Erdős Problem #394

declared status 'open'. Formalized: yes. Let $t_k(n)$ denote the least $m$ such that$$n\mid m(m+1)(m+2)\cdots (m+k-1).$$Is it true that$$\sum_{n\leq x}t_2(n)\ll \frac{x^2}{(\log x)^c}$$for some $c>0$? Is it true that, for $k\geq 2$,$$\sum_{n\leq x}t_{k+1}(n) =o\left(\sum_{n\leq x}t_k(n)\right)?$$ Current best: This was proved by Erd\H{o}s and Hall [ErHa78], who proved that in fact$$\sum_{n\leq x}t_2(n)\ll \frac{\log\log\log x}{\log\log x}x^2.$$Erd\H{o}s and Hall conjecture that the sum is $o(x^2/(\log x)^c)$ for any $c<\log 2$. Since $t_2(p)=p-1$ for prime $p$ it is trivial that$$\sum_{n\leq x}t_2(n)\gg \frac{x^2}{\log x}.$$Erd\H{o}s and Hall [ErHa78] also note that $t_{n-1}(n!)=2$ and $t_{n-2}(n!)\ll n$, which $n=2^r$ shows is the best possible. They ask about the behaviour of $t_{n-3}(n!)$ and also ask ask whether, for infinitely many $n$,$$t_k(n!)< t_{k-1}(n!)-1$$for all $1\leq k<n$. Prize: no. OEIS: A344005. Tags: number theory.

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