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

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

Offset

1,2

Comments

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

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

Example

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

Mathematica

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

Prog

(PARI) a(n)=(2*n-1)*(2*n-1)!/n \\ Charles R Greathouse IV, Apr 21 2015
(Magma) [(2*n-1)*Factorial(2*n-1)/n: n in [1..20]]; // Vincenzo Librandi, Apr 22 2015

Crossrefs

Cf. A126074, A293211.

Sequence in context: A064756 A197077 A093708 * A012061 A012168 A012133

Adjacent sequences: A052142 A052143 A052144 * A052146 A052147 A052148

Keyword

nonn,easy

Author

N. J. A. Sloane, Jan 23 2000

Status

approved