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

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

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• nice

Comments

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

Examples

			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 + ...
		

Links

• 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).

Crossrefs

Cf. A000984, A006480, A008978, A178529, A000897, A002894, A002897, A006480, A008979, A186420, A188662.

Row 4 of A187783.

Related to diagonal of rational functions: A268545-A268555.

Programs

Magma

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

Maple

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

Mathematica

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 *)

Maxima

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

PARI

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

Python

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

Formula

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