A295431 -id:A295431 - cp's OEIS frontend

A347854 a(n) = (6n)!/((3n)!(2n)!) * (n/2)!/(3*n/2)!.

1, 40, 4620, 622336, 89237148, 13236695040, 2005604901300, 308350245273600, 47913489552349980, 7505566011722039296, 1183237138556438547120, 187495217080545878999040, 29836408028165719837829700, 4764790302634058161217077248

Offset: 0

Peter Bala, Sep 16 2021

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.

			a(11) - a(1) = 187495217080545878999040 - 40 = (2^3)*(5^3)*(11^3)*140867931690868429 == 0 (mod 11^3).

Peter Bala, Some integer ratios of factorials
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.

Cf. A295431, A347855, A347856, A347857, A347858.

seq(binomial(6*n,2*n)*binomial(4*n,n)/binomial(3*n/2,n), n = 0..13);

from math import factorial
from sympy import factorial2
def A347854(n): return int((factorial(6*n)*factorial2(n)<Chai Wah Wu, Aug 10 2023

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)

A347855 a(n) = (4n)!/((2n)!(n)!) (n/3)!/(4*n/3)!.

Original entry on oeis.org

1, 9, 189, 4620, 120285, 3241134, 89237148, 2493521172, 70429218525, 2005604901300, 57481750139814, 1656023714623980, 47913489552349980, 1391243084942932620, 40519970408738302020, 1183237138556438547120

Offset: 0

Peter Bala, Sep 17 2021

Fractional factorials are defined using the Gamma function; for example, (n/3)! := Gamma(1 + n/3). 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. Here we are essentially considering the sequence (u(n/3))n>=0. The sequence is conjectured to be integral.

			Congruence: a(11) - a(1) = 1656023714623980 - 9 = (3^2)*7*(11^3)*17* 1161713471 == 0 (mod 11^3).

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.

Cf. A295431, A347854, A347855, A347856, A347857, A347858.

seq( (4*n)!/((2*n)!*(n)!) * GAMMA(1+n/3)/GAMMA(1+4*n/3), n = 0..15);

Table[Binomial[4n,2n] Binomial[2n,n]/Binomial[4 n/3,n],{n,0,20}] (* Harvey P. Dale, Apr 09 2022 *)

a(n) = binomial(4*n,2*n)*binomial(2*n,n)/binomial(4*n/3,n).
a(3*n) = A295431(n).
D-finite with recurrence -n*(n-1)*(n-2)*(2*n-3)*a(n) + 216*(4*n-11)*(4*n-1)*(4*n-5)*(4*n-7)*a(n-3).
Asymptotics: a(n) ~ 1/(2*sqrt(Pi*n))*2^(10*n/3)*3^n as n -> infinity.
O.g.f.: A(x) = hypergeom([11/12, 7/12, 5/12, 1/12], [2/3, 1/2, 1/3], 27648*x^3) + 9*x*hypergeom([11/12, 5/4, 5/12, 3/4], [5/6, 4/3, 2/3], 27648*x^3) + 189*x^2*hypergeom([19/12, 13/12, 5/4, 3/4], [7/6, 5/3, 4/3], 27648*x^3) 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.

A364173 a(n) = (9n)!(2n)!(3n/2)!/((9n/2)!(4n)!(3n)!*n!).

Original entry on oeis.org

1, 128, 43758, 17039360, 7012604550, 2976412336128, 1288415796384780, 565399665327996928, 250622090889055155270, 111950839825145979207680, 50312973039218473430585508, 22723567527558510746926055424, 10304958075870392958137083227804

Offset: 0

Peter Bala, Jul 13 2023

A295440, defined by A295440(n) = (18*n)!*(4*n)!*(3*n)! / ((9*n)!*(8*n)!*(6*n)!*(2*n)!), is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 10). Here we are essentially considering the sequence {A295440(n/2) : n >= 0}. Fractional factorials are defined in terms of the gamma function; for example, (3*n/2)! := Gamma(1 + 3*n/2).
This sequence is only conjecturally an integer sequence.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.

