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If I undderstand the problem correctly one case would be that all bricks are translates of each other in which case a tessalation can be summarized by the lattice of centers of bricks. Up to reordering coordinates, generators for this lattice can be chosen to be the columns of a triangular matrix M $M$ and two which differ by an integer triangular basis change are equivalent. In this case s(M)=dt(M)/L(M) $s(M)=dt(M)/L(M)$ where t(M) $t(M)$ is the trace of the matrix $M$ and L(M) $L(M)$ is the minimum 1-norm of a nonzero lattice element.

Claim: s(d+1)/(d+1) >= $s(d+1)/(d+1) \geq s(d)/d + 1/21/2$.

Thus s(d) >= d(d+1)/2$s(d) \geq d(d+1)/2$.

Proof: If M is obtained by adding a new vector v with diagonal entry r (and dimension) to N then up to translation by the lattice spanned by columns of N, the projection of v to the space spanned by N has 1-norm at least L(N)/2 $L(N)/2$ and s(M)/(d+1)=(t(N)+r)/min{L(N), L(N)/2 $s(M)/(d+1)=(t(N)+r)/\min(L(N),L(N)/2 + r} >= ) \geq s(N)/d + 1/21/2$, with equality if r=L(N)/2. $r=L(N)/2$.

There is a large space of lattices achieving this bound.
One simple solution has diagonal entries (brick side edge lengths) of (d/2)(2,1,1,...,1) and first off diagonal of (d/2)(1,1,...,1).

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If I undderstand the problem correctly one case would be that all bricks are translates of each other in which case a tessalation can be summarized by the lattice of centers of bricks. Up to reordering coordinates, generators for this lattice can be chosen to be the columns of a triangular matrix M and two which differ by an integer triangular basis change are equivalent. In this case s(M)=dt(M)/L(M) where t(M) is the trace of the matrix and L(M) is the minimum 1-norm of a nonzero lattice element.

Claim: s(d+1)/(d+1) >= s(d)/d + 1/2. Thus s(d) >= d(d+1)/2. Proof: If M is obtained by adding a new vector v with diagonal entry r (and dimension) to N then up to translation by the lattice spanned by columns of N, the projection of v to the space spanned by N has 1-norm at least L(N)/2 and s(M)/(d+1)=(t(N)+r)/min{L(N), L(N)/2 + r} >= s(N)/d + 1/2, with equality if r=L(N)/2.

There is a large space of lattices achieving this bound.
One simple solution has diagonal entries (brick side lengths) of (d/2)(2,1,1,...,1) and first off diagonal of (d/2)(1,1,...,1).