A275652 -id:A275652 - cp's OEIS frontend

A364303 Square array read by ascending antidiagonals: T(n,k) = [x^k] (1 - x)^(2k) Legendre_P(n*k, (1 + x)/(1 - x)) for n, k >= 0.

1, 1, -2, 1, 0, 6, 1, 4, -6, -20, 1, 10, 36, 0, 70, 1, 18, 300, 400, 90, -252, 1, 28, 1050, 11440, 4900, 0, 924, 1, 40, 2646, 77616, 485100, 63504, -1680, -3432, 1, 54, 5544, 316540, 6370650, 21841260, 853776, 0, 12870, 1, 70, 10296, 972400, 42031990, 554822268, 1022041020, 11778624, 34650, -48620

Offset: 0

Peter Bala, Jul 19 2023

The first row of the table is a signed version of the central binomial coefficients A000984. The central binomial coefficients 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 (see Meštrović, equation 39). 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  -2      6      -20          70          -252    ...  (-1)^k*A000984(k)
  1 |  1   0     -6        0          90             0    ...  A245086
  2 |  1   4     36      400        4900         63504    ...  A002894
  3 |  1  10    300    11440      485100      21841260    ...  A275652
  4 |  1  18   1050    77616     6370650     554822268    ...  A275653
  5 |  1  28   2646   316540    42031990    5921058528    ...  A275654
  6 |  1  40   5544   972400   189290920   39089615040    ...  A275655
  7 |  1  54  10296  2484000   665091000  188907932304    ...  A364304
  8 |  1  70  17550  5567380  1960044750  732012601320    ...  A364305

R. Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.
Wikipedia, Dixon's identity

Cf. A000984 (row 0 unsigned), A245086 (row 1), A002894 (row 2), A275652 (row 3), A275653 (row 4), A275654 (row 5), A275655 (row 6), A364304 (row 7), A364305 (row 8).

Cf. A364113, A364298

T(n,k) := coeff(series( (1 - x)^(2*k) * LegendreP(n*k, (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, k-i)^2 * binomial((n-2)*k+i-1, i).
For n >= 2, T(n,k) = binomial((n-1)*k, k)^2 * hypergeom([a, b, b], [1 + a - b, 1 + a - b], 1), where a = (n - 3)*k and b = -k.
For n >= 3, T(n,k) = ((n - 1)*k)! * ((n + 1)*k/2)! * ((n - 3)*k/2)! / ( ((n - 1)*k/2)!^2 * k!^2 *  ((n - 3)*k)! ) by Dixon's 3F2 summation theorem, where fractional factorials are defined in terms of the gamma function.

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).

A275653 a(n) = binomial(4n,2n)binomial(3n,2*n).

Original entry on oeis.org

1, 18, 1050, 77616, 6370650, 554822268, 50199951984, 4664758248000, 442077195513690, 42533571002422500, 4141601026094832300, 407220411993767798400, 40363606408574136870000, 4028061310168832261158176, 404311537318239680601595200, 40785601782042745410592271616

Offset: 0

Peter Bala, Aug 04 2016

Right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(n+k)*binomial(4*n + k,4*n - k)*binomial(2*k,k)* binomial(2*n - k,n) = binomial(4*n,2*n)* binomial(3*n,2*n).
We also have Sum_{k = 0..4*n} (-1)^(n+k)*binomial(4*n + k,4*n - k)*binomial(2*k,k)*binomial(2*n - k,n) = binomial(4*n,2*n)* binomial(3*n,2*n).
Compare with the identities
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).
Sum_{k = 0..n} (-1)^(n+k)*binomial(6*n + k,6*n - k)* binomial(2*k,k)*binomial(2*n - k,n) = binomial(6*n,3*n)* binomial(2*n,n) = A275655(n)
Sum_{k = 0..n} (-1)^(n+k)*binomial(8*n + k,8*n - k)* binomial(2*k,k)*binomial(2*n - k,n) = binomial(8*n,4*n)* binomial(5*n,2*n)*binomial(2*n,n)/binomial(6*n,3*n).
See also A275652, A275654 and A275655.

