A186360 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)<... .

0, 1, 3, 10, 42, 215, 1306, 9203, 73896, 666449, 6672426, 73447207, 881720276, 11465066353, 160533297198, 2408198818951, 38533084860528, 655081834141121, 11791682879883154, 224044379597455367, 4480916680834220172, 94099620668706861137, 2070196606209604069110

Offset

0,3

Comments

a(n) = Sum(k*A186358(n,k), k=0..n).

Links

Alois P. Heinz, Table of n, a(n) for n = 0..300

Emeric Deutsch and Sergi Elizalde, Cycle up-down permutations, arXiv:0909.5199 [math.CO], 2009; and also, Australas. J. Combin. 50 (2011), 187-199.

Formula

a(n) = n!*Sum(E(j-1)/j!, j=1..n), where E(i) = A000111(i) are the Euler (or up-down) numbers.

E.g.f.: -log(1-sin z)/(1-z).

a(n) ~ n! * (-log(1-sin(1))). - Vaclav Kotesovec, Oct 08 2013

Example

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.

Maple

g := -ln(1-sin(z))/(1-z): gser := series(g, z = 0, 25): seq(factorial(n)*coeff(gser, z, n), n = 0 .. 22);

Mathematica

CoefficientList[Series[-Log[1-Sin[x]]/(1-x), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Oct 08 2013 *)

Crossrefs

Cf. A000111, A186358.

Sequence in context: A030964 A263823 A030867 * A352856 A007680 A232606

Adjacent sequences: A186357 A186358 A186359 * A186361 A186362 A186363

Keyword

nonn

Author

Emeric Deutsch, Feb 28 2011

Status

approved