the numbers
> > >> of the following quotients of quivers:
> > >>
> > >> Associate to a category with n morphisms and k objects
> > >> the quiver with k vertices corresponding to the objects
> > >> and a(X,Y) arrows directed from the object X to the object
> > >> Y if there are a(X,Y) morphisms from X into Y.
> > >>
> > >> Given two morphism g:X-->Y, f:Y-->Z with composition
> > >> h=f o g:X-->Z, put the relation gf=h on the quiver algebra
> > >>
> > >> The resulting quotient algebra has dimension n and a basis
> > >> given by the (simple) arrows.
> > >>
> > >> This leads to the following "algorithm" for enumerating
> > >> all categories with n morphisms and k objects:
> > >>
> > >> (a) enumerate all quivers (directed graphs) with k vertices and
> > >> n oriented edges.
> > >>
> > >> (b) Associate the following "weight" to such a quiver as follows :
> > >>
> > >> (b1)given a triplet of vertices X,Y,Z,
> > >> set w(X,Y,Z)=(a(X,Z))^{a(X,Y) \cdot a(Y,Z)}
> > >> (this counts the number of ways which associate a morphism
> > >> X-->Z to a composition X-->Y-->Z) where a(U,V) denotes the
> > >> number of oriented arrows starting at U and ending at W.
> > >>
> > >> (b2) associate to a given quiver the weight
> > >> \prod_{X,Y,Z} w(X,Y,Z)
> > >> where the product is over all triplets of vertices.
> > >>
> > >> The total sum of such weighted quivers yields then the solution.
> > >>
> > >> Roland Bacher
> > >>
> > >>
> > >>
> > >>
> > >>
> > >>
> > >> On Wed, Nov 22, 2006 at 04:20:32PM -0500, franktaw at netscape.netwrote:
> > >> > How many categories are there?
> > >> >
> > >> > First, how many categories are there with n morphisms and k
> objects?
> > >> > This table starts:
> > >> >
> > >> > 1
> > >> > 2  1
> > >> > 7  3 1
> > >> > 35 16 3 1
> > >> >
> > >> > The first column is A058129, the number of monoids; the main
> diagonal
> > >> > is all 1's.  I am not
> > >> > 100% certain of the 16 in the final row.
> > >> >
> > >> > Taking the row sums, we get:
> > >> >
> > >> > 1,3,11,55
> > >> >
> > >> > the number of categories with n morphisms.  This is probably not in
> > >> the
> > >> > OEIS (only
> > >> > A001776 is possible - other matches become less than A058129).  The
> > >> > inverse Euler
> > >> > transform,
> > >> >
> > >> > 1,2,8,41
> > >> >
> > >> > is the number of connected categories with n morphisms; this is
> > >> > likewise probably not
> > >> > in the OEIS (only A052447 is possible).
> > >> >
> > >> > Can somebody generate more data?
> > >> >
> > >> > Franklin T. Adams-Watters
> > >> >
> > >> > A category is a collection of objects and morphisms; each morphism
> is
> > >> > from one object
> > >> > to another (not necessarily different) object.  Where the
> destination
> > >> > of one morphism
> > >> > is the source of a second, their composition is defined;
> composition
> > >> is
> > >> > associative where
> > >> > it is defined.  Each object has an identity morphism, which
> connects
> > >> it
> > >> > to itself; this
> > >> > is an identity when composed with morphisms coming in and with
> > >> > morphisms going
> > >> > out.
> > >> >
> > >>
> ________________________________________________________________________
> > >> > Check Out the new free AIM(R) Mail -- 2 GB of storage and
> > >> > industry-leading spam and email virus protection.
> > >> >
> > >>
> > >
> > >
>
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From jvospost3 at gmail.com  Sun Nov 26 22:27:11 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 26 Nov 2006 13:27:11 -0800
Subject: A124080 10 times triangular numbers
In-Reply-To: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
References: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
Message-ID: <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>

I wondered too.  It intersects the much more interesting
 A000537 <http://www.research.att.com/%7Enjas/sequences/A000537>
 Sum of first n cubes; or n-th triangular number squared.  at 100.

Not that all of my roughly 1 kilophi sequences and comments are interesting,
either.

-- Jonathan

On 11/26/06, zak seidov <zakseidov at yahoo.com> wrote:
>
> Neil, Zerinvary, seqfans,
> Why this sequence?!
> Thanks, Zak
>
> %I A124080
> %S A124080
> 0,10,30,60,100,150,210,280,360,450,550,660,780,910,1050,1200,1360,1530,
> %T A124080
> 1710,1900,2100,2310,2530,2760,3000,3250,3510,3780,4060,4350,4650,4960,
> %U A124080
> 5280,5610,5950,6300,6660,7030,7410,7800,8200,8610,9030,9460,9900,10350
> %N A124080 10 times triangular numbers.
> %F A124080 a(n)=10*C(n,2), n>=1
> %p A124080 [seq(10*binomial(n,2) ,n=1..51)];
> %Y A124080 Cf. A028895, A046092, A045943, A002378,
> A028896, A024966, A033996, A027468.
> %K A124080 easy,nonn,new
> %O A124080 0,2
> %A A124080 Zerinvary Lajos
> (zerinvarylajos(AT)yahoo.com), Nov 24 2006
>
>
>
>
>
> ____________________________________________________________________________________
> Do you Yahoo!?
> Everyone is raving about the all-new Yahoo! Mail beta.
> http://new.mail.yahoo.com
>
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From jvospost3 at gmail.com  Sun Nov 26 23:43:48 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 26 Nov 2006 14:43:48 -0800
Subject: A124080 10 times triangular numbers
In-Reply-To: <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>
References: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
	 <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>
Message-ID: <5542af940611261443g32cd3bf3lf2174cfc2d42e716@mail.gmail.com>

On the other hand, every sequence is related to some interesting sequence...
I just submitted by form the following:

 NEW SEQUENCE FROM Jonathan Vos Post


%I A000001
%S A000001 11, 29, 31, 59, 61, 101, 149, 151, 211, 281, 359, 449, 659,
661, 911, 1049, 1051, 1201, 1361, 1531, 1709, 1901, 2099, 2309, 2311,
2531, 2999, 3001
%N A000001 Primes of the form A124080 (10 times triangular numbers), +1
or -1.
%C A000001 The j, such that A124080(j)-1 is prime or A124080(j)+1 is
prime, where repetition means a twin prime, are 1, 2, 2, 3, 3, 4, 5, 5,
6, 7, 8, 9, 11, 11, 13, 14, 14, 15, 16, 17, 18, 19, 20, 21, 22, 24, 24.
%F A000001 {A124080(j)-1 when prime} U {A124080(j)+1 when prime} =
{i = 10*T(j)-1 such that i is prime} U {i = 10*T(j)+1 such that i is
prime} where T(j) = A000217(j) = j*(j+1)/2.
%e A000001 a(1) = A124080(1)+1 = (10*T(1)) - 1 = 10*(1*(1+1)/2) + 1 =
10+1 = 11 is prime.
a(2) = A124080(2)-1 = (10*T(2))-1 = 10*(2*(2+1)/2) - 1 = 30-1 = 29 is
prime.
a(3) = A124080(2)+1 = (10*T(2))+1 = 10*(2*(2+1)/2) + 1 = 30+1 = 31 is
prime.
%Y A000001 Cf. A000040, A000217, A028895, A046092, A045943, A002378,
A028896, A024966, A033996, A027468.
%O A000001 1,1
%K A000001 ,easy,nonn,
%A A000001 Jonathan Vos Post (jvospost2 at yahoo.com
<http://us.f551.mail.yahoo.com/ym/Compose?To=jvospost2 at yahoo.com&YY=86457&y5beta=yes&y5beta=yes&order=down&sort=date&pos=0&view=a&head=b>),
Nov 26 2006
RH
RA 192.20.225.32
RU
RI



On 11/26/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> I wondered too.  It intersects the much more interesting
>  A000537 <http://www.research.att.com/%7Enjas/sequences/A000537>
>  Sum of first n cubes; or n-th triangular number squared.  at 100.
>
> Not that all of my roughly 1 kilophi sequences and comments are
> interesting, either.
>
> -- Jonathan
>
> On 11/26/06, zak seidov <zakseidov at yahoo.com> wrote:
> >
> > Neil, Zerinvary, seqfans,
> > Why this sequence?!
> > Thanks, Zak
> >
> > %I A124080
> > %S A124080
> > 0,10,30,60,100,150,210,280,360,450,550,660,780,910,1050,1200,1360,1530,
> > %T A124080
> > 1710,1900,2100,2310,2530,2760,3000,3250,3510,3780,4060,4350,4650,4960,
> > %U A124080
> > 5280,5610,5950,6300,6660,7030,7410,7800,8200,8610,9030,9460,9900,10350
> > %N A124080 10 times triangular numbers.
> > %F A124080 a(n)=10*C(n,2), n>=1
> > %p A124080 [seq(10*binomial(n,2) ,n=1..51)];
> > %Y A124080 Cf. A028895, A046092, A045943, A002378,
> > A028896, A024966, A033996, A027468.
> > %K A124080 easy,nonn,new
> > %O A124080 0,2
> > %A A124080 Zerinvary Lajos
> > (zerinvarylajos(AT)yahoo.com), Nov 24 2006
> >
> >
> >
> >
> >
> > ____________________________________________________________________________________
> > Do you Yahoo!?
> > Everyone is raving about the all-new Yahoo! Mail beta.
> > http://new.mail.yahoo.com
> >
>
>
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From njas at research.att.com  Mon Nov 27 03:23:49 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Sun, 26 Nov 2006 21:23:49 -0500 (EST)
Subject: A124080 10 times triangular numbers
References: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
    <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>
Message-ID: <200611270223.VAA41276@fry.research.att.com>

JVP, you submitted this sequence:
%N A000001 Primes of the form A124080 (10 times triangular numbers), +1
or -1.
I will use it, but I do not find it interesting.  Why?  Because it 
is derivative.  It is based on a sequence of little interest - acceptable
only because it was sent in by someone who perhaps does not know
what is worth being included in the database - and is a step further 
down the ladder towards the basement.

