emails about div. chains
N. J. A. Sloane
njas at research.att.com
Tue May 4 16:34:00 CEST 2004
apparently some people have been posting to the math fun list
when they should have been posting to the seqfan list
let's keep the sequence stuff on the seq fan list , Ok?
and when you have a nice sequence, please send it to the OEIS
by filling out the "Submit New Sequence or Comment" web page, OK?
Don't expect that i will go though all the postings and all
the submissions to try to figure out what was posted and not submitted,
and what was submitted and not posted. OK?
At the last count i was getting over 350 emails a day. The associate
editrors are supposed to help, not make my life more complicated, OK?
Here are all the postings about divisor chains that i saved, from
both mailing lists. They have been summarized in sequences A094096-A094099.
They are not in order and there are duplicates, since they arrived
from several sources with multiple copies.
I will send this to both lists, to bring everyone up to date
unless you are interested in divosor chains, delete this.
NJAS
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Cc: Paul Vaderlind <paul at math.su.se>,
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Subject: [math-fun] Divisor chains
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The following might be an acceptable sequence
for Neil Sloane's OEIS, but someone needs to
do some work
1 2 3 4 5 6 7 8 9 10 11 12 13 14 ...
1 1 1 1 1 1 1 5 4 1 2 5 5 4 ...
It arises from a problem I got recently from
Paul Vaderlind.
If the sequence of numbers from 1 to 37,
is arranged so that each term is a divisor
of the sum of preceding ones, starting
37, 1, ...
what is the next term?
The answer is either 2 or 19. P'r'aps
I won't spoil your fun by pointing out
which. But I will spoil it by asking:
How do you know that there is such a
`divisor chain' ?
The sequence is (my present state of knowledge
of) the number of divisor chains of length n.
Here are the ones I found
1 2 1 3 1 2 4 2 3 1
5 1 2 4 3 6 2 4 3 5 1 7 1 2 5 3 6 4
8 2 5 3 6 4 7 1
8 4 2 7 3 1 5 6
8 4 2 7 3 6 5 1
8 4 3 5 1 7 2 6
8 4 6 3 7 2 5 1
9 1 2 4 8 6 5 7 3
9 1 2 6 3 7 4 8 5
9 3 4 8 6 5 7 2 1
9 3 6 2 1 7 4 8 5
10 2 4 8 3 9 6 7 1 5
11 1 2 7 3 8 4 9 5 10 6
11 1 4 8 6 10 5 9 2 7 3
12 2 1 5 10 3 11 4 8 7 9 6
12 2 7 3 8 4 9 5 10 6 11 1
12 3 5 10 2 8 4 11 1 7 9 6
12 4 8 3 9 6 2 11 5 10 7 1
12 4 8 6 10 5 9 2 7 3 11 1
13 1 2 8 3 9 4 10 5 11 6 12 7
13 1 2 8 6 10 4 11 5 12 9 3 7
13 1 2 8 6 10 4 11 5 3 9 12 7
13 1 2 8 12 4 10 5 11 3 9 3 7
13 1 2 8 12 4 10 5 11 6 9 3 7
14 2 8 6 10 4 11 5 3 9 12 7 13 1
14 2 8 6 10 4 11 5 12 9 3 7 13 1
14 2 8 12 4 10 5 11 6 9 3 7 13 1
14 2 8 12 9 3 4 13 1 11 7 6 10 5
Enough sins of omission, to say nothing of
commission, for today. Please check and extend.
Any ideas for proving anything? R.
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From math-fun-bounces+njas=research.att.com at mailman.xmission.com Mon May 3 16:53:57 2004
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Date: Mon, 3 May 2004 13:53:20 -0700 (PDT)
From: Eugene Salamin <gene_salamin at yahoo.com>
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Subject: Re: [math-fun] Divisor chains
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You seem to be considering only those chains for which the first
element is the largest. Otherwise we have chains such as (2 1 3), (2 4
3 1), (2 4 3 1 5), (3 1 4 2 5), or (4 1 5 2 3).
Gene
--- Richard Guy <rkg at cpsc.ucalgary.ca> wrote:
> The following might be an acceptable sequence
> for Neil Sloane's OEIS, but someone needs to
> do some work
>
> 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ...
> 1 1 1 1 1 1 1 5 4 1 2 5 5 4 ...
>
> It arises from a problem I got recently from
> Paul Vaderlind.
>
> If the sequence of numbers from 1 to 37,
> is arranged so that each term is a divisor
> of the sum of preceding ones, starting
> 37, 1, ...
> what is the next term?
>
> The answer is either 2 or 19. P'r'aps
> I won't spoil your fun by pointing out
> which. But I will spoil it by asking:
> How do you know that there is such a
> `divisor chain' ?
>
> The sequence is (my present state of knowledge
> of) the number of divisor chains of length n.
