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%I #2 Mar 30 2012 18:37:20
%S 1,1,6,60,803,13071,244917,5101603,115451307,2794682082,71579132742,
%T 1924722618873,54022011952266,1575777019075715,47606721776494443,
%U 1485688929610479498,47790055655273649449,1581727833458617151379
%N G.f. satisfies: A(x/A(x)^3) = G(x) where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.
%F G.f. satisfies: A(x) = 1 + A(x)^3*Series_Reversion[x/A(x)^3].
%F G.f. satisfies: A( (x*(1-x)^2)/A(x*(1-x)^2)^3 ) = 1/(1-x).
%F G.f. satisfies: A( (x/(1+x)^3)/A(x/(1+x)^3)^3 ) = 1 + x.
%e G.f.: A(x) = 1 + x + 6*x^2 + 60*x^3 + 803*x^4 + 13071*x^5 +...
%e A(x/A(x)^3) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 +...+ A001764(n)*x^n +...
%o (PARI) {a(n)=local(A=1+x, F=sum(k=0, n, binomial(3*k+1, k)/(3*k+1)*x^k)+x*O(x^n)); for(i=0, n, A=subst(F, x, serreverse(x/(A+x*O(x^n))^3))); polcoeff(A, n)}
%o (PARI) {a(n)=local(A=1+x);for(i=1,n,A=1+A^3*serreverse(x/(A+x*O(x^n))^3)); polcoeff(A,n)}
%Y Cf. A168479 (cube), A168448 (variant), A001764.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Dec 06 2009
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