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Will Sawin
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I'll use Christian Remling's notation: We want to show that we can always find an empty interval of size $\epsilon_n$ among any set of $n$ runners.

The Masked Avenger gives an argument for $\epsilon_n =1/n$, and suggests that $\epsilon_n=2/n$ may be close to optimal by an argument I can't follow.

Assuming the lonely runner conjecture for $n+1$, we get an argument for $\epsilon_n =2/n+1$ - just add one more runner, randomly placed, and wait until they're lonely.

In terms of actually provable statements, I was only able to get a very slight improvement - $\epsilon_n = 1/n + c/n^{3/2}$ for some $c$. For this argument, we may assume without loss of generality that the speeds are integers. Consider each runner's position on the circle as a unit complex number function of time, and add them up. This gives a periodic function whose Fourier series $\hat{f}(n)$ is $0$ if $n$ is not the speed of the runner and, if $n$ is the speed fo a runner, the starting position of that runner. So $\hat{f}(n)$ has $L_2$ norm $\sqrt{n}$, and $f(t)$ has $L_2$ norm $\sqrt{n}$. This means at some time it must take a value at least $\sqrt{n}$. We can easily see that if all the gaps are at most $1/n+\delta$, each runner can be no further than $n/\delta$ from a perfectly even position, meaning summing over the runners gives $\sqrt{n}\geq n^2/\delta$, with some constant thrown in there, which gives our result.

On the other hand we get $\epsilon_n < 2/\sqrt{\log n}$ from Christian Remling's argument. So we have an exponential gap to close.

Will Sawin
  • 148.4k
  • 9
  • 324
  • 563