A276100 -id:A276100 - cp's OEIS frontend

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

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).

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

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

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)

A352651 a(n) = ( binomial(5n,2n)binomial(5n/2,2n)binomial(2n,n)^2 ) / binomial(5n/2,n)^2.

Original entry on oeis.org

1, 12, 378, 14700, 629850, 28540512, 1341310320, 64676424384, 3178603964250, 158529793422000, 7999466594747628, 407514796591710600, 20924507330066816112, 1081581197431986720000, 56225684939117297889600, 2937292879652230377427200, 154108110471294720105987930

Offset: 0

Peter Bala, Mar 25 2022

We write x! as shorthand for Gamma(x+1) and binomial(x,y) as shorthand for x!/(y!*(x-y)!) = Gamma(x+1)/(Gamma(y+1)*Gamma(x-y+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. Soundararajan gives many examples of two-parameter families of integral factorial ratio sequences of height 2.
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. See A276098 and the cross references for further examples of this type.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k. The case n = k = 1 is easily proved.
More generally, for an integer N not equal to 0 or 1, the height 2 factorial ratio sequence whose n-th term is given by ( binomial(N*n,2*n)* binomial(N*n/2,2*n)* binomial(2*n,n)^2 )/binomial(N*n/2,n)^2 is conjectured to be integral and satisfy the same supercongruences. This is the case N = 5. See A352652 (N = 7)

			Examples of supercongruences:
a(2*7) - a(2) = 56225684939117297889600 - 378 = 2*(3^3)*(7^4)*6553*411473* 160830097 == 0 (mod 7^4).
a(13) - a(1) = 1081581197431986720000 - 12 = (2^2)*3*(13^3)* 41024927834622467 == 0 (mod 13^3)

Peter Bala, Some integer ratios of factorials
J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, 2007, arXiv:0709.1977v1 [math.NT], 2007; J. London Math. Soc., 79, Issue 2, (2009), 422-444.
K. Soundararajan, Integral factorial ratios: irreducible examples with height larger than 1, Phil. Trans. R. Soc.A378: 2018044, 2019.

Cf. A008978, A262732, A275652, A276098, A276100, A276101, A276102, A352652.

a := n -> if n = 0 then 1 elif n = 1 then 12 else
5*(3*n - 2)*(3*n - 4)*(5*n - 1)*(5*n - 3)*(5*n - 7)*(5*n - 9)/(n^2*(n - 1)^2*(3*n - 1)*(3*n - 5))*a(n-2) end if:
seq(a(n), n = 0..20);

from math import factorial
from sympy import factorial2
def A352651(n): return int(factorial(5*n)*factorial2(3*n)**2//factorial(3*n)//factorial2(5*n)//factorial(n)**2//factorial2(n)) # Chai Wah Wu, Aug 08 2023

a(n) = (5*n)!*(3*n/2)!^2/( (3*n)!*(5*n/2)!*n!^2*(n/2)! ).
a(n) = 3*Sum_{k = 0..n} (-1)^(n+k)*binomial(5*n,n-k)*binomial(3*n+k-1,k)^2 for n >= 1 (this formula shows the sequence is integral).
a(n) = 3*Sum_{k = 0..n} binomial(2*n-k-2,n-k)*binomial(3*n-1,k)^2 for n >= 1.
a(n) = 3 * [x^n] ( (1 - x)^(2*n) * P(3*n-1,(1 + x)/(1 - x)) ) for n >= 1, where P(n,x) denotes the n-th Legendre polynomial.
a(n) ~ (sqrt(3)/Pi)*(5^n)^(5/2)*( 1/(2*n) - 2/(15*n^2) + 4/(225*n^3) + O(1/n^4) ).
a(n) = A008978(n)/A275652(n).
a(n) = binomial(3*n/2,n)*A262732(n).
a(n) = 3*(-1)^n*binomial(5*n,n)*hypergeom([-n, 3*n, 3*n], [1, 4*n+1], 1) for n >= 1.
a(n) = 5*(3*n-2)*(3*n-4)*(5*n-1)*(5*n-3)*(5*n-7)*(5*n-9)/(n^2*(n-1)^2*(3*n- 1)*(3*n-5)) * a(n-2) with a(0) = 1 and a(1) = 12.
a(p) == 12 (mod p^3) for prime p >= 5.
O.g.f.: A(x) = hypergeom([1/10, 3/10, 7/10, 9/10, 1/3, 2/3], [1/6, 5/6, 1/2, 1/2, 1], (5^5)*x^2) + 12*x*hypergeom([3/5, 4/5, 6/5, 7/5, 5/6, 7/6], [2/3, 4/3, 3/2, 3/2, 1], (5^5)*x^2).

A352652 a(n) = ( binomial(7n,2n)binomial(7n/2,2n)binomial(2n,n)^2 ) / binomial(7n/2,n)^2.

Original entry on oeis.org

1, 30, 2860, 343200, 45643500, 6435891280, 942422020540, 141696569678400, 21724714133822700, 3381208130986900500, 532553441617598475360, 84695057996350934903680, 13578009523892192555221500, 2191530567314796197691108600, 355765014009052303028935320000

Offset: 0

Peter Bala, Apr 03 2022

We write x! as shorthand for Gamma(x+1) and binomial(x,y) as shorthand for x!/(y!*(x-y)!) = Gamma(x+1)/(Gamma(y+1)*Gamma(x-y+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. Soundararajan gives many examples of two-parameter families of integral factorial ratio sequences of height 2. Usually, it is 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. See A276098 and the cross references for further examples of factorial ratios of this type.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k. The case n = k = 1 is easily proved.

