A052145 a(n) = (2n-1)*(2n-1)!/n.

1, 9, 200, 8820, 653184, 73180800, 11564467200, 2451889440000, 671854030848000, 231125690776780800, 97537253236899840000, 49549698749529538560000, 29829250083328819200000000, 20999962511521107738624000000, 17094073187896757112117657600000

Offset: 1

Author

N. J. A. Sloane, Jan 23 2000

Keywords

• nonn
• easy

Comments

This is the number of permutations of 2n letters having a cycle of length n. - Marko Riedel, Apr 21 2015

Examples

			For n=2, there are 9 permutations of [4] = { 1, 2, 3, 4 } which have a cycle of length 2: each of the 4*3/2 = 6 transpositions, plus the 3 different possible products of two transpositions. - _M. F. Hasler_, Apr 21 2015
		

References

• R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 7.68(d).

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..225
• Math Stackexchange, How many permutations

Crossrefs

Cf. A126074, A293211.

Programs

Magma

[(2*n-1)*Factorial(2*n-1)/n: n in [1..20]]; // Vincenzo Librandi, Apr 22 2015

Mathematica

Table[(2 n - 1) (2 n - 1)! / n, {n, 30}] (* Vincenzo Librandi, Apr 22 2015 *)

PARI

a(n)=(2*n-1)*(2*n-1)!/n \\ Charles R Greathouse IV, Apr 21 2015

Formula

a(n) = 2*m*m!/(m+1) where m=2n-1.

a(n) = A126074(2n,n). - Alois P. Heinz, Apr 21 2017

a(n) = A293211(2n,n). - Alois P. Heinz, Oct 11 2017