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a(n) = Sum_{k=0..floor(n/3)} (n-3*k)^n.
4

%I #15 Apr 16 2022 09:39:55

%S 1,1,4,27,257,3157,47385,839928,17168097,397517868,10283523826,

%T 293950435376,9200707298514,312972062711525,11496160809184370,

%U 453507273286873875,19122193920118268355,858239308206920523297,40851073484808464297979

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

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

%t a[0] = 1; a[n_] := Sum[(n-3*k)^n, {k, 0, Floor[n/3]}]; Array[a, 20, 0] (* _Amiram Eldar_, Apr 16 2022 *)

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

%o (PARI) my(N=20, x='x+O('x^N)); Vec(sum(k=0, N, (k*x)^k/(1-(k*x)^3)))

%Y Cf. A352982, A353014, A353017.

%K nonn,easy

%O 0,3

%A _Seiichi Manyama_, Apr 16 2022