Cf. A001448, A002894, A005809, A275652, A275654, A275655.

seq((4*n)!*(3*n)!/(n!*(2*n)!^3), n = 0..20);

Table[Binomial[4 n, 2 n] Binomial[3 n, 2 n], {n, 0, 15}] (* Michael De Vlieger, Aug 07 2016 *)

a(n) = (4*n)!*(3*n)!/(n!*(2*n)!^3).
a(n) = A001448(n) * A005809(n).
Recurrence: a(n) = 3*(3*n - 1)*(3*n - 2)*(4*n - 1)*(4*n - 3)/(n^2*(2*n - 1)^2) * a(n-1).
a(n) = [x^n] ((1 + x)^2/(1 - x))^(2*n) * [x^n] (1 + x)^(3*n) = [x^n] G(x)^(6*n) where G(x) = 1 + 3*x + 38*x^2 + 1150*x^3 + 47099^x^4 + 2264968*x^5 + 120311611*x^6 + ... appears to have integer coefficients.
exp( Sum_{n >= 1} a(n)*x^n/n ) = F(x)^6, where F(x) = 1 + 3*x + 92*x^2 + 4579*x^3 + 282605*x^4 + 19698991*x^5 + 1484923315*x^6 + ... appears to have integer coefficients.
a(n) ~ sqrt(3/2)*108^n/(2*Pi*n). - Ilya Gutkovskiy, Aug 07 2016
From Peter Bala, Mar 23 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(3*n-k-1,n-k)*binomial(4*n,k)^2.
a(n) = [x^n] (1 - x)^(2*n) * P(4*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A275652.
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)

A275655 a(n) = binomial(6n,3n)binomial(2n,n).

Original entry on oeis.org

1, 40, 5544, 972400, 189290920, 39089615040, 8385425017200, 1847301025078080, 415026659401497000, 94660194875011205440, 21850091031597537252544, 5092815839064962373499680, 1196622940864849837505171824, 283073284848591452381449360000

Offset: 0

Peter Bala, Aug 04 2016

Right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(n+k)*binomial(6*n + k,6*n - k)*binomial(2*k,k) *binomial(2*n - k,n) = binomial(6*n,3*n)*binomial(2*n,n).
We also note that Sum_{k = 0..6*n} (-1)^(n+k)*binomial(6*n + k,6*n - k)*binomial(2*k,k)*binomial(2*n - k,n) = binomial(6*n,3*n)*binomial(2*n,n).
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 A275654.

Cf. A000984, A002894, A066802, A275652, A275653, A275654.

seq((6*n)!*(2*n)!/((3*n)!*n!)^2, n = 0..20);

Table[Binomial[6 n, 3 n] Binomial[2 n, n], {n, 0, 13}] (* Michael De Vlieger, Aug 07 2016 *)

a(n) = (6*n)!*(2*n)!/((3*n)!*n!)^2.
a(n) = A066802(n) * A000984(n).
Recurrence: a(n) = 16*(2*n - 1)^2*(6*n - 1)*(6*n - 5)/(n^2*(3*n - 1)*(3*n - 2)) * a(n-1).
a(n) = [x^(3*n)] (1 + x)^(6*n) * [x^n] (1 + x)^(2*n) = [x^n] G(x)^(8*n) where G(x) = 1 + 5*x + 159*x^2 + 11690*x^3 + 1160817*x^4 + 135123516*x^5 + 17357714116*x^6 + ... appears to have integer coefficients.
exp( Sum_{n >= 1} a(n)*x^n/n ) = F(x)^8, where F(x) = 1 + 5*x + 359*x^2 + 42270*x^3 + 6182313*x^4 + 1021669966*x^5 + 182605696304*x^6 + ... appears to have integer coefficients.
a(n) ~ 256^n/(sqrt(3)*Pi*n). - Ilya Gutkovskiy, Aug 07 2016
From Peter Bala, Mar 23 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(5*n-k-1,n-k)*binomial(6*n,k)^2.
a(n) = [x^n] (1 - x)^(2*n) * P(6*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A275652.
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)

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

A365026 a(n) = (5n)!(9n/2)!(n/2)! / ((2n)!^2 (5n/2)!^2 n!).

Original entry on oeis.org

1, 126, 79380, 65523780, 60634147860, 59774707082376, 61346313465418800, 64736852770959042240, 69724035322703253191700, 76277370761329867481375100, 84482032811073922526904281880, 94508142285721995026811874069200, 106599928449546340546215262030974000

Offset: 0

Peter Bala, Aug 17 2023

Fractional factorials are defined in terms of the gamma function; for example, (9*n/2)! = Gamma(1 + 9*n/2).

Row 2 of A365025.

