Apollonian gasket 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 


*

*Is there anywhere this will fail ?

*Can it be more simplified to get $R_n$ in terms $n$ 

*Is there any faster way to calculate for large numbers as $n=10^8$

 A: 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.
