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a(n) = Sum_{k=0..floor(n/2)} k^(2*n).
5

%I #20 Apr 16 2022 09:34:11

%S 1,0,1,1,257,1025,535538,4799354,4338079554,69107159370,

%T 96470431101379,2401809362313955,4798267740520031875,

%U 172076350440523281571,466164803742660494432996,22761346686115003736962100,80340572151131167125889902852

%N a(n) = Sum_{k=0..floor(n/2)} k^(2*n).

%H Seiichi Manyama, <a href="/A352983/b352983.txt">Table of n, a(n) for n = 0..240</a>

%F G.f.: Sum_{k>=0} (k^2 * x)^(2 * k) / (1 - k^2 * x).

%F a(n) ~ exp(3 + (-1)^n) * (n/2)^(2*n) / (exp(4) - 1). - _Vaclav Kotesovec_, Apr 14 2022

%t a[0] = 1; a[n_] := Sum[k^(2*n), {k, 0, Floor[n/2]}]; Array[a, 17, 0] (* _Amiram Eldar_, Apr 13 2022 *)

%o (PARI) a(n) = sum(k=0, n\2, k^(2*n));

%o (PARI) my(N=20, x='x+O('x^N)); Vec(sum(k=0, N, (k^2*x)^(2*k)/(1-k^2*x)))

%Y Cf. A249459, A352984.

%K nonn,easy

%O 0,5

%A _Seiichi Manyama_, Apr 13 2022