A276099 -id:A276099 - cp's OEIS frontend

A276098 a(n) = (7n)!(3/2n)!/((7n/2)!(3n)!(2n)!).

1, 48, 6006, 860160, 130378950, 20392706048, 3254013513660, 526470648692736, 86047769258554950, 14173603389190963200, 2349023203055914140756, 391249767795614684282880, 65434374898388743460014620, 10981406991821583404677201920

Offset: 0

Peter Bala, Aug 22 2016

Let a > b be nonnegative integers. The ratio of factorials (2*a*n)!*(b*n)!/( (a*n)!*(2*b*n)!*((a - b)*n)! ) is known to be an integer for n >= 0 (see, for example, Bober, Theorem 1.1). We have the companion result: 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 n >= 0. This is the case a = 3, b = 1. Other cases include A091496 (a = 2, b = 0), A091527 (a = 1, b = 0), A262732 (a = 2, b = 1), A262733 (a = 3, b = 2) and A276099 (a = 4, b = 2).

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

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. A046521, A091496, A091527, A262732, A262733, A276099.

seq(simplify((7*n)!*(3/2*n)!/((7*n/2)!*(3*n)!*(2*n)!)), n = 0..20);

from math import factorial
from sympy import factorial2
def A276098(n): return int((factorial(7*n)*factorial2(3*n)<<(n<<1))//factorial2(7*n)//factorial(3*n)//factorial(n<<1)) # Chai Wah Wu, Aug 10 2023

a(n) = Sum_{k = 0..2*n} binomial(7*n, 2*n - k)*binomial(3*n + k - 1, k).
a(n) = Sum_{k = 0..n} binomial(10*n, 2*n - 2*k)*binomial(3*n + k - 1, k).
Recurrence: a(n) = 28*(7*n - 1)*(7*n - 3)*(7*n - 5)*(7*n - 9)*(7*n - 11)*(7*n - 13)/(3*n*(n - 1)*(2*n - 1)*(2*n - 3)*(3*n - 1)*(3*n - 5)) * a(n-2).
a(n) ~ 1/sqrt(4*Pi*n) * (7^7/3^3)^(n/2).
O.g.f. A(x) = Hypergeom([13/14, 11/14, 9/14, 5/14, 3/14, 1/14], [5/6, 3/4, 1/2, 1/4, 1/6], (7^7/3^3)*x^2) + 48*x*Hypergeom([10/7, 9/7, 8/7, 6/7, 5/7, 4/7], [5/4, 4/3, 3/2, 3/4, 2/3], (7^7/3^3)*x^2).
a(n) = [x^(2*n)] H(x)^n, where H(x) = (1 + x)^7/(1 - x)^3.
It follows that the o.g.f. A(x) equals 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.
Let F(x) = (1/x)*Series_Reversion( x*sqrt((1 - x)^3/(1 + x)^7) ) and put G(x) = 1 + x*d/dx(log(F(x))). Then A(x^2) = (G(x) + G(-x))/2.

A091527 a(n) = ((3n)!/n!^2)(Gamma(1+n/2)/Gamma(1+3n/2)).

Original entry on oeis.org

1, 4, 30, 256, 2310, 21504, 204204, 1966080, 19122246, 187432960, 1848483780, 18320719872, 182327718300, 1820797698048, 18236779032600, 183120225632256, 1842826521244230, 18581317012684800, 187679234340049620, 1898554215471513600, 19232182592635611060

Offset: 0

Michael Somos, Jan 18 2004

Sequence terms are given by [x^n] ( (1 + x)^(k+2)/(1 - x)^k )^n for k = 1. See the crossreferences for related sequences obtained from other values of k. - Peter Bala, Sep 29 2015

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 n >= 0. This is the case a = 1, b = 0. - Peter Bala, Aug 28 2016

