A211420 -id:A211420 - cp's OEIS frontend

A211419 a(n) = (6n)!(2n)!/((4n)!(3n)!*n!).

1, 10, 198, 4420, 104006, 2521260, 62300700, 1560167752, 39457579590, 1005490725148, 25776935824948, 664048851069240, 17175945353271068, 445775181599116600, 11602978540817349240, 302767701121286251920, 7917664916276259668550, 207452338901630123085180

Offset: 0

Peter Bala, Apr 10 2012

This sequence is the particular case a = 3, b = 2 of the following result (see Bober, Theorem 1.2): Let a, b be nonnegative integers with a > b and gcd(a,b) = 1. Then (2*a*n)!*(b*n)!/((a*n)!*(2*b*n)!*((a-b)*n)!) is an integer for all integer n >= 0. Other cases include A061162 (a = 3, b = 1), A211420 (a = 4, b = 1), A211421 (a = 4, b = 3) and A061163 (a = 5, b = 1).
This is the case m = 3n in Catalan's formula (2m)!*(2n)!/(m!*(m+n)!*n!) - see Umberto Scarpis in References. - Bruno Berselli, Apr 27 2012
Sequence terms are given by the coefficient of x^n in the expansion of ((1 + x)^(k+2)/(1 - x)^k)^n when k = 4. See the cross references for related sequences obtained from other values of k. - Peter Bala, Sep 29 2015

Umberto Scarpis, Sui numeri primi e sui problemi dell'analisi indeterminata in Questioni riguardanti le matematiche elementari, Nicola Zanichelli Editore (1924-1927, third Edition), page 11.

Seiichi Manyama, Table of n, a(n) for n = 0..500
Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion
J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977v1 [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. A061162, A061163, A182400, A211420, A211421.

Cf. A000984 (k=0), A091527 (k=1), A001448 (k=2), A262732 (k=3), A262733 (k=5), A211421 (k=6), A262738.

[Factorial(6*n) * Factorial(2*n) / (Factorial(4*n) * Factorial(3*n) * Factorial(n)): n in [0..20]]; // Vincenzo Librandi, May 03 2018

A211419 := n-> (6*n)!*(2*n)!/((4*n)!*(3*n)!*n!):
seq(A211419(n), n=0..20);
# Using the o.g.f. from Karol A. Penson and Jean-Marie Maillard:
u := 27*x-1: c := (u^3*((3*x*u)^(1/2)*(12+81*x)-u^2+216*x-7))^(1/3):
gf := ((c^2-2*c*u+27*u*(7-81*x)*x-4*u)/(6*c*u^2))^(1/2):
ser := series(gf, x, 8); # Peter Luschny, May 03 2018
ogf := hypergeom([1/6, 1/2, 5/6], [1/4, 3/4], 27*z): ser := series(ogf, z, 20):
seq(coeff(ser, z, n), n = 0..17);  # Peter Luschny, Feb 22 2024

Table[(6 n)!*(2 n)!/((4 n)!*(3 n)!*n!), {n, 0, 16}] (* Michael De Vlieger, Oct 04 2015 *)
CoefficientList[Series[Sqrt[(4 + 7290 x^2 - 59049 x^3 + 2 (8 + 3 Sqrt[3] Sqrt[x (-1 + 27 x)^7 (4 + 27 x)^2] - 27 x (-2 + 27 x) (-17 + 27 x (19 + 27 x (-11 + 27 x))))^(1/3) + (8 + 3 Sqrt[3] Sqrt[x (-1 + 27 x)^7 (4 + 27 x)^2] - 27 x (-2 + 27 x) (-17 + 27 x (19 + 27 x (-11 + 27 x))))^(2/3) - 27 x (11 + 2 (8 + 3 Sqrt[3] Sqrt[x (-1 + 27 x)^7 (4 + 27 x)^2] - 27 x (-2 + 27 x) (-17 + 27 x (19 + 27 x (-11 + 27 x))))^(1/3)))/(6 (1 - 27 x)^2 (8 + 3 Sqrt[3] Sqrt[x (-1 + 27 x)^7 (4 + 27 x)^2] - 27 x (-2 + 27 x) (-17 + 27 x (19 + 27 x (-11 + 27 x))))^(1/3))],{x,0,16}],x] (* Karol A. Penson and Jean-Marie Maillard, May 02 2018 *)

a(n) = (6*n)!*(2*n)!/((4*n)!*(3*n)!*n!);
vector(30, n, a(n-1)) \\ Altug Alkan, Oct 02 2015

