A233045 1^m + 2^m + ... + m^m (mod m) for primary pseudoperfect numbers m.

1, 1, 1, 1, 5797, 272753965, 8749232767, 1045741078641946876220133713545

Offset: 1

Author

Jonathan Sondow, Dec 10 2013

Keywords

• more
• nonn

Comments

A031971(m) (mod m) for m in A054377 = 2, 6, 42, 1806, 47058, 2214502422, 52495396602, 8490421583559688410706771261086. The known values of m for which 1^m + 2^m + ... + m^m == 1 (mod m) are m = 1, 2, 6, 42, 1806.

For any m and prime p | m, use Sum_{j=1..m} j^m == -m/p (mod p) if p-1 | m or == 0 (mod p) otherwise (see Lemma 3 in Grau et al.) and the Chinese Remainder Theorem.

Examples

			The 1st primary pseudoperfect number is 2, and 1^2 + 2^2 = 5 == 1 (mod 2), so a(1) = 1.
		

Links

• J. M. Grau, A. M. Oller-Marcen, and J. Sondow, On the congruence 1^m + 2^m + ... + m^m == n (mod m) with n|m, arXiv:1309.7941 [math.NT].

Crossrefs

Cf. A031971, A054377, A231409.

Programs

Mathematica

ps={2, 6, 42, 1806, 47058, 2214502422,52495396602, 8490421583559688410706771261086}; fa = FactorInteger; VonStaudt[n_] := Mod[n - Sum[If[IntegerQ[n/(fa[n][[i, 1]] - 1)], n/fa[n][[i, 1]], 0], {i, Length[fa[n]]}], n]; Table[VonStaudt[ps[[i]]], {i, 1, 8}]

Formula

a(n) = 1 for n = 1, 2, 3, 4.