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Number of cycles with 2 alternating runs in all permutations of [n] (it is assumed that the smallest element of the cycle is in the first position).
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%I #15 Jul 26 2022 13:19:01

%S 0,0,0,1,7,42,267,1900,15263,137494,1375195,15127656,181532895,

%T 2359929682,33039019643,495585302836,7929364861759,134799202682670,

%U 2426385648353595,46101327318849376,922026546377249663,19362557473922767210,425976264426301927195

%N Number of cycles with 2 alternating runs in all permutations of [n] (it is assumed that the smallest element of the cycle is in the first position).

%C a(n) = Sum_{k>=0} k*A187244(n,k).

%H Alois P. Heinz, <a href="/A187246/b187246.txt">Table of n, a(n) for n = 0..450</a>

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

%F a(n) ~ (5/4-exp(1)+exp(2)/4) * n! = 0.378982196273617... * n!. - _Vaclav Kotesovec_, Mar 15 2014

%F D-finite with recurrence a(n) +(-n-3)*a(n-1) +(3*n-1)*a(n-2) +2*(-n+2)*a(n-3)=0. - _R. J. Mathar_, Jul 26 2022

%e a(4)=7 because each of the following permutations of {1,2,3,4} has 1 cycle with 2 alternating runs: (132)(4), (142)(3), (143)(2), (1)(243), (1243), (1342), and (1432); the remaining 17 permutations have none.

%p g := (1/4*(3+2*z+exp(2*z)-4*exp(z)))/(1-z): gser := series(g, z = 0, 25): seq(factorial(n)*coeff(gser, z, n), n = 0 .. 22);

%p # second Maple program:

%p b:= proc(n) option remember; expand(

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

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

%p end:

%p a:= n-> (p-> add(coeff(p, x, i)*i, i=0..degree(p)))(b(n)):

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

%t CoefficientList[Series[(3+2*x+E^(2*x)-4*E^(x))/(4*(1-x)), {x, 0, 20}], x]* Range[0, 20]! (* _Vaclav Kotesovec_, Mar 15 2014 *)

%Y Cf. A187244.

%K nonn

%O 0,5

%A _Emeric Deutsch_, Mar 07 2011