A friend of mine asked me to prove or disprove that:
$$ \left\lceil\frac{2}{2^{1/n}-1}\right\rceil=\left\lfloor\frac{2n}{\ln 2}\right\rfloor \forall n \in \mathbb{Z^+} $$
First of all, I run a python code to numerically compute the values upto $5000$ in order to conjecture about the truth of proposition. I got a positive result. Then I tried plotting to get a rough idea about the behavior of functions, the conjecture seemed to be true. Nonetheless, I couldn't prove it. Now I am really frustrated to know that the first counter example is $777,451,915,729,368$.
Is an existential proof possible without brute force?
I read about this problem some time ago...
You can easily see that the difference $$\frac{2n}{\ln 2} - \frac{2}{2^{\frac{1}{n}}-1} $$ is increasing and getting closer to one. If $\frac{2n}{\ln 2}$ is smaller than a certain integer $k$, but close enough to $k$ so that $\frac{2}{2^{1/n}-1}$ is greater than $k-1$, then you have found a counterexample. But asking for a $n$ such that $\frac{2n}{\ln 2}$ is well enough approximated by $k$ is the same as asking for a good enough rational approximation $\frac{k}{n}$ of $\frac{2}{\ln 2}$. The convergents of the continued fraction of $\frac{2}{\ln 2}$ are the best approximations you can get. You just have to calculate them until you reach a counterexample, and you'll probably have to compute just tens of convergents, not billions. $777,451,915,729,368$ should be the denominator of one of these convergents.