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The motivation of this question comes from number theory (I add the tag number theory for this reason, in that it is possible that someone with a number-theoretic background has already thought about this question for similar reason and has already come up with an answer).

Let $r,s$ be positive integers, where $s$ is bigger then $r$. For a sub-lattice $\Lambda \subseteq \{(x_i) \in \mathbb{Z}^s:\sum_{i=1}^{s}x_i=0\}$ of rank $r$, we call a vector $\lambda \in \Lambda$ special if there is another vector $\lambda^{*} \in \Lambda$ such that the negative coordinates (with respect to the standard basis in $\mathbb{Z}^s$) of $\lambda$ and $\lambda^{*}$ coincide with exactly the same values, and for all coordinates where $\lambda$ is positive then $\lambda^{*}$ has 0 in that coordinate (and now it follows the same property exchanging $\lambda$ and $\lambda^{*}$ from the already listed properties). I denote by $\Lambda^{*}$ the set of special vectors in $\Lambda$. I denote by $w(\Lambda)$ the number of distinct Hamming weights of vectors in $\Lambda^{*}$ (the Hamming weight of a vector is the number of non-zero entries of it).

Question: Can $w(\Lambda)$ be bounded purely in function of $r$ (independently of $s$)? Or is it possible to create a family of $\Lambda$ of rank $r$, with bigger and bigger s, with $w(\Lambda)$ arbitrarily large?

Remarks: 1) It is very easy to create lattices of rank 2, so r=2, with at least $\frac{\sqrt s}{2}$ different Hamming weights.

2)On the other hand the question above is positively answered for r=2, so the condition of doing it only on special vectors add some constraint.

3)I have not finished the argument, but I would not be surprised if the answer is positive also for r=3 (I have done a subdivision in 2 cases and in one case I can prove it). But I do not have a good feeling of what happens for r large. I apologize if there is a trivial counter-example.

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