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a(n) = Sum_{k = 0..n} binomial(8*n, n-k)^2 * binomial(6*n+k-1, k).
a(n) = [x^n] (1 - x)^(2*n) * P(8*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial.
a(n) = (5/12)*(5*n-1)*(5*n-2)*(5*n-3)*(5*n-4)*(8*n-1)*(8*n-3)*(8*n-5)*(8*n-7)/((4*n-1)*(4*n-3)*(6*n-1)*(6*n-5)*n^2*(2*n-1)^2) * a(n-1) with a(0) = 1.
a(n) ~ c^n * sqrt(5)/(4*Pi*n), where c = (2^2)*(5^5)/(3^3).
a(n) = binomial(8*n,2*n)*binomial(5*n,n)*binomial(2*n,n)/binomial(4*n,n) = A001449(n) * A211421(n).
a(p) == a(1) (mod p^3) for all primes p >= 5.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k.
a(n) = G(x)^(10*n), where the power series G(x) = 1 + 7*x + 412*x^2 + 55524*x^3 + 10088066*x^4 + 2146473322*x^5 + 503731865112*x^6 + ... appears to have integer coefficients
exp(Sum_{n >= 1} a(n)*x^n/n) = F(x)^10, where the power series F(x) = 1 + 7*x + 902*x^2 + 191779*x^3 + 50706776*x^4 + 15153397742*x^5 + 4898289306180*x^6 + ... appears to have integer coefficients.
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