A249489 a(n) = [x^n/n!] Sum_{k=0..n} cosh(k*x)^k.

1, 0, 9, 0, 12070, 0, 126447741, 0, 5100496997940, 0, 562605048135059545, 0, 138523311740417986721274, 0, 66543520389763227261554370645, 0, 56664734898911130799849838608991176, 0, 79610326854782816434044397510470501877041

Offset

0,3

Links

Vaclav Kotesovec, Table of n, a(n) for n = 0..220

Formula

a(n) = Sum_{k=0..n} Sum_{j=0..k} binomial(k, j) * k^n * (k-2*j)^n / 2^k. [Explicitly stating Vaclav's formula in Mma program - Paul D. Hanna, Oct 15 2018]

If n is even, then a(n) ~ c * (1-2*r)^n * n^(2*n) / (2^n * exp(n) * (r*(1-r))^(n/2)), where r = 0.0832217201995176507819192648878903254298041... is the root of the equation (r/(1-r))^(1-2*r) = exp(-2), and c = 2.09233700490262732901066903251002074102409436600891921766318742438...

Mathematica

Flatten[{1, Table[Sum[Sum[Binomial[k, j] * k^n*(k-2*j)^n / 2^k, {j, 0, k}], {k, 0, n}], {n, 1, 20}]}]

Prog

(PARI) {a(n)=n!*polcoeff(sum(k=0, n, cosh(k*x+x*O(x^n))^k), n)}
for(n=0, 20, print1(a(n), ", "))
(PARI) {a(n) = sum(k=0, n, sum(j=0, k, binomial(k, j) * k^n*(k-2*j)^n / 2^k ))}
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Oct 15 2018, using Vaclav's formula.

Crossrefs

Cf. A224899, A249459, A245322, A221077, A221078.

Sequence in context: A191564 A067479 A308803 * A166734 A155783 A257097

Adjacent sequences: A249486 A249487 A249488 * A249490 A249491 A249492

Keyword

nonn

Author

Vaclav Kotesovec, Oct 30 2014

Extensions

Name clarified by Paul D. Hanna, Oct 15 2018

Status

approved