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a(n) = Sum_{k=0..floor(n/3)} k^n.
5

%I #22 Nov 02 2022 10:29:56

%S 1,0,0,1,1,1,65,129,257,20196,60074,179196,17312754,68711380,

%T 273234810,31605701625,156925970179,780248593545,105443761093411,

%U 628709267031321,3752628871164355,580964060390826448,4043844561787569140,28170468954985342384

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

%H Seiichi Manyama, <a href="/A352982/b352982.txt">Table of n, a(n) for n = 0..458</a>

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

%t a[0] = 1; a[n_] := Sum[k^n, {k, 0, Floor[n/3]}]; Array[a, 24, 0] (* _Amiram Eldar_, Apr 13 2022 *)

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

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

%o (Magma) [(&+[k^n: k in [0..Floor(n/3)]]): n in [0..40]]; // _G. C. Greubel_, Nov 01 2022

%o (SageMath) [sum( k^n for k in range((n//3)+1)) for n in range(41)] # _G. C. Greubel_, Nov 01 2022

%Y Cf. A089072, A352945, A352981.

%K nonn,easy

%O 0,7

%A _Seiichi Manyama_, Apr 13 2022