Problem Statement
Let $r(n)$ count the number of $d_1,d_2$ such that $d_1\mid n$ and $d_2\mid n$ and $d_1<d_2<2d_1$. Is it true that, for every $\epsilon>0$,\[r(n) < \epsilon \tau(n)\]for almost all $n$, where $\tau(n)$ is the number of divisors of $n$?
Categories:
Number Theory Divisors
Progress
This is false - indeed, for any constant $K>0$ we have $r(n)>K\tau(n)$ for a positive density set of $n$. Kevin Ford has observed this follows from the negative solution to [448]: the Cauchy-Schwarz inequality implies\[r(n)+\tau(n)\geq \tau(n)^2/\tau^+(n)\]where $\tau^+(n)$ is as defined in [448], and the negative solution to [448] implies the right-hand side is at least $(K+1)\tau(n)$ for a positive density set of $n$. (This argument is given for an essentially identical problem by Hall and Tenenbaum [HaTe88], Section 4.6.)See also [448].
Source: erdosproblems.com/449 | Last verified: January 15, 2026