Blumenthal's problem. Szekeres [Sz41] showed that\[\alpha_{2^n+1}> \pi \left(1-\frac{1}{n}+\frac{1}{n(2^n+1)^2}\right)\]and\[\alpha_{2^n}\leq \pi\left(1-\frac{1}{n}\right).\]Erdős and Szekeres [ErSz60] showed that\[\alpha_{2^n}=\alpha_{2^n-1}= \pi\left(1-\frac{1}{n}\right),\]and suggested that perhaps $\alpha_{N}=\pi(1-1/n)$ for $2^{n-1}<N\leq 2^n$. This was disproved by Sendov [Se92].

Sendov [Se93] provided the definitive answer, proving that $\alpha_N=\pi(1-1/n)$ for $2^{n-1}+2^{n-3}<N\leq 2^n$ and $\alpha_N=\pi(1-\frac{1}{2n-1})$ for $2^{n-1}<N\leq 2^{n-1}+2^{n-3}$.

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This page was last edited 16 October 2025.