%I #13 Oct 23 2020 10:44:30
%S 1,8,156,4696,186406,9053640,515875660,33585910968,2453913830097,
%T 198609146859416,17630476159933080,1703025192274201272,
%U 177846105338917975896,19968484152350242447288
%N Self-convolution 8th power of A113674, where a(n) = A113674(n+1)/(n+1).
%C From _Vaclav Kotesovec_, Oct 23 2020: (Start)
%C 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 c(k) tends to A238223*exp(1) = 0.592451670452494179138706... if k tends to infinity.
%C (End)
%H Vaclav Kotesovec, <a href="/A113668/b113668.txt">Table of n, a(n) for n = 0..330</a>
%F G.f. satisfies: A(x) = (1 + x*d/dx[x*A(x)] )^8.
%F a(n) ~ c * 8^n * n! * n^(7/8), where c = 0.6523348263871879460325... - _Vaclav Kotesovec_, Oct 23 2020
%o (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)}
%Y Cf. A113674, A113662, A113663, A113664, A113665, A113666, A113667, A338377.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Nov 04 2005