A276098 -id:A276098 - cp's OEIS frontend

A091496 a(n) = ((5n)!/(n!(2n)!))(Gamma(1+n/2)/Gamma(1+5*n/2)).

1, 16, 630, 28672, 1385670, 69206016, 3528923580, 182536110080, 9540949030470, 502682972323840, 26651569523959380, 1420217179365703680, 75998432812419471900, 4081125953526124511232, 219813190240007470094520, 11869871068877664049692672, 642409325786050322446410310

Offset: 0

Michael Somos, Jan 15 2004

nonn

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 = 2, b = 0. - Peter Bala, Aug 28 2016

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

G. C. Greubel, Table of n, a(n) for n = 0..500
Peter Bala, Some integer ratios of factorials

Cf. A061163(n)=a(2n), A061162, A091527, A262732, A262733, A276098, A276099.

Table[((5 n)!/(n! (2 n)!)) (Gamma[1 + n/2]/Gamma[1 + 5 n/2]), {n, 0, 14}] (* or *)
Table[Sum[Binomial[6 n, 2 n - 2 k] Binomial[n + k - 1, k], {k, 0, n}], {n, 0, 14}] (* or *)
Table[Sum[Binomial[5 n, k] Binomial[3 n - k - 1, 2 n - k], {k, 0, 2 n}], {n, 0, 14}] (* Michael De Vlieger, Aug 28 2016 *)

a(n)=16^n*sum(i=0,2*n,binomial(i-1+(n-1)/2,i))

from math import factorial
from sympy import factorial2
def A091496(n): return int((factorial(5*n)*factorial2(n)<<(n<<1))//(factorial(n)*factorial(n<<1)*factorial2(5*n))) # Chai Wah Wu, Aug 10 2023

n*(n-1)*(2*n-1)*(2*n-3)*a(n) = 20*(5*n-1)*(5*n-3)*(5*n-7)*(5*n-9)*a(n-2).
From Peter Bala, Aug 22 2016: (Start)
a(n) = Sum_{k = 0..2*n} binomial(5*n, k)*binomial(3*n - k - 1, 2*n - k).
a(n) = Sum_{k = 0..n} binomial(6*n, 2*n - 2*k)*binomial(n + k - 1, k).
a(n) ~ 5^(5*n/2)/(2*sqrt(Pi*n)).
O.g.f. A(x) = Hypergeom([9/10, 7/10, 3/10, 1/10], [3/4, 1/2, 1/4], 3125*x^2) + 16*x*Hypergeom([7/5, 6/5, 4/5, 3/5], [5/4, 3/2, 3/4], 3125*x^2).
a(n) = [x^(2*n)] H(x)^n, where H(x) = (1 + x)^5/(1 - x). Cf. A061162 and A262732.
It follows that the o.g.f. 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.
exp(Sum_{n >= 1} a(n)*x^n/n) = 1 + 16*x + 443*x^2 + 15280*x^3 + 591998*x^4 + 24635360*x^5 + 1075884051*x^6 + ... has integer coefficients.
Let F(x) = 1/x*Series_Reversion( x*sqrt((1 - x)/(1 + x)^5) ) and put G(x) = 1 + x*d/dx(log(F(x))). Then A(x) satisfies A(x^2) = (G(x) + G(-x))/2. (End)
O.g.f. denoted by h(x), satisfies algebraic equation of order 10: -800000*x^4 + 81*x^2 - 25000*x^3*h(x) - 25*x^2*(1400000*x^2 - 123)*h(x)^2 + 8*x*(178125*x^2 - 32)*h(x)^3 + (-31250000*x^4 + 22500*x^2 + 4)*h(x)^4 + 32*x*(137500*x^2 - 19)*(3125*x^2 - 1)*h(x)^5 + 12*(3125*x^2 - 1)*(3125*x^2 + 3)*h(x)^6 + 800*x*(3125*x^2 - 1)^2*h(x)^7 + 96*(3125*x^2 - 1)^2*h(x)^8 + 64*(3125*x^2 - 1)^3*h(x)^10 = 0. - Karol A. Penson, Apr 30 2025

A352373 a(n) = [x^n] ( 1/((1 - x)^2*(1 - x^2)) )^n for n >= 1.

