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G.f. A(x) satisfies: A(x) = 1 / (1 - 2*x) + x * (1 - 2*x)^2 * A(x)^4.
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%I #16 Nov 26 2021 04:35:20

%S 1,3,12,66,460,3681,31848,289176,2714044,26103468,255876048,

%T 2546717454,25666830724,261407935366,2686191839232,27815564456544,

%U 289960011573212,3040424427011492,32046741183678288,339345854532800136,3608307717155678256,38511520730570169033

%N G.f. A(x) satisfies: A(x) = 1 / (1 - 2*x) + x * (1 - 2*x)^2 * A(x)^4.

%C Second binomial transform of A002293.

%F a(n) = Sum_{k=0..n} binomial(n,k) * binomial(4*k,k) * 2^(n-k) / (3*k+1).

%F a(n) = 2^n*F([1/4, 1/2, 3/4, -n], [2/3, 1, 4/3], -2^7/3^3), where F is the generalized hypergeometric function. - _Stefano Spezia_, Nov 22 2021

%F a(n) ~ 2^(n - 10) * 155^(n + 3/2) / (sqrt(Pi) * n^(3/2) * 3^(3*n + 3/2)). - _Vaclav Kotesovec_, Nov 26 2021

%t nmax = 21; A[_] = 0; Do[A[x_] = 1/(1 - 2 x) + x (1 - 2 x)^2 A[x]^4 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%t Table[Sum[Binomial[n, k] Binomial[4 k, k] 2^(n - k)/(3 k + 1), {k, 0, n}], {n, 0, 21}]

%o (PARI) a(n) = sum(k=0, n, binomial(n,k)*binomial(4*k,k)*2^(n-k)/(3*k+1)); \\ _Michel Marcus_, Nov 23 2021

%Y Cf. A002293, A064613, A346646, A346762, A349582, A349584, A349590, A349591.

%K nonn

%O 0,2

%A _Ilya Gutkovskiy_, Nov 22 2021