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A233045
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1^m + 2^m + ... + m^m (mod m) for primary pseudoperfect numbers m.
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2
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OFFSET
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1,5
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COMMENTS
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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.
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LINKS
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FORMULA
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a(n) = 1 for n = 1, 2, 3, 4.
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EXAMPLE
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The 1st primary pseudoperfect number is 2, and 1^2 + 2^2 = 5 == 1 (mod 2), so a(1) = 1.
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MATHEMATICA
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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}]
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CROSSREFS
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KEYWORD
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more,nonn
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AUTHOR
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STATUS
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approved
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