A292716 a(n) = [x^n] 1/(1 - n*x - n*x^2/(1 - n*x - n*x^2/(1 - n*x - n*x^2/(1 - n*x - n*x^2/(1 - ...))))), a continued fraction.

1, 1, 6, 54, 672, 10625, 203256, 4554697, 116842496, 3373056027, 108134200000, 3809118341028, 146170521796608, 6066719073261639, 270692733123460480, 12917478278285156250, 656311833287586742272, 35364920064570086779227, 2014028255250518880457728, 120852950097737555898105210

Offset: 0

Author

Ilya Gutkovskiy, Sep 21 2017

Keywords

• nonn

Comments

Also coefficient of x^n in the expansion of 1/(n+1) * (1 + n*x + n*x^2)^(n+1). - Seiichi Manyama, May 06 2019

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..381

Crossrefs

Main diagonal of A107267.

Cf. A001006, A247496.

Programs

Mathematica

Table[SeriesCoefficient[1/(1 - n x + ContinuedFractionK[-n x^2, 1 - n x, {i, 1, n}]), {x, 0, n}], {n, 0, 19}]
Table[SeriesCoefficient[2/(1 - n x + Sqrt[1 + n x ((n - 4) x - 2)]), {x, 0, n}], {n, 0, 19}]
Table[n! SeriesCoefficient[E^(n x) Hypergeometric0F1Regularized[2, n x^2], {x, 0, n}], {n, 0, 19}]
Flatten[{1, Table[Sum[Binomial[k+1, n-k+1] * Binomial[n, k] * n^k / (k+1), {k, 0, n}], {n, 1, 20}]}] (* Vaclav Kotesovec, May 05 2019 *)

PARI

{a(n) = polcoef((1+n*x+n*x^2)^(n+1)/(n+1), n)} \\ Seiichi Manyama, May 06 2019

Formula

a(n) = [x^n] 2/(1 - n*x + sqrt(1 + n*x*((n - 4)*x - 2))).

a(n) = n! * [x^n] exp(n*x)*BesselI(1,2*sqrt(n)*x)/(sqrt(n)*x), for n > 0.

a(n) = A107267(n,n).

a(n) ~ exp(2*sqrt(n) - 2) * n^(n - 3/4) / (2*sqrt(Pi)). - Vaclav Kotesovec, May 05 2019