A187249 Number of cycles with at most 2 alternating runs in all permutations of [n] (it is assumed that the smallest element of the cycle is in the first position).
0, 1, 3, 11, 48, 248, 1504, 10560, 84544, 761024, 7610496, 83715968, 1004592640, 13059706368, 182835893248, 2742538406912, 43880614526976, 745970446991360, 13427468045910016, 255121892872421376, 5102437857448689664, 107151195006423007232, 2357326290141307207680
Offset: 0
Keywords
Examples
a(3)=11 because in (1)(2)(3), (1)(23), (12)(3), (13)(2), (123), and (132) all cycles have at most 2 alternating runs.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..450
Crossrefs
Cf. A187247.
Programs
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Maple
g := (1/4*(exp(2*z)-1+2*z))/(1-z): gser := series(g, z = 0, 25): seq(factorial(n)*coeff(gser, z, n), n = 0 .. 22); # second Maple program: b:= proc(n) option remember; expand( `if`(n=0, 1, add(b(n-j)*binomial(n-1, j-1)* `if`(j=1, x, (j-1)!+2^(j-2)*(x-1)), j=1..n))) end: a:= n-> (p-> add(coeff(p, x, i)*i, i=0..n))(b(n)): seq(a(n), n=0..30); # Alois P. Heinz, Apr 15 2017 -
Mathematica
CoefficientList[Series[(E^(2*x)-1+2*x)/(4*(1-x)), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Mar 15 2014 *)
Formula
E.g.f.: g(z) = (1/4)[exp(2z) - 1 +2z]/(1-z).
a(n) ~ (exp(2)+1)/4 * n! = 2.09726402473266... * n!. - Vaclav Kotesovec, Mar 15 2014
D-finite with recurrence a(n) +(-n-2)*a(n-1) +2*(n-1)*a(n-2)=0. - R. J. Mathar, Jul 26 2022
Comments