Paolo Xausa, Table of n, a(n) for n = 0..300

Cf. A275652, A365025, A365027.

seq( simplify((5*n)!*(9*n/2)!*(n/2)! / ((2*n)!^2 * (5*n/2)!^2 * n!)), n = 0..15);

A365026[n_]:=(5n)!(9n/2)!(n/2)!/((2n)!^2(5n/2)!^2n!);Array[A365026,15,0] (* Paolo Xausa, Oct 05 2023 *)

from math import factorial
from sympy import factorial2
def A365026(n): return int(factorial(5*n)*factorial2(9*n)*factorial2(n)//((factorial2(5*n)*factorial(n<<1))**2*factorial(n))) # Chai Wah Wu, Aug 24 2023

a(n) = Sum_{j = 0..2*n} binomial(5*n, 2*n-j)^2 * binomial(n+j-1, j).
P-recursive: (5*n-2)*(5*n-4)*(5*n-6)*(5*n-8)*(2*n)^2*(2*n-1)^2*(2*n-2)^2*(2*n-3)^2*a(n)= 9*(9*n-2)*(9*n-4)*(9*n-6)*(9*n-8)*(9*n-10)*(9*n-12)*(9*n-14)*(9*n-16)*(5*n-1)*(5*n-3)*(5*n-7)*(5*n-9)*a(n-2) with a(0) = 1 and a(1) = 126.
a(n) ~ c^n * 3*sqrt(5)/(20*Pi*n), where c = (3^9)/(2^4).
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all integers n and r.
a(n) = [x^n] G(x)^(9*n), where the power series G(x) = 1 + 14*x + 2744*x^2 + 1130724*x^3 + 615596785*x^4 + 388901411712*x^5 + 269588153179744*x^6 + ... appears to have integer coefficients.
exp( Sum_{n >= 1} a(n)*x^n/n ) = F(x)^9, where the power series F(x) = 1 + 14*x + 4508*x^2 + 2489004*x^3 + 1728415009*x^4 + 1362984972918*x^5 + 1165343050808188*x^6 + ... appears to have integer coefficients.

A365027 a(n) = (7n)!(13n/2)!(n/2)! / ((3n)!^2 (7n/2)!^2 n!).

Original entry on oeis.org

1, 1716, 20612592, 328206021000, 5876083665270000, 112210544802995673216, 2232092469681027490937400, 45670179632369542491712236480, 953926390279492216468973361270000, 20241460048032081192591594667805420400, 434878619369192244460121948456800558766592

Offset: 0

Peter Bala, Aug 18 2023

Fractional factorials are defined in terms of the gamma function; for example, (13*n/2)! = Gamma(1 + 13*n/2).

Row 3 of A365025.

Paolo Xausa, Table of n, a(n) for n = 0..200

Cf. A275652, A365025, A365026.

seq( simplify((7*n)!*(13*n/2)!*(n/2)! / ((3*n)!^2 * (7*n/2)!^2 * n!)), n = 0..15);

A365027[n_]:=(7n)!(13n/2)!(n/2)!/((3n)!^2(7n/2)!^2n!);Array[A365027,10,0] (* Paolo Xausa, Oct 05 2023 *)

from math import factorial
from sympy import factorial2
def A365027(n): return int(factorial(7*n)*factorial2(13*n)*factorial2(n)//((factorial2(7*n)*factorial(3*n))**2*factorial(n))) # Chai Wah Wu, Aug 24 2023

a(n) = Sum_{j = 0..3*n} binomial(7*n, 3*n-j)^2 * binomial(n+j-1, j).
a(n) ~ c^n * sqrt(91)/(42*Pi*n), where c = sqrt(13)*(13/3)^6.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all integers n and r.
a(n) = [x^n] G(x)^(78*n), where the power series G(x) = 1 + 22*x + 94622*x^2 + 821626080*x^3 + 9321370449728*x^4 + 122052794640882192*x^5 + 1748115226331150054950*x^6 + ... appears to have integer coefficients.
exp( Sum_{n >= 1} a(n)*x^n/n ) = F(x)^78, where the power series F(x) = 1 + 22*x + 132374*x^2 + 1405498512*x^3 + 18873219886000*x^4 + 288319543590164888*x^5 + 4779239354183722040470*x^6 + ... appears to have integer coefficients.

A364304 a(n) = (7n)!(9n/2)!(5n/2)!/((5n)!(7n/2)!^2*n!^2).

