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

1,3

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

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

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 *)

Crossrefs

Cf. A059375 (rotations are counted as different).

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

Sequence in context: A012422 A122767 A260321 * A300045 A091472 A156518

Adjacent sequences: A094044 A094045 A094046 * A094048 A094049 A094050

Keyword

nonn

Author

_Matthijs Coster_, Apr 29 2004

Extensions

Better definition from _Joel B. Lewis_, Jun 30 2007

Formula and further terms from _Max Alekseyev_, Feb 10 2008

Status

approved