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
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
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