A082368 a(n) = (4*n-1)! / (n! * n! * n! * (n-1)! * 3!).

1, 105, 15400, 2627625, 488864376, 96197645544, 19688264481600, 4148378852099625, 893864677761055000, 196056702961398759480, 43627992869961630486720, 9825387560922608865863400, 2235197406895366368301560000, 512889830640524227455318600000

Offset: 1

Author

John A. Trono (jtrono(AT)smcvt.edu), May 10 2003

Keywords

• easy
• nonn

Comments

Number of combinations that are possible when placing teams ranked #1 to #4*N in a single elimination tournament where there are four columns of N teams (as in the NCAA Men's Division-1 basketball tournament that is played in March) and each column is a separate regional tournament that produces one of the four semi-finalists. (The teams in the columns appear in sorted order and the relative positions of the four columns is irrelevant.)

Number of ways of dividing 4n labeled items into 4 unlabeled boxes with n items in each box. - Dan Parrish, Apr 09 2015

Examples

			8 ranked teams (n=2) in a four region, single elimination tournament generates 105 different possible tournament orderings, where the teams in each region are ordered from best to worst. (Teams would be matched up from top to bottom and continue towards the middle two for other matchups, when more than two teams are listed in each column.) 105 tournaments is too many to list here. As this formula applies to single elimination tournaments, this enumeration formula really only makes sense when n is even.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..300

Crossrefs

Row 4 of A060540.

Cf. A000984, A005809, A005810, A008977.

Programs

Magma

[Factorial(4*n-1) / (Factorial(n)*Factorial(n)* Factorial(n)*Factorial(n-1)*6): n in [1..15]]; // Vincenzo Librandi, Jun 16 2017

Maple

[seq(binomial(4*n,n)*binomial(3*n,n)*binomial(2*n,n)/24,n=1..17)]; # Zerinvary Lajos, Jun 25 2006

Mathematica

Table[(4 n)! / (4! n!^4), {n, 30}] (* Vincenzo Librandi, Jun 16 2017 *)

PARI

a(n)=(4*n)!/(4!*n!^4) \\ Charles R Greathouse IV, Apr 09 2015

Formula

a(n) = binomial(4*n,n)*binomial(3*n,n)*binomial(2*n,n)/24. - Zerinvary Lajos, Jun 25 2006

a(n) = (4n)!/(4!*n!^4). - Dan Parrish, Apr 09 2015

From Robert Israel, Apr 09 2015: (Start)

a(n) = Gamma(2*n+1/2)*Gamma(n+1/2)*64^n/(24*Pi*(n!)^3).

a(n+1) = 8*(2*n+1)*(4*n+1)*(4*n+3)*a(n)/(n+1)^3.

G.f.: g(x) = x*hypergeom([1,5/4,3/2,7/4],[2,2,2],256*x) satisfies

x^4*(256*x-1)*g''''(x) + 5*x^3*(384*x-1)*g'''(x) + 4*x^2*(780*x-1)*g''(x) + 840*x^2*g'(x) = 0. (End)

From Karol A. Penson, Dec 31 2023: (Start)

a(n) = Integral_{x=0..256} x^n*W(x) dx, n>=0, where W(x) = x^(1/4)*hypergeometric3F2([1/4, 1/4, 1/4], [1/2, 3/4], x/256)/(96*Gamma(3/4)^4) - sqrt(x)*hypergeometric3F2([1/2, 1/2, 1/2], [3/4, 5/4], x/256)/(96*Pi^2) + Gamma(3/4)^4*x^(3/4)*hypergeometric3F2([3/4, 3/4, 3/4], [5/4, 3/2], x/256)/(768*Pi^4) is positive and unimodal on x = [0, 256]. It has a single maximum at approximately x = 31, and it goes to zero with W'(x) diverging, at both x = 0 and x = 256. This integral representation as the n-th power moment of the positive function W(x) on the interval [0, 256] is unique, as W(x) is the solution of the Hausdorff moment problem. (End)

a(n) = A008977(n)/24. - Vaclav Kotesovec, Feb 14 2024

Extensions

More terms from Zerinvary Lajos, Jun 25 2006