A187251 Number of permutations of [n] having no cycle with 3 or more alternating runs (it is assumed that the smallest element of a cycle is in the first position).

1, 1, 2, 6, 22, 94, 460, 2532, 15420, 102620, 739512, 5729192, 47429896, 417429800, 3888426512, 38192416048, 394239339792, 4264424937488, 48212317486112, 568395755184224, 6973300915138656, 88860103591344864, 1174131206436335296, 16061756166912244800

Offset: 0

Author

Emeric Deutsch, Mar 08 2011

Keywords

• nonn

Comments

a(n) = A187250(n,0).

It appears that a(n) = A216964(n,1), for n>0. - Michel Marcus, May 17 2013.

The above comment is correct. Let b(n) be the n-th element of the first column of the triangle in A216964. By definition, b(n) is the number of permutations of [n] with no cyclic valleys. Recall that alternating runs of permutations are monotonically increasing or decreasing subsequences. In other words, b(n) is the number of permutations of [n] with the restriction that every cycle has at most two alternating runs, so b(n) = A187251(n) = a(n). - Shi-Mei Ma, May 18 2013.

Examples

			a(4)=22 because only the permutations 3421=(1324) and 4312=(1423) have cycles with more than 2 alternating runs.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..528
• Veronica Bitonti, Bishal Deb, and Alan D. Sokal, Thron-type continued fractions (T-fractions) for some classes of increasing trees, arXiv:2412.10214 [math.CO], 2024. See p. 56.

Crossrefs

Cf. A001519, A187245, A187248, A187250, A216964.

Row sums of A344855.

Programs

Maple

g := exp((2*z-1+exp(2*z))*1/4): gser := series(g, z = 0, 28): seq(factorial(n)*coeff(gser, z, n), n = 0 .. 23);
# second Maple program:
a:= proc(n) option remember; `if`(n=0, 1, add(
      a(n-j)*binomial(n-1, j-1)*ceil(2^(j-2)), j=1..n))
    end:
seq(a(n), n=0..23);  # Alois P. Heinz, May 30 2021

Mathematica

nmax = 20; CoefficientList[Series[E^((2*x-1+E^(2*x))/4), {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Apr 17 2020 *)

Maxima

a(n):=n!*sum(2^(n-2*k)*sum(binomial(k,j)*stirling2(n-k+j,j)*j!/(n-k+j)!,j,0,k)/k!,k,1,n); /* Vladimir Kruchinin, Apr 25 2011 */

PARI

x='x+O('x^66); Vec(serlaplace(exp( (2*x-1+exp(2*x))/4 ))) /* Joerg Arndt, Apr 26 2011 */

PARI

lista(m) = {P = x*y; for (n=1, m, M = subst(P, x, 1); M = subst(M, y, 1); print1(polcoeff(M, 0, q), ", "); P = (n*q+x*y)*P + 2*q*(1-q)*deriv(P, q)+ 2*x*(1-q)*deriv(P, x)+ (1-2*y+q*y)*deriv(P, y); ); } \\ (adapted from PARI prog in A216964) \\ Michel Marcus, May 17 2013

Formula

E.g.f.: exp( (2*z-1+exp(2*z))/4 ).

For n>=1: a(n)=n!*sum(k=1..n, 2^(n-2*k)*sum(j=0..k, binomial(k,j)*stirling2(n-k+j,j)*j!/(n-k+j)!)/k!); [From Vladimir Kruchinin, Apr 25 2011]

G.f.: 1/Q(0) where Q(k) = 1 - x*k - x/(1 - x*(k+1)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 07 2013

G.f.: 1/Q(0) where Q(k) = 1 - x*(2*k+1) - m*x^2*(k+1)/Q(k+1) and m=1 (continued fraction); setting m=2 gives A004211, m=4 gives A124311 without signs. - Sergei N. Gladkovskii, Sep 26 2013

G.f.: T(0)/(1-x), where T(k) = 1 - x^2*(k+1)/( x^2*(k+1) - (1-x-2*x*k)*(1-3*x-2*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 25 2013

Sum_{k=0..n} binomial(n,k) * a(k) * a(n-k) = A007405(n). - Vaclav Kotesovec, Apr 17 2020

a(n) = Sum_{j=1..n} a(n-j)*binomial(n-1,j-1)*ceiling(2^(j-2)) for n > 0, a(0) = 1. - Alois P. Heinz, May 30 2021