Apolloniana

Golden Satellite

To see again-- click on the image. To see dimensions -- press any key.

Intriguing properties:
1. Fibonacci and Lucas! The sequence b_i of the curvatures of the yellow disks is:

...14, 6, 3, 2, 2, 3, 6, 14, 35, 90, 234, ...
Supprisingly, it may be obtained from Lucas and Fibonacci sequences. Below, the first column represents Fibonacci numbers (F), the second -- Lucas numbers (L). Consider every other number in each sequence, as marked by red characters. They form a zig-zag sequence: 2, 1, 3, 2, 7, 5, 18, ...

F L

0 2
1 1
1 3
2 4
3 7
5 11
8 18
13 29
21 47
34 123
55 199
89 322
... ...
Call this an "underground sequence s_k. Next, multiply every two consecutive red numbers to form a "product" sequence:

s_i: ... 2 1 3 2 7 5 18 13 ...
b_i: ... 2 3 6 14 35 90 234 ...

The resulting sequence turns out to be the original sequence (b_i) of the curvatures of the disk in the in the chain!
Another fact: The sequence of curvatures nmay be obtained from the odd subsequence of the Fibonacci numbers (the red entries) by simply adding 1 to each entry: $$ \begin{array}{ccccccccc} F:\ &1 & 2 & 5 & 13 & 34 & 89 \\ & \downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow\\ \mathbf b: \ &2 & 3 & 6 & 14 & 35 & 90 & \end{array} $$ And yet another observation: the odd Fibonacci numbers are hypothenuses of a subfamily of the Pythagorean triples. Here are some: $$ 0^2+1^2 = \mathbf 1^2 \qquad 1^2+1^2 = \mathbf 2^2 \qquad 3^2+4^2 = \mathbf 5^2 \qquad 5^2+12^2 = \mathbf{13}^2 \qquad 16^2+30^2 = \mathbf{34}^2 $$ The justification of these facts may be found in here and here.
2. Golden ratio $\varphi$. The geometric proportions of the above figure contain the Golden ratio. To see it, press any key. It will show the golden rectangle and the golden proportion (toggle with any key). The suggested identity: $$ \frac{1}{2} + \frac{1}{2} + \frac{1}{3} + \frac{1}{6} + \frac{1}{14}\ldots = \frac{1+\sqrt{5}}{2} \equiv \varphi $$
3. Recurrence. The sequence of curvatures follows a nonhomogeneous recurrence:
b_n+1 = 3b_n − b_n-1 −1
The underground sequence (s_i) satisfies
f_n+1 = A·f_n − f_n-1
where the constant A alternates between the values A = 5, 1.

Thanks to Christian Rose who spotted some numerical typos in the original page