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A113668
Self-convolution 8th power of A113674, where a(n) = A113674(n+1)/(n+1).
9
1, 8, 156, 4696, 186406, 9053640, 515875660, 33585910968, 2453913830097, 198609146859416, 17630476159933080, 1703025192274201272, 177846105338917975896, 19968484152350242447288
OFFSET
0,2
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
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
PROG
(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)}
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Nov 04 2005
STATUS
approved