Let $E\subset \mathbb{R}^n$ be a cross-polytope:
$$E= \left\lbrace x : \frac{|x\_1|}{q\_1}+\cdots+\frac{|x\_n|}{q\_n}\leq 1 \right\rbrace, $$
where $q\_1,\dots,q\_n$ are positive integers. I am interested in estimating the number $L$ of lattice points in $E$.


**Conjecture.** 
$$L:=\# \left(E \cap \mathbb{Z}^n\right)>\mathrm{vol}(E)\,.$$


It's obviously true for $n=1$, and very easy to verify (using Pick's formula) for $n=2$. I don't know how to prove it for $n\geq 3$ without some additional constraints.


According to (generalized) Blichfeldt's theorem, there exists $x\in\mathbb{R}^n$ such that
$E+x$ has more lattice points than $\mathrm{vol}(E)$. This suggests the following


**Conjecture (strong version).** For any $x\in\mathbb{R}^n$
$$L\geq \#\left((E+x)\cap\mathbb{Z}^n\right)\,.$$


I don't know how to prove (or disprove) this version even for $n=2$.



Lattice points in cross-polytopes
The conjecture fails for $n=3$ and $(q\_1,q\_2,q\_3) = (9,10,10)$,
when ${\rm vol}(E) = (2^3/3!) 9 \cdot 10 \cdot 10 = 1200$ but
$ \#(E \cap {\bf Z}^3) = 1199 $. In general,
if $(q\_1,q\_2,q\_3) = (k,k+1,k+1)$ then
$$
\#(E \cap {\bf Z}^3) - {\rm vol}(E) = -\frac23 k^2 + \frac{16}{3} k + 5
$$
which is negative for $k \geq 9$, and increasingly so as $k$ grows.
This was found experimentally, but is not hard to prove: the
integer points are those for which
$\left| x\_1 \right| + \left| x\_2 \right| + \left| x\_3 \right| \leq k$
(i.e. the lattice points inside the $(k,k,k)$ octahedron),
together with the $4(k+1)$ points with $x\_1=0$ and
$\left| x\_2 \right| + \left| x\_3 \right| = k+1$.
By a standard induction there are $\frac43 k^3 + 2k^2 + \frac83 k + 1$
integer solutions of
$\left| x\_1 \right| + \left| x\_2 \right| + \left| x\_3 \right| \leq k$, etc.


The other examples with $n=3$ and $q\_1 \leq q\_2 \leq q\_3 \leq 20$ are
$(6,17,18)$, $(7,13,14)$, $(8,15,16)$, $(9,17,18)$, $(9,19,19)$, $(10,19,20)$
with a shortfall of $-1$, $-17/3$, $-17$, $-31$, $-1$, $-143/3$ respectively,
and suggesting similar generalizations, all with two of the $q\_i$
related by a simple ratio. The OP commented "That is quite surprising.
I assumed that is better to look for coprime $q\_i$ for counterexamples",
and I confess that this surprised me too (though the proof for
$(k,k+1,k+1)$ explains the mechanism). Curiously the conjecture
seems to be true when the $q\_i$ are pairwise coprime; in fact
an exhaustive search of the region $q\_i \leq 256$ found no
counterexamples to the stronger inequality
$$
\#(E \cap {\bf Z}^3) - {\rm vol}(E) > q\_1.
$$
It's not easy even to find examples where the two sides are nearly equal.
The smallest ratios that the search turned up all fit the pattern
$$
(q\_1,q\_2,q\_3) = (2n^2-3n, 2n^2-2n-1, 2n^2-n-1)
$$
with 
$$
\#(E \cap {\bf Z}^3) - {\rm vol}(E) = 2n^2-2n+5 = q\_1 + O(q\_1^{1/2})
$$
(any $n>1$). I have not tried to prove that this persists for all $n$,
though I guess it won't be too hard $-$ not as easy as the $(k,k+1,k+1)$
explanation but much easier than proving the inequality for all
pairwise coprime $q\_1,q\_2,q\_3$.


Still, even in the pairwise coprime case one soon find
counterexamples with $n=3$ to the "Conjecture (strong version)"
that $E$ has at least as many lattice points than $E+x$ for any
$x \in {\bf R}^n$. We can even take $x = (\frac12, 0, 0)$,
and then test the conjecture by comparing the count of integer points
in the $(2q\_1,q\_2,q\_3)$ octahedron with twice the $(q\_1,q\_2,q\_3)$ count.
If the former is larger then we have a counterexample, and this happens
already for $(q\_1,q\_2,q\_3) = (5,2,3)$, when $E$ has $49$ integral points
but $E+(\frac12,0,0)$ has $50$. There are $30$ more such examples
with $\max\_i q\_i \leq 16$. (As it happens $(2,3,5)$ is the $n=2$
case of $(2n^2-3n, 2n^2-2n-1, 2n^2-n-1)$, and larger $n$ account for
some of the other counterexamples.)


In the previous edit of this answer I gave some needlessly convoluted
**gp** code for computing the difference between the number $L$ of
integral points in $E$ and the volume $\frac43 q\_1 q\_2 q\_3$ of $E$.
Here's a much simpler and faster program, which takes time about $q\_1 q\_2$
and negligible space, whereas the earlier code (using generating functions)
took time and space $q\_1 q\_2 q\_3$.



```
{
L3(q1,q2,q3) =
  sum(x1=0,q1,if(x1>0,2,1) *
    sum(x2=0,q2+(-q2*x1)\q1,if(x2>0,2,1) *
      (1 + 2*(q3+(-q3*(q2*x1+q1*x2))\(q1*q2)))
    )
  )
}
d3(q1,q2,q3) = L3(q1,q2,q3) - (4/3)*q1*q2*q3

```