J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc., 79, Issue 2, (2009), 422-444.

Cf. A276100, A276101, A276102, A295431, A295440, A347854, A347855, A347856, A347857, A347858, A364172 - A364185.

seq( simplify((9*n)!*(2*n)!*(3*n/2)!/((9*n/2)!*(4*n)!*(3*n)!*n!)) , n = 0..15);

a(n) ~ c^n * 1/sqrt(4*Pi*n), where c = (3^7)/(2^3) * sqrt(3) = 473.4993895191418....

a(n) = 108*(9*n - 1)*(9*n - 5)*(9*n - 7)*(9*n - 11)*(9*n - 13)*(9*n - 17)/(n*(n - 1)*(4*n - 1)*(4*n - 3)*(4*n - 5)*(4*n - 7))*a(n-2) for n >= 2 with a(0) = 1 and a(1) = 128.

A211417 Integral factorial ratio sequence: a(n) = (30n)!n!/((15n)!(10n)!(6*n)!).

Original entry on oeis.org

1, 77636318760, 53837289804317953893960, 43880754270176401422739454033276880, 38113558705192522309151157825210540422513019720, 34255316578084325260482016910137568877961925210286281393760

Offset: 0

Peter Bala, Apr 11 2012

The integrality of this sequence can be used to prove Chebyshev's estimate C(1)*x/log(x) <= #{primes <= x} <= C(2)*x/log(x), for x sufficiently large; the constant C(1) = 0.921292... and C(2) = 1.105550.... Chebyshev's approach used the related step function floor(x) -floor(x/2) -floor(x/3) -floor(x/5) +floor(x/30). See A182067.
This sequence is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin.
The o.g.f. sum {n >= 0} a(n)*z^n is a generalized hypergeometric series of type 8F7 (see Bober, Table 2, Entry 31) and is an algebraic function of degree 483840 over the field of rational functions Q(z) (see Rodriguez-Villegas). Bober remarks that the monodromy group of the differential equation satisfied by the o.g.f. is W(E_8), the Weyl group of the E_8 root system.
See the Bala link for the proof that a(n), n = 0,1,2..., is an integer.
Congruences: a(p^k) == a(p^(k-1)) ( mod p^(3*k) ) for any prime p >= 5 and any positive integer k (write a(n) as C(30*n,15*n)*C(15*n,5*n)/C(6*n,n) and use equation 39 in Mestrovic, p. 12). More generally, the congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) may hold for any prime p >= 5 and any positive integers n and k. Cf. A295431. - Peter Bala, Jan 24 2020

N. J. A. Sloane, Table of n, a(n) for n = 0..50
Peter Bala, Proof of the integrality of A211417 and A211418
Frits Beukers, Hypergeometric functions, how special are they?, Notices Amer. Math. Soc. 61 (2014), no. 1, 48--56. MR3137256
J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc., 79, Issue 2, (2009), 422-444.
Florian Fürnsinn and Sergey Yurkevich, Algebraicity of hypergeometric functions with arbitrary parameters, arXiv:2308.12855 [math.CA], 2023.
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
Fernando Rodriguez Villegas, Integral ratios of factorials and algebraic hypergeometric functions, arXiv:math.NT/0701362, 2007.
Fernando Rodriguez Villegas, Mixed Hodge numbers and factorial ratios, arXiv:1907.02722 [math.NT], 2019.
K. Soundararajan, Integral Factorial Ratios, arXiv:1901.05133 [math.NT], 2019.
Wadim Zudilin, Integer-valued factorial ratios, MathOverflow question 26336, 2010.

Cf. A182067, A211418, A061162, A061163, A061164, A091496, A091527, A112292, A182400, A211419, A211420, A211421, A276100, A262733, A295431.

