A229244 Number of n-permutations such that at least one cycle has size ceiling(n/2).

1, 1, 3, 9, 40, 200, 1260, 8820, 72576, 653184, 6652800, 73180800, 889574400, 11564467200, 163459296000, 2451889440000, 39520825344000, 671854030848000, 12164510040883200, 231125690776780800, 4644631106519040000, 97537253236899840000, 2154334728240414720000, 49549698749529538560000, 1193170003333152768000000

Offset: 1

Author

Geoffrey Critzer, Sep 17 2013

Keywords

• nonn

Examples

			a(4) = 9 because we have:
1: (1)(2)(4,3)
2: (1)(3,2)(4)
3: (1)(4,2)(3)
4: (2,1)(3)(4)
5: (2,1)(4,3)
6: (3,1)(2)(4)
7: (3,1)(4,2)
8: (4,1)(2)(3)
9: (4,1)(3,2).
		

Crossrefs

Cf. A110468.

Programs

Mathematica

f[n_]:=If[EvenQ[n],Binomial[n,n/2](n/2-1)!((n/2)!-(n/2-1)!)+n!/2/(n/2)^2,Binomial[n,Ceiling[n/2]]Floor[n/2]!^2]; Table[f[n],{n,1,25}]

Formula

For odd n, a(2m+1)= binomial(2m+1,m+1)*m!^2.

For even n, a(2m) = binomial(2m,m)*(m-1)!*(m!-(m-1)!) + (2m)!/(2*m^2).

Conjecture: (n+1)*a(n) +(-3*n+1)*a(n-1) -(n-2)*(n^2-2*n-1)*a(n-2) +(n-2)*(n-3)^2*a(n-3)=0. - R. J. Mathar, May 23 2014