The o.g.f. Sum_{n >= 1} a(n)*z^n is algebraic over the field of rational functions Q(z) (see Rodriguez-Villegas).
From Peter Bala, Sep 29 2015: (Start)
a(n) = Sum_{i = 0..n} binomial(6*n, i) * binomial(5*n-i-1, n-i).
a(n) = [x^n] ( (1 + x)^6/(1 - x)^4 )^n.
O.g.f. exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 10*x + 149*x^2 + 2630*x^3 + 51002*x^4 + ... has integer coefficients and equals 1/x * series reversion of x*(1 - x)^4/ (1 + x)^6. See A262738. (End)
a(n) ~ 27^n/sqrt(2*Pi*n). - Ilya Gutkovskiy, Jul 31 2016
O.g.f.: sqrt((4 + 7290*x^2 - 59049*x^3 + 2*(8 + 3*sqrt(3)*sqrt(x*(-1 + 27*x)^7*(4 + 27*x)^2) - 27*x*(-2 + 27*x)*(-17 + 27*x*(19 + 27*x*(-11 + 27*x))))^(1/3) + (8 + 3*sqrt(3)*sqrt(x*(-1 + 27*x)^7*(4 + 27*x)^2) - 27*x*(-2 + 27*x)*(-17 + 27*x*(19 + 27*x*(-11 + 27*x))))^(2/3) - 27*x*(11 + 2*(8 + 3*sqrt(3)*sqrt(x*(-1 + 27*x)^7*(4 + 27*x)^2) - 27*x*(-2 + 27*x)*(-17 + 27*x*(19 + 27*x*(-11 + 27*x))))^(1/3)))/(6*(1 - 27*x)^2*(8 + 3*sqrt(3)*sqrt(x*(-1 + 27*x)^7*(4 + 27*x)^2) - 27*x*(-2 + 27*x)*(-17 + 27*x*(19 + 27*x*(-11 + 27*x))))^(1/3))). - Karol A. Penson and Jean-Marie Maillard, May 02 2018
Right-hand side of the binomial sum identity: Sum_{k = 0..2*n} (-1)^(n+k) * binomial(6*n, 2*n+k) * binomial(2*n, k) = (6*n)!*(2*n)!/((4*n)!*(3*n)!*n!). - Peter Bala, Jan 19 2020
a(n) = 6*(6*n - 1)*(2*n - 1)*(6*n - 5)*a(n-1)/(n*(4*n - 1)*(4*n - 3)). - Neven Sajko, Jul 19 2023
From Peter Luschny, Feb 22 2024: (Start)
a(n) = 4^n*(Gamma(3*n + 1/2)/Gamma(2*n + 1/2))/Gamma(n + 1).
O.g.f.: hypergeom([1/6, 1/2, 5/6], [1/4, 3/4], 27*z). (End)
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] 1/((1-x)^(n+1) * (1-2*x)^(4*n)).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(6*n,k) * binomial(2*n-k,n-k).
a(n) = Sum_{k=0..n} 2^k * binomial(4*n+k-1,k) * binomial(2*n-k,n-k).
a(n) = 4^n * binomial((6*n-1)/2,n).
a(n) = [x^n] 1/(1-4*x)^((4*n+1)/2).
a(n) = [x^n] (1+4*x)^((6*n-1)/2). (End)

A211421 Integral factorial ratio sequence: a(n) = (8n)!(3n)!/((6n)!(4n)!*n!).

