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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.)

Steven
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  • 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. – Steven Oct 02 '14 at 06:26
  • See : http://arxiv.org/abs/math/0505036 ( giant step method ) HTH – Adam Jan 20 '15 at 19:07
  • @Adam, I've just now started looking into that article and I'm running into something I don't understand on definition 2. If $S$ is a set of complex numbers and $T$ is a set of complex numbers, how can $B(S,2^{-n})$ be a proper subset of $T$, when $B$ is defined to be a set containing only the squares of real numbers? – Steven Jan 29 '15 at 00:03
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    z0=0 (critical value) not z0=c – Adam Jan 30 '15 at 14:58
  • @Adam You're correct. I have fixed the post accordingly. However, that doesn't answer my question, does it? I mean, I still don't understand Definition 2 from the article you linked. – Steven Feb 02 '15 at 16:45
  • @Adam Nevermind. I misunderstood some of it. I understand it now. – Steven Feb 02 '15 at 22:07
  • Have you made a spreadsheet to calculate $f(m,n)$ for small values? It would be good to post the array as somebody might recognize the numbers, like you did with the Catalans. If you copy from Excel and paste into Word, you can do a global replace of ^t to & and ^p to \ and be close to a MathJax array. – Ross Millikan Feb 03 '15 at 03:12
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    See OEIS A$137560$ and OEIS A$137867$. – Lucian Feb 03 '15 at 06:53
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    The Mandelbrot set is a contender for "most complicated object in the universe", so I wouldn't expect much simplification in the formulas. – Gerry Myerson Feb 03 '15 at 08:15
  • @GerryMyerson I always found that "most complicated object in the universe" business to be a bit funny. It is a compact and connected subset of the plane, after all. (Sorry about the ping a year and a half later, but I couldn't resist.) – Mark McClure Jun 30 '16 at 00:14
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    @Mark, just goes to show how complicated compact connected subsets of the plane can be, I suppose. – Gerry Myerson Jun 30 '16 at 02:51

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