A276098 a(n) = (7*n)!*(3/2*n)!/((7*n/2)!*(3*n)!*(2*n)!).

1, 48, 6006, 860160, 130378950, 20392706048, 3254013513660, 526470648692736, 86047769258554950, 14173603389190963200, 2349023203055914140756, 391249767795614684282880, 65434374898388743460014620, 10981406991821583404677201920

Offset: 0

Author

Peter Bala, Aug 22 2016

Keywords

• nonn
• easy

Comments

Let a > b be nonnegative integers. The ratio of factorials (2*a*n)!*(b*n)!/( (a*n)!*(2*b*n)!*((a - b)*n)! ) is known to be an integer for n >= 0 (see, for example, Bober, Theorem 1.1). We have the companion result: Let a > b be nonnegative integers. Then the ratio of factorials ((2*a + 1)*n)!*((b + 1/2)*n)!/(((a + 1/2)*n)!*((2*b + 1)*n)!*((a - b)*n)!) is an integer for n >= 0. This is the case a = 3, b = 1. Other cases include A091496 (a = 2, b = 0), A091527 (a = 1, b = 0), A262732 (a = 2, b = 1), A262733 (a = 3, b = 2) and A276099 (a = 4, b = 2).

References

• R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Links

• Peter Bala, Some integer ratios of factorials
• J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc., 79, Issue 2, (2009), 422-444.

Crossrefs

Cf. A046521, A091496, A091527, A262732, A262733, A276099.

Programs

Maple

seq(simplify((7*n)!*(3/2*n)!/((7*n/2)!*(3*n)!*(2*n)!)), n = 0..20);

Python

from math import factorial
from sympy import factorial2
def A276098(n): return int((factorial(7*n)*factorial2(3*n)<<(n<<1))//factorial2(7*n)//factorial(3*n)//factorial(n<<1)) # Chai Wah Wu, Aug 10 2023

Formula

a(n) = Sum_{k = 0..2*n} binomial(7*n, 2*n - k)*binomial(3*n + k - 1, k).

a(n) = Sum_{k = 0..n} binomial(10*n, 2*n - 2*k)*binomial(3*n + k - 1, k).

Recurrence: a(n) = 28*(7*n - 1)*(7*n - 3)*(7*n - 5)*(7*n - 9)*(7*n - 11)*(7*n - 13)/(3*n*(n - 1)*(2*n - 1)*(2*n - 3)*(3*n - 1)*(3*n - 5)) * a(n-2).

a(n) ~ 1/sqrt(4*Pi*n) * (7^7/3^3)^(n/2).

O.g.f. A(x) = Hypergeom([13/14, 11/14, 9/14, 5/14, 3/14, 1/14], [5/6, 3/4, 1/2, 1/4, 1/6], (7^7/3^3)*x^2) + 48*x*Hypergeom([10/7, 9/7, 8/7, 6/7, 5/7, 4/7], [5/4, 4/3, 3/2, 3/4, 2/3], (7^7/3^3)*x^2).

a(n) = [x^(2*n)] H(x)^n, where H(x) = (1 + x)^7/(1 - x)^3.

It follows that the o.g.f. A(x) equals the diagonal of the bivariate rational generating function 1/2*( 1/(1 - t*H(sqrt(x))) + 1/(1 - t*H(-sqrt(x))) ) and hence is algebraic by Stanley 1999, Theorem 6.33, p. 197.

Let F(x) = (1/x)*Series_Reversion( x*sqrt((1 - x)^3/(1 + x)^7) ) and put G(x) = 1 + x*d/dx(log(F(x))). Then A(x^2) = (G(x) + G(-x))/2.