> In geometry, a [kissing number][1] is defined as the number of
> non-overlapping unit spheres that can be arranged such that they each
> touch another given unit sphere.

Let $\tau_n$ be the kissing number in $n$ dimension.  
Kabatiansky-Levenshtein (1978) proved the following *asymptotic* upper bound: $$\tau_n \le 2^{0.401n(1+o(1))} = (1.32\dots)^{n(1+o(1))}$$

**Question**: What is the smallest $\alpha$ such that $\tau_n \le \alpha^n$, for all $n$?  

By using volume, we can prove that $\tau_n \le \frac{Vol(B(3))-Vol(B(1))}{Vol(B(1))}=3^n-1$, so $\alpha \le 3$.

Now $\tau_2 = 6$, so $\alpha \ge \sqrt 6 \simeq 2.45$. Moreover, for $n \le 24$, $e^{\ln(\tau_n)/n} \le \sqrt 6$. Is it true that $\alpha = \sqrt 6$?  
  

This post is motivated by [arXiv:1710.00285][2], Section 5.  


  [1]: https://en.wikipedia.org/wiki/Kissing_number_problem
  [2]: https://arxiv.org/pdf/1710.00285.pdf