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%I #44 Aug 02 2021 02:29:48
%S 0,1,9,98,1300,20515,376761,7907396,186884496,4914341925,142364319625,
%T 4505856912854,154718778284148,5729082486784839,227584583172284625,
%U 9654782997596059912,435659030617933827136,20836030169620907691465
%N a(n) = Sum_{k=1..n-1} k^n.
%C n^3 divides a(n) for n in A121707.
%C It appears that p^(3k-1) divides a(p^k) for all integer k > 1 and prime p > 2:
%C for prime p > 2, p^2 divides a(p), p^5 divides a(p^2) and p^8 divides a(p^3).
%C Additionally, p^3 divides a(3p) for prime p > 2.
%C For prime p > 3, p divides a(p+1) and p^3 divides a(2p+1);
%C for prime p > 5, p divides a(3p+1) and p^3 divides a(4p+1);
%C for prime p > 7, p divides a(5p+1) and p^3 divides a(6p+1):
%C 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.
%C p divides a((p+1)/2) for primes in A002145: primes of the form 4n+3, n >= 1.
%C p^2 divides a((p+1)/2) for primes in A007522: primes of the form 8n+7, n >= 0.
%C n*(2*n+1) divides a(2*n+1) for n >= 1. - _Franz Vrabec_, Dec 20 2020
%H Seiichi Manyama, <a href="/A121706/b121706.txt">Table of n, a(n) for n = 1..386</a>
%F a(n) = Sum(k^n, k=1..n) - n^n = A031971(n) - A000312(n) for n > 1.
%F a(n) = zeta(-n) - zeta(-n, n).
%p A121706 := proc(n)
%p (bernoulli(n+1,n)-bernoulli(n+1))/(n+1) ;
%p end proc: # _R. J. Mathar_, May 10 2013
%t Table[Sum[k^n,{k,1,n-1}],{n,1,35}]
%o (PARI) a(n)=sum(k=1,n-1,k^n) \\ _Charles R Greathouse IV_, May 09 2013
%o (PARI) a(n)=subst(sumformal('x^n),'x,n-1) \\ _Charles R Greathouse IV_, May 09 2013
%Y Cf. A121707, A031971, A000312, A002145, A007522.
%K nonn
%O 1,3
%A _Alexander Adamchuk_, Aug 16 2006
%E Edited by _M. F. Hasler_, Jul 22 2019
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