Problem Statement
Are there infinitely many $n$ such that there exists some $t\geq 2$ and distinct integers $a_1,\ldots,a_t\geq 1$ 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?
Categories:
Number Theory
Progress
Related to [260].In [Er88c] Erdős notes that Cusick had a simple proof that there do exist infinitely many such $n$. Erdős does not record what this was, but a later paper by Borwein and Loring [BoLo90] provides the following proof: for every positive integer $m$ and $n=2^{m+1}-m-2$ we have\[\frac{n}{2^n}=\sum_{n<k\leq n+m}\frac{k}{2^k}.\]Tengely, Ulas, and Zygadlo [TUZ20] have verified that all $n\leq 10000$ have the required property.
In [Er88c] Erdős weakens the second question to asking for the existence of a rational $x$ which has two solutions.
Source: erdosproblems.com/261 | Last verified: January 14, 2026