A001448 -id:A001448 - cp's OEIS frontend

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

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.

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 = 2, b = 1. - 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..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)

A211421 Integral factorial ratio sequence: a(n) = (8n)!(3n)!/((6n)!(4n)!*n!).

Original entry on oeis.org

1, 14, 390, 12236, 404550, 13777764, 478273692, 16825310040, 597752648262, 21397472070260, 770557136489140, 27884297395587240, 1013127645555452700, 36935287875280348776, 1350441573221798941560, 49498889739033621986736, 1818284097150186829038150

Offset: 0

Peter Bala, Apr 10 2012

This sequence is the particular case a = 4, b = 3 of the following result (see Bober, Theorem 1.2): let a, b be nonnegative integers with a > b and GCD(a,b) = 1. Then (2*a*n)!*(b*n)!/((a*n)!*(2*b*n)!*((a-b)*n)!) is an integer for all integer n >= 0. Other cases include A061162 (a = 3, b = 1), A211419 (a = 3, b = 2) and A211420 (a = 4, b = 1).

Sequence terms are given by the coefficient of x^n in the expansion of ( (1 + x)^(k+2)/(1 - x)^k )^n when k = 6. See the cross references for related sequences obtained from other values of k. - Peter Bala, Sep 29 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..600
Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion
J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, 2007, arXiv:0709.1977v1 [math.NT], 2007; J. London Math. Soc., Vol. 79, Issue 2 (2009), 422-444.
F. Rodriguez-Villegas, Integral ratios of factorials and algebraic hypergeometric functions, arXiv:math/0701362 [math.NT], 2007.

Cf. A061162, A061163, A211419, A211420.

Cf. A000984 (k = 0), A091527 (k = 1), A001448 (k = 2), A262732 (k = 3), A211419 (k = 4), A262733 (k = 5), A262740.

[Factorial(8*n)*Factorial(3*n)/(Factorial(6*n)*Factorial(4*n)*Factorial(n)): n in [0..20]]; // Vincenzo Librandi, Aug 01 2016

#A211421
a := n -> (8*n)!*(3*n)!/((6*n)!*(4*n)!*n!);
seq(a(n), n = 0..16);

Table[(8 n)!*(3 n)!/((6 n)!*(4 n)!*n!), {n, 0, 15}] (* Michael De Vlieger, Oct 04 2015 *)

a(n) = (8*n)!*(3*n)!/((6*n)!*(4*n)!*n!);
vector(30, n, a(n-1)) \\ Altug Alkan, Oct 02 2015

The o.g.f. sum {n >= 1} a(n)*z^n is algebraic over the field of rational functions Q(z) (see Rodriguez-Villegas).
From Peter Bala, Sep 29 2015: (Start)
a(n) = Sum_{i = 0..n} binomial(8*n,i) * binomial(7*n-i-1,n-i).
a(n) = [x^n] ( (1 + x)^8/(1 - x)^6 )^n.
a(0) = 1 and a(n) = 2*(8*n - 1)*(8*n - 3)*(8*n - 5)*(8*n - 7)/( n*(6*n - 1)*(6*n - 3)*(6*n - 5) ) * a(n-1) for n >= 1.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 14*x + 293*x^2 + 7266*x^3 + 197962*x^4 + 5726364*x^5 + ... has integer coefficients and equals 1/x * series reversion of x*(1 - x)^6/(1 + x)^8. See A262740. (End)
a(n) ~ 2^(10*n)*27^(-n)/sqrt(2*Pi*n). - Ilya Gutkovskiy, Jul 31 2016
a(n) = (2^n/n!)*Product_{k = 3*n..4*n-1} (2*k + 1). - Peter Bala, Feb 26 2023
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] 1/((1-x)^(n+1) * (1-2*x)^(6*n)).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(8*n,k) * binomial(2*n-k,n-k).
a(n) = Sum_{k=0..n} 2^k * binomial(6*n+k-1,k) * binomial(2*n-k,n-k).
a(n) = 4^n * binomial((8*n-1)/2,n).
a(n) = [x^n] 1/(1-4*x)^((6*n+1)/2).
a(n) = [x^n] (1+4*x)^((8*n-1)/2). (End)

A241477 Triangle read by rows, number of orbitals classified with respect to the first zero crossing, n>=1, 1<=k<=n.

