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 [ 3 ] , 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 self-similar 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 bottom-left triangle x 1 is made up of 4 triangles and that this tessellated x 1 is self-similar to the tessellated triangles x 2 and x 3 . It then follows that | S 2 | = 4 ( 3 ) + 1 = 13 . For n 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.

Counting Number of Triangles of S n
| S n |