I seem to have to play the role of the singing teacher
at times:  so be it - this song was off-key!

Neil





From jvospost3 at gmail.com  Mon Nov 27 04:10:47 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Sun, 26 Nov 2006 19:10:47 -0800
Subject: A124080 10 times triangular numbers
In-Reply-To: <200611270223.VAA41276@fry.research.att.com>
References: <20061126195832.67293.qmail@web38206.mail.mud.yahoo.com>
	 <5542af940611261327s79a6b2d5uc133675de46fbaf1@mail.gmail.com>
	 <200611270223.VAA41276@fry.research.att.com>
Message-ID: <5542af940611261910x10195634teeac87bdf4bfd1e7@mail.gmail.com>

As usual, you're right, Neil.

I was trying to be nice to the newcomer, by showing him the collaborative
nature of the endeavor. I'm sure that his enthusiasm can be made fruitful.
I've certainly erred in judgment, many times, and you and your associate
editors have been splendid in guiding me towards better work.

Sometimes, as they sing: "you've got to be cruel to be kind."

Hence the "probation" keyword and, one of my useful suggestions to you and
your board, the "less" keyword.

Sorry for where that road leads, which I help pave with good intentions.

-- Jonathan

On 11/26/06, N. J. A. Sloane <njas at research.att.com> wrote:
>
> JVP, you submitted this sequence:
> %N A000001 Primes of the form A124080 (10 times triangular numbers), +1
> or -1.
> I will use it, but I do not find it interesting.  Why?  Because it
> is derivative.  It is based on a sequence of little interest - acceptable
> only because it was sent in by someone who perhaps does not know
> what is worth being included in the database - and is a step further
> down the ladder towards the basement.
>
> I seem to have to play the role of the singing teacher
> at times:  so be it - this song was off-key!
>
> Neil
>
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From pauldhanna at juno.com  Mon Nov 27 04:58:25 2006
From: pauldhanna at juno.com (Paul D. Hanna)
Date: Sun, 26 Nov 2006 22:58:25 -0500
Subject: Missing Table?
Message-ID: <20061126.225826.332.1.pauldhanna@juno.com>

Seqfans,
     Where is the rectangular table referred to by these sequences?
 
A040027  Main diagonal of an array of binomial recurrence coefficients.  
 0, 1, 1, 3, 9, 31, 121, 523, 2469, 12611, 69161, 404663,  
A045501  Second diagonal in table of binomial recurrence coefficients.  
 1, 1, 4, 14, 54, 233, 1101, 5625, 30846, 
A045499  Third diagonal in table of binomial recurrence coefficients. 
 1, 1, 5, 20, 85, 400, 2046, 11226, 65676, 407787, 2675410, 
A045500  Fourth diagonal in table of binomial recurrence coefficients. 
 1, 1, 6, 27, 125, 635, 3488, 20425, 126817, 831915, 5744784, 41618459 
(AUTHOR of these sequences:  H. W. Gould) 
 
The closest thing I can find is Emeric Deutsch's recent triangle A124496,

in which these sequences appear as columns. 
 
The lower half of the table (I guess) must look like this:
1, ...
1, 1, ...
1, 1, 3, ...
1, 1, 4, 9, ...
1, 1, 5, 14, 31, ...
1, 1, 6, 20, 54, 121, ...
1, 1, 7, 27, 85, 233, 523, ...
1, 1, 8, 35, 125, 400, 1101, 2469, ...
1, 1, 9, 44, 175, 635, 2046, 5625, 12611, ...
 
Is the complete table in the OEIS?  
If not, it certainly should be. 
     Paul
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From mathar at strw.leidenuniv.nl  Mon Nov 27 10:47:35 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Mon, 27 Nov 2006 10:47:35 +0100
Subject: Missing Table?
Message-ID: <200611270947.kAR9lZBJ001619@amer.strw.leidenuniv.nl>


The master table of A040027 could also be A046936..but this does not fit the
2nd, 3rd and 4th diagonal as well as A124496 does.

RJM

> From seqfan-owner at ext.jussieu.fr  Mon Nov 27 05:03:23 2006
> Return-Path: <seqfan-owner at ext.jussieu.fr>
> To: Seqfan at ext.jussieu.fr
> Cc: gould at math.wvu.edu
> Date: Sun, 26 Nov 2006 22:58:25 -0500
> Subject: Missing Table?
> From: "Paul D. Hanna" <pauldhanna at juno.com>
> 
> Seqfans,
>      Where is the rectangular table referred to by these sequences?
>  
> A040027  Main diagonal of an array of binomial recurrence coefficients.  
>  0, 1, 1, 3, 9, 31, 121, 523, 2469, 12611, 69161, 404663,  
> A045501  Second diagonal in table of binomial recurrence coefficients.  
>  1, 1, 4, 14, 54, 233, 1101, 5625, 30846, 
> A045499  Third diagonal in table of binomial recurrence coefficients. 
>  1, 1, 5, 20, 85, 400, 2046, 11226, 65676, 407787, 2675410, 
> A045500  Fourth diagonal in table of binomial recurrence coefficients. 
>  1, 1, 6, 27, 125, 635, 3488, 20425, 126817, 831915, 5744784, 41618459 
> (AUTHOR of these sequences:  H. W. Gould) 
>  
> The closest thing I can find is Emeric Deutsch's recent triangle A124496,
> 
> in which these sequences appear as columns. 
>  
> The lower half of the table (I guess) must look like this:
> 1, ...
> 1, 1, ...
> 1, 1, 3, ...
> 1, 1, 4, 9, ...
> 1, 1, 5, 14, 31, ...
> 1, 1, 6, 20, 54, 121, ...
> 1, 1, 7, 27, 85, 233, 523, ...
> 1, 1, 8, 35, 125, 400, 1101, 2469, ...
> 1, 1, 9, 44, 175, 635, 2046, 5625, 12611, ...
>  
> Is the complete table in the OEIS?  
> If not, it certainly should be. 
>      Paul





From tanyakh at TanyaKhovanova.com  Mon Nov 27 17:05:13 2006
From: tanyakh at TanyaKhovanova.com (Tanya Khovanova)
Date: Mon, 27 Nov 2006 08:05:13 -0800
Subject: practical numbers
Message-ID: <200611270805.AA74449328@TanyaKhovanova.com>


Hello seqfans,

There is some confusion with practical numbers. Wikipedia article on practical numbers: http://en.wikipedia.org/wiki/Practical_number
gives the following definition of a practical number:
A practical number or panarithmic number is a positive integer n such that all preceding positive integers are a sum of distinct divisors of n.

At the same time it refers to the sequence A005153 which gives a slightly different definition:
Practical numbers (first definition): all k <= sigma(n) are sums of distinct divisors of n. Also called panarithmic numbers.

Browsing other websites (including my own :-) ) I've found that the wiki definition is used more often. Moreover, my calculation with the my definition exactly matches the sequence A005153, though they are supposed to be different.

Please, clarify my confusion.

Tanya Khovanova






From mathar at strw.leidenuniv.nl  Mon Nov 27 17:54:41 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Mon, 27 Nov 2006 17:54:41 +0100
Subject: practical numbers
Message-ID: <200611271654.kARGsfOX022242@amer.strw.leidenuniv.nl>

> From seqfan-owner at ext.jussieu.fr  Mon Nov 27 17:07:28 2006
> Date: Mon, 27 Nov 2006 08:05:13 -0800
> From: "Tanya Khovanova" <tanyakh at tanyakhovanova.com>
> Reply-To: <tanyakh at tanyakhovanova.com>
> To: <seqfan at ext.jussieu.fr>
> Subject: practical numbers
> ...
> There is some confusion with practical numbers. Wikipedia article on practical numbers: http://en.wikipedia.org/wiki/Practical_numbe
> r
> gives the following definition of a practical number:
> A practical number or panarithmic number is a positive integer n such that all preceding positive integers are a sum of distinct div
> isors of n.
> 
> At the same time it refers to the sequence A005153 which gives a slightly different definition:
> Practical numbers (first definition): all k <= sigma(n) are sums of distinct divisors of n. Also called panarithmic numbers.
> 
> Browsing other websites (including my own :-) ) I've found that the wiki definition is used more often. Moreover, my calculation wit
> h the my definition exactly matches the sequence A005153, though they are supposed to be different.
> 
> Please, clarify my confusion.
> ...

The A005153 definition [with reference to the sigma(n)] is also used
in
http://www.dm.unipi.it/gauss-pages/melfi/public_html/articoli/jnt.ps
but the same author uses in his web page
http://members.unine.ch/giuseppe.melfi/pratica.html
just the other definition.
see also http://citeseer.ist.psu.edu/285.html
http://arxiv.org/abs/math.NT/0404555

B. M Stewart in Am J. Math 76 (4) (1954) p 779, available from
http://www.jstor.org, quotes Srinivasan [Current Science (1948), pp 179] with
the definition without the sigma.

E J Scourfield in J. Number Theory 62 (1) (1997) p 163 uses the definition
including the sigma:
http://www.emis.de/projects/EULER/detail?ide=1997saiaentidividens&matchno=7&matchtotal=18&q=cr%3AE*+Saias+

I have to stop here. The recent IAU resolution on what a useful definition of a
"planet" might be, in particular the status of Pluto, be has already caused
enough confusion, and we're still recovering from this.

