A276098 -id:A276098 - cp's OEIS frontend

A276100 a(n) = (15n)!(n/2)!/((15n/2)!(5n)!(3*n)!).

1, 114688, 77636318760, 62505037015810048, 53837289804317953893960, 48066503353826060675410034688, 43880754270176401422739454033276880, 40671547154451909281150562260837340282880, 38113558705192522309151157825210540422513019720

Offset: 0

Peter Bala, Aug 22 2016

Fractional factorials are defined in terms of the gamma function, for example, (n/2)! := gamma(n/2 + 1).
This is only conjecturally an integer sequence. The similarly defined sequence (15*n)!*floor(n/2)!/(floor(15*n/2)!*(5*n)!*(3*n)!) = A211418(15*n) is integral.
Let u(n) = (30*n)!*n!/((15*n)!*(10*n)!*(6*n)!) = A211417(n). This sequence of ratios of factorials is integral and was used by Chebyshev in his estimate of the number of primes less than or equal to a fixed integer n. The three sequences u(1/2*n), u(1/3*n) and u(1/5*n) also appear to be integral (checked up to n = 200). This is the sequence u(1/2*n). See A276101( u(1/3*n) ) and A276102( u(1/5*n) ).
The generating function for u(n) is Hypergeom([29/30, 23/30, 19/30, 17/30, 13/30, 11/30, 7/30, 1/30], [4/5, 3/5, 2/5, 1/5, 2/3, 1/3, 1/2], (2^14*3^9*5^5)*x) and is known to be algebraic - see Rodriguez-Villegas. Are the generating functions for u(1/2*n), u(1/3*n) and u(1/5*n) also algebraic?

Carauleanu Marc, Table of n, a(n) for n = 0..111
P. Bala, Some integer ratios of factorials
F. Rodriguez-Villegas, Integral ratios of factorials and algebraic hypergeometric functions, arXiv:math/0701362 [math.NT], 2007.

Cf. A211417, A211418, A276101, A276102.

A211417 := proc(n)
(30*n)!*(n)!/((15*n)!(10*n)!(6*n)!);
end proc:
seq(simplify(A211417(1/2*n)), n = 0..10);

Table[(15 n)!*(n/2)!/((15 n/2)!*(5 n)!*(3 n)!), {n, 0, 8}] (* Michael De Vlieger, Aug 28 2016 *)

O.g.f.: A(x) = Hypergeom([29/30, 23/30, 19/30, 17/30, 13/30, 11/30, 7/30, 1/30,], [4/5, 3/5, 2/5, 1/5, 2/3, 1/3, 1/2], (2^14*3^9*5^5)*x^2) + 114688*x*Hypergeom([22/15, 19/15, 17/15, 16/15, 14/15, 13/15, 11/15, 8/15], [13/10, 11/10, 9/10, 7/10, 7/6, 5/6, 3/2], (2^14*3^9*5^5)*x^2).

a(n) ~ (2^14*3^9*5^5)^(n/2)/sqrt(30*Pi*n).

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)

A276101 a(n) = (10n)!(n/3)!/((5n)!(10n/3)!(2*n)!).

Original entry on oeis.org

1, 1458, 9723402, 77636318760, 665145965903562, 5915482311008318958, 53837289804317953893960, 497704257299202369371725086, 4653371135224869009103021872330, 43880754270176401422739454033276880

Offset: 0

Peter Bala, Aug 22 2016

Fractional factorials are defined in terms of the gamma function, for example, (n/3)! := gamma(n/3 + 1).
This is only conjecturally an integer sequence. The similarly defined sequence (10*n)!*floor(n/3)!/((5*n)!*floor(10*n/3)!*(2*n)!) = A211418(10*n) is integral.
Let u(n) = (30*n)!*n!/((15*n)!*(10*n)!*(6*n)!) = A211417(n). The three sequences u(1/2*n), u(1/3*n) and u(1/5*n) appear to be integral (checked up to n = 200). This is the sequence u(1/3*n). See A276100 ( u(1/2*n) ) and A276102 ( u(1/5*n) ).
The generating function for u(n) is Hypergeom([29/30, 23/30, 19/30, 17/30, 13/30, 11/30, 7/30, 1/30], [4/5, 3/5, 2/5, 1/5, 2/3, 1/3, 1/2], (2^14*3^9*5^5)*x) and is known to be algebraic. Are the generating functions for u(1/2*n), u(1/3*n) and u(1/5*n) also algebraic?

