A307091 a(n) = Sum_{k=0..floor(n/2)} (-1)^k * binomial(n,2*k)^2.

1, 1, 0, -8, -34, -74, 0, 736, 3334, 7606, 0, -80464, -372436, -864772, 0, 9400192, 43976774, 103061158, 0, -1137528688, -5355697084, -12623082284, 0, 140697113792, 665238165916, 1574005263676, 0, -17663830073504, -83769667651816, -198760191043784, 0

Offset

0,4

Links

Seiichi Manyama, Table of n, a(n) for n = 0..1879

Formula

a(4*n+2) = 0 for n >= 0.

From Peter Bala, Mar 17 2023: (Start)

n*(n-1)*(6*n^2-24*n+23)a(n) = 4*(n-1)*(2*n-3)*(3*n^2-9*n+4)*a(n-1) - 4*(3*n^2-9*n+4)*(2*n-3)^2*a(n-2) - 8*(n-2)*(2*n-3)*(3*n^2-9*n+4)*a(n-3) - 4*(n-2)*(n-3)*(6*n^2-12*n+5)*a(n-4) with a(0) = 1, a(1) = 1, a(2) = 0 and a(3) = -8.

a(n) = hypergeom([(1-n)/2, (1-n)/2, -n/2, -n/2], [1/2, 1/2, 1], -1).

Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for positive integers n and r and all primes p >= 3. (End)

Mathematica

Table[Sum[(-1)^k*Binomial[n, 2*k]^2, {k, 0, Floor[n/2]}], {n, 0, 30}] (* Vaclav Kotesovec, Mar 24 2019 *)
Table[HypergeometricPFQ[{1/2 - n/2, 1/2 - n/2, -n/2, -n/2}, {1/2, 1/2, 1}, -1], {n, 0, 30}] (* Vaclav Kotesovec, Mar 24 2019 *)

Prog

(PARI) {a(n) = sum(k=0, n\2, (-1)^k*binomial(n, 2*k)^2)}

Crossrefs

Central coefficients of number triangle A307090.

Cf. A000984, A119358, A119363.

Sequence in context: A301966 A196143 A373356 * A024847 A154516 A212744

Adjacent sequences: A307088 A307089 A307090 * A307092 A307093 A307094

Keyword

sign,easy

Author

Seiichi Manyama, Mar 24 2019

Status

approved