A082368 -id:A082368 - cp's OEIS frontend

A008977 a(n) = (4*n)!/(n!)^4.

1, 24, 2520, 369600, 63063000, 11732745024, 2308743493056, 472518347558400, 99561092450391000, 21452752266265320000, 4705360871073570227520, 1047071828879079131681280, 235809301462142612780721600, 53644737765488792839237440000, 12309355935372581458927646400000

Offset: 0

N. J. A. Sloane

Number of paths of length 4*n in an n X n X n X n grid from (0,0,0,0) to (n,n,n,n).
a(n) occurs in Ramanujan's formula 1/Pi = (sqrt(8)/9801) * Sum_{n>=0} (4*n)!/(n!)^4 * (1103 + 26390*n)/396^(4*n). - Susanne Wienand, Jan 05 2013
a(n) is the number of ballot results that lead to a 4-way tie when 4*n voters each cast three votes for three out of four candidates vying for 3 slots on a county commission; each of these ballot results give 3*n votes to each of the four candidates. - Dennis P. Walsh, May 02 2013
a(n) is the constant term of (X + Y + Z + 1/(X*Y*Z))^(4*n). - Mark van Hoeij, May 07 2013
In Narumiya and Shiga on page 158 the g.f. is given as a hypergeometric function. - Michael Somos, Aug 12 2014
Diagonal of the rational function R(x,y,z,w) = 1/(1-(w+x+y+z)). - Gheorghe Coserea, Jul 15 2016

			a(13)=52!/(13!)^4=53644737765488792839237440000 is the number of ways of dealing the four hands in Bridge or Whist. - _Henry Bottomley_, Oct 06 2000
a(1)=24 since, in a 4-voter 3-vote election that ends in a four-way tie for candidates A, B, C, and D, there are 4! ways to arrange the needed vote sets {A,B,C}, {A,B,D}, {A,C,D}, and {B,C,D} among the 4 voters. - _Dennis P. Walsh_, May 02 2013
G.f. = 1 + 24*x + 2520*x^2 + 369600*x^3 + 63063000*x^4 + 11732745024*x^5 + ...

T. D. Noe, Table of n, a(n) for n=0..100
R. M. Dickau, Paths through a 4-D lattice.
S. Hassani, J.-M. Maillard, and N. Zenine, On the diagonals of rational functions: the minimal number of variables (unabridged version), arXiv:2502.05543 [math-ph], 2025. See pp. 41.
Timothy Huber, Daniel Schultz, and Dongxi Ye, Ramanujan-Sato series for 1/pi, Acta Arith. (2023) Vol. 207, 121-160. See p. 11.
Markus Kuba and Alois Panholzer, Lattice paths and the diagonal of the cube, arXiv:2411.03930 [math.CO], 2024.
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
Michaël Moortgat, The Tamari order for D^3 and derivability in semi-associative Lambek-Grishin Calculus, 15th Workshop: Computational Logic and Applications (CLA 2020).
N. Narumiya and H. Shiga, The mirror map for a family of K3 surfaces induced from the simplest 3-dimensional reflexive polytope, Proceedings on Moonshine and related topics (Montréal, QC, 1999), 139-161, CRM Proc. Lecture Notes, 30, Amer. Math. Soc., Providence, RI, 2001. MR1877764 (2002m:14030).

Cf. A000984, A006480, A008978, A178529, A000897, A002894, A002897, A006480, A008979, A186420, A188662.
Row 4 of A187783.
Related to diagonal of rational functions: A268545-A268555.