> Here are the ones I found
>
> 1 2 1 3 1 2 4 2 3 1
>
> 5 1 2 4 3 6 2 4 3 5 1 7 1 2 5 3 6 4
>
> 8 2 5 3 6 4 7 1
> 8 4 2 7 3 1 5 6
> 8 4 2 7 3 6 5 1
> 8 4 3 5 1 7 2 6
> 8 4 6 3 7 2 5 1
>
> 9 1 2 4 8 6 5 7 3
> 9 1 2 6 3 7 4 8 5
> 9 3 4 8 6 5 7 2 1
> 9 3 6 2 1 7 4 8 5
>
> 10 2 4 8 3 9 6 7 1 5
>
> 11 1 2 7 3 8 4 9 5 10 6
> 11 1 4 8 6 10 5 9 2 7 3
>
> 12 2 1 5 10 3 11 4 8 7 9 6
> 12 2 7 3 8 4 9 5 10 6 11 1
> 12 3 5 10 2 8 4 11 1 7 9 6
> 12 4 8 3 9 6 2 11 5 10 7 1
> 12 4 8 6 10 5 9 2 7 3 11 1
>
> 13 1 2 8 3 9 4 10 5 11 6 12 7
> 13 1 2 8 6 10 4 11 5 12 9 3 7
> 13 1 2 8 6 10 4 11 5 3 9 12 7
> 13 1 2 8 12 4 10 5 11 3 9 3 7
> 13 1 2 8 12 4 10 5 11 6 9 3 7
>
> 14 2 8 6 10 4 11 5 3 9 12 7 13 1
> 14 2 8 6 10 4 11 5 12 9 3 7 13 1
> 14 2 8 12 4 10 5 11 6 9 3 7 13 1
> 14 2 8 12 9 3 4 13 1 11 7 6 10 5
>
> Enough sins of omission, to say nothing of
> commission, for today. Please check and extend.
> Any ideas for proving anything? R.
>
>
> _______________________________________________
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> math-fun at mailman.xmission.com
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From math-fun-bounces+njas=research.att.com at mailman.xmission.com Tue May 4 02:03:52 2004
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From: Chris Landauer <cal at rushg.aero.org>
Date: Mon, 3 May 2004 23:05:20 -0700 (PDT)
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Subject: [math-fun] penultimate divisor counts output
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hihi, all -
yes, well, there were some typos - now the program counts them, so i don't
have to
there will be a few more numbers tomorrow (hopefully, n=37 will appear), but
for now, here is what i have
1 chains of length 1, 1 anchored, 1 cyclic, 1 both
1 chains of length 2, 1 anchored, 0 cyclic, 0 both
2 chains of length 3, 1 anchored, 2 cyclic, 1 both
2 chains of length 4, 1 anchored, 0 cyclic, 0 both
4 chains of length 5, 1 anchored, 2 cyclic, 1 both
5 chains of length 6, 1 anchored, 0 cyclic, 0 both
7 chains of length 7, 1 anchored, 3 cyclic, 1 both
7 chains of length 8, 5 anchored, 0 cyclic, 0 both
24 chains of length 9, 4 anchored, 5 cyclic, 4 both
22 chains of length 10, 3 anchored, 0 cyclic, 0 both
29 chains of length 11, 2 anchored, 6 cyclic, 2 both
39 chains of length 12, 8 anchored, 0 cyclic, 0 both
67 chains of length 13, 4 anchored, 6 cyclic, 4 both
55 chains of length 14, 6 anchored, 0 cyclic, 0 both
386 chains of length 15, 47 anchored, 147 cyclic, 47 both
235 chains of length 16, 44 anchored, 1 cyclic, 0 both
312 chains of length 17, 6 anchored, 22 cyclic, 6 both
347 chains of length 18, 37 anchored, 2 cyclic, 0 both
451 chains of length 19, 6 anchored, 27 cyclic, 6 both
1319 chains of length 20, 166 anchored, 165 cyclic, 0 both
5320 chains of length 21, 462 anchored, 519 cyclic, 462 both
3220 chains of length 22, 232 anchored, 0 cyclic, 0 both
4489 chains of length 23, 372 anchored, 516 cyclic, 372 both
20237 chains of length 24, 2130 anchored, 2021 cyclic, 0 both
36580 chains of length 25, 1589 anchored, 1912 cyclic, 1589 both
52875 chains of length 26, 9093 anchored, 506 cyclic, 0 both
197103 chains of length 27, 20896 anchored, 45658 cyclic, 20896 both
216562 chains of length 28, 20314 anchored, 514 cyclic, 0 both
(for a sequence to be both cyclic and anchored, the first element must
both divide n*(n+1)/2 and be equal to n, so n must be odd)
one more tomorrow,
cal
Chris Landauer
Aerospace Integration Science Center
The Aerospace Corporation
cal at aero.org
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From seqfan-owner at ext.jussieu.fr Tue May 4 03:32:32 2004
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From: David Wasserman <dwasserm at earthlink.com>
Subject: Re: Divisor chains
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>From the results that have been posted, I expect that there are many divisor chains for n = 37, but no one has exhibited any. So here's two of them:
37 1 2 4 11 5 3 7 14 6 15 35 28 8 22 33 21 36 18 34 20 12 31 13 32 16 29 17 30 10 25 23 26 24 27 9 19
37 1 2 4 11 5 3 7 14 6 15 35 28 8 22 33 21 36 18 34 20 12 31 13 32 16 29 17 30 10 25 23 26 24 27 9 19
I found these in about 20 minutes of computer time.
