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A347856
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a(n) = (6*n)!/((4*n)!*n!) * (n/2)!/(3*n/2)!.
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18
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1, 20, 990, 56576, 3432198, 215147520, 13768454700, 893768826880, 58626071754822, 3876225891958784, 257898242928604740, 17245961314149335040, 1158088115444301759900, 78040346182201091555328
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OFFSET
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0,2
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COMMENTS
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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.
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LINKS
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FORMULA
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a(n) = binomial(6*n,4*n)*binomial(2*n,n)/binomial(3*n/2,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.
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EXAMPLE
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Congruence: a(11) - a(1) = 17245961314149335040 - 20 = (2^2)*5*(11^3)*103*6289876694707 == 0 (mod 11^3).
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MAPLE
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seq( (6*n)!/((4*n)!*n!) * GAMMA(1+n/2)/GAMMA(1+3*n/2), n = 0..12);
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PROG
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(Python)
from math import factorial
from sympy import factorial2
def A347856(n): return int((factorial(6*n)*factorial2(n)<<n)//(factorial(n<<2)*factorial(n)*factorial2(3*n))) # Chai Wah Wu, Aug 10 2023
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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