Richard





From franktaw at netscape.net  Mon Nov 27 18:04:03 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 27 Nov 2006 12:04:03 -0500
Subject: practical numbers
In-Reply-To: <200611270805.AA74449328@TanyaKhovanova.com>
References: <200611270805.AA74449328@TanyaKhovanova.com>
Message-ID: <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com>

The definitions are equivalent.  The condition cited in the Wikipedia 
article, to make everything up to n a sum of the divisors of n, is 
sufficient to make everything up to sigma(n) be such a sum.

Franklin T. Adams-Watters


-----Original Message-----
From: tanyakh at TanyaKhovanova.com

Hello seqfans,

There is some confusion with practical numbers. Wikipedia article on 
practical
numbers: http://en.wikipedia.org/wiki/Practical_number
gives the following definition of a practical number:
A practical number or panarithmic number is a positive integer n such 
that all
preceding positive integers are a sum of distinct divisors of n.

At the same time it refers to the sequence A005153 which gives a 
slightly
different definition:
Practical numbers (first definition): all k <= sigma(n) are sums of 
distinct
divisors of n. Also called panarithmic numbers.

Browsing other websites (including my own :-) ) I've found that the 
wiki
definition is used more often. Moreover, my calculation with the my 
definition
exactly matches the sequence A005153, though they are supposed to be 
different.

Please, clarify my confusion.

Tanya Khovanova


________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From davidwwilson at comcast.net  Mon Nov 27 18:30:31 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Mon, 27 Nov 2006 12:30:31 -0500
Subject: practical numbers
References: <200611270805.AA74449328@TanyaKhovanova.com> <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com>
Message-ID: <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p>

Assuming Franklin's observation to be correct, why not

%C A005153 Also, n such that all k <= n are sums of distinct divisors of n.

Some poorly-thought-out questions before I head off to work:

Are there not-quite-practical numbers, that is, numbers such that n-1 is the 
only number < n which is not the sum of distinct divisors of n?

If not, is there some interesting bound f(n) < n such that if all k <= f(n) 
are sums of distinct divisors of n, then n is also a sum of distinct 
divisors of n?

----- Original Message ----- 
From: <franktaw at netscape.net>
To: <tanyakh at TanyaKhovanova.com>; <seqfan at ext.jussieu.fr>
Sent: Monday, November 27, 2006 12:04 PM
Subject: Re: practical numbers


> The definitions are equivalent.  The condition cited in the Wikipedia 
> article, to make everything up to n a sum of the divisors of n, is 
> sufficient to make everything up to sigma(n) be such a sum.
>
> Franklin T. Adams-Watters






From davidwwilson at comcast.net  Mon Nov 27 18:42:56 2006
From: davidwwilson at comcast.net (David Wilson)
Date: Mon, 27 Nov 2006 12:42:56 -0500
Subject: practical numbers
References: <200611270805.AA74449328@TanyaKhovanova.com> <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com> <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p>
Message-ID: <000601c7124b$7c8b0040$6501a8c0@yourxhtr8hvc4p>

Simce any number n is a repeated sum of any of its divisors, "sum of 
distinct divisors" seems like overkill.

Anyway, another question: What is the smallest number a(n) that is a sum of 
its distinct divisors in n ways? 






From franktaw at netscape.net  Mon Nov 27 20:12:37 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 27 Nov 2006 14:12:37 -0500
Subject: practical numbers
In-Reply-To: <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p>
References: <200611270805.AA74449328@TanyaKhovanova.com> <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com> <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p>
Message-ID: <8C8E06F230775CD-7BC-A6B4@FWM-M18.sysops.aol.com>

To answer the second question first, it suffices to show that every 
number up to and including ceiling(n/2) is a sum of the divisors of n.  
Except for n=3, that can be floor(n/2) instead.  (If n is odd, 2 is not 
a sum; if n is even, every m with n/2 <= m < n is a sum n/2 plus a 
number less than n/2.)  This answers the first question, essentially in 
the negative: 3 is the only such number.

More interesting, perhaps: what about numbers such that every number up 
to n, with only 1 exception, are the sum of divisors of n?  (Note that 
this has to be n in this case, not sigma(n), since the possible sums 
are symmetric with respect to sigma(n) - if a set S of divisors sums to 
m, then the complement of S, with respect to the set of divisors of n, 
sums to sigma(n) - m.)

70 is one such number, skipping only 4; I think 70p for any prime 5 <= 
p < 70 does the same.  945 is another, skipping 2.  (945 is the 
smallest odd abundant number.)  I think that these values (70, 350, 
490, 770, 910, 945) are the only ones up to 1000.

The number skipped is always even; in fact, it always one less than a 
prime divisor p of n (and 1 more than the sum of the divisors of a 
practical divisor of n - which implies that that practical divisor is a 
square or twice a square).  Except for n=3, there must be another prime 
divisor q of n with p < q < 2p.

Franklin T. Adams-Watters


-----Original Message-----
From: davidwwilson at comcast.net

Assuming Franklin's observation to be correct, why not?
?
%C A005153 Also, n such that all k <= n are sums of distinct divisors 
of n.?
?
Some poorly-thought-out questions before I head off to work:?
?
Are there not-quite-practical numbers, that is, numbers such that n-1 
is the only number < n which is not the sum of distinct divisors of n??
?
If not, is there some interesting bound f(n) < n such that if all k <= 
f(n) are sums of distinct divisors of n, then n is also a sum of 
distinct divisors of n??
?
----- Original Message ----- From: <franktaw at netscape.net>?
To: <tanyakh at TanyaKhovanova.com>; <seqfan at ext.jussieu.fr>?
Sent: Monday, November 27, 2006 12:04 PM?
Subject: Re: practical numbers?
?
> The definitions are equivalent. The condition cited in the Wikipedia 
 > article, to make everything up to n a sum of the divisors of n, is > 
sufficient to make everything up to sigma(n) be such a sum.?
>?
> Franklin T. Adams-Watters?
?


________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
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From franktaw at netscape.net  Mon Nov 27 20:25:04 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Mon, 27 Nov 2006 14:25:04 -0500
Subject: practical numbers
In-Reply-To: <000601c7124b$7c8b0040$6501a8c0@yourxhtr8hvc4p>
References: <200611270805.AA74449328@TanyaKhovanova.com> <8C8E05D2CDF68E9-658-31C6@FWM-M04.sysops.aol.com> <001401c71249$bd6a4780$6501a8c0@yourxhtr8hvc4p> <000601c7124b$7c8b0040$6501a8c0@yourxhtr8hvc4p>
Message-ID: <8C8E070DFC523C7-7BC-A7AB@FWM-M18.sysops.aol.com>

I don't think "sum of distinct divisors" is really overkill.  If you 
leave out the "distinct", you'll get people complaining "but every 
number is a sum of its divisors" - or worse, misunderstanding.

A096356.  (Which should really be edited to say "proper divisors" 
instead of just "divisors".  Adding an initial 1 at offset 0 would also 
make sense to me.)

Franklin T. Adams-Watters


-----Original Message-----
From: davidwwilson at comcast.net

Simce any number n is a repeated sum of any of its divisors, "sum of 
distinct divisors" seems like overkill.?
?
Anyway, another question: What is the smallest number a(n) that is a 
sum of its distinct divisors in n ways? ?


________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From mathar at strw.leidenuniv.nl  Mon Nov 27 21:46:49 2006
From: mathar at strw.leidenuniv.nl (Richard Mathar)
Date: Mon, 27 Nov 2006 21:46:49 +0100
Subject: Re^2: practical numbers
Message-ID: <200611272046.kARKknOV015522@amer.strw.leidenuniv.nl>


> From seqfan-owner at ext.jussieu.fr  Mon Nov 27 18:44:49 2006
> From: "David Wilson" <davidwwilson at comcast.net>
> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
> Subject: Re: practical numbers
> ...
> Anyway, another question: What is the smallest number a(n) that is a sum of 
> its distinct divisors in n ways? 

For each n we get the "trivial" solution where the sum consists of n itself
as the only term. Further solutions, where the sum contains more than one term
(all of which are distinct divisors), are listed below in the format
n [list of terms that sum to n]:

6 [1, 2, 3]
12 [1, 2, 3, 6]
12 [2, 4, 6]
18 [1, 2, 6, 9]
18 [3, 6, 9]
20 [1, 4, 5, 10]
24 [1, 2, 3, 4, 6, 8]
24 [1, 2, 3, 6, 12]
24 [2, 4, 6, 12]
24 [1, 3, 8, 12]
24 [4, 8, 12]
28 [1, 2, 4, 7, 14]
30 [1, 3, 5, 6, 15]
30 [2, 3, 10, 15]
30 [5, 10, 15]
36 [2, 3, 4, 6, 9, 12]
36 [2, 3, 4, 9, 18]
36 [1, 2, 6, 9, 18]
36 [3, 6, 9, 18]
36 [1, 2, 3, 12, 18]
36 [2, 4, 12, 18]
36 [6, 12, 18]
40 [1, 2, 4, 5, 8, 20]
40 [1, 4, 5, 10, 20]
40 [2, 8, 10, 20]
42 [1, 6, 14, 21]
42 [7, 14, 21]


n=6 has 1 extra solution, n=12 or 18 have 2 extra solutions, n=20 has one,
n=24 has five etc.

