A211421 -id:A211421 - cp's OEIS frontend

A262740 O.g.f. exp( Sum_{n >= 1} A211421(n)*x^n/n ).

1, 14, 293, 7266, 197962, 5726364, 172662765, 5367187226, 170772853790, 5534640052292, 182070248073826, 6063785526898644, 204055962203476788, 6927718839334775608, 236994877398511998717, 8161492483543100398410, 282705062046649346154006, 9843330120848835962213940

Offset: 0

Peter Bala, Sep 29 2015

O.g.f. is 1/x * the series reversion of x*(1 - x)^k/(1 + x)^(k+2) at k = 6. See the cross references for related sequences obtained from other values of k.

Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion

Cf. A211421, A000108 (k = 0), A007297 (k = 1), A066357 (k = 2), A262737 (k = 3), A262738 (k = 4), A262739 (k = 5).

#A262740
A262740 := proc (n) option remember; if n = 0 then 1 else add(1/k!*(8*k)!/(4*k)!*(3*k)!/(6*k)!*A262740(n-k), k = 1 .. n)/n end if; end proc:
seq(A262740(n), n = 0..17);

a(n) = sum(k=0, n, binomial(8*(n+1),k)*binomial(7*(n+1)-k-2,(n+1)-k-1))/(n+1); \\ Altug Alkan, Oct 03 2015

a(n-1) = 1/n * Sum_{i = 0..n-1} binomial(8*n,i)*binomial(7*n-i-2,n-i-1).
O.g.f.: A(x) = exp ( Sum_{n >= 1} 1/n! * (8*n)!/(4*n)! * (3*n)!/(6*n)!*x^n/n ) = 1 + 14*x + 293*x^2 + 7266*x^3 + ....
1 + x*A'(x)/A(x) is the o.g.f. for A211421.
O.g.f. is the series reversion of x*(1 - x)^6/(1 + x)^8.
a(0) = 1 and for n >= 1, a(n) = 1/n * Sum {k = 1..n} 1/k! * (8*k)!/(4*k)! * (3*k)!/(6*k)!*a(n-k).

A001448 a(n) = binomial(4n,2n) or (4n)!/((2n)!(2n)!).

Original entry on oeis.org

1, 6, 70, 924, 12870, 184756, 2704156, 40116600, 601080390, 9075135300, 137846528820, 2104098963720, 32247603683100, 495918532948104, 7648690600760440, 118264581564861424, 1832624140942590534, 28453041475240576740, 442512540276836779204, 6892620648693261354600

Offset: 0

N. J. A. Sloane

Corollary 8 in Chapman et alia says: "For n>=1, there are binomial(4n,2n) binary sequences of length 4n+1 with the property that for all j, the j-th occurrence of 10 appears in positions 4j+1 and 4j+2 or later (if it exists at all)." - Peter Luschny, Nov 21 2011

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

			a(n) = (1/Pi)*Integral_{x=0..4} x^(2n)/sqrt(4-(x-2)^2) dx. - _Paul Barry_, Sep 17 2010
G.f. = 1 + 6*x + 70*x^2 + 924*x^3 + 12870*x^4 + 184756*x^5 + 2704156*x^6 + ...

T. D. Noe, Table of n, a(n) for n = 0..100
R. J. Chapman, T. Y. Chowa, A. Khetana, D. P. Moulton and R. J. Waters, Simple formulas for lattice paths avoiding certain periodic staircase boundaries, Journal of Combinatorial Theory, Series A 116 (2009) 205-214.
Maciej Dziemianczuk, On Directed Lattice Paths With Additional Vertical Steps, arXiv preprint arXiv:1410.5747 [math.CO], 2014.
Quy Nhan, A finite sum of binomial coefficients, MathStackExchange, 2025-06-21.
Tito Piezas III, Fermat primes and Binomial sums, tpiezas, A Collection of Algebraic IdentitiesJune 3, 2012.
K. H. Pilehrood and T. H. Pilehrood, Jacobi Polynomials and Congruences Involving Some Higher-Order Catalan Numbers and Binomial Coefficients, J. Int. Seq. 18 (2015) 15.11.7.
Ricardo A. Podestá, New identities for binary Krawtchouk polynomials, binomial coefficients and Catalan numbers, arXiv:1603.09156 [math.CO], 2016.
Renzo Sprugnoli, Sums of reciprocals of the central binomial coefficients, INTEGERS 6 (2006) #A27.

