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G.f. A(x) satisfies A(x) = Product_{k>=1} (1 + x^k*A(x)^(2*k)).
2

%I #6 Sep 26 2023 12:09:25

%S 1,1,3,13,64,340,1903,11053,65992,402508,2497207,15709873,99980007,

%T 642535004,4164018953,27181480712,178559253274,1179546465168,

%U 7830695860690,52216823047741,349584244515573,2348869478981267,15833924106623011,107057382854642578,725829177205070854

%N G.f. A(x) satisfies A(x) = Product_{k>=1} (1 + x^k*A(x)^(2*k)).

%F A(x) satisfies QPochhammer(-1, x*A(x)^2) = 2*A(x).

%F a(n) ~ c * d^n / n^(3/2), where d = 7.2188305975020061051473056449576894316519... and c = 0.2182691546096422371919544994005940622002...

%t nmax = 30; A[_] = 0; Do[A[x_] = Product[1 + x^k*A[x]^(2*k), {k, 1, nmax}] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%t (* The constants {d,c}: *) {1/r, 1/(2*Sqrt[Pi*(1/s^2 + 2*r^2*s*Derivative[0, 2][QPochhammer][-1, r*s^2])])} /. FindRoot[{2*s == QPochhammer[-1, r*s^2], r*s*Derivative[0, 1][QPochhammer][-1, r*s^2] == 1}, {r, 1/8}, {s, 1}, WorkingPrecision -> 120]

%Y Cf. A171802, A181315.

%K nonn

%O 0,3

%A _Vaclav Kotesovec_, Sep 26 2023