A242006 G.f. satisfies: 2*A(x) = 1 + x + A(x*A(x)^4).

1, 1, 4, 54, 1156, 32917, 1149264, 47083228, 2203193792, 115647869941, 6721947019280, 428408002112146, 29705584153315352, 2226637865972203345, 179445974881472237440, 15475783832452270534780, 1422388135341144845327744, 138817119057328298887651613

Offset

0,3

Comments

In general, if g.f. satisfies: 2*A(x) = 1 + x + A(x*A(x)^q), then a(n) ~ c(q) * q^n * n^(n + (q-1)/(2*q)*log(2)) / (exp(n) * (log(2))^n), where c(q) is a constant independent on n.

Links

Vaclav Kotesovec, Table of n, a(n) for n = 0..310

Formula

a(n) ~ c * 4^n * n^(n + 3/8*log(2)) / (exp(n) * (log(2))^n), where c = 0.25046401546046145...

Mathematica

nmax = 17; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[2 A[x] - (1 + x + A[x A[x]^4]) + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
sol /. Rule -> Set;
a /@ Range[0, nmax] (* Jean-François Alcover, Nov 01 2019 *)

Prog

(PARI) {a(n)=local(A=1+x); for(i=1, n, A = 1+x + subst(A, x, x*A^4 +x*O(x^n)) - A); polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))

Crossrefs

Cf. A242003 (q=1), A242004 (q=2), A242005 (q=3).

Sequence in context: A346291 A258846 A093983 * A259063 A138459 A111161

Adjacent sequences: A242003 A242004 A242005 * A242007 A242008 A242009

Keyword

nonn

Author

Vaclav Kotesovec, Aug 11 2014

Extensions

Name corrected by Vaclav Kotesovec and Paul D. Hanna, Aug 15 2014

Status

approved