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Number of excedances in all cycle-up-down permutations of {1,2,...,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 #24 Jan 31 2023 12:13:48

%S 0,0,1,4,20,100,568,3480,23552,172280,1369600,11687996,107154944,

%T 1046876220,10891574272,119977808752,1397745975296,17147547551920,

%U 221248695107584,2992325932512948,42370119842398208,626422742409351380,9659199474524225536,154997830582262971784

%N Number of excedances in all cycle-up-down permutations of {1,2,...,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 a(n) = Sum_{k>=0} k*A186368(n,k).

%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 E.g.f.: [z(1+sin z)-cos z * log(sec z + tan z)]/[2 cos z *(1-sin z)].

%F a(n) ~ n! * n^2 * (2/Pi)^(n+2) * (1 - log(n)/n). - _Vaclav Kotesovec_, Aug 23 2014

%e a(3) = 4 because the permutations (1)(2)(3), (1)(23), (12)(3), (13)(2), (132) have a total of 0+1+1+1+1=4 excedances.

%p g := ((z*(1+sin(z))-cos(z)*ln(sec(z)+tan(z)))*1/2)/(cos(z)*(1-sin(z))): gser := series(g, z = 0, 30): seq(factorial(n)*coeff(gser, z, n), n = 0 .. 22);

%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}];

%t Table[Sum[k T[n, k], {k, 0, n/2}], {n, 0, 22}] (* _Jean-François Alcover_, Aug 07 2018 *)

%Y Cf. A186368.

%K nonn

%O 0,4

%A _Emeric Deutsch_, Feb 28 2011