[Factorial(4*n)/Factorial(n)^4: n in [0..20]]; // Vincenzo Librandi, Aug 13 2014

Maple
```
A008977 := n->(4*n)!/(n!)^4;
```

Table[(4n)!/(n!)^4,{n,0,16}] (* Harvey P. Dale, Oct 24 2011 *)
a[ n_] := If[ n < 0, 0, (4 n)! / n!^4]; (* Michael Somos, Aug 12 2014 *)
a[ n_] := SeriesCoefficient[ HypergeometricPFQ[ {1/4, 2/4, 3/4}, {1, 1}, 256 x], {x, 0, n}]; (* Michael Somos, Aug 12 2014 *)

A008977(n):=(4*n)!/(n!)^4$ makelist(A008977(n),n,0,20); /* Martin Ettl, Nov 15 2012 */

a(n) = (4*n)!/n!^4; \\ Gheorghe Coserea, Jul 15 2016

from math import factorial
def A008977(n): return factorial(n<<2)//factorial(n)**4 # Chai Wah Wu, Mar 15 2023

a(n) = A139541(n)*(A001316(n)/A049606(n))^3. - Reinhard Zumkeller, Apr 28 2008
Self-convolution of A178529, where A178529(n) = (4^n/n!^2) * Product_{k=0..n-1} (8*k + 1)*(8*k + 3).
G.f.: hypergeom([1/8, 3/8], [1], 256*x)^2. - Mark van Hoeij, Nov 16 2011
a(n) ~ 2^(8*n - 1/2) / (Pi*n)^(3/2). - Vaclav Kotesovec, Mar 07 2014
G.f.: hypergeom([1/4, 2/4, 3/4], [1, 1], 256*x). - Michael Somos, Aug 12 2014
From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(2*n,n)*binomial(3*n,n)*binomial(4*n,n) = ( [x^n](1 + x)^(2*n) ) * ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) = [x^n](F(x)^(24*n)), where F(x) = 1 + x + 29*x^2 + 2246*x^3 + 239500*x^4 + 30318701*x^5 + 4271201506*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008978, A008979, A186420 and A188662. (End)
0 = (x^2-256*x^3)*y''' + (3*x-1152*x^2)*y'' + (1-816*x)*y' - 24*y, where y is the g.f. - Gheorghe Coserea, Jul 15 2016
From Peter Bala, Jul 17 2016: (Start)
a(n) = Sum_{k = 0..3*n} (-1)^(n+k)*binomial(4*n,n + k)* binomial(n + k,k)^4.
a(n) = Sum_{k = 0..4*n} (-1)^k*binomial(4*n,k)*binomial(n + k,k)^4. (End)
E.g.f.: 3F3(1/4,1/2,3/4; 1,1,1; 256*x). - Ilya Gutkovskiy, Jan 23 2018
From Peter Bala, Feb 16 2020: (Start)
a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k - apply Mestrovic, equation 39, p. 12.
a(n) = [(x*y*z)^n] (1 + x + y + z)^(4*n). (End)
D-finite with recurrence n^3*a(n) -8*(4*n-3)*(2*n-1)*(4*n-1)*a(n-1)=0. - R. J. Mathar, Aug 01 2022
a(n) = 24*A082368(n). - R. J. Mathar, Jun 21 2023

A060540 Square array read by antidiagonals downwards: T(n,k) = (nk)!/(k!^nn!), (n>=1, k>=1), the number of ways of dividing nk labeled items into n unlabeled boxes with k items in each box.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 10, 15, 1, 1, 35, 280, 105, 1, 1, 126, 5775, 15400, 945, 1, 1, 462, 126126, 2627625, 1401400, 10395, 1, 1, 1716, 2858856, 488864376, 2546168625, 190590400, 135135, 1, 1, 6435, 66512160, 96197645544, 5194672859376, 4509264634875, 36212176000, 2027025, 1

Offset: 1

Henry Bottomley, Apr 02 2001

The Copeland link gives the associations of this entry with the operator calculus of Appell Sheffer polynomials, the combinatorics of simple set partitions encoded in the Faa di Bruno formula for composition of analytic functions (formal Taylor series), the Pascal matrix, and the geometry of the n-dimensional simplices (hypertriangles, or hypertetrahedra). These, in turn, are related to simple instances of the application of the exponential formula / principle / schema giving the number of not-necessarily-connected objects composed from an ensemble of connected objects. - Tom Copeland, Jun 09 2021

			Array begins:
  1,   1,       1,          1,             1,                 1, ...
  1,   3,      10,         35,           126,               462, ...
  1,  15,     280,       5775,        126126,           2858856, ...
  1, 105,   15400,    2627625,     488864376,       96197645544, ...
  1, 945, 1401400, 2546168625, 5194672859376, 11423951396577720, ...
  ...

