How to calculate number of triangles and points after dividing a triangle n times? When having a triangle and dividing it n times, how to get the number of triangles and points?
      a
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   f-----d
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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.

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