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

Michael De Vlieger, Table of n, a(n) for n = 0..985
Peter Bala, Some integer ratios of factorials
Paul Barry, On the Central Coefficients of Bell Matrices, J. Int. Seq. 14 (2011) # 11.4.3.
Karl Dilcher, Armin Straub, and Christophe Vignat, Identities for Bernoulli polynomials related to multiple Tornheim zeta functions, arXiv:1903.11759 [math.NT], 2019. See p. 11.
I. M. Gessel, A short proof of the Deutsch-Sagan congruence for connected non crossing graphs, arXiv preprint arXiv:1403.7656 [math.CO], 2014.
Bernhard Heim and Markus Neuhauser, Asymptotic Distribution of the Zeros of recursively defined Non-Orthogonal Polynomials, arXiv:2107.05013 [math.CA], 2021.
W. Mlotkowski and K. A. Penson, Probability distributions with binomial moments, arXiv preprint arXiv:1309.0595 [math.PR], 2013.

Cf. A061162(n) = a(2n), A007297, A000984 (k = 0), A001448 (k = 2), A262732 (k = 3), A211419 (k = 4), A262733 (k = 5), A211421 (k = 6), A276098, A276099.

a := n -> 4^n * `if`(n<2, 1, (2*(n+1)*binomial((3*n-1)/2, n + 1))/(n-1)):
seq(a(n), n=0..18); # Peter Luschny, Feb 03 2020

Table[((3 n)!/n!^2) Gamma[1 + n/2]/Gamma[1 + 3 n/2], {n, 0, 18}] (* Michael De Vlieger, Oct 02 2015 *)
Table[4^n Sum[Binomial[k - 1 + (n - 1)/2, k], {k, 0, n}], {n, 0, 18}] (* Michael De Vlieger, Aug 28 2016 *)

B(x):=(-1/3+(2/3)*sqrt(1+9*x)*sin((1/3)*asin((2+27*x+54*x^2)/2/(1+9*x)^(3/2))))/x-1;
taylor(x*diff(B(x),x)/B(x),x,0,10); /* Vladimir Kruchinin, Oct 02 2015 */

a(n)=4^n*sum(i=0,n,binomial(i-1+(n-1)/2,i))

vector(30, n, sum(k=0, n, binomial(3*n-3, k)*binomial(2*n-k-3, n-k-1))) \\ Altug Alkan, Oct 04 2015

from math import factorial
from sympy import factorial2
def A091527(n): return int((factorial(3*n)*factorial2(n)<Chai Wah Wu, Aug 10 2023

D-finite with recurrence n*(n - 1)*a(n) = 12*(3*n - 1)*(3*n - 5)*a(n-2).
From Peter Bala, Sep 29 2015: (Start)
a(n) = Sum_{i = 0..n} binomial(3*n,i) * binomial(2*n-i-1,n-i).
a(n) = [x^n] ( (1 + x)^3/(1 - x) )^n.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 4*x + 23*x^2 + 156*x^3 + 1162*x^4 + 9192*x^5 + ... is the o.g.f. for A007297 (but with an offset of 0). (End)
a(n) = (n+1)*A078531(n). [Barry, JIS (2011)]
G.f.: x*B'(x)/B(x), where x*B(x)+1 is g.f. of A007297. - Vladimir Kruchinin, Oct 02 2015
From Peter Bala, Aug 22 2016: (Start)
a(n) = Sum_{k = 0..floor(n/2)} binomial(4*n,n-2*k)*binomial(n+k-1,k).
O.g.f.: A(x) = Hypergeom([5/6, 1/6], [1/2], 108*x^2) + 4*x*Hypergeom([4/3, 2/3], [3/2], 108*x^2).
The o.g.f. is the diagonal of the bivariate rational function 1/(1 - t*(1 + x)^3/(1 - x)) and hence is algebraic by Stanley 1999, Theorem 6.33, p. 197. (End)
a(n) ~ 2^n*3^(3*n/2)/sqrt(2*Pi*n). - Ilya Gutkovskiy, Aug 22 2016
a(n) = 4^n*2*(n+1)*binomial((3*n-1)/2, n+1)/(n-1) for n >= 2. - Peter Luschny, Feb 03 2020
From Peter Bala, Mar 04 2022: (Start)
The o.g.f. A(x) satisfies the algebraic equation (1 - 108*x^2)*A(x)^3 - A(x) = 8*x. Cf. A244039.
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.
Conjecture: the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for primes p >= 5 and positive integers n and k. (End)
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] 1/((1-x)^(n+1) * (1-2*x)^n).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(3*n,k) * binomial(2*n-k,n-k).
a(n) = Sum_{k=0..n} 2^k * binomial(n+k-1,k) * binomial(2*n-k,n-k).
a(n) = 4^n * binomial((3*n-1)/2,n).
a(n) = [x^n] 1/(1-4*x)^((n+1)/2).
a(n) = [x^n] (1+4*x)^((3*n-1)/2). (End)