Bisection of A000984. Cf. A002458, A066357, A000984 (k = 0), A091527 (k = 1), A262732 (k = 3), A211419 (k = 4), A262733 (k = 5), A211421 (k = 6).

Cf. A370100, A370101.

[Factorial(4*n)/(Factorial(2*n)*Factorial(2*n)): n in [0..20]]; // Vincenzo Librandi, Sep 13 2011

Maple
```
A001448 := n-> binomial(4*n,2*n) ;
```

Table[Binomial[4n,2n],{n,0,20}] (* Harvey P. Dale, Apr 26 2014 *)
a[ n_] := If[ n < 0, 0, HypergeometricPFQ[ {-2 n, -2 n}, {1}, 1]]; (* Michael Somos, Oct 22 2014 *)

a(n)=binomial(4*n,2*n) \\ Charles R Greathouse IV, Sep 13 2011

from math import comb
def A001448(n): return comb(n<<2,n<<1) # Chai Wah Wu, Aug 10 2023

a(n) = A000984(2*n).
Using Stirling's formula in sequence A000142 it is easy to get the asymptotic expression a(n) ~ 16^n / sqrt(2 * Pi * n). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
From Wolfdieter Lang, Dec 13 2001: (Start)
a(n) = 2*A001700(2*n-1) = (2*n+1)*C(2*n), n >= 1, C(n) := A000108(n) (Catalan).
G.f.: (1-y*((1+4*y)*c(y)-(1-4*y)*c(-y)))/(1-(4*y)^2) with y^2=x, c(y) = g.f. for A000108 (Catalan). (End)
a(n) ~ 2^(-1/2)*Pi^(-1/2)*n^(-1/2)*2^(4*n)*{1 - (1/16)*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Jun 11 2002
a(n) = (1/Pi)*Integral_{x=-2..2} (2+x)^(2*n)/sqrt((2-x)*(2+x)) dx. Peter Luschny, Sep 12 2011
G.f.: (1/2) * (1/sqrt(1+4*sqrt(x)) + 1/sqrt(1-4*sqrt(x))). - Mark van Hoeij, Oct 25 2011
Sum_{n>=1} 1/a(n) = 16/15 + Pi*sqrt(3)/27 - 2*sqrt(5)*log(phi)/25, [T. Trif, Fib Quart 38 (2000) 79] with phi=A001622. - R. J. Mathar, Jul 18 2012
D-finite with recurrence n*(2*n-1)*a(n) -2*(4*n-1)*(4*n-3)*a(n-1)=0. - R. J. Mathar, Dec 02 2012
G.f.: sqrt((1 + sqrt(1-16*x))/(2*(1-16*x))) = 1 + 6*x/(G(0)-6*x), where G(k) = 2*x*(4*k+3)*(4*k+1) + (2*k+1)*(k+1) - 2*x*(k+1)*(2*k+1)*(4*k+5)*(4*k+7)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jun 30 2013
a(n) = hypergeom([1-2*n,-2*n],[2],1)*(2*n+1). - Peter Luschny, Sep 22 2014
From Michael Somos, Oct 22 2014: (Start)
0 = a(n)*(+65536*a(n+2) - 16896*a(n+3) + 858*a(n+4)) + a(n+1)*(-3584*a(n+2) + 1176*a(n+3) - 66*a(n+4)) + a(n+2)*(+14*a(n+2) - 14*a(n+3) + a(n+4)) for all n in Z.
0 = a(n)^2*(+196608*a(n+1)^2 - 40960*a(n+1)*a(n+2) + 2100*a(n+2)^2) + a(n)*a(n+1)*(-12288*a(n+1)^2 + 2840*a(n+1)*a(n+2) - 160*a(n+2)^2) + a(n+1)^2*(+180*a(n+1)^2 - 48*a(n+1)*a(n+2) + 3*a(n+2)^2) for all n in Z. (End)
