The sequence can be extended to negative values of n:
a(-n) = Sum_{k} binomial(-n,k)*binomial(-n+k,k)*binomial(-n-1,k)*binomial(-n-1+k,k) = a(n), since binomial(-m,k) = (-1)^k*binomial(m+k-1,k) for nonnegative k.
a(n) = Sum_{k = 1..n} binomial(n,k)^2*binomial(n+k-1,k-1)^2.
a(n) = (1/2)*Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k-1,k)^2 for n >= 1.
a(n) = hypergeom([n, n+1, -n, -n+1], [1, 1, 1], 1) for n >= 1.
a(n) = (1/2)*hypergeom([n, n, -n, -n], [1, 1, 1], 1) for n >= 1.
a(n) = (1/4)*binomial(2*n,n)^2*hypergeom([-n+1, -n+1, -n, -n], [1, -2*n+1, -2*n+1], 1) for n >= 1
a(n) = [x^n*y^(n-1)*z^n*t^(n-1)] 1/((1 - x - y)(1 - z - t) - x*y*z*t).
a(n) = [x^n] 1/(1 - x) * P(n-1,(1 + x)/(1 - x))^2 = [x^(n-1)] 1/(1 - x) * P(n,(1 + x)/(1 - x))^2 for n >= 1, where P(n,x) denotes the n-th Legendre polynomial. Cf.
A005259(n) = [x^n] 1/(1 - x) * P(n,(1 + x)/(1 - x))^2.
(n + 1)^3*(2*n - 1)(3*n^2 - 3*n + 1)*a(n+1) = 2*(102*n^6 - 68*n^4 + 21*n^2 - 3)*a(n) - (n - 1)^3*(2*n + 1)*(3*n^2 + 3*n + 1)*a(n-1) with a(0) = 0 and a(1) = 1.
The Gauss congruences a(n*p^r) == a(n*p^(r-1)) (mod p^r) hold for all primes p and all positive integers n and r. It follows that the expansion of exp( Sum_{n >= 1} a(n)/n*x^n ) = 1 + x + 7*x^2 + 91*x^3 + 1544*x^4 + 30448*x^5 + 661506*x^6 + 15377010*x^7 + ... has integer coefficients.
a(n) = n^2*hypergeom([1 - n, 1 - n, 1 + n, 1 + n], [1, 2, 2], 1). -
Peter Luschny, Apr 17 2022
a(n) = Sum_{k = 0..n-1} (n-k)/(n+k) * binomial(n,k)^2 * binomial(n+k,k)^2.
a(n) ~ (1 + sqrt(2))^(4*n) / (2^(9/4) * Pi^(3/2) * n^(3/2)). -
Vaclav Kotesovec, Mar 29 2023
a(n) = (1/3)*Sum_{k = 0..n} binomial(n,k)*binomial(n+k-1,k)*f(k) for n >= 1, where f(n) =
A000172(n) denotes the n-th Franel number. Cf.
A005259(n) = Sum_{k = 0..n} binomial(n,k)*binomial(n+k,k)*f(k) (the second Strehl identity). -
Peter Bala, Jun 26 2023
a(n) = Sum_{0 <= j <= k <= n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)* binomial(n+j-1, j)^2*binomial(n-1, k-j) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*
A108625(n-1, k). Cf. Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*
A108625(n, k) =
A005259(n). -
Peter Bala, Sep 22 2024