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RIchard Guy wrote:
<< How do you know that there is such a `divisor chain' ?
It took quite a while for me to realize that for {1,2,...,n}, if we
let M = floor(n/2), the following permutations always work:
For n even: M+1, 1, M+2, 2, ..., n-1, M-1, n, M.
For n odd: M+1, 1, M+2, 2, ..., M-1, n-1, M, n
The proof is both magical and completely obvious.
E.g.,
for n = 10: 6, 1, 7, 2, 8, 3, 9, 4, 10, 5;
for n = 37: 19, 1, 20, 2, ..., 35, 17, 36, 18, 37.
Still don't know how to show there exists a 37-chain beginning
with 37, 1. But since the final term must divide 37*19, the third
term must be 2 in any such chain.
D'uh!
--Dan
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From seqfan-owner at ext.jussieu.fr Tue May 4 06:16:45 2004
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Subject: Re: [seqfan] Re: Divisor chains
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In about 6 hours of computing time, I've got 2283 chains for n=37 so far and
clearly many more to go.
There are 1572 chains that begin with 37,1,2,4,11.
-- Chuck
----- Original Message -----
From: "David Wasserman" <dwasserm at earthlink.com>
To: <seqfan at ext.jussieu.fr>
Sent: Tuesday, May 04, 2004 3:30 AM
Subject: [seqfan] Re: Divisor chains
> From the results that have been posted, I expect that there are many
divisor chains for n = 37, but no one has exhibited any. So here's two of
them:
>
> 37 1 2 4 11 5 3 7 14 6 15 35 28 8 22 33 21 36 18 34 20 12 31 13 32 16 29
17 30 10 25 23 26 24 27 9 19
>
> 37 1 2 4 11 5 3 7 14 6 15 35 28 8 22 33 21 36 18 34 20 12 31 13 32 16 29
17 30 10 25 23 26 24 27 9 19
>
> I found these in about 20 minutes of computer time.
>
Delivered-To: njas at research.att.com
From: Chris Landauer <cal at rushg.aero.org>
Date: Mon, 3 May 2004 16:25:02 -0700 (PDT)
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Subject: [math-fun] divisor chains are strange
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R.K.Guy wrote (approximately)
a division chain for n is the sequence of numbers from 1 to n,
arranged so that each term is a divisor of the sum of the preceding ones
i wrote a program to compute them so i could look at them
here are the counts up to n=27
(more are pending; this is the simplest program i could think of, and it is
not very smart)
1 of length 1
1 of length 2
1 of length 3
1 of length 4
2 of length 5
2 of length 6
3 of length 7
6 of length 8
14 of length 9
12 of length 10
15 of length 11
17 of length 12
29 of length 13
24 of length 14
178 of length 15
128 of length 16
156 of length 17
140 of length 18
182 of length 19
507 of length 20
2210 of length 21
1636 of length 22
2272 of length 23
9455 of length 24
17437 of length 25
30202 of length 26
102292 of length 27
there are two curious phenomena i notice by looking at the chains themselves:
there are very limited choices for the first and last element
(1) first elements seem to be all large values (mostly, but not always, all
consecutive values from some point up to the largest)
(2) last elements seem to be extremely limited (mostly, but not always, no
more than 3 different choices)
i'll put the entire file, with all chains, counts, and the program, out on the
web when the program finishes (unless somebody disputes the counts, in which
case i'll post only the program on the web instead - it is only 160 lines of C
code - and we can figure out what i did wrong)
more soon,
cal
Chris Landauer
Aerospace Integration Science Center
The Aerospace Corporation
cal at aero.org
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From math-fun-bounces+njas=research.att.com at mailman.xmission.com Mon May 3 19:30:01 2004
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From: Chris Landauer <cal at rushg.aero.org>
Date: Mon, 3 May 2004 16:31:48 -0700 (PDT)
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Subject: [math-fun] divisor chain last elements - duh
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hihi, all -
i wrote that the last elements of these divisor chains were very limited -
well, they have to divide n*(n+1)/2 and be no larger than n, and there are a
lot of small primes
more later,
cal
Chris Landauer
Aerospace Integration Science Center
The Aerospace Corporation
cal at aero.org
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From math-fun-bounces+njas=research.att.com at mailman.xmission.com Mon May 3 20:36:42 2004
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From: Chris Landauer <cal at rushg.aero.org>
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Subject: [math-fun] divisor chains counts fixed
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hihi, all -
oops - my program did not count sequences in which one of the terms was equal
to the sum of the previous terms (< instead of <= in the loop condition 8-()
anyway, here are the revised counts
1 chains of length 1
1 chains of length 2
2 chains of length 3
2 chains of length 4
4 chains of length 5
5 chains of length 6
7 chains of length 7
7 chains of length 8
24 chains of length 9
22 chains of length 10
29 chains of length 11
39 chains of length 12
67 chains of length 13
55 chains of length 14
386 chains of length 15
235 chains of length 16
312 chains of length 17
347 chains of length 18
451 chains of length 19
1319 chains of length 20
5320 chains of length 21
3220 chains of length 22
4489 chains of length 23
20237 chains of length 24
36580 chains of length 25
52875 chains of length 26
197103 chains of length 27
more soon,
cal
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From seqfan-owner at ext.jussieu.fr Mon May 3 21:56:29 2004
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"Vaderlind, Paul -- Paul Vaderlind" <paul at matematik.su.se>,
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Subject: Re: Divisor chains
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On Mon, 3 May 2004, Richard Guy wrote:
> The following might be an acceptable sequence
> for Neil Sloane's OEIS, but someone needs to
> do some work
>
> 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ...
