login
Number of permutations of [n] having only cycles with at least 3 alternating runs (it is assumed that the smallest element of the cycle is in the first position).
4

%I #15 Jun 16 2018 13:35:21

%S 1,0,0,0,2,16,104,688,5116,44224,438560,4851136,58603544,764606016,

%T 10715043104,160692920256,2570016145680,43678554864128,

%U 786135111482112,14936420654180864,298733557232591136,6273502048592506112,138018105454095739008,3174423293668325353216

%N Number of permutations of [n] having only cycles with at least 3 alternating runs (it is assumed that the smallest element of the cycle is in the first position).

%C a(n) = A187247(n,0).

%H Alois P. Heinz, <a href="/A187248/b187248.txt">Table of n, a(n) for n = 0..450</a>

%F E.g.f.: g(z) = exp[(1-2z-exp(2z))/4]/(1-z).

%F a(n) ~ exp((-1-exp(2))/4) * n!. - _Vaclav Kotesovec_, Mar 18 2014

%e a(4)=2 because we have (1423) and (1324).

%p g := exp((1-2*z-exp(2*z))*1/4)/(1-z): gser := series(g, z = 0, 25): seq(factorial(n)*coeff(gser, z, n), n = 0 .. 23);

%p # second Maple program:

%p a:= proc(n) option remember;

%p `if`(n=0, 1, add(a(n-j)*binomial(n-1, j-1)*

%p `if`(j=1, 0, (j-1)!-2^(j-2)), j=1..n))

%p end:

%p seq(a(n), n=0..30); # _Alois P. Heinz_, Apr 15 2017

%t a[n_] := a[n] = If[n == 0, 1, Sum[a[n-j]*Binomial[n-1, j-1]* If[j == 1, 0, (j-1)! - 2^(j-2)], {j, 1, n}]];

%t Table[a[n], {n, 0, 30}] (* _Jean-François Alcover_, Jun 16 2018, after _Alois P. Heinz_ *)

%Y Cf. A187247.

%K nonn

%O 0,5

%A _Emeric Deutsch_, Mar 07 2011