cp's OEIS Frontend

a(n) = Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i)^2 * binomial(4*n, 2*n+i).

Compare with Dixon's identity: Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i)^3 = (3*n)!/n!^3.

a(n) = A005810(n)^2.

P-recursive: a(n) = 16 * ( (4*n - 1)*(4*n - 2)*(4*n - 3)/(3*n*(3*n - 1)*(3*n - 2)) )^2 * a(n-1) with a(0) = 1.

a(n) ~ c^n * 2/(3*Pi*n), where c = (2^16)/(3^6).

a(n) = [x^n] G(x)^(16*n), where the power series G(x) = 1 + x + 9*x^2 + 225*x^3 + 7525*x^4 + 295228*x^5 + 12787152*x^6 + 592477457*x^7 + 28827755219*x^8 + ... appears to have integer coefficients.

exp( Sum_{n > = 1} a(n)*x^n/n ) = F(x)^16, where the power series F(x) = 1 + x + 25*x^2 + 1033*x^3 + 53077*x^4 + 3081944*x^5 + 193543624*x^6 + 12835533333*x^7 + 886092805699*x^8 + ... appears to have integer coefficients.

The supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3^r)) hold for all primes p >= 5 and all positive integers n and r.

a(n) = [x^(3*n)] ( (1 - x)^(6*n)*Legendre_P(2*n, (1 + x)/(1 - x)) ). - Peter Bala, Aug 14 2023

a(n) = Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i)^2 * binomial(6*n, 3*n+i).

Compare with Dixon's identity: Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i)^3 = (3*n)!/n!^3.

a(n) = (-1)^n*binomial(6*n,2*n) * hypergeom([-2*n, -2*n, -4*n], [1, 2*n+1], 1).

P-recursive: a(n) = (15/4)*(5*n-1)*(5*n-2)*(5*n-3)*(5*n-4)*(6*n-1)*(6*n-5)/((4*n-1)^2*(4*n-3)^2*n^2) * a(n-1) with a(0) = 1.

a(n) ~ c^n * sqrt(5)/(4*Pi*n), where c = (3^3)*(5^5)/(2^8).

a(n) = [x^n] G(x)^(10*n), where the power series G(x) = 1 + 5*x + 208*x^2 + 19960*x^3 + 2580710*x^4 + 390721786*x^5 + 65243160516*x^6 + 11646611942100*x^7 + 2182248792056787*x^8 + ... appears to have integer coefficients.

exp( Sum_{n > = 1} a(n)*x^n/n ) = F(x)^10, where the power series F(x) = 1 + 5*x + 458*x^2 + 69285*x^3 + 13037740*x^4 + 2773287786*x^5 + 638122182196*x^6 + 155077758079485*x^7 + 39234250338228617*x^8 + ... appears to have integer coefficients.

Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3^r)) hold for all primes p >= 5 and all positive integers n and r [added Oct 11 2024: follows from Meštrović, Section 6, equation 39, since a(n) = binomial(6*n, 3*n)*binomial(5*n, n)* binomial(2*n, n)/binomial(4*n, n)].

a(n) = (-1)^n * [x^(4*n)] (1 - x)^(8*n)*Legendre_P(2*n, (1 + x)/(1 - x)). - Peter Bala, Aug 14 2023