A Series of spiral bays in the Mandelbrot set

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

figure 1 (1k)

I assume (without proof) that if fc(z)=z2+c, and fcn=fcofco...fc, then a bay (circle) number n corresponds to a fixed point of fcn.

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:

figure 2 (12k)

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

a(c) = inf { | fcn(0) | : n = 1,2,...},
index(c) = smallest n providing a(c) = | fcn(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:

figure 3 (1k)

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

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:

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.

Note:
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)
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