> 1 1 1 1 1 1 1 5 4 1 2 5 5 4 ...
>
I compute up to n = 19 the following number of divisor chains:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
1 1 1 1 1 1 1 5 4 3 2 8 4 6 47 44 6 37 6
which differs from your values at 10, 12, 13, 14.
It's interesting that a(15) > a(16). One might think powers of 2 would
give larger values.
If anyone cares to check my chains from 10 to 16 here they are:
10 2 4 8 6 5 7 3 9 1
10 2 4 8 3 9 6 7 1 5
10 2 6 3 7 4 8 5 9 1
11 1 4 8 6 10 5 9 2 7 3
11 1 2 7 3 8 4 9 5 10 6
12 3 5 10 2 1 11 4 8 7 9 6
12 2 7 1 11 3 9 5 10 4 8 6
12 2 7 3 8 4 9 5 10 6 11 1
12 4 8 3 9 6 2 11 5 10 7 1
12 4 8 6 10 5 9 2 7 3 11 1
12 6 9 3 2 8 4 11 5 10 7 1
12 3 5 10 2 8 4 11 1 7 9 6
12 2 1 5 10 3 11 4 8 7 9 6
13 1 2 8 6 10 4 11 5 3 9 12 7
13 1 2 8 6 10 4 11 5 12 9 3 7
13 1 2 8 12 4 10 5 11 6 9 3 7
13 1 2 8 3 9 4 10 5 11 6 12 7
14 2 8 6 10 4 11 5 3 9 12 7 13 1
14 2 8 6 10 4 11 5 12 9 3 7 13 1
14 2 8 3 9 12 4 13 1 11 7 6 10 5
14 2 8 3 9 4 10 5 11 6 12 7 13 1
14 2 8 12 4 10 5 11 6 9 3 7 13 1
14 2 8 12 9 3 4 13 1 11 7 6 10 5
15 3 2 1 7 14 6 4 13 5 10 8 11 9 12
15 5 10 6 12 8 14 7 11 2 9 1 4 13 3
15 5 4 3 9 12 2 10 6 11 1 13 7 14 8
15 5 4 2 13 1 8 12 3 7 14 6 9 11 10
15 5 10 3 11 4 12 2 1 9 6 13 7 14 8
15 5 10 3 11 4 6 9 7 14 12 8 13 1 2
15 3 6 4 14 7 1 10 5 13 2 8 11 9 12
15 3 2 4 12 9 5 10 6 11 1 13 7 14 8
15 1 2 9 3 10 4 11 5 12 6 13 7 14 8
15 5 10 3 11 1 9 6 12 4 2 13 7 14 8
15 5 4 12 9 3 2 10 6 11 1 13 7 14 8
15 1 4 2 11 3 9 5 10 12 6 13 7 14 8
15 1 8 12 6 2 11 5 10 14 7 13 4 9 3
15 5 10 2 1 11 4 6 9 7 14 12 8 13 3
15 3 6 4 14 7 1 2 13 5 10 8 11 9 12
15 5 10 6 3 13 4 7 9 12 14 1 11 2 8
15 5 10 2 4 9 3 12 6 11 1 13 7 14 8
15 5 4 2 13 3 6 1 7 14 10 8 11 9 12
15 5 10 3 11 1 9 2 4 12 6 13 7 14 8
15 5 10 6 3 13 4 14 7 11 8 12 9 1 2
15 5 4 2 13 1 8 6 9 7 14 12 3 11 10
15 5 10 6 9 3 4 13 1 11 7 12 2 14 8
15 5 10 6 2 1 13 4 14 7 11 8 12 9 3
15 5 1 7 2 6 3 13 4 14 10 8 11 9 12
15 5 10 2 4 12 6 9 3 11 1 13 7 14 8
15 5 4 2 13 3 14 7 1 8 12 6 9 11 10
15 5 10 6 12 4 2 9 3 11 1 13 7 14 8
15 5 4 2 13 3 6 8 7 9 12 14 1 11 10
15 5 10 6 12 8 14 7 11 2 1 13 4 9 3
15 5 10 6 3 13 4 14 7 11 8 2 1 9 12
15 3 9 1 2 10 4 11 5 12 6 13 7 14 8
15 5 10 6 9 3 8 14 7 11 2 1 13 4 12
15 1 2 3 7 14 6 4 13 5 10 8 11 9 12
15 1 8 12 9 5 10 6 11 7 14 2 4 13 3
15 5 10 6 3 13 4 8 2 11 7 14 1 9 12
15 5 10 6 9 3 12 4 2 11 1 13 7 14 8
15 5 10 2 1 11 4 12 3 9 6 13 7 14 8
15 5 2 11 3 6 14 7 9 8 10 1 13 4 12
15 5 10 3 11 4 8 14 2 6 13 7 1 9 12
15 5 10 6 2 1 13 4 14 7 11 8 3 9 12
15 3 9 1 14 6 4 13 5 7 11 8 12 2 10
15 3 6 12 9 5 10 4 2 11 1 13 7 14 8
15 1 8 3 9 6 2 11 5 10 14 7 13 4 12
15 5 4 12 9 3 8 14 2 6 13 7 1 11 10
15 3 2 10 5 7 14 8 1 13 6 4 11 9 12
15 5 10 6 3 13 4 14 7 1 2 8 11 9 12
