0,3

Compare to the identities:

(1) F(x) = exp( Integral 1 + x*F(x) dx ) when F(x) = 1/(1-x),

(2) G(x) = exp( Integral x*G(x)^3 dx ) when G(x) = 1/(1-3*x^2/2)^(1/3).

In general, if e.g.f. satisfies: A(x) = exp( Integral(1 + x*A(x)^p) dx ), p>1, and the constant of integration is zero, then A(x) = (1/p + (p-1)/(exp(p*x)*p) - x)^(-1/p), and a(n) ~ n! * p^(n+1/p) / (Gamma(1/p) * n^(1-1/p)* (1+LambertW((p-1)*exp(-1)))^(n+2/p)). - Vaclav Kotesovec, Jul 16 2014

E.g.f.: 3^(1/3)*exp(x)/(exp(3*x) - 3*exp(3*x)*x + 2)^(1/3). - Vaclav Kotesovec, Jan 05 2014

a(n) ~ 3^(n+5/6) * n^(n-1/6) * Gamma(2/3) / (sqrt(2*Pi) * exp(n) * (1+LambertW(2*exp(-1)))^(n+2/3)). - Vaclav Kotesovec, Jan 05 2014

E.g.f.: A(x) = 1 + x + 2*x^2/2! + 10*x^3/3! + 70*x^4/4! + 614*x^5/5! + 6694*x^6/6! + ...

such that, by definition,

log(A(x)) = x + x^2/2! + 6*x^3/3! + 36*x^4/4! + 288*x^5/5! + 2970*x^6/6! + 36612*x^7/7! + ...

Related expansions:

x*A(x)^3 = x + 6*x^2/2! + 36*x^3/3! + 288*x^4/4! + 2970*x^5/5! + 36612*x^6/6! + ...

A(x)^3 = 1 + 3*x + 12*x^2/2! + 72*x^3/3! + 594*x^4/4! + 6102*x^5/5! + 75006*x^6/6! + ...

CoefficientList[Series[3^(1/3)*E^x/(E^(3*x) - 3*E^(3*x)*x + 2)^(1/3), {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Jan 05 2014 *)

(PARI) {a(n)=local(A=1+x); for(i=1, n, A=exp(intformal(1+x*A^3)+x*O(x^n))); n!*polcoeff(A, n)}

for(n=0, 30, print1(a(n), ", "))

Cf. A212913, A245266, A245267.

Sequence in context: A051405 A005568 A036075 * A123881 A289680 A089845

Adjacent sequences: A212911 A212912 A212913 * A212915 A212916 A212917

Paul D. Hanna, May 30 2012

approved