Original entry on oeis.org

2, 12, 74, 484, 3252, 22260, 154352, 1080612, 7621526, 54071512, 385454940, 2758690636, 19810063392, 142662737376, 1029931873824, 7451492628260, 54013574117106, 392188079586468, 2851934621212598, 20766924805302984, 151403389181347160, 1105047483656041080

Offset: 1

Peter Bala, Mar 14 2022

Suppose n identical objects are distributed in 3*n labeled baskets, 2*n colored white and n colored black. White baskets can contain any number of objects (or be empty), while black baskets must contain an even number of objects (or be empty). a(n) is the number of distinct possible distributions.
Number of nonnegative integer solutions to n = x_1 + x_2 + ... + x_(2*n) + 2*y_1 + 2*y_2 + ... + 2*y_n.
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.
Calculation suggests that, in fact, stronger congruences may hold.
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.
More generally, let r and s be integers and define a sequence (a(r,s;n))n>=1 by a(r,s;n) = [x^n] ( (1 + x)^r * (1 - x)^s )^n.
Conjecture: for each r and s the above supercongruences hold for the sequence (a(r,s;n))n>=1.
The present sequence is the case r = -1 and s = -3. Other cases include A000984 (r = 2, s = 0), A001700 with offset 1 (r = 0, s = -1), A002003 (r = 1, s = -1), A091527 (r = 3, s = -1), A119259 (r = 2, s = -1), A156894 (r = 1, s = -2), A165817 (r = 0, s = -2), A234839 (r = 1, s = 2), A348410 (r = -1, s = -2) and A351857 (r = -2, s = -4).

			n = 2: 12 distributions of 2 identical objects in 4 white and 2 black baskets
             White         Black
   1)   (0) (0) (0) (0)   [2] [0]
   2)   (0) (0) (0) (0)   [0] [2]
   3)   (2) (0) (0) (0)   [0] [0]
   4)   (0) (2) (0) (0)   [0] [0]
   5)   (0) (0) (2) (0)   [0] [0]
   6)   (0) (0) (0) (2)   [0] [0]
   7)   (1) (1) (0) (0)   [0] [0]
   8)   (1) (0) (1) (0)   [0] [0]
   9)   (1) (0) (0) (1)   [0] [0]
  10)   (0) (1) (1) (0)   [0] [0]
  11)   (0) (1) (0) (1)   [0] [0]
  12)   (0) (0) (1) (1)   [0] [0]
Examples of supercongruences:
a(7) - a(1) = 154352 - 2 = 2*(3^2)*(5^2)*(7^3) == 0 (mod 7^3);
a(2*11) - a(2) = 1105047483656041080 - 12 = (2^2)*3*(11^3)*13*101*103*2441* 209581 == 0 (mod 11^3).

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

Paolo Xausa, Table of n, a(n) for n = 1..1000

Cf. A000984, A001448, A001700, A002003, A091527, A119259, A156894, A165817, A211419, A211421, A234839, A262733, A276098, A348410, A351856, A351857.

seq(add( binomial(3*n-2*k-1,n-2*k)*binomial(n+k-1,k), k = 0..floor(n/2)), n = 1..25);

nterms=25;Table[Sum[Binomial[3n-2k-1,n-2k]Binomial[n+k-1,k],{k,0,Floor[n/2]}],{n,nterms}] (* Paolo Xausa, Apr 10 2022 *)

a(n) = Sum_{k = 0..floor(n/2)} binomial(3*n-2*k-1,n-2*k)*binomial(n+k-1,k).
a(n) = Sum_{k = 0..n} (-1)^k*binomial(4*n-k-1,n-k)*binomial(n+k-1,k).
a(n) = binomial(4*n-1,n)*hypergeom([n, -n], [1-4*n], -1).
48*n*(n-1)*(3*n-1)*(3*n-2)*(93*n^3-434*n^2+668*n-339)*a(n) = 12*(n-1)*(21762*n^6-134199*n^5+323805*n^4-386685*n^3+237728*n^2-70336*n+7680)*a(n-1) + 5*(5*n-9)*(5*n-8)*(5*n-7)*(5*n-6)*(93*n^3-155*n^2+79*n-12)*a(n-2) with a(1) = 2 and a(2) = 12.
The o.g.f. A(x) = 2*x + 12*x^2 + 74*x^3 + ... is the diagonal of the bivariate rational function x*t/(1 - t/((1 - x)^2*(1 - x^2))) and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.
A(x) = x*d/dx(log(F(x))), where F(x) = (1/x)*Series_Reversion( x*(1 - x)^2*(1 - x^2) ).
a(n) ~ sqrt(4 + sqrt(6)) * (13/4 + 31*sqrt(6)/18)^n / (2*sqrt(5*Pi*n)). - Vaclav Kotesovec, Mar 15 2022

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

Original entry on oeis.org

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

Offset: 0

Peter Bala, Jul 31 2023

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

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

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

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

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

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

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.

