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a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-k,k) * a(n-2*k-1).
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%I #6 Apr 07 2022 09:25:03

%S 1,1,1,3,6,13,34,84,230,653,1893,5794,18080,58345,193761,657959,

%T 2295398,8177305,29775086,110676222,419169483,1617868052,6353518921,

%U 25376986471,103017630200,424704411564,1777458163195,7546547411488,32490058003914,141774055915497,626739661952337

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

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

%t a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - k, 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 + x A[x/(1 - x^2)]/(1 - x^2)^2 + O[x]^(nmax + 1) // Normal,nmax + 1]; CoefficientList[A[x], x]

%Y Cf. A040027, A172383, A352865.

%K nonn

%O 0,4

%A _Ilya Gutkovskiy_, Apr 06 2022