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
KEYWORD
sign,easy
AUTHOR
Seiichi Manyama, Mar 24 2019
STATUS
approved