A113668 Self-convolution 8th power of A113674, where a(n) = A113674(n+1)/(n+1).

1, 8, 156, 4696, 186406, 9053640, 515875660, 33585910968, 2453913830097, 198609146859416, 17630476159933080, 1703025192274201272, 177846105338917975896, 19968484152350242447288

Offset: 0

Author

Paul D. Hanna, Nov 04 2005

Keywords

• nonn

Comments

From Vaclav Kotesovec, Oct 23 2020: (Start)

In general, for k>=1, if g.f. satisfies: A(x) = (1 + x*d/dx[x*A(x)] )^k, then a(n) ~ c(k) * k^n * n! * n^((k-1)/k), where c(k) is a constant (dependent only on k).

c(k) tends to A238223*exp(1) = 0.592451670452494179138706... if k tends to infinity.

(End)

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..330

Crossrefs

Cf. A113674, A113662, A113663, A113664, A113665, A113666, A113667, A338377.

Programs

PARI

{a(n)=local(A=1+x*O(x^n));for(i=1,n, A=(1+x*deriv(x*A))^8);polcoeff(A,n,x)}

Formula

G.f. satisfies: A(x) = (1 + x*d/dx[x*A(x)] )^8.

a(n) ~ c * 8^n * n! * n^(7/8), where c = 0.6523348263871879460325... - Vaclav Kotesovec, Oct 23 2020