How to find examples of periodic points of the (complex) exponential-function $z \to \exp(z)$? Background: By considering the question  which asks whether a certain summation-method $\mathfrak M$ for the (extremely divergent!) sum $\mathfrak M: S(z)=z + e^z + e^{e^z}+e^{e^{e^z}} + ...$ might be valid, I had the idea, that a good test for the validity of the summation-method would be to compare one result with cases, where the sum can be given on some standard way. 
For instance, if $z$ is a fixpoint $p$  of $\exp(z)$  having  $\exp(p)=p$ then $S(p) = p + p+ p+ p+...$ which could possibly been evaluated using the $\zeta(0)$ - definition.
But this is possibly no good test, a better one were, if for instance $p$ were a $2$- periodic point where the sign changes, (let's denote a fixpoint $p_1$ and a $2$-periodic point $p_2$) such that $\exp(p_2)=-p_2$ and the series becomes $S(p_2)=p_2 - p_2 + p_2 - ... + ...$ which can then be assumed to be the Cesarosum of the alternating series of a constant.              
First steps of my own approach: This led to the idea to find such $2-periodic$ points to have example cases. 
What I've done so far was to screen segments of the complex plane , say a square around some $z_0$ of size $2x2$ in steps of $1/10$ and find local minima. Then take each local minima of the error ($|z_0 - e^{e^z}|$  as new center of a new square with $0.2 x 0.2$ size, and iterate this two or three times until the error is small enough and then use Newton-iteration to finally find a point $p_2$ (hopefully meeting the requirements for Cesaro-summability!).       
This gave by tedious manual screening an initial set of $2$-periodic points. But all so far found $2$-periodic points were on the real-positive halfplane and so not suited for Cesaro-summation of $S(p_2)=p_2-p_2+p_2-...$ with alternating signs.        
Here is a picture which included  a couple more points which I found by extrapolating linear trends in subsets of the found ones. See my initial answer here
 
Question 1:
So my first question is now for ideas,
 - 1) how could I improve the search-routine? (Perhaps avoiding that manual screening at all) 
The extrapolation-idea is very useful and allows to find infinitely many more $2$-periodic points completely automatically, but it still needs an initial manual screening to get at least $3$ points as germ for the extrapolation.     
Question 2:
The picture contain no pairs of $2$-periodic points having alternating signs on the real part.
I've tried to find analytically better solutions, of to even prove there are no such points, but arrived nowhere definite so far.
 - 2) Are there really no $2$-periodic points with negative real part? 
Question 3 & 4:
If there are indeed no such cases, what about $3$-periodic points or in general $n$-periodic points?
 - 3) Are there $n$-periodic points with negative real part?
 - 4) Is there any analytical method known how to find $n$-periodic points without manual screening? (For the finding of $1$-periodic (or "fix"-) points we have the Lambert W-function. For generalizing the Lambert W I had found a scheme giving power series, but which have extremely small range of convergence and I think are thus useless here)
 A: The structure of the periodic points of the exponential map is well-understood. Observe that the real axis is invariant and contains no periodic points; its preimages contain the horizontal lines at imaginary parts that are integer multiplies of $\pi$.
Consider the strips
$$ S_k := \{ a + ib\colon (2k-1)\pi < b < (2k+1)\pi\}.$$
Every periodic point $z$ has an "itinerary" $(k_n)_{n=0}^{\infty}$, defined by
$$ f^n(z) \in S_{k_n}.$$
(Here $f$ is the exponential map and $f^n$ its $n$-th iterate.)
Clearly the itinerary is a periodic sequence whose period divides the period of $z$.
It is known that every periodic sequence is realised by a periodic point. When $(k_n)_{n=0}^{\infty}$ is not the sequence $k_n \equiv 0$ for all $n$, this periodic point is unique and, in particular, has the same period as $(k_n)$.
For the sequence defined by $k_n\equiv 0$, there are two fixed points, one in the upper half-plane and one in the lower half-plane, which are complex conjugates of each other.
The periodic point for a given itinerary can be obtained by backwards iteration, as you mention. That is, let $L_k\colon \mathbb{C}\setminus (-\infty,0]\to S_k$ denote the inverse of the map $f|_{S_k}$. If $z\notin\mathbb{R}$, then define
$$ z_m := L_{k_0}(L_{k_1}(\dots(L_{k_m})\dots)).$$
Then the sequence $z_m$ will converge to a periodic point with the desired itinerary, and this point is unique, except for the exceptional case $k_n\equiv 0$, where it depends on whether $\operatorname{Im} z$ is positive or negative.
It is also known that every periodic point as above, except for the case $k_n\equiv 0$ has a periodic curve to $\infty$ attached to it, periodic of the same period. This curve is called a "Devaney hair" or "external ray", and all points except the periodic endpoint converge to infinity under iteration.
I think that all of this can already be found in the paper by Devaney and Krych (Dynamics of $\exp(z)$, ETDS, 1984). They use a somewhat different convention for itineraries, but the results are equivalent.
A: As you may remember, you need to solve in general the $p$-th auxilliary exponential equation to find the fixed points first. The first auxilliary is $f(z)=z$, with $f=exp$. This is solved by using Lambert's $W$ function as $z_k=-W_k(-1)$, $k\in\mathbb{Z}$. None of these is stable as $|f'(z_k)|>1$. The second auxilliary is $f^{(2)}(z)=z$ or $e^{e^z}=z$. This cannot be solved in terms of elementary functions, so you need to use Newton's method. Once you find a solution $z_0$, the fixed points will be $z_0$ and $f(z_0)$. You can test then for the period, by using the modulus of the derivative of the multiplier, as given by Shell, as $|(f^{(2)})'(z_0)|$. If that's less than one, then it will be a 2-cycle, with limits $z_0$ and $f(z_0)$.
In general you'll have to solve the $p$-th auxiliary $f^{(p)}(z)=z$, for which you can still use Newton's method, but as the composition becomes more and more entangled, it will progressively be slower and may even fail. If you get a point $z_0$, then all $\{z_0,f(z_0),f^{(2)}(z_0),\ldots,f^{(p-1)}(z_0)\}$ will be fixed points, which you can check using again the modulus of the multipler $|(f^{(p)})'(z_0)|$.
With that said, it's unknown really whether there are points of period $p$ for arbitrary $p$ in both tetration of the exponential and/or regular tetration. You can only check individual values. In one of my papers I extract the fixed points using functions which generalize $W$, but the check is still the same. They all have to go through Shell's multiplier to check for the modulus. You can get a rough idea on where to look, by graphing the iterates of the exponential like this.
