For $n\in\Bbb N$, consider $n$ points $x_1,\ldots, x_n$ in the unit square $Q=[0,1]^2$. Let $f(x_1,\ldots, x_n)$ denote the minimal total edge length of a tree with $x_1,\ldots, x_n$ as vertices. Let $$ L_n=\sup_{x_1,\ldots, x_n\in Q}f(x_1,\ldots,x_n).$$ Then in Prove that there exists a graph with these points such that G is connected, it was asked to show that $$\tag1\forall n\ge 3\colon L_n\le c\sqrt n $$ is true for $c=10$ and false for $c=1$.
In my answer (see there), an inductive argument based on subdividing the square showed that $(1)$ is true for $c=2\sqrt 2\approx 2.828$, and has infinitely many counterexamples for $c=1$.
In a later addendum to my answer, I showed that $$ \tag2\exists \ell_0\colon \forall n\colon L_n<\ell_0+c\sqrt n$$ holds for any $c>\frac 4{\sqrt{2\sqrt 3}}\approx 2.149$ and dared to
Conjecture. $$\inf\{\,c\in\Bbb R\mid (2) \text{ is true}\,\}=\frac{4}{\sqrt{2\sqrt 3}}. $$
I dared to do so because the constant stems from the factor $\frac{\pi}{2\sqrt 3}$ known from the densest circle packing.
My question for you us: Is my conjecture true?
Silly me! After a good night's rest, this direction is of course the trivial part!
For $m\gg 0$, the lattice $\frac 1m\Bbb Z[\frac{1+i\sqrt 3}2]$ has $\frac {2m}{\sqrt 3}+O(1)$ rows of $m+O(1)$ points each in the unit square, so that's $n=\frac 2{\sqrt 3}m² +O(m)$ points with minimal distance $\frac 1m$ between them. Any tree with these vertices certainly has edge length $$\ell =\frac{n-1}m=\frac 2{\sqrt 3}m+O(1) =\frac{\sqrt 2}{\sqrt[4]3}\sqrt n+O(1)=\frac 2{\sqrt{2\sqrt 3}}\sqrt n+O(1).$$ I suppose, a spurious factor of $2$ somehow slipped into the calculations leading to the conjecture.