A178575 Number of permutations of {1,2,...,3n} whose cycle lengths are all divisible by 3.

1, 2, 160, 62720, 68992000, 163235072000, 710399033344000, 5129081020743680000, 57096929922918645760000, 927825111247427993600000000, 21095031729321522862489600000000, 648714415740095471067280179200000000, 26246985260844262759382156050432000000000

Offset: 0

Author

Geoffrey Critzer, Dec 23 2010

Keywords

• nonn

Examples

			a(1) = 2 because we have (123) and (132).
		

References

• Herbert S. Wilf, Generatingfunctiontology, page 209

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..150
• H. Crane and P. McCullagh, Reversible Markov structures on divisible set partitions, Journal of Applied Probability, 52(3), 2015.

Crossrefs

Cf. A001818, A258878.

Programs

Maple

a:= n-> factorial(3*n)*(mul(1+3*i, i = 1 .. n-1))/(factorial(n)*3^n): seq(a(n), n = 0 .. 11);

Mathematica

Table[(-1)^(n/3) Binomial[-1/3,n/3]n!,{n,0,30,3}]

PARI

v=Vec(serlaplace(1/(1-x^3+O(x^50))^(1/3))); vector(#v\3,n,v[3*n-2])

Formula

a(n) = (-1)^(n/3)*binomial(-1/3,n/3)*n!.

E.g.f.: 1/(1-x^3)^(1/3).

a(n) = ((3*n)!/(n!*3^n))*Product_{i=1..n-1} (1+3*i) (from the Wilf reference).

a(n) ~ (3*n)! / (Gamma(1/3) * n^(2/3)). - Vaclav Kotesovec, Jun 15 2015

D-finite with recurrence: a(n) = (3*n-1)*(3*n-2)^2*a(n-1), a(0)=1. - Georg Fischer, Jul 02 2021 (from the 3rd formula)