A121706 a(n) = Sum_{k=1..n-1} k^n.

0, 1, 9, 98, 1300, 20515, 376761, 7907396, 186884496, 4914341925, 142364319625, 4505856912854, 154718778284148, 5729082486784839, 227584583172284625, 9654782997596059912, 435659030617933827136, 20836030169620907691465

Offset

1,3

Comments

n^3 divides a(n) for n in A121707.

It appears that p^(3k-1) divides a(p^k) for all integer k > 1 and prime p > 2:

for prime p > 2, p^2 divides a(p), p^5 divides a(p^2) and p^8 divides a(p^3).

Additionally, p^3 divides a(3p) for prime p > 2.

For prime p > 3, p divides a(p+1) and p^3 divides a(2p+1);

for prime p > 5, p divides a(3p+1) and p^3 divides a(4p+1);

for prime p > 7, p divides a(5p+1) and p^3 divides a(6p+1):

It appears that p divides a((2k+1)p+1) for integer k >= 0 and prime p > 2k+3, and p^3 divides a(2kp+1) for integer k > 0 and prime p > 2k+2.

p divides a((p+1)/2) for primes in A002145: primes of the form 4n+3, n >= 1.

p^2 divides a((p+1)/2) for primes in A007522: primes of the form 8n+7, n >= 0.

n*(2*n+1) divides a(2*n+1) for n >= 1. - Franz Vrabec, Dec 20 2020

Links

Seiichi Manyama, Table of n, a(n) for n = 1..386

Formula

a(n) = Sum(k^n, k=1..n) - n^n = A031971(n) - A000312(n) for n > 1.

a(n) = zeta(-n) - zeta(-n, n).

Maple

A121706 := proc(n)
    (bernoulli(n+1, n)-bernoulli(n+1))/(n+1) ;
end proc: # R. J. Mathar, May 10 2013

Mathematica

Table[Sum[k^n, {k, 1, n-1}], {n, 1, 35}]

Prog

(PARI) a(n)=sum(k=1, n-1, k^n) \\ Charles R Greathouse IV, May 09 2013
(PARI) a(n)=subst(sumformal('x^n), 'x, n-1) \\ Charles R Greathouse IV, May 09 2013

Crossrefs

Cf. A121707, A031971, A000312, A002145, A007522.

Sequence in context: A024115 A066557 A289214 * A306567 A061817 A083835

Adjacent sequences: A121703 A121704 A121705 * A121707 A121708 A121709

Keyword

nonn

Author

Alexander Adamchuk, Aug 16 2006

Extensions

Edited by M. F. Hasler, Jul 22 2019

Status

approved