The lattice polytop $[0,n_1]\times[0,n_2]\times\dots\times[0,n_{d-1}]\times[0,1]$ contains $(n_1+1)(n_2+1)\cdots(n_{d-1}+1)2$ integral points on the boundary and no integral points in its interior. Its number of vertices, $2^d$, is however bounded by a function depending only on its dimension $d$. Does there exist a sequence of convex $d-$dimensional lattice-polytops without interior lattice points and more and more vertices?

Remarks: (1) The answer is no in dimension $2$.

(2) This question is motivated by a result of Lagarias-Ziegler who showed that the volume (and thus the number of vertices) of a convex $d-$dimensional lattice polytop is bounded if it contains exactly $k>0$ interior lattice points. If no sequence as above exist, then the condition on the existence of $k$ interior lattice points can perhaps be modified into a condition on the number of integral vertices (which has to be sufficiently large) and integral boundary points.

maximalhollow lattice polytopes in a fixed dimension, where "maximal hollow" means "not properly contained in a larger hollow lattice polytope" (Averkov, Wagner, Weismantel; Nill & Ziegler). – Joseph O'Rourke Feb 9 '12 at 19:07