Original entry on oeis.org

1, 14, 390, 12236, 404550, 13777764, 478273692, 16825310040, 597752648262, 21397472070260, 770557136489140, 27884297395587240, 1013127645555452700, 36935287875280348776, 1350441573221798941560, 49498889739033621986736, 1818284097150186829038150

Offset: 0

Peter Bala, Apr 10 2012

This sequence is the particular case a = 4, b = 3 of the following result (see Bober, Theorem 1.2): let a, b be nonnegative integers with a > b and GCD(a,b) = 1. Then (2*a*n)!*(b*n)!/((a*n)!*(2*b*n)!*((a-b)*n)!) is an integer for all integer n >= 0. Other cases include A061162 (a = 3, b = 1), A211419 (a = 3, b = 2) and A211420 (a = 4, b = 1).

Sequence terms are given by the coefficient of x^n in the expansion of ( (1 + x)^(k+2)/(1 - x)^k )^n when k = 6. See the cross references for related sequences obtained from other values of k. - Peter Bala, Sep 29 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..600
Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion
J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, 2007, arXiv:0709.1977v1 [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. A061162, A061163, A211419, A211420.

Cf. A000984 (k = 0), A091527 (k = 1), A001448 (k = 2), A262732 (k = 3), A211419 (k = 4), A262733 (k = 5), A262740.

[Factorial(8*n)*Factorial(3*n)/(Factorial(6*n)*Factorial(4*n)*Factorial(n)): n in [0..20]]; // Vincenzo Librandi, Aug 01 2016

#A211421
a := n -> (8*n)!*(3*n)!/((6*n)!*(4*n)!*n!);
seq(a(n), n = 0..16);

Table[(8 n)!*(3 n)!/((6 n)!*(4 n)!*n!), {n, 0, 15}] (* Michael De Vlieger, Oct 04 2015 *)

a(n) = (8*n)!*(3*n)!/((6*n)!*(4*n)!*n!);
vector(30, n, a(n-1)) \\ Altug Alkan, Oct 02 2015

The o.g.f. sum {n >= 1} a(n)*z^n is algebraic over the field of rational functions Q(z) (see Rodriguez-Villegas).
From Peter Bala, Sep 29 2015: (Start)
a(n) = Sum_{i = 0..n} binomial(8*n,i) * binomial(7*n-i-1,n-i).
a(n) = [x^n] ( (1 + x)^8/(1 - x)^6 )^n.
a(0) = 1 and a(n) = 2*(8*n - 1)*(8*n - 3)*(8*n - 5)*(8*n - 7)/( n*(6*n - 1)*(6*n - 3)*(6*n - 5) ) * a(n-1) for n >= 1.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 14*x + 293*x^2 + 7266*x^3 + 197962*x^4 + 5726364*x^5 + ... has integer coefficients and equals 1/x * series reversion of x*(1 - x)^6/(1 + x)^8. See A262740. (End)
a(n) ~ 2^(10*n)*27^(-n)/sqrt(2*Pi*n). - Ilya Gutkovskiy, Jul 31 2016
a(n) = (2^n/n!)*Product_{k = 3*n..4*n-1} (2*k + 1). - Peter Bala, Feb 26 2023
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] 1/((1-x)^(n+1) * (1-2*x)^(6*n)).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(8*n,k) * binomial(2*n-k,n-k).
a(n) = Sum_{k=0..n} 2^k * binomial(6*n+k-1,k) * binomial(2*n-k,n-k).
a(n) = 4^n * binomial((8*n-1)/2,n).
a(n) = [x^n] 1/(1-4*x)^((6*n+1)/2).
a(n) = [x^n] (1+4*x)^((8*n-1)/2). (End)

A061162 a(n) = (6n)!n!/((3n)!(2n)!^2).

Original entry on oeis.org

1, 30, 2310, 204204, 19122246, 1848483780, 182327718300, 18236779032600, 1842826521244230, 187679234340049620, 19232182592635611060, 1980665038436368775400, 204826599735691440534300, 21255328931341321610645544, 2212241139727064219063537016

Offset: 0

Richard Stanley, Apr 17 2001

According to page 781 of the cited reference the generating function F(x) for a(n) is algebraic but not obviously so and the minimal polynomial satisfied by F(x) is quite large.

