OPEN
Are there infinitely many $n$ such that there exists some $t\geq 2$ and $a_1,\ldots,a_t\geq 1$ such that
\[\frac{n}{2^n}=\sum_{1\leq k\leq t}\frac{a_k}{2^{a_k}}?\]
Is this true for all $n$? Is there a rational $x$ such that
\[x = \sum_{k=1}^\infty \frac{a_k}{2^{a_k}}\]
has at least $2^{\aleph_0}$ solutions?
Related to
[260].
In [Er88c] Erdős notes that Cusick had a simple proof that there do exist infinitely many such $n$. Erdődoes not record what this was, but Kovač has provided the following proof: for every positive integer $m$ and $n=2^{m+1}-m-2$ we have
\[\frac{n}{2^n}=\sum_{n<k\leq n+m}\frac{k}{2^k}.\]
