A178575 - OEIS

%I #35 Oct 10 2022 10:45:26

%S 1,2,160,62720,68992000,163235072000,710399033344000,

%T 5129081020743680000,57096929922918645760000,

%U 927825111247427993600000000,21095031729321522862489600000000,648714415740095471067280179200000000,26246985260844262759382156050432000000000

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

%D Herbert S. Wilf, Generatingfunctiontology, page 209

%H Alois P. Heinz, <a href="/A178575/b178575.txt">Table of n, a(n) for n = 0..150</a>

%H H. Crane and P. McCullagh, <a href="https://doi.org/10.1239/jap/1445543836">Reversible Markov structures on divisible set partitions</a>, Journal of Applied Probability, 52(3), 2015.

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

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

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

%F a(n) ~ (3*n)! / (Gamma(1/3) * n^(2/3)). - _Vaclav Kotesovec_, Jun 15 2015

%F 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)

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

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

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

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

%Y Cf. A001818, A258878.

%K nonn

%O 0,2

%A _Geoffrey Critzer_, Dec 23 2010