# Density of numbers with multiple factors near square root

Fix constants $$1\leq \alpha<\beta$$. What is the density of the set of positive integers $$n$$ with at least two factors between $$\alpha\sqrt{n}$$ and $$\beta\sqrt{n}$$?

(I am specifically interested when $$\alpha=\sqrt{2}$$ and $$\beta=\sqrt{3}$$, and I am hoping the density is zero. I am not an expert in this field, so apologies in advance.)

Even one such factor gives you zero density. Indeed, if $$d \mid n$$, $$\alpha \sqrt{n} \le d \le \beta \sqrt{n}$$ then $$\frac{1}{\beta}\sqrt{n} \le \frac{n}{d} \le \frac{1}{\alpha}\sqrt{n}$$ therefore $$n$$ is a product of two numbers not greater than $$\max(\frac{1}{\alpha}, \beta)\sqrt{n}$$. Therefore amount of desired numbers in the interval $$[0, N]$$ is not greater than amount of numbers in the multiplicative table of size $$\max(\frac{1}{\alpha}, \beta)\sqrt{N} \times \max(\frac{1}{\alpha}, \beta)\sqrt{N}$$. And this number is $$o(\sqrt{N}^2) = o(N)$$, see this MO question.
• Unless I missed something, the other question you link to, deals with the whole multiplication table. What if the subtables around $\sqrt{N}$ have relatively few products overlapping? Then the number of distinct products there would be $O(N)$ rather than $o(N)$. Is it is easy to complete the argument? – Yaakov Baruch Jan 10 '19 at 11:14
• The whole table is $o(N^2)$. – Yaakov Baruch Jan 10 '19 at 11:16
• @YaakovBaruch no, as I wrote, the whole table is $o(\sqrt{N}^2)=o(N)$ as needed. – Aleksei Kulikov Jan 10 '19 at 11:21