a(n) = [x^n] ( (1 + x)^4/(1 - x)^2 )^n; exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 6*x + 53*x^2 + 554*x^3 + ... = Sum_{n >= 0} A066357(n+1)*x^n. - Peter Bala, Jun 23 2015
a(n) = Sum_{i = 0..n} binomial(4*n,i) * binomial(3*n-i-1,n-i). - Peter Bala, Sep 29 2015
a(n) = A000984(n)*Product_{j=0..n} (2^j/(j!*(2*j-1)!!))*A068424(n, j)^2, with A068424 the falling factorial. See (5.4) in Podestá link. - Michel Marcus, Mar 31 2016
a(n) = GegenbauerC(2*n, -2*n, -1). - Peter Luschny, May 07 2016
a(n) = [x^n] 1/sqrt(1 - 4*x)^(2*n+1). - Ilya Gutkovskiy, Oct 10 2017
a(n) is the n-th moment of the positive weight function w(x) on (0,16), i.e. a(n) = Integral_{x=0..16} x^n*w(x) dx, n = 0,1,..., where w(x) = (1/(2*Pi))/(sqrt(4 - sqrt(x))*x^(3/4)). The function w(x) is the solution of the Hausdorff moment problem and is unique. - Karol A. Penson, Mar 06 2018
a(n) = (16^n*(Beta(2*n - 1/2, 1/2) - Beta(2*n - 1/2, 3/2)))/Pi. - Peter Luschny, Mar 06 2018
E.g.f.: hypergeom([1/4,3/4],[1/2,1],16*x). - Karol A. Penson, Mar 08 2018
From Peter Bala, Feb 16 2020: (Start)
a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k.
a(n) = [(x*y)^(2*n)] (1 + x + y)^(4*n). (End)
a(n) = (2^n/n!)*Product_{k = n..2*n-1} (2*k + 1). - Peter Bala, Feb 26 2023
a(n) = Sum_{k = 0..2*n} binomial(2*n+k-1, k). - Peter Bala, Nov 02 2024
Sum_{n>=0} (-1)^n/a(n) = 16/17 + 4*sqrt(34)*(sqrt(17)-2)*arctan(sqrt(2/(sqrt(17)-1)))/(289*sqrt(sqrt(17)-1)) + 2*sqrt(34)*(sqrt(17)+2)*log((sqrt(sqrt(17)+1)-sqrt(2))/(sqrt(sqrt(17)+1)+sqrt(2)))/(289*sqrt(sqrt(17)+1)) (Sprugnoli, 2006, Theorem 3.8, p. 11; Piezas, 2012). - Amiram Eldar, Nov 03 2024
For n >= 1, a(n) = Sum_{k=1}^n a(n-k) * A337350(n) = Sum_{k=1}^n a(n-k) * a(k) * (8k + 1) / (8k^2 + 2k - 1).  For proof, see the Quy Nhan link. - Lucas A. Brown, Jun 26 2025
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] 1/((1-x)^(n+1) * (1-2*x)^(2*n)).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(4*n,k) * binomial(2*n-k,n-k).
a(n) = Sum_{k=0..n} 2^k * binomial(2*n+k-1,k) * binomial(2*n-k,n-k).
a(n) = 4^n * binomial((4*n-1)/2,n).
a(n) = [x^n] (1+4*x)^((4*n-1)/2). (End)

A091527 a(n) = ((3n)!/n!^2)(Gamma(1+n/2)/Gamma(1+3n/2)).