15 5 4 3 9 12 8 14 2 6 13 7 1 11 10
16 4 10 15 3 2 5 11 6 12 7 13 8 14 9 1
16 2 6 12 9 5 10 15 3 13 7 14 8 1 11 4
16 8 12 9 5 10 6 11 7 14 2 4 13 3 15 1
16 8 12 9 5 10 15 3 13 7 1 11 2 14 6 4
16 8 3 9 6 2 11 5 10 14 7 13 4 12 15 1
16 8 12 9 15 10 5 3 13 7 14 2 6 1 11 4
16 8 12 9 5 10 15 3 13 7 14 2 6 1 11 4
16 2 6 4 14 3 15 10 7 11 8 12 9 13 5 1
16 8 6 15 9 2 14 5 3 13 7 1 11 10 12 4
16 4 2 11 3 9 5 10 12 6 13 7 14 8 15 1
16 2 9 3 10 4 11 5 12 6 13 7 14 8 15 1
16 2 9 3 15 5 10 12 6 13 7 14 8 1 11 4
16 8 12 6 2 11 5 10 14 7 13 4 9 3 15 1
16 8 2 13 3 14 7 9 12 6 15 5 10 1 11 4
16 8 2 13 1 5 15 3 7 14 6 9 11 10 12 4
16 2 6 12 9 15 10 5 3 13 7 14 8 1 11 4
16 8 12 3 13 2 9 7 14 6 15 5 10 1 11 4
16 2 9 3 15 5 10 6 11 1 13 7 14 8 12 4
16 8 2 13 3 6 1 7 14 5 15 9 11 10 12 4
16 8 6 3 11 4 12 5 13 2 10 15 7 14 9 1
16 8 12 3 13 4 14 5 15 9 11 10 6 7 1 2
16 8 12 3 13 4 14 7 11 2 15 5 10 6 9 1
16 8 12 9 15 5 13 6 14 7 3 2 10 1 11 4
16 8 3 9 12 2 10 5 13 6 14 7 15 1 11 4
16 8 3 9 6 2 11 5 10 14 7 13 1 15 12 4
16 4 10 15 3 8 14 7 11 2 6 12 9 13 5 1
16 2 3 7 14 6 4 13 5 10 8 11 9 12 15 1
16 8 4 14 3 15 10 7 11 2 6 12 9 13 5 1
16 2 9 3 10 5 15 6 11 1 13 7 14 8 12 4
16 8 2 13 3 14 7 9 12 6 10 5 15 1 11 4
16 8 12 9 3 2 10 5 13 6 14 7 15 1 11 4
16 8 2 13 1 10 5 11 6 9 3 14 7 15 12 4
16 8 12 3 13 2 9 7 14 6 10 5 15 1 11 4
16 8 12 9 15 10 5 3 13 7 1 11 2 14 6 4
16 4 10 15 9 2 14 7 11 8 12 6 3 13 5 1
16 8 4 14 6 12 10 7 11 2 15 3 9 13 5 1
16 2 9 3 10 8 12 5 13 6 14 7 15 1 11 4
16 4 10 6 12 8 14 7 11 2 15 3 9 13 5 1
16 8 12 3 13 4 14 7 11 2 10 5 15 6 9 1
16 4 10 15 3 6 2 14 7 11 8 12 9 13 5 1
16 2 9 3 10 5 15 12 6 13 7 14 8 1 11 4
16 4 2 11 3 9 15 12 6 13 7 14 8 10 5 1
16 8 12 6 14 4 10 7 11 2 15 3 9 13 5 1
16 8 12 9 15 6 11 7 14 2 4 13 3 10 5 1
From seqfan-owner at ext.jussieu.fr Mon May 3 23:10:30 2004
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To: "Richard Guy" <rkg at cpsc.ucalgary.ca>,
"Math Fun" <math-fun at mailman.xmission.com>, <seqfan at ext.jussieu.fr>
Cc: "Vaderlind, Paul -- Paul Vaderlind" <paul at matematik.su.se>,
"Paul Vaderlind" <paul at math.su.se>,
"Loren Larson" <lllarsson at earthlink.net>
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Subject: Re: [seqfan] Divisor chains
Date: Mon, 3 May 2004 23:13:44 -0400
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Wrote a litte recursive Maple code to solve this problem. My output agrees
with Richards for n from 1 to 9 and 11. For n=10 I find 3 chains (Richard's
+ 2), for n=12 I find 8 chains (Richard's + 3), for n=13 I find 4 (Richard's
4'th n=13 chain includes "3" twice), for n=14 I find 6 chains (Richard's +
2). Other than the errant chain for 13, the chains Richard provides appear
to be correct.