Maple program to create that one (not checked!):

a := proc(n)
	local findx,d,f,found,i,count ;
	count :=0 :
	# decompose n into all sums that are candidates for the decomposition
	d := combinat[partition](n) ;
	# take each sum indivdually
	for findx from 1 to nops(d) do
		f := op(findx,d) ;
		# start with the assumption that it works
		found := true ;
		for i from 2 to nops(f) do
			# if two terms are equal in the Maple list, 
			# we flag this case as "not admitted"
			if op(i,f) = op(i-1,f) then
				found := false ;
				break ;
			fi ;
		od ;
		for i from 1 to nops(f) do
			# if any term is not a divisor,
			# we flag this case as "not admitted"
			if n mod op(i,f) <> 0 then
				found := false ;
				break ;
			fi ;
		od ;
		# If this sum seems to work and has more than one term, we list it
		if found and nops(f) > 1 then
			printf("%a %a\n",n,f) ;
			count := count+1 ;
		fi ;
	od ;
	RETURN(count) ;
end:

# loop over the most fundamental cases..
for n from 1 to 47 do
	a(n) ;
od ;

# Richard Mathar





From zbi74583 at boat.zero.ad.jp  Tue Nov 28 07:18:47 2006
From: zbi74583 at boat.zero.ad.jp (kohmoto)
Date: Tue, 28 Nov 2006 15:18:47 +0900
Subject: A038260
Message-ID: <025d01c712b5$16134290$977fd7dc@FMC57937B6BE00>

    Neil wrote on comment line of A038260.

    >Surely there are smaller members! njas, Nov 07 2006



    Why do you think that?

    I searched the ?seeds? for ?regular? Rational Amicable Number again.

       .

    And {11*19, 239 } is the smallest one.           .

    So, I think that at least it is the first terms of regular Rational 
Amicable Pair.

            .

    Such a Diophantine equation which is high degree and is involved with 
Sigma function has many conditions  , so the solution must have many prime 
factors for having good property, hence it becomes to have many digits.



    I guess that 2^2*7*3*5^2*13*31*139*277*3877*11*19 is the smallest and if 
an irregular example exists then it has around 16 digits.



    Yasutoshi


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From Annette.Warlich at t-online.de  Wed Nov 29 08:38:17 2006
From: Annette.Warlich at t-online.de (Gottfried Helms)
Date: Wed, 29 Nov 2006 08:38:17 +0100
Subject: Curious binomial-identity /A002720
In-Reply-To: <4568BB64.9050703@indiana.edu>
References: <456898F9.8020608@t-online.de> <45689C81.2020302@t-online.de> <4568BB64.9050703@indiana.edu>
Message-ID: <456D38E9.8030900@t-online.de>

Am 25.11.2006 22:53 schrieb Joseph Biberstine:
> The example you gave has an off-by-one row index as follows:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n+1)
> 
> Note that the row index is incremented.  Mathematica simplifies this
> fraction to LaguerreL[-(n+1), 1]/LaguerreL[n+1, -1].  This is not equal
> to Exp[1], but interestingly seems to approach Exp[1] monotonically from
> below as n approaches infinity.
> 
> I'm sure you meant:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n)
> 
> Mathematica simplifies this to LaguerreL[-(n+1), 1]/LaguerreL[n, -1].
> Although I can't get it to explicitly resolve that this is exactly
> Exp[1], the values do seem to bear this truth out.


I think I got it. As usual I like to write that problem in
matrix-mode.
So with

     V(x)~ * Pf = B

defined as the (infinite) matrix product as set of formal powerseries
in x:

                     1/0!   .     .     .
                     1/1!  1/0!
                     1/2!  2/1!  1/0!
                     1/3!  3/2!  3/1!  1/0!
                 .....
                   -------------------------------
[1 x x^2 x^3...]  = [ b0   b1    b2     b3 .... ]

we have, using the apostroph ' for derivative wrt x:

   b0  =            exp(x)      /0!
   b1  =  x*  [ x  *exp(x) ]'   /1!
   b2 =   x^2*[ x^2*exp(x) ]''  /2!
   b3 =   x^3*[ x^3*exp(x) ]''' /3!
 ...
   bk =   x^k*[ x^k*exp(x) ]'(k)' /k!

Expanding the derivatives gives

(1):

   b0 *0! / exp(x) =   1
   b1 *1! / exp(x) =   1 + x

   b2 *2! / exp(x) =   (2 + 4x + x^2)
                   =   2!/0! + 2*2!/1!  + 1*2!/2! x^2

   b3 *3! / exp(x) =   (6 + 18x + 9x^2 + x^3)
                        3! + 3*3!/1! x + 3*3!/2! x^2 + 1 *3!/3! x^3

   b4 *4! / exp(x) =   (24   + 96x  + 72x^2   + 16x^3 + 1x^4)
                       (1*4! + 4*4!/1! x + 6*4!/2!x^2 + 4*4!/3!x^3 + 1*4!/4!*x^4)



and the rhs written as matrix, where also the k! of the k'th row is cancelled , gives

(2):
   b0 / exp(x)      =   1                              *  [1,  x,  x^2,  x^3, ....]~
   b1 / exp(x)      =   1/0!  1/1!
   b2 / exp(x)      =   1/0!  2/1!  1/2!
   b3 / exp(x)      =   1/0!  3/1!  3/2!   1/3!
     ...            =     ....

If we now consider the result when letting x->1 we can further use
the horizontal symmetry of the binomial-coefficients to reorder
the rows and just get the matrix Pf again as coefficients

(3):

 [b0 b1 b2 b3...]~ /exp(1) =    1/0!                            * [1, 1, 1, 1, 1, ...]~
                                1/1!   1/0!
                                1/2!   2/1!   1/0!
                                1/3!   3/2!   3/1!   1/0!


 so

   V(1)~ * Pf = B~
 or
           Pf~ * V(1) = B
 and
           Pf  * V(1) = B / exp(x)

The systematic of the derivation of matrix Pf as resultant
of the derivative-proceeding in (1) and (2) needs now a better
systematic proof, but I think, that could be done...


--------------------------------------------------

One could proceed one further step and state

 for a current row and column m of Pf we have

(4):
    V(x)~ * Pf[0..inf,m] = exp(x) * Pf[m,0..m] * x^(2m) * V(1/x)

or (Pari:)

   suminf( k=0, x^k*Pf[k+1,m+1]) = exp(x) * x^(2*m) *  sum(k=0,m,Pf[m+1,k+1]/x^k)

where for a row r and col c the entry Pf[r+1,c+1] may be replaced by

      Pf[r+1,c+1] = ch(r,c) / (r-c)!          // for r>=c
                  = 0                         // for r<c

where ch(a,b) is the binomial-coefficient.

----------------

The common notation for the above observation may be given as:

    oo            x^k               m    m              x^(m-k)
   sum   ch(k,m)*------   = exp(x) x  * sum   ch(m,k)* ---------
    k=m          (k-m)!                 k=0              (m-k)!

                where ch(a,b) is the binomial-coefficient.


Gottfried Helms






From bowerc at usa.net  Wed Nov 29 10:49:58 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 29 Nov 2006 01:49:58 -0800
Subject: Categories
Message-ID: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>


I have one more row to add

------ Original Message ------
From: franktaw at netscape.net
To: seqfan at ext.jussieu.fr
Subject: Categories

> How many categories are there?
> 
> First, how many categories are there with n morphisms and k objects?  
> This table starts:
> 
>  1
>  2  1
>  7  3 1
> 35 16 3 1
> 
228 77 20 3 1

> The first column is A058129, the number of monoids; the main diagonal 
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
> 
> Taking the row sums, we get:
> 
> 1,3,11,55
329
> 
> the number of categories with n morphisms.  This is probably not in the 
> OEIS (only
> A001776 is possible - other matches become less than A058129).

While the 329 is eerily close to A001776's 330, EULER(A058129) is a
lower limit for this sequence and we have 2982 vs 2345 for #6 there.
>  The 
> inverse Euler
> transform,
> 
> 1,2,8,41
258
> 
> is the number of connected categories with n morphisms; this is 
> likewise probably not
> in the OEIS (only A052447 is possible).
No longer possible
> 
> Can somebody generate more data?
> 
> Franklin T. Adams-Watters
> 

Christian

PS

I don't know if it will be feasible to collect enough data to do
this, but the columns of the triangle converge and that convergence
would make an interesting sequence in and of itself.








From zakseidov at yahoo.com  Wed Nov 29 13:05:28 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 29 Nov 2006 04:05:28 -0800 (PST)
Subject: Difference between neighbor square and cube
Message-ID: <566064.14049.qm@web38215.mail.mud.yahoo.com>

Just submitted (with no autoreply).
My Q is:
Is the Conjectured list (see %C).. 
old hat or what?
Thanks, Zak

%I A125643
%S A125643
0,0,1,1,4,8,9,16,25,27,36,49,64,64,81,100,121,125,144,169,196,216,225,256,289,324,343,361,400,441,484,512,529,576,625,676,729,729,784,841,900,961,1000,1024,1089,1156,1225,1296,1331,1369,1444,1521,1600,1681,1728,1764,1849,1936,2025,2116,2197
%N A125643 Squares and cubes (with repetitions)
%C A125643 Cf. A002760 Squares and cubes (without
repetitions).
Conjectured list of numbers not appeared as difference
between neighbor terms:
5,6,10,14,16,22,23,29,32,34,42,44,46,50,52,54,58,62,64,66,70,72,78,82,84,86,88,90,92,93,94,96,98,102,105,110,112,114,117,118,120,122,124,126,130,132,134,136,140,141,142,144,153,156,158,160,162,164,165,166,172,176,178,179,182,188,190,192,194
%Y A125643 A002760
%O A125643 0
%K A125643 ,nonn,
%A A125643 Zak Seidov  (zakseidov at gmail.com), Nov 29 2006


 
____________________________________________________________________________________
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Everyone is raving about the all-new Yahoo! Mail beta.
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From maxale at gmail.com  Wed Nov 29 13:37:32 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 29 Nov 2006 04:37:32 -0800
Subject: Difference between neighbor square and cube
In-Reply-To: <566064.14049.qm@web38215.mail.mud.yahoo.com>
References: <566064.14049.qm@web38215.mail.mud.yahoo.com>
Message-ID: <d3dac270611290437x23c81f2fq6fb37e5d662b801d@mail.gmail.com>

Zak,

Your conjecture is related to Mordell curves of the form y^2 = x^3 + n, see
http://mathworld.wolfram.com/MordellCurve.html
http://tnt.math.metro-u.ac.jp/simath/MORDELL/

The latter link can help to resolve the most difficult case of whether
a given number is the difference of a square and a cube. In
particular, for n=5 and n=-5 the corresponding equation y^2 = x^3 + n
has no positive solutions, meaning that 5 is not a difference of
positive square and cube (not necessary neighbors, so this is even
stronger than needed by conjecture).
It is also easy to see that 5 is not the difference of two cubes. And
5 can be represented in an unique way as the difference of two
squares: 5 = 3^2 - 2^2 but 2^2 and 3^2 are not neighbors in A125643
since they are separated by 2^3.
Therefore, 5 indeed never appears as the difference between neighbor
terms of A125643.

