1

We call it an ugly number if a positive integer can be written as $2^a3^b$, where $a,b$ are non-negative integers. Prove that the interval $(n^2,(n+1)^2),n\in\mathbb{N}^*$ has at most $2$ such numbers.

I tried to count the number in the interval $[1,N]$ of such numbers by inclusion and exclusion principle, $$ s(N)=\sum_{k\ge1,(k,2)=(k,3)=1}\mu(k)\left[\frac{N}{k}\right], $$ but it's very hard to make proper estimations.

Though the problem maybe a simple task, but it had really bothered me for a long period. Could you offer me some hints so that I can proceed? Thanks for your help.

Sakup2485
  • 467
  • 1
    Surely the first thing to do would be a simple search. How far have you checked this? Is there any other reason to imagine that the claim is true? – lulu Apr 12 '25 at 11:32
  • If you are ignoring "ugly" squares as not being in any $(n^2,(n+1)^2)$ then the only small examples of $2$ such numbers are ${2^1 3^0,2^0 3^1}$, ${2^1 3^1,2^3 3^0}$, ${2^1 3^2,2^3 3^1}$, ${2^0 3^3,2^5 3^0}$. – Henry Apr 12 '25 at 12:11
  • I suspect that proving what you're stating would involve something similar to what I wrote in another comment, i.e., using this answer which states "we can use the [Rhin bound][1] (page 160: $|µ_1\log 2+µ_2\log 3|\geq H^{-13.3}$ with $H=max(|µ_1|,|µ_2|)$)". – John Omielan Apr 12 '25 at 14:23

0 Answers0