This sequence is the particular case a = 3, b = 1 of the following result (see Bober, Theorem 1.2): let a, b be nonnegative integers with a > b and gcd(a,b) = 1. Then (2*a*n)!*(b*n)!/((a*n)!*(2*b*n)!*((a-b)*n)!) is an integer for all integer n >= 0. Other cases include A211419 (a = 3, b = 2), A211420 (a = 4, b = 1) and A211421 (a = 4, b = 3) and A061163 (a = 5, b = 1). The o.g.f. Sum_{n >= 1} a(n)*z^n is algebraic over the field of rational functions Q(z) (see Rodriguez-Villegas). - Peter Bala, Apr 10 2012

M. Kontsevich and D. Zagier, Periods, in Mathematics Unlimited - 2001 and Beyond, Springer, Berlin, 2001, pp. 771-808.
R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Harry J. Smith, Table of n, a(n) for n = 0..100
J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, 2007, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc., Vol. 79, Issue 2, (2009) 422-444.
M. Kontsevich and D. Zagier, Periods, Institut des Hautes Etudes Scientifiques 2001 IHES/M/01/22, p. 11.
W. Mlotkowski and Karol A. Penson, Probability distributions with binomial moments, arXiv preprint arXiv:1309.0595 [math.PR], 2013.
F. Rodriguez-Villegas, Integral ratios of factorials and algebraic hypergeometric functions, arXiv:math/0701362 [math.NT], 2007.

Cf. A061163, A061164, A007297, A091527, A262732.

See also A091527, A211419, A211420, A211421.

A061162 := n->(6*n)!*n!/((3*n)!*(2*n)!^2);

a[n_] := 16^n Gamma[3 n + 1/2]/(Gamma[n + 1/2] Gamma[2 n + 1]);
Table[a[n], {n, 0, 14}] (* Peter Luschny, Mar 01 2018 *)

{ for (n=0, 100, write("b061162.txt", n, " ", (6*n)!*n!/((3*n)!*(2*n)!^2)) ) } \\ Harry J. Smith, Jul 18 2009

a(n) ~ 1/2*Pi^(-1/2)*n^(-1/2)*2^(2*n)*3^(3*n)*{1 - 1/72*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Jun 11 2002
n*(2*n-1)*a(n) -6*(6*n-1)*(6*n-5)*a(n-1)=0. - R. J. Mathar, Oct 26 2014
From Peter Bala, Aug 21 2016: (Start)
a(n) = Sum_{k = 0..2*n} binomial(6*n, k)*binomial(4*n - k - 1, 2*n - k).
a(n) = Sum_{k = 0..n} binomial(8*n, 2*n - 2*k)*binomial(2*n + k - 1, k).
O.g.f. A(x) = Hypergeom([5/6, 1/6], [1/2], 108*x).
a(n) = [x^(2*n)] H(x)^n, where H(x) = (1 + x)^6/(1 - x)^2. Cf. A091496 and A262732. It follows that the o.g.f. A(x) for this sequence is the diagonal of the bivariate rational generating function 1/2*( 1/(1 - t*H(sqrt(x))) + 1/(1 - t*H(-sqrt(x))) ) and hence is algebraic by Stanley 1999, Theorem 6.33, p. 197.
Let G(x) = 1/x * series reversion( x*(1 - x)/(1 + x)^3 ) = 1 + 4*x + 23*x^2 + 156*x^3 + 1162*x^4 + ..., essentially the o.g.f. for A007297. Then A(x^2) equals the even part of 1 + x*(d/dx log(G(x))).
exp(Sum_{n >= 1} a(n)*x^n/n) = F(x), where F(x) = 1 + 30*x + 1605*x^2 + 107218*x^3 + 8043114*x^4 + 647773116*x^5 + 54730094637*x^6 + ... has integer coefficients since F(x^2) = G(x)*G(-x). Furthermore, F(x)^(1/6) = 1 + 5*x + 205*x^2 + 12328*x^3 + 874444*x^4 + 68022261*x^5 + 5613007167*x^6 + ... appears to have all integer coefficients. (End)
a(n) is the n-th moment of the positive weight function w(x) on x = (0,108), i.e.: a(n) = Integral_{x=0..108} x^n*w(x) dx, n >= 0, where w(x) = sqrt(3)*(1 + sqrt(1 - x/108))^(2/3)/(12*2^(1/3)*Pi*x^(5/6)*sqrt(1 - x/108)) + 2^(4/3)*sqrt(3)/(864*Pi*x^(1/6)*(1 + sqrt(1 - x/108))^(2/3)*sqrt(1 - x/108)). The weight function w(x) is singular at x=0 and at x=108 and is the solution of the Hausdorff moment problem. This solution is unique. - Karol A. Penson, Mar 01 2018
a(n) = 2^(4*n)*binomial(-n-1/2, 2*n). - Ira M. Gessel, Jan 04 2025

