A186367 Number of cycles 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(3)<... .

1, 3, 10, 38, 165, 812, 4478, 27408, 184529, 1356256, 10809786, 92892928, 856329253, 8430600960, 88292571934, 980197173248, 11499036105537, 142147625652224, 1846872283846922, 25161923756064768, 358706981125488581, 5340498034862030848

Offset: 1

Author

Emeric Deutsch, Feb 28 2011

Keywords

• nonn

Examples

			a(3) = 10 because the cycle-up-down permutations (1)(2)(3), (12)(3), (13)(2), (1)(23), and (132), have a total of 3+2+2+2+1=10 cycles.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..200
• E. Deutsch and S. Elizalde, Cycle up-down permutations, arXiv:0909.5199 [math.CO], 2009.

Crossrefs

Cf. A186366.

Programs

Maple

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

Mathematica

Rest[CoefficientList[Series[-Log[1-Sin[x]]/(1-Sin[x]), {x, 0, 20}], x]* Range[0, 20]!] (* Vaclav Kotesovec, Oct 02 2013 *)

PARI

x='x+O('x^30); Vec(serlaplace(-log(1-sin(x))/(1-sin(x)))) \\ G. C. Greubel, Aug 30 2018

Formula

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

a(n) = Sum_{k=1..n} k*A186366(n,k).

a(n) ~ n!*n*2^(n+3)/Pi^(n+2)*(2*log(n/Pi) + 2*gamma + 3*log(2) - 2), where gamma is the Euler-Mascheroni constant (A001620). - Vaclav Kotesovec, Oct 02 2013