# How many unique vertices in octahedron based sphere approximation

Using a triangular facet approximation of a sphere based on Sphere Generation by Paul Bourke.

We take an octahedron and bisect the edges of its facets to form 4 triangles from each triangle.

 /\ /\ /\ / \ / \ /__\ / \ / \ /\ /\ / \ /______\ /__\/__\ / \ /\ /\ /\ /\ /\ / \ / \ / \ /__\/__\/__\ / \ / \ / \ /\ /\ /\ /\ /______________\ /______\/______\ /__\/__\/__\/__\ 0th generation 1st generation 2nd generation 

This happens for every face so 8 times for the first generation.

Between each generation new vertices are pushed to the surface of the sphere.

The number of facets will be $(4^\mathrm{generations}) \cdot 8$

Some facets will share vertices: in the $0^{th}$ generation there are 6 unique vertices (it's an octahedron).

How many unique vertices will there be in the $N^{th}$ generation?

The key is the Euler characteristic formula $$V-E+F = 2.$$
You've already calculated that $F=8\cdot 4^N$. You also know that all of the facets are triangles; that means that each triangle has three half-edges or $3F=2E$. That means $$2V - F = 4$$ or $$V = 2 + 4^{N+1}.$$
Observe that each vertex is shared by $6$ faces, except the six vertices of the original octahedron, which are always shared by only $4$ faces in each generation. Each face has $3$ vertices. The number of faces in generation $N$ is $8 \cdot 4^{N}$. Putting this together, we see that the number of vertices in that generation is $$\frac{3 \cdot 8 \cdot 4^{N} - 4 \cdot 6}{6} + 6 = 4^{N+1} + 2$$