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%I #17 Nov 07 2020 06:00:51
%S 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,
%T 1,799,1468,1351,840,420,140,21,1,6392,11760,10820,6692,3185,1176,266,
%U 28,1,57527,105905,97458,60058,28098,10479,2856,462,36,1,575270,1059306,975140,599640,278500,103593,30345,6210,750,45,1
%N Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k up-down cycles (0<=k<=n). 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 n!.
%C T(n,0) = A186359(n).
%C Sum_{k=0..n} k * T(n,k) = A186360(n).
%H Emeric Deutsch and Sergi Elizalde, <a href="http://arxiv.org/abs/0909.5199">Cycle up-down permutations</a>, arXiv:0909.5199 [math.CO], 2009; and <a href="https://ajc.maths.uq.edu.au/pdf/50/ajc_v50_p187.pdf">also</a>, Australas. J. Combin. 50 (2011), 187-199.
%F E.g.f.: (1-sin z)^(1-t)/(1-z).
%F 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 up-down cycles (marked by t), and number of cycles that are not up-down (marked by s) is given by H(t,s,z) = (1-sin z)^(s-t)/(1-z)^s.
%e T(3,0)=1 because we have (123).
%e T(4,2)=7 because we have (1)(243), (142)(3), (132)(4), (13)(24), (12)(34), (143)(2), and (14)(23).
%e Triangle starts:
%e 1;
%e 0,1;
%e 0,1,1;
%e 1,1,3,1;
%e 4,6,7,6,1;
%e 19,35,30,25,10,1;
%p G := (1-sin(z))^(1-t)/(1-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], t, j), j = 0 .. n) end do; # yields sequence in triangular form
%t m = maxExponent = 11;
%t (CoefficientList[# + O[t]^m, t] Range[0, m-1]!&) /@ CoefficientList[(1 - Sin[z])^{1-t}/(1-z) + O[z]^m, z] // Flatten (* _Jean-François Alcover_, Aug 07 2018 *)
%Y Cf. A186359, A168360.
%K nonn,tabl
%O 0,9
%A _Emeric Deutsch_, Feb 20 2011
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