A262732 a(n) = (1/n!) * (5n)!/(5n/2)! * (3n/2)!/(3n)!.

Original entry on oeis.org

1, 8, 126, 2240, 41990, 811008, 15967980, 318636032, 6421422150, 130395668480, 2663825039876, 54684895150080, 1127155102890908, 23311847679590400, 483537022180231320, 10054732930602762240, 209536624110664757830, 4375058594685417160704, 91505601042318156186900

Offset: 0

Peter Bala, Sep 29 2015

Sequence terms are given by the coefficient of x^n in the expansion of ( (1 + x)^(k+2)/(1 - x)^k )^n when k = 3. See the cross references for related sequences obtained from other values of k.

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

Michael De Vlieger, Table of n, a(n) for n = 0..751
Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion
Peter Bala, Some integer ratios of factorials
Peter Bala, A supercongruence for A262732

Cf. A000984 (k = 0), A091527 (k = 1), A001448 (k = 2), A211419 (k = 4), A262733 (k = 5), A211421 (k = 6), A262737, A276098, A276099.

Cf. A115293.

a := n -> 1/n! * (5*n)!/GAMMA(1 + 5*n/2) * GAMMA(1 + 3*n/2)/(3*n)!:
seq(a(n), n = 0..18);

Table[1/n!*(5 n)!/(5 n/2)!*(3 n/2)!/(3 n)!, {n, 0, 18}] (* or *)
Table[Sum[Binomial[8 n, n - 2 k] Binomial[3 n + k - 1, k], {k, 0, Floor[n/2]}], {n, 0, 18}] (* Michael De Vlieger, Aug 28 2016 *)

a(n) = sum(k=0, n, binomial(5*n,k)*binomial(4*n-k-1,n-k));
vector(30, n, a(n-1)) \\ Altug Alkan, Oct 03 2015

from math import factorial
from sympy import factorial2
def A262732(n): return int((factorial(5*n)*factorial2(3*n)<Chai Wah Wu, Aug 10 2023

