I found this somewhere on the web:
Theorem. The total number of vertices for $n$ squares that share exactly one common vertex is given by the formula $f(n) = 3n + 1$.
Proof. Each of the $n$ squares has $3$ vertices that are not shared with any other square; this gives $3n$ unshared vertices in all. In addition there is a single vertex that is shared among all the squares. So the total number of vertices is $3n + 1$.
So I've drawn the above theorem for $n=4$ ($4$ squares): And we get $3*4+1=13$, the total number of vertices should be $13$. So this is correct based on the drawing (we have to count every edge that share vertices drawn with a black dot):
the red dots are vertices that are not shared.
Does this have any connections to the Collatz Conjecture? Since the odd case in standard C.C. is $3n + 1$. Are there any graph visualization of the C. C. when it comes to vertices of graphs like this except for the typical Collatz-trees that we see on wikipedia or books. I hope this is not to vague a question. Where can I find information about this in geometric terms and visualizations of all the details around graphs with squares and connections like this?
If its too a vague question, I want some ideas for what kinds of question except the number of vertices we can ask by the geometry of this problem.
It looks like the theorem statement "shares exactly one" is causing confusion. Your counterexample has the center vertex, which is shared among all 4 squares, however in addition to this there are four vertices which are shared among 2 squares. Hence 13-4=9.
This breaks the conditional statement. Perhaps the theorem would've been better read had it been written "The total number of vertices for $n$ squares that share exactly 1 common vertex where no other vertices are shared is given by the function $f(n)=3n+1$."
This is someone explained in the proof, where each square has 3 vertices which are not shared by any other square, hence $3n$ vertices with +1 as the only shared vertex.