PROVED
This has been solved in the affirmative.
A minimal cut of a graph is a minimal set of vertices whose removal disconnects the graph. Let $c(n)$ be the maximum number of minimal cuts a graph on $n$ vertices can have.
Does $c(n)^{1/n}\to \alpha$ for some $\alpha <2$?
Asked by Erdős and Nešetřil, who also ask whether $c(3m+2)=3^m$. Seymour observed that $c(3m+2)\geq 3^m$, as seen by the graph of $m$ independent paths of length $4$ joining two vertices.
Solved by Bradač
[Br24], who proved that $\alpha=\lim c(n)^{1/n}$ exists and\[\alpha \leq 2^{H(1/3)}=1.8899\cdots,\]where $H(\cdot)$ is the binary entropy function. Seymour's construction proves that $\alpha\geq 3^{1/3}=1.442\cdots$. Bradač conjectures that this lower bound is the true value of $\alpha$.
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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 #150, https://www.erdosproblems.com/150, accessed 2026-01-16