%I #8 Nov 21 2021 05:16:32
%S 1,5,65,1145,23185,509005,11782465,283138545,6996125985,176633573205,
%T 4536739406465,118166489152745,3113854691067185,82864654201672605,
%U 2223776891616904065,60113561634017675745,1635364503704652830785,44739382956328846263205,1230059816693141938275265
%N G.f. A(x) satisfies: A(x) = (1 + 4 * x * A(x)^3) / (1 - x).
%F a(0) = 1; a(n) = a(n-1) + 4 * Sum_{i=0..n-1} Sum_{j=0..n-i-1} a(i) * a(j) * a(n-i-j-1).
%F a(n) = Sum_{k=0..n} binomial(n+2*k,3*k) * binomial(3*k,k) * 4^k / (2*k+1).
%F a(n) ~ sqrt((1 + (1 + 1/phi^(2/3) + phi^(2/3))^3/2) / (2*Pi)) / (6 * n^(3/2) * (1 + 3/phi^(1/3) - 3*phi^(1/3))^n), where phi = A001622 is the golden ratio. - _Vaclav Kotesovec_, Nov 21 2021
%t nmax = 18; A[_] = 0; Do[A[x_] = (1 + 4 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] + 4 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, 18}]
%t Table[Sum[Binomial[n + 2 k, 3 k] Binomial[3 k, k] 4^k/(2 k + 1), {k, 0, n}], {n, 0, 18}]
%o (PARI) a(n) = sum(k=0, n, binomial(n+2*k,3*k) * binomial(3*k,k) * 4^k / (2*k+1)) \\ _Andrew Howroyd_, Nov 20 2021
%Y Cf. A001764, A133305, A346626, A348912, A349516.
%K nonn
%O 0,2
%A _Ilya Gutkovskiy_, Nov 20 2021