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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.)

DCT
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1 Answers1

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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.

Aleksei Kulikov
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  • @YaakovBaruch you are right that we probably have not the whole table (that is why I wrote “not greater” in my answer) but nevertheless we definitely can’t have more elements than in the whole table since our set is a subset of the elements of the table. – Aleksei Kulikov Jan 10 '19 at 11:14
  • 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
  • Apologies for editing my comment and pushing it below yours... – Yaakov Baruch Jan 10 '19 at 11:15
  • 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
  • Yes, you are right. Nice proof. – Yaakov Baruch Jan 10 '19 at 11:26