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A347854 a(n) = (6*n)!/((3*n)!*(2*n)!) * (n/2)!/(3*n/2)!. 18
1, 40, 4620, 622336, 89237148, 13236695040, 2005604901300, 308350245273600, 47913489552349980, 7505566011722039296, 1183237138556438547120, 187495217080545878999040, 29836408028165719837829700, 4764790302634058161217077248 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
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)!*n!/((6*n)!*(4*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 1). See A295431. It is known that u(n) is integral and satisfies the congruences u(n*p) == u(n) ( mod p^3 ) for prime p >= 5 and any positive integer n (Zudilin, Section 5); the o.g.f. Sum_{n >= 0} u(n)*x^n is algebraic over Q(x) (Rodriguez-Villegas).
Here we are essentially considering the sequence ( u(n/2) )n>=0. The sequence is conjectured to be integral.
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.
Wadim Zudilin, Congruences for q-binomial coefficients, arXiv:1901.07843 [math.NT], 2019.
FORMULA
a(n) = binomial(6*n,2*n)*binomial(4*n,n)/binomial(3*n/2,n).
a(2*n) = A295431(n).
a(2*n) = 24*(12*n - 1)*(12*n - 5)*(12*n - 7)*(12*n - 11)/( n*(2*n - 1)*(3*n - 1)*(3*n - 2) )*a(2*n-2);
a(2*n+1) = 96*(12*n + 1)*(12*n - 1)*(12*n + 5)*(12*n - 5)/( n*(2*n + 1)*(6*n + 1)*(6*n - 1) )*a(2*n-1).
Asymptotics: a(n) ~ 32^n/sqrt(6*Pi*n) * 3^(3*n/2) as n -> infinity.
O.g.f.: A(x) = hypergeom([1/12, 5/12, 7/12, 11/12], [1/3, 1/2, 2/3], 27648*x^2) + 40*x*hypergeom([11/12, 13/12, 7/12, 17/12], [3/2, 5/6, 7/6], 27648*x^2) is conjectured to be algebraic over Q(x).
Conjectural: a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k.
From Karol A. Penson, Feb 17 2023: (Start)
An integral representation of a(n) as the n-th power moment of the weight function W(x) is given by a(n) = Integral_{x=0..96*sqrt(3)} x^n*W(x), where W(x) = W_1(x) + W_2(x) + W_3(x) + W_4(x) and the functions W_n(x) are:
W_1(x) = sqrt(2)*3^(3/4)*hypergeom([1/12, 5/12, 7/12, 3/4], [1/6, 1/2, 2/3], x^2/27648)*Gamma(3/4)/(18*sqrt(Pi)*x^(5/6)*Gamma(2/3)*Gamma(7/12)).
W_2(x) = sqrt(2)*cos((5*Pi)/12)*Gamma(2/3)*csc(Pi/12)*Gamma(3/4)*3^(1/4)* hypergeom([5/12, 3/4, 11/12, 13/12], [1/2, 5/6, 4/3], x^2/27648)/(2304*Pi^(3/2)* Gamma(11/12)*x^(1/6)).
W_3(x) = cos((5*Pi)/12)*3^(1/4)*Gamma(11/12)*x^(1/6)*hypergeom([7/12, 11/12, 13/12, 5/4], [2/3, 7/6, 3/2], x^2/27648)/(3456*sqrt(Pi)*Gamma(2/3)*Gamma(3/4)).
W_4(x) = 7*sin((5*Pi)/12)*Gamma(2/3)*Gamma(7/12)*3^(3/4)*x^(5/6)*hypergeom([11/12, 5/4,17/12, 19/12], [4/3, 3/2, 11/6], x^2/27648))/(1327104*Pi^(3/2)*Gamma(3/4)).
The function W(x) is positive on the support x = (0..96*sqrt(3)) and is singular at both endpoints of the support. The function W(x) is unique as it is the solution of the Hausdorff moment problem. (End)
EXAMPLE
a(11) - a(1) = 187495217080545878999040 - 40 = (2^3)*(5^3)*(11^3)*140867931690868429 == 0 (mod 11^3).
MAPLE
seq(binomial(6*n, 2*n)*binomial(4*n, n)/binomial(3*n/2, n), n = 0..13);
PROG
(Python)
from math import factorial
from sympy import factorial2
def A347854(n): return int((factorial(6*n)*factorial2(n)<<n)//(factorial(3*n)*factorial(n<<1)*factorial2(3*n))) # Chai Wah Wu, Aug 10 2023
CROSSREFS
Sequence in context: A059948 A229604 A045502 * A259461 A229692 A270053
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Sep 16 2021
STATUS
approved

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Last modified February 21 09:51 EST 2024. Contains 370228 sequences. (Running on oeis4.)