Seiichi Manyama, Antidiagonals n = 1..50, flattened (first 20 antidiagonals from Harry J. Smith)
Tom Copeland, Calculus, Combinatorics, and Geometry Underlying OEIS A060540, and the Exponential Formula, 2021.
Nattawut Phetmak and Jittat Fakcharoenphol, Uniformly Generating Derangements with Fixed Number of Cycles in Polynomial Time, Thai J. Math. (2023) Vol. 21, No. 4, 899-915. See pp. 901, 914.
Elena L. Wang and Guoce Xin, On Ward Numbers and Increasing Schröder Trees, arXiv:2507.15654 [math.CO], 2025. See p. 13.

Rows include A000012, A001700, A060542, A082368, A322252.
Columns k=1..10 give A000012, A001147, A025035, A025036, A025037, A025038, A025040, A025041, A025042.
Main diagonal is A057599.
Related to A057599, see also A096126 and A246048.
Cf. A060358, A361948 (includes row/col 0).
Cf. A000217, A000292, A000332, A000389, A000579, A000580, A007318, A036040, A099174, A133314, A132440, A135278 (associations in Copeland link).

T[n_, k_] := (n*k)!/(k!^n*n!);
Table[T[n-k+1, k], {n, 1, 10}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Jun 29 2018 *)

{ i=0; for (m=1, 20, for (n=1, m, k=m - n + 1; write("b060540.txt", i++, " ", (n*k)!/(k!^n*n!))); ) } \\ Harry J. Smith, Jul 06 2009

T(n,k) = (n*k)!/(k!^n*n!) = T(n-1,k)*A060543(n,k) = A060538(n,k)/k!.

T(n,k) = Product_{j=2..n} binomial(j*k-1,k-1). - M. F. Hasler, Aug 22 2014

Definition reworded by M. F. Hasler, Aug 23 2014

A322252 a(0) = 1 and a(n) = (5n)!/(5!n!^5) for n > 0.

Original entry on oeis.org

1, 1, 945, 1401400, 2546168625, 5194672859376, 11423951396577720, 26478825654361766400, 63805953776276649848625, 158421985022100255941485000, 402789797982510165934296910320, 1044048983553856888083223814102400, 2749848597736878877579660426025283000

Offset: 0

Seiichi Manyama, Nov 30 2018

nonn

Row 5 of A060540.

Cf. A000142, A100734, A008978, A060542, A082368.

[1] cat [Factorial(5*n)/(120*Factorial(n)^5):n in [1..12]]; // Marius A. Burtea, Feb 18 2020

a[n_]:=(5*n)!/(5!*n!^5); Array[a, 20] (* or *) CoefficientList[Series[HypergeometricPFQ[{1/5, 2/5, 3/5, 4/5}, {1, 1, 1}, 3125 x]/(120 x) , {x, 0, 20}], x] (* Stefano Spezia, Dec 01 2018 *)

O.g.f.: F({1/5, 2/5, 3/5, 4/5}, {1, 1, 1}, 3125*x)/(120*x), where F is the generalized hypergeometric function. - Stefano Spezia, Dec 01 2018

a(n) = (1/5!)*A008978(n) for n >= 1. - Peter Bala, Feb 18 2020

A370294 G.f.: exp(Sum_{k>=1} (4k)!/(4!k!^4) * x^k/k).

Original entry on oeis.org

1, 1, 53, 5186, 663444, 98703235, 16179000550, 2837251240021, 522937525075783, 100134345595461824, 19762585810520535829, 3997199042964419204924, 825055790810846248226675, 173231819660726985218760834, 36906136513918240767383588700, 7962139696794640558535530147729

Offset: 0

Vaclav Kotesovec, Feb 14 2024

nonn

Cf. A008977, A082368, A229452, A370295, A333042.

