Positivity of coefficients of Ehrhart polynomial of n-Tetrahedron

A set of positive integers $d_1, \dots, d_n$ describe two n-dimensional closed lattice tetrahedron: $$T = \left\{ (x_1, \dots, x_n) \in \mathbb{R}^n: \sum_{i=1}^n \frac{x_i}{d_i} \leq 1 \textrm{ and all } x_i \geq 0 \right\}.$$ Let $L_T(t)$ be the Ehrhart polynomial of $T$, i.e. the number of integer lattice points in $tT$.

When are all coefficients of $L_{T}(t)$ positive?

• Let me take a look at the case $n=2$ to get an idea of what one might expect. In this case the triangle $tT$ in question has vertices $$(td_1,0),(0,td_2),(0,0).$$ Pick's formula gives the number of points as $${\rm Area}+\frac 12\sharp({\rm pts ~on~the~boundary})+1 =\frac 12t^2 d_1d_2 + \frac 12t(\gcd(d_1,d_2)+d_1+d_2)+1$$ so the coefficients are all positive. – Lev Borisov Feb 20 '16 at 20:18
• To add some information: 1. Pommersheim has given an explicit formula for $n=3$ based on Dedekind sums. I can post the polynomial on request. 2. In the case that $d_1 = \dots = d_n$, then the coefficients are the absolute value of the Stirling numbers of the first kind and therefore always positive. – Sebastian Schlecht Feb 21 '16 at 8:16
• @LevBorisov: Thanks for the start! Any idea how to proceed from there? This paper of Beck gives a description of all coefficients via the Residue Theorem, however I cannot derive anything about the coefficient signs. – Sebastian Schlecht Feb 25 '16 at 8:39
• $T_2$ does not have integer vertices, so it does not have an Ehrhart polynomial. Let $M$ be the least common multiple of $d_1,\dots,d_n$. You could ask for each $0\leq j<M$ whether the polynomial $p_j(t)=L_{T_2}(tM+j)$ has positive coefficients. – Richard Stanley Feb 25 '16 at 13:38
• I agree with Lev that this is a difficult problem, already for $n=3$. As Sebastian mentioned, one can compute these polynomials through Dedekind sums (and there generalized versions, for $n \ge 5$), but those are delicate... – matthias beck Feb 25 '16 at 18:21