Original entry on oeis.org

1, 4, 30, 256, 2310, 21504, 204204, 1966080, 19122246, 187432960, 1848483780, 18320719872, 182327718300, 1820797698048, 18236779032600, 183120225632256, 1842826521244230, 18581317012684800, 187679234340049620, 1898554215471513600, 19232182592635611060

Offset: 0

Michael Somos, Jan 18 2004

Sequence terms are given by [x^n] ( (1 + x)^(k+2)/(1 - x)^k )^n for k = 1. See the crossreferences for related sequences obtained from other values of k. - Peter Bala, Sep 29 2015

Let a > b be nonnegative integers. Then the ratio of factorials ((2*a + 1)*n)!*((b + 1/2)*n)!/(((a + 1/2)*n)!*((2*b + 1)*n)!*((a - b)*n)!) is an integer for n >= 0. This is the case a = 1, b = 0. - Peter Bala, Aug 28 2016

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Michael De Vlieger, Table of n, a(n) for n = 0..985
Peter Bala, Some integer ratios of factorials
Paul Barry, On the Central Coefficients of Bell Matrices, J. Int. Seq. 14 (2011) # 11.4.3.
Karl Dilcher, Armin Straub, and Christophe Vignat, Identities for Bernoulli polynomials related to multiple Tornheim zeta functions, arXiv:1903.11759 [math.NT], 2019. See p. 11.
I. M. Gessel, A short proof of the Deutsch-Sagan congruence for connected non crossing graphs, arXiv preprint arXiv:1403.7656 [math.CO], 2014.
Bernhard Heim and Markus Neuhauser, Asymptotic Distribution of the Zeros of recursively defined Non-Orthogonal Polynomials, arXiv:2107.05013 [math.CA], 2021.
W. Mlotkowski and K. A. Penson, Probability distributions with binomial moments, arXiv preprint arXiv:1309.0595 [math.PR], 2013.

Cf. A061162(n) = a(2n), A007297, A000984 (k = 0), A001448 (k = 2), A262732 (k = 3), A211419 (k = 4), A262733 (k = 5), A211421 (k = 6), A276098, A276099.

a := n -> 4^n * `if`(n<2, 1, (2*(n+1)*binomial((3*n-1)/2, n + 1))/(n-1)):
seq(a(n), n=0..18); # Peter Luschny, Feb 03 2020

Table[((3 n)!/n!^2) Gamma[1 + n/2]/Gamma[1 + 3 n/2], {n, 0, 18}] (* Michael De Vlieger, Oct 02 2015 *)
Table[4^n Sum[Binomial[k - 1 + (n - 1)/2, k], {k, 0, n}], {n, 0, 18}] (* Michael De Vlieger, Aug 28 2016 *)

B(x):=(-1/3+(2/3)*sqrt(1+9*x)*sin((1/3)*asin((2+27*x+54*x^2)/2/(1+9*x)^(3/2))))/x-1;
taylor(x*diff(B(x),x)/B(x),x,0,10); /* Vladimir Kruchinin, Oct 02 2015 */

a(n)=4^n*sum(i=0,n,binomial(i-1+(n-1)/2,i))

vector(30, n, sum(k=0, n, binomial(3*n-3, k)*binomial(2*n-k-3, n-k-1))) \\ Altug Alkan, Oct 04 2015

from math import factorial
from sympy import factorial2
def A091527(n): return int((factorial(3*n)*factorial2(n)<Chai Wah Wu, Aug 10 2023

