Infinite exponential representation of real numbers I was thinking about infinite exponential representation of real numbers (like $2=e^{e^{-e^{-e^{e^{-e^{e^{e^{-e^{-e^{-e^{-e^{-e^{e^{-e^{e^{e^{-e^{e^{\cdot^{\cdot^{\cdot}}}}}}}}}}}}}}}}}}}}}$.
The sequence of signs before exponents can be obtained by repeated application of $\ln|x|$ to $2$ and taking a sign of each result. It seems that this gives an almost 1-1 correspondence between $\mathbb{R}$ and the set of infinite sequences of signs (or positive and negative 1's) , except that $0$ has two representations $\pm e^{-e^{e^{e^{\cdot^{\cdot^{\cdot}}}}}}$ (this is also true for every number that has representation as a finite exponent tower ending with $0$) and sequences $\pm e^{e^{e^{\cdot^{\cdot^{\cdot}}}}}$ diverge to $\pm \infty$. 
Has this representation been studied? Does any algebraic number (except $0$, $1$ and $-1$) have an eventually periodic representation? What we can say about frequency of each sign in representation of a particular number (say, $2$)? (first several hundreds of elements suggest that $-1$ appears two times more often than $1$). Are there arbitrarily long runs of the same sign?
 A: One must be careful with convergence of such "infinite representations".  Almost any
positive real $x_0$ is the start of a sequence $x_i$ of positive reals such that 
$x_i = e^{x_{i+1}}$ if $x_i > 1$ or $e^{-x_{i+1}}$ if $x_i < 1$; thus $x_{i+1} = |\ln x_i|$.
It's only "almost any" because the sequence would end if some $x_i = 0$. There is a fixed point, $\text{LambertW}(1) = 0.5671432904$, which is a value $x$ such that $\ln x = -x$.
This would correspond to a "representation" $\text{LambertW}(1) =\exp(-\exp(-\exp(-\ldots)))$.
There's a 2-cycle consisting of $a = 0.2698741377$ and $b = 1.309799586$, corresponding to
"representations" $a = \exp(-\exp(\exp(-\exp(\ldots)))$ and $b = \exp(\exp(-\exp(\exp(\exp(\ldots))))$.  I would expect numbers with periodic representations to be transcendental, as they correspond to solutions of non-algebraic equations.
A: First, even though I think this is a fun question,
it's not really research mathematics and I'm not sure it belongs on mathoverflow.
(You know that some really smart people answer questions on math.stackexchange, right?)
As was noted in Robert's answer, one is investigating the sequence $x_{n+1} = | \log(x_n)|$, which makes
sense for all $x_0$ outside some countable set. Moreover, the periodic points
will (surely) be transcendental, and it will be impossible to prove this fact
(except for periods of length $1$).
So what should one expect for the number of $+$ and $-$ signs?
Here is a heuristic description of what happens, with the caveat that I have made no attempt to be rigorous (to do so, I would compare this with the probability theory of the 
Gauss–Kuzmin distribution).
First, choose $x_0$ in $[0,\infty]$ according to some probability measure
$f_0(x)$, with cumulative probability distribution $F_0(x)$.
Let $f_n(x)$ be the distribution of $x_n$.
By definition, $f_n(x)$ must satisfy the following equation:
$$\int^{b}_{a} f_n(x) dx = \int_{e^a}^{e^b} f_{n-1}(x) dx + \int_{e^{-b}}^{e^{-a}} f_{n-1}(x) dx$$
for all $b \ge a \ge 0$.
If $F_n(x) = \int^{x}_{0}f_n(t)dt$ is the cumulative distribution function of $x_n$, then this equation becomes:
$$F_n(b) - F_n(a) = F_{n-1}(e^b) - F_{n-1}(e^a) +  F_{n-1}(e^{-a}) - F_{n-1}(e^{-b}).$$
Letting $a = 0$, one obtains:
$$F_n(z) = F_{n-1}(e^z) - F_{n-1}(e^{-z}).$$
Given some basic assumption on $F_0(z)$, the sequence
of functions $F_n(z)$  converges to the unique increasing function $F(z)$
such that
$$F(z) = F(e^z) - F(e^{-z}).$$
The independence of $F(z)$ on $F_0(z)$ implies that this function 
should describe the cumulative distribution function of $x_n$ for $n$ sufficiently large
for almost all initial values $x_0$.
By choosing random functions $F_0(z)$, one can estimate that
$$F(1) \simeq 0.6518\ldots$$
Since the sign in the exponential is determined by whether $x_n > 1$ or not,
this implies that the ratio of $-$ signs to $+$ signs (for almost all initial values, which presumably includes $x_0 = 2$)
 is  roughly $1.872$ to $1$. This seems to confirm what you observed experimentally.
Moreover, since the function $F(z)$ is strictly increasing,
it follows by Kolmogorov's zero-one law that, for almost all initial values $x_0$, that there are arbitrarily long
runs of $+$ signs, $-$ signs, etc.

 Edit: To make this completely rigorous, one can
make $\mathrm{R}^{+}$ a compact measure space given by the measure
specified by $\mu([a,b]) = F(b) - F(a)$. 
The function $T:=|\log(x)|$ (modified so that $T(1) = 1$) is then measure preserving and (as is relatively easy to check) ergodic.
The claims then follow from the Birkhoff Ergodic theorem, for suitable choices of 
test function $f$ (like the step function which is zero for $x < 1$ and one for $x > 1$).
(BTW, I may have added a few more decimal digits above than was really justified.) 