A365025 Square array read by antidiagonals: T(n, k) := (k/2)!/k! * ((2n+1)k)! * ((2n+1/2)k)! / ( (nk)!^2 ((n+1/2)*k)!^2 ) for n, k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 10, 1, 1, 126, 300, 1, 1, 1716, 79380, 11440, 1, 1, 24310, 20612592, 65523780, 485100, 1, 1, 352716, 5318784900, 328206021000, 60634147860, 21841260, 1, 1, 5200300, 1368494343216, 1552041334596844, 5876083665270000, 59774707082376, 1022041020, 1

Offset: 0

Peter Bala, Aug 17 2023

Fractional factorials are defined in terms of the gamma function; for example, ((2*n+1/2)*k)! = Gamma(1 + (2*n+1/2)*k).
Given two sequences of integers c = (c_1, c_2, ..., c_K) and d = (d_1, d_2, ..., d_L) where c_1 + ... + c_K = d_1 + ... + d_L  we can define the factorial ratio sequence u_k(c, d) = (c_1*k)!*(c_2*k)!* ... *(c_K*k)!/ ( (d_1*k)!*(d_2*k)!* ... *(d_L*k)! ) and ask whether it is integral for all k >= 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 half-integers. See A276098 for further examples of this type.
Each row sequence of the present table is an integral factorial ratio sequence of height 2.
Conjecture: each row sequence of the table satisfies the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r.

			 Square array begins:
 n\k|  0      1               2                    3                      4
  - + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |  1      1               1                    1                      1  ...
  1 |  1     10             300                11440                 485100  ...
  2 |  1    126           79380             65523780            60634147860  ...
  3 |  1   1716        20612592         328206021000       5876083665270000  ...
  4 |  1  24310      5318784900     1552041334596844  510031828417402714500  ...
  5 |  1 352716   1368494343216  7108360304262169344 ...

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.
K. Soundararajan, Integral factorial ratios: irreducible examples with height larger than 1, Phil. Trans. Royal Soc., A378: 2018044, 2019.
Wikipedia, Dixon's identity

Cf. A275652 (row 1), A365026 (row 2), A365027 (row 3).

Cf. A276098, A364506, A364509, A364513, A364518.

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

from itertools import count, islice
from math import factorial
from sympy import factorial2
def A365025_T(n,k): return int(factorial2(k)*factorial(r:=((m:=n<<1)+1)*k)*factorial2(((m<<1)+1)*k)//((factorial(n*k)*factorial2(r))**2*factorial(k)))
def A365025_gen(): # generator of terms
    for n in count(0):
        yield from (A365025_T(n-k,k) for k in range(n+1))
A365025_list = list(islice(A365025_gen(),20)) # Chai Wah Wu, Aug 24 2023

T(n,k) = Sum_{j = 0..n*k} binomial((2*n+1)*k, n*k-j)^2 * binomial(k+j-1, j).
T(n,k) = binomial((2*n+1)*k,n*k)^2 * hypergeom([k, -n*k, -n*k], [1 + (n+1)*k, 1 + (n+1)*k], 1) = (k/2)!/k! * ((2*n+1)*k)! * ((2*n+1/2)*k)! / ( (n*k)!^2 * ((n+1/2)*k)!^2 ) by Dixon's 3F2 summation theorem.
T(n,k) = [x^(n*k)] ( (1 - x)^(2*n*k) * Legendre_P((2*n+1)*k, (1 + x)/(1 - x)) ).
T(n,k) = k!!*((2*n+1)*k)!*((4*n+1)*k)!!/(k!*((n*k)!*((2*n+1)*k)!!)^2). - Chai Wah Wu, Aug 24 2023

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.

A364518 Square array read by ascending antidiagonals: T(n,k) = [x^(2*k)] ( (1 + x)^(n+2)/(1 - x)^(n-2) )^k for n, k >= 0.

