A Mersenne number is a number of the form $2^k-1$ for some $k \in \mathbb{N}$. Consider the set of $2^n-1$ products of Mersenne numbers $$M_n=\left\{ \prod_{k\in S} (2^k-1) : S \subseteq [n], S\neq \emptyset\right\}.$$

**Question:** What is the minimum $r \in \mathbb{N}$ for which there exists $\alpha_1,\dots, \alpha_r \in \mathbb{R}$ such that for all $S \subseteq [n]$ there exists $T \subseteq [r]$ for which $\sum_{a \in T} \alpha_a = \prod_{k\in S} (2^k-1)$?

**Known so far: A linear lower bound.**

One can prove that for any $S \neq R \subseteq [n]$, it holds that $\prod_{k\in S} (2^k-1) \neq \prod_{k\in R} (2^k-1)$, so $M_n$ consists of $2^n-1$ distinct numbers. Since subset sums of a set of $r$ real numbers can take at most $2^r$ values, it follows that $r \geq n$. Can we prove a superlinear lower bound?

Our intuition is that there should be a superlinear lower bound, because the numbers in $M_n$ are somewhat "spread out." For example, they range from 1 to $\sim2^{n^2}$, and if you place the elements of $M_n$ in strictly increasing order $$1=m_1 <\dots < m_{2^n-1}\sim 2^{n^2},$$ then it is not hard to prove that $\frac{m_{i+1}}{m_{i}}\leq 3$ for all $i=1,\dots, 2^n-2$.

**Where this question comes from**

We arrived at this question because a superlinear lower bound would give us new lower bounds on the "stabilizer rank" of $n$ copies of the quantum T-state, which would have important implications for classical simulation of quantum circuits.

To avoid posting an XY question, I briefly mention that the above question is a reduction from the following: We would like a lower bound on $c \in \mathbb{N}$ for which there exists a $\{0, \pm 1, \pm i\}$-valued matrix $A \in \mathbb{C}^{2^n-1 \times c}$ for which $M_n$, viewed as a $2^n-1$-dimensional vector, lies in the image of $A$. A lower bound for the main question of this post would imply a lower bound for the question in this paragraph, but not necessarily the other way around.