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A303291 G.f. A(x) satisfies: 3 = Sum_{n>=0} (2/3)^n * (1 + x)^(n^2) / A(x)^n.

%I A303291 #14 Oct 14 2020 11:08:15
%S A303291 1,5,70,3170,252160,27705800,3806286820,621124623740,116766042046000,
%T A303291 24783363325335440,5854493683431121840,1522701357625214096240,
%U A303291 432347094526718807347480,133078785461406479045306360,44145742694332046133435657280,15702781293109570148744738306240,5962874290966165187708554294296880,2407878412120285331813837276575565360
%N A303291 G.f. A(x) satisfies: 3 = Sum_{n>=0} (2/3)^n * (1 + x)^(n^2) / A(x)^n.
%H A303291 Paul D. Hanna, <a href="/A303291/b303291.txt">Table of n, a(n) for n = 0..200</a>
%F A303291 G.f.: 3 = 1/(1 - q/(3/2*A(x) - q*(q^2-1)/(1 - q^5/(3/2*A(x) - q^3*(q^4-1)/(1 - q^9/(3/2*A(x) - q^5*(q^6-1)/(1 - q^13/(3/2*A(x) - q^7*(q^8-1)/(1 - ...))))))))), where q = (1+x), a continued fraction due to a partial elliptic theta function identity.
%F A303291 G.f.: 3 = Sum_{n>=0} (2/3)^n * (1+x)^n / A(x)^n * Product_{k=1..n} (3*A(x) - 2*(1+x)^(4*k-3)) / (3*A(x) - 2*(1+x)^(4*k-1)), due to a q-series identity.
%F A303291 a(n) ~ 2^(2*n + 2 + log(3/2)/8) * n^n / (3^(log(3/2)/8 + 7/2) * exp(n) * log(3/2)^(2*n + 1)). - _Vaclav Kotesovec_, Oct 14 2020
%e A303291 G.f.: A(x) = 1 + 5*x + 70*x^2 + 3170*x^3 + 252160*x^4 + 27705800*x^5 + 3806286820*x^6 + 621124623740*x^7 + 116766042046000*x^8 + ...
%e A303291 such that A = A(x) satisfies:
%e A303291 3 = 1 + (1+x)/(3*A/2) + (1+x)^4/(3*A/2)^2 + (1+x)^9/(3*A/2)^3 + (1+x)^16/(3*A/2)^4 + (1+x)^25/(3*A/2)^5 + (1+x)^36/(3*A/2)^6 + (1+x)^49/(3*A/2)^7 + ...
%o A303291 (PARI) /* Find A(x) that satisfies the continued fraction: */
%o A303291 {a(n) = my(A=[1], q=1+x, CF=1); for(i=1, n, A=concat(A, 0); m=#A; for(k=0, m, CF = 1/(1 - q^(4*m-4*k+1)/(3/2*Ser(A) - q^(2*m-2*k+1)*(q^(2*m-2*k+2) - 1)*CF)) ); A[#A] = Vec(CF)[#A]/6 ); A[n+1]}
%o A303291 for(n=0, 30, print1(a(n), ", "))
%Y A303291 Cf. A303290, A303292.
%K A303291 nonn
%O A303291 0,2
%A A303291 _Paul D. Hanna_, Apr 22 2018