Original entry on oeis.org

1, 1, -2, 1, 0, 6, 1, 6, -10, -20, 1, 16, 70, 0, 70, 1, 30, 630, 924, 198, -252, 1, 48, 2310, 28672, 12870, 0, 924, 1, 70, 6006, 204204, 1385670, 184756, -4420, -3432, 1, 96, 12870, 860160, 19122246, 69206016, 2704156, 0, 12870, 1, 126, 24310, 2704156, 130378950, 1848483780, 3528923580, 40116600, 104006, -48620

Offset: 0

Peter Bala, Aug 07 2023

Compare with A364303 and A364519.
Given two sequences of integers c = (c_1, c_2, ..., c_K) and d = (d_1, d_2, ..., d_L), where c_1 + ... + c_K = d_1 + ... + d_L, we can define the factorial ratio sequence u_n(c, d) = (c_1*n)!*(c_2*n)!* ... *(c_K*n)!/ ( (d_1*n)!*(d_2*n)!* ... *(d_L*n)! ) and ask whether it is integral for all n >= 0. The integer L - K is called the height of the sequence. Bober completed the classification of integral factorial ratio sequences of height 1 (see A295431). Soundararajan gives many examples of two-parameter families of integral factorial ratio sequences of height 2.
Each row of the present table is an integral factorial ratio sequence of height 1. It is usually assumed that the c's and d's are integers but here some of the c's and d's are half-integers. See A276098 and the cross references there for further examples of this type.
It is known that the unsigned version of row 0 (the central binomial numbers A000984) and row 2 satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r. We conjecture that all the row sequences of the table satisfy the same supercongruences.

			 Square array begins:
 n\k|  0   1      2        3           4             5
  - + - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |  1  -2      6      -20          70          -252   ... see A000984
  1 |  1   0    -10        0         198             0   ... see A211419
  2 |  1   6     70      924       12870        184756   ... A001448
  3 |  1  16    630    28672     1385670      69206016   ... A091496
  4 |  1  30   2310   204204    19122246    1848483780   ... A061162
  5 |  1  48   6006   860160   130378950   20392706048   ... A276098
  6 |  1  70  12870  2704156   601080390  137846528820   ... A001448 bisected
  7 |  1  96  24310  7028736  2149374150  678057476096   ... A276099

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

Cf. A000984 (row 0 unsigned), A211419 (row 1 unsigned without 0's), A001448 (row 2), A091496 (row 3), A061162 (row 4), A276098 (row 5), A001448 bisected (row 6), A276099 (row 7).

Cf. A330843, A364303, A364506, A364509, A364513, A364519.

T(n,k) = add( binomial((n+2)*k, j)*binomial(n*k-j-1, 2*k-j), j = 0..2*k):
# display as a square array
seq(print(seq(T(n, k), k = 0..10)), n = 0..10);
# display as a sequence
seq(seq(T(n-k, k), k = 0..n), n = 0..10);

T(n,k) = sum(j = 0, 2*k, binomial((n+2)*k, j)*binomial(n*k-j-1, 2*k-j));
lista(nn) = for( n=0, nn, for (k=0, n, print1(T(n-k, k), ", "))); \\ Michel Marcus, Aug 13 2023

T(n,k) = Sum_{j = 0..2*k} binomial((n+2)*k, j)*binomial(n*k-j-1, 2*k-j).
T(2,k) = binomial(4*k,2*k).
For n >= 3, T(n,k) = binomial(n*k-1,2*k) * hypergeom([-(n+2)*k, -2*k], [1 - n*k], -1) except when (n,k) = (3,1).
For n >= 2, T(n,k) = ((n+2)*k)!*((n-2)*k/2)!/(((n+2)*k/2)!*((n-2)*k)!*(2*k)!) by Kummer's Theorem.
T(n,k) = [x^k] (1 - x)^(2*k) * Chebyshev_T(n*k, (1 + x)/(1 - x)).
T(n,k) = Sum_{j = 0..k} binomial(2*n*k, 2*j)*binomial((n-1)*k-j-1, k-j).
For n >= 3, T(n,k) = binomial((n-1)*k-1,k) * hypergeom([-n*k, -k, -n*k + 1/2], [1 - (n-1)*k, 1/2], 1).
The row generating functions are algebraic functions over the field of rational functions Q(x).

