How to calculate number of triangles and points after dividing a triangle n times?

Question

When having a triangle and dividing it n times, how to get the number of triangles and points?

      a
     / \
    /   \
   f-----d
  / \   / \
 /   \ /   \
c-----e-----b 

edit: This triangle is part of an icosahedron as base geometry for a sphere, left it out due to jitter information. Then the assumptions took place.

Answer 1

Each division multiplies the number of small triangles by $4$, and you start with one triangle at $0$ divisions, so the number of triangles after $n$ divisions is $4^n$.

The number of distinct horizontal sides of small triangles after $n$ divisions is

$$\sum_{k=1}^{2^n}k=\frac{2^n(2^n+1)}2=2^{n-1}(2^n+1)\;.$$

There are the same number of distinct sides of small triangles parallel to each of the non-horizontal sides of the original triangle, so there are altogether $3\cdot2^{n-1}(2^n+1)$ distinct sides of small triangles after $n$ divisions.

Now let $p_n$ be the number of vertices after $n$ divisions; clearly $p_0=3$. If $n\ge 1$, there are $3\cdot 2^{n-2}(2^{n-1}+1)$ small sides after $n-1$ divisions, and the $n$-th division adds one vertex for each of those sides, so

$$p_n=p_{n-1}+3\cdot2^{n-2}(2^{n-1}+1)\;.$$

Thus,

$$\begin{align*} p_n&=p_0+\sum_{k=1}^n3\cdot 2^{k-2}(2^{k-1}+1)\\\\ &=3+3\sum_{k=1}^n\left(2^{2k-3}+2^{k-2}\right)\\\\ &=3+\frac32\left(\sum_{k=0}^{n-1}4^k+\sum_{k=0}^{n-1}2^k\right)\\\\ &=3+\frac32\left(\frac{4^n-1}3+2^n-1\right)\\\\ &=2^{2n-1}+3\cdot2^{n-1}+1\;. \end{align*}$$