A347856 a(n) = (6*n)!/((4*n)!*n!) * (n/2)!/(3*n/2)!.

1, 20, 990, 56576, 3432198, 215147520, 13768454700, 893768826880, 58626071754822, 3876225891958784, 257898242928604740, 17245961314149335040, 1158088115444301759900, 78040346182201091555328

Offset: 0

Author

Peter Bala, Sep 17 2021

Keywords

• nonn
• easy

Comments

Fractional factorials are defined using the Gamma function; for example, (n/2)! := Gamma(1 + n/2).

The sequence defined by u(n) = (12*n)!/((8*n)!*(2*n)!) * n!/(3*n)! is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 3). See A295433. Here we are essentially considering the sequence (u(n/2))n>=0. The sequence is conjectured to be integral.

Examples

			Congruence: a(11) - a(1) = 17245961314149335040 - 20 = (2^2)*5*(11^3)*103*6289876694707 == 0 (mod 11^3).
		

Links

• J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc., Vol. 79, Issue 2 (2009), 422-444.
• F. Rodriguez-Villegas, Integral ratios of factorials and algebraic hypergeometric functions, arXiv:math/0701362 [math.NT], 2007.

Crossrefs

Cf. A295433, A347854 - A347858.

Programs

Maple

seq( (6*n)!/((4*n)!*n!) * GAMMA(1+n/2)/GAMMA(1+3*n/2), n = 0..12);

Python

from math import factorial
from sympy import factorial2
def A347856(n): return int((factorial(6*n)*factorial2(n)<Chai Wah Wu, Aug 10 2023

Formula

a(n) = binomial(6*n,4*n)*binomial(2*n,n)/binomial(3*n/2,n).

a(2*n) = A295433(n).

a(2*n) = 54*(6*n-1)*(6*n-5)*(12*n-1)*(12*n-5)*(12*n-7)*(12*n-11)/(n*(2*n-1)*(8*n-1)*(8*n-3)*(8*n-5)*(8*n-7))*a(2*n-2);

a(2*n+1) = 216*(9*n^2-1)*(144*n^2-1)*(144*n^2-5^2)/(n*(2*n+1)*(64*n^2-1)*(64*n^2-3^2))*a(2*n-1).

Asymptotics: a(n) ~ 1/(2*sqrt(n*Pi)) *3^(9*n/2)/2^n as n -> infinity.

O.g.f. A(x) = hypergeom([1/12, 1/6, 5/12, 7/12, 5/6, 11/12], [1/8, 3/8, 1/2, 5/8, 7/8], (19683/4)*x^2) + 20*x*hypergeom([17/12, 13/12, 11/12, 7/12, 4/3, 2/3], [11/8, 9/8, 7/8, 5/8, 3/2], (19683/4)*x^2) is conjectured to be algebraic over Q(x).

Conjectural congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k.