A219550 Sum(m^p, m=1..p-1)/p as p runs through the odd primes.

3, 260, 53823, 12942210875, 11901444483396, 25627001801054931008, 55413915436873048932459, 490667517005738962388828685983, 48588952813858892791005036793649985985124, 303307728036900627681487165427498812641117375, 158544898951978777519612048992784361843596346824881328548

Offset: 1

Author

Jonathan Sondow, Dec 04 2012

Keywords

• nonn

Comments

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.

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

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

Examples

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

Links

• K. MacMillan and J. Sondow, Proofs of power sum and binomial coefficient congruences via Pascal's identity, Amer. Math. Monthly, 118 (2011), 549-551.
• J. Sondow and K. MacMillan, Reducing the Erdos-Moser equation 1^n + 2^n + ... + k^n = (k+1)^n modulo k and k^2, Integers 11 (2011), #A34.

Crossrefs

Cf. A055030, A294507.

Programs

Mathematica

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