The Mandelbrot set has ---
1 main cardioid of period 1.
1 main bulb of period 2.
3 bulbs of period 3 (one of them the needle minibrot).
5 bulbs of period 4? (two of them minibrots on the period 3 branches).

I'd like this to be a sequence in OEIS. How many bulbs of period 5, 6, 7, 8, and so on are there?

  • 2
    $\begingroup$ M. Lutzky: Counting hyperbolic components of the Mandelbrot set. Physics Letters A Volume 177, Issues 4-5, 21 June 1993, Pages 338-340. $\endgroup$
    – Adam
    Jul 31, 2016 at 20:09

1 Answer 1


A so-called period $n$ bulb is a maximal, open, connected set $U$ in the complex plane with the property that for every $c\in U$ the corresponding function $f_c(z)=z^2+c$ has an attractive cycle of period $n$. For each such $U$, there is exactly one point $c\in U$ such that $f_c$ has a super-attracting cycle. A cycle is super-attracting if and only the critical point zero is in that cycle. Thus, the period $n$ bulbs correspond to solutions of the equation (in the variable $c$) $f_c^n(0)=0$. Put another way, we're looking for the roots of the polynomial $f_c^n$.

Using this, we can show that there are actually 6 period 4 bulbs. We can see this with the following little bit of Mathematica code:

f[c_][z_] = z^2 + c;
Factor[Nest[f[c], 0, 4]]

(* Out: c (1 + c) (1 + 2 c^2 + 3 c^3 + 3 c^4 + 3 c^5 + c^6) *)

That Nest[f[c],0,4] expresses $f_c^4$. We factor it since we want the roots. The roots of the smaller factors satisfy $f_c^4(0)=0$ but actually yield super-attractive orbits of smaller degree. The actual super-periodic parameters of period 4 arise from the irreducible factor of degree 6. Thus, there are 6 of them.

We can push this further to find how many bulbs of higher degree there are:

Table[Max[Exponent[List @@ Factor[Nest[f[c], 0, k]], c]], {k, 10}]

(* Out: {1, 1, 3, 6, 15, 27, 63, 120, 252, 495} *)

You can plug this into OEIS to find that the sequence is already there!


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