# $\tan (n) > n$ for infinitely many positive integers

I heard the following problem is open:

$\tan(n ) > n$ for infinitely many positive integers in radians.

Does anyone know if it is still open or if any progress has been made on this problem?

• @Raskolnikov Not all positive integers $n$, but for infinitely many positive integers one. Commented Jul 27, 2013 at 4:27
• mathworld.wolfram.com/TancFunction.html Commented Jul 27, 2013 at 4:31
• See the paper by Bellamy et al. here: math.udel.edu/~lazebnik/papers/tan_n.pdf I saw this referred to on OEIS.org in relation to the sequence A088306 after posting the less interesting sequence A249836 myself, on purely recreational grounds.
– user203812
Commented Dec 29, 2014 at 20:04

You are needing $n$ such that there is an integer $k$ with, roughly:
$$(2k+1)\pi- 2\epsilon_n < 2n < (2k+1)\pi$$ with $\epsilon_n =\arctan\frac{1}{n}$.
Dividing by $2(2k+1)$, that means we need a $n$ so that:
$$\frac{\pi}2 - \frac{\epsilon_n}{(2k+1)}<\frac{n}{2k+1} < \frac{\pi}2$$ So if we are seeking odd convergents in the continued fraction expansion for $\frac{\pi}{2}$ with an odd denominator, and the next coefficient is reasonably large. The first odd convergent with odd denominator is $\frac{1}{1}$. The next is $\frac{344}{219}$, giving $n=344$, but it's not quite good enough, since $\tan 344 = 227.5\dots$. (The next coefficient is $1$, so this is not entirely a surprise.)
Going to have to go further out to find $n$, which, for me, is going to require a much better approximation for $\pi$ to get the calculations to work.