I had asked this on reddit and someone suggested that I try here:
Assuming that the pattern in the image below continues infinitely, how much would each generation of circles have to decrease to avoid any overlap/collision. In my example the diameter of each circle is 1/3 of its parent circle. As you can see that's not small enough and overlap occurs quickly.
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Define the sequence of $\theta_n$ to be,
\begin{cases} 0 \ , & \text{if $n=0$} \\ \cfrac{3\pi}{4}\ , & \text{if $n=1$} \\ \pi \ , & \text{$n \gt 1$} \end{cases}
Visually speaking, this is the sequence that tries to cross back on the original circle the "hardest". The exceptions are $\theta_0$ and $\theta_1$. The initial angle has to be $0$. The second angle can't be $\pi$ or it automatically hits the circle. The other angles are $\pi$ because we want the angle that takes the most direct route to the circle, but ends up just touching the edge of the first circle.
Then, as noted by Mark McClure, we have a representation with complex numbers,
$$(1) \quad \left|\sum_{k=0}^{\infty} (1+r) \cdot r^k e^{\theta_k i}\right| > 1$$
Because $\theta_n=\pi$ for $n \gt 1$, we should analyze the sum starting with $k=2$.
$e^{\pi i}=-1$, so we immediately have,
$$(2) \quad \sum_{k=2}^{\infty} (1+r) \cdot r^k e^{\theta_k i}=-\sum_{k=2}^{\infty} (1+r)r^k=\cfrac{r^2 \cdot (r+1)}{r-1}$$
So, if we substitute the value found in $(2)$ into $(1)$, we get,
$$(3) \quad \left|(1+r)+(1+r) \cdot r \cdot \cfrac{-\sqrt{2}+\sqrt{2}i}{2}+\cfrac{r^2 \cdot (r+1)}{r-1}\right| > 1$$
We can expand this using algebra, I prefer equalities,
$$(3.9) \quad \sqrt{(2-\sqrt{2}) \cdot r^4+2\sqrt{2} \cdot r^2-(2+\sqrt{2}) \cdot r+1} \cdot \cfrac{r+1}{1-r} = 1$$
More conducive for manipulations, we have,
$$(4) \quad \left((2-\sqrt{2}) \cdot r^4+2\sqrt{2} \cdot r^2-(2+\sqrt{2}) \cdot r+1\right) \cdot \cfrac{(r+1)^2}{(1-r)^2} = 1$$
This is quite an odd equation. First of all, here's a plot of $(3.9)$ against $1$.
Numerically, there are three solutions, which is unexpected. We have,
$$r=0$$ $$r=\phi=\cfrac{\sqrt{5}-1}{2}=0.618...$$ and then, this... $$r=\cfrac{-1}{3}-\cfrac{2\cdot(4+3\sqrt{2})}{3 \cdot \left(26+9\sqrt{2}+3\sqrt{3 \cdot (114+76\sqrt{2})} \right)^{1/3}}+\cfrac{1}{3} \cdot \left(26+9\sqrt{2}+3\sqrt{3 \cdot (114+76\sqrt{2})} \right)^{1/3}=0.1660561895...$$
I just find it odd two of the solutions are so simple, and then the third one is so complex.
It seems that there are three non-negative solutions: the complicated $r\approx0.166$ illustrated above, $r=0$ (which has a simple interpretation, as the partial sums are all 1), and $r=1/\varphi$ the reciprocal of the golden ratio. In this final case, the small appended circles actually cross the larger circle and touch it on the other side. Here's an illustration where most of the circles are highly translucent while the path that leads to just touching is highlighted:
My first guess was $r=1/(2+\sqrt{2})$, which is exactly the correct scaling factor for the octagasket:
That leads to a picture like the following, however:
In this picture, the largest central circle has radius 1 and the 8 next largest circles just tangent to it have radius $r$. Each subsequent child circle has radius $r$ times the radius of the parent. As you can see, the clusters are just touching (as expected) but we haven't dealt with the overlap with the original circle appropriately.
There are three arrows on the diagram indicating the problem explicitly: we have vectors $v_1$, $v_2$, and $v_3$ of lengths $1+r$, $(1+r)r$, and $(1+r)r^2$ each of which is a rotation of the previous through an angle which is a multiple of $\pi/4$. Assuming we never step backwards, the center of each circle can be determined by a sum of such vectors. In order to avoid overlap, we must ensure that the sum of such vectors always has magnitude larger than 1. We see that this doesn't happen for this choice of $r$.
We can write this semi-nicely using complex numbers. We seek a value of $r$ such that $$\left|\sum_{k=0}^{\infty} (1+r)r^k e^{\theta_k i}\right| > 1,$$ for any sequence $(\theta_k)_{k=1}^{\infty}$ where each $\theta_k$ is a multiple of $\pi/4$ and we're not allowed to double back (that is, $\theta_{k+1} \neq \theta_k \pm\pi$). It's not too hard to program this on a computer to find a value of $r$ where we just have contact. Doing so, I found $$r\approx 0.16605618952678$$ which leads to the following picture:
Here's a zoom:
I don't know if there's a closed form. I did examine the sequence of rotations involve, since a simple pattern might lead to a closed form. I didn't notice any obvious such pattern, though.