CoefficientList[Series[Exp[Sum[(4*k)!/(4!*k!^4)*x^k/k, {k, 1, 20}]], {x, 0, 20}], x]
CoefficientList[Series[Exp[x*HypergeometricPFQ[{1, 1, 5/4, 3/2, 7/4}, {2, 2, 2, 2}, 256*x]], {x, 0, 20}], x]

G.f. A(x) = G(x)^(1/24), where G(x) is the g.f. for A333042.

a(n) ~ c * 4^(4*n)/n^(5/2), where c = exp(HypergeometricPFQ[{1, 1, 5/4, 3/2, 7/4}, {2, 2, 2, 2}, 1] / 256) / (24*sqrt(2)*Pi^(3/2)) = 0.005320414767134132512371690902604699480645296829596277834542636529157577...

A361948 Array read by ascending antidiagonals. A(n, k) = Product_{j=0..k-1} binomial((j + 1)*n - 1, n - 1) if n >= 1, and A(0, k) = 1 for all k.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 10, 15, 1, 1, 1, 1, 35, 280, 105, 1, 1, 1, 1, 126, 5775, 15400, 945, 1, 1, 1, 1, 462, 126126, 2627625, 1401400, 10395, 1, 1, 1, 1, 1716, 2858856, 488864376, 2546168625, 190590400, 135135, 1, 1

Offset: 0

Peter Luschny, Apr 13 2023

Row n gives the leading coefficients of the set partition polynomials of type n. The sequence of these polynomial sequences starts: A097805, A048993, A156289, A291451, A291452, ...

			Array A(n, k) starts:
  [0] 1, 1,   1,       1,           1,                 1, ...
  [1] 1, 1,   1,       1,           1,                 1, ...
  [2] 1, 1,   3,      15,         105,               945, ...  A001147
  [3] 1, 1,  10,     280,       15400,           1401400, ...  A025035
  [4] 1, 1,  35,    5775,     2627625,        2546168625, ...  A025036
  [5] 1, 1, 126,  126126,   488864376,     5194672859376, ...  A025037
  [6] 1, 1, 462, 2858856, 96197645544, 11423951396577720, ...  A025038
.
Triangle A(n-k, k) starts:
  [0] 1;
  [1] 1, 1;
  [2] 1, 1,  1;
  [3] 1, 1,  1,   1;
  [4] 1, 1,  3,   1,   1;
  [5] 1, 1, 10,  15,   1, 1;
  [6] 1, 1, 35, 280, 105, 1, 1;

Cyril Banderier, Philippe Marchal, and Michael Wallner, Rectangular Young tableaux with local decreases and the density method for uniform random generation (short version), arXiv:1805.09017 [cs.DM], 2018.
Antoine Chambert-Loir, Combinatorics of partitions, blog post 2024.
Alexander Karpov, Generalized knockout tournament seedings, International Journal of Computer Science in Sport, vol. 17(2), 2018.

Cf. A060540 (subarray), A370407 (antidiagonal sums, row sums).
Cf. A001147 (row 2), A025035 (row 3), A025036 (row 4), A025037 (row 5), A025038 (row 6), A025039 (row 7), A025040 (row 8), A025041 (row 9).
Cf. A088218 (column 2), A060542 (column 3), A082368 (column 4), A322252 (column 5), A057599 (main diagonal).
Cf. A097805, A048993, A156289, A291451, A291452.

A := (n, k) -> mul(binomial((j + 1)*n - 1, n - 1), j = 0..k-1):
seq(seq(A(n-k, k), k = 0..n), n = 0..9);
# Alternative, using recursion:
A := proc(n, k) local P; P := proc(n, k) option remember;
if n = 0 then return x^k*k! fi; if k = 0 then 1 else add(binomial(n*k, n*j)*
P(n,k-j)*x, j=1..k) fi end: coeff(P(n, k), x, k) / k! end:
seq(print(seq(A(n, k), k = 0..5)), n = 0..6);
# Alternative, using exponential generating function:
egf := n -> ifelse(n=0, 1, exp(x^n/n!)): ser := n -> series(egf(n), x, 8*n):
row := n -> local k; seq((n*k)!*coeff(ser(n), x, n*k), k = 0..6):
for n from 0 to 6 do [n], row(n) od;  # Peter Luschny, Aug 15 2024