Max


On 11/29/06, zak seidov <zakseidov at yahoo.com> wrote:
> Just submitted (with no autoreply).
> My Q is:
> Is the Conjectured list (see %C)..
> old hat or what?
> Thanks, Zak
>
> %I A125643
> %S A125643
> 0,0,1,1,4,8,9,16,25,27,36,49,64,64,81,100,121,125,144,169,196,216,225,256,289,324,343,361,400,441,484,512,529,576,625,676,729,729,784,841,900,961,1000,1024,1089,1156,1225,1296,1331,1369,1444,1521,1600,1681,1728,1764,1849,1936,2025,2116,2197
> %N A125643 Squares and cubes (with repetitions)
> %C A125643 Cf. A002760 Squares and cubes (without
> repetitions).
> Conjectured list of numbers not appeared as difference
> between neighbor terms:
> 5,6,10,14,16,22,23,29,32,34,42,44,46,50,52,54,58,62,64,66,70,72,78,82,84,86,88,90,92,93,94,96,98,102,105,110,112,114,117,118,120,122,124,126,130,132,134,136,140,141,142,144,153,156,158,160,162,164,165,166,172,176,178,179,182,188,190,192,194
> %Y A125643 A002760
> %O A125643 0
> %K A125643 ,nonn,
> %A A125643 Zak Seidov  (zakseidov at gmail.com), Nov 29 2006
>
>
>
> ____________________________________________________________________________________
> Do you Yahoo!?
> Everyone is raving about the all-new Yahoo! Mail beta.
> http://new.mail.yahoo.com
>





From zakseidov at yahoo.com  Wed Nov 29 14:44:52 2006
From: zakseidov at yahoo.com (zak seidov)
Date: Wed, 29 Nov 2006 05:44:52 -0800 (PST)
Subject: Difference between neighbor square and cube
In-Reply-To: <d3dac270611290437x23c81f2fq6fb37e5d662b801d@mail.gmail.com>
Message-ID: <948080.43313.qm@web38204.mail.mud.yahoo.com>

Max,
thanks a lot for your (excellent as usual) reply!!

Using it I found:
A054504(*
Numbers n such that Mordell's equation y^2 = x^3 + n
has no integral solutions. *)=
{6, 7, 11, 13, 14, 20, 21, 23, 29, 32, 34, 39, 42, 45,
46, 47, 51, 53, 58, 59, 60, 61, 62, 66, 67, 69, 70,
74, 75, 77, 78, 83, 84, 85, 86, 87, 88, 90, 93, 95,
96, 102, 103, 104, 109, 110, 111, 114, 115, 116, 118,
123, 124, 130, 133, 135, 137, 139, 140, 146, 147, 149,
153, 155};

A081121=(*
Numbers n such that Mordell's equation y^2 = x^3 - n
has no integral solutions. *){
3, 5, 6, 9, 10, 12, 14, 16, 17, 21, 22, 24, 29, 30,
31, 32, 33, 34, 36, 37, 38, 41, 42, 43, 46, 50, 51,
52, 57, 58, 59, 62, 65, 66, 68, 69, 70, 73, 75, 77,
78, 80, 82, 84, 85, 86, 88, 90, 91, 92, 93, 94, 96,
97, 98, 99};

Intersection[A081121,A054504]
ND={6,14,21,29,32,34,42,46,51,58,59,62,66,69,70,75,77,78,84,85,86,88,90,93,96}

And ND list gives 
(Absolute values of) Non-difference between square and
cube (not only neighbor as in %C A125643).

Hence any number in ID should be in %C A125643.
But 21 is in ND and absent in %C A125643!

Indeed in A125643,
we have:
..,64,81,100,121,125,...,
and 121-100=11^2-10^2=21!!

21 is not difference between neighbor square and cube,
but it is difference between neighbor terms in
A125643.

As a result: my subject 'Difference between neighbor
square and cube' is not correct,
and in %C A125643 we have 
'Difference between neighbor square/cube and
square/cube'.

Sorry and thanks to Max & all seqfans,
Zak  


--- "Max A." <maxale at gmail.com> wrote:

> Zak,
> 
> Your conjecture is related to Mordell curves of the
> form y^2 = x^3 + n, see
> http://mathworld.wolfram.com/MordellCurve.html
> http://tnt.math.metro-u.ac.jp/simath/MORDELL/
> 
> The latter link can help to resolve the most
> difficult case of whether
> a given number is the difference of a square and a
> cube. In
> particular, for n=5 and n=-5 the corresponding
> equation y^2 = x^3 + n
> has no positive solutions, meaning that 5 is not a
> difference of
> positive square and cube (not necessary neighbors,
> so this is even
> stronger than needed by conjecture).
> It is also easy to see that 5 is not the difference
> of two cubes. And
> 5 can be represented in an unique way as the
> difference of two
> squares: 5 = 3^2 - 2^2 but 2^2 and 3^2 are not
> neighbors in A125643
> since they are separated by 2^3.
> Therefore, 5 indeed never appears as the difference
> between neighbor
> terms of A125643.
> 
> Max
> 
> 
> On 11/29/06, zak seidov <zakseidov at yahoo.com> wrote:
> > Just submitted (with no autoreply).
> > My Q is:
> > Is the Conjectured list (see %C)..
> > old hat or what?
> > Thanks, Zak
> >
> > %I A125643
> > %S A125643
> >
>
0,0,1,1,4,8,9,16,25,27,36,49,64,64,81,100,121,125,144,169,196,216,225,256,289,324,343,361,400,441,484,512,529,576,625,676,729,729,784,841,900,961,1000,1024,1089,1156,1225,1296,1331,1369,1444,1521,1600,1681,1728,1764,1849,1936,2025,2116,2197
> > %N A125643 Squares and cubes (with repetitions)
> > %C A125643 Cf. A002760 Squares and cubes (without
> > repetitions).
> > Conjectured list of numbers not appeared as
> difference
> > between neighbor terms:
> >
>
5,6,10,14,16,22,23,29,32,34,42,44,46,50,52,54,58,62,64,66,70,72,78,82,84,86,88,90,92,93,94,96,98,102,105,110,112,114,117,118,120,122,124,126,130,132,134,136,140,141,142,144,153,156,158,160,162,164,165,166,172,176,178,179,182,188,190,192,194
> > %Y A125643 A002760
> > %O A125643 0
> > %K A125643 ,nonn,
> > %A A125643 Zak Seidov  (zakseidov at gmail.com), Nov
> 29 2006
> >
> >
> >
> >
>
____________________________________________________________________________________
> > Do you Yahoo!?
> > Everyone is raving about the all-new Yahoo! Mail
> beta.
> > http://new.mail.yahoo.com
> >
> 



 
____________________________________________________________________________________
Do you Yahoo!?
Everyone is raving about the all-new Yahoo! Mail beta.
http://new.mail.yahoo.com





From jvospost3 at gmail.com  Wed Nov 29 19:52:44 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 29 Nov 2006 10:52:44 -0800
Subject: Categories
In-Reply-To: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
References: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
Message-ID: <5542af940611291052l1a3b2654k7467715f3073dfde@mail.gmail.com>

Thank you, Christian.  That looks right.

Note that the partial sums of the first column are A118601 Number of monoids
(semigroups with identity) of order <=n.  This first column also equals the
row sums of A058137, which partitions the monoids of order n by the number k
of idempotents.






On 11/29/06, Christian G. Bower <bowerc at usa.net> wrote:
>
>
> I have one more row to add
>
> ------ Original Message ------
> Received: Wed, 22 Nov 2006 01:21:01 PM PST
> From: franktaw at netscape.net
> To: seqfan at ext.jussieu.fr
> Subject: Categories
>
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> >  1
> >  2  1
> >  7  3 1
> > 35 16 3 1
> >
> 228 77 20 3 1
>
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
> > Taking the row sums, we get:
> >
> > 1,3,11,55
> 329
> >
> > the number of categories with n morphisms.  This is probably not in the
> > OEIS (only
> > A001776 is possible - other matches become less than A058129).
>
> While the 329 is eerily close to A001776's 330, EULER(A058129) is a
> lower limit for this sequence and we have 2982 vs 2345 for #6 there.
> >  The
> > inverse Euler
> > transform,
> >
> > 1,2,8,41
> 258
> >
> > is the number of connected categories with n morphisms; this is
> > likewise probably not
> > in the OEIS (only A052447 is possible).
> No longer possible
> >
> > Can somebody generate more data?
> >
> > Franklin T. Adams-Watters
> >
>
> Christian
>
> PS
>
> I don't know if it will be feasible to collect enough data to do
> this, but the columns of the triangle converge and that convergence
> would make an interesting sequence in and of itself.
>
>
>
>
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From franktaw at netscape.net  Wed Nov 29 20:09:00 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Wed, 29 Nov 2006 14:09:00 -0500
Subject: Categories
In-Reply-To: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
References: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
Message-ID: <8C8E200F6031B3A-794-8706@FWM-M14.sysops.aol.com>

Thanks, Christian.

