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A121706 a(n) = Sum_{k=1..n-1} k^n. 5

%I

%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 = {35, 55, 77, 95, 115, 119, 143, 155, 161,...} = 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 prime p = {7, 11, 19, 23, 31, 43, 47, 59, 67, 71,...} = A002145: primes of the form 4n+3, n >= 1.

%C p^2 divides a((p+1)/2) for prime p = {7, 23, 31, 47, 71, 79, 103, 127,...} = A007522: primes of the form 8n+7, n >= 0.

%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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Last modified February 25 10:39 EST 2020. Contains 332228 sequences. (Running on oeis4.)