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

Cf. A211417, A276100, A276102.

A211417 := proc(n)
(30*n)!*(n)!/((15*n)!(10*n)!(6*n)!);
end proc:
seq(simplify(A211417(1/3*n)), n = 0..10);

Table[(10*n)!*(n/3)!/((5*n)!*(10*n/3)!*(2*n)!) // FullSimplify, {n, 0, 9}] (* Jean-François Alcover, Nov 27 2017 *)

O.g.f. A(x) = Hypergeom([29/30, 23/30, 19/30, 17/30, 13/30, 11/30, 7/30, 1/30], [4/5, 3/5, 2/5, 1/5, 2/3, 1/3, 1/2], (2^14*3^9*5^5)*x^3) + 1458*x*Hypergeom([29/30, 23/30, 17/30, 11/30, 13/10, 11/10, 9/10, 7/10], [17/15, 14/15, 11/15, 8/15, 5/6, 4/3, 2/3], (2^14*3^9*5^5)*x^3) + 9723402*x^2*Hypergeom([ 49/30, 43/30, 37/30, 31/30, 13/10, 11/10, 9/10, 7/10], [22/15, 19/15, 16/15, 13/15, 7/6, 5/3, 4/3],(2^14*3^9*5^5)*x^3).

a(n) ~ (2^14*3^9*5^5)^(n/3)/(sqrt(20*Pi*n)).

A276102 a(n) = (6n)!(n/5)!/((3n)!(2n)!(6*n/5)!).

Original entry on oeis.org

1, 50, 8250, 1636250, 349456250, 77636318760, 17672894531250, 4089765214843750, 957711284472656250, 226280605806640625000, 53837289804317953893960, 12880759628253295898437500

Offset: 0

Peter Bala, Aug 22 2016

Fractional factorials are defined in terms of the gamma function, for example, (n/5)! := gamma(n/5 + 1).
This is only conjecturally an integer sequence. The similarly defined sequence (6*n)!*floor(n/5)!/((3*n)!*(2*n)!*floor(6*n/5)!) = A211418(6*n) is integral.
Let u(n) = (30*n)!*n!/((15*n)!*(10*n)!*(6*n)!) = A211417(n). The three sequences u(1/2*n), u(1/3*n) and u(1/5*n) appear to be integral (checked up to n = 200). This is the sequence u(1/5*n). See A276100 ( u(1/2*n) ) and A276101 ( u(1/3*n) ).
The generating function for u(n) is Hypergeom([29/30, 23/30, 19/30, 17/30, 13/30, 11/30, 7/30, 1/30], [4/5, 3/5, 2/5, 1/5, 2/3, 1/3, 1/2], (2^14*3^9*5^5)*x) and is algebraic - see Rodriguez-Villegas. Are the generating functions for u(1/2*n), u(1/3*n) and u(1/5*n) also algebraic?
The o.g.f. for this sequence can be expressed as a sum of 5 generalized hypergeometric functions of type 8F7.

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

Cf. A211417, A276100, A276101.

A211417 := proc(n)
(30*n)!*(n)!/((15*n)!(10*n)!(6*n)!);
end proc:
seq(simplify(A211417(1/5*n)), n = 0..10);

Table[(6*n)!*(n/5)!/((3*n)!*(2*n)!*(6*n/5)!) // FullSimplify, {n, 0, 11}] (* Jean-François Alcover, Nov 27 2017 *)

a(n) ~ (2^14*3^9*5^5)^(n/5)/sqrt(12*Pi*n).

