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a(0) = 1; a(n) = Sum_{k=0..floor(n/4)} binomial(n-3*k,k) * a(k).
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%I #5 Mar 02 2022 08:45:07

%S 1,1,1,1,2,3,4,5,7,10,14,19,26,36,50,69,96,136,196,285,417,614,909,

%T 1349,2002,2968,4394,6493,9572,14074,20639,30189,44049,64123,93151,

%U 135080,195599,282915,408884,590658,853080,1232168,1780190,2573059,3721103

%N a(0) = 1; a(n) = Sum_{k=0..floor(n/4)} binomial(n-3*k,k) * a(k).

%F G.f. A(x) satisfies: A(x) = A(x^4/(1 - x)) / (1 - x).

%t a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - 3 k, k] a[k], {k, 0, Floor[n/4]}]; Table[a[n], {n, 0, 44}]

%t nmax = 44; A[_] = 1; Do[A[x_] = A[x^4/(1 - x)]/(1 - x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%Y Cf. A092684, A180184, A352039.

%K nonn

%O 0,5

%A _Ilya Gutkovskiy_, Mar 01 2022