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A255839 G.f. satisfies: A(x) = exp( Sum_{n>=1} [Sum_{k=0..3*n} binomial(3*n,k)^2 * x^k] / A(x)^n * x^n/n ).

%I A255839 #7 Mar 08 2015 21:06:19
%S A255839 1,1,9,18,64,172,477,1368,3681,10485,28701,80829,225090,632160,
%T A255839 1778553,5010948,14181849,40161357,114151716,324873027,926918784,
%U A255839 2649218580,7585705665,21758756931,62508649059,179859399129,518234494662,1495275239115,4319808231645,12495043092609,36183457564425
%N A255839 G.f. satisfies: A(x) = exp( Sum_{n>=1} [Sum_{k=0..3*n} binomial(3*n,k)^2 * x^k] / A(x)^n * x^n/n ).
%C A255839 Compare the definition of this sequence to G(x) = exp( Sum_{n>=1} [Sum_{k=0..2*n} binomial(2*n,k)^2 * x^k] / G(x)^n * x^n/n ), which is satisfied by the rational function: G(x) = (1+x^2)^2*(1+x^3)/((1-x)*(1-x^2)).
%H A255839 Paul D. Hanna, <a href="/A255839/b255839.txt">Table of n, a(n) for n = 0..100</a>
%e A255839 G.f.: A(x) = 1 + x + 9*x^2 + 18*x^3 + 64*x^4 + 172*x^5 + 477*x^6 +...
%e A255839 where
%e A255839 log(A(x)) = (1 + 3^2*x + 3^2*x^2 + x^3)/A(x) * x +
%e A255839 (1 + 6^2*x + 15^2*x^2 + 20^2*x^3 + 15^2*x^4 + 6^2*x^5 + x^6)/A(x)^2 * x^/2 +
%e A255839 (1 + 9^2*x + 36^2*x^2 + 84^2*x^3 + 126^2*x^4 + 126^2*x^5 + 84^2*x^6 + 36^2*x^7 + 9^2*x^8 + x^9)/A(x)^3 * x^3/3 +
%e A255839 (1 + 12^2*x + 66^2*x^2 + 220^2*x^3 + 495^2*x^4 + 792^2*x^5 + 924^2*x^6 + 792^2*x^7 + 495^2*x^8 + 220^2*x^9 + 66^2*x^10 + 12^2*x^11 + x^12)/A(x)^4 * x^4/4 +...
%e A255839 which involves the squares of the coefficients in (1 + x)^(3*n).
%o A255839 (PARI) /* By Definition: */
%o A255839 {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, sum(k=0, min(3*m,n-m), binomial(3*m,k)^2 * x^k) / (A +x*O(x^n))^m * x^m/m)+x*O(x^n))); polcoeff(A, n)}
%o A255839 for(n=0, 40, print1(a(n), ", "))
%Y A255839 Cf. A248876, A200537, A251687, A251688, A005721.
%K A255839 nonn
%O A255839 0,3
%A A255839 _Paul D. Hanna_, Mar 07 2015