A037224 Number of permutations p of {1,2,3...,n} that are fixed points under the operation of first reversing p, then taking the inverse.

1, 0, 0, 2, 2, 0, 0, 12, 12, 0, 0, 120, 120, 0, 0, 1680, 1680, 0, 0, 30240, 30240, 0, 0, 665280, 665280, 0, 0, 17297280, 17297280, 0, 0, 518918400, 518918400, 0, 0, 17643225600, 17643225600, 0, 0, 670442572800, 670442572800, 0, 0, 28158588057600

Offset: 1

Author

Miklos SZABO (mike(AT)ludens.elte.hu)

Keywords

• nonn
• easy

Comments

A122670 is an essentially identical sequence.

Also the number of rotationally symmetric solutions to non-attacking rooks problem on an n X n board.

Reversal of a permutation reflects the associated permutation matrix through an axis parallel to its sides, while inversion reflects the matrix through its main diagonal. The composition of these two operations is rotation by 90 degrees, and so permutations fixed by this composition correspond to rotationally symmetric rook diagrams by taking the associated permutation matrix. [Ian Duff, Mar 09 2007 and Joel B. Lewis, Jun 10 2009]

Equivalently, the number of permutations fixed by first inverting and then reversing. We may also replace "reversing" with "complementing" in the preceding sentences, where the complement of (w(1), ..., w(n)) is (n + 1 - w(1), ..., n + 1 - w(n)). [Joel B. Lewis, Jun 10 2009]

Examples

			Let p be the permutation {11,1,9,3,7,5,8,6,10,4,12,2} of {1,2,3,..,12}. Then the reverse Rp of p is {2,12,4,10,6,8,5,7,3,9,1,11} and the inverse IRp of Rp is {11,1,9,3,7,5,8,6,10,4,12,2}. Thus p counts as one of the a(12)=120 fixed-points for n=12.
		

Links

• C. Bebeacua, T. Mansour, A. Postnikov and S. Severini, On the X-rays of permutations, arXiv:math/0506334 [math.CO], 2005.
• M. Szabo, Non-attacking Queens Problem Page

Crossrefs

Cf. A001813, A033148, A032522, A037223, A122670.

Programs

Maple

a:= n-> `if`(irem(n, 4, 'm')>1, 0,
        `if`(m=0, 1, (2*m-1)! * 2/(m-1)!)):
seq(a(n), n=1..99);  # Alois P. Heinz, Jan 21 2011

Mathematica

{1}~Join~Table[If[MemberQ[{0, 1}, Mod[n, 4]], (2 # - 1)!*2/(# - 1)! &[Floor[n/4]], 0], {n, 2, 44}] (* Michael De Vlieger, Oct 05 2016 *)

PARI

a(n)=
{
    if ( n%4>=2, return(0) );
    n = n\4;
    if ( n==0, return(1) );
    return( (2*n-1)!*2/(n-1)! );
}
vector(55,n,a(n)) /* Joerg Arndt, Jan 21 2011 */

Formula

a(4n) = a(4n+1) = (2n-1)!*2/(n-1)!, a(4n+2) = a(4n+3) = 0.

Extensions

Edited by N. J. A. Sloane, Jun 12 2009, incorporating comments from John W. Layman, Sep 17 2004