When I start searching systematically for spirals in the Mandelbrot set (see trick1), I used this figure:

I assume (without proof) that if f_{c}(z)=z^{2}+c, and
f_{c}^{n}=f_{c}of_{c}o...f_{c},
then a bay (circle) number n corresponds to a fixed point of f_{c}^{n}.

Later, I discovered that my series is the same as the one represented in figure 34 on
page 61 of the book **The Beauty of Fractals** by Peitgen & Richter.
Fractint implemented this drawing mode under the name of 'bof 61'; so here is a figure I
obtained with fractint:

And here is the definition that **The Beauty of Fractals** gives:

a(c) = inf { | f_{c}^{n}(0) | : n = 1,2,...},

index(c) = smallest n providing a(c) = | f_{c}^{n}(0) |.

The book claims that *" index(c) introduces a Fibonacci partition
in M. "*

The "Fibonacciness" of the figure is, that the index of any region is the sum of the indexes of its two neighbor regions.

I reconsidered both figures to come up with this revisited figure:

This is the **first form** of the series of the main bays (circles) on the top half
counted counterclockwise arround the belly (cardioid) of the Mandelbrot set
(as an exception, the number 1 corresponds to the cardioid itself).

step 0: 1 2 step 1: 1 3 2 step 2: 1 4 3 5 2 step 3: 1 5 4 7 3 8 5 7 2

in this form there is no unique series, as on every step a new term
is inserted between two existing terms.

Nevertheless, I will have to come back to this form later.

Here is the **second form** of the series: I write down the new terms
of every step in the order that I obtained them:

(step 0) (step 1) (step 2) (step 3) 1 2 3 4 5 5 7 8 7 ...

To simplify the formula that describes how a term is obtained, it is best to start indexing with -1:

index u | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |

value a(u) | 1 | 2 | 3 | 4 | 5 | 5 | 7 | 8 | 7 | 6 | 9 | 11 | 10 | 11 | 13 | 12 | 9 |

- a(-1) = 1; a(0) = 2; a(1) = 3
- a(2
^{k-1}) = a(-1) + a(2^{k-2}) , if u is a power of 2; else: - a(u) = a( i(u) ) + a( i(u+1) ), where i(u) = { i(u/2) if 2|u; [u/2] else } , and [ ] is the integer part of a number.

The recursive formula for i(u) can well been described if u is written in base 2:

as long as the result is pair, cut the last zero; finally cut the last 1 too.
If nothing is left, we have i(1)=0.

**How often does a given number n appear in the series a(u)?**

I will prove that for n>2 it is phi(n)/2 times, where phi is the Euler function (number of positive integers <n relatively prime to n).

**Proof:**

If a number n>2 appears in the series, it is in some step k of the first form, so n = a + b. (a and b are successive terms in step k).

Looking at step k-1, one of the numbers a or b is still there (the greater one), the other number ( b or a) is the sum of (a or b) and of another number (the difference). Lets say, a<b with b = a + c. (It is impossible to have a zero anywhere in a series!) So step (k-1) has the successive terms { a, c }.

We will iterate, like in Euclid's algorithm to calculate the greatest common divisor of two numbers! The greatest common divisor of a and b, (a,b) is the same as (a,c).

Necessarily, we arrive at step 0 with the numbers 1 and 2, and so:

- the greatest common divisor of our initial (a,b) is 1, and
- with a given couple (a,b) with (a,b)=1, there is exactly one occurrence of successive terms ...a,b,... or ...b,a,... in exactly one step k.

But (a,b) = (a,a+b) = (a,n); so there are phi(n) ways to choose a, for
a given n; this number phi must be divided by 2 because "n=a+b"
and "n=b+a" count as one solution.

QED.

If we consider also the bays on the lower half of the cardioid, then there are phi(n) circles of index n arround the cardioid for every n > 1.

Created: 1997-03-30; Bugfix: 1997-04-21

**Your comments are welcome!** Mail to Patrick Hahn (phahn@vo.lu)

Homepage: http://www2.vo.lu/homepages/phahn