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a(0) = 1; a(n) = Sum_{k=0..floor(n/3)} binomial(n-2*k,k) * a(k).
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%I #6 Mar 02 2022 08:44:07

%S 1,1,1,2,3,4,6,9,13,20,32,51,82,133,215,346,555,886,1408,2231,3528,

%T 5572,8797,13892,21950,34707,54919,86958,137761,218339,346178,549073,

%U 871261,1383243,2197542,3494019,5560580,8858687,14128865,22560717,36067022,57725840

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

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

%t a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - 2 k, k] a[k], {k, 0, Floor[n/3]}]; Table[a[n], {n, 0, 41}]

%t nmax = 41; A[_] = 1; Do[A[x_] = A[x^3/(1 - x)]/(1 - x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%Y Cf. A092684, A102547, A352041.

%K nonn

%O 0,4

%A _Ilya Gutkovskiy_, Mar 01 2022