a(n) = (1/C(2n,n))*sum {k = 0..n} C(2k,k)*C(4k,2k)*C(2n-2k,n-k)*C(4n-4k,2n-2k).
Recurrence relation:
a(0) = 1, a(1) = 12, n^2*a(n) = 4*(8*n^2-8*n+3)*a(n-1) - 256*(n-1)^2*a(n-2).
Congruences:
For odd prime p, a(m*p^r) = a(m*p^(r-1)) (mod p^r) for any m,r in N.
a(n) ~ 16^n/(Pi*sqrt(Pi*n)) * (log(n) + gamma + 6*log(2)), where gamma is the Euler-Mascheroni constant (
A001620). -
Vaclav Kotesovec, Oct 11 2013
a(n) = sum {k = 0..n} 4^(n-k) C(2k,k)^2*C(2n-2k,n-k). -
Tito Piezas III, Dec 12 2014
a(n) = hypergeom([1/2,1/2,n+1],[1,n+3/2],1)*2^(5*n+1)*n!/((2*n+1)!!*Pi) -
G. A. Edgar, Dec 10 2016
a(n) = binomial(4*n,2*n)*hypergeom([1/4,3/4,-n,-n], [1,1/4-n,3/4-n], 1). -
Peter Luschny, May 14 2020
a(n) = 16^n*Sum_{k=0..n} (-1)^k*binomial(-1/2, k)^2*binomial(n, k).
a(n) = 16^n*hypergeom([1/2, 1/2, -n], [1, 1], 1). (End)
