Discrete disjoint covering of integer lattices Is there a covering of $\mathbb{Z}^n$ by disjoint translates of the basis-and-origin 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.
 A: 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 basis-and-origin 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 $S$).  
A: 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,
  $$ \begin{bmatrix} 1 & 0 & 0 & 2\\\ 0 & 1 & 0 & 3\\\
          0 & 0 & 1 & 4\\\ 0 & 0 & 0 & 5\end{bmatrix},
  \qquad \begin{bmatrix} 1 & 0 & 0 & 1\\\ 0 & 1 & 0 & 0\\\
          0 & 0 & 1 & 1\\\ 0 & 0 & 0 & 5\end{bmatrix}. $$
