OFFSET

1,2

COMMENTS

R. Mestrovic conjectures that, for n >= 3, 2*sum(j = 1, n) j^n == -n^2 (mod n^3) implies that n is prime. See proposition 2.2(ii) - the conjecture is proved to be equivalent to Giuga's conjecture.

This proves the first half of Mestrovic's conjecture and hence the first half of Giuga's conjecture : If n>2 is prime, a(n)=0. Proof: [1] [(n-k)^n + k^n] mod n^3 with n odd becomes [2] n^2*k^(n-1) mod n^3. k^(n-1) == 1 mod n if n is prime, so we can write k^(n-1) = 1 + jn. Place this back into [2], we find that [2] is just n^2 mod n^3. From [1] we have (n-1)*n^2 = -n^2 mod n^3. Q.E.D. The second half of Mestrovic's conjecture involves proving that if n is composite then a(n)<>0. It is easy enough to prove a(4k+2)<>0 by noticing that the residue must be even on the RHS, but the residue on the LHS is odd. a(8k+4)<>0 by a similar method. - Jon Perry, Dec 01 2012

LINKS

T. D. Noe, Table of n, a(n) for n = 1..1000

R. Mestrovic, A congruence modulo n^3 involving two consecutive sums of powers and its applications, arXiv:1211.4570 [math.NT], 2012.

R. Mestrovic, On a Congruence Modulo n^3 Involving Two Consecutive Sums of Powers, Journal of Integer Sequences, Vol. 17 (2014), 14.8.4.

Eric Weisstein's World of Mathematics, Giuga's Conjecture.

Eric Weisstein's World of Mathematics, Agoh's Conjecture.

Eric Weisstein's World of Mathematics, Fermat's Little Theorem.

Eric Weisstein's World of Mathematics, Giuga Number.

Eric Weisstein's World of Mathematics, Bernoulli Number.

Eric Weisstein's World of Mathematics, Carmichael Number.

EXAMPLE

a(2) = 6 because 2^2 + 2(1^2 + 2^2) = 14 and 14 = 6 mod 2^3.

MATHEMATICA

Table[Mod[n^2 + 2*Sum[j^n, {j, 1, n}], n^3], {n, 100}] (* T. D. Noe, Nov 29 2012 *)

PROG

(PARI) a(n) = (2*sum(j=1, n, j^n) + n^2) % n^3;

CROSSREFS

KEYWORD

nonn

AUTHOR

Michel Marcus, Nov 22 2012

STATUS

approved