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%I #2 Mar 30 2012 18:37:20
%S 1,3,21,217,2895,46479,857670,17619348,394066449,9445681950,
%T 239946999264,6407385578778,178774882463450,5188026867995184,
%U 156036783823130184,4850255971984578744,155467140310522090338
%N G.f. satisfies: A(x/A(x)) = G(x)^3 where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.
%F G.f. satisfies: A(x) = [1 + A(x)*Series_Reversion(x/A(x))]^3.
%F G.f. satisfies: A( (x*(1-x)^2)/A(x*(1-x)^2) ) = 1/(1-x)^3.
%F G.f. satisfies: A( (x/(1+x)^3)/A(x/(1+x)^3) ) = (1 + x)^3.
%F Self-convolution cube of A168478.
%e G.f.: A(x) = 1 + 3*x + 21*x^2 + 217*x^3 + 2895*x^4 + 46479*x^5 +...
%e A(x/A(x)) = 1 + 3*x + 12*x^2 + 55*x^3 + 273*x^4 + 1428*x^5 +...+ A001764(n+1)*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^3, x, serreverse(x/(A+x*O(x^n))))); polcoeff(A, n)}
%o (PARI) {a(n)=local(A=1+x);for(i=1,n,A=(1+A*serreverse(x/(A+x*O(x^n))))^3); polcoeff(A,n)}
%Y Cf. A168478, A168449 (variant), A001764.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Dec 06 2009
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