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

Original entry on oeis.org

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)

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)

A275652 a(n) = binomial(3n,3n/2)binomial(2n,n)binomial(5n/2,n/2)/binomial(n,n/2).

Original entry on oeis.org

1, 10, 300, 11440, 485100, 21841260, 1022041020, 49128552000, 2408829328620, 119918393838100, 6042249840712800, 307438844121252480, 15770112362658517500, 814459593645444166560, 42308586942403276440000, 2208850973597860123741440, 115825519836558228435979500

Offset: 0

Peter Bala, Aug 04 2016

Right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(n+k)* binomial(3*n+k,3*n-k)*binomial(2*k,k)*binomial(2*n-k,n) = binomial(3*n,3*n/2)*binomial(2*n,n)*binomial(5*n/2,n/2)/binomial(n,n/2).
We also have Sum_{k = 0..3*n} (-1)^k*binomial(3*n+k,3*n-k)* binomial(2*k,k)*binomial(2*n-k,n) = binomial(3*n,3*n/2)*binomial(2*n,n)* binomial(5*n/2,n/2) /binomial(n,n/2).
Compare with Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n+k,2*n-k)* binomial(2*k,k)*binomial(2*n-k,n) = binomial(2*n,n)^2 = A002894(n). See also A275653, A275654 and A275655.

Cf. A002894, A245086, A275653, A275654, A275655, A276098, A276100, A276101, A276102, A352651, A352652, A365025.

seq(simplify(factorial(3*n)*factorial(n/2)*factorial(5*n/2)/(factorial(n)^3*factorial(3*n/2)^2)), n = 0 .. 20);

Table[Binomial[3 n, 3 n/2] Binomial[2 n, n] Binomial[5 n/2, n/2] / Binomial[n, n/2], {n, 0, 16}] (* Michael De Vlieger, Aug 07 2016 *)

a(n) = sum(k = 0, n, binomial(2*n-k-1,n-k)*binomial(3*n,k)^2); \\ Michel Marcus, Apr 21 2022

from math import factorial
from sympy import factorial2
def A275652(n): return int(factorial(3*n)*factorial2(5*n)*factorial2(n)//factorial2(3*n)**2//factorial(n)**3) # Chai Wah Wu, Aug 08 2023

a(n) = (3*n)!*(5*n/2)!*(n/2)!/((3*n/2)!^2*n!^3).
Recurrence: a(n) = 5*(3*n - 1)*(3*n - 5)*(5*n - 2)*(5*n - 4)*(5*n - 6)*(5*n - 8)/(n^2*(n - 1)^2*(3*n - 2)*(3*n - 4)) * a(n-2).
a(n) = [x^n] G(x)^(5*n), where G(x) = 1 + 2*x + 12*x^2 + 184*x^3 + 3811*x^4 + 92796*x^5 + 2497358*x^6 + ... appears to have integer coefficients.
exp(Sum_{n >= 1} a(n)*x^n/n) = F(x)^5, where F(x) = 1 + 2*x + 32*x^2 + 824*x^3 + 26291*x^4 + 947506*x^5 + 36934522*x^6 + ... appears to have integer coefficients.
a(n) ~ sqrt(5/3)*5^(5*n/2)/(2*Pi*n). - Ilya Gutkovskiy, Aug 07 2016
From Peter Bala, Mar 22 2022: (Start)
For n >= 1, a(n) = (5/3)*binomial(m*n,2*n)*binomial(m*n/2,2*n)*binomial(2*n,n)^2/ binomial(m*n/2,n)^2 at m = -1. See A352651 for the case m = 1.
a(n) = Sum_{k = 0..n} binomial(2*n-k-1,n-k)*binomial(3*n,k)^2.
a(n) = [x^n] (1 - x)^(2*n) * P(3*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A245086.
a(p) == a(1) (mod p^3) for prime p >= 5.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k. (End)
Row 1 of A365025. - Peter Bala, Aug 18 2023

A275654 a(n) = (5n)!/((3n)!n!^2) ((3n/2)!(7n/2)!)/(5n/2)!^2.

