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
Robin Visser, Table of n, a(n) for n = 1..800
O. Aichholzer, Counting Triangulations - Olympics, 2006.
V. Kaibel and G. M. Ziegler, Counting Lattice Triangulations, arXiv:math/0211268 [math.CO], 2002.
PROG
(SageMath) # Implements the recursive formula by V. Kaibel and G. Ziegler
def a(n):
g, ans = [[0 for i in range(n+1)] for j in range(n+1)], 0
for (A, B) in ((x, y) for x in range(n+1) for y in range(n+1)):
g[A][B] = binomial((3*A+B-1)/2, A)*binomial((A+3*B-1)/2, B)
for (a, b) in ((x, y) for x in range(A+1) for y in range(B+1)):
if ((a+b)%2==1) and (a+b < A+B):
g[A][B]+=g[a][b]*binomial((3*A+B-3*a-b)/2-1, A-a)*binomial((A+3*B-a-3*b)/2-1, B-b)
for (B, A) in ((x, y) for x in range(n+1) for y in range(x)):
if ((A+B)%2==1):
ans+=g[A][B]*binomial(2*n-(3*A+B+1)/2, n-A)*binomial(2*n-(A+3*B+1)/2, n-B)
return binomial(2*n, n)^2 + 2*ans # Robin Visser, May 29 2025
CROSSREFS
KEYWORD
nonn
AUTHOR
John Kieffer, Dec 06 2017
EXTENSIONS
More terms from Robin Visser, May 29 2025
STATUS
approved