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a(n) = Sum_{k=0..floor(n/2)} (n-2*k)^n.
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%I #40 Apr 04 2024 10:57:18

%S 1,1,4,28,272,3369,50816,903856,18522624,429746905,11135257600,

%T 318719062236,9987013488640,340037795872369,12500401969233920,

%U 493467700789897408,20819865970795610112,934939160745193002321,44523294861684890664960

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

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

%F Conjecture: a(n) = (1 - 2^n)*zeta(-n) - (2^n)*zeta(-n, n/2 + 1) for n > 0, where the bivariate zeta function is the Hurwitz zeta function. - _Velin Yanev_, Mar 25 2024

%F a(n) ~ n^n / (1 - exp(-2)). - _Vaclav Kotesovec_, Mar 25 2024

%t a[0] = 1; a[n_] := Sum[(n-2*k)^n, {k, 0, Floor[n/2]}]; Array[a, 20, 0] (* _Amiram Eldar_, Apr 16 2022 *)

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

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

%Y Cf. A352981, A353013, A353016.

%K nonn,easy

%O 0,3

%A _Seiichi Manyama_, Apr 16 2022