introduction and new sequence

[Robert G. Wilson v]

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-
>