%I #16 Oct 30 2017 10:28:20
%S 1,1,1,1,1,4,1,10,5,1,20,40,1,35,175,61,1,56,560,768,1,84,1470,4996,
%T 1385,1,120,3360,22720,24320,1,165,6930,81730,214445,50521,1,220,
%U 13200,248512,1288320,1152512,1,286,23595,665236,5986695,12989678,2702765,1,364,40040,1610752,23063040,98169344,76477440
%N 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)).
%C 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) < ... .
%C Row n has 1+floor(n/2) entries.
%C Sum of entries in row n is A000111(n+1) (the Euler or up-down numbers).
%H E. Deutsch and S. Elizalde, <a href="https://arxiv.org/abs/0909.5199">Cycle up-down permutations</a>, arXiv:0909.5199 [math.CO], 2009.
%F T(n,1) = (1/6)n(n-1)(n+1) = A000292(n-1).
%F T(2n,n) = A000111(2n) (the Euler numbers).
%F Sum(k*T(n,k),k>=0) = A186369(n).
%F E.g.f.: G(t,z)=[sec(z*sqrt(t)) + tan(z*sqrt(t))]^{1/sqrt(t)}/cos(z*sqrt(t)).
%e T(n,0)=1, the identity permutation.
%e T(4,2)=5 because we have (12)(34), (13)(24), (1324), (1423), and (14)(23).
%e Triangle starts:
%e 1;
%e 1;
%e 1,1;
%e 1,4;
%e 1,10,5;
%e 1,20,40;
%e 1,35,175,61;
%p 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
%t 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 *)
%Y Cf. A000111, A000292, A186369.
%K nonn,tabf
%O 0,6
%A _Emeric Deutsch_, Feb 28 2011