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
MAPLE
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
KEYWORD
nonn
AUTHOR
Alexander Adamchuk, Aug 16 2006
EXTENSIONS
Edited by M. F. Hasler, Jul 22 2019
STATUS
approved