[Factorial(30*n)*Factorial(n)/(Factorial(15*n)*Factorial(10*n)*Factorial(6*n)): n in [0..10]]; // Vincenzo Librandi, Oct 03 2015

Table[(30 n)!*n!/((15 n)!*(10 n)!*(6 n)!), {n, 0, 5}] (* Michael De Vlieger, Oct 02 2015 *)

a(n) = (30*n)!*n!/((15*n)!*(10*n)!*(6*n)!);
vector(10, n, a(n-1)) \\ Altug Alkan, Oct 02 2015

a(n) ~ 2^(14*n-1) * 3^(9*n-1/2) * 5^(5*n-1/2) / sqrt(Pi*n). - Vaclav Kotesovec, Aug 30 2016

A364513 Square array read by ascending antidiagonals: T(n,k) = [x^k] (1 - x)^(2k) Legendre_P(n*k-1, (1 + x)/(1 - x)) for n, k >= 0.

Original entry on oeis.org

1, 1, -2, 1, -2, 6, 1, 0, 0, -20, 1, 4, 0, 16, 70, 1, 10, 126, 0, 0, -252, 1, 18, 594, 4900, 0, -252, 924, 1, 28, 1716, 44200, 209950, 0, 0, -3432, 1, 40, 3900, 205920, 3640210, 9513504, 0, 4800, 12870, 1, 54, 7650, 685216, 27386100, 317678760, 447103440, 0, 0, -48620, 1, 70, 13566, 1847560, 133501500, 3861534768, 28782923400, 21558808128, 0, -100100, 184756

Offset: 0

Peter Bala, Jul 31 2023

Compare with A364303.
Given two sequences of integers c = (c_1, c_2, ..., c_K) and d = (d_1, d_2, ..., d_L), where c_1 + ... + c_K = d_1 + ... + d_L, we can define the factorial ratio sequence u_n(c, d) = (c_1*n)!*(c_2*n)!* ... *(c_K*n)!/ ( (d_1*n)!*(d_2*n)!* ... *(d_L*n)! ) and ask whether it is integral for all n >= 0. The integer L - K is called the height of the sequence. Bober completed the classification of integral factorial ratio sequences of height 1 (see A295431). Soundararajan gives many examples of two-parameter families of integral factorial ratio sequences of height 2.
Each row of the present table is an integral factorial ratio sequence of height 2. It is usually assumed that the c's and d's are integers but here some of the c's and d's are half-integers. See A276098 and the cross references there for further examples of this type.

			 Square array begins:
 n\k|  0    1       2         3            4              5
  - + - - - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |  1   -2       6       -20           70           -252   ... (see A000984)
  1 |  1   -2       0        16            0           -252   ...  A364514
  2 |  1    0       0         0            0              0
  3 |  1    4     126      4900       209950        9513504   ...  (1/3)*A352651
  4 |  1   10     594     44200      3640210      317678760   ...  A364515
  5 |  1   18    1716    205920     27386100     3861534768   ...  (3/5)*A352652
  6 |  1   28    3900    685216    133501500    27583083528   ...  A364516
  7 |  1   40    7650   1847560    494944450   140625140040   ...  A364517

J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc. (2) 79 2009, 422-444.
K. Soundararajan, Integral factorial ratios: irreducible examples with height larger than 1, Phil. Trans. Royal Soc., A378: 2018044, 2019.
Wikipedia, Dixon's identity

Cf. A000984 (row 0 unsigned), A276098, A295431, A352651 (3*row 3), A352652 ((5/3)*row 5), A364303, A364506, A364509, A364514 (row 1), A364515 (row 4), A364516 (row 6), A364517 (row 7).

T(n,k) := coeff(series( (1 - x)^(2*k) * LegendreP(n*k-1, (1 + x)/(1 - x)), x, 11), x, k):
# display as a square array
seq(print(seq(T(n, k), k = 0..10)), n = 0..10);
# display as a sequence
seq(seq(T(n-k, k), k = 0..n), n = 0..10);

