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Triangle read by rows: T(n,k) is the number of cycle-up-down permutations of {1,2,...,n} having k fixed points (0 <= k <= n). 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) < ... .
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%I #11 Nov 03 2019 19:41:32

%S 1,0,1,1,0,1,1,3,0,1,5,4,6,0,1,15,25,10,10,0,1,71,90,75,20,15,0,1,341,

%T 497,315,175,35,21,0,1,1945,2728,1988,840,350,56,28,0,1,12135,17505,

%U 12276,5964,1890,630,84,36,0,1,84091,121350,87525,40920,14910,3780,1050,120,45,0,1

%N Triangle read by rows: T(n,k) is the number of cycle-up-down permutations of {1,2,...,n} having k fixed points (0 <= k <= n). 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 Sum of entries in row n is A000111(n+1) (the Euler or up-down numbers).

%C T(n,0) = A186364(n).

%C T(n,k) = T(n-k,0)*binomial(n,k).

%C Sum_{k=0..n} k*T(n,k) = A186365(n).

%H E. Deutsch and S. Elizalde, <a href="http://arxiv.org/abs/0909.5199v1"> Cycle up-down permutations</a>, arXiv:0909.5199v1 [math.CO].

%F E.g.f. = exp((x-1)z)/(1-sin z).

%F The trivariate e.g.f. H(t,s,z) of the cycle-up-down permutations of {1,2,...,n} with respect to size (marked by z), number of cycles (marked by t), and number of fixed points (marked by x) is given by H(t,x,z) = exp((x-1)*t*z)/(1-sin(z))^t.

%e T(3,1)=3 because we have (1)(23), (12)(3), and (13)(2).

%e T(4,2)=6 because we have (1)(2)(34), (1)(23)(4), (1)(24)(3), (12)(3)(4), (13)(2)(4), and (14)(2)(3).

%e Triangle starts:

%e 1;

%e 0, 1;

%e 1, 0, 1;

%e 1, 3, 0, 1;

%e 5, 4, 6, 0, 1;

%e 15, 25, 10, 10, 0, 1;

%p G := exp((x-1)*z)/(1-sin(z)): 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], x, j), j = 0 .. n) end do; # yields sequence in triangular form

%t T[n_, k_] := T[n, k] = If[k == 0, SeriesCoefficient[Exp[-x]/(1 - Sin[x]), {x, 0, n}] n!, T[n - k, 0] Binomial[n, k]];

%t Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jul 24 2018 *)

%Y Cf. A186364, A186365.

%K nonn,tabl

%O 0,8

%A _Emeric Deutsch_, Feb 28 2011