I think we need to look at one more table:  the number of connected
categories with n morphisms and k objects.  This starts:

  1
  2
  7 1
 35 6
228 28 2

(Row n has length ceiling(n/2).)

The table of the number of categories that I started with is a two-
dimensional Euler transform of this table.

The downward-sloping diagonal sums of this table are 1,3,15,???.
Shifting this left and taking the Euler transform gives the limiting
values Christian is referring to; starting with b(0):

1,3,21,???

For the table above, a(2n-1,n) has a purely algebraic or graph-
theoretic interpretation.  It is the number of connected
anti-transitive relations on n objects (meaning that if a R b and
b R c, then NOT (a R c)); equivalently, the number of bipartite
oriented trees, where each edge origin is in the same part.

If I haven't made any mistakes, this sequence starts:

1,1,2,3,6,10

This is not enough data to determine whether it is in the OEIS.

Franklin T. Adams-Watters


-----Original Message-----
From: bowerc at usa.net

I have one more row to add

------ Original Message ------

> How many categories are there?
>
> First, how many categories are there with n morphisms and k objects?
> This table starts:
>
>  1
>  2  1
>  7  3 1
> 35 16 3 1
>
228 77 20 3 1

> The first column is A058129, the number of monoids; the main diagonal
> is all 1's.  I am not
> 100% certain of the 16 in the final row.
>
> Taking the row sums, we get:
>
> 1,3,11,55
329
>
> the number of categories with n morphisms.  This is probably not in 
the
> OEIS (only
> A001776 is possible - other matches become less than A058129).

While the 329 is eerily close to A001776's 330, EULER(A058129) is a
lower limit for this sequence and we have 2982 vs 2345 for #6 there.
>  The
> inverse Euler
> transform,
>
> 1,2,8,41
258
>
> is the number of connected categories with n morphisms; this is
> likewise probably not
> in the OEIS (only A052447 is possible).
No longer possible
>
> Can somebody generate more data?
>
> Franklin T. Adams-Watters
>

Christian

PS

I don't know if it will be feasible to collect enough data to do
this, but the columns of the triangle converge and that convergence
would make an interesting sequence in and of itself.

________________________________________________________________________
Check Out the new free AIM(R) Mail -- 2 GB of storage and 
industry-leading spam and email virus protection.






From jvospost3 at gmail.com  Wed Nov 29 20:23:47 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Wed, 29 Nov 2006 11:23:47 -0800
Subject: Categories
In-Reply-To: <8C8E200F6031B3A-794-8706@FWM-M14.sysops.aol.com>
References: <461kkCJW77882S21.1164793798@cmsweb21.cms.usa.net>
	 <8C8E200F6031B3A-794-8706@FWM-M14.sysops.aol.com>
Message-ID: <5542af940611291123l4f9c3385l283f3b2243dab1c2@mail.gmail.com>

The issue of "connected" in the underlying functional graph, as a subset of
the set of all digraphs, is further revealed by the distinction (definitions
at MathWorld) of Weakly Conncted versus Strongly Conncted Digraphs.

Every functional digraph is the disjoint union of connected digraphs.
That's why there is a recursion in the enumeration of F(n) = functional
graphs on n points. F(n) is equal or greater to the number of functional
graphs on n-1 points, unioned with an isolated point looping to itself, plus
the number of functional graphs on n-2 points unioned with any of the 3
functional graphs on 3 points, ..., throwing out any double or multiple
counts by equivalence class on isomorphisms of the unlabelled digraphs.

Note that MathWorld's correct definition of functional graph, and its
Mathematica, is identical to what I'd called endofunctions in earlier
posting of this thread.  Hence the enumeration problem that I solved for
"prime" endofunctions is identical to a subset of your category enumeration
problem, relating to the number of categories that are not unions nor
categorical products of smaller categories.

However, when I referred to 2-categories before, I should more correctly
have said Bicategories, the difference being based on whether associativity
is strict or up to 2-isomorphism.  Your enumeration of categories is more
formally explained in terms of Cat, the category of categories and functors,
but this is not the right venue for me to go into excruciating details.

On 11/29/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
> Thanks, Christian.
>
> I think we need to look at one more table:  the number of connected
> categories with n morphisms and k objects.  This starts:
>
>   1
>   2
>   7 1
> 35 6
> 228 28 2
>
> (Row n has length ceiling(n/2).)
>
> The table of the number of categories that I started with is a two-
> dimensional Euler transform of this table.
>
> The downward-sloping diagonal sums of this table are 1,3,15,???.
> Shifting this left and taking the Euler transform gives the limiting
> values Christian is referring to; starting with b(0):
>
> 1,3,21,???
>
> For the table above, a(2n-1,n) has a purely algebraic or graph-
> theoretic interpretation.  It is the number of connected
> anti-transitive relations on n objects (meaning that if a R b and
> b R c, then NOT (a R c)); equivalently, the number of bipartite
> oriented trees, where each edge origin is in the same part.
>
> If I haven't made any mistakes, this sequence starts:
>
> 1,1,2,3,6,10
>
> This is not enough data to determine whether it is in the OEIS.
>
> Franklin T. Adams-Watters
>
>
> -----Original Message-----
> From: bowerc at usa.net
>
> I have one more row to add
>
> ------ Original Message ------
> Received: Wed, 22 Nov 2006 01:21:01 PM PST
>
> > How many categories are there?
> >
> > First, how many categories are there with n morphisms and k objects?
> > This table starts:
> >
> >  1
> >  2  1
> >  7  3 1
> > 35 16 3 1
> >
> 228 77 20 3 1
>
> > The first column is A058129, the number of monoids; the main diagonal
> > is all 1's.  I am not
> > 100% certain of the 16 in the final row.
> >
> > Taking the row sums, we get:
> >
> > 1,3,11,55
> 329
> >
> > the number of categories with n morphisms.  This is probably not in
> the
> > OEIS (only
> > A001776 is possible - other matches become less than A058129).
>
> While the 329 is eerily close to A001776's 330, EULER(A058129) is a
> lower limit for this sequence and we have 2982 vs 2345 for #6 there.
> >  The
> > inverse Euler
> > transform,
> >
> > 1,2,8,41
> 258
> >
> > is the number of connected categories with n morphisms; this is
> > likewise probably not
> > in the OEIS (only A052447 is possible).
> No longer possible
> >
> > Can somebody generate more data?
> >
> > Franklin T. Adams-Watters
> >
>
> Christian
>
> PS
>
> I don't know if it will be feasible to collect enough data to do
> this, but the columns of the triangle converge and that convergence
> would make an interesting sequence in and of itself.
>
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and
> industry-leading spam and email virus protection.
>
>
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From njas at research.att.com  Wed Nov 29 22:34:56 2006
From: njas at research.att.com (N. J. A. Sloane)
Date: Wed, 29 Nov 2006 16:34:56 -0500 (EST)
Subject: OEIS on vacation in December
Message-ID: <200611292134.QAA53261@fry.research.att.com>


This means:
- do send in sequences in published works or 
   that are important for your work
- do send important corrections, extensions, updates
- do send EDITED versions

- don't send in sequences that you made up (unless
   they are really beautiful - and if they
   are base dependent or involve primes 
   they are probably not)
- don't send very minor corrections

I will be doing occasional updates, but I won't be
working on the OEIS 24/7 !

Thanks!

Neil





From maxale at gmail.com  Wed Nov 29 22:45:38 2006
From: maxale at gmail.com (Max A.)
Date: Wed, 29 Nov 2006 13:45:38 -0800
Subject: Difference between neighbor square and cube
In-Reply-To: <948080.43313.qm@web38204.mail.mud.yahoo.com>
References: <d3dac270611290437x23c81f2fq6fb37e5d662b801d@mail.gmail.com>
	 <948080.43313.qm@web38204.mail.mud.yahoo.com>
Message-ID: <d3dac270611291345n2ad5dae4id319f3bce8fca71c@mail.gmail.com>

On 11/29/06, zak seidov <zakseidov at yahoo.com> wrote:

> Intersection[A081121,A054504]
> ND={6,14,21,29,32,34,42,46,51,58,59,62,66,69,70,75,77,78,84,85,86,88,90,93,96}

That's A110223. But Robert G. Wilson posed as "conjectured".
I believe this sequence can be confirmed/verified using the
information about Mordell curves from
http://tnt.math.metro-u.ac.jp/simath/MORDELL/

Max





From bowerc at usa.net  Thu Nov 30 00:42:31 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 29 Nov 2006 15:42:31 -0800
Subject: Categories
Message-ID: <623kkCXpf3086S27.1164843751@cmsweb27.cms.usa.net>

ftaw
> I think we need to look at one more table:  the number of connected
> categories with n morphisms and k objects.  This starts:

>   1
>   2
>   7 1
>  35 6
> 228 28 2

> (Row n has length ceiling(n/2).)
...
> For the table above, a(2n-1,n) has a purely algebraic or graph-
> theoretic interpretation.  It is the number of connected
> anti-transitive relations on n objects (meaning that if a R b and
> b R c, then NOT (a R c)); equivalently, the number of bipartite
> oriented trees, where each edge origin is in the same part.
> 
> If I haven't made any mistakes, this sequence starts:
> 
> 1,1,2,3,6,10
> 
> This is not enough data to determine whether it is in the OEIS.

