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a(n) = Sum_{k=0..floor(n/3)} (n-3*k)^(3*k).
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%I #12 Apr 16 2022 09:39:28

%S 1,1,1,1,2,9,28,66,190,946,4441,16650,67069,380795,2220697,11142307,

%T 58133022,380165427,2581541092,15919859932,101602799146,758173118356,

%U 5826902270129,42158185020684,316416126945385,2656178496077301,22725296418141937,187568834724460765

%N a(n) = Sum_{k=0..floor(n/3)} (n-3*k)^(3*k).

%F G.f.: Sum_{k>=0} x^k / (1 - (k * x)^3).

%t a[0] = 1; a[n_] := Sum[(n-3*k)^(3*k), {k, 0, Floor[n/3]}]; Array[a, 30, 0] (* _Amiram Eldar_, Apr 16 2022 *)

%o (PARI) a(n) = sum(k=0, n\3, (n-3*k)^(3*k));

%o (PARI) my(N=40, x='x+O('x^N)); Vec(sum(k=0, N, x^k/(1-(k*x)^3)))

%Y Cf. A352946, A353015.

%K nonn,easy

%O 0,5

%A _Seiichi Manyama_, Apr 16 2022