

A186358


Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k updown cycles (0<=k<=n). A cycle (b(1), b(2), ...) is said to be updown if, when written with its smallest element in the first position, it satisfies b(1)<b(2)>b(3)<... .


4



1, 0, 1, 0, 1, 1, 1, 1, 3, 1, 4, 6, 7, 6, 1, 19, 35, 30, 25, 10, 1, 114, 210, 190, 125, 65, 15, 1, 799, 1468, 1351, 840, 420, 140, 21, 1, 6392, 11760, 10820, 6692, 3185, 1176, 266, 28, 1, 57527, 105905, 97458, 60058, 28098, 10479, 2856, 462, 36, 1, 575270, 1059306, 975140, 599640, 278500, 103593, 30345, 6210, 750, 45, 1
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OFFSET

0,9


COMMENTS

Sum of entries in row n is n!.
Sum_{k=0..n} k * T(n,k) = A186360(n).


LINKS



FORMULA

E.g.f.: (1sin z)^(1t)/(1z).
The trivariate e.g.f. H(t,s,z) of the permutations of {1,2,...,n} with respect to size (marked by z), number of updown cycles (marked by t), and number of cycles that are not updown (marked by s) is given by H(t,s,z) = (1sin z)^(st)/(1z)^s.


EXAMPLE

T(3,0)=1 because we have (123).
T(4,2)=7 because we have (1)(243), (142)(3), (132)(4), (13)(24), (12)(34), (143)(2), and (14)(23).
Triangle starts:
1;
0,1;
0,1,1;
1,1,3,1;
4,6,7,6,1;
19,35,30,25,10,1;


MAPLE

G := (1sin(z))^(1t)/(1z): Gser := simplify(series(G, z = 0, 16)): for n from 0 to 10 do P[n] := sort(expand(factorial(n)*coeff(Gser, z, n))) end do: for n from 0 to 10 do seq(coeff(P[n], t, j), j = 0 .. n) end do; # yields sequence in triangular form


MATHEMATICA

m = maxExponent = 11;
(CoefficientList[# + O[t]^m, t] Range[0, m1]!&) /@ CoefficientList[(1  Sin[z])^{1t}/(1z) + O[z]^m, z] // Flatten (* JeanFrançois Alcover, Aug 07 2018 *)


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AUTHOR



STATUS

approved



