So here is the question: If we look at the Sierpinski triangle (left column of attached image) and think about how many triangle's it takes to make the shape at each iteration we can get the sequence $\frac{3^n -1}{2}$ for $n \in N$. If we look at another shape that we'll call the Inverted Sierpinski triangle (right column of attached image) where we take the center most triangle and turn it into 4 triangles we can find a sequence to give use the number of triangles as $3n+1$ for $n \in N$. Both sequences diverge to infinity. But do they go to the same infinity? e.g. Which shape has more triangles after an infinite number of iterations?
Now I think that they go to different infinity's here is my reasoning. If I look at the sequence $\{(\frac{3^n -1}{2})- (3n+1)\}= \{0,0,6,30,108...\}$ this seems to suggest to me that after the second iteration of drawing the shapes I can take the triangles in the Inverted Sierpinski triangle and match them 1-1 with some of the triangles in the Sierpinski triangle but I'll have left overs after every iteration. Is it enough to claim that because this series $\{(\frac{3^n -1}{2})- (3n+1)\}$ diverges that the two shapes contain a different infinity of triangles? If not how would one show this?
PS: If the second shape has an actual name can someone point that out as well?
PPS: I didn't name the second shape it comes from the classmate who got us started in thinking about the problem.