OFFSET
1,1
COMMENTS
As stated by Kaibel and Ziegler, the number of unimodular triangulations of [0,1]x[0,n] is (2n)!/(n!*n!). This gives a(1)=6.
No formula for a(n) is known. Aichholzer computed a(n) for n<=15.
Kaibel and Ziegler computed a(n) for n<=375. Aichholzer also computed the number of unimodular triangulations of [0,m]x[0,n] for m=3,4,5 and various n, and Kaibel-Ziegler extended these calculations to m=6.
REFERENCES
V. Kaibel and G. Ziegler, "Counting lattice triangulations," London Math. Soc. Lecture Notes Series, Vol. 307, pp. 277-307, 2003.
LINKS
O. Aichholzer, Counting Triangulations - Olympics, 2006.
V. Kaibel and G. M. Ziegler, Counting Lattice Triangulations, arXiv:math/0211268 [math.CO], 2002.
CROSSREFS
KEYWORD
nonn,more
AUTHOR
John Kieffer, Dec 06 2017
STATUS
approved