Erdős Problem #460

declared status 'open'. Formalized: no. Let $a_0=n$ and $a_1=1$, and in general $a_k$ is the least integer $>a_{k-1}$ for which $(n-a_k,n-a_i)=1$ for all $1\leq i<k$. Does$$\sum_{i}\frac{1}{a_i}\to \infty$$as $n\to \infty$? What about if we restrict the sum to those $i$ such that $n-a_j$ is divisible by some prime $\leq a_j$, or the complement of such $i$? Current best: This question arose in work of Eggleton, Erd\H{o}s, and Selfridge, who could prove that $a_k <k^{2+o(1)}$ for $k$ large enough depending on $n$, but conjectured that in fact $a_k\ll k\log k$ is true. Prize: no. Tags: number theory.

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