T(n,k) = Sum_{i = 0..k} binomial(n*k-1, k-i)^2 * binomial((n-2)*k+i-2, i).
T(n,1) = 1 for all n and for n >= 2 and k >= 1, T(n,k) = binomial((k*n-1), k)^2 * hypergeom([a, b, b], [1 + a - b, 1 + a - b], 1), where a = (n - 2)*k - 1 and b = -k.
For n >= 3 and k >= 1, T(n,k) = ((n*k - 1))! * ( ((n+2)*k - 1)/2 )! *  ( ((n-2)*k - 1)/2 )! / ( k!^2 *  ((n-2)*k - 1)! * ((n*k - 1)/2)!^2 ) by Dixon's 3F2 summation theorem, where fractional factorials are defined in terms of the gamma function.
For n >= 3 and k >= 1, T(n,k) = (n-2)/n * ((n+2)*k)!*(n*k/2)!^2 / ( ((n+2)*k/2)! * (n*k)! * ((n-2)*k/2)! * k!^2 ).
The central binomial numbers A000984, row 1 unsigned, satisfy the supercongruences A000984(n*p^r) == A000984(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r. We conjecture that each row sequence of the table satisfies the same supercongruences.

A304126 a(n) = (6n)!(4n)!/((2n)!(3n)!(5n)!).

Original entry on oeis.org

1, 12, 308, 8976, 276276, 8767512, 283728368, 9307523160, 308397041460, 10297160887440, 345907798472808, 11677249143964768, 395812039652176368, 13462467839121604380, 459228293024549285160, 15704603616059963888976, 538244114669755948787508, 18482918763808824442733616

Offset: 0

Karol A. Penson, May 07 2018

nonn

From Bober's reference, s. Theorem 1.2: Eq.(9) for a=3, b=2 gives a(n).

J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc. (2) 79 (2009), 422-444.

Cf. A295431.

seq((6*n)!*(4*n)!/((2*n)!*(3*n)!*(5*n)!),n=0..17);
# Alternative (to illustrate the constant C):
k := [10,15,30,45,50]/60: j := [12,24,36,48,60]/60: C := 110592/3125:
a := n -> C^n*mul(pochhammer(k[i],n)/pochhammer(j[i],n), i=1..5): # Peter Luschny, May 07 2018

a[n_] := (2^(10 n) Gamma[1/2 + 2 n] Gamma[1/2 + 3 n])/(Pi Gamma[1 + 5 n]);
Table[a[n], {n, 0, 17}] (* Peter Luschny, May 07 2018 *)
Table[((6n)!(4n)!)/((2n)!(3n)!(5n)!),{n,0,20}] (* Harvey P. Dale, Jul 26 2019 *)

