A338472 (1 + Sum_{k(even)=2..p-1} 2*k^(p-1))/p as p runs through the odd primes.

3, 109, 14519, 2024592291, 1536463613637, 2449395996564189425, 4686662617019462175259, 33724155827962966577589860263, 2606282943971359343146382147809434583605, 15159042500551578738018590862773479717960671, 6576976543997974825092367662248938303820921894460988333

Offset: 1

Author

Davide Rotondo, Oct 29 2020

Keywords

• nonn

Comments

Conjecture: (1 + Sum_{k(even)=2..p-1} 2*k^(p-1))/p is an integer iff p is an odd prime.

Crossrefs

Cf. A055030.

Programs

Mathematica

a[n_] := Module[{p = Prime[n + 1]}, (1 + 2 * Sum[k^(p - 1), {k, 2, p - 1, 2}])/p]; Array[a, 11] (* Amiram Eldar, Oct 29 2020 *)

PARI

a(n) = my(p=prime(n+1)); (1 + sum(k=1, (p-1)\2, 2*(2*k)^(p-1)))/p; \\ Michel Marcus, Oct 29 2020