D-finite with recurrence n*(n - 1)*a(n) = 12*(3*n - 1)*(3*n - 5)*a(n-2).
From Peter Bala, Sep 29 2015: (Start)
a(n) = Sum_{i = 0..n} binomial(3*n,i) * binomial(2*n-i-1,n-i).
a(n) = [x^n] ( (1 + x)^3/(1 - x) )^n.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 4*x + 23*x^2 + 156*x^3 + 1162*x^4 + 9192*x^5 + ... is the o.g.f. for A007297 (but with an offset of 0). (End)
a(n) = (n+1)*A078531(n). [Barry, JIS (2011)]
G.f.: x*B'(x)/B(x), where x*B(x)+1 is g.f. of A007297. - Vladimir Kruchinin, Oct 02 2015
From Peter Bala, Aug 22 2016: (Start)
a(n) = Sum_{k = 0..floor(n/2)} binomial(4*n,n-2*k)*binomial(n+k-1,k).
O.g.f.: A(x) = Hypergeom([5/6, 1/6], [1/2], 108*x^2) + 4*x*Hypergeom([4/3, 2/3], [3/2], 108*x^2).
The o.g.f. is the diagonal of the bivariate rational function 1/(1 - t*(1 + x)^3/(1 - x)) and hence is algebraic by Stanley 1999, Theorem 6.33, p. 197. (End)
a(n) ~ 2^n*3^(3*n/2)/sqrt(2*Pi*n). - Ilya Gutkovskiy, Aug 22 2016
a(n) = 4^n*2*(n+1)*binomial((3*n-1)/2, n+1)/(n-1) for n >= 2. - Peter Luschny, Feb 03 2020
From Peter Bala, Mar 04 2022: (Start)
The o.g.f. A(x) satisfies the algebraic equation (1 - 108*x^2)*A(x)^3 - A(x) = 8*x. Cf. A244039.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k.
Conjecture: the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for primes p >= 5 and positive integers n and k. (End)
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] 1/((1-x)^(n+1) * (1-2*x)^n).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(3*n,k) * binomial(2*n-k,n-k).
a(n) = Sum_{k=0..n} 2^k * binomial(n+k-1,k) * binomial(2*n-k,n-k).
a(n) = 4^n * binomial((3*n-1)/2,n).
a(n) = [x^n] 1/(1-4*x)^((n+1)/2).
a(n) = [x^n] (1+4*x)^((3*n-1)/2). (End)

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

Original entry on oeis.org

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)

A262732 a(n) = (1/n!) * (5n)!/(5n/2)! * (3n/2)!/(3n)!.

Original entry on oeis.org

1, 8, 126, 2240, 41990, 811008, 15967980, 318636032, 6421422150, 130395668480, 2663825039876, 54684895150080, 1127155102890908, 23311847679590400, 483537022180231320, 10054732930602762240, 209536624110664757830, 4375058594685417160704, 91505601042318156186900

Offset: 0

Peter Bala, Sep 29 2015

Sequence terms are given by the coefficient of x^n in the expansion of ( (1 + x)^(k+2)/(1 - x)^k )^n when k = 3. See the cross references for related sequences obtained from other values of k.

Let a > b be nonnegative integers. Then the ratio of factorials ((2*a + 1)*n)!*((b + 1/2)*n)!/(((a + 1/2)*n)!*((2*b + 1)*n)!*((a - b)*n)!) is an integer for n >= 0. This is the case a = 2, b = 1. - Peter Bala, Aug 28 2016

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Michael De Vlieger, Table of n, a(n) for n = 0..751
Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion
Peter Bala, Some integer ratios of factorials
Peter Bala, A supercongruence for A262732

Cf. A000984 (k = 0), A091527 (k = 1), A001448 (k = 2), A211419 (k = 4), A262733 (k = 5), A211421 (k = 6), A262737, A276098, A276099.

Cf. A115293.

a := n -> 1/n! * (5*n)!/GAMMA(1 + 5*n/2) * GAMMA(1 + 3*n/2)/(3*n)!:
seq(a(n), n = 0..18);

Table[1/n!*(5 n)!/(5 n/2)!*(3 n/2)!/(3 n)!, {n, 0, 18}] (* or *)
Table[Sum[Binomial[8 n, n - 2 k] Binomial[3 n + k - 1, k], {k, 0, Floor[n/2]}], {n, 0, 18}] (* Michael De Vlieger, Aug 28 2016 *)

a(n) = sum(k=0, n, binomial(5*n,k)*binomial(4*n-k-1,n-k));
vector(30, n, a(n-1)) \\ Altug Alkan, Oct 03 2015

from math import factorial
from sympy import factorial2
def A262732(n): return int((factorial(5*n)*factorial2(3*n)<Chai Wah Wu, Aug 10 2023

