In geometry, a kissing number 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 $n$ dimensions. Kabatiansky and Levenshtein proved the following asymptotic upper bound (ppi1518, mr514023, 1978): $$\tau_n \le 2^{0.401n(1+o(1))} = (1.32\dots)^{n(1+o(1))}$$

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, Section 5.

In geometry, a kissing number 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 $n$ dimensions. Kabatiansky and Levenshtein proved the following asymptotic upper bound (ppi1518, mr514023, 1978): $$\tau_n \le 2^{0.401n(1+o(1))} = (1.32\dots)^{n(1+o(1))}$$

Question: What is the smallest $\alpha$ such that $\tau_n \le \alpha^n$, for all $n$?
($\alpha := \min_{n \ge 1} \tau_n^{1/n}$)

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 $n \le 24$, $\tau_n^{1/n} \le \sqrt 6$. Is it true that $\alpha = \sqrt 6$?

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

In geometry, a kissing number 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, Section 5.

In geometry, a kissing number 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 $n$ dimensions. Kabatiansky and Levenshtein proved the following asymptotic upper bound (ppi1518, mr514023, 1978): $$\tau_n \le 2^{0.401n(1+o(1))} = (1.32\dots)^{n(1+o(1))}$$

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, Section 5.