OPEN

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 $i$?

#460: [ErGr80]