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G.f.: A(x) = exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^3 * x^k / A(x)^k] * x^n/n ).
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%I #5 Mar 30 2012 18:37:32

%S 1,1,2,5,10,22,58,150,392,1097,3139,9069,26903,81299,248305,768521,

%T 2407340,7607947,24248690,77906841,251995121,820096599,2684160567,

%U 8830103123,29183369411,96865043941,322780531149,1079491353973,3622338207474,12193038599714,41161594789286

%N G.f.: A(x) = exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^3 * x^k / A(x)^k] * x^n/n ).

%C Compare to g.f. G(x) = (1+x^2)/(1-x-x^3) that satisfies:

%C G(x) = exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^2 * x^k / G(x)^k] * x^n/n ).

%e G.f.: A(x) = 1 + x + 2*x^2 + 5*x^3 + 10*x^4 + 22*x^5 + 58*x^6 +...

%e where

%e log(A(x)) = (1 + x/A(x))*x + (1 + 2^3*x/A(x) + x^2/A(x)^2)*x^2/2 +

%e (1 + 3^3*x/A(x) + 3^3*x^2/A(x)^2 + x^3/A(x)^3)*x^3/3 +

%e (1 + 4^3*x/A(x) + 6^3*x^2/A(x)^2 + 4^3*x^3/A(x)^3 + x^4/A(x)^4)*x^4/4 +

%e (1 + 5^3*x/A(x) + 10^3*x^2/A(x)^2 + 10^3*x^3/A(x)^3 + 5^3*x^4/A(x)^4 + x^5/A(x)^5)*x^5/5 +...

%e more explicitly,

%e log(A(x)) = x + 3*x^2/2 + 10*x^3/3 + 19*x^4/4 + 46*x^5/5 + 162*x^6/6 + 477*x^7/7 + 1371*x^8/8 +...

%o (PARI) {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^3*x^j/(A+x*O(x^n))^j)*x^m/m))); polcoeff(A, n, x)}

%K nonn

%O 0,3

%A _Paul D. Hanna_, Oct 31 2011