%I #5 Jul 05 2015 17:19:24
%S 1,1,4,37,586,13612,424621,16827976,815866699,47093387797,
%T 3170897237125,245127016240321,21482473673228266,2112385883734692910,
%U 231062843227493844112,27913223028923592662539,3701041353685453743060265,535729316331363978105167557,84263588534262286958390813305
%N E.g.f. A(x) satisfies: A( Integral 1/A(x)^3 dx ) = exp(x).
%F E.g.f. satisfies: A(x) = exp( Series_Reversion( Integral 1/A(x)^3 dx ) ).
%F E.g.f. A(x) such that A(x/3)^3 is the e.g.f. of A233335.
%e E.g.f.: A(x) = 1 + x + 4*x^2/2! + 37*x^3/3! + 586*x^4/4! + 13612*x^5/5! + 424621*x^6/6! +...
%e where log(A(x)) = Series_Reversion( Integral 1/A(x)^3 dx ):
%e log(A(x)) = x + 3*x^2/2! + 27*x^3/3! + 432*x^4/4! + 10206*x^5/5! + 323919*x^6/6! +...+ 3^(n-1)*A214645(n)*x^n/n! +...
%e and
%e A(x/3)^3 = 1 + x + 2*x^2/2! + 7*x^3/3! + 38*x^4/4! + 292*x^5/5! + 2975*x^6/6! +...+ A233335(n)*x^n/n! +...
%o (PARI) {a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=exp(serreverse(intformal(1/A^3+x*O(x^n))))); n!*polcoeff(A, n)}
%o for(n=0, 30, print1(a(n), ", "))
%Y Cf. A233335, A214645.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jul 05 2015