A347858 a(n) = (9*n)!/((3*n)!*(2*n)!) * (n/2)!/(9*n/2)!.

1, 512, 1021020, 2399141888, 6016814703900, 15626259253952512, 41477110789150966020, 111745115394167694950400, 304331361887290342345862940, 835666006020766806513664655360, 2309513382640863232775760738593520, 6416034331756986890806503962421755904

Offset: 0

Author

Peter Bala, Sep 19 2021

Keywords

• nonn
• easy

Comments

Fractional factorials are defined using the Gamma function; for example, (9*n/2)! := Gamma(1 + 9*n/2).

The sequence defined by u(n) = (18*n)!*(n)!/((4*n)!*(6*n)!*(9*n)!) is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 7). See A295437. 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) = 6416034331756986890806503962421755904 - 512 = (2^9)*(11^3)*7207*1306363809854375553476366323 == 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. A295437, A347854, A347855, A347856, A347857.

Programs

Maple

seq( (9*n)!/((3*n)!*(2*n)!) * GAMMA(1+n/2)/GAMMA(1+9*n/2), n = 0..11);

Python

from math import factorial
from sympy import factorial2
def A347858(n): return int((factorial(9*n)*factorial2(n)<<(n<<2))//(factorial(3*n)*factorial(n<<1)*factorial2(9*n))) # Chai Wah Wu, Aug 10 2023

Formula

a(n) = binomial(9*n,3*n)*binomial(6*n,2*n)/binomial(9*n/2,4*n).

a(2*n) = A295437(n).

a(2*n) = 72*(18*n-1)*(18*n-5)*(18*n-7)*(18*n-11)*(18*n-13)*(18*n-17)/(n*(2*n-1)*(3*n-1)*(3*n-2)*(4*n-1)*(4*n-3))*a(2*n-2);

a(2*n+1) = 18432*(81*n^2-1)*(81*n^2-4)*(81*n^2-16)/(n*(2*n+1)*(16*n^2-1)*(36*n^2-1))*a(2*n-1).

Asymptotics: a(n) ~ 1/(2*sqrt(3*Pi*n)) * 2916^n as n -> infinity.

O.g.f.: A(x) = hypergeom([1/18, 5/18, 7/18, 11/18, 13/18, 17/18], [1/4, 1/3, 1/2, 2/3, 3/4], (2^4)*(3^12)*x^2) + 512*x*hypergeom([5/9, 7/9, 8/9, 10/9, 11/9, 13/9], [3/4, 5/6, 7/6, 5/4, 3/2], (2^4)*(3^12)*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.