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A349535 G.f. A(x) satisfies A(x) = 1 / ((1 - 4 * x) * (1 - x * A(x)^2)).

%I A349535 #13 Aug 01 2023 09:33:32
%S A349535 1,5,31,232,2007,19213,196912,2114360,23467039,266945359,3094963919,
%T A349535 36433302124,434269638196,5230582604476,63560612650120,
%U A349535 778281652963504,9593248679529551,118938874458415987,1482248642350235245,18557299563105608336
%N A349535 G.f. A(x) satisfies A(x) = 1 / ((1 - 4 * x) * (1 - x * A(x)^2)).
%H A349535 Seiichi Manyama, <a href="/A349535/b349535.txt">Table of n, a(n) for n = 0..887</a>
%F A349535 a(n) = 4^n + Sum_{i=0..n-1} Sum_{j=0..n-i-1} a(i) * a(j) * a(n-i-j-1).
%F A349535 a(n) = Sum_{k=0..n} binomial(n+k,2*k) * binomial(3*k,k) * 4^(n-k) / (2*k+1).
%F A349535 a(n) = 4^n*F([1/3, 2/3, -n, 1+n], [1/2, 1, 3/2], -3^3/2^6), where F is the generalized hypergeometric function. - _Stefano Spezia_, Nov 21 2021
%F A349535 a(n) ~ 273^(1/4) * (59 + 3*sqrt(273))^(n + 1/2) / (9 * sqrt(Pi) * n^(3/2) * 2^(3*n + 5/2)). - _Vaclav Kotesovec_, Nov 22 2021
%t A349535 nmax = 19; A[_] = 0; Do[A[x_] = 1/((1 - 4 x) (1 - x A[x]^2)) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
%t A349535 a[n_] := a[n] = 4^n + 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, 19}]
%t A349535 Table[Sum[Binomial[n + k, 2 k] Binomial[3 k, k] 4^(n - k)/(2 k + 1), {k, 0, n}], {n, 0, 19}]
%Y A349535 Cf. A001764, A104498, A199475, A349533, A349534.
%K A349535 nonn
%O A349535 0,2
%A A349535 _Ilya Gutkovskiy_, Nov 21 2021