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A219540
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a(n) = (n^2 + 2*(Sum_{j = 1..n} j^n)) (mod n^3).
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1
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0, 6, 0, 20, 0, 26, 0, 200, 567, 950, 0, 1276, 0, 2058, 2475, 784, 0, 3750, 0, 4932, 6615, 4114, 0, 2264, 13125, 3718, 13851, 588, 0, 8050, 0, 19488, 25047, 34102, 1225, 23700, 0, 38266, 41067, 18664, 0, 46622, 0, 53724, 44550, 34914, 0, 72496, 103243, 123750
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OFFSET
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1,2
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COMMENTS
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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
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LINKS
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EXAMPLE
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a(2) = 6 because 2^2 + 2(1^2 + 2^2) = 14 and 14 = 6 mod 2^3.
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MATHEMATICA
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Table[Mod[n^2 + 2*Sum[j^n, {j, 1, n}], n^3], {n, 100}] (* T. D. Noe, Nov 29 2012 *)
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PROG
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(PARI) a(n) = (2*sum(j=1, n, j^n) + n^2) % n^3;
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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