A201546 The number of permutations of {1,2,...,2n} that contain a cycle of length greater than n.

1, 14, 444, 25584, 2342880, 312888960, 57424792320, 13869128448000, 4264876094976000, 1627055289796608000, 754132445894209536000, 417405110861381271552000, 271933770461631065948160000, 205985221930119691492392960000, 179512031423815845458883379200000

Offset

1,2

Comments

(2n)!*(H(2n)-H(n)) where H(n) is the harmonic number. See "One Hundred Prisoners" in Wikipedia reference below.

References

Richcard Stanley, Enumerative Combinatorics Volume 1 second edition, Cambridge Univ. Press, 2011, page 194 solution 119.

Links

Alois P. Heinz, Table of n, a(n) for n = 1..100

Wikipedia, Random permutation statistics

Formula

The E.g.f. for the number of n-permutations that have a cycle of length greater than k is G(x)=1/(1-x) - exp(A(x)) where A(x)=Sum_{j=1..k}x^j/j.

a(n)/(2n)! is the coefficient of x^(2n) in G(x).

a(n) = Sum_{k=n+1..2n} C(2n,k)*(k-1)!*(2n-k)! - Geoffrey Critzer, Jun 01 2013

a(n) ~ sqrt(Pi)*log(2)*2^(2*n+1)*n^(2*n+1/2)/exp(2*n). - Vaclav Kotesovec, Sep 29 2013

Example

a(2)=14 because there are 6 permutations of {1,2,3,4} that contain a cycle of length four and 8 that contain a cycle of length three. 6+8=14.

Maple

a:= proc(n) option remember; `if`(n<2, n,
      (6-12*n+8*n^2)*a(n-1)-4*(n-1)^2*(2*n-3)^2*a(n-2))
    end:
seq(a(n), n=1..20);  # Alois P. Heinz, Jun 13 2013

Mathematica

Drop[Table[a = Sum[x^j/j, {j, 1, n}]; Coefficient[Series[1/(1 - x) - Exp[a], {x, 0, 40}], x^(2 n)]*(2 n)!, {n, 1, 40}], -20]

Crossrefs

Sequence in context: A033815 A187358 A103916 * A305115 A282245 A319096

Adjacent sequences: A201543 A201544 A201545 * A201547 A201548 A201549

Keyword

nonn

Author

Geoffrey Critzer, Dec 02 2011

Status

approved