O.g.f.: hypergeom([1/6,1/4,1/2,3/4,5/6],[1/5,2/5,3/5,4/5],110592*z/3125).
a(n) ~ sqrt(10)*sqrt(1/n)*(32248627200000*n^4 - 1657221120000*n^3 + 42581376000*n^2 + 12688032960*n - 680136331)*3125^(-n)*110592^n/(161243136000000*n^4*sqrt(Pi)).
Integral representation as the n-th moment of the positive function V(x) on x = (0, 110592/3125), i.e. in Maple notation: a(n) = int(x^n*V(x),x=0..110592/3125), where V(x) = 2^(1/3)*hypergeom([1/6, 11/30, 17/30, 23/30, 29/30], [1/3, 5/12, 2/3, 11/12], 3125*x*(1/110592))/(12*Pi*x^(5/6))+sqrt(2)*hypergeom([1/4, 9/20, 13/20, 17/20, 21/20], [5/12, 1/2, 3/4, 13/12], 3125*x*(1/110592))/(16*Pi*x^(3/4)) +cos((1/5)*Pi)*cos(2*Pi*(1/5))*hypergeom([1/2, 7/10, 9/10, 11/10, 13/10], [2/3, 3/4, 5/4, 4/3], 3125*x*(1/110592))/(16*Pi*sqrt(x))+11*sqrt(2)*hypergeom([3/4, 19/20, 23/20, 27/20, 31/20], [11/12, 5/4, 3/2, 19/12], 3125*x*(1/110592))/ (2048*Pi*x^(1/4))+247*2^(2/3)*hypergeom([5/6, 31/30, 37/30, 43/30, 49/30], [13/12, 4/3, 19/12, 5/3], 3125*x*(1/110592))/(110592*Pi*x^(1/6)). The function V(x) is singular at both edges of its support and is U-shaped. The function V(x) is unique as it is the solution of the Hausdorff moment problem.
a(n) = (2^(10*n)*Gamma(1/2 + 2*n)*Gamma(1/2 + 3*n))/(Pi*Gamma(1 + 5*n)). - Peter Luschny, May 07 2018
D-finite with recurrence 5*n*(5*n-4)*(5*n-3)*(5*n-2)*(5*n-1)*a(n) -96*(6*n-5)*(6*n-1)*(4*n-3)*(2*n-1)*(4*n-1)*a(n-1)=0. - R. J. Mathar, Jul 27 2022
From Karol A. Penson, Nov 18 2022: (Start)
O.g.f. = h(z) satisfies the following  algebraic equation of order 16:
268435456*z - 759375 + (-8657043456*z + 32400000)*h(z) + (113508352000*z - 590625000)*h(z)^2 + (-737869824000*z + 5850000000)*h(z)^3 + (1099511627776*z^2 + 2023598080000*z - 32039062500)*h(z)^4 + (-22265110462464*z^2 + 2487091200000*z + 72500000000)*h(z)^5 + (160030481448960*z^2 - 30842368000000*z + 172265625000)*h(z)^6 + (-234827336908800*z^2 + 57507840000000*z - 1437500000000)*h(z)^7 + (-2334117593088000*z^2 + 82630400000000*z + 1970214843750)*h(z)^8 + (8806025134080000*z^2 - 497664000000000*z + 7031250000000)*h(z)^9 + (25649407252758528*z^3 + 2525390438400000*z^2 + 659840000000000*z - 21240234375000)*h(z)^10 + (519400496868360192*z^3 - 44030125670400000*z^2 + 1244160000000000*z - 11718750000000)*h(z)^11 + (405781638178406400*z^3 + 80263249920000000*z^2 - 5508000000000000*z + 82397460937500)*h(z)^12 + (6763027302973440000*z^3 - 573308928000000000*z^2 + 16200000000000000*z - 152587890625000)*h(z)^14 + (149587343098087735296*z^4 - 16907568257433600000*z^3 + 716636160000000000*z^2 - 13500000000000000*z + 95367431640625)*h(z)^16=0. (End)

A295432 a(n) = (12n)!(3n)!(2n)!/((6n)!(6n)!(4n)!*n!).

Original entry on oeis.org

1, 462, 579462, 825012300, 1240292449350, 1922257130698212, 3037843525927941276, 4866407206411278522456, 7873830606510488097067590, 12837724391783995395083457780, 21058175422921386792478888300212, 34712605923460405627194955063564200

Offset: 0

Gheorghe Coserea, Nov 23 2017

nonn

Cf. A295431.

Array[(12 #)!*(3 #)!*(2 #)!/((6 #)!*(6 #)!*(4 #)!*#!) &, 12, 0] (* Michael De Vlieger, Nov 23 2017 *)
CoefficientList[ Series[ HypergeometricPFQ[{1/12, 5/12, 7/12, 11/12}, {1/6, 1/2, 5/6}, 1728 x], {x, 0, 11}], x] (* Robert G. Wilson v, Nov 23 2017 *)

G.f.: hypergeom([1/12, 5/12, 7/12, 11/12], [1/6, 1/2, 5/6], 1728*x).
a(n) = a(n-1)*6*(12*n - 1)*(12*n - 5)*(12*n - 7)*(12*n - 11)/(n*(2*n - 1)*(6*n - 1)*(6*n - 5)). - Neven Sajko, Jul 22 2023
a(n) ~ 2^(6*n - 1) * 3^(3*n) / sqrt(Pi*n). - Vaclav Kotesovec, Jul 11 2025

A295433 a(n) = (12n)!n!/((8n)!(3n)!(2*n)!).

