When does Mandelbrot's quadratic recurrence have explicit solutions? According to Mathworld's Quadratic Map, there is no general explicit solution to the following quadratic map :
$$ z_{n+1} = z_n^2 + c ~, ~z_0 = 0$$
However, if $c=0$, the solution is quite obvious. 
So I want to know if there are known explicit solutions for some other values of $c$ and with different $z_0$ value. 
(Actually, what I need to know is when $c=1/4$ and $z_0 \in (-1/2,1/2)$, but just got curious about other cases too. $c$ and $z_0$ may range complex numbers.)
 A: Let's make a substitution $y_{n}=z_n+a$, then $$y_{n+1}=y_n^2-2ay_n+a^2+a+c.$$
Selecting $a$ to eliminate the free term, i.e. $a^2+a+c=0$ we can rewrite the recurrence as:
$$
\frac{y_{n+1}}{2a}=-2a\,\frac{y_{n}}{2a}\left(1-\frac{y_{n}}{2a}\right)
$$
This is a logistic recurrence of the form $x_{n+1}=rx_n(1-x_n)$ with $r=-2a$. According to Mathworld's Logistic Map, exact solutions (with real $r$) are only known for $r=\pm2,4$. They are of the form $$x_n=\frac12\left(1-f(r^nf^{-1}(1-2x_0))\right)$$
with some function $f$, $f^{-1}$ is its inverse ($f=e^x$ for $r=2$ and $\cos x$ for $r=4$).
For $c=\frac14$ we have $a=-\frac12$ and $r=1$, so it is not one of them. Explicitly, the only known good values for $c$ are $0$ and $-2$. If Wolfram's conjecture about logistic recurrence is correct they are the only ones (with real $r$). Another known value with exact solutions is $c=i$ (with complex $r$). For generic $r$ the dynamics is chaotic.
A: For $z_0=0$, the given recurrence is related to the definition of the Mandelbrot Set which is a quite complicated set. We have an explicit formula for the recurrence only in a "few" special cases. For example if $z_0=0$ and $c\in\{0,-1/2,1/2,-2,i\}$ the recurrence is easy to solve. See Misiurewicz points.
For $c=1/4$ and $z_0 \in [-1/2,1/2]$ the recurrence is bounded and it has limit $1/2$. One can show this property without finding an explicit formula. By induction, we have that if $|z_0|\leq 1/2$ then $|z_n|\leq 1/2$ for all $n\geq 1$:
$$|z_{n}|=|z_{n-1}|^2+1/4\leq |1/2|^2+1/4=1/2.$$
Moreover for $c=1/4$, if $z_0\in \mathbb{R}$ then the sequence is non-decreasing:
$$z_n=z_{n-1}^2+1/4=z_{n-1}+\left(z_{n-1}-1/2\right)^2\geq z_{n-1}.$$
Hence for $z_0\in [-1/2,1/2]$, the sequence $(z_n)_n$ has a limit, which is $1/2$, the unique root of the equation $z^2-1/4=z$.