Original entry on oeis.org

1, 0, 2, 2, 2, 2, 0, 4, 0, 2, 6, 12, 4, 2, 6, 0, 12, 0, 4, 0, 4, 20, 60, 12, 12, 12, 4, 20, 0, 40, 0, 12, 0, 8, 0, 10, 70, 280, 40, 60, 36, 24, 40, 10, 70, 0, 140, 0, 40, 0, 24, 0, 20, 0, 28, 252, 1260, 140, 280, 120, 120, 120, 60, 140, 28, 252, 0, 504, 0

Offset: 1

Peter Luschny, Apr 23 2014

For the combinatorial definitions see A232500. An orbital w over n sectors has its first zero crossing at k if k is the smallest j such that the partial sum(1<=i<=j, w(i))) = 0, where w(i) are the jumps of the orbital represented by -1, 0, 1.

			[1], [ 1]
[2], [ 0,  2]
[3], [ 2,  2,  2]
[4], [ 0,  4,  0,  2]
[5], [ 6, 12,  4,  2,  6]
[6], [ 0, 12,  0,  4,  0, 4]
[7], [20, 60, 12, 12, 12, 4, 20]

Row sums: A056040.

Cf. A232500.

A241477 := proc(n, k)
  if n = 0 then 1
elif k = 0 then 0
elif irem(n, 2) = 0 and irem(k, 2) = 1 then 0
elif k = 1 then (n-1)!/iquo(n-1,2)!^2
else 2*(n-k)!*(k-2)!/iquo(k,2)/(iquo(k-2,2)!*iquo(n-k,2)!)^2
  fi end:
for n from 1 to 9 do seq(A241477(n, k), k=1..n) od;

T[n_, k_] := Which[n == 0, 1, k == 0, 0, Mod[n, 2] == 0 && Mod[k, 2] == 1,  0, k == 1, (n-1)!/Quotient[n-1, 2]!^2, True, 2*(n-k)!*(k-2)!/Quotient[k, 2]/(Quotient[k-2, 2]!*Quotient[n-k, 2]!)^2];
Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 20 2018, from Maple *)

def A241477_row(n):
    if n == 0: return [1]
    Z = [0]*n; T = [0] if is_odd(n) else []
    for i in (1..n//2): T.append(-1); T.append(1)
    for p in Permutations(T):
        i = 0; s = p[0]
        while s != 0: i += 1; s += p[i];
        Z[i] += 1
    return Z
for n in (1..9): A241477_row(n)

If n is even and k is odd then T(n, k) = 0 else if k = 1 then T(n, 1) = A056040(n-1) else T(n, k) = 2*A057977(k-2)*A056040(n-k).
T(n, n) = A241543(n).
T(n+1, 1) = A126869(n).
T(2*n, 2*n) = |A002420(n)|.
T(2*n+1, 1) = A000984(n).
T(2*n+1, n+1) = A241530(n).
T(2*n+2, 2) = A028329(n).
T(4*n, 2*n) = |A010370(n)|.
T(4*n, 4*n) = |A024491(n)|.
T(4*n+1, 1) = A001448(n).
T(4*n+1, 2*n+1) = A002894(n).

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)

A023834 Sum of exponents in prime-power factorization of C(4n,2n).