Original entry on oeis.org

1, 990, 3432198, 13768454700, 58626071754822, 257898242928604740, 1158088115444301759900, 5274648294443835666191640, 24274428407219984638641697350, 112600884656246368465947735787860, 525586819747900180429387794022656948, 2465690419601453784410118364323664257000

Offset: 0

Gheorghe Coserea, Nov 23 2017

nonn

Cf. A295431, A495432.

Array[(12 #)!*#!/((8 #)!*(3 #)!*(2 #)!) &, 12, 0] (* Michael De Vlieger, Nov 23 2017 *)
CoefficientList[ Series[ HypergeometricPFQ[{1/12, 1/6, 5/12, 7/12, 5/6, 11/12}, {1/8, 3/8, 1/2, 5/8, 7/8}, 19683 x], {x, 0, 10}], x] (* Robert G. Wilson v, Nov 23 2017 *)

G.f.: 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).
a(n) ~ 3^(9*n) / (sqrt(Pi*n) * 2^(2*n + 3/2)). - Vaclav Kotesovec, Oct 06 2021
From  Karol A. Penson, Jun 13 2025: (Start)
a(n) = Integral_{x=0..19683/4} x^n*W(x)*dx, where W(x) is the Meijer G-function which in the Maple notation is equal to (sqrt(2)/(19683*sqrt(Pi)))*MeijerG([[],[0,-1/2,-7/8,-5/8,-3/8,-1/8]],[[-11/12,-5/6,-7/12,-5/12,-1/6,-1/12],[]],4*x/19683).
W(x) can be represented in terms of 6 different generalized hypergeometric functions of type 6F5 of argument 4*x/19683. W(x) is a positive, U-shaped function on x=(0,19683/4), and is singular at x=0 and at x=19683/4. Thus a(n) is a positive definite sequence. This representation is unique as W(x) is the solution of the Hausdorff moment problem. (End)

A295437 a(n) = (18n)!n!/((9n)!(6n)!(4*n)!).

Original entry on oeis.org

1, 1021020, 6016814703900, 41477110789150966020, 304331361887290342345862940, 2309513382640863232775760738593520, 17900786683992090777583802478540374204100, 140773522690118901080808611014971776285838687000

Offset: 0

Gheorghe Coserea, Nov 23 2017

nonn

Cf. A295431.

Array[(18 #)!*#!/((9 #)!*(6 #)!*(4 #)!) &, 8, 0] (* Michael De Vlieger, Nov 23 2017 *)
CoefficientList[ Series[ HypergeometricPFQ[{1/18, 5/18, 7/18, 11/18, 13/18, 17/18}, {1/4, 1/3, 1/2, 2/3, 3/4}, 8503056 x], {x, 0, 8}], x] (* Robert G. Wilson v, Nov 23 2017 *)

G.f.: hypergeom([1/18, 5/18, 7/18, 11/18, 13/18, 17/18], [1/4, 1/3, 1/2, 2/3, 3/4], 8503056*x).

a(n) ~ 2^(4*n - 3/2) * 3^(12*n - 1/2) / sqrt(Pi*n). - Vaclav Kotesovec, Jul 11 2025

A364172 a(n) = (6n)!(n/3)!/((3n)!(2n)!(4*n/3)!).

Original entry on oeis.org

1, 45, 6237, 1021020, 178719453, 32427545670, 6016814703900, 1133540594837892, 215925912619400925, 41477110789150966020, 8019784929635201045862, 1558875476359831844951100, 304331361887290342345862940, 59629409730107012112361325820