Original entry on oeis.org

1, 54, 10296, 2484000, 665091000, 188907932304, 55737530929080, 16888537352985408, 5218680924762089400, 1637124203403474142500, 519752205290081232622296, 166620892958456148158454144, 53846423260084127389865311800

Offset: 0

Peter Bala, Jul 21 2023

Fractional factorials are defined in terms of the gamma function; for example, (9*n/2)! = gamma(1 + 9*n/2).

Paolo Xausa, Table of n, a(n) for n = 0..300

Row 7 of A364303.

Cf. A275652, A275654.

seq( simplify((7*n)!*(9*n/2)!*(5*n/2)!/((5*n)!*(7*n/2)!^2*n!^2)), n = 0..12);

A364304[n_]:=(7n)!(9n/2)!(5n/2)!/((5n)!(7n/2)!^2n!^2);Array[A364304,15,0] (* Paolo Xausa, Oct 06 2023 *)

from math import factorial
from sympy import factorial2
def A364304(n): return int(factorial(7*n)*factorial2(9*n)*factorial2(5*n)//factorial(5*n)//factorial2(7*n)**2//factorial(n)**2) # Chai Wah Wu, Aug 10 2023

a(n) = Sum_{k = 0..n} binomial(7*n, n - k)^2 * binomial(5*n + k - 1, k).
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(7*n + k, 7*n - k)*binomial(2*k, k)*binomial(2*n - k, n).
Conjecture: a(n) =  Sum_{k = 0..7*n} (-1)^k * binomial(7*n + k, 7*n - k)* binomial(2*k, k)*binomial(2*n - k, n)
a(n) = [x^n] (1 - x)^(2*n) * P(7*n, (1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial.
a(n) = [x^n] G(x)^(9*n), where G(x) = 1 + 6*x + 266*x^2 + 27104*x^3 + 3726380*x^4 + 600232416*x^5 + 106662768380*x^6 + ... appears to have integer coefficients.
exp( Sum_{n > = 1} a(n)*x^n/n ) = F(x)^9, where F(x) = 1 + 6*x + 590*x^2 + 95468*x^3 + 19200692*x^4 + 4364084760*x^5 + 1072849548644*x^6 + ... appears to have integer coefficients.
a(p) == a(1) (mod p^3).
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.
P-recursive: a(n) =  9*(9*n-2)*(9*n-4)*(9*n-6)*(9*n-8)*(9*n-10)*(9*n-12)*(9*n-14)*(9*n-16)*(7*n-1)*(7*n-3)*(7*n-5)*(7*n-9)*(7*n-11)*(7*n-13)/((7*n-2)*(7*n-4)*(7*n-6)*(7*n-8)*(7*n-10)*(7*n-12)*(5*n-1)*(5*n-3)*(5*n-5)*(5*n-7)*(5*n-9)*n^2*(n-1)) * a(n-2) with a(0) = 1 and a(1) = 54.
a(n) ~ c^n * 3*sqrt(7)/(14*Pi*n), where c = (3^9)/(5^3) * sqrt(5).

cp's OEIS Frontend

A364303 Square array read by ascending antidiagonals: T(n,k) = [x^k] (1 - x)^(2*k) * Legendre_P(n*k, (1 + x)/(1 - x)) for n, k >= 0.

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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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A275653 a(n) = binomial(4*n,2*n)*binomial(3*n,2*n).

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A275655 a(n) = binomial(6*n,3*n)*binomial(2*n,n).

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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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A365026 a(n) = (5*n)!*(9*n/2)!*(n/2)! / ((2*n)!^2 * (5*n/2)!^2 * n!).

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A364303 Square array read by ascending antidiagonals: T(n,k) = [x^k] (1 - x)^(2k) Legendre_P(n*k, (1 + x)/(1 - x)) for n, k >= 0.

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.

A275653 a(n) = binomial(4n,2n)binomial(3n,2*n).

A275655 a(n) = binomial(6n,3n)binomial(2n,n).

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.

A365026 a(n) = (5n)!(9n/2)!(n/2)! / ((2n)!^2 (5n/2)!^2 n!).

A365027 a(n) = (7n)!(13n/2)!(n/2)! / ((3n)!^2 (7n/2)!^2 n!).

A364304 a(n) = (7n)!(9n/2)!(5n/2)!/((5n)!(7n/2)!^2*n!^2).