Recurrent sequence in Ford Circle I have been working on this question for quite some time now but couldn't solve it. If anyone can help or give me some insight into the problem, I would very much appreciate it.
So imagine 2 touching circles that are also tangent to a horizontal line, it's possible to add another circle in the space between them, then more circles can be added through generation (say each generation add the maximum circles that are available to the spaces given the condition that it must be tangent to other circles and the horizontal line). This is basically Ford circles and the circles can be added continuously and infinitely through generations.
What I'm trying to do is to find the sequence of the length of the radius of the added circles after each generation.
The fact that there are many branches to consider but I'm looking specifically at the fastest sequence that the radii are changing the fastest. I've figured out that the Radius of the circle that is tangent to the nearest generation would have the smallest radius within the circles being added within that generation.
Here's a diagram I made for the problem
enter image description here
Say a radius R of any circles within generation n, would be the smallest if it's tangent to the circle from generation n-1 and n-2
We can find the relationship of the elements of the sequence of circles that have the fastest decreasing rate
$$
\frac{1}{\sqrt{r_{n}}} = \frac{1}{\sqrt{r_{n-1}}} + \frac{1}{\sqrt{r_{n-2}}}
$$
given that $r_0$ is defined (say 30cm), what is the length of the radius of the $n^{th}$ radius. or at which $n$ number that the circle would have the radius of, or close to $1.00 *10^-5cm$ for example
I found this paper that addresses exactly what I need (I assume) but my poor Math knowledge is not good enough to understand. It seems to be a recurrence relation or sequence I think
This is my first post so I very much apologize for any mistake that I made. I'm not a native speaker either so there might be some English issues, please tell me and I'll clarify if needed.
Thanks a lot
 A: Let's simplify the notation somewhat and express this in terms of the curvature $k=1/r$. Then your equation becomes $\sqrt{k_3}=\sqrt{k_1}+\sqrt{k_2}$. And for definiteness, let's assume that $k_1<k_2$. Now, on the next iteration there will be two circles generated. Since you want the smallest radius, we want
$$\sqrt{k_4}=\sqrt{k_3}+\max(\sqrt{k_1},\sqrt{k_2})=\sqrt{k_3}+\sqrt{k_2}$$
Now, repeating this process we'll find that
$$\sqrt{k_n}=\sqrt{k_{n-1}}+\sqrt{k_{n-2}}$$
If we now let $f_n=\sqrt{k_n}$ you can see that we have the classic Fibonacci recursion, but not Fibonacci sequence, because we do not have the right initial conditions. I have previously shown how to solve that problem here. In your case, this would be expressed as
$$
f_n = \left(f_1 - \frac{f_0}{2}\right) \frac{\varphi^n - \psi^n}{\varphi-
 \psi} + \frac{f_0}{2} (\varphi^n + \psi^n)
$$
or
$$
\sqrt{k_n} = \left(\sqrt{k_2} - \frac{\sqrt{k_1}}{2}\right) \frac{\varphi^n - \psi^n}{\varphi-
 \psi} + \frac{\sqrt{k_1}}{2} (\varphi^n + \psi^n); \quad n\ge3
$$
where $\varphi=(1+\sqrt{5})/2$ is the golden ratio and $\psi=1-\varphi=-1/\varphi$.
