A365025 -id:A365025 - 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

A364506 Square array read by ascending antidiagonals: T(n,k) = (2k)!/k! ( (2nk)! * ((2n+1)k)! )/( (nk)!^2 ((n+1)*k)!^2 ).

Original entry on oeis.org

1, 1, 2, 1, 6, 6, 1, 40, 90, 20, 1, 350, 5880, 1680, 70, 1, 3528, 594594, 1101100, 34650, 252, 1, 38808, 75088728, 1299170600, 229265400, 756756, 924, 1, 453024, 10861066216, 2066315135040, 3164045050530, 50678855040, 17153136, 3432, 1, 5521230, 1721929279200, 3943172216808000

Offset: 0

Peter Bala, Jul 27 2023

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.
Each row sequence of the present table is an integral factorial ratio sequence of height 2.
It is known that both row 0, the central binomial numbers, and row 1, the de Bruijn numbers, 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 ...
  1 |  1     6        90           1680              34650            756756 ...
  2 |  1    40      5880        1101100          229265400       50678855040 ...
  3 |  1   350    594594     1299170600      3164045050530  8188909171581600 ...
  4 |  1  3528  75088728  2066315135040  63464046079757400  ...
  5 |  1 38808  ...

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

A000984 (row 0), A006480 (row 1), A364507 (row 2), A364508 (row 3). Cf. A364303, A364509, A365025.

# display as a square array
T(n,k) := (2*k)!/k! * ( (2*n*k)! * ((2*n+1)*k)! )/((n*k)!^2 * ((n+1)*k)!^2):
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 = -k..k} (-1)^i * binomial(2*k, k+i) * binomial(2*n*k, n*k+i)^2 (shows that the table entries are integers).
For n >= 1, T(n,k) = (-1)^k * binomial(2*n*k, (n+1)*k)^2 * hypergeom([-2*k, -(n+1)*k, -(n+1)*k], [1 + (n-1)*k, 1 + (n-1)*k], 1) = (2*k)!/k! * ( (2*n*k)! * ((2*n+1)*k)! )/( (n*k)!^2 * ((n+1)*k)!^2 ) by Dixon's 3F2 summation theorem.
T(n,k) = (-1)^k * [x^((n + 1)*k)] ( (1 - x)^(2*(n+1)*k) * Legendre_P(2*n*k, (1 + x)/(1 - x)) ). - Peter Bala, Aug 15 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.

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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A364506 Square array read by ascending antidiagonals: T(n,k) = (2*k)!/k! * ( (2*n*k)! * ((2*n+1)*k)! )/( (n*k)!^2 * ((n+1)*k)!^2 ).

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

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Maple

Mathematica

Python

Formula

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

A364506 Square array read by ascending antidiagonals: T(n,k) = (2k)!/k! ( (2nk)! * ((2n+1)k)! )/( (nk)!^2 ((n+1)*k)!^2 ).

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!).