a(n) = Sum_{i = 0..n} binomial(5*n,i) * binomial(4*n-i-1,n-i).
a(n) = [x^n] ( (1 + x)^5/(1 - x)^3 )^n.
D-finite with recurrence a(n) = 20*(5*n - 1)*(5*n - 3)*(5*n - 7)*(5*n - 9)/( n*(3*n - 1)*(3*n - 3)*(3*n - 5) ) * a(n-2).
The o.g.f. exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 8*x + 95*x^2 + 1336*x^4 + ... has integer coefficients and equals (1/x) * (series reversion of x*(1 - x)^3/(1 + x)^5). See A262737.
a(n) ~ 2^n*3^(-3*n/2)*5^(5*n/2)/sqrt(2*Pi*n). - Ilya Gutkovskiy, Jul 31 2016
From Peter Bala, Aug 22 2016: (Start)
a(n) = Sum_{k = 0..floor(n/2)} binomial(8*n,n - 2*k) * binomial(3*n + k - 1,k).
O.g.f.: A(x) = Hypergeom([9/10, 7/10, 3/10, 1/10], [5/6, 1/2, 1/6], (12500/27)*x^2) + 8*x*Hypergeom([7/5, 6/5, 4/5, 3/5], [4/3, 3/2, 2/3], (12500/27)*x^2).
The o.g.f. is the diagonal of the bivariate rational function 1/(1 - t*(1 + x)^5/(1 - x)^3) and hence is algebraic by Stanley 1999, Theorem 6.33, p. 197. (End)
From Karol A. Penson, Apr 26 2018: (Start)
Integral representation of a(n) as the n-th moment of a positive function w(x) on the support (0, sqrt(12500/27)):
a(n) = Integral_{x=0..sqrt(12500/27)} x^n*w(x) dx,
where w(x) = sqrt(5)*2^(3/5)*csc((1/5)*Pi)*sin((1/10)*Pi)*hypergeom([1/10, 4/15, 3/5, 14/15], [1/5, 2/5, 4/5], 27*x^2*(1/12500))/(10*Pi*x^(4/5)) + sqrt(5)*2^(4/5)*csc(2*Pi*(1/5))*sin(3*Pi*(1/10))*hypergeom([3/10, 7/15, 4/5, 17/15], [2/5, 3/5, 6/5], 27*x^2*(1/12500))/(50*Pi*x^(2/5)) + sqrt(5)*2^(1/5)*csc(2*Pi*(1/5))*sin(3*Pi*(1/10))*x^(2/5)*hypergeom([7/10, 13/15, 6/5, 23/15], [4/5, 7/5, 8/5], 27*x^2*(1/12500))/(625*Pi) + 11*sqrt(5)*2^(2/5)*csc((1/5)*Pi)*sin((1/10)*Pi)*x^(4/5)*hypergeom([9/10, 16/15, 7/5, 26/15], [6/5, 8/5, 9/5], 27*x^2*(1/12500))/(50000*Pi). The function w(x) involves four different hypergeometric functions of type 4F3. The function w(x) is singular at both ends of the support. It is the solution of the Hausdorff moment problem and as such it is unique. (End)
From Peter Bala, Sep 15 2021: (Start)
a(n) = [x^n] (1 + 4*x)^((5*n-1)/2) = 4^n*binomial((5*n-1)/2,n).
a(p) == a(1) (mod p^3) for prime p >= 5.
More generally, we conjecture that a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) for prime p >= 5 and positive integers n and k. (End)
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] 1/((1-x)^(n+1) * (1-2*x)^(3*n)).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(5*n,k) * binomial(2*n-k,n-k).
a(n) = Sum_{k=0..n} 2^k * binomial(3*n+k-1,k) * binomial(2*n-k,n-k).
a(n) = [x^n] 1/(1-4*x)^((3*n+1)/2). (End)

A262733 a(n) = (1/n!) * (7n)!/(7n/2)! * (5n/2)!/(5n)!.

Original entry on oeis.org

1, 12, 286, 7680, 217350, 6336512, 188296108, 5670567936, 172459427910, 5284842700800, 162922160580036, 5047099485847552, 156983503897469340, 4899363753956474880, 153349672416272587800, 4811846645261721927680, 151316978279502571401798, 4767566079229070105640960

Offset: 0

Peter Bala, Sep 29 2015

Sequence terms are given by the coefficient of x^n in the expansion of ( (1 + x)^(k+2)/(1 - x)^k )^n when k = 5. See the cross references for related sequences obtained from other values of k.

