The number of points in diameters defined by a subdivided hexagon

Question

Just as in the image, imagine that we have $n$ nested hexagons which have subdivided sides just as in the image i.e. the first inner hexagon has no subdivisions, it is just a regular hexagon, the second hexagon nested has its sides divided into two with a point on the center and so on. Hence in general the $n$th hexagon has its edges subdivided into $n$ equal parts.

Now, from the very outer hexagon, from each subdivision point as well as from each vertex we draw a diameter through the center of the hexagon. Does there exist a rule that determines the number of points these diameters must go through if we number points on one edge as $p_0,...,p_n$ where $p_1$ and $p_n$ would be the vertices of the hexagon itself.

Edit: in the image we have a diameter that goes through two subdivision points as well as the center.

Answer 1

If we just look at $\frac16$ of the hexagon, we get a triangle formed by the center and one of the sides. It's an equilateral triangle, but from the point of view of the intersection points, it might as well be a right triangle: map the center of the triangle to $(0,0)$ and the side to the line segment between $(n,0)$ and $(0,n)$. The subdivision points on the side are the points of the form $(x,n-x)$ for $1 \le x \le n-1$. The subdivision points inside the hexagon are the other lattice points in that triangle.

The line connecting $(x,n-x)$ to $(0,0)$ contains $\gcd(x,n-x) = \gcd(x,n)$ lattice points, counting $(0,0)$ but not $(x,n-x)$ itself (or vice versa). The corresponding diameter of the hexagon contains $2\gcd(x,n)-1$ subdivision points, counting the center: we subtract $1$ to avoid counting the center twice.

If the points on the side are numbered $p_0, p_1, \dots, p_n$, then $p_k$ corresponds to $(k,n-k)$ and so the diameter through $p_k$ contains $2\gcd(k,n)-1$ subdivision points. In the example, we are looking at $p_2$ and $n=6$, so we get $2\gcd(2,6)-1 = 2 \cdot 2 - 1 = 3$ points, as desired.