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a(n) = Sum_{k = 1..n} (-1)^(n+k) * k^3 * binomial(n,k)^2.
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%I #19 Nov 03 2023 07:12:18

%S 0,1,4,-36,-96,450,1080,-3920,-8960,28350,63000,-182952,-399168,

%T 1093092,2354352,-6177600,-13178880,33474870,70887960,-175518200,

%U -369512000,896251356,1877859984,-4478082336,-9345563136,21971267500,45700236400,-106148523600,-220159900800

%N a(n) = Sum_{k = 1..n} (-1)^(n+k) * k^3 * binomial(n,k)^2.

%C Compare with the alternating binomial sum evaluation Sum_{k = 0..2*n+1} (-1)^(n+k+1) * k^2 * binomial(2*n+1,k)^2. = (2*n+1)^2 * binomial(2*n,n) = A294486(n).

%H Winston de Greef, <a href="/A361719/b361719.txt">Table of n, a(n) for n = 0..3271</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant">Apéry's constant</a>

%F a(n) = (-1)^((n-1)*(n+2)/2) * n*floor((n+1)/2)^2 * binomial(n, floor(n/2)) = (-1)^((n-1)*(n+2)/2) * n*floor((n+1)/2)^2 * A001405(n).

%F a(2*n) = (-1)^(n+1) * 2*n^3 * binomial(2*n,n).

%F a(2*n+1) = (-1)^n * (n+1)*(2*n+1)^2 * binomial(2*n,n).

%F a(n) = (-1)^(n+1) * n^2 * hypergeom([2, 1 - n, 1 - n], [1, 1], -1).

%F P-recursive: (2*n^2 - 5*n + 4)*(n - 2)*(n - 1)^3*a(n) = 2*n^2*(3*n - 5)*(n - 2)*a(n-1) - 4*n^2*(2*n^2 - n + 1)*(n - 1)^2*a(n-2) with a(0) = 0 and a(1) = 1.

%F 5*Sum_{n >= 1} 1/a(2*n) = zeta(3), a result due to Markov (1890), rediscovered by Apéry (1979). - _Peter Bala_, Oct 24 2023

%p seq(add( (-1)^(n+k)*k^3*binomial(n,k)^2, k = 0..n ), n = 0..20);

%o (PARI) a(n) = (-1)^((n-1)*(n+2)/2) * n*((n+1)\2)^2 * binomial(n, n\2) \\ _Winston de Greef_, Mar 24 2023

%o (Python)

%o from math import comb

%o def A361719(n): return (-((m:=n>>1)+1)*n**2*comb(n-1,m) if n&2 else ((m:=n>>1)+1)*n**2*comb(n-1,m)) if n&1 else (((m:=n>>1)**3<<1)*comb(n,m) if n&2 else -((m:=n>>1)**3<<1)*comb(n,m)) # _Chai Wah Wu_, Mar 24 2023

%Y Cf. A000984, A001405, A002117, A294486.

%K sign,easy

%O 0,3

%A _Peter Bala_, Mar 24 2023