This makes the sequence up to 14 look like this:
1, 1, 1, 1, 1, 1, 1, 5, 4, 3, 2, 8, 4, 6
>From there on the values start getting much larger (here is n=15 to n=24):
47, 44, 6, 37, 6, 166, 462, 232, 372, 2130
If anyone is interested I'm happy to share the code. I'm waiting on n=25
right now. The trend is generally upward so I suspect there are MANY chains
for n=25.
Given how many chains there are for these numbers, I would expect the number
of chains for n=37 to be pretty high, although for some primes (such as 17
and 19) there are small numbers of chains. Richard do you suggest n=37
because you believe there to be very few chains?
-- Chuck
----- Original Message -----
From: "Richard Guy" <rkg at cpsc.ucalgary.ca>
To: "Math Fun" <math-fun at mailman.xmission.com>; <seqfan at ext.jussieu.fr>
Cc: "Vaderlind, Paul -- Paul Vaderlind" <paul at matematik.su.se>; "Paul
Vaderlind" <paul at math.su.se>; "Loren Larson" <lllarsson at earthlink.net>
Sent: Monday, May 03, 2004 4:20 PM
Subject: [seqfan] Divisor chains
> The following might be an acceptable sequence
> for Neil Sloane's OEIS, but someone needs to
> do some work
>
> 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ...
> 1 1 1 1 1 1 1 5 4 1 2 5 5 4 ...
>
> It arises from a problem I got recently from
> Paul Vaderlind.
>
> If the sequence of numbers from 1 to 37,
> is arranged so that each term is a divisor
> of the sum of preceding ones, starting
> 37, 1, ...
> what is the next term?
>
> The answer is either 2 or 19. P'r'aps
> I won't spoil your fun by pointing out
> which. But I will spoil it by asking:
> How do you know that there is such a
> `divisor chain' ?
>
> The sequence is (my present state of knowledge
> of) the number of divisor chains of length n.
> Here are the ones I found
>
> 1 2 1 3 1 2 4 2 3 1
>
> 5 1 2 4 3 6 2 4 3 5 1 7 1 2 5 3 6 4
>
> 8 2 5 3 6 4 7 1
> 8 4 2 7 3 1 5 6
> 8 4 2 7 3 6 5 1
> 8 4 3 5 1 7 2 6
> 8 4 6 3 7 2 5 1
>
> 9 1 2 4 8 6 5 7 3
> 9 1 2 6 3 7 4 8 5
> 9 3 4 8 6 5 7 2 1
> 9 3 6 2 1 7 4 8 5
>
> 10 2 4 8 3 9 6 7 1 5
>
> 11 1 2 7 3 8 4 9 5 10 6
> 11 1 4 8 6 10 5 9 2 7 3
>
> 12 2 1 5 10 3 11 4 8 7 9 6
> 12 2 7 3 8 4 9 5 10 6 11 1
> 12 3 5 10 2 8 4 11 1 7 9 6
> 12 4 8 3 9 6 2 11 5 10 7 1
> 12 4 8 6 10 5 9 2 7 3 11 1
>
> 13 1 2 8 3 9 4 10 5 11 6 12 7
> 13 1 2 8 6 10 4 11 5 12 9 3 7
> 13 1 2 8 6 10 4 11 5 3 9 12 7
> 13 1 2 8 12 4 10 5 11 3 9 3 7
> 13 1 2 8 12 4 10 5 11 6 9 3 7
>
> 14 2 8 6 10 4 11 5 3 9 12 7 13 1
> 14 2 8 6 10 4 11 5 12 9 3 7 13 1
> 14 2 8 12 4 10 5 11 6 9 3 7 13 1
> 14 2 8 12 9 3 4 13 1 11 7 6 10 5
>
> Enough sins of omission, to say nothing of
> commission, for today. Please check and extend.
> Any ideas for proving anything? R.
