# Is there a simplification for the coefficients generated with the Mandelbrot iteration rule?

The Mandelbrot Set is obtained using the equation $z_n=z_{n-1}^2+c$ for some constant $c \in \mathbb{C}$ with $z_0=0$. Therefore, $z_1=c$, $z_2=c^2+c$, $z_3=c^4+2c^3+c^2+c$, etc.

I have a function which generates the $n$th coefficient for $z_m$: $f(m,n)$, where $m \in \mathbb{N} \land n \in \mathbb{N},$ such that

$$z_{m}=f(m,n)c^{n+1}+f(m,n-1)c^{n}+...+f(m,1)c^2+f(m,0)c^1$$

However, $f$ is piecewise-defined, recursive, and even involves some summations; It's a tad complicated.

$$f(m,n) := \begin{cases} 0,& \mbox{if} \quad m = 0 \lor n \geq 2^{m-1} \\ 1,& \mbox{if} \quad n = 0 \land m \neq 0 \\ \sum_{i=0}^{n-1} \left( f(m-1,i)f(m-1,n-1-i) \right) ,& otherwise\\ \end{cases}$$

On a side note, if $m \gt n$,

$$f(m,n) = \frac{\left( 2n \right)!}{(n+1)!n!}$$

which generates Catalan Numbers, but that doesn't have much to do with my question: is there any way to simplify this function? (Preferably in a way that eliminates the recursion. Also, note that I haven't proven any of this. I probably could, but I haven't and don't plan to. I just want a simplification of $f$.)

Edit:

Per request, here is a table of $f(m,n)$ for $0\leq m\leq 5$ and $0\leq n\lt 2^4$

$$\begin{cases} &n_0&n_1&n_2&n_3&n_4&n_5&n_6&n_7&n_8&n_9&n_{10}&n_{11}&n_{12}&n_{13}&n_{14}&n_{15}\\ m_0:&0\\ m_1:&1\\ m_2:&1&1\\ m_3:&1&1&2&1\\ m_4:&1&1&2&5&6&6&4&1\\ m_5:&1&1&2&5&14&26&44&69&94&114&116&94&60&28&8&1\\ \end{cases}$$ (The extra zeros have been removed for readability. Technically, each row should have an infinite amount of zeros on the end.)

• I would really like some sort of resolution to this. I would add a bounty to it, but I don't really have enough rep for the bounty to be worth it. Oct 2 '14 at 6:26
• z0=0 (critical value) not z0=c
• See OEIS A$137560$ and OEIS A$137867$. Feb 3 '15 at 6:53