a(n) = Sum_{i = 0..n} binomial(5*n,i) * binomial(4*n-i-1,n-i).
a(n) = [x^n] ( (1 + x)^5/(1 - x)^3 )^n.
D-finite with recurrence a(n) = 20*(5*n - 1)*(5*n - 3)*(5*n - 7)*(5*n - 9)/( n*(3*n - 1)*(3*n - 3)*(3*n - 5) ) * a(n-2).
The o.g.f. exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 8*x + 95*x^2 + 1336*x^4 + ... has integer coefficients and equals (1/x) * (series reversion of x*(1 - x)^3/(1 + x)^5). See A262737.
a(n) ~ 2^n*3^(-3*n/2)*5^(5*n/2)/sqrt(2*Pi*n). - Ilya Gutkovskiy, Jul 31 2016
From Peter Bala, Aug 22 2016: (Start)
a(n) = Sum_{k = 0..floor(n/2)} binomial(8*n,n - 2*k) * binomial(3*n + k - 1,k).
O.g.f.: A(x) = Hypergeom([9/10, 7/10, 3/10, 1/10], [5/6, 1/2, 1/6], (12500/27)*x^2) + 8*x*Hypergeom([7/5, 6/5, 4/5, 3/5], [4/3, 3/2, 2/3], (12500/27)*x^2).
The o.g.f. is the diagonal of the bivariate rational function 1/(1 - t*(1 + x)^5/(1 - x)^3) and hence is algebraic by Stanley 1999, Theorem 6.33, p. 197. (End)
From Karol A. Penson, Apr 26 2018: (Start)
Integral representation of a(n) as the n-th moment of a positive function w(x) on the support (0, sqrt(12500/27)):
a(n) = Integral_{x=0..sqrt(12500/27)} x^n*w(x) dx,
where w(x) = sqrt(5)*2^(3/5)*csc((1/5)*Pi)*sin((1/10)*Pi)*hypergeom([1/10, 4/15, 3/5, 14/15], [1/5, 2/5, 4/5], 27*x^2*(1/12500))/(10*Pi*x^(4/5)) + sqrt(5)*2^(4/5)*csc(2*Pi*(1/5))*sin(3*Pi*(1/10))*hypergeom([3/10, 7/15, 4/5, 17/15], [2/5, 3/5, 6/5], 27*x^2*(1/12500))/(50*Pi*x^(2/5)) + sqrt(5)*2^(1/5)*csc(2*Pi*(1/5))*sin(3*Pi*(1/10))*x^(2/5)*hypergeom([7/10, 13/15, 6/5, 23/15], [4/5, 7/5, 8/5], 27*x^2*(1/12500))/(625*Pi) + 11*sqrt(5)*2^(2/5)*csc((1/5)*Pi)*sin((1/10)*Pi)*x^(4/5)*hypergeom([9/10, 16/15, 7/5, 26/15], [6/5, 8/5, 9/5], 27*x^2*(1/12500))/(50000*Pi). The function w(x) involves four different hypergeometric functions of type 4F3. The function w(x) is singular at both ends of the support. It is the solution of the Hausdorff moment problem and as such it is unique. (End)
From Peter Bala, Sep 15 2021: (Start)
a(n) = [x^n] (1 + 4*x)^((5*n-1)/2) = 4^n*binomial((5*n-1)/2,n).
a(p) == a(1) (mod p^3) for prime p >= 5.
More generally, we conjecture that a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) for prime p >= 5 and positive integers n and k. (End)
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] 1/((1-x)^(n+1) * (1-2*x)^(3*n)).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(5*n,k) * binomial(2*n-k,n-k).
a(n) = Sum_{k=0..n} 2^k * binomial(3*n+k-1,k) * binomial(2*n-k,n-k).
a(n) = [x^n] 1/(1-4*x)^((3*n+1)/2). (End)

A262733 a(n) = (1/n!) * (7n)!/(7n/2)! * (5n/2)!/(5n)!.