For n>1 it's A122086
http://www.research.att.com/~njas/sequences/A122086
A sequence which could use a nicer description and formula such as
a(n) = A000055(n) - A000081(n/2) where a sequence evaluated at a
noninteger is 0.
http://www.research.att.com/~njas/sequences/A000055
http://www.research.att.com/~njas/sequences/A000081

Christian








From bowerc at usa.net  Thu Nov 30 01:36:23 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Wed, 29 Nov 2006 16:36:23 -0800
Subject: Categories
Message-ID: <209kkdajX8954S10.1164846983@cmsweb10.cms.usa.net>

I have row 6 now:

> How many categories are there?
>
> First, how many categories are there with n morphisms and k objects?
> This table starts:
>
>  1
>  2  1
>  7  3 1
> 35 16 3 1
>
228 77 20 3 1
2237 485 111 21 3 1

> Taking the row sums, we get:
>
> 1,3,11,55
329
2858

>  The
> inverse Euler
> transform,
>
> 1,2,8,41
258
2407

> I think we need to look at one more table:  the number of connected
> categories with n morphisms and k objects.  This starts:

>   1
>   2
>   7 1
>  35 6
> 228 28 2
2237 159 11
31559 ?   ? 3

> (Row n has length ceiling(n/2).)
> 
> The table of the number of categories that I started with is a two-
> dimensional Euler transform of this table.
> 
> The downward-sloping diagonal sums of this table are 1,3,15,???.
77
> Shifting this left and taking the Euler transform gives the limiting
> values Christian is referring to; starting with b(0):
> 
> 1,3,21,???
132








From franktaw at netscape.net  Thu Nov 30 14:53:38 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 30 Nov 2006 08:53:38 -0500
Subject: Categories
In-Reply-To: <623kkCXpf3086S27.1164843751@cmsweb27.cms.usa.net>
References: <623kkCXpf3086S27.1164843751@cmsweb27.cms.usa.net>
Message-ID: <8C8E29E12708888-5D8-4EB@WEBMAIL-RA17.sysops.aol.com>

A typo there; that should be a(n) = 2*A000055(n) - A000081(n/2).

Franklin T. Adams-Watters


-----Original Message-----
From: bowerc at usa.net

...
For n>1 it's A122086
http://www.research.att.com/~njas/sequences/A122086
A sequence which could use a nicer description and formula such as
a(n) = A000055(n) - A000081(n/2) where a sequence evaluated at a
noninteger is 0.
http://www.research.att.com/~njas/sequences/A000055
http://www.research.att.com/~njas/sequences/A000081

Christian





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From jvospost3 at gmail.com  Thu Nov 30 19:34:51 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 30 Nov 2006 10:34:51 -0800
Subject: Categories
In-Reply-To: <8C8E29E12708888-5D8-4EB@WEBMAIL-RA17.sysops.aol.com>
References: <623kkCXpf3086S27.1164843751@cmsweb27.cms.usa.net>
	 <8C8E29E12708888-5D8-4EB@WEBMAIL-RA17.sysops.aol.com>
Message-ID: <5542af940611301034t3bba13fdnadc86411facc535d@mail.gmail.com>

a(n) = 2*A000055(n) - A000081(n/2) = (2* Number of trees with n unlabeled
nodes) - Number of rooted trees with n/2 nodes (or connected functions with
a fixed point)

where a(n) is downwards sloping diagonal sums of main category table, or
what?

It isn't until we look at bicategories that we are forced to look at the
symmetries and structure of the individual categories being enumerated here,
as objects in the higher categories, and have to enumerate the morphisms of
the functions. A (finite, concrete) category whose underlying directed graph
has no symmetry (i.e. automorphism group of order 1) can only map to itself
or an identical copy of itself, point to point, loop to loop, arc to arc.
But any cycle of length >1 in that directed graph allows morphisms to other
directed graphs.  The directed graph C_n, a cycle of length n, can thus map
to the same graph but mapping each point to the next, and each arc to the
next, and so forth. That requires then analysis of cycles in directed
graphs, as by de Bruijn in 1952 and more deeply understood since then. This
is another enumeration problem, unnecessary to completing the answer to
Frank's original question, and more tied to my endofunctions analysis. Sorry
if my interruptions distracted you from the straightforward question.  Good
work, guys!

Funny how Category Theory folks tend to be addicted to theory for its own
sake, and less willing to "get their hands dirty" with anything as old
fashioned as enumeration.  OEIS helps us focus here, by demanding the
integer sequences!

On 11/30/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
> A typo there; that should be a(n) = 2*A000055(n) - A000081(n/2).
>
> Franklin T. Adams-Watters
>
>
> -----Original Message-----
> From: bowerc at usa.net
>
> ...
> For n>1 it's A122086
> http://www.research.att.com/~njas/sequences/A122086
> A sequence which could use a nicer description and formula such as
> a(n) = A000055(n) - A000081(n/2) where a sequence evaluated at a
> noninteger is 0.
> http://www.research.att.com/~njas/sequences/A000055
> http://www.research.att.com/~njas/sequences/A000081
>
> Christian
>
>
>
>
>
> ________________________________________________________________________
> Check Out the new free AIM(R) Mail -- 2 GB of storage and
> industry-leading spam and email virus protection.
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>
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From all at abouthugo.de  Thu Nov 30 21:28:06 2006
From: all at abouthugo.de (Hugo Pfoertner)
Date: Thu, 30 Nov 2006 21:28:06 +0100
Subject: Difference between neighbor square and cube
References: <948080.43313.qm@web38204.mail.mud.yahoo.com>
Message-ID: <456F3ED6.D0619322@abouthugo.de>

zak seidov wrote:
> 
> Max,
> thanks a lot for your (excellent as usual) reply!!
> 
> Using it I found:
> A054504(*
> Numbers n such that Mordell's equation y^2 = x^3 + n
> has no integral solutions. *)=
> {6, 7, 11, 13, 14, 20, 21, 23, 29, 32, 34, 39, 42, 45,
> 46, 47, 51, 53, 58, 59, 60, 61, 62, 66, 67, 69, 70,
> 74, 75, 77, 78, 83, 84, 85, 86, 87, 88, 90, 93, 95,
> 96, 102, 103, 104, 109, 110, 111, 114, 115, 116, 118,
> 123, 124, 130, 133, 135, 137, 139, 140, 146, 147, 149,
> 153, 155};
> 
> A081121=(*
> Numbers n such that Mordell's equation y^2 = x^3 - n
> has no integral solutions. *){
> 3, 5, 6, 9, 10, 12, 14, 16, 17, 21, 22, 24, 29, 30,
> 31, 32, 33, 34, 36, 37, 38, 41, 42, 43, 46, 50, 51,
> 52, 57, 58, 59, 62, 65, 66, 68, 69, 70, 73, 75, 77,
> 78, 80, 82, 84, 85, 86, 88, 90, 91, 92, 93, 94, 96,
> 97, 98, 99};
> 
> Intersection[A081121,A054504]
> ND={6,14,21,29,32,34,42,46,51,58,59,62,66,69,70,75,77,78,84,85,86,88,90,93,96}
> 
> And ND list gives
> (Absolute values of) Non-difference between square and
> cube (not only neighbor as in %C A125643).
> 
> Hence any number in ID should be in %C A125643.
> But 21 is in ND and absent in %C A125643!
> 
> Indeed in A125643,
> we have:
> ..,64,81,100,121,125,...,
> and 121-100=11^2-10^2=21!!
> 
> 21 is not difference between neighbor square and cube,
> but it is difference between neighbor terms in
> A125643.
> 
> As a result: my subject 'Difference between neighbor
> square and cube' is not correct,
> and in %C A125643 we have
> 'Difference between neighbor square/cube and
> square/cube'.
> 
> Sorry and thanks to Max & all seqfans,
> Zak

Also closely related:

http://www.research.att.com/~njas/sequences/A087285
http://www.research.att.com/~njas/sequences/A087286
http://www.research.att.com/~njas/sequences/A088017

Hugo





From bowerc at usa.net  Thu Nov 30 22:29:20 2006
From: bowerc at usa.net (Christian G. Bower)
Date: Thu, 30 Nov 2006 13:29:20 -0800
Subject: Categories
Message-ID: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>



------ Original Message ------
From: "Jonathan Post" <jvospost3 at gmail.com>
To: "franktaw at netscape.net" <franktaw at netscape.net>Cc: bowerc at usa.net,
seqfan at ext.jussieu.fr, jvospost2 at yahoo.com
Subject: Re: Categories

> a(n) = 2*A000055(n) - A000081(n/2) = (2* Number of trees with n unlabeled
> nodes) - Number of rooted trees with n/2 nodes (or connected functions with
> a fixed point)
> 
> where a(n) is downwards sloping diagonal sums of main category table, or
> what?

Actually it's the "corner" values from this table

>   *1*
>   2
>   7 *1*
>  35 6
> 228 28 *2*
> 2237 159 11
> 31559 ?   ? *3*

The downward sloping sums were an aide in calculating the convergent
sequence of the columns

The interpretation of these corners comes from the fact we have a
saturated category. It has 2n-1 elements, n of which are object
identities. It's connected, so it has a minimum of n-1 "crossover"
morphisms, it also has a maximum of n-1 cm's since there are only
n-1 elements left. Any connected graph with n points and n-1 edges is
a tree. The associativity of categories states that if we have an
a-b morphism (an edge on the graph) and a b-c we must have an a-c.
But if we have all 3, it's no longer a tree, hence we never have the
a-b and the b-c, hence each point has all edges either pointing in or
pointing out, hence the biparite graph analogy, hence a little
combinatorial magic.

...
> Funny how Category Theory folks tend to be addicted to theory for its > own
> sake, and less willing to "get their hands dirty" with anything as old
> fashioned as enumeration.  OEIS helps us focus here, by demanding the
> integer sequences!
> 

I try to stay out of philosophical stuff here, but sometimes it's too
hard to resist. I figure I'm guilty of spending too much time counting
stuff and not taking the theory in. Most of the world would not give a
hoot about either, so I'm glad there are folk spending time on what I
would not give any time to.

