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a(n) = (2n-1)*(2n-1)!/n.
4

%I #34 Sep 08 2022 08:44:59

%S 1,9,200,8820,653184,73180800,11564467200,2451889440000,

%T 671854030848000,231125690776780800,97537253236899840000,

%U 49549698749529538560000,29829250083328819200000000,20999962511521107738624000000,17094073187896757112117657600000

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

%C This is the number of permutations of 2n letters having a cycle of length n. - _Marko Riedel_, Apr 21 2015

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

%H Alois P. Heinz, <a href="/A052145/b052145.txt">Table of n, a(n) for n = 1..225</a>

%H Math Stackexchange, <a href="http://math.stackexchange.com/questions/1241917/">How many permutations</a>

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

%F a(n) = A126074(2n,n). - _Alois P. Heinz_, Apr 21 2017

%F a(n) = A293211(2n,n). - _Alois P. Heinz_, Oct 11 2017

%e 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

%t Table[(2 n - 1) (2 n - 1)! / n, {n, 30}] (* _Vincenzo Librandi_, Apr 22 2015 *)

%o (PARI) a(n)=(2*n-1)*(2*n-1)!/n \\ _Charles R Greathouse IV_, Apr 21 2015

%o (Magma) [(2*n-1)*Factorial(2*n-1)/n: n in [1..20]]; // _Vincenzo Librandi_, Apr 22 2015

%Y Cf. A126074, A293211.

%K nonn,easy

%O 1,2

%A _N. J. A. Sloane_, Jan 23 2000