A187245 Number of permutations of [n] having no cycle with 2 alternating runs (it is assumed that the smallest element of the cycle is in the first position).

1, 1, 2, 5, 17, 78, 463, 3315, 27164, 247975, 2492539, 27422698, 328607417, 4266367567, 59686293284, 895068242601, 14320843215019, 243467476610732, 4382635181281015, 83272415871044649, 1665465961530365026, 34974843092354081119, 769445564105823722109

Offset: 0

Author

Emeric Deutsch, Mar 07 2011

Keywords

• nonn

Comments

a(n) = A187244(n,0).

Examples

			a(3)=5 because we have among the 6 permutations of {1,2,3} only 312=(132) has a cycle with 2 alternating runs.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..450

Crossrefs

Cf. A187244.

Programs

Maple

g := exp((4*exp(z)-exp(2*z)-3-2*z)*1/4)/(1-z): gser := series(g, z = 0, 25): seq(factorial(n)*coeff(gser, z, n), n = 0 .. 22);
# second Maple program:
a:= proc(n) option remember;
      `if`(n=0, 1, add(a(n-j)*binomial(n-1, j-1)*
      `if`(j=1, 1, (j-1)!-(2^(j-2)-1)), j=1..n))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Apr 15 2017

Mathematica

CoefficientList[Series[E^((4*E^x-E^(2*x)-3-2*x)/4)/(1-x), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Mar 15 2014 *)

Formula

E.g.f.: g(z)=exp[(4exp(z)-exp(2z)-3-2z)/4]/(1-z).

a(n) ~ exp(exp(1)-exp(2)/4-5/4) * n! = 0.68455780023755436... * n!. - Vaclav Kotesovec, Mar 15 2014