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

%I A178575 #35 Oct 10 2022 10:45:26
%S A178575 1,2,160,62720,68992000,163235072000,710399033344000,
%T A178575 5129081020743680000,57096929922918645760000,
%U A178575 927825111247427993600000000,21095031729321522862489600000000,648714415740095471067280179200000000,26246985260844262759382156050432000000000
%N A178575 Number of permutations of {1,2,...,3n} whose cycle lengths are all divisible by 3.
%D A178575 Herbert S. Wilf, Generatingfunctiontology, page 209
%H A178575 Alois P. Heinz, <a href="/A178575/b178575.txt">Table of n, a(n) for n = 0..150</a>
%H A178575 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 A178575 a(n) = (-1)^(n/3)*binomial(-1/3,n/3)*n!.
%F A178575 E.g.f.: 1/(1-x^3)^(1/3).
%F A178575 a(n) = ((3*n)!/(n!*3^n))*Product_{i=1..n-1} (1+3*i) (from the Wilf reference).
%F A178575 a(n) ~ (3*n)! / (Gamma(1/3) * n^(2/3)). - _Vaclav Kotesovec_, Jun 15 2015
%F A178575 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 A178575 a(1) = 2 because we have (123) and (132).
%p A178575 a:= n-> factorial(3*n)*(mul(1+3*i, i = 1 .. n-1))/(factorial(n)*3^n): seq(a(n), n = 0 .. 11);
%t A178575 Table[(-1)^(n/3) Binomial[-1/3,n/3]n!,{n,0,30,3}]
%o A178575 (PARI) v=Vec(serlaplace(1/(1-x^3+O(x^50))^(1/3))); vector(#v\3,n,v[3*n-2])
%Y A178575 Cf. A001818, A258878.
%K A178575 nonn
%O A178575 0,2
%A A178575 _Geoffrey Critzer_, Dec 23 2010