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G.f. A(x) satisfies A(x) = (1 + 2 * x * A(x)^3) / (1 - x).
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%I #12 Jul 24 2023 09:22:36

%S 1,3,21,201,2217,26535,335001,4391553,59203137,815580507,11430639165,

%T 162470033625,2336381642649,33930648153615,496935405133617,

%U 7331179445170689,108846406625097729,1625145134034548019,24385673680861258533,367546405595389076649,5561980053932228243529

%N G.f. A(x) satisfies A(x) = (1 + 2 * x * A(x)^3) / (1 - x).

%H Seiichi Manyama, <a href="/A348912/b348912.txt">Table of n, a(n) for n = 0..828</a>

%F a(0) = 1; a(n) = a(n-1) + 2 * Sum_{i=0..n-1} Sum_{j=0..n-i-1} a(i) * a(j) * a(n-i-j-1).

%F a(n) ~ sqrt(-50 + 30*sqrt(3) + (22 - 12*sqrt(3))*(2*(sqrt(3) - 1))^(1/3) + (2*(sqrt(3) - 1))^(2/3)*(-11 + 7*sqrt(3)))/(4*sqrt(3*Pi)*(-1 + sqrt(3))^(3/2) * n^(3/2) * (1 + (3*(-1 + sqrt(3))^(1/3))/2^(2/3) - 3/(2*(-1 + sqrt(3)))^(1/3))^n). - _Vaclav Kotesovec_, Nov 04 2021

%F a(n) = Sum_{k=0..n} 2^k * binomial(n,k) * binomial(n+2*k+1,n) / (n+2*k+1). - _Seiichi Manyama_, Jul 24 2023

%t nmax = 20; A[_] = 0; Do[A[x_] = (1 + 2 x A[x]^3)/(1 - x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%t a[0] = 1; a[n_] := a[n] = a[n - 1] + 2 Sum[Sum[a[i] a[j] a[n - i - j - 1], {j, 0, n - i - 1}], {i, 0, n - 1}]; Table[a[n], {n, 0, 20}]

%o (PARI) a(n) = sum(k=0, n, 2^k*binomial(n, k)*binomial(n+2*k+1, n)/(n+2*k+1)); \\ _Seiichi Manyama_, Jul 24 2023

%Y Cf. A001764, A103210, A153231, A346626.

%K nonn

%O 0,2

%A _Ilya Gutkovskiy_, Nov 03 2021