>
>
From seqfan-owner at ext.jussieu.fr Mon May 3 23:21:18 2004
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To: "Richard Guy" <rkg at cpsc.ucalgary.ca>,
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"Paul Vaderlind" <paul at math.su.se>,
"Loren Larson" <lllarsson at earthlink.net>
References: <Pine.LNX.4.44.0405031403560.12751-100000 at csl> <000c01c43185$d159a120$0200a8c0 at plastereddragon>
Subject: Re: [seqfan] Divisor chains
Date: Mon, 3 May 2004 23:25:00 -0400
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Sorry for the double-post... for n=25 there are 1589 chains. First 25 terms
are therefore:
1, 1, 1, 1, 1, 1, 1, 5, 4, 3, 2, 8, 4, 6, 47, 44, 6, 37, 6, 166, 462, 232,
372, 2130, 1589
-- Chuck
----- Original Message -----
From: "Chuck Seggelin" <seqfan at plastereddragon.com>
To: "Richard Guy" <rkg at cpsc.ucalgary.ca>; "Math Fun"
<math-fun at mailman.xmission.com>; <seqfan at ext.jussieu.fr>
Cc: "Vaderlind, Paul -- Paul Vaderlind" <paul at matematik.su.se>; "Paul
Vaderlind" <paul at math.su.se>; "Loren Larson" <lllarsson at earthlink.net>
Sent: Monday, May 03, 2004 11:13 PM
Subject: Re: [seqfan] Divisor chains
> Wrote a litte recursive Maple code to solve this problem. My output
agrees
> with Richards for n from 1 to 9 and 11. For n=10 I find 3 chains
(Richard's
> + 2), for n=12 I find 8 chains (Richard's + 3), for n=13 I find 4
(Richard's
> 4'th n=13 chain includes "3" twice), for n=14 I find 6 chains (Richard's +
> 2). Other than the errant chain for 13, the chains Richard provides
appear
> to be correct.
>
> This makes the sequence up to 14 look like this:
>
> 1, 1, 1, 1, 1, 1, 1, 5, 4, 3, 2, 8, 4, 6
>
> From there on the values start getting much larger (here is n=15 to n=24):
>
> 47, 44, 6, 37, 6, 166, 462, 232, 372, 2130
>
> If anyone is interested I'm happy to share the code. I'm waiting on n=25
> right now. The trend is generally upward so I suspect there are MANY
chains
> for n=25.
>
> Given how many chains there are for these numbers, I would expect the
number
> of chains for n=37 to be pretty high, although for some primes (such as 17
> and 19) there are small numbers of chains. Richard do you suggest n=37
> because you believe there to be very few chains?
>
> -- Chuck
>
> ----- Original Message -----
> From: "Richard Guy" <rkg at cpsc.ucalgary.ca>
> To: "Math Fun" <math-fun at mailman.xmission.com>; <seqfan at ext.jussieu.fr>
> Cc: "Vaderlind, Paul -- Paul Vaderlind" <paul at matematik.su.se>; "Paul
> Vaderlind" <paul at math.su.se>; "Loren Larson" <lllarsson at earthlink.net>
> Sent: Monday, May 03, 2004 4:20 PM
> Subject: [seqfan] Divisor chains
>
>
> > The following might be an acceptable sequence
> > for Neil Sloane's OEIS, but someone needs to
> > do some work
> >
> > 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ...
> > 1 1 1 1 1 1 1 5 4 1 2 5 5 4 ...
> >
> > It arises from a problem I got recently from
> > Paul Vaderlind.
> >
> > If the sequence of numbers from 1 to 37,
> > is arranged so that each term is a divisor
> > of the sum of preceding ones, starting
> > 37, 1, ...
> > what is the next term?
> >
> > The answer is either 2 or 19. P'r'aps
> > I won't spoil your fun by pointing out
> > which. But I will spoil it by asking:
> > How do you know that there is such a
> > `divisor chain' ?
> >
> > The sequence is (my present state of knowledge
> > of) the number of divisor chains of length n.
> > Here are the ones I found
> >
> > 1 2 1 3 1 2 4 2 3 1
> >
> > 5 1 2 4 3 6 2 4 3 5 1 7 1 2 5 3 6 4
> >
> > 8 2 5 3 6 4 7 1
> > 8 4 2 7 3 1 5 6
> > 8 4 2 7 3 6 5 1
> > 8 4 3 5 1 7 2 6
> > 8 4 6 3 7 2 5 1
> >
> > 9 1 2 4 8 6 5 7 3
> > 9 1 2 6 3 7 4 8 5
> > 9 3 4 8 6 5 7 2 1
> > 9 3 6 2 1 7 4 8 5
> >
> > 10 2 4 8 3 9 6 7 1 5
> >
> > 11 1 2 7 3 8 4 9 5 10 6
> > 11 1 4 8 6 10 5 9 2 7 3
> >
> > 12 2 1 5 10 3 11 4 8 7 9 6
> > 12 2 7 3 8 4 9 5 10 6 11 1
> > 12 3 5 10 2 8 4 11 1 7 9 6
> > 12 4 8 3 9 6 2 11 5 10 7 1
> > 12 4 8 6 10 5 9 2 7 3 11 1
> >
> > 13 1 2 8 3 9 4 10 5 11 6 12 7
> > 13 1 2 8 6 10 4 11 5 12 9 3 7
> > 13 1 2 8 6 10 4 11 5 3 9 12 7
> > 13 1 2 8 12 4 10 5 11 3 9 3 7
> > 13 1 2 8 12 4 10 5 11 6 9 3 7
> >
> > 14 2 8 6 10 4 11 5 3 9 12 7 13 1
> > 14 2 8 6 10 4 11 5 12 9 3 7 13 1
> > 14 2 8 12 4 10 5 11 6 9 3 7 13 1
> > 14 2 8 12 9 3 4 13 1 11 7 6 10 5
> >
> > Enough sins of omission, to say nothing of
> > commission, for today. Please check and extend.
