A349116 a(n) = Sum_{m=1..n} (Sum_{k=1..m} (Sum_{j=1..k} j^n)).

1, 7, 57, 605, 7990, 126378, 2331462, 49183590, 1168343775, 30871159429, 898302280271, 28547663567787, 983905660878140, 36556999252315556, 1456715662183096092, 61973464872507939132, 2803744852868605501965, 134413716665674685292183, 6807005707226954237422477

Offset

1,2

Links

Vaclav Kotesovec, Table of n, a(n) for n = 1..200

Hjalmar Rosengren, Identity with binomial coefficients and Stirling numbers of the second kind, answer to question on MathOverflow, 2026.

Formula

a(n) ~ c * n^n, where c = 1/(1 - 3/exp(1) + 3/exp(2) - 1/exp(3)) = 3.959134481...

a(n) = Sum_{j=1..n} binomial(n-j+2,2)*j^n (due to Hjalmar Rosengren). - Mikhail Kurkov, Mar 22 2026

Mathematica

Table[Sum[Sum[Sum[j^n, {j, 1, k}], {k, 1, m}], {m, 1, n}], {n, 1, 20}]
Table[(Zeta[-2-n] - HurwitzZeta[-2-n, 1+n] + (3+2*n)*(HurwitzZeta[-1-n, 1+n] - Zeta[-1-n]) + (1+n)*(2+n)*(Zeta[-n] - HurwitzZeta[-n, 1+n]))/2, {n, 1, 20}] (* Vaclav Kotesovec, Mar 23 2026 *)

Prog

(PARI) a(n) = sum(j=1, n, binomial(n-j+2, 2)*j^n) \\ Mikhail Kurkov, Mar 22 2026

Crossrefs

Cf. A215084, A349117.

Sequence in context: A122649 A051846 A231540 * A337556 A390063 A337022

Adjacent sequences: A349113 A349114 A349115 * A349117 A349118 A349119

Keyword

nonn

Author

Vaclav Kotesovec, Nov 08 2021

Status

approved