A061163 a(n) = (10n)!n!/((5n)!(4n)!*(2n)!).

Original entry on oeis.org

1, 630, 1385670, 3528923580, 9540949030470, 26651569523959380, 75998432812419471900, 219813190240007470094520, 642409325786050322446410310, 1892390644737640220059489996260

Offset: 0

Richard Stanley, Apr 17 2001

According to page 781 of the cited reference the generating function F(x) for a(n) is algebraic but not obviously so and the minimal polynomial satisfied by F(x) is quite large.
This sequence is the particular case a = 5, b = 1 of the following result (see Bober, Theorem 1.2): let a, b be nonnegative integers with a > b and GCD(a,b) = 1. Then (2*a*n)!*(b*n)!/((a*n)!*(2*b*n)!*((a-b)*n)!) is an integer for all integer n >= 0. Other cases include A061162 (a = 3, b = 1), A211419 (a = 3, b = 2) and A211420(a = 4, b = 1) and A211421 (a = 4, b = 3). The o.g.f. Sum_{n >= 1} a(n)*z^n is algebraic over the field of rational functions Q(z) (see Rodriguez-Villegas). - Peter Bala, Apr 10 2012
Continuing the comment above: This is case n = 4 of the array of sequences
  A(n, k) = 4^(n*k)*(Gamma((n + 1)*k + 1/2)/Gamma(k + 1/2)) / Gamma(n * k + 1). See the cross-references for other cases. - Peter Luschny, Feb 21 2024

M. Kontsevich and D. Zagier, Periods, in Mathematics Unlimited - 2001 and Beyond, Springer, Berlin, 2001, pp. 771-808.

J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, 2007, arXiv:0709.1977 [math.NT], 2007; Jour. of the London Math. Soc., Vol. 79, Issue 2, 422-444.
M. Kontsevich and D. Zagier, Periods, Institut des Hautes Etudes Scientifiques 2001 IHES/M/01/22, p. 11.
F. Rodriguez-Villegas, Integral ratios of factorials and algebraic hypergeometric functions, arXiv:math/0701362 [math.NT], 2007.

Cf. A061164, A211419, A211421.

Cf. A000012 (n=0), A001448 (n=1), A061162 (n=2), A211420 (n=3), this sequence (n=4).

A061163 := n->(10*n)!*n!/((5*n)!*(4*n)!*(2*n)!);
# Alternative:
A := (n, k) -> 4^(n*k)*(GAMMA((n + 1)*k + 1/2)/GAMMA(k + 1/2))/GAMMA(n*k + 1):
seq(A(4, k), k = 0..9);  # Peter Luschny, Feb 21 2024

Table[(10n)! n!/((5n)!(4n)!(2n)!),{n,0,10}] (* Harvey P. Dale, Oct 24 2022 *)

n*(4*n-3)*(2*n-1)*(4*n-1)*a(n) -10*(10*n-9)*(10*n-7)*(10*n-3)*(10*n-1)*a(n-1)=0. - R. J. Mathar, Oct 26 2014
O.g.f. is a generalized hypergeometric function 4F3([1/10, 3/10, 7/10, 9/10], [1/4, 1/2, 3/4], 5^5*z). - Karol A. Penson, Apr 13 2022
From Karol A. Penson, Feb 21 2024: (Start)
(O.g.f.(z))^2 satisfies the algebraic equation of order 15, in which the powers of (O.g.f.(z))^2 are multiplied by polynomials p(n, z) with integer coefficients, in the form: Sum_{n = 0..15} p(n, z)*(O.g.f.(z))^(2*n) = 0.
Here is the list of orders, in the variable z, of all polynomials p(n, z) for n=0..15: 9,9,9,9,9,10,10,10,10,10,10,11,11,11,11,11,12. For example p(15, z) = 2^50*(5^5*z-1)^12. (End)
a(n) ~ 5^(5*n) / (2^(3/2) * sqrt(Pi*n)). - Vaclav Kotesovec, Aug 27 2024

