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A258338
Ternary ménage problem: number of seating arrangements for n opposite-sex couples around a circular table such that no spouses and no triples of the same sex seat next to each other. Seats are labeled.
4
0, 8, 84, 3456, 219120, 19281600, 2324085120, 370554347520, 74897768655360, 18761274367718400, 5708008284647961600, 2072453585852572876800, 885341762559654194995200, 439630143301970662603161600, 251099117378080818090596352000, 163464570058143774978660630528000
OFFSET
1,2
COMMENTS
Conjecture: (a(n)/n!^2)^(1/n) ~ (3+sqrt(5))/2. - Vaclav Kotesovec, May 29 2015
LINKS
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.
FORMULA
a(n) = A114939(n) * 4 * n.
MATHEMATICA
a[1] = 0;
a[n_] := n! Sum[(-1)^j (n-j)! SeriesCoefficient[ SeriesCoefficient[ Tr[ MatrixPower[{{0, 1, 0, y^2, 0, 0}, {z y^2, 0, 1, 0, y^2, 0}, {z y^2, 0, 0, 0, y^2, 0}, {0, 1, 0, 0, 0, z}, {0, 1, 0, y^2, 0, z}, {0, 0, 1, 0, y^2, 0}}, 2n]], {y, 0, 2n}], {z, 0, j}], {j, 0, n}];
Array[a, 16] (* Jean-François Alcover, Dec 03 2018, from 1st PARI program *)
PROG
(PARI) { a(n) = if(n<2, 0, n! * sum(j=0, n, (-1)^j * (n-j)! *polcoeff( polcoeff( trace([0, 1, 0, y^2, 0, 0; z*y^2, 0, 1, 0, y^2, 0; z*y^2, 0, 0, 0, y^2, 0; 0, 1, 0, 0, 0, z; 0, 1, 0, y^2, 0, z; 0, 0, 1, 0, y^2, 0]^(2*n)), 2*n, y) , j, z)) ); }
(PARI) { a(n) = if(n<2, 0, n! * polcoeff( serlaplace( polcoeff( trace([-y, z*y, z, 0, z*y, -y; -y, (z - 1)*y, 0, (z - 1)*y^2, z*y, -y; 0, (z - 1)*y, 0, (z - 1)*y^2, 0, -y; -y, 0, z - 1, 0, (z - 1)*y, 0; -y, z*y, z - 1, 0, (z - 1)*y, -y; -y, z*y, 0, z*y^2, z*y, -y]^n), n, y) )/(1-z) + O(z^(n+1)), n, z) ) }
CROSSREFS
Cf. A114939 (counts up to rotations and reflections)
Sequence in context: A350264 A113376 A205311 * A245197 A281340 A298283
KEYWORD
nonn
AUTHOR
Max Alekseyev, May 27 2015
STATUS
approved