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

%I A082368 #31 Feb 14 2024 10:33:05
%S A082368 1,105,15400,2627625,488864376,96197645544,19688264481600,
%T A082368 4148378852099625,893864677761055000,196056702961398759480,
%U A082368 43627992869961630486720,9825387560922608865863400,2235197406895366368301560000,512889830640524227455318600000
%N A082368 a(n) = (4*n-1)! / (n! * n! * n! * (n-1)! * 3!).
%C A082368 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.)
%C A082368 Number of ways of dividing 4n labeled items into 4 unlabeled boxes with n items in each box. - _Dan Parrish_, Apr 09 2015
%H A082368 Vincenzo Librandi, <a href="/A082368/b082368.txt">Table of n, a(n) for n = 1..300</a>
%F A082368 a(n) = binomial(4*n,n)*binomial(3*n,n)*binomial(2*n,n)/24. - _Zerinvary Lajos_, Jun 25 2006
%F A082368 a(n) = (4n)!/(4!*n!^4). - _Dan Parrish_, Apr 09 2015
%F A082368 From _Robert Israel_, Apr 09 2015: (Start)
%F A082368 a(n) = Gamma(2*n+1/2)*Gamma(n+1/2)*64^n/(24*Pi*(n!)^3).
%F A082368 a(n+1) = 8*(2*n+1)*(4*n+1)*(4*n+3)*a(n)/(n+1)^3.
%F A082368 G.f.: g(x) = x*hypergeom([1,5/4,3/2,7/4],[2,2,2],256*x) satisfies
%F A082368 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)
%F A082368 From _Karol A. Penson_, Dec 31 2023: (Start)
%F A082368 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)
%F A082368 a(n) = A008977(n)/24. - _Vaclav Kotesovec_, Feb 14 2024
%e A082368 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.
%p A082368 [seq(binomial(4*n,n)*binomial(3*n,n)*binomial(2*n,n)/24,n=1..17)]; # _Zerinvary Lajos_, Jun 25 2006
%t A082368 Table[(4 n)! / (4! n!^4), {n, 30}] (* _Vincenzo Librandi_, Jun 16 2017 *)
%o A082368 (PARI) a(n)=(4*n)!/(4!*n!^4) \\ _Charles R Greathouse IV_, Apr 09 2015
%o A082368 (Magma) [Factorial(4*n-1) / (Factorial(n)*Factorial(n)* Factorial(n)*Factorial(n-1)*6): n in [1..15]]; // _Vincenzo Librandi_, Jun 16 2017
%Y A082368 Row 4 of A060540.
%Y A082368 Cf. A000984, A005809, A005810, A008977.
%K A082368 easy,nonn
%O A082368 1,2
%A A082368 John A. Trono (jtrono(AT)smcvt.edu), May 10 2003
%E A082368 More terms from _Zerinvary Lajos_, Jun 25 2006