A361719 a(n) = Sum_{k = 1..n} (-1)^(n+k) * k^3 * binomial(n,k)^2.

0, 1, 4, -36, -96, 450, 1080, -3920, -8960, 28350, 63000, -182952, -399168, 1093092, 2354352, -6177600, -13178880, 33474870, 70887960, -175518200, -369512000, 896251356, 1877859984, -4478082336, -9345563136, 21971267500, 45700236400, -106148523600, -220159900800

Offset

0,3

Comments

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).

Links

Winston de Greef, Table of n, a(n) for n = 0..3271

Wikipedia, Apéry's constant

Formula

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).

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

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

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

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.

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

Maple

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

Prog

(PARI) a(n) = (-1)^((n-1)*(n+2)/2) * n*((n+1)\2)^2 * binomial(n, n\2) \\ Winston de Greef, Mar 24 2023
(Python)
from math import comb
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

Crossrefs

Cf. A000984, A001405, A002117, A294486.

Sequence in context: A193833 A193183 A152760 * A016826 A190318 A193874

Adjacent sequences: A361716 A361717 A361718 * A361720 A361721 A361722

Keyword

sign,easy

Author

Peter Bala, Mar 24 2023

Status

approved