cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

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

Views

Author

Alexander Adamchuk, Aug 16 2006

Keywords

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

Crossrefs

Programs

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).

Extensions

Edited by M. F. Hasler, Jul 22 2019