Imagine covering a triangle ${S}_{0}$ with four congruent triangles ${x}_{1}$, ${x}_{2}$, ${x}_{3}$ and ${x}_{4}$ and that it is done in such a way that for any side of ${S}_{0}$ , three of the four triangles share the same vertex at the midpoint. It then follows that these four smaller triangles cover ${S}_{0}$ leaving no gaps or overlaps. Call the set of these four a tessellation ${S}_{1}$ of ${S}_{0}$ . If a mental picture of this is not evident, attempt to sketch it on paper. A tessellation of ${S}_{0}$ with more than just four triangles is easily created. The approach taken will require to think about the triangle whose vertices are the three midpoints of the sides of ${S}_{0}$ as not being tessellated, so name it ${x}_{4}$. Then take ${x}_{i}$, where $i\in \left[3\right]$, and tessellate it the same way ${S}_{0}$ was tessellated. A new tessellation ${S}_{2}$ of ${S}_{0}$ is obtained and is known as the second iteration of Sierpinski's Triangle. To obtain the third iteration of Sierpinski's Triangle ${S}_{3}$, continue to tessallate the triangles which tessellate ${x}_{i}$ with the exception of the middle fourth triangle. This approach is a process of iteration which can be continued indefinitely where every step of this process produces many triangles which are selfsimilar to each other. This makes it feasible to count the triangles of ${S}_{2}$, ${S}_{3}$, ${S}_{4}$, and ${S}_{5}$. For example, sketch ${S}_{2}$ and observe that the tessellation of the bottomleft triangle ${x}_{1}$ is made up of $4$ triangles and that this tessellated ${x}_{1}$ is selfsimilar to the tessellated triangles ${x}_{2}$ and ${x}_{3}$. It then follows that $\left{S}_{2}\right=4\left(3\right)+1=13$. For $n\ge 6$, a general forumula should be sought to enumerate the elements of ${S}_{n}$. The enumeration results for any iterates of Sierpinski's triangle can then be checked below.