Original entry on oeis.org

1, 12, 286, 7680, 217350, 6336512, 188296108, 5670567936, 172459427910, 5284842700800, 162922160580036, 5047099485847552, 156983503897469340, 4899363753956474880, 153349672416272587800, 4811846645261721927680, 151316978279502571401798, 4767566079229070105640960

Offset: 0

Peter Bala, Sep 29 2015

Sequence terms are given by the coefficient of x^n in the expansion of ( (1 + x)^(k+2)/(1 - x)^k )^n when k = 5. See the cross references for related sequences obtained from other values of k.

let a > b be nonnegative integers. Then the ratio of factorials ((2*a + 1)*n)!*((b + 1/2)*n)!/(((a + 1/2)*n)!*((2*b + 1)*n)!*((a - b)*n)!) is an integer for n >= 0. This is the case a = 3, b = 2. - Peter Bala, Aug 28 2016

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion
Peter Bala, Some integer ratios of factorials

Cf. A000984 (k = 0), A091527 (k = 1), A001448 (k = 2), A262732 (k = 3), A211419 (k = 4), A211421 (k = 6), A262739, A276098, A276099.

a := n -> 1/n! * (7*n)!/GAMMA(1 + 7*n/2) * GAMMA(1 + 5*n/2)/(5*n)!:
seq(a(n), n = 0..18);

Table[1/n!*(7 n)!/(7 n/2)!*(5 n/2)!/(5 n)!, {n, 0, 17}] (* Michael De Vlieger, Oct 04 2015 *)

a(n) = sum(k=0, n, binomial(7*n,k)*binomial(6*n-k-1,n-k));
vector(30, n, a(n-1)) \\ Altug Alkan, Oct 03 2015

from math import factorial
from sympy import factorial2
def A262733(n): return int((factorial(7*n)*factorial2(5*n)<Chai Wah Wu, Aug 10 2023

a(n) = [x^n] ( (1 + x)^7/(1 - x)^5 )^n.
a(n) = Sum_{i = 0..n} binomial(7*n,i) * binomial(6*n-i-1,n-i).
a(n) = 28*(7*n - 1)*(7*n - 3)*(7*n - 9)*(7*n - 11)*(7*n - 13) / ( n*(5*n - 1)*(5*n - 3)*(5*n - 5)*(5*n - 7)*(5*n - 9) ) * a(n-2).
The o.g.f. exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 12*x + 215*x^2 + 4564*x^3 + 106442*x^4 + ... has integer coefficients and equals 1/x * series reversion of x*(1 - x)^5/(1 + x)^7. See A262739.
a(n) ~ 2^n*5^(-5*n/2)*7^(7*n/2)/sqrt(2*Pi*n). - Ilya Gutkovskiy, Jul 31 2016
From Peter Bala, Aug 22 2016: (Start)
a(n) = Sum_{k = 0..floor(n/2)} binomial(12*n,n - 2*k) * binomial(5*n + k - 1,k).
O.g.f.: A(x) = Hypergeom([13/14, 11/14, 9/14, 5/14, 3/14, 1/14], [9/10, 7/10, 3/10, 1/2, 1/10], (2^2*7^7/5^5)*x^2) + 12*x*Hypergeom([10/7, 9/7, 8/7, 6/7, 5/7, 4/7], [7/5, 6/5, 4/5, 3/2, 3/5], (2^2*7^7/5^5)*x^2).
The o.g.f. is the diagonal of the bivariate rational function 1/(1 - t*(1 + x)^7/(1 - x)^5) and hence is algebraic by Stanley 1999, Theorem 6.33, p. 197. (End)
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] 1/((1-x)^(n+1) * (1-2*x)^(5*n)).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(7*n,k) * binomial(2*n-k,n-k).
a(n) = Sum_{k=0..n} 2^k * binomial(5*n+k-1,k) * binomial(2*n-k,n-k).
a(n) = 4^n * binomial((7*n-1)/2,n).
a(n) = [x^n] 1/(1-4*x)^((5*n+1)/2).
a(n) = [x^n] (1+4*x)^((7*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

A211420 a(n) = (8n)!n!/((4n)!(3n)!(2*n)!).