Original entry on oeis.org

1, 28, 2646, 316540, 42031990, 5921058528, 866486466720, 130220534668224, 19958454291525750, 3105489721784166640, 489023391870111994896, 77758775451291032116200, 12464212878673327376454304, 2011515147856766922731424000

Offset: 0

Peter Bala, Aug 04 2016

Right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(n+k)*binomial(5*n + k,5*n - k)*binomial(2*k,k)*binomial(2*n - k,n) = (5*n)!/((3*n)!*n!^2) * ((3*n/2)!*(7*n/2)!)/(5*n/2)!^2.
We also have Sum_{k = 0..5*n} (-1)^k*binomial(5*n + k,5*n - k)* binomial(2*k,k) *binomial(2*n - k,n) = (5*n)!/((3*n)!*n!^2) * ((3*n/2)!*(7*n/2)!)/(5*n/2)!^2.
Compare with Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n + k,2*n - k)*binomial(2*k,k)*binomial(2*n - k,n) = binomial(2*n,n)^2 = A002894(n). See also A275652, A275653 and A275655.

Cf. A275652, A275653, A275655, A276098, A276100, A276101, A276102, A352651, A352652.

seq(simplify(factorial(3*n/2)*factorial(5*n)*factorial(7*n/2)/(factorial(n)^2*factorial(3*n)*factorial(5*n/2)^2)), n = 0 .. 20);

Table[(5 n)!/((3 n)! n!^2) ((3 n/2)! (7 n/2)!)/(5 n/2)!^2, {n, 0, 13}] (* Michael De Vlieger, Aug 07 2016 *)

a(n) = sum(k = 0, n, binomial(4*n-k-1,n-k)*binomial(5*n,k)^2); \\ Michel Marcus, Apr 21 2022

from math import factorial
from sympy import factorial2
def A275654(n): return int(factorial(5*n)*factorial2(3*n)*factorial2(7*n)//factorial(3*n)//factorial(n)**2//factorial2(5*n)**2) # Chai Wah Wu, Aug 08 2023

a(n) = (3*n/2)!*(5*n)!*(7*n/2)!/(n!^2*(3*n)!*(5*n/2)!^2).
Recurrence: 3*a(n)*n^2*(n - 1)^2*(3*n - 1)*(3*n - 5)*(5*n - 2)*(5*n - 4)*(5*n - 6)*(5*n - 8) = 7*(5*n - 1)*(5*n - 3)*(5*n - 7)*(5*n - 9)*(7*n - 2)*(7*n - 4)*(7*n - 6)*(7*n - 8)*(7*n - 10)*(7*n - 12)*a(n-2).
a(n) = [x^n] G(x)^(7*n) where G(x) = 1 + 4*x + 85*x^2+ 4220*x^3 + 283285*x^4 + 22308156*x^5 + 1939419083*x^6 + ... appears to have integer coefficients.
exp( Sum_{n >= 1} a(n)*x^n/n ) = F(x)^7, where F(x) = 1 + 4*x + 197*x^2 + 15840*x^3 + 1580819*x^4 + 178220584*x^5 + 21729476664*x^6 + ... appears to have integer coefficients.
a(n) ~ 7^(7*n/2+1/2)/(2*sqrt(5)*Pi*3^(3*n/2)*n). - Ilya Gutkovskiy, Aug 07 2016
From Peter Bala, Mar 23 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(4*n-k-1,n-k)*binomial(5*n,k)^2.
For n >= 1, a(n) = (7/5)*binomial(m*n,2*n)*binomial(m*n/2,2*n)* binomial(2*n,n)^2/binomial(m*n/2,n)^2 at m = -3. Se A352651 for the case m = 1.
a(n) = [x^n] (1 - x)^(2*n) * P(5*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A275652.
a(p) == a(1) (mod p^3) for primes p >= 5.
Conjecture: The supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k. (End)

A276099 a(n) = (9n)!(5/2n)!/((9n/2)!(5n)!(2n)!).

