Is there a covering of $\mathbb{Z}^n$ by disjoint translates of the basisandorigin minimal integer $n$simplex? By haphazard I have such coverings for $\mathbb{Z}$, $\mathbb{Z}^2$ and $\mathbb{Z}^3$, where the wanted translations are lattices spanned by $\{2\}$, $\{(2,1),(1,2)\}$, and $\{(1,1,1),(1,1,1),(1,1,1)\}$, but rhyme nor reason can I see in this sequence of families to extend.


show 2 more comments 
Let $S$ be the set of integer points $(x_1,x_2,\dots,x_n)$ satisfying $$x_1+2x_2+3x_3+\dots+nx_n \equiv 0 \mod n+1,$$ and $T$ be the basisandorigin simplex as described in Ben's comment. Then translates of $T$ by $S$ disjointly cover $\mathbb{Z}^n$ (since decreasing the $x_i$ coordinate by $1$ changes the left hand side of the above relation by $i$, for any point not in $S$ there's exactly one direction we can move in to reach $S$). 


For fixed dimension $n$ there is an algorithm to find all such lattice tilings. Namely, let $S_n$ be the set of all $n\times n$ matrices $A$ of determinant $n+1$ that are in Hermite normal form over $\mathbb{Z}$. If the columns of $A$ are $v_1,\dots,v_n$, then there are $n$ nonzero integer column vectors $u_1,\dots,u_n$ for which there exist $0\leq a_i<1$ satisfying $\sum a_i v_i=u_i$. If the determinant of the matrix $M$ with columns $u_i$ is $\pm 1$, then the translates by the lattice generated by $v_1,\dots,v_n$ of the origin and the vectors $u_i$ gives a tiling of $\mathbb{Z}^n$. By a unimodular integral change of basis we can convert the $u_i$'s to the unit coordinate vectors. This construction gives all the desired lattice tilings, and is easy to implement algorithmically. For $n=4$ there are exactly two Hermite normal forms such that $\det M=\pm 1$, namely, $$ \begin{bmatrix} 1 & 0 & 0 & 2\\\ 0 & 1 & 0 & 3\\\ 0 & 0 & 1 & 4\\\ 0 & 0 & 0 & 5\end{bmatrix}, \qquad \begin{bmatrix} 1 & 0 & 0 & 1\\\ 0 & 1 & 0 & 0\\\ 0 & 0 & 1 & 1\\\ 0 & 0 & 0 & 5\end{bmatrix}. $$ 


$\{(0,0), (1,0), (0,1)\}$
. – Ben Barber May 8 '13 at 7:21