Original entry on oeis.org

0, 2, 3, 5, 6, 6, 7, 11, 9, 11, 12, 12, 13, 15, 14, 15, 15, 16, 15, 18, 17, 20, 22, 22, 22, 23, 22, 23, 22, 23, 25, 28, 25, 25, 29, 28, 28, 32, 30, 32, 31, 30, 31, 34, 34, 34, 34, 36, 35, 37, 34, 36, 38, 38, 38, 40, 38, 41, 42, 42, 40, 44, 46, 44, 43, 43, 44, 46, 42, 46, 47, 47, 46, 48, 47, 50, 51, 50

Offset: 0

Clark Kimberling

nonn

Ivan Neretin, Table of n, a(n) for n = 0..10000

Cf. A001222, A001448, A023816, A022559.

Join[{0}, Table[Total[FactorInteger[Binomial[4 n, 2 n]][[All, 2]]], {n, 77}]] (* Ivan Neretin, Oct 26 2017 *)

a(n) = bigomega(binomial(4*n, 2*n)); \\ Amiram Eldar, Jun 11 2025

From Amiram Eldar, Jun 11 2025: (Start)
a(n) = A001222(A001448(n)).
a(n) = A023816(2*n).
a(n) = A022559(4*n) - 2*A022559(2*n). (End)

Offset changed to 0 and a(0) prepended by Amiram Eldar, Jun 11 2025

A214282 Largest Euler characteristic of a downset on an n-dimensional cube.

Original entry on oeis.org

1, 1, 1, 3, 6, 10, 15, 35, 70, 126, 210, 462, 924, 1716, 3003, 6435, 12870, 24310, 43758, 92378, 184756, 352716, 646646, 1352078, 2704156, 5200300, 9657700, 20058300, 40116600, 77558760, 145422675, 300540195, 601080390, 1166803110, 2203961430, 4537567650, 9075135300, 17672631900

Offset: 1

Terence Tao, Jul 09 2012

nonn

An m-downset is a set of subsets of 1..m such that if S is in the set, so are all subsets of S. The Euler characteristic of a downset is the number of sets in the downset with an even cardinality, minus the number with an odd cardinality.

			G.f. = x + x^2 + x^3 + 3*x^4 + 6*x^5 + 10*x^6 + 15*x^7 + 35*x^8 + ...

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Terry Tao, Optimal bounds for an alternating sum on a downset, 2012.

Cf. A214283.

a214282 n = a007318 (n - 1) (a004524 (n - 1))
-- Reinhard Zumkeller, Jul 14 2012

Table[{Binomial[n - 1, n/2], Binomial[n, n/2], Binomial[n + 1, n/2 + 1], Binomial[n + 2, n/2 + 2]}, {n, 0, 28, 4}] (* Alonso del Arte, Jul 09 2012 *)

a(n)=binomial(n-1,if(n%2,(n+1)\4*2,n/2)) \\ Charles R Greathouse IV, Jul 09 2012

{a(n) = if( n<1, 0, vecmax( Vec((1 - x)^(n-1))))}; /* Michael Somos, Apr 21 2014 */

from math import comb
def A214282(n): return comb(n-1, (n+1>>1)&(-1^(n&1))) # Chai Wah Wu, Jan 31 2024

a(n) = binomial(n - 1, n/2) when n is even, a(n) = binomial(n - 1, (n + 1)/2) when n is 3 mod 4, and a(n) = binomial(n - 1, (n - 1)/2) when n is 1 mod 4.
a(2n) = A001700(n-1). a(4n+1) = A001448(n). a(4n+3) = A186231(n).
a(n) = A214283(n) + A001405(n). - Reinhard Zumkeller, Jul 14 2012
a(n) = A007318(n-1, A004524(n-1)). - Reinhard Zumkeller, Jul 14 2012
a(n+1) = A000108([n/2])*A215495(n). - M. F. Hasler, Aug 25 2012
A214282(n) - A214283(n) is A056040(n) if n is even and A056040(n)/((n+1)/2) otherwise. - Peter Luschny, Jul 08 2016

A061163 a(n) = (10n)!n!/((5n)!(4n)!*(2n)!).