A[n_, k_] := Product[Binomial[n (j + 1) - 1, n - 1], {j, 0, k - 1}]; Table[A[n - k, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Apr 13 2023 *)

def Arow(n, size):
    if n == 0: return [1] * size
    return [prod(binomial((j + 1)*n - 1, n - 1) for j in range(k)) for k in range(size)]
for n in range(7): print(Arow(n, 7))
# Alternative, using exponential generating function:
def SetPolyLeadCoeff(m, n):
    x, z = var("x, z")
    if m == 0: return 1
    w = exp(2 * pi * I / m)
    o = sum(exp(z * w ** k) for k in range(m)) / m
    t = exp(x * (o - 1)).taylor(z, 0, m*n)
    p = factorial(m*n) * t.coefficient(z, m*n)
    return p.leading_coefficient(x)
for m in range(7):
    print([SetPolyLeadCoeff(m, k) for k in range(6)])

A(n, k) = (1/k!) * [x^k] P(n, k), where P(n, k) = k!*x^k if n = 0 and otherwise 1 if k = 0 and otherwise Sum_{j=1..k} binomial(n*k, n*j)*P(n, k-j)*x.

A(n, k) = (n*k)!*[x^(n*k)] exp(x^n/n!) for n >= 1. - Peter Luschny, Aug 15 2024

A368650 a(n) = (6n + 1)!(9n + 1)!/((2n)!(3n)!((5n + 1)!)^2).

Original entry on oeis.org

1, 2940, 27511848, 324265486545, 4234842288963000, 58626067532977225512, 842744763083824037236800, 12437726604034570811549435040, 187171833825593326056635733697560, 2859197188199406875783449346275416000, 44198453917285616202092687086145825181264, 689863061309915307698539343386922516078167200

Offset: 0

Karol A. Penson, Jan 02 2024

nonn

a(n) can be rigorously proven to be an integer for n>=0.

Cf. A304126, A368545, A082368, A113424.

seq((6*n + 1)!*(9*n + 1)!/((2*n)!*(3*n)!*((5*n + 1)!)^2), n=0..12);

G.f.: hypergeometric10F9([2/9, 1/3, 4/9, 5/9, 2/3, 7/9, 5/6, 8/9, 10/9, 7/6], [2/5, 2/5, 3/5, 3/5, 4/5, 4/5, 1, 6/5, 6/5], (167365651248*z)/9765625).
O.g.f.: hypergeometric10F10([2/9, 1/3, 4/9, 5/9, 2/3, 7/9, 5/6, 8/9, 10/9, 7/6], [2/5, 2/5, 3/5, 3/5, 4/5, 4/5, 1, 1, 6/5, 6/5], (167365651248*z)/9765625).
a(n) = Integral_{x=0..167365651248/9765625} x^n*W(x) dx, n>=0, where W(x) = (78125*MeijerG([[], [-3/5, -3/5, -2/5, -2/5, -1/5, -1/5, 0, 0, 1/5, 1/5]], [[1/6, 1/9, -1/9, -1/6, -2/9, -1/3, -4/9, -5/9, -2/3, -7/9], []], (9765625*x)/167365651248))/(2066242608*Pi). MeijerG is the Meijer G - function. W(x) can be represented as a sum of 10 hypergeometric functions of type 10F9. W(x) can be proven to be a positive function in the interval [0, 167365651248/9765625]. W(x) is singular at x=0 and monotonically decreases to zero at x = 167365651248/9765625. This integral representation as the n-th power moment of the positive function  W(x) in the interval [0, 167365651248/9765625] is unique, as W(x) is the solution of the Hausdorff moment problem.

A377218 Expansion of the o.g.f. A(x) defined by [x^n] A(x)^(24n) = (4n)!/n!^4 for n >= 0.

