Erdős Problem #261

declared status 'open'. Formalized: no. 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? Current best: Erd\H{o}s 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. Prize: no. Tags: number theory.

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