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A199257 G.f. satisfies: A(x) = exp( Sum_{n>=1} [Sum_{k=0..2*n} C(2*n,k)^2 * x^k * A(x)^k]* x^n/n ).

%I A199257 #8 Mar 30 2012 18:37:32
%S A199257 1,1,5,18,86,408,2075,10787,57655,313643,1733450,9700574,54867895,
%T A199257 313145033,1801150861,10430094658,60758092753,355795743385,
%U A199257 2093295146379,12367548160650,73346850194969,436486017193373,2605656191324094,15599323024019360,93634195155551584
%N A199257 G.f. satisfies: A(x) = exp( Sum_{n>=1} [Sum_{k=0..2*n} C(2*n,k)^2 * x^k * A(x)^k]* x^n/n ).
%F A199257 G.f. satisfies: A(x) = G(x*A(x)) where A(x/G(x)) = G(x) = (1-x+x^2)*(1+x^2)^2/(1-x)^2.
%F A199257 G.f.: A(x) = (1/x)*Series_Reversion( x*(1-x)^2/((1-x+x^2)*(1+x^2)^2) ).
%e A199257 G.f.: A(x) = 1 + x + 5*x^2 + 18*x^3 + 86*x^4 + 408*x^5 + 2075*x^6 +...
%e A199257 such that A(x) = G(x*A(x)) where G(x) = (1-x+x^2)*(1+x^2)^2/(1-x)^2:
%e A199257 G(x) = 1 + x + 4*x^2 + 5*x^3 + 9*x^4 + 12*x^5 + 16*x^6 + 20*x^7 + 24*x^8 +...
%e A199257 ...
%e A199257 Let A = x*A(x), then the logarithm of the g.f. A(x) equals the series:
%e A199257 log(A(x)) = (1 + 2^2*A + A^2)*x +
%e A199257 (1 + 4^2*A + 6^2*A^2 + 4^2*A^3 + A^4)*x^2/2 +
%e A199257 (1 + 6^2*A + 15^2*A^2 + 20^2*A^3 + 15^2*A^4 + 6^2*A^5 + A^6)*x^3/3 +
%e A199257 (1 + 8^2*A + 28^2*A^2 + 56^2*A^3 + 70^2*A^4 + 56^2*A^5 + 28^2*A^6 + 8^2*A^7 + A^8)*x^4/4 +
%e A199257 (1 + 10^2*A + 45^2*A^2 + 120^2*A^3 + 210^2*A^4 + 252^2*A^5 + 210^2*A^6 + 120^2*A^7 + 45^2*A^8 + 10^2*A^9 + A^10)*x^5/5 +...
%e A199257 which involves the squares of binomial coefficients C(2*n,k).
%o A199257 (PARI) {a(n)=local(A=1+x); A=1/x*serreverse(x*(1-x)^2/((1-x+x^2)*(1+x^2)^2+x*O(x^n))); polcoeff(A, n)}
%o A199257 (PARI) {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, sum(k=0, 2*m, binomial(2*m, k)^2 *x^k*A^k) *x^m/m)+x*O(x^n))); polcoeff(A, n)}
%Y A199257 Cf. A197601, A199248.
%K A199257 nonn
%O A199257 0,3
%A A199257 _Paul D. Hanna_, Nov 04 2011