A352651 -id:A352651 - cp's OEIS frontend

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

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)

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.

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.

cp's OEIS Frontend

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

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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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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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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.

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.

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