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 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). 6
 1, 1, 2, 6, 22, 94, 460, 2532, 15420, 102620, 739512, 5729192, 47429896, 417429800, 3888426512, 38192416048, 394239339792, 4264424937488, 48212317486112, 568395755184224, 6973300915138656, 88860103591344864, 1174131206436335296, 16061756166912244800 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 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. LINKS Alois P. Heinz, Table of n, a(n) for n = 0..528 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 EXAMPLE a(4)=22 because only the permutations 3421=(1324) and 4312=(1423) have cycles with more than 2 alternating runs. 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 *) PROG (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 CROSSREFS Cf. A001519, A187245, A187248, A187250, A216964. Row sums of A344855. Sequence in context: A030453 A001861 A049526 * A193763 A301385 A093793 Adjacent sequences:  A187248 A187249 A187250 * A187252 A187253 A187254 KEYWORD nonn AUTHOR Emeric Deutsch, Mar 08 2011 STATUS approved

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Last modified November 28 06:42 EST 2021. Contains 349401 sequences. (Running on oeis4.)
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