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

Original entry on oeis.org

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

Offset: 0

Peter Bala, Aug 07 2023

Compare with A364303 and A364518.
Given two sequences of integers c = (c_1, c_2, ..., c_K) and d = (d_1, d_2, ..., d_L), where c_1 + ... + c_K = d_1 + ... + d_L, we can define the factorial ratio sequence u_n(c, d) = (c_1*n)!*(c_2*n)!* ... *(c_K*n)!/ ( (d_1*n)!*(d_2*n)!* ... *(d_L*n)! ) and ask whether it is integral for all n >= 0. The integer L - K is called the height of the sequence. Bober completed the classification of integral factorial ratio sequences of height 1 (see A295431).
Each row of the present table is an integral factorial ratio sequence of height 1. It is usually assumed that the c's and d's are integers but here some of the c's and d's are half-integers. See A276098 and the cross references there for further examples of this type.
It is known that A005810, the unsigned version of row 1, satisfies the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r. We conjecture that each row sequence of the table satisfies the same supercongruences.

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

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

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

Cf. A330843, A364303, A364518.

T(n,k) := add( binomial((n+3)*k, j)*binomial(n*k-j-1, 3*k-j), j = 0..3*k):
# display as a square array
seq(print(seq(T(n, k), k = 0..10)), n = 0..10);
# display as a sequence
seq(seq(T(n-k, k), k = 0..n), n = 0..10);

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

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

A364520 a(n) = (7n)!(n/2)!/((7n/2)!(n)!(3n)!).

Original entry on oeis.org

1, 64, 12012, 2621440, 608435100, 146028888064, 35794148650260, 8901646138474496, 2237242000722428700, 566823049100850626560, 144520856111821003326512, 37036782455383679028953088, 9531607276865293630675462980, 2461693334077582876433071472640

Offset: 0

Peter Bala, Aug 09 2023

Row 4 of A364519.

Paolo Xausa, Table of n, a(n) for n = 0..400

Cf. A276098, A364519.

seq( simplify((7*n)!*(n/2)!/((7*n/2)!*(n)!*(3*n)!)),  n = 0..15);

A364520[n_]:=(7n)!(n/2)!/((7n/2)!n!(3n)!);Array[A364520,15,0] (* Paolo Xausa, Oct 05 2023 *)

from math import factorial
from sympy import factorial2
def A364520(n): return int((factorial(7*n)*factorial2(n)<<(3*n))//(factorial2(7*n)*factorial(n)*factorial(3*n))) # Chai Wah Wu, Aug 13 2023

a(n) = [x^(3*n)] ( (1 + x)^7/(1 - x) )^n.
a(n) = Sum_{j = 0..3*n} binomial(7*n, j)*binomial(4*n-j-1, 3*n-j).
a(n) = binomial(4*n-1, 3*n) * hypergeom([-7*n, -3*n], [1 - 4*n], -1) for n >= 2.
a(n) ~ c^n * 1/sqrt(6*Pi*n) where c = (14/3)^3*sqrt(7).
P-recursive: a(n) = 448*(7*n-1)*(7*n-3)*(7*n-5)*(7*n-9)*(7*n-11)*(7*n-13)/(3*n*(3*n-1)*(3*n-2)*(3*n-3)*(3*n-4)*(3*n-5)) * a(n-2) with a(0) = 1 and a(1) = 64.
The generating function is an algebraic function over the field of rational functions Q(x).

cp's OEIS Frontend

A091496 a(n) = ((5*n)!/(n!*(2*n)!))*(Gamma(1+n/2)/Gamma(1+5*n/2)).

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A352373 a(n) = [x^n] ( 1/((1 - x)^2*(1 - x^2)) )^n for n >= 1.

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

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

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A365025 Square array read by antidiagonals: T(n, k) := (k/2)!/k! * ((2*n+1)*k)! * ((2*n+1/2)*k)! / ( (n*k)!^2 * ((n+1/2)*k)!^2 ) for n, k >= 0.

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

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A091496 a(n) = ((5n)!/(n!(2n)!))(Gamma(1+n/2)/Gamma(1+5*n/2)).

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

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.

A365025 Square array read by antidiagonals: T(n, k) := (k/2)!/k! * ((2n+1)k)! * ((2n+1/2)k)! / ( (nk)!^2 ((n+1/2)*k)!^2 ) for n, k >= 0.

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

A364520 a(n) = (7n)!(n/2)!/((7n/2)!(n)!(3n)!).