Offset: 0

Peter Bala, Jul 12 2023

Fractional factorials are defined in terms of the gamma function; for example, (n/3)! := Gamma(n/3 + 1).
Given two sequences of numbers c = (c_1, c_2, ..., c_K) and d = (d_1, d_2, ..., d_L) where c_1 + ... + c_K = d_1 + ... + d_L  we can define the factorial ratio sequence u_n(c, d) = (c_1*n)!*(c_2*n)!* ... *(c_K*n)!/ ( (d_1*n)!*(d_2*n)!* ... *(d_L*n)! ) and ask whether it is integral for all n >= 0. The integer L - K is called the height of the sequence. Bober completed the classification of integral factorial ratio sequences of height 1. For a list of the 52 sporadic integral factorial ratio sequences see A295431.
It is usually assumed that the c's and d's are integers but here we allow for some of the c's and d's to be rational numbers.
A295437, defined by A295437(n) = (18*n)!*n! / ((9*n)!*(6*n)!*(4*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). Here we are essentially considering the sequence {A295437(n/3) : n >= 0}. This sequence is only conjecturally an integer sequence.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.

Peter Bala, Some integer ratios of factorials
J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc., 79, Issue 2, (2009), 422-444.

Cf. A276100, A276101, A276102, A295431, A295437, A347854, A347855, A347856, A347857, A347858, A364173 - A364185.

seq( simplify((6*n)!*(n/3)!/((3*n)!*(2*n)!*(4*n/3)!)), n = 0..15);

Table[Product[36*(6*k - 5)*(6*k - 1)/(k*(3*k + n)), {k, 1, n}], {n, 0, 20}] (*  Vaclav Kotesovec, Jul 13 2023 *)

a(n) ~ 2^(4*n/3 - 3/2) * 3^(4*n) / sqrt(Pi*n). - Vaclav Kotesovec, Jul 13 2023

a(n) = 5832*(6*n - 1)*(6*n - 5)*(6*n - 7)*(6*n - 11)*(6*n - 13)*(6*n - 17)/(n*(n - 1)*(n - 2)*(2*n - 3)*(4*n - 3)*(4*n - 9))*a(n-3) for n >= 3 with a(0) = 1, a(1) = 45 and a(2) = 6237.

cp's OEIS Frontend

A347854 a(n) = (6*n)!/((3*n)!*(2*n)!) * (n/2)!/(3*n/2)!.

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A347855 a(n) = (4*n)!/((2*n)!*(n)!) * (n/3)!/(4*n/3)!.

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A364173 a(n) = (9*n)!*(2*n)!*(3*n/2)!/((9*n/2)!*(4*n)!*(3*n)!*n!).

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A211417 Integral factorial ratio sequence: a(n) = (30*n)!*n!/((15*n)!*(10*n)!*(6*n)!).

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A364513 Square array read by ascending antidiagonals: T(n,k) = [x^k] (1 - x)^(2*k) * Legendre_P(n*k-1, (1 + x)/(1 - x)) for n, k >= 0.

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A304126 a(n) = (6*n)!*(4*n)!/((2*n)!*(3*n)!*(5*n)!).

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A295432 a(n) = (12*n)!*(3*n)!*(2*n)!/((6*n)!*(6*n)!*(4*n)!*n!).

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A295433 a(n) = (12*n)!*n!/((8*n)!*(3*n)!*(2*n)!).

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A347854 a(n) = (6n)!/((3n)!(2n)!) * (n/2)!/(3*n/2)!.

A347855 a(n) = (4n)!/((2n)!(n)!) (n/3)!/(4*n/3)!.

A364173 a(n) = (9n)!(2n)!(3n/2)!/((9n/2)!(4n)!(3n)!*n!).

A211417 Integral factorial ratio sequence: a(n) = (30n)!n!/((15n)!(10n)!(6*n)!).

A364513 Square array read by ascending antidiagonals: T(n,k) = [x^k] (1 - x)^(2k) Legendre_P(n*k-1, (1 + x)/(1 - x)) for n, k >= 0.

A304126 a(n) = (6n)!(4n)!/((2n)!(3n)!(5n)!).

A295432 a(n) = (12n)!(3n)!(2n)!/((6n)!(6n)!(4n)!*n!).

A295433 a(n) = (12n)!n!/((8n)!(3n)!(2*n)!).

A295437 a(n) = (18n)!n!/((9n)!(6n)!(4*n)!).

A364172 a(n) = (6n)!(n/3)!/((3n)!(2n)!(4*n/3)!).