|
%I #7 Dec 12 2014 05:04:40
%S 3,5,7,11,15,19,23,25,27,31,35,39,43,45,47,51,55,59,63,65,67,71,75,79,
%T 83,85,87,91,95,99,103,105,107,111,115,119,123,125,127,131,135,139,
%U 143,145,147,151,155,159,163,165,167,171,175,179,183,185,187,191,195,199
%N Numbers n such that 25 divides Sum[ Prime[k]^n, {k,1,n}].
%C a(n) up to a(7) = 23 coincides with A007665[n+1] = Tower of Hanoi with 5 pegs. It appears that a(n) includes all A007665[n] = {1, 3, 5, 7, 11, 15, 19, 23, 27, 31, 39, 47, 55, 63, 71, 79, 87, 95, 103, 111, 127, 143, 159, 175, 191, 207, 223, 239, 255, 271, 287, 303, 319, 335, 351, 383, 415, 447, 479, 511, 543, 575, 607, 639, 671, 703, 735, 767, 799, ...} except A007665[1] = 1.
%C Primes in this sequence include 5 and all primes of the form 4k+3, A002145[n]. Terms include all numbers of the form 10k+5 (with nonnegative k), A017329[n].
%e There are 25 primes p < 100, p(n) = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.
%e a(1) = because 25 divides Sum[p(n)^3,{n,1,25}] = 2^3 + 3^3 + ... + 89^3 + 97^3 = A098999[25] and does not divide Sum[p(n)^1,{n,1,25}] = A007504[25] and Sum[p(n)^2,{n,1,25}] = A024450[25].
%e The next a(2) = 5 because 25 divides Sum[p(n)^5,{n,1,25}] = A122103[25] and does not divide Sum[p(n)^4,{n,1,25}] = A122102[25].
%t Select[Range[300],IntegerQ[Sum[ Prime[k]^#1, {k,1,25}]/25]&]
%o (PARI) for(n=1,100,if(sum(k=1,25,prime(k)^n)%25==0,print1(n,",")));
%o print;print("Alternative method not using primes:");
%o for(n=1,100,m=(n-1)%6;print1((n-m)*3+(n-m+if(m>1,(m-1)*12-1,m*6-1))/3,",")) \\ _K. Spage_, Oct 23 2009
%Y Cf. A007504, A024450, A098999, A122102, A122103, A007665.
%Y Cf. A002144, A002145, A017329.
%K nonn
%O 1,1
%A _Alexander Adamchuk_, Aug 21 2006, Sep 18 2006, Sep 21 2006
|
