A060706 For n >= 1 a(n) is the size of the conjugacy class in the symmetric group S_(4n) consisting of permutations whose cycle decomposition is a product of n disjoint 4-cycles.

1, 6, 1260, 1247400, 3405402000, 19799007228000, 210384250804728000, 3692243601622976400000, 99579809935771673508000000, 3910499136177753618659160000000, 214428309633170941925556379440000000

Offset: 0

Author

Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 21 2001

Keywords

• nonn

Comments

a(n) is the number of ways to seat 4n bridge players at n circular tables with four players at each table. - Geoffrey Critzer, Dec 17 2011

Links

• Harry J. Smith, Table of n, a(n) for n = 0..100
• Bishal Deb and Alan D. Sokal, Higher-order Stirling cycle and subset triangles: Total positivity, continued fractions and real-rootedness, arXiv:2507.18959 [math.CO], 2025. See p. 7.

Crossrefs

Cf. A000142. A001147, A052502, A052504.

Programs

Maple

for n from 0 to 20 do printf(`%d,`,(4*n)! / (n! * 4^n)) od:

Mathematica

nn = 40; a = x^4/4;f[list_] := Select[list, # > 0 &];
f[Range[0, nn]! CoefficientList[Series[Exp[a], {x, 0, nn}], x]]  (* Geoffrey Critzer, Dec 17 2011 *)

PARI

{ for (n=0, 100, write("b060706.txt", n, " ", (4*n)! / (n! * 4^n)); ) } \\ Harry J. Smith, Jul 09 2009

Formula

a(n) = (4n)! / (n! * 4^n). Recursion: a(0) = 1, a(1) = 6, for n >= 2 a(n) = a(n-1) * C(4n - 1, 3)* 6 = a(n-1)*(4n-1)*(4n-2)*(4n-3). Using Stirling's formula in A000142 we have a(n) ~ 2 * 64^n * (n/e)^(3n).

E.g.f.: exp(x^4/4). - Geoffrey Critzer, Dec 17 2011

Write the generating function for this sequence in the form A(x) = sum_{n>=0} a(n)* x^(3*n+1)/(3*n+1)!. Then A'(x)*( 1 - A(x)^3) = 1, consequently A(x) is a root of z^4 - 4*z + 4*x with A(0) = 0. Cf. A052502. - Peter Bala, Jan 02 2015

Extensions

More terms from James Sellers, Apr 23 2001