Lets define following function $f : \mathbb{N} \rightarrow \mathbb{N}$
First consider arbitraty four points on plane $\mathbb{R}^{2}$, that are corner of some square. We write then $f(1)=4$
Later we mark each middle point between any pair of previous set of points. Now we count number of new points, and forget about the past ones this will be our number $f(2)$ (i claim that $f(2)=5$)
We continue that procedure.
Find formula for $f(k)$
If it is impossible find lower bound of that function.
I tried to apply some theory about acting groups on set, because in general for $f(k)$ we have at most ${f(k)\choose 2}$ points, but we must be aware of repetition. The repetition comes from symmetries and that is why i seriously thinking about usege of group theory.
Generation 3 forms a square with the centre removed, that can go at integer coordinates from $-1$ to $1$.
Double coordinates at every stage, I think you get an integer square, with point of even coordinates removed.
If generation $2n-1$ runs from $-m$ to $m$, then generation $2n$ runs from $-2m$ to $2m$.
If generation $2n$ runs from $-m$ to $m$ then generation $2n+1$ runs from $1-2m$ to $2m-1$ because the outer layer is wiped out, the midpoints are all from previous generations.
I think the square sides are either $(2^{gen}+10)/6$ or $(2^{gen}+14)/6$, and the removed square of previous points respectively $(2^{gen}+4)/12$ and $(2^{gen}+20)/12$.
That gives the following sequence. The first two are different, but from the third generation on they are the difference of two squares mentioned above.
$$4,5,8,16,40,120,408,...$$