SOLVED
This has been resolved in some other way than a proof or disproof.
Let $N\geq 1$. How many $A\subseteq \{1,\ldots,N\}$ are there such that $\sum_{n\in A}\frac{1}{n}=1$?
It was not even known for a long time whether this is $2^{cN}$ for some $c<1$ or $2^{(1+o(1))N}$. In fact the former is true, and the correct value of $c$ is now known.
- Steinerberger [St24] proved the relevant count is at most $2^{0.93N}$;
- Independently, Liu and Sawhney [LiSa24] gave both upper and lower bounds, proving that the count is\[2^{(0.91\cdots+o(1))N},\]where $0.91\cdots$ is an explicit number defined as the solution to a certain integral equation;
- Again independently this same asymptotic was proved (with a different proof) by Conlon, Fox, He, Mubayi, Pham, Suk, and Verstraëte [CFHMPSV24], who prove more generally, for any $x\in \mathbb{Q}_{>0}$, a similar expression for the number of $A\subseteq \{1,\ldots,N\}$ such that $\sum_{n\in A}\frac{1}{n}=x$;
- The above papers all appeared within weeks of each other in 2024; in 2017 a similar question (with $\leq 1$ rather than $=1$) was asked on MathOverflow by Mikhail Tikhomirov and proofs of the correct asymptotic were sketched by Lucia, RaphaelB4, and js21.
See also
[362].
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Additional thanks to: Zachary Chase
When referring to this problem, please use the original sources of Erdős. If you wish to acknowledge this website, the recommended citation format is:
T. F. Bloom, Erdős Problem #297, https://www.erdosproblems.com/297, accessed 2026-04-11
