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a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/2)} 2^k * a(k) * a(n-2*k-1).
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%I #6 Mar 01 2022 07:28:39

%S 1,1,1,3,5,11,21,59,117,283,597,1467,3125,7387,16149,39931,87541,

%T 207643,463061,1107515,2473909,5819739,13132437,31080571,70236533,

%U 164315035,373572693,875121339,1991869237,4639482331,10599986709,24765957371,56617082101

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

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

%t a[0] = 1; a[n_] := a[n] = Sum[2^k a[k] a[n - 2 k - 1], {k, 0, Floor[(n - 1)/2]}]; Table[a[n], {n, 0, 32}]

%t nmax = 32; A[_] = 0; Do[A[x_] = 1/(1 - x A[2 x^2]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%Y Cf. A000621, A015083, A352007, A352008.

%K nonn

%O 0,4

%A _Ilya Gutkovskiy_, Feb 28 2022