Number of circles in Apollonian fractals

Question

I am studying Apollonian fractals.

According to this web page, number of circles in Apollonian fractals is $2+3^{(n+1)}$, which n means stage.

How can I prove it? I guess the number of the circles each stage depends on the initial conditions. Am I wrong?

Answer 1

You start with three circles initially, that are not going to be counted initially.

Then you add the inner and outer tangent circles : $a_0=2$.

From then on every circle from stage $n$ is going to be surrounded by three new circles at stage $n+1$, therefore the amount of circles added at stage $n+1$ is $$a_{n+1}=3a_n$$

Given that you start $a_0=2$ it follows that $a_n=2\cdot 3\cdot\ldots\cdot 3=2\cdot3^{n}$

Now the total number of circles at stage $n$ is $$a_0+\ldots+a_n=2\cdot(3^0+\ldots+3^n)$$ which is $$2\cdot\frac{3^{n+1}-1}{3-1}$$that is $3^{n+1}-1$.

Finally, add in the three initial circles: $3^{n+1}-1+3=3^{n+1}+2$.

Answer 2

If you are happy that the number of new circles at each stage is $2\times 3^{n}$ then the total is obtained by summing a geometrical progression.

$$3+2(1+3+3^2+...+3^{n})=3+2(\frac {3^{n+1}-1}{3-1})$$$$=2+3^{n+1}.$$