Original entry on oeis.org

1, 96, 24310, 7028736, 2149374150, 678057476096, 218191487357116, 71184392021606400, 23459604526110889542, 7791432263086689484800, 2603575153867220801823060, 874329826463740757819785216, 294822072977645830504963830300

Offset: 0

Peter Bala, Aug 22 2016

Let a > b be nonnegative integers. The ratio of factorials (2*a*n)!*(b*n)!/( (a*n)!*(2*b*n)!*((a - b)*n)! ) is known to be an integer for all integer n >= 0 (see, for example, Bober, Theorem 1.1). We have the companion result: 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 all integer n >= 0. This is the case a = 4, b = 2. Other cases include A091496 (a = 2, b = 0), A091527 (a = 1, b = 0), A262732 (a = 2, b = 1), A262733 (a = 3, b = 2) and A276098 (a = 3, b = 1).

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

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. A091496, A091527, A262732, A262733, A276098.

seq(simplify((9*n)!*(5/2*n)!/((9*n/2)!*(5*n)!*(2*n)!)), n = 0..20);

Table[((9n)!(5/2 n)!)/((9 n/2)!(5n)!(2n)!),{n,0,15}] (* Harvey P. Dale, May 21 2024 *)

from math import factorial
from sympy import factorial2
def A276099(n): return int((factorial(9*n)*factorial2(5*n)<<(n<<1))//factorial2(9*n)//factorial(5*n)//factorial(n<<1)) # Chai Wah Wu, Aug 10 2023

a(n) = Sum_{k = 0..2*n} binomial(9*n, k)*binomial(7*n - k - 1, 2*n - k).
a(n) = Sum_{k = 0..n} binomial(14*n, 2*n - 2*k)*binomial(5*n + k - 1, k).
a(n) ~ 1/sqrt(4*Pi*n) * (3^18/5^5)^(n/2).
O.g.f. A(x) = Hypergeom([17/18, 13/18, 11/18, 7/18, 5/18, 1/18, 5/6, 1/6], [9/10, 7/10, 3/10, 1/10, 3/4, 1/4, 1/2], (3^18/5^5)*x^2) + 96*x*Hypergeom([13/9, 11/9, 10/9, 8/9, 7/9, 5/9, 4/3, 2/3], [7/5, 6/5, 4/5, 3/5, 5/4, 3/4, 3/2], (3^18/5^5)*x^2).
a(n) = [x^(2*n)] H(x)^n, where H(x) = (1 + x)^9/(1 - x)^5.
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 F(x) = (1/x)*Series_Reversion( x*sqrt((1 - x)^5/(1 + x)^9) ) and put G(x) = 1 + x*d/dx(log(F(x))). Then A(x^2) = (G(x) + G(-x))/2.

cp's OEIS Frontend

A276100 a(n) = (15*n)!*(n/2)!/((15*n/2)!*(5*n)!*(3*n)!).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

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

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Mathematica

Maxima

PARI

PARI

Python

Formula

A276101 a(n) = (10*n)!*(n/3)!/((5*n)!*(10*n/3)!*(2*n)!).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A276102 a(n) = (6*n)!*(n/5)!/((3*n)!*(2*n)!*(6*n/5)!).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Python

Formula

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

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Formula

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

Views

Author

Keywords

A276100 a(n) = (15n)!(n/2)!/((15n/2)!(5n)!(3*n)!).

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

A276101 a(n) = (10n)!(n/3)!/((5n)!(10n/3)!(2*n)!).

A276102 a(n) = (6n)!(n/5)!/((3n)!(2n)!(6*n/5)!).

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

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

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

A275652 a(n) = binomial(3n,3n/2)binomial(2n,n)binomial(5n/2,n/2)/binomial(n,n/2).

A275654 a(n) = (5n)!/((3n)!n!^2) ((3n/2)!(7n/2)!)/(5n/2)!^2.

A276099 a(n) = (9n)!(5/2n)!/((9n/2)!(5n)!(2n)!).