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

A364519 Square array read by ascending antidiagonals: T(n,k) = [x^(3*k)] ( (1 + x)^(n+3)/(1 - x)^(n-3) )^k for n, k >= 0.

Original entry on oeis.org

1, 1, 0, 1, -4, -20, 1, 0, 28, 0, 1, 20, -84, -220, 924, 1, 64, 924, 0, 1820, 0, 1, 140, 12012, 48620, 16796, -15504, -48620, 1, 256, 60060, 2621440, 2704156, 0, 134596, 0, 1, 420, 204204, 29745716, 608435100, 155117520, -3801900, -1184040, 2704156, 1, 640, 554268, 187432960, 15628090140, 146028888064, 9075135300, 0, 10518300, 0

Offset: 0

Peter Bala, Aug 07 2023

Compare with A364303 and A364518.
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).
Each row of the present table is an integral factorial ratio sequence of height 1. 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.
It is known that A005810, the unsigned version of row 1, satisfies the supercongruences u(n*p^r) == u(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.

			Square array begins:
 n\k| 0    1       2          3             4                5
  - + - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  0 | 1    0     -20          0           924                0  ... see A066802
  1 | 1   -4      28       -220          1820           -15504  ... see A005810
  2 | 1    0     -84          0         16796                0
  3 | 1   20     924      48620       2704156        155117520  ... A066802
  4 | 1   64   12012    2621440     608435100     146028888064  ... A364520
  5 | 1  140   60060   29745716   15628090140    8480843582640  ... A211420

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. (2) 79 2009, 422-444.
Wikipedia, Hypergeometric function

Cf. A066802 (row 3, also row 0 unsigned and without 0's), A005810 (row 1 unsigned), A364520 (row 4), A211420 (row 5).

Cf. A330843, A364303, A364518.

T(n,k) := add( binomial((n+3)*k, j)*binomial(n*k-j-1, 3*k-j), j = 0..3*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(j = 0, 3*k, binomial((n+3)*k, j)*binomial(n*k-j-1, 3*k-j));
lista(nn) = for( n=0, nn, for (k=0, n, print1(T(n-k, k), ", "))); \\ Michel Marcus, Aug 13 2023

T(n,k) = Sum_{j = 0..3*k} binomial((n+3)*k, j)*binomial(n*k-j-1, 3*k-j).
For n >= 3, T(n,k) = binomial(n*k-1,3*k) * hypergeom([-(n+3)*k, -3*k], [1 - n*k], -1) = ((n+3)*k)!*((n-3)*k/2)!/(((n+3)*k/2)!*((n-3)*k)!*(3*k)!) by Kummer's Theorem.
The row generating functions are algebraic functions over the field of rational functions Q(x).

cp's OEIS Frontend

A211419 a(n) = (6*n)!*(2*n)!/((4*n)!*(3*n)!*n!).

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A211421 Integral factorial ratio sequence: a(n) = (8*n)!*(3*n)!/((6*n)!*(4*n)!*n!).

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A061162 a(n) = (6n)!n!/((3n)!(2n)!^2).

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A061163 a(n) = (10n)!*n!/((5n)!*(4n)!*(2n)!).

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

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A364519 Square array read by ascending antidiagonals: T(n,k) = [x^(3*k)] ( (1 + x)^(n+3)/(1 - x)^(n-3) )^k for n, k >= 0.

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A211419 a(n) = (6n)!(2n)!/((4n)!(3n)!*n!).

A211421 Integral factorial ratio sequence: a(n) = (8n)!(3n)!/((6n)!(4n)!*n!).

A061163 a(n) = (10n)!n!/((5n)!(4n)!*(2n)!).

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