Let $q_n$ denote the $n$-th square-free number. By Chinese remainder theorem (see this post), it is not difficult to show that there is arbitrarily large gap between two consecutive square-free numbers, i.e., $\limsup_{n\to\infty}(q_{n+1}-q_n)=\infty$. How to prove the stronger bound? $$\limsup_{n\to\infty}\frac{q_{n+1}-q_n}{\log n/\log\log n}\geq \frac{1}{2}.$$ I don't how to get started, thanks for any help.
Edit: This is an exercise (Exercise 2.20) from A.J. Hildebrand's An introduction to analytic number theory.
I was curious about how large that ratio could stay. I had it print out only when the ratio exceeded 1.5. Does not seem to be dying out. The later part of that exercise says prove the limsup is at least $\pi^2/12 \approx 0.822467$ From what I am seeing, a limsup between 1 and 2 seems a good bet. Probably hard to prove, though.
First, just when the ratio increases:
Now, ratio larger than 1.5
Worth noting that the actual $q_{n+1} - q_n$ are quite small, near the end of this output just 9 or 10. Eh, shorter output, I told it to keep going forever and just print when the gap size increases. Here it is so far:
This last version is extended at https://oeis.org/A020754 and https://oeis.org/A020754/b020754.txt