Show that there are no even natural numbers between $\frac{\pi}{10^{-k}}$ and $\frac{\pi}{\arctan (10^{-k})}$ for integer $k \ge 2$ Show that there are no even natural numbers between $\frac{\pi}{10^{-k}}$ and $\frac{\pi}{\arctan (10^{-k})}$ for integer $k \ge 2$. I encountered this while solving a physics problem and I am not even sure if it is true or not.
I tried to bound the difference of the two using the fact that $x > \arctan (x)$ for $x > 0$, but still there can be cases like $\frac{\pi}{10^{-k}} = n + 0.999999999$ for some integer $n$, regardless of how small the bound is.
Can anyone help?
EDIT : I have found Galperin's paper and it seems it is still left as a conjecture. All sources are in the comments of the first answer.
 A: We have $$ \frac{\pi}{\arctan (x)}-\frac{\pi}{x}=\frac{\pi}{x}\left(\frac{x}{\arctan(x)}-1\right)=\frac{\pi}{x}\left(\frac{x^2}{3}+O(x^4)\right)=\frac{\pi}{3}x + O(x^3) $$
Then the difference is around $\frac{\pi}{3}x \approx 10^{-k}$
To find an integer inbetween, then, roughly amounts to find, after $k$ digits in the decimal expansion of $\pi$, a sequence of at least $k$ consecutive nines, for some $k$.
This seems very unlikely. But also very difficult to prove. I don't think probabilistic approaches, assuming $\pi$ is normal, would lead anywhere.
Edit: as pointed out in the comments, the paper "Playing pool with π (the number π from a billiard point of view)" studies this (sec 10, question 1). It basically agrees with the previous paragraphs (the property "holds if and only if the string of $2N$ first decimal
digits of $\pi$ contains $N − 1$ nines in its right half"). It concedes that "modern mathematics is powerless to answer it" but it also conjectures -with almost certainity- that it's not true. Some probabilistic analysis is provided.
