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a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/2)} 3^k * a(k) * a(n-2*k-1).
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%I #6 Mar 01 2022 07:28:34

%S 1,1,1,4,7,19,40,178,379,1237,2941,10378,24628,78928,198820,813550,

%T 1971907,6587245,16980079,61488286,155573011,515316037,1363261084,

%U 4937498686,12796438252,42078038668,113153315824,390012381346,1036020692356,3379994401042,9240830253940

%N a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/2)} 3^k * a(k) * a(n-2*k-1).

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

%t a[0] = 1; a[n_] := a[n] = Sum[3^k a[k] a[n - 2 k - 1], {k, 0, Floor[(n - 1)/2]}]; Table[a[n], {n, 0, 30}]

%t nmax = 30; A[_] = 0; Do[A[x_] = 1/(1 - x A[3 x^2]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%Y Cf. A000621, A015084, A352006, A352008.

%K nonn

%O 0,4

%A _Ilya Gutkovskiy_, Feb 28 2022