Original entry on oeis.org

1, 1, 29, 2246, 239500, 30318701, 4271201506, 647359627557, 103476937050223, 17223017775652625, 2959285397777331751, 521687007046376376544, 93932798602803741121051, 17215649571517858590782737, 3203146941738318544432065500, 603763082812549420389330837978, 115095760617137117019641563685386

Offset: 0

Peter Bala, Oct 20 2024

Compare with A000984(n) = [x^n] (1 + x)^(2*n) = (2*n)!/n!^2.
The central binomial coefficients A000984(n) satisfy the supercongruences u(n*p^k) == u(n*p^(k-1)) (mod p^(3*k)) for all primes p >= 5 and positive integers n and k.
More generally, for positive integers r and s, the sequence {u(r,s; n) : n >= 0} defined by u(r,s; n) = [x^(s*n)] (1 + x)^(r*n) = binomial(r*n, s*n) satisfies the same supercongruences (Meštrović, Section 6, equation 39).
Conjecture: for positive integers r and s, the sequence {v(r,s; n) : n >= 0} defined by v(r,s; n) = [x^(s*n)] A(x)^(r*n) also satisfies the same supercongruences.

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

Cf. A008977, A082368, A333042, A370294, A377217, A377219.

Order := 25:
E(x) := exp(add((4*n)!/n!^4 * x^n/n, n = 1..25)):
solve(series(x*E(x),x) = y, x):
convert(%, polynom):
g := taylor(y/%, y = 0, 25):
seq(coeftayl(g^(1/24), y = 0,  n), n = 0..20);

O.g.f.: A(x) = ( x/(x * series_reversion(E(x)))^(1/24), where E(x) = exp(Sum_{n >= 1} (4*n)!/n!^4 *x^n/n) is the o.g.f. of A333042.

A181386 Tetrahedron of terms C(r,n,m) representing the number of ways of choosing m disjoint subsets of r members from an original set of n members.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 1, 1, 4, 6, 4, 1, 1, 3, 1, 1, 1, 1, 5, 10, 10, 5, 1, 1, 6, 3, 1, 1, 1, 1, 1, 1, 6, 15, 20, 15, 6, 1, 1, 10, 15, 1, 4, 1, 1, 1, 1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 15, 45, 15, 1, 10, 1, 1, 1, 1, 1, 1, 1, 8, 28, 56, 70, 56, 28, 8, 1, 1, 21, 105, 105, 1, 20

Offset: 1

Frank M Jackson, Oct 16 2010

The start index for r is 1 but the start index for m and n is 0. For each value of r, the triangle T_r(n,m) has row n containing 1 + floor(n/r) terms.
From Frank M Jackson, Nov 20 2010: (Start)
C(r,mr,m) = C(r,mr-1,m-1).
C(1,m,m) = A000012, C(2,2m,m) = A001147,
C(3,3m,m), ..., C(10,10m,m) = A025035, ..., A025042.
C(2,26,10) = 150738274937250 and represents the number of possible plugboard settings for a WWII German Enigma Enciphering Machine.
C(r,2r,2) = A001700, C(r,3r,3) = A060542, C(r,4r,4) = A082368.
C(r,n,m) = C(r,mr-1,m-1)*binomial(n,rm),
  and applied recursively gives the identity
C(r,n,m) = Binomial(n,r*m) * Product_{p=1..m} Binomial(r*(m-p+1)-1,r-1).
(End)
C(2,26,10) = A266365(10), where 26 is the size of the alphabet. - Jonathan Sondow, Dec 29 2015

			r=1, C(1,n,m) is
  1
  1, 1
  1, 2,  1
  1, 3,  3,  1
  1, 4,  6,  4, 1
  1, 5, 10, 10, 5, 1
r=2, C(2,n,m) is
  1
  1
  1,  1
  1,  3
  1,  6,  3
  1, 10, 15
r=3, C(3,n,m) is
  1
  1
  1
  1,  1
  1,  4
  1, 10

C(1,n,m) = T_1(n,m) = A007318, C(2,n,m) = T_2(n,m) = A100861, and C(2,26,m) = A266365.

