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G.f. satisfies: A(x) = theta_3( x*A(x) )^2, where theta_3(x) is Jacobi's theta_3 function.
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%I #27 Nov 16 2023 11:04:55

%S 1,4,20,112,676,4312,28704,197600,1397060,10090676,74152456,552666448,

%T 4167528000,31736182776,243698432960,1884809367456,14668777816708,

%U 114789815231560,902661488046900,7129068237647408,56524456978032904,449752267499647104

%N G.f. satisfies: A(x) = theta_3( x*A(x) )^2, where theta_3(x) is Jacobi's theta_3 function.

%F G.f. A(x) satisfies:

%F (1) A(x) = (1 + 2*Sum_{n>=1} (x*A(x))^(n^2) )^2.

%F (2) A(x) = 1 + 4*Sum_{n>=1} (x*A(x))^n / (1 + (x*A(x))^(2*n)).

%F (3) A(x) = Product_{n>=1} (1 - (-x)^n*A(x)^n)^2 / (1 + (-x)^n*A(x)^n)^2.

%F (4) A( x/theta_3(x)^2 ) = theta_3(x)^2.

%F (5) A(x) = (1/x)*Series_Reversion(x/theta_3(x)^2), where theta_3(x) = 1 + 2*Sum_{n>=1} x^(n^2).

%F a(n) = [x^n] theta_3(x)^(2*n+2) / (n+1).

%F a(n) ~ c * d^n / n^(3/2), where d = 8.54148362320612002563896433934021488424489314523756456892173912667254... and c = 1.2437677914754786190190604348779334425700766084860016245397106832001... - _Vaclav Kotesovec_, Nov 16 2023

%e G.f.: A(x) = 1 + 4*x + 20*x^2 + 112*x^3 + 676*x^4 + 4312*x^5 + 28704*x^6 +...

%e Given g.f. A(x), let q = x*A(x), then by a q-series identity:

%e A(x) = 1 + 4*q/(1+q^2) + 4*q^2/(1+q^4) + 4*q^3/(1+q^6) + 4*q^4/(1+q^8) +...

%e A(x) = (1 + 2*q + 2*q^4 + 2*q^9 + 2*q^16 + 2*q^25 +...)^2.

%e ...

%e Illustrate a(n) = [x^n] theta_3(x)^(2*n+2) / (n+1) by the following table of coefficients in powers theta_3(x)^(2*n+2) for n>=0:

%e n=0: [(1), 4, 4, 0, 4, 8, 0, 0, 4, 4, 8, 0, 0, 8, 0, 0,...];

%e n=1: [1, (8), 24, 32, 24, 48, 96, 64, 24, 104, 144, 96, 96, 112,...];

%e n=2: [1, 12, (60), 160, 252, 312, 544, 960, 1020, 876, 1560, 2400,...];

%e n=3: [1, 16, 112, (448), 1136, 2016, 3136, 5504, 9328, 12112,...];

%e n=4: [1, 20, 180, 960, (3380), 8424, 16320, 28800, 52020, 88660,...];

%e n=5: [1, 24, 264, 1760, 7944, (25872), 64416, 133056, 253704,...];

%e n=6: [1, 28, 364, 2912, 16044, 64792, (200928), 503360, ...];

%e n=7: [1, 32, 480, 4480, 29152, 140736, 525952, (1580800), ...]; ...

%e where the coefficients in parenthesis form the initial terms of this sequence:

%e A = [1/1, 8/2, 60/3, 448/4, 3380/5, 25872/6, 200928/7, 1580800/8, ...].

%t CoefficientList[1/x * InverseSeries[Series[x/EllipticTheta[3, 0, x]^2, {x, 0, 25}], x], x] (* _Vaclav Kotesovec_, Nov 16 2023 *)

%t (* Calculation of constants {d,c}: *) {1/r, s/Sqrt[Pi*(1 + 4 * r^2 * s^(3/2) * Derivative[0, 0, 2][EllipticTheta][3, 0, r*s])]} /. FindRoot[{s == EllipticTheta[3, 0, r*s]^2, 2*r*Sqrt[s]*Derivative[0, 0, 1][EllipticTheta][3, 0, r*s] == 1}, {r, 1/8}, {s, 3/2}, WorkingPrecision -> 70] (* _Vaclav Kotesovec_, Nov 16 2023 *)

%o (PARI) {a(n)=local(A=1+x);for(i=1,n,A=1+4*sum(m=1,n,(x*A)^m/(1+(x*A+x*O(x^n))^(2*m))));polcoeff(A,n)}

%o (PARI) {a(n)=local(A=1+x);for(i=1,n,A=(1+2*sum(m=1,sqrtint(n+1),(x*A+x*O(x^n))^(m^2)))^2);polcoeff(A,n)}

%o (PARI) {a(n)=local(A=1+x);for(i=1,n,A=prod(m=1,n,(1-(-x)^m*A^m)/(1+(-x)^m*A^m +x*O(x^n)))^2);polcoeff(A,n)}

%o for(n=0,30,print1(a(n),", "))

%Y Cf. A166952.

%K nonn

%O 0,2

%A _Paul D. Hanna_, May 14 2012