Superattracting period-n orbits in the Mandelbrot set

Question

So I'm trying to find the number of superattracting period-$n$ orbits in the family $z\rightarrow z^2+c$ for $n = 1,2,3,4,5,6$.

I think I found an algorithm to compute this.

$0 \rightarrow c \rightarrow c^2 + c \rightarrow (c^2+c)^2+c \rightarrow ((c^2+c)^2+c)^2+c\rightarrow (((c^2+c)^2+c)^2+c)^2 + c \rightarrow ((((c^2+c)^2+c)^2+c)^2 + c)^2 + c$

So I know that Period 1 has a superattracting point at $c = 0$. To find the superattracting orbit for the Period-2 bulb, you solve for $c^2 + c = 0$, which yields $c = 0,-1$. How would one solve for $c$ for the higher periods? Would you have to use Mathematica?

Answer 1

Methods :

• all centers for given period $p$
• one center $c_p$ for given period p near given point c

For all centers

• solving $F_p(0,c) = 0 $ or $G_p(0,c) = 0$
  • using Matrices
  • The iterated refinement Newton method

...

Answer 2

Mu-ency's enumeration of features page details combinatorial methods for finding the number of superattracting period-$n$ orbits:

We start by considering the lemniscates as polynomials set equal to zero, and solving them, i.e. "finding the roots". Each root gives the nucleus of some mu-atom. Since the $n$th lemniscate is of order $2^{n-1}$, it follows that for each period $n$ there are $2^{n-1}$ roots.

However, a root of period $3$ (for example) also shows up as a root of period $6$, $9$, etc. but we don't count it that way. It only counts as a root of period $3$. So, from $2^{n-1}$ we must subtract the number of roots of all lower periods that are divisors of $n$. This gives the following formula for the number of Mu-Atoms of period $n$: $$N(n) = 2^{n-1} - \sum_{0 < f < n \\n \equiv 0 \mod f} N(f)$$

... This is Sloane's sequence A000740.