Flatten[Table[{n!/((n-r*m)!*m!*r!^m)}, {r, 1, 50}, {n, 0, 50}, {m, 0, Floor[n/r]}]]

C(r,n,m) = n!/((n-r*m)!*m!*(r!)^m).

A368692 a(n) = (12n + 6)!(6n + 9)!/(108(4n + 2)!(2n + 3)!((6*n + 5)!)^2).

Original entry on oeis.org

14, 563108, 54231252075, 6700034035890000, 928978310614152999200, 137569863175651804211692560, 21253098849879053645154605945160, 3381375421559384124434964404229384000, 549714622911935710495977183989400234273000

Offset: 0

Karol A. Penson, Jan 03 2024

nonn

According to A. Adolphson and S. Sperber, "On the integrality of hypergeometric series whose coefficients are factorial ratios", ArXiv: 2001.03296, s.page 14, first equation after Eq.(7.4): for any two integers K, L, the ratios (3*K)!*(3*L)!/(K!*L!*((K+L)!)^2) are proven to be integers. 108*a(n) results from K = 4*n+2 and L = 2*n+3, n>=0. It is conjectured here that a(n) are integers.

A. Adolphson and S. Sperber, On the integrality of hypergeometric series whose coefficients are factorial ratios, Acta Arithmetica 200 (2021), no.1, 39-59.

Cf. A368650, A304126, A368545, A082368, A113424, A368875.

seq((12*n + 6)!*(6*n + 9)!/(108*(4*n + 2)!*(2*n + 3)!*((6*n + 5)!)^2),n=0..9);

G.f.:  14*hypergeometric8F7([7/12, 2/3, 5/6, 11/12, 13/12, 17/12, 13/6, 7/3], [1, 7/6, 4/3, 3/2, 3/2, 5/3, 11/6], 186624*z).
E.g.f.: 14*hypergeometric8F8([7/12, 2/3, 5/6, 11/12, 13/12, 17/12, 13/6, 7/3], [1, 1, 7/6, 4/3, 3/2, 3/2, 5/3, 11/6], 186624*z).
a(n) = Integral_{x=0..186624} x^n*W(x) dx, n>=0, where W(x) = (1/(20736*Pi))*MeijerG([[], [0, 0, 1/6, 1/3, 1/2, 1/2, 2/3, 5/6]], [[-5/12, -1/3, -1/6, -1/12, 1/12, 5/12, 7/6, 4/3], []], x/186624). MeijerG is the Meijer G - function.  W(x) can be represented as an expression containing the sum of 4 generalized hypergeometric functions of type 8F7. W(x) is a positive function in the interval [0, 186624], is singular at x=0 and monotonically decreases to zero at x = 186624. This integral representation as the n-th power moment of the positive function  W(x) in the interval [0, 186624] is unique, as W(x) is the solution of the Hausdorff moment problem.
Let b(n) = Gamma(7+ 12*n)/(6*Gamma(2 + 2*n)*Gamma(3 + 4*n)*Gamma(6 + 6*n)), then a(n) = b(n) * A272399(n+2). - Peter Luschny, Jan 06 2024

A368875 a(n) = 24(3n + 1)!/(n!*((n + 2)!)^2).

Original entry on oeis.org

6, 16, 105, 1008, 12012, 164736, 2494206, 40646320, 701149020, 12655450080, 237026033790, 4577828250240, 90739095674400, 1838979005667840, 37993593597567210, 798259862714284080, 17022152442879594780, 367791659430639444000, 8040845154302354844450

Offset: 0

Karol A. Penson, Jan 08 2024

nonn

According to A. Adolphson and S. Sperber (see Links), see page 14, second equation after Eq.(7.4): for any two integers K, L, the ratios (3*K+1)!*(3*L+1)!/(K!*L!*((K+L+1)!)^2) are proven to be integers. Here a(n) results from K = 1 and L = n, n >= 0.

