A094047 Number of seating arrangements of n couples around a round table (up to rotations) so that each person sits between two people of the opposite sex and no couple is seated together.

0, 0, 2, 12, 312, 9600, 416880, 23879520, 1749363840, 159591720960, 17747520940800, 2363738855385600, 371511874881100800, 68045361697964851200, 14367543450324474009600, 3464541314885011705344000, 946263209467217020194816000, 290616691739323132839591936000

Offset: 1

Author

Matthijs Coster, Apr 29 2004

Keywords

• nonn

Comments

Also, the number of Hamiltonian directed circuits in the crown graph of order n.

Or the number of those 3 X n Latin rectangles (cf. A000186) the second row of which is a full cycle. - Vladimir Shevelev, Mar 22 2010

References

• V. S. Shevelev, Reduced Latin rectangles and square matrices with equal row and column sums, Diskr.Mat.(J. of the Akademy of Sciences of Russia) 4(1992),91-110.

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..253
• M. A. Alekseyev, Weighted de Bruijn Graphs for the Menage Problem and Its Generalizations. Lecture Notes in Computer Science 9843 (2016), 151-162. doi:10.1007/978-3-319-44543-4_12, arXiv:1510.07926, [math.CO], 2015-2016.
• H. M. Taylor, A problem on arrangements, Mess. Math., 32 (1902), 60ff. [Annotated scanned copy]
• Eric Weisstein's World of Mathematics, Crown Graph
• Eric Weisstein's World of Mathematics, Hamiltonian Cycle

Crossrefs

Cf. A059375 (rotations are counted as different).

Cf. A000179, A000186, A114939, A137729, A306496.

Programs

Maple

A094047 := proc(n)
    if n < 3 then
        0;
    else
        (-1)^n*2*(n-1)!+n!*add( (-1)^j*(n-j-1)!*binomial(2*n-j-1,j),j=0..n-1) ;
    end if;
end proc: # R. J. Mathar, Nov 02 2015

Mathematica

Join[{0},Table[(-1)^n 2(n-1)!+n!Sum[(-1)^j (n-j-1)!Binomial[2n-j-1,j],{j,0,n-1}],{n,2,20}]] (* Harvey P. Dale, Mar 07 2012 *)

Formula

For n>1, a(n) = (-1)^n * 2 * (n-1)! + n! * Sum_{j=0..n-1} (-1)^j * (n-j-1)! * binomial(2*n-j-1,j). - Max Alekseyev, Feb 10 2008

a(n) = A059375(n) / (2*n) = A000179(n) * (n-1)!.

Conjecture: a(n) +(-n^2+2*n-3)*a(n-1) -(n-2)*(n^2-3*n+5)*a(n-2) -3*(n-2)*(n-3)*a(n-3) +(n-2)*(n-3)*(n-4)*a(n-4)=0. - R. J. Mathar, Nov 02 2015

Conjecture: (-n+2)*a(n) +(n-1)*(n^2-3*n+3)*a(n-1) +(n-2)*(n-1)*(n^2-3*n+3)*a(n-2) +(n-2)*(n-3)*(n-1)^2*a(n-3)=0. - R. J. Mathar, Nov 02 2015

a(n) = (n-1) * (n * (a(n-1) + a(n-2)) - 4 * (-1)^n * (n-3)!) for n > 3. - Seiichi Manyama, Jan 18 2020

a(n) = 2 * A306496(n). - Alois P. Heinz, Jun 19 2022

Extensions

Better definition from Joel B. Lewis, Jun 30 2007

Formula and further terms from Max Alekseyev, Feb 10 2008