Original entry on oeis.org

1, 630, 1385670, 3528923580, 9540949030470, 26651569523959380, 75998432812419471900, 219813190240007470094520, 642409325786050322446410310, 1892390644737640220059489996260

Offset: 0

Richard Stanley, Apr 17 2001

According to page 781 of the cited reference the generating function F(x) for a(n) is algebraic but not obviously so and the minimal polynomial satisfied by F(x) is quite large.
This sequence is the particular case a = 5, b = 1 of the following result (see Bober, Theorem 1.2): let a, b be nonnegative integers with a > b and GCD(a,b) = 1. Then (2*a*n)!*(b*n)!/((a*n)!*(2*b*n)!*((a-b)*n)!) is an integer for all integer n >= 0. Other cases include A061162 (a = 3, b = 1), A211419 (a = 3, b = 2) and A211420(a = 4, b = 1) and A211421 (a = 4, b = 3). The o.g.f. Sum_{n >= 1} a(n)*z^n is algebraic over the field of rational functions Q(z) (see Rodriguez-Villegas). - Peter Bala, Apr 10 2012
Continuing the comment above: This is case n = 4 of the array of sequences
  A(n, k) = 4^(n*k)*(Gamma((n + 1)*k + 1/2)/Gamma(k + 1/2)) / Gamma(n * k + 1). See the cross-references for other cases. - Peter Luschny, Feb 21 2024

M. Kontsevich and D. Zagier, Periods, in Mathematics Unlimited - 2001 and Beyond, Springer, Berlin, 2001, pp. 771-808.

J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, 2007, arXiv:0709.1977 [math.NT], 2007; Jour. of the London Math. Soc., Vol. 79, Issue 2, 422-444.
M. Kontsevich and D. Zagier, Periods, Institut des Hautes Etudes Scientifiques 2001 IHES/M/01/22, p. 11.
F. Rodriguez-Villegas, Integral ratios of factorials and algebraic hypergeometric functions, arXiv:math/0701362 [math.NT], 2007.

Cf. A061164, A211419, A211421.

Cf. A000012 (n=0), A001448 (n=1), A061162 (n=2), A211420 (n=3), this sequence (n=4).

A061163 := n->(10*n)!*n!/((5*n)!*(4*n)!*(2*n)!);
# Alternative:
A := (n, k) -> 4^(n*k)*(GAMMA((n + 1)*k + 1/2)/GAMMA(k + 1/2))/GAMMA(n*k + 1):
seq(A(4, k), k = 0..9);  # Peter Luschny, Feb 21 2024

Table[(10n)! n!/((5n)!(4n)!(2n)!),{n,0,10}] (* Harvey P. Dale, Oct 24 2022 *)

n*(4*n-3)*(2*n-1)*(4*n-1)*a(n) -10*(10*n-9)*(10*n-7)*(10*n-3)*(10*n-1)*a(n-1)=0. - R. J. Mathar, Oct 26 2014
O.g.f. is a generalized hypergeometric function 4F3([1/10, 3/10, 7/10, 9/10], [1/4, 1/2, 3/4], 5^5*z). - Karol A. Penson, Apr 13 2022
From Karol A. Penson, Feb 21 2024: (Start)
(O.g.f.(z))^2 satisfies the algebraic equation of order 15, in which the powers of (O.g.f.(z))^2 are multiplied by polynomials p(n, z) with integer coefficients, in the form: Sum_{n = 0..15} p(n, z)*(O.g.f.(z))^(2*n) = 0.
Here is the list of orders, in the variable z, of all polynomials p(n, z) for n=0..15: 9,9,9,9,9,10,10,10,10,10,10,11,11,11,11,11,12. For example p(15, z) = 2^50*(5^5*z-1)^12. (End)
a(n) ~ 5^(5*n) / (2^(3/2) * sqrt(Pi*n)). - Vaclav Kotesovec, Aug 27 2024

A337389 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt((1+(k-4)x+sqrt(1-2(k+4)x+((k-4)x)^2)) / (2 * (1-2(k+4)x+((k-4)*x)^2))).

