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A187246
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).
3
0, 0, 0, 1, 7, 42, 267, 1900, 15263, 137494, 1375195, 15127656, 181532895, 2359929682, 33039019643, 495585302836, 7929364861759, 134799202682670, 2426385648353595, 46101327318849376, 922026546377249663, 19362557473922767210, 425976264426301927195
OFFSET
0,5
COMMENTS
a(n) = Sum_{k>=0} k*A187244(n,k).
LINKS
FORMULA
E.g.f.: g(z)=(1/4)[3+2z+exp(2z)-4exp(z)]/(1-z).
a(n) ~ (5/4-exp(1)+exp(2)/4) * n! = 0.378982196273617... * n!. - Vaclav Kotesovec, Mar 15 2014
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
EXAMPLE
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.
MAPLE
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);
# 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, 1, (j-1)!+(2^(j-2)-1)*(x-1)), j=1..n)))
end:
a:= n-> (p-> add(coeff(p, x, i)*i, i=0..degree(p)))(b(n)):
seq(a(n), n=0..30); # Alois P. Heinz, Apr 15 2017
MATHEMATICA
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 *)
CROSSREFS
Cf. A187244.
Sequence in context: A152239 A152240 A221794 * A278152 A366222 A332082
KEYWORD
nonn
AUTHOR
Emeric Deutsch, Mar 07 2011
STATUS
approved