If $\delta'(n)$ is the density of integers which have exactly one divisor in $(n,2n)$ then is it true that $\delta'(n)=o(\delta(n))$?

SOLVED

Let $\delta(n)$ denote the density of integers which are divisible by some integer in $(n,2n)$. What is the growth rate of $\delta(n)$?

If $\delta'(n)$ is the density of integers which have exactly one divisor in $(n,2n)$ then is it true that $\delta'(n)=o(\delta(n))$?

Besicovitch [Be34] proved that $\liminf \delta(n)=0$. Erdős [Er35] proved that $\delta(n)=o(1)$. Erdős [Er60] proved that $\delta(n)=(\log n)^{-\alpha+o(1)}$ where
\[\alpha=1-\frac{1+\log\log 2}{\log 2}=0.08607\cdots.\]
This estimate was refined by Tenenbaum [Te84], and the true growth rate of $\delta(n)$ was determined by Ford [Fo08] who proved
\[\delta(n)\asymp \frac{1}{(\log n)^\alpha(\log\log n)^{3/2}}.\]

Among many other results in [Fo08], Ford also proves that the second conjecture is false, and more generally that if $\delta_r(n)$ is the density of integers with exactly $r$ divisors in $(n,2n)$ then $\delta_r(n)\gg_r\delta(n)$.