A. Adolphson and S. Sperber, On the integrality of hypergeometric series whose coefficients are factorial ratios, arXiv:2001.03296 [math.NT], 2020.
A. Adolphson and S. Sperber, On the integrality of hypergeometric series whose coefficients are factorial ratios, Acta Arithmetica 200 (2021), no.1, 39-59.

Cf. A368650, A304126, A368545, A082368, A113424, A368692.

seq(24*(3*n + 1)!/(n!*((n + 2)!)^2),n=0..17);

Table[24*(3*n + 1)!/(n!*((n + 2)!)^2),{n,0,16}] (* James C. McMahon, Jan 08 2024 *)

def a(n): return (24 * (n + 1) * (n + 2) * gamma(3*n + 2)) / gamma(n + 3)^3
print([a(n) for n in range(19)])  # Peter Luschny, Jan 09 2024

G.f.:   6*hypergeometric3F2([2/3, 1, 4/3], [3, 3], 27*z).
G.f.:   -(hypergeometric2F1([-4/3, -2/3], [1], 27*z) - 1)/(3*z^2) + 8/z.
E.g.f.: 6*hypergeometric3F3([2/3, 1, 4/3], [3, 3, 1], 27*z).
a(n) = Integral_{x=0..27} x^n*W(x) dx, n >= 0, where
  W(x) = (243*2^(2/3)*Gamma(5/6)*Gamma(2/3)*hypergeometric2F1([-4/3, -4/3], [1/3], x/27)) / (16*Pi^(5/2)*x^(1/3)) - (3*sqrt(3)*2^(1/3)*x^(1/3)* hypergeometric2F1([-2/3, -2/3], [5/3], x/27))/(2*sqrt(Pi)*Gamma(5/6)* Gamma(2/3)).
W(x) is a positive function in the interval [0, 27], is singular at x = 0 with the singularity x^(-1/3), and monotonically decreases to zero at x = 27, with W'(x) tending to zero at x = 27. This integral representation as the n-th power moment of the positive function W(x) in the interval [0, 27] is unique, as W(x) is the solution of the Hausdorff moment problem.

cp's OEIS Frontend

A008977 a(n) = (4*n)!/(n!)^4.

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A060540 Square array read by antidiagonals downwards: T(n,k) = (n*k)!/(k!^n*n!), (n>=1, k>=1), the number of ways of dividing nk labeled items into n unlabeled boxes with k items in each box.

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A322252 a(0) = 1 and a(n) = (5*n)!/(5!*n!^5) for n > 0.

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A370294 G.f.: exp(Sum_{k>=1} (4*k)!/(4!*k!^4) * x^k/k).

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A361948 Array read by ascending antidiagonals. A(n, k) = Product_{j=0..k-1} binomial((j + 1)*n - 1, n - 1) if n >= 1, and A(0, k) = 1 for all k.

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A368650 a(n) = (6*n + 1)!*(9*n + 1)!/((2*n)!*(3*n)!*((5*n + 1)!)^2).

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A377218 Expansion of the o.g.f. A(x) defined by [x^n] A(x)^(24*n) = (4*n)!/n!^4 for n >= 0.

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A181386 Tetrahedron of terms C(r,n,m) representing the number of ways of choosing m disjoint subsets of r members from an original set of n members.

A060540 Square array read by antidiagonals downwards: T(n,k) = (nk)!/(k!^nn!), (n>=1, k>=1), the number of ways of dividing nk labeled items into n unlabeled boxes with k items in each box.

A322252 a(0) = 1 and a(n) = (5n)!/(5!n!^5) for n > 0.

A370294 G.f.: exp(Sum_{k>=1} (4k)!/(4!k!^4) * x^k/k).

A368650 a(n) = (6n + 1)!(9n + 1)!/((2n)!(3n)!((5n + 1)!)^2).

A377218 Expansion of the o.g.f. A(x) defined by [x^n] A(x)^(24n) = (4n)!/n!^4 for n >= 0.

A368692 a(n) = (12n + 6)!(6n + 9)!/(108(4n + 2)!(2n + 3)!((6*n + 5)!)^2).

A368875 a(n) = 24(3n + 1)!/(n!*((n + 2)!)^2).