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For each complex number $c$, define $P_{0}(c)=0$ and $P_{n+1}(c) = (P_{n}(c))^{2} + c$ . The Mandelbrot set is the set of complex numbers c for which $|P_{n}(c)|$ stays bounded as $n\rightarrow \infty$.

$P_{n}(c)$ is a polynomial in $c$. Donald D. Cross noticed in 2005 that the leading terms settle down
$p_{0} + p_{1}c + p_{2} c^{2} + p_{3} c^{3} + ... = c + c^{2} + 2c^{3} + 5c^{4} + ...$ and submitted them to the Online Encyclopedia of Integer Sequences (A000108). It was subsequently noticed that these are the terms for the generating function $P(c)$ of the Catalan numbers, which is defined by: $P(c) = cP(c)^{2} + 1$. Indeed, $p_{n} = \sum_{i=1}^{n-1}p_{i}p_{n-i}$.

Philippe Flajolet and Robert Sedgewick mention this connection with enthusiasm in their definitive textbook Analytic Combinatorics, 2009, available online for free. See pages 535-537 and also the link to the theta function is discussed on page 328-330. Curiously, there is no mention of the Mandelbrot set in Richard Stanley's 2015 book Catalan Numbers.

My question is whether $P(c)$ can be used to define the Mandelbrot set. Is it the case that $|P(c)|$ is bounded if and only if $|P_{n}(c)|$ stays bounded as $n\rightarrow \infty$ ? Is the Mandelbrot set simply the set for which $P(c)$ is bounded?

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    $\begingroup$ What do you mean? $P(c) = (1 - \sqrt{1-4c})/(2c)$ is an analytic function (except for a branch cut on $[1/4, \infty)$). It doesn't do anything special on the Mandelbrot set. $\endgroup$ – Robert Israel Aug 8 '16 at 1:31
  • $\begingroup$ Robert, that helps, thank you. So does that mean $P(c)$ is well defined for all c except on the branch cut? $\endgroup$ – Andrius Kulikauskas Aug 8 '16 at 7:19
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    $\begingroup$ I am intrigued by this question, but I don't yet have anything intelligent to contribute, except that a search on "mandelbrot catalan" shows that someone else noticed this in 1999: mathforum.org/kb/message.jspa?messageID=162350 $\endgroup$ – Nathan Reading Aug 8 '16 at 16:42
  • $\begingroup$ @AndriusKulikauskas : I'm puzzled as to why you're asking this here since I already answered your question when you asked it on FOM back in May. cs.nyu.edu/pipermail/fom/2016-May/019822.html $\endgroup$ – Timothy Chow Aug 8 '16 at 16:53
  • $\begingroup$ @NathanReading thank you for that link! Yes, I think I'm asking the same question. The answer they got is helpful for me. $\endgroup$ – Andrius Kulikauskas Aug 9 '16 at 11:04

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