> > Any ideas for proving anything? R.
> >
> >
>
>
From math-fun-bounces+njas=research.att.com at mailman.xmission.com Tue May 4 00:47:38 2004
Delivered-To: njas at research.att.com
From: Chris Landauer <cal at rushg.aero.org>
Date: Mon, 3 May 2004 21:49:09 -0700 (PDT)
Message-Id: <200405040449.i444n9795633 at calamari.aero.org>
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Subject: [math-fun] counts for related divisor chains
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hihi, all -
several people have reminded me that there are some sequences related
to the divisor chains
one of them must start with n
one of them must start with a divisor of n*(n+1)/2
(so it can form a cycle)
actually, if the requirement is that each cyclic permutation
of the sequence is also a divisor chain,
then there are none (except the trivial one 1 for n=1),
since not all integers start divisor chains
for any given n>1
all of these subsequent counts were done by manual editing of the sequence
output from the first program, so typos are quie possible
the first counts are when the sequence must start with n
(i call these anchored):
1 chains of length 1
1 chains of length 2 @
1 chains of length 3 *
1 chains of length 4 @
1 chains of length 5 *
1 chains of length 6 @
1 chains of length 7 *
5 chains of length 8 @
4 chains of length 9 *
3 chains of length 10 @
2 chains of length 11
8 chains of length 12
4 chains of length 13 *
6 chains of length 14
47 chains of length 15
44 chains of length 16 @
6 chains of length 17 *
37 chains of length 18 @
6 chains of length 19
166 chains of length 20
462 chains of length 21
232 chains of length 22 @
372 chains of length 23
2130 chains of length 24
1589 chains of length 25 *
9093 chains of length 26 @
(* = n and (n+1)/2 are prime powers, n odd)
(@ = n/2 and n+1 are prime powers, n even)
the second case is when the first element must divide n*(n+1)/2
(i call these cyclic):
1 chains of length 1
1 chains of length 2
2 chains of length 3
1 chains of length 4
2 chains of length 5
0 chains of length 6
3 chains of length 7
0 chains of length 8
5 chains of length 9
0 chains of length 10
6 chains of length 11
0 chains of length 12
6 chains of length 13
0 chains of length 14
145 chains of length 15
0 chains of length 16
22 chains of length 17
2 chains of length 18
27 chains of length 19
165 chains of length 20
57 chains of length 21
0 chains of length 22
516 chains of length 23
2021 chains of length 24
1912 chains of length 25
453 chains of length 26
the third case is when both conditions must hold:
1 chains of length 1
1 chains of length 2
1 chains of length 3
1 chains of length 4
1 chains of length 5
0 chains of length 6
1 chains of length 7
0 chains of length 8
4 chains of length 9
0 chains of length 10
2 chains of length 11
0 chains of length 12
4 chains of length 13
0 chains of length 14
47 chains of length 15
0 chains of length 16
6 chains of length 17
0 chains of length 18
6 chains of length 19
0 chains of length 20
0 chains of length 21
0 chains of length 22
372 chains of length 23
0 chains of length 24
1589 chains of length 25
0 chains of length 26
more later,
cal
Chris Landauer
Aerospace Integration Science Center
The Aerospace Corporation
cal at aero.org
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From seqfan-owner at ext.jussieu.fr Tue May 4 03:32:32 2004
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Date: Tue, 4 May 2004 00:30:16 -0700
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From: David Wasserman <dwasserm at earthlink.com>
Subject: Re: Divisor chains
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>From the results that have been posted, I expect that there are many divisor chains for n = 37, but no one has exhibited any. So here's two of them:
37 1 2 4 11 5 3 7 14 6 15 35 28 8 22 33 21 36 18 34 20 12 31 13 32 16 29 17 30 10 25 23 26 24 27 9 19
37 1 2 4 11 5 3 7 14 6 15 35 28 8 22 33 21 36 18 34 20 12 31 13 32 16 29 17 30 10 25 23 26 24 27 9 19
I found these in about 20 minutes of computer time.
From math-fun-bounces+njas=research.att.com at mailman.xmission.com Tue May 4 03:48:25 2004
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Date: Tue, 4 May 2004 00:47:45 -0700 (GMT-07:00)
From: Daniel Asimov <dasimov at earthlink.net>
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RIchard Guy wrote:
<< How do you know that there is such a `divisor chain' ?
It took quite a while for me to realize that for {1,2,...,n}, if we
let M = floor(n/2), the following permutations always work:
For n even: M+1, 1, M+2, 2, ..., n-1, M-1, n, M.
For n odd: M+1, 1, M+2, 2, ..., M-1, n-1, M, n
The proof is both magical and completely obvious.
E.g.,
for n = 10: 6, 1, 7, 2, 8, 3, 9, 4, 10, 5;
for n = 37: 19, 1, 20, 2, ..., 35, 17, 36, 18, 37.
Still don't know how to show there exists a 37-chain beginning
with 37, 1. But since the final term must divide 37*19, the third
term must be 2 in any such chain.
D'uh!
--Dan
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