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Sum(m^p, m=1..p-1)/p as p runs through the odd primes.
2

%I #20 Nov 05 2017 11:52:39

%S 3,260,53823,12942210875,11901444483396,25627001801054931008,

%T 55413915436873048932459,490667517005738962388828685983,

%U 48588952813858892791005036793649985985124,303307728036900627681487165427498812641117375,158544898951978777519612048992784361843596346824881328548

%N Sum(m^p, m=1..p-1)/p as p runs through the odd primes.

%C Always an integer: for an elementary proof that Sum(m^k,m=1..p-1)/p is an integer if p is prime and p-1 does not divide k (and a discussion of other proofs), see MacMillan and Sondow 2011. Applications are in Sondow and MacMillan 2011.

%C For (Sum(m^(p-1), m=1..p-1)+1)/p as p runs through the primes, see A055030.

%C For Sum(m^p, m=1..p-1) / p^2 as p runs through the odd primes, see A294507.

%H K. MacMillan and J. Sondow, <a href="http://arxiv.org/abs/1011.0076">Proofs of power sum and binomial coefficient congruences via Pascal's identity</a>, Amer. Math. Monthly, 118 (2011), 549-551.

%H J. Sondow and K. MacMillan, <a href="http://www.integers-ejcnt.org/l34/l34.pdf">Reducing the Erdos-Moser equation 1^n + 2^n + ... + k^n = (k+1)^n modulo k and k^2</a>, Integers 11 (2011), #A34.

%e a(1) = (1^3 + 2^3)/3 = (1 + 8)/3 = 3.

%t Array[Sum[m^#, {m, # - 1}]/# &@ Prime@ # &, 11, 2] (* _Michael De Vlieger_, Nov 04 2017 *)

%Y Cf. A055030, A294507.

%K nonn

%O 1,1

%A _Jonathan Sondow_, Dec 04 2012