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A186368
Triangle read by rows: T(n,k) is the number of cycle-up-down permutations of {1,2,...,n} having k excedances (0<=k<=floor(n/2)).
1
1, 1, 1, 1, 1, 4, 1, 10, 5, 1, 20, 40, 1, 35, 175, 61, 1, 56, 560, 768, 1, 84, 1470, 4996, 1385, 1, 120, 3360, 22720, 24320, 1, 165, 6930, 81730, 214445, 50521, 1, 220, 13200, 248512, 1288320, 1152512, 1, 286, 23595, 665236, 5986695, 12989678, 2702765, 1, 364, 40040, 1610752, 23063040, 98169344, 76477440
OFFSET
0,6
COMMENTS
A permutation is said to be cycle-up-down if it is a product of up-down cycles. A cycle (b(1), b(2), ...) is said to be up-down if, when written with its smallest element in the first position, it satisfies b(1) < b(2) > b(3) < ... .
Row n has 1+floor(n/2) entries.
Sum of entries in row n is A000111(n+1) (the Euler or up-down numbers).
LINKS
E. Deutsch and S. Elizalde, Cycle up-down permutations, arXiv:0909.5199 [math.CO], 2009.
FORMULA
T(n,1) = (1/6)n(n-1)(n+1) = A000292(n-1).
T(2n,n) = A000111(2n) (the Euler numbers).
Sum(k*T(n,k),k>=0) = A186369(n).
E.g.f.: G(t,z)=[sec(z*sqrt(t)) + tan(z*sqrt(t))]^{1/sqrt(t)}/cos(z*sqrt(t)).
EXAMPLE
T(n,0)=1, the identity permutation.
T(4,2)=5 because we have (12)(34), (13)(24), (1324), (1423), and (14)(23).
Triangle starts:
1;
1;
1,1;
1,4;
1,10,5;
1,20,40;
1,35,175,61;
MAPLE
G := (sec(z*sqrt(t))+tan(z*sqrt(t)))^(1/sqrt(t))/cos(z*sqrt(t)): Gser := simplify(series(G, z = 0, 16)): for n from 0 to 13 do P[n] := sort(expand(factorial(n)*coeff(Gser, z, n))) end do: for n from 0 to 13 do seq(coeff(P[n], t, k), k = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form
MATHEMATICA
T[n_, k_] := n! SeriesCoefficient[(Sec[z*Sqrt[t]] + Tan[z*Sqrt[t]])^(1/ Sqrt[t])/Cos[z*Sqrt[t]], {z, 0, n}, {t, 0, k}]; Table[T[n, k], {n, 0, 13}, {k, 0, n/2}] // Flatten (* Jean-François Alcover, Oct 30 2017 *)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, Feb 28 2011
STATUS
approved