			Examples of supercongruences:
a(11) - a(1) = 84695057996350934903680 - 30 = 2*(5^2)*(11^3)*23*593* 3671*5693*4464799 == 0 (mod 11^3)
a(2*7) - a(2) = 355765014009052303028935320000 - 2860 = (2^2)*5*(7^3)*11* 269*3307*375101*14129010228023 == 0 (mod 7^3)

Peter Bala, Some integer ratios of factorials
J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, 2007, arXiv:0709.1977v1 [math.NT], 2007; J. London Math. Soc., 79, Issue 2, (2009), 422-444.
K. Soundararajan, Integral factorial ratios: irreducible examples with height larger than 1, Phil. Trans. R. Soc. A378: 2018044, 2019.

Cf. A275654, A276098, A276100, A276101, A276102, A352651.

a := n -> if n = 0 then 1 elif n = 1 then 30 else
7*(5*n-2)*(5*n-4)*(5*n-6)*(5*n-8)*(7*n-1)*(7*n-3)*(7*n-5)*(7*n-9)*(7*n-11)*(7*n-13)/(3*n^2*(n-1)^2*(3*n-2)*(3*n-4)*(5*n-1)*(5*n- 3)*(5*n -7)*(5*n-9)) *a(n-2) end if:
seq(a(n), n = 0..20);

from math import factorial
from sympy import factorial2
def A352652(n): return int(factorial(7*n)*factorial2(5*n)**2//factorial(5*n)//factorial2(7*n)//factorial2(3*n)//factorial(n)**2) # Chai Wah Wu, Aug 08 2023

a(n) = (5/3)*Sum_{k = 0..n} (-1)^(n+k)*binomial(7*n,n-k)*binomial(5*n+k-1,k)^2 for n >= 1 (this formula shows 3*a(n) is integral; how to show a(n) is integral?).
a(n) = (5/3)*Sum_{k = 0..n} binomial(4*n-k-2,n-k)*binomial(5*n-1,k)^2 for n >= 1.
a(n) = (7*n)!*(5*n/2)!^2/((5*n)!*(7*n/2)!*(3*n/2)!*n!^2!).
a(n) = (5/3) * [x^n] ( (1 - x)^(2*n) * P(5*n-1,(1 + x)/(1 - x)) ) for n >= 1, where P(n,x) denotes the n-th Legendre polynomial.
a(n) = (5/3)*(-1)^n*binomial(7*n,n)*hypergeom([-n, 5*n, 5*n], [1, 6*n+1], 1) for n >= 1.
a(n) ~ sqrt(15)/Pi * 7^(7*n/2)/3^(3*n/2) * ( 1/(6*n) - 29/(945*n^2) + 841/(297675*n^3) + O(1/n^4) ).
a(n) = 7*(5*n-2)*(5*n-4)*(5*n-6)*(5*n-8)*(7*n-1)*(7*n-3)*(7*n-5)*(7*n-9)*(7*n-11)*(7*n-13)/(3*n^2*(n-1)^2*(3*n-2)*(3*n-4)*(5*n-1)*(5*n- 3)*(5*n -7)*(5*n-9)) * a(n-2) with a(0) = 1 and a(1) = 30.
a(n)*A275654(n) = (7*n)!/(n!^4*(3*n)!) = A071549(n)/A006480(n).
a(p) == 30 (mod p^3) for all primes p >= 5.

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.

A364183 a(n) = (12n)!(2n)!(n/2)!/((6n)!(4n)!(7n/2)!n!).

Original entry on oeis.org

1, 4224, 76488984, 1626105446400, 36856530424884600, 864687003650148532224, 20728451893251973782071160, 504292670666772382512278667264, 12401082728528113445556802226795640, 307453669544695584297743425538327838720, 7671567513095586883562392061857092727662984

Offset: 0

Peter Bala, Jul 13 2023

A295479, defined by A295479(n) = (24*n)!*(4*n)!*n! / ((12*n)!*(8*n)!*(7*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 49). Here we are essentially considering the sequence {A295479(n/2) : n >= 0}. Fractional factorials are defined in terms of the gamma function; for example, (7*n/2)! := Gamma(1 + 7*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, A295479, A347854, A347855, A347856, A347857, A347858, A364173 - A364185.

seq( simplify((12*n)!*(2*n)!*(n/2)!/((6*n)!*(4*n)!*(7*n/2)!*n!)), n = 0..15);

a(n) ~ c^n * 1/sqrt(14*Pi*n), where c = (2^15)*(3^6)/(7^4) * sqrt(7).

a(n) = 1327104*(12*n - 1)*(12*n - 5)*(12*n - 7)*(12*n - 11)*(12*n - 13)*(12*n - 17)*(12*n - 19)*(12*n - 23)/(7*n*(n - 1)*(7*n - 2)*(7*n - 4)*(7*n - 6)*(7*n - 8)*(7*n - 10)*(7*n - 12))*a(n-2) with a(0) = 1 and a(1) = 4224.

cp's OEIS Frontend

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

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

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

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

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A352651 a(n) = ( binomial(5*n,2*n)*binomial(5*n/2,2*n)*binomial(2*n,n)^2 ) / binomial(5*n/2,n)^2.

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

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

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

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)!).

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

A352651 a(n) = ( binomial(5n,2n)binomial(5n/2,2n)binomial(2n,n)^2 ) / binomial(5n/2,n)^2.

A352652 a(n) = ( binomial(7n,2n)binomial(7n/2,2n)binomial(2n,n)^2 ) / binomial(7n/2,n)^2.

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

A364183 a(n) = (12n)!(2n)!(n/2)!/((6n)!(4n)!(7n/2)!n!).