Trying to Resolve One Recursion with Two Solutions

Question

Background: I recently answered a question about the sequence of minimum Ford circles on each successive iteration here. I then asked myself the related question about the maximum circles on each iteration. Although this recursion formula was simpler, I found two different solutions by different methods.

Problem: I considered a more general problem than the Ford circles consisting of two arbitrary kissing circles sitting on a straight line (see the first figure below). However, for the analysis we consider only those circles that touch the straight line. This allows us to use the degenerate form of Descartes' theorem (since one curvature is zero). Thus, $\sqrt{k_3}=\sqrt{k_2}+\sqrt{k_1}$. The next circle of largest radius would be

$$ \sqrt{k_4}=\sqrt{k_3}+\min(\sqrt{k_1},\sqrt{k_2}) $$

Since $k_1$ is always the smallest curvature, we find that $$ \sqrt{k_n}=\sqrt{k_{n-1}}+\sqrt{k_1} $$

If we define $f_n=\sqrt{k_n}$ we have $f_n=f_{n-1}+f_1$, which we can solve by induction, such that

$$ \begin{align} &f_4=f_3+f_1=f_2+2f_1\\ &f_5=f_4+2f_1=f_2+3f_1\\ &...\\ &f_n=f_2+(n-2)f_1\\ \end{align} $$

The second figure below shows five levels of Apollonian circles for the same problem. The above result is in agreement with the result of circles labeled $k=81, 169, 289, 441, 625,...$

Solution by equation: Consider that we can eliminate the constant $f_1$ by down-indexing the equation and subtracting, hence

$$ \begin{align} &f_n=f_{n-1}+f_1\\ &f_{n-1}=f_{n-2}+f_1\\ &--------\\ &f_n=2f_{n-1}-f_{n-2}\\ \end{align} $$

Notice that this form will give the same induction solution as before.

A solution for the generalized Fibonacci sequence $f_n=af_{n-1}+bf_{n-2}$ with arbitrary initial conditions $f_{0,1}$ can be found here. The solution can expressed in various ways. The following one suits us best:

$$ f_n=\left(f_1-\frac{af_0}{2}\right) \frac{\alpha^n-\beta^n}{\alpha-\beta}+a\frac{f_0}{2} \frac{\alpha^n+\beta^n}{\alpha+\beta} $$

where $\alpha,\beta=(a\pm\sqrt{a^2+4b})/2$.

For the present problem we can show that $\alpha,\beta=1\pm0$. This may look like we have a zero divide, but note that

$$ \begin{align} \frac{\alpha^n-\beta^n}{\alpha-\beta}&=\sum_{k=1}^n\alpha^{n-k}\beta^{k-1}\\ &=n \text{ for } \alpha=\beta=1 \end{align} $$

Thus we end up with

$$ f_n=n(f_1-f_0)+f_0 $$

Since we know $f_1$ and $f_2$ but not $f_0$, we can show from above that $f_0=2f_1-f_2$ with this final result

$$ f_n=(n-1)f_2-(n-2)f_1 $$

This is clearly not the same as the induction solution are therein lies my problem.

EDIT: In response to comment by @jimjim I have added a colorized image of the packing by levels.

Answer 1

Note that if your original equation $\ f_n=f_{n-1}+f_1\ $ held for $\ n=2\ $ (presumably it doesn't necessarily) then you'd have $\ f_2=2f_1\ $ and both solutions would reduce to $\ f_n=nf_1\ $. If, however, that equation doesn't hold for $\ n=2\ $, then your derived equation $\ f_n=2f_{n-1}-f_{n-2}\ $ won't hold for $\ n=3\ $.

Since it holds for $\ n>3\ $, however, you can write its solution as $$ f_n=(f_3-f_2)n+3f_2-2f_3\ . $$ But your original equation for $\ n=3\ $ gives you $\ f_3=f_2+f_1\ $, and substituting this for $\ f_3\ $ in the above equation gives $$ f_n=(n-2)f_1+f_2\ , $$ the same as your original solution.