A347857 a(n) = (6*n)!/((3*n)!*(2*n)!) * (3*n/2)!/(5*n/2)!.

1, 24, 1386, 91392, 6374082, 458625024, 33660501840, 2504739913728, 188276393811330, 14262358156247040, 1087036407409838886, 83262603737872465920, 6403774152656100209808, 494217792537649867653120

Offset: 0

Author

Peter Bala, Sep 18 2021

Keywords

• nonn
• easy

Comments

Fractional factorials are defined using the Gamma function; for example, (3*n/2)! := Gamma(1 + 3*n/2). The sequence defined by u(n) = (12*n)!*(3*n)!/((6*n)!*(4*n)!*(5*n)!) is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 5). See A295435. 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) = 83262603737872465920 - 24 = (2^3)*3*(11^3)*2369413*1100068993 == 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. A295435, A347854, A347855, A347856, A347858.

Programs

Maple

seq( (6*n)!/((3*n)!*(2*n)!) * GAMMA(1+3*n/2)/GAMMA(1+5*n/2), n = 0..12);

Python

from math import factorial
from sympy import factorial2
def A347857(n): return int((factorial(6*n)*factorial2(3*n)<Chai Wah Wu, Aug 10 2023

Formula

a(n) = binomial(6*n,3*n)*binomial(3*n,2*n)/binomial(5*n/2,n).

a(2*n) = A295435(n).

a(2*n) = (216/5)*(3*n-1)*(3*n-2)*(12*n-1)*(12*n-5)*(12*n-7)*(12*n-11)/(n*(2*n-1)*(5*n-1)*(5*n-2)*(5*n-3)*(5*n-4))*a(2*n-2);

a(2*n+1) = (864/5)*(36*n^2-1)*(144*n^2-1)*(144*n^2-25)/( n*(2*n+1)*(100*n^2-1)*(100*n^2-9))*a(2*n-1).

Asymptotics: a(n) ~ sqrt(3/(10*Pi*n))* 32^n * 3^(9*n/2)/5^(5*n/2) as n -> infinity.

O.g.f. A(x) = hypergeom([1/12, 1/3, 5/12, 7/12, 2/3, 11/12], [1/5, 2/5, 1/2, 3/5, 4/5], (2^10*3^9/5^5)*x^2) + 24*x*hypergeom([7/12, 5/6, 11/12, 13/12, 7/6, 17/12], [7/10, 9/10, 11/10, 13/10, 3/2], (2^10*3^9/5^5)*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.