Original entry on oeis.org

1, 140, 60060, 29745716, 15628090140, 8480843582640, 4697400936504900, 2638798257262351800, 1497753729733989900060, 856840435680656569701776, 493243073668546377605912560, 285369375758780754651194529300, 165789876049841088844342275759300

Offset: 0

Peter Bala, Apr 10 2012

This sequence is the particular case a = 4, 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), A211421(a = 4, b = 3) and A061163 (a = 5, b = 1).

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

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.
F. Rodriguez-Villegas, Integral ratios of factorials and algebraic hypergeometric functions, arXiv:math/0701362 [math.NT], 2007.

Cf. A061162, A061163, A211419, A211421.

a := n -> (2^(6*n)*GAMMA(4*n + 1/2))/(GAMMA(n + 1/2)*GAMMA(3*n + 1)):
seq(a(n), n = 0..12); # Peter Luschny, Jul 11 2023

Table[ 2^(6*n) * Gamma[4*n + 1/2] / (Gamma[n + 1/2] * Gamma[3*n + 1]), {n, 0, 12}] (* James C. McMahon, Feb 24 2024 *)

The o.g.f. Sum_{n >= 1} a(n)*z^n is algebraic over the field of rational functions Q(z) (see Rodriguez-Villegas).
D-finite with recurrence: 3*(3*n-1)*(2*n-1)*(3*n-2)*n*a(n) - 8*(8*n-3)*(8*n-1)*(8*n-7)*(8*n-5)*a(n-1) = 0. - Georg Fischer, Nov 30 2022
From Peter Bala, Jul 10 2023: (Start)
a(n) = Sum_{k = 0..3*n} binomial(8*n, k)*binomial(5*n-k-1, 3*n-k).
a(n) = [x^(3*n)] F(x)^n, where F(x) = (1 + x)^8/(1 - x)^2.
It follows that the o.g.f. A(x) for this sequence is the diagonal of the bivariate rational generating function 1/3*( 1/(1 - t*F(x^(1/3))) +  1/(1 - t*F(w*x^(1/3))) +  1/(1 - t*F(w^2*x^(1/3))) ), where w = exp(2*Pi*i/3), and hence A(x), as stated above, is algebraic over Q(x) by Stanley 1999, Theorem 6.33, p. 197. (End)
From Karol A. Penson, Feb 23 2024: (Start)
O.g.f.: hypergeometric4F3([1/8, 3/8, 5/8, 7/8], [1/3, 1/2, 2/3], (2^14*z)/27). (O.g.f.(z))^2 satisfies the algebraic equation of order 16, 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..16} 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..16: 8,8,8,8,9,9,9,9,10,10,10,11,11,11,11,11,12. For example p(14, z) = 6*(2^13*z + 31*27)*(2^14*z - 27)^10. (End)
a(n) ~ 2^(14*n - 1/2) / (3^(3*n + 1/2) * sqrt(Pi*n)). - Vaclav Kotesovec, Aug 27 2024

cp's OEIS Frontend

A262740 O.g.f. exp( Sum_{n >= 1} A211421(n)*x^n/n ).

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A001448 a(n) = binomial(4n,2n) or (4*n)!/((2*n)!*(2*n)!).

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A091527 a(n) = ((3*n)!/n!^2)*(Gamma(1+n/2)/Gamma(1+3n/2)).

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

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A262732 a(n) = (1/n!) * (5*n)!/(5*n/2)! * (3*n/2)!/(3*n)!.

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A262733 a(n) = (1/n!) * (7*n)!/(7*n/2)! * (5*n/2)!/(5*n)!.

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A001448 a(n) = binomial(4n,2n) or (4n)!/((2n)!(2n)!).

A091527 a(n) = ((3n)!/n!^2)(Gamma(1+n/2)/Gamma(1+3n/2)).

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

A262732 a(n) = (1/n!) * (5n)!/(5n/2)! * (3n/2)!/(3n)!.

A262733 a(n) = (1/n!) * (7n)!/(7n/2)! * (5n/2)!/(5n)!.

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

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

A211420 a(n) = (8n)!n!/((4n)!(3n)!(2*n)!).