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User: John Kieffer

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John Kieffer has authored 1 sequences.

A296165 a(n) is the number of unimodular triangulations of [0,2]x[0,n].

Original entry on oeis.org

6, 64, 852, 12170, 182132, 2801708, 43936824, 698607816, 11224598424, 181815529916, 2964167665340, 48580814410080, 799696199314500, 13212398835196240, 218976668040908248, 3639020246503687098, 60616163842958990268, 1011775545312594580868, 16918718677672553292440, 283368129709983000763876
Offset: 1

Author

John Kieffer, Dec 06 2017

Keywords

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.

Crossrefs

Second column of array A082640.

Programs

  • 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

Extensions

More terms from Robin Visser, May 29 2025