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a(n) = 1 + Sum_{k=0..floor((n-1)/2)} a(k) * a(n-2*k-1).
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%I #6 Feb 28 2022 08:10:04

%S 1,2,3,6,11,21,40,78,151,294,572,1115,2172,4234,8252,16088,31361,

%T 61140,119191,232370,453010,883167,1721768,3356675,6543988,12757830,

%U 24871992,48489172,94531974,184294706,359291464,700456240,1365573493,2662252082,5190190005

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

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

%t a[n_] := a[n] = 1 + Sum[a[k] a[n - 2 k - 1], {k, 0, Floor[(n - 1)/2]}]; Table[a[n], {n, 0, 34}]

%t nmax = 34; A[_] = 0; Do[A[x_] = 1/((1 - x) (1 - x A[x^2])) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%Y Cf. A000621, A007317, A351973.

%K nonn

%O 0,2

%A _Ilya Gutkovskiy_, Feb 26 2022