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%I #19 Nov 07 2020 05:26:38
%S 0,1,3,10,42,215,1306,9203,73896,666449,6672426,73447207,881720276,
%T 11465066353,160533297198,2408198818951,38533084860528,
%U 655081834141121,11791682879883154,224044379597455367,4480916680834220172,94099620668706861137,2070196606209604069110
%N Number of up-down cycles in all permutations of {1,2,...,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 a(n) = Sum(k*A186358(n,k), k=0..n).
%H Alois P. Heinz, <a href="/A186360/b186360.txt">Table of n, a(n) for n = 0..300</a>
%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 a(n) = n!*Sum(E(j-1)/j!, j=1..n), where E(i) = A000111(i) are the Euler (or up-down) numbers.
%F E.g.f.: -log(1-sin z)/(1-z).
%F a(n) ~ n! * (-log(1-sin(1))). - _Vaclav Kotesovec_, Oct 08 2013
%e a(3) = 10 because the permutations (1)(2)(3), (12)(3), (13)(2), (1)(23), (123), and (132) have a total of 3 + 2 + 2 + 2 + 0 + 1 = 10 up-down cycles.
%p g := -ln(1-sin(z))/(1-z): gser := series(g, z = 0, 25): seq(factorial(n)*coeff(gser, z, n), n = 0 .. 22);
%t CoefficientList[Series[-Log[1-Sin[x]]/(1-x), {x, 0, 20}], x]* Range[0, 20]! (* _Vaclav Kotesovec_, Oct 08 2013 *)
%Y Cf. A000111, A186358.
%K nonn
%O 0,3
%A _Emeric Deutsch_, Feb 28 2011
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