A122124 Numbers n such that 25 divides Sum[ Prime[k]^n, {k,1,n}].

3, 5, 7, 11, 15, 19, 23, 25, 27, 31, 35, 39, 43, 45, 47, 51, 55, 59, 63, 65, 67, 71, 75, 79, 83, 85, 87, 91, 95, 99, 103, 105, 107, 111, 115, 119, 123, 125, 127, 131, 135, 139, 143, 145, 147, 151, 155, 159, 163, 165, 167, 171, 175, 179, 183, 185, 187, 191, 195, 199

Offset: 1

Author

Alexander Adamchuk, Aug 21 2006, Sep 18 2006, Sep 21 2006

Keywords

• nonn

Comments

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.

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].

Examples

			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}.
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].
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].
		

Crossrefs

Cf. A007504, A024450, A098999, A122102, A122103, A007665.

Cf. A002144, A002145, A017329.

Programs

Mathematica

Select[Range[300],IntegerQ[Sum[ Prime[k]^#1, {k,1,25}]/25]&]

PARI

for(n=1,100,if(sum(k=1,25,prime(k)^n)%25==0,print1(n,",")));
print;print("Alternative method not using primes:");
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