Let $\tau(n)$ count the divisors of $n$ and $\tau^+(n)$ count the number of $k$ such that $n$ has a divisor in $[2^k, 2^{k+1})$. Is it true that, for all $\epsilon > 0$, $$\tau^+(n) < \epsilon \cdot \tau(n)$$ for almost all $n$?

Let $\tau(n)$ count the divisors of $n$ and $\tau^+(n)$ count the number of $k$ such that $n$ has a divisor in $[2^k, 2^{k+1})$. Is it true that, for all $\epsilon > 0$, $$\tau^+(n) < \epsilon \cdot \tau(n)$$ for almost all $n$?

This is false, and was disproved by Erdős and Tenenbaum [ErTe81], who showed that in fact the upper density of the set of such $n$ is $\asymp \epsilon^{1-o(1)}$ (where the $o(1)$ in the exponent $\to 0$ as $\epsilon \to 0$). A more precise result was proved by Hall and Tenenbaum [HaTe88] (see Section 4.6), who showed that the upper density is $\ll \epsilon \log(2/\epsilon)$. Hall and Tenenbaum further prove that $\tau^+(n)/\tau(n)$ has a distribution function. Erdős and Graham also asked whether there is a good inequality known for $\sum_{n \leq x} \tau^+(n)$. This was provided by Ford [Fo08] who proved $$\sum_{n \leq x} \tau^+(n) \asymp x\frac{(\log x)^{1-\alpha}}{(\log\log x)^{3/2}}$$ where $$\alpha = 1-\frac{1+\log\log 2}{\log 2} = 0.08607\cdots.$$