Original entry on oeis.org

1, 1, 2, 1, 3, 6, 1, 4, 19, 20, 1, 5, 34, 141, 70, 1, 6, 51, 328, 1107, 252, 1, 7, 70, 587, 3334, 8953, 924, 1, 8, 91, 924, 7123, 34904, 73789, 3432, 1, 9, 114, 1345, 12870, 89055, 372436, 616227, 12870, 1, 10, 139, 1856, 20995, 184756, 1135005, 4027216, 5196627, 48620

Offset: 0

Seiichi Manyama, Aug 25 2020

			Square array begins:
    1,    1,     1,     1,      1,      1, ...
    2,    3,     4,     5,      6,      7, ...
    6,   19,    34,    51,     70,     91, ...
   20,  141,   328,   587,    924,   1345, ...
   70, 1107,  3334,  7123,  12870,  20995, ...
  252, 8953, 34904, 89055, 184756, 337877, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=0..5 give A000984, A082758, A337390, A245926, A001448, A243946.
Main diagonal gives A337388.
Cf. A337369, A337419.

T[n_, k_] := Sum[If[k == 0, Boole[n == j], k^(n - j)] * Binomial[2*j, j] * Binomial[2*n, 2*j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, Aug 25 2020 *)

{T(n, k) = sum(j=0, n, k^(n-j)*binomial(2*j, j)*binomial(2*n, 2*j))}

T(n,k) = Sum_{j=0..n} k^(n-j) * binomial(2*j,j) * binomial(2*n,2*j).
T(0,k) = 1, T(1,k) = k+2 and n * (2*n-1) * (4*n-5) * T(n,k) = (4*n-3) * (4*(k+4)*n^2-6*(k+4)*n+k+6) * T(n-1,k) - (k-4)^2 * (n-1) * (2*n-3) * (4*n-1) * T(n-2,k) for n > 1. - Seiichi Manyama, Aug 28 2020
For fixed k > 0, T(n,k) ~ (2 + sqrt(k))^(2*n + 1/2) / sqrt(8*Pi*n). - Vaclav Kotesovec, Aug 31 2020
Conjecture: the k-th column entries, k >= 0, are given by [x^n] ( (1 + (k-2)*x + x^2)*(1 + x)^2/(1 - x)^2 )^n. This is true for k = 0 and k = 4. - Peter Bala, May 03 2022

A370101 a(n) = Sum_{k=0..n} binomial(4n,k) binomial(4*n-k-1,n-k).

Original entry on oeis.org

1, 7, 97, 1519, 25089, 427007, 7408897, 130287871, 2313945089, 41409732607, 745530884097, 13488086405119, 245014271688705, 4465915098890239, 81637668328243201, 1496095489290731519, 27477504726883368961, 505627095685486608383, 9320167322334416338945

Offset: 0

Seiichi Manyama, Feb 10 2024

nonn

Cf. A001448, A370100, A370102.
Cf. A119259, A370097.
Cf. A365846.

Table[Sum[Binomial[4n,k]Binomial[4n-k-1,n-k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Dec 09 2024 *)
Table[Sum[2^k*(-1)^(n-k)*Binomial[4*n, k], {k, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Jul 31 2025 *)

a(n) = sum(k=0, n, binomial(4*n, k)*binomial(4*n-k-1, n-k));

a(n) = [x^n] ( (1+x)^4/(1-x)^3 )^n.
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x*(1-x)^3/(1+x)^4 ). See A365846.
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(4*n,k). - Seiichi Manyama, Jul 31 2025
a(n) ~ 2^(9*n + 3/2) / (7 * sqrt(Pi*n) * 3^(3*n - 1/2)). - Vaclav Kotesovec, Jul 31 2025
a(n) = Sum_{k=0..n} 2^k * binomial(3*n+k-1,k). - Seiichi Manyama, Aug 01 2025
a(n) = [x^n] 1/((1-x) * (1-2*x)^(3*n)). - Seiichi Manyama, Aug 09 2025

A370102 a(n) = Sum_{k=0..n} binomial(4n,k) binomial(5*n-k-1,n-k).

