a(n) = (9*n)!/((7*n)!*n!^2) = binomial(9*n,2*n)* binomial(2*n,n).
a(n) = [x^n](1 + x)^(8*n) * [x^n] (1 + x)^(9*n).
It appears that a(n) = [x^n] F(x)^(72*n), where F(x) = 1 + x + 56*x^2 + 7700*x^3 + 1422008*x^4 + 307144278*x^5 + 73118586828*x^6 + ... has all integer coefficients. Cf.
A273628 and
A008979.
Recurrence: 7*n^2*(7*n - 1)*(7*n - 2)*(7*n - 3)*(7*n - 4)*(7*n - 5)*(7*n - 6)*a(n) = 9*(9*n - 1)*(9*n - 2)*(9*n - 3)*(9*n - 4)*(9*n - 5)*(9*n - 6)*(9*n - 7)*(9*n - 8)*a(n-1).
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) *
A108625(8*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). -
Peter Bala, Oct 15 2024