Still, I find it jarring any time I read a math paper and they write
about several things that can be represented as sequences, yet all they
give is a mathematical description, rarely do they cite the OEIS
reference or even give me the first n terms leaving me to calculate them
myself and hope I didn't make a mistake. If I wrote a paper, I'd
include that stuff, not just because I'm an OEIS nerd, but also because
I would want people to be able to find my paper who didn't know they
were looking for it.

Just a little more self indulgence here with my list of semigroup pipe
dreams. After doing sequences on semigroups
http://www.research.att.com/~njas/sequences/A027851
and monoids
http://www.research.att.com/~njas/sequences/A058129
I knew categories was one of the next logical steps, but I held off
because I knew it would be difficult. Perhaps I can spur someone else's
interest in related problems that may prove even more difficult.

There's transformation semigroups, the semigroup version of permutation
groups
http://www.research.att.com/~njas/sequences/A000638
I've seen TSs written about, but no attempts to enumerate them.
(Note, a TS is any set of endofunctions closed under composition)
There TS's close cousins which could be called:
transformation monoids
relation semigroups
relation monoids
(Composition of relations is such that (a,c) is in R1(R2) iff there
is b such that (a,b) in R1 and (b,c) in R2)
Then there are the generalizations of categories to semigroups and
groupoids. Note I mean the groupoid that is a set and closed operation
http://www.research.att.com/~njas/sequences/A001329
Not the group like category groupoid which might also be interesting
to count and not too difficult; oh terminology.
Anyway, with the semigroup and groupoid categories, there comes the
issue of treating objects and morphims. The ordinary (monoid like)
categories avoid this problem by having object identities among the
morphisms that can be associatied with the objects themselves. Without
that restriction, it's possible to have objects untouched by any
morphism or two structures with identical (or isomorphic) multiplication
tables, but not isomorphic when you take the objects into account.
Basically many of the same counting problems one has with multigraphs
(and probably the same solutions too.)








From jvospost3 at gmail.com  Thu Nov 30 22:53:42 2006
From: jvospost3 at gmail.com (Jonathan Post)
Date: Thu, 30 Nov 2006 13:53:42 -0800
Subject: Categories
In-Reply-To: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>
References: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>
Message-ID: <5542af940611301353m486a4788wff2e4eb539b76943@mail.gmail.com>

I strongly agree with Christian.  My doctoral dissertation involved taking
the Krohn-Rhodes decomposition of the semigroup of differential operators of
a system of nonlinear differential equations that had been in the
mathematical Biology literature for decades, and routinely dismissed as not
capable of closed-form solution. Marvin Minsky (who had been the PhD advisor
of the eponymous John Rhodes) thought my work was the best use yet of his
student's theorem.  The great Stan Ulam (coinventor of cellular automata,
the H-bomb, and the nuclear pulse rocket) also thought it was cool. But only
half a dozen biologists in theb world knew that much about semigroups, and
some of them wrote to me from Edinburgh, Scotland, and places in the USSR.
Ulam died before we could coauthor; Semigroup Forum accepted my paper with
delight for a special issue, but wanted it reformatted, and I was laid off
from the company where I worked, and they erased on only machine-readable
version of the paper.  Equations were harder to typeset those decades
ago...  But yes, lots of so-called professionals go out of their way to
avoid citing the OEIS, or give any useful examples.  Also yes: semigroup
enumerations are worthwhile.  Also, you nailed the differences between those
and categories.

See for instance:

  A118581 <http://www.research.att.com/~njas/sequences/A118581> Number of
nonisomorphic *semigroups* of order <= n.

  A118099 <http://www.research.att.com/~njas/sequences/A118099> Number of
inverse *semigroups* of order <= n.

  A118100 <http://www.research.att.com/~njas/sequences/A118100>

Number of commutative *semigroups* of order <= n.

  A118601 <http://www.research.att.com/~njas/sequences/A118601> Number of
monoids (*semigroups* with identity) of order <= n.
  A113534 <http://www.research.att.com/~njas/sequences/A113534> Ascending
descending base exponent transform of the flipped tribonacci substitution (
A092782 <http://www.research.att.com/~njas/sequences/A092782>).
with its semigroup reference, and some others.

-- Jonathan


On 11/30/06, Christian G. Bower <bowerc at usa.net> wrote:
>
>
>
> ------ Original Message ------
> Received: Thu, 30 Nov 2006 10:34:55 AM PST
> From: "Jonathan Post" <jvospost3 at gmail.com>
> To: "franktaw at netscape.net" <franktaw at netscape.net>Cc: bowerc at usa.net,
> seqfan at ext.jussieu.fr, jvospost2 at yahoo.com
> Subject: Re: Categories
>
> > a(n) = 2*A000055(n) - A000081(n/2) = (2* Number of trees with n
> unlabeled
> > nodes) - Number of rooted trees with n/2 nodes (or connected functions
> with
> > a fixed point)
> >
> > where a(n) is downwards sloping diagonal sums of main category table, or
> > what?
>
> Actually it's the "corner" values from this table
>
> >   *1*
> >   2
> >   7 *1*
> >  35 6
> > 228 28 *2*
> > 2237 159 11
> > 31559 ?   ? *3*
>
> The downward sloping sums were an aide in calculating the convergent
> sequence of the columns
>
> The interpretation of these corners comes from the fact we have a
> saturated category. It has 2n-1 elements, n of which are object
> identities. It's connected, so it has a minimum of n-1 "crossover"
> morphisms, it also has a maximum of n-1 cm's since there are only
> n-1 elements left. Any connected graph with n points and n-1 edges is
> a tree. The associativity of categories states that if we have an
> a-b morphism (an edge on the graph) and a b-c we must have an a-c.
> But if we have all 3, it's no longer a tree, hence we never have the
> a-b and the b-c, hence each point has all edges either pointing in or
> pointing out, hence the biparite graph analogy, hence a little
> combinatorial magic.
>
> ...
> > Funny how Category Theory folks tend to be addicted to theory for its >
> own
> > sake, and less willing to "get their hands dirty" with anything as old
> > fashioned as enumeration.  OEIS helps us focus here, by demanding the
> > integer sequences!
> >
>
> I try to stay out of philosophical stuff here, but sometimes it's too
> hard to resist. I figure I'm guilty of spending too much time counting
> stuff and not taking the theory in. Most of the world would not give a
> hoot about either, so I'm glad there are folk spending time on what I
> would not give any time to.
>
> Still, I find it jarring any time I read a math paper and they write
> about several things that can be represented as sequences, yet all they
> give is a mathematical description, rarely do they cite the OEIS
> reference or even give me the first n terms leaving me to calculate them
> myself and hope I didn't make a mistake. If I wrote a paper, I'd
> include that stuff, not just because I'm an OEIS nerd, but also because
> I would want people to be able to find my paper who didn't know they
> were looking for it.
>
> Just a little more self indulgence here with my list of semigroup pipe
> dreams. After doing sequences on semigroups
> http://www.research.att.com/~njas/sequences/A027851
> and monoids
> http://www.research.att.com/~njas/sequences/A058129
> I knew categories was one of the next logical steps, but I held off
> because I knew it would be difficult. Perhaps I can spur someone else's
> interest in related problems that may prove even more difficult.
>
> There's transformation semigroups, the semigroup version of permutation
> groups
> http://www.research.att.com/~njas/sequences/A000638
> I've seen TSs written about, but no attempts to enumerate them.
> (Note, a TS is any set of endofunctions closed under composition)
> There TS's close cousins which could be called:
> transformation monoids
> relation semigroups
> relation monoids
> (Composition of relations is such that (a,c) is in R1(R2) iff there
> is b such that (a,b) in R1 and (b,c) in R2)
> Then there are the generalizations of categories to semigroups and
> groupoids. Note I mean the groupoid that is a set and closed operation
> http://www.research.att.com/~njas/sequences/A001329
> Not the group like category groupoid which might also be interesting
> to count and not too difficult; oh terminology.
> Anyway, with the semigroup and groupoid categories, there comes the
> issue of treating objects and morphims. The ordinary (monoid like)
> categories avoid this problem by having object identities among the
> morphisms that can be associatied with the objects themselves. Without
> that restriction, it's possible to have objects untouched by any
> morphism or two structures with identical (or isomorphic) multiplication
> tables, but not isomorphic when you take the objects into account.
> Basically many of the same counting problems one has with multigraphs
> (and probably the same solutions too.)
>
>
>
>
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From franktaw at netscape.net  Thu Nov 30 22:54:39 2006
From: franktaw at netscape.net (franktaw at netscape.net)
Date: Thu, 30 Nov 2006 16:54:39 -0500
Subject: Categories
In-Reply-To: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>
References: <944kkdVCu7440S30.1164922160@cmsweb30.cms.usa.net>
Message-ID: <8C8E2E14509A6B2-934-A97D@FWM-M10.sysops.aol.com>

I'd like to put my plug in for more application in math.

I heard of a case some years ago where a paper was published about some 
kind of mathematical object (I think it was a kind of topological 
space, but it really doesn't matter).  Three or four more papers were 
published, establishing more and more properties for this type of 
object - until finally it was proved that they don't exist!  This 
wouldn't have happened if somebody had asked for an example at an early 
stage.

There is a strong tendency in mathematics to start at the end.  The 
researcher pursues a line of thought, which eventually leads to a 
spiffy proof.  The proof is then published, with no hint of the process 
by which it was reached.  This is a disservice to anybody who might use 
a similar approach to solve some other problem.  It is especially a 
disservice when presented to students.

On a more personal level, I find when looking a math paper, I want to 
know how this relates to problems that I am already interested in or at 
least familiar with.  If I can't get an answer to that, I have a hard 
time maintaining any interest.

Franklin T. Adams-Watters

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