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 n >= 0. This is the case a = 3, b = 2. - Peter Bala, Aug 28 2016

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

Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion
Peter Bala, Some integer ratios of factorials

Cf. A000984 (k = 0), A091527 (k = 1), A001448 (k = 2), A262732 (k = 3), A211419 (k = 4), A211421 (k = 6), A262739, A276098, A276099.

a := n -> 1/n! * (7*n)!/GAMMA(1 + 7*n/2) * GAMMA(1 + 5*n/2)/(5*n)!:
seq(a(n), n = 0..18);

Table[1/n!*(7 n)!/(7 n/2)!*(5 n/2)!/(5 n)!, {n, 0, 17}] (* Michael De Vlieger, Oct 04 2015 *)

a(n) = sum(k=0, n, binomial(7*n,k)*binomial(6*n-k-1,n-k));
vector(30, n, a(n-1)) \\ Altug Alkan, Oct 03 2015

from math import factorial
from sympy import factorial2
def A262733(n): return int((factorial(7*n)*factorial2(5*n)<Chai Wah Wu, Aug 10 2023

a(n) = [x^n] ( (1 + x)^7/(1 - x)^5 )^n.
a(n) = Sum_{i = 0..n} binomial(7*n,i) * binomial(6*n-i-1,n-i).
a(n) = 28*(7*n - 1)*(7*n - 3)*(7*n - 9)*(7*n - 11)*(7*n - 13) / ( n*(5*n - 1)*(5*n - 3)*(5*n - 5)*(5*n - 7)*(5*n - 9) ) * a(n-2).
The o.g.f. exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 12*x + 215*x^2 + 4564*x^3 + 106442*x^4 + ... has integer coefficients and equals 1/x * series reversion of x*(1 - x)^5/(1 + x)^7. See A262739.
a(n) ~ 2^n*5^(-5*n/2)*7^(7*n/2)/sqrt(2*Pi*n). - Ilya Gutkovskiy, Jul 31 2016
From Peter Bala, Aug 22 2016: (Start)
a(n) = Sum_{k = 0..floor(n/2)} binomial(12*n,n - 2*k) * binomial(5*n + k - 1,k).
O.g.f.: A(x) = Hypergeom([13/14, 11/14, 9/14, 5/14, 3/14, 1/14], [9/10, 7/10, 3/10, 1/2, 1/10], (2^2*7^7/5^5)*x^2) + 12*x*Hypergeom([10/7, 9/7, 8/7, 6/7, 5/7, 4/7], [7/5, 6/5, 4/5, 3/2, 3/5], (2^2*7^7/5^5)*x^2).
The o.g.f. is the diagonal of the bivariate rational function 1/(1 - t*(1 + x)^7/(1 - x)^5) and hence is algebraic by Stanley 1999, Theorem 6.33, p. 197. (End)
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] 1/((1-x)^(n+1) * (1-2*x)^(5*n)).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(7*n,k) * binomial(2*n-k,n-k).
a(n) = Sum_{k=0..n} 2^k * binomial(5*n+k-1,k) * binomial(2*n-k,n-k).
a(n) = 4^n * binomial((7*n-1)/2,n).
a(n) = [x^n] 1/(1-4*x)^((5*n+1)/2).
a(n) = [x^n] (1+4*x)^((7*n-1)/2). (End)

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

Original entry on oeis.org

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

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).

cp's OEIS Frontend

A276098 a(n) = (7*n)!*(3/2*n)!/((7*n/2)!*(3*n)!*(2*n)!).

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A091527 a(n) = ((3*n)!/n!^2)*(Gamma(1+n/2)/Gamma(1+3n/2)).

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A262732 a(n) = (1/n!) * (5*n)!/(5*n/2)! * (3*n/2)!/(3*n)!.

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A262733 a(n) = (1/n!) * (7*n)!/(7*n/2)! * (5*n/2)!/(5*n)!.

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

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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.

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A276098 a(n) = (7n)!(3/2n)!/((7n/2)!(3n)!(2n)!).

A091527 a(n) = ((3n)!/n!^2)(Gamma(1+n/2)/Gamma(1+3n/2)).

A262732 a(n) = (1/n!) * (5n)!/(5n/2)! * (3n/2)!/(3n)!.

A262733 a(n) = (1/n!) * (7n)!/(7n/2)! * (5n/2)!/(5n)!.

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