introduction and new sequence

Andrew Plewe aplewe at sbcglobal.net
Wed Aug 11 02:09:00 CEST 2004


At lunch today I noticed I'd made some errors in the values in my first
email (I had worked it out by hand and wanted a couple of days to look
over my calculations).  It looks like you've found and corrected them.
Thanks for your assistance, and for the generating function and extra
terms.

	-Andrew Plewe-

-----Original Message-----
From: Robert G. Wilson v [mailto:rgwv at rgwv.com] 
Sent: Tuesday, August 10, 2004 4:44 PM
To: Andrew Plewe
Cc: seqfan
Subject: Re: introduction and new sequence


Et al,

	I quickly noticed the connection between this problem and the
number of times a 
divisor divides a factorial. Thus the algorithm is to find the prime
factors of n, 
and then figure out difference between the indexes of n! and (n-1)!. As
an 
example, for a(100) its prime divisors are just two and five. 2^97 |
100! and 2^95 
| 99! thus the indexes for two @ 100 are just 96 and 97. 5^24 | 100! and
5^22 | 99! thus the indexes for five @ 100 are just 23 and 24. a(100) =
96+97+23+24 = 240.

%I A000001
%S A000001
1,1,5,1,6,1,18,7,10,1,24,1,13,9,54,1,31,1,39,12,21,1,73,11,25,36,53,1,
%T A000001
47,1,145,18,34,13,100,1,37,21,120,1,64,1,85,51,44,1,200,15,70,26,101,1,
%U A000001
125,18,165,30,56,1,153,1,59,69,363,20,101,1,135,35,94,1,274,1,73,70,149
%N A000001 Sum of index values of the prime factors (with multiplicity)
for n.
%C A000001 Let P be equal to the set of prime factors of the positive
integers, 
counted with multiplicity. Order the members of this set into subsets
such that 
each prime has its own set with an index value assigned to each instance
of the 
prime. Therefore, P = {{2_1, 2_2,..2_i}, {3_1, 3_2,..3_j}, . . {p_1,
p_2,..p_x}}. 
In generating the sequence, each indexed instance of a prime can only be
used once.
%e A000001 2 = 2_1, thus a(2)=1
%e A000001 3 = 3_1, thus a(3)=1
%e A000001 4 = 2_2 * 2_3, thus a(4)=5
%e A000001 5 = 5_1, thus a(5)=1
%e A000001 6 = 2_4 * 3_2, thus a(6)=6
%e A000001 7 = 7_1, thus a(7)=1
%e A000001 8 = 2_5 * 2_6 * 2_7, thus a(8)=5+6+7=18, etc.
%t A000001 PrimeFactors[n_Integer] := Flatten[ Table[ # [[1]], {1}] & /@

FactorInteger[n]]; f[n_, p_] := Block[{t = 0, q = p}, While[s =
Floor[n/q]; t = t 
+ s; s > 0, q *= p]; t]; g[n_] := Block[{s = 0, pf = PrimeFactors[n], k
= 1}, l = 
Length[pf]; While[k <= l, s = s + Sum[i, {i, f[n - 1, pf[[k]]] + 1, f[n,

pf[[k]]]}]; k++ ]; s]; Table[g[n], {n, 2, 75}]
%Y A000001 Cf. A123456.
%O A000001 2,3
%K A000001 nonn
%A A000001 Andrew Plewe (aplewe at sbcglobal.net), Aug 10 2004
%E A000001 Edited and extended by RGWv (rgwv at rgwv.com), Aug 10 2004


Andrew Plewe wrote:
> Hello, I've recently joined the list and I'd like to introduce myself.
> My name is Andrew Plewe, I work as a computer programmer in
California.
> I'm interested in number theory, specifically factoring and primes.
The
> sequences I've submitted to the OEIS originate from my explorations of
> number theory.
> 
> 
> Here is one such sequence.  If anyone has suggestions/corrections feel
> free to email them to me at aplewe at sbcglobal.net.  I plan to submit
this
> sequence in a couple of days:
> 
> 
> The sequence:
> 1, 1, 5, 1, 6, 1, 18, 7, 10, 1, 24, 1, 13, 9, 54, 1, 31, 1, 34, 12,
21,
> 1, 73, 11, 25, 36, 53, 1, 47, 1, 145, 18, 34, 13, 100, 1, 37, 21, 120,
> 1, 54, 1, 85, 51, 44, 1, 200, 15, 70
> 
> 
> Definition:
> Sum of index values of the prime factors (counted with multiplicity)
for
> each positive integer greater than two.
> 
> 
> Comment:
> herein p_x denotes prime p with index value x.
> 
> This sequence is generated as follows:
> 
> P = the set of prime factors of the positive integers greater than
two,
> counted with multiplicity.  Order the members of this set into subsets
> such that each prime has its own set with an index value assigned to
> each instance of the prime.  Therefore, P = {{2_1, 2_2,..2_x}, {3_1,
> 3_2,..3_x}, . . {p_1, p_2,..p_x}}.  In generating the sequence, each
> indexed instance of a prime can only be used once.  Starting with
> integers i = 2 to i = 8, we proceed as follows:
> 
> i = 2_1,              n = 1
> i = 3_1,              n = 1
> i = 4 = 2_2 * 2_3,    n = 2 + 3 = 5
> i = 5_1,              n = 1
> i = 6 = 3_2 * 2_4,    n = 2 + 4 = 6
> i = 7_1,              n = 1
> i = 2_5 * 2_6 * 2_7,  n = 5 + 6 + 7 = 18
> 
> and etc.
> 
> i = 2 to i = 50 are shown.
> 
>     -Andrew Plewe-
> 






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