Apollonian gasket

Question

Okay , is there a way to find the radius of the nth circle in a apollonian gasket ..

Something like this

Its like simple case of apollonian gasket ..

I found from descartes' theorem

$R_n = 2\cdot\sqrt{ R_{n-1}\cdot R_1 + R_n-1\cdot R_2 + R_1\cdot R_2} +R_{n-1} + R_1 +R_2 $

I have the values of $R_1,R_2,R_3,R_4$ ..

where $R_n$ is the curvature and not the radius So I have this question

1. Is there anywhere this will fail ?
2. Can it be more simplified to get $R_n$ in terms $n$
3. Is there any faster way to calculate for large numbers as $n=10^8$

Answer 1

Sure, there is a way.

To simplify the formula, I will relabel your circles as follows.

• Let $S_a$ and $S_b$ be your circle $C_1$ and $C_2$.
• Let $S_0, S_1, S_2, \ldots$ be your circle $C_3, C_4, C_5, \ldots$.
• Given $C_1, C_2, C_3$, there are two possible configurations of $C_4$.
  Let $S_{-1}$ be the other possible configuration of $C_4$ differ from $S_{1}$.

Let $r_n$ be the radius of circle $S_n$ where $n = a, b$ or $\ge -1$ and $\rho_n = \frac{1}{r_n}$.

Recall $S_a$ is the outer circle. If we apply Descartes circle theorem to circles $S_a, S_b, S_{n}, S_{n\pm 1}$ for $n \ge 0$, we have

$$ 2( \rho_a^2 + \rho_b^2 + \rho_n^2 + \rho_{n\pm 1}^2 ) = (-\rho_a + \rho_b + \rho_n + \rho_{n\pm 1})^2$$

So $\rho_{n\pm 1}$ are the two roots of a quadratic equation.

$$\rho^2 - 2(-\rho_a + \rho_b + \rho_n ) \rho + \left(2(\rho_a^2 + \rho_b^2 + \rho_n^2 ) - (-\rho_a + \rho_b + \rho_n)^2\right) = 0\tag{*1}$$ This implies $$\rho_{n+1}-2\rho_n + \rho_{n-1} = 2A \quad\text{ where }\quad A = \rho_b - \rho_a = \frac{1}{r_b} - \frac{1}{r_a}$$

This is a linear recurrence relation in $\rho_n$ and its general solution has the form $$\rho_{n} = A n^2 + B n + C$$ In particular, we have $$\begin{cases} \rho_1 &= A + B + C\\ \rho_0 &= C\\ \rho_{-1} &= A - B + C\\ \end{cases} \implies \begin{cases} B &= \frac12\left( \rho_1 - \rho_{-1} \right)\\ C &= \rho_0 \end{cases}$$ This leads to

$$\rho_n = (\rho_b - \rho_a) n^2 + \frac12 (\rho_1 - \rho_{-1}) n + \rho_0$$

You can obtain the values of $\rho_{\pm 1}$ by solving the quadratic equation $(*1)$ for $n = 0$. To match your diagram, $\rho_1$ should be the smaller one of the two roots.