Original entry on oeis.org

1, 8, 128, 2312, 44032, 864008, 17282432, 350353928, 7172939776, 147972367880, 3070951360128, 64044689834760, 1341056098444288, 28176478479561992, 593725756425591680, 12542160174109922312, 265525958014053580800, 5632170795392966388744

Offset: 0

Seiichi Manyama, Feb 10 2024

Cf. A006318, A103885, A365847, A370098, A370099.

Cf. A001448, A370100, A370101.

seq(simplify(binomial(5*n-1, n)*hypergeom([-n, -4*n], [1 - 5*n], -1)), n = 0..20); # Peter Bala, Jul 29 2024

a(n) = sum(k=0, n, binomial(4*n, k)*binomial(5*n-k-1, n-k));

a(n) = [x^n] ( (1+x)^4/(1-x)^4 )^n.
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x*(1-x)^4/(1+x)^4 ). See A365847.
From Peter Bala, Jul 20 2024: (Start)
a(n) = binomial(5*n-1, n)*hypergeom([-n, -4*n], [1 - 5*n], -1).
For n >=1, a(n) = (4/3) * [x^n] S(x)^(3*n) = (4/5) * [x^n] (1/S(-x))^(5*n), where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the sequence of large Schröder numbers A006318.
n*(4*n - 3)*(2*n - 1)*(4*n - 1)*(85*n^4 - 510*n^3 + 1138*n^2 - 1119*n + 409)*a(n) = 2*(29665*n^8 - 237320*n^7 + 794282*n^6 - 1443212*n^5 + 1544750*n^4 - 987560*n^3 + 363568*n^2 - 69168*n + 5040)*a(n-1) + (n - 2)*(4*n - 7)*(2*n - 3)*(4*n - 5)*(85*n^4 - 170*n^3 + 118*n^2 - 33*n + 3)*a(n-2) with a(0) = 1 and a(1) = 8.
The Gauss congruences hold: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and r. (End)
a(n) ~ (349 + 85*sqrt(17))^n / (17^(1/4) * sqrt(Pi*n) * 2^(5*n - 1/2)). - Vaclav Kotesovec, Aug 08 2024
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] (1-x)^(n-1)/(1-2*x)^(4*n).
a(n) = Sum_{k=0..n} 2^k * binomial(4*n,k) * binomial(n-1,n-k).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(4*n+k-1,k) * binomial(n-1,n-k). (End)

cp's OEIS Frontend

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

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A211421 Integral factorial ratio sequence: a(n) = (8*n)!*(3*n)!/((6*n)!*(4*n)!*n!).

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A241477 Triangle read by rows, number of orbitals classified with respect to the first zero crossing, n>=1, 1<=k<=n.

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

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A023834 Sum of exponents in prime-power factorization of C(4n,2n).

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A214282 Largest Euler characteristic of a downset on an n-dimensional cube.

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

A211421 Integral factorial ratio sequence: a(n) = (8n)!(3n)!/((6n)!(4n)!*n!).

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

A061163 a(n) = (10n)!n!/((5n)!(4n)!*(2n)!).

A337389 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt((1+(k-4)x+sqrt(1-2(k+4)x+((k-4)x)^2)) / (2 * (1-2(k+4)x+((k-4)*x)^2))).

A370101 a(n) = Sum_{k=0..n} binomial(4n,k) binomial(4*n-k-1,n-k).

A370102 a(n) = Sum_{k=0..n} binomial(4n,k) binomial(5*n-k-1,n-k).