Calculating the Number of Vertices in Koch Snowflake I need help in trying to calculate the number of vertices in the different stages of the Koch flake. I know that in Stage $0$ there are $3$ vertices & in Stage $1$ there are $12$ vertices. I noticed that $3$ new vertices appeared on each edge in the Stage $1$ flake and I think that for every new edge that comes out there will be $3$ more vertices. However, I do not know how to write this in an equation. Any help will be greatly appreciated. 
 A: First, note that in every approximation of the von Koch curve, there are exactly as many edges as there are vertices.  You can think of the vertices as marks along the closed curve which partition that curve into segments.  Every segments needs a "starting" point and an "ending" point (as we traverse the curve counterclockwise, for example), and the ending point of the $k$-th segment serves as the starting point of the $(k+1)$-st segment.  Since the curve is a closed loop, the endpoint of the last segment is the starting point of the first segment.  Thus there is a one-to-one correspondence between segments and starting points (we can uniquely specify a segment by determining its starting point, or uniquely specify a point by determining the segment which starts at that point).  Therefore there are as many edges as vertices in each approximation.
(The above is a little hand-wavy and is meant to be informal, but it can be made rigorous if we require.  Note also that this argument implies that if we have any any closed curve made up of segments, then the number of segments and the number of vertices will be the same—in this sense, there is nothing special about approximations to the snowflake.)
To pass from the $n$-th approximation of the von Koch snowflake to the $(n+1)$-st, we replace every edge with four edges.  Hence the number of edges in the $(n+1)$-st approximation is four times the number of edges in the $n$-th approximation.  If we let $E_n$ denote the number of edges in the $n$-th approximation, this means that
$$ E_{n+1} = 4 E_{n}. $$
Using the fact that the number of edges and vertices are the same, this implies that
$$ V_{n+1} = 4 V_{n}, $$
where $V_n$ denotes the number of vertices in the $n$-th approximation of the von Koch curve.
This actually allows us to determine the number of vertices by recursion:  since the $0$-th approximation of the von Koch snowflake is an equilateral triangle, it has three vertices and three edges.  Hence $V_0 = 3$.  Thus we have
\begin{align}
V_0 &= 3 \\
V_{n+1} &= 4 V_n.
\end{align}
We can then show by induction (if nothing else) that
$$ \boxed{V_n = 3\cdot 4^n,} $$
which gives a closed-form expression for the number of vertices in the $n$-th approximation of the von Koch snowflake.
