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%I #34 Jul 30 2023 08:46:13
%S 1,3,15,225,6003,223029,10403175,577700889,37009173207,2679339499305,
%T 216031850406327,19187294118006057,1861057604220294591,
%U 195742656849628038465,22192660352433291780159,2698458809215198981964481,350326879575505922875480047,48370384900519379918253881361,7078145146554395463373624118319,1094300840117324691452685873392145
%N G.f. A(x) satisfies: 2 = Sum_{n>=0} (1/2^n) * (1+x)^(n^2) / A(x)^n.
%H Paul D. Hanna, <a href="/A303290/b303290.txt">Table of n, a(n) for n = 0..200</a>
%F G.f.: 2 = 1/(1 - q/(2*A(x) - q*(q^2-1)/(1 - q^5/(2*A(x) - q^3*(q^4-1)/(1 - q^9/(2*A(x) - q^5*(q^6-1)/(1 - q^13/(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 G.f.: 2 = Sum_{n>=0} (1+x)^n/(2^n*A(x)^n) * Product_{k=1..n} (2*A(x) - (1+x)^(4*k-3)) / (2*A(x) - (1+x)^(4*k-1)), due to a q-series identity.
%F a(n) ~ c * 2^(2*n) * n^n / (exp(n) * log(2)^(2*n)), where c = 0.339650521725496... - _Vaclav Kotesovec_, Oct 06 2020
%e G.f.: A(x) = 1 + 3*x + 15*x^2 + 225*x^3 + 6003*x^4 + 223029*x^5 + 10403175*x^6 + 577700889*x^7 + 37009173207*x^8 + 2679339499305*x^9 + 216031850406327*x^10 + ...
%e such that A = A(x) satisfies:
%e 2 = 1 + (1+x)/(2*A) + (1+x)^4/(2*A)^2 + (1+x)^9/(2*A)^3 + (1+x)^16/(2*A)^4 + (1+x)^25/(2*A)^5 + (1+x)^36/(2*A)^6 + (1+x)^49/(2*A)^7 + (1+x)^64/(2*A)^8 + ...
%o (PARI) /* Find A(x) that satisfies the continued fraction: */
%o {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)/(2*Ser(A) - q^(2*m-2*k+1)*(q^(2*m-2*k+2) - 1)*CF)) ); A[#A] = Vec(CF)[#A]/2 );A[n+1]}
%o for(n=0,30,print1(a(n),", "))
%Y Cf. A303291, A303292, A303058.
%Y Cf. A325286.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Apr 20 2018
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