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%I #32 Nov 08 2021 09:14:14
%S 0,1,6,46,470,6035,93436,1695036,35277012,828707925,21693441550,
%T 626254969978,19766667410282,677231901484775,25031756512858200,
%U 992872579254244088,42066929594261568840,1896157095455962952169,90601933352843530354170,4574495282686422755339734,243359175218492577008763726
%N a(n) = sum of the sums of the k first n-th powers.
%C First term a(0) may be computed as 1 by starting the inner sum at j=0 and taking the convention 0^0 = 1.
%H Vincenzo Librandi, <a href="/A215084/b215084.txt">Table of n, a(n) for n = 0..200</a>
%F a(n) = Sum_{k=1..n} Sum_{j=1..k} j^n.
%F a(n) = Sum_{k=1..n} H_k^{-n} where H_k^{-n} is the k-th harmonic number of order -n.
%F a(n) = Sum_{k=1..n} (B(n+1, k+1) - B(n+1, 1))/(n+1), where B(n, x) are the Bernoulli polynomials. - _Daniel Suteu_, Jun 25 2018
%F G.f.: Sum_{k>=1} k^k*x^k/(1 - k*x)^2. - _Ilya Gutkovskiy_, Oct 11 2018
%F a(n) ~ c * n^n, where c = 1/(1 - 2*exp(-1) + exp(-2)) = 2.50265030107711874333... - _Vaclav Kotesovec_, Nov 06 2021
%e a(3) = (1^3) + (1^3 + 2^3) + (1^3 + 2^3 + 3^3) = (1^3 + 1^3 + 1^3) + (2^3 + 2^3) + (3^3) = 3 * 1^3 + 2 * 2^3 + 1 * 3^3 = 46. - _David A. Corneth_, Jun 27 2018
%t Table[Sum[Sum[j^n, {j, 1, k}], {k, 0, n}], {n, 0, 20}]
%t a[n_] := (n+1)*HarmonicNumber[-1, -n] - HarmonicNumber[n, -n-1] + (n+1)*HarmonicNumber[n, -n]; Table[a[n], {n, 0, 20}] (* _Jean-François Alcover_, Mar 05 2013 *)
%t Table[Total[Accumulate[Range[n]^n]],{n,0,20}] (* _Harvey P. Dale_, Mar 29 2020 *)
%o (PARI) a(n) = sum(k=1, n, sum(j=1, k, j^n)); \\ _Michel Marcus_, Jun 25 2018
%o (PARI) a(n) = sum(i=1, n, (n+1-i) * i^n); \\ _David A. Corneth_, Jun 27 2018
%Y Row sums of A215083.
%Y Cf. A086787, A349116.
%K nonn
%O 0,3
%A _Olivier Gérard_, Aug 02 2012
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