Continuing logarithm $\log(\log(\dots\log(z)))$ I don't know the best way to describe it in technical terms, but what is the result of a continuing logarithm of $z$, for example:
$$\log(\log(\dots\log(z)))$$
Where it is taking the logarithm of the logarithm and so on, for an infinite amount of times?
How would this type of thing behave? Does it converge, go off to infinity or an infinitesimal? Does the resulting behavior depend on whether $z$ is imaginary or real? Positive or negative?
 A: If $\Im(z)$ (the imaginary part of $z$) is greater than or equal to zero, then this converges to roughly $z_\infty = 0.318152 + 1.33724 i$.  
If $\Im(z)<0$ it converges to  $z_\infty = 0.318152 - 1.33724 i$.
This of course relies on the usual branch cut for $\log z$.  
The exceptional cases are any cases where a finite number of itetations lands on $1$ (or starting with $z=0$.  These include $1, e, e^e$, and so forth.
However, there are isolated points for which the iterated log neither goes to infinity nor converges.  For example, for any $z$ such that  $$e^z = \log z \neq z$$
the iterated log oscillates between $z$ and $\log z$. I think there are such points; for example, there is an unstable 2-cycle fixed point at roughly
$$
z= 0.883998 + 6.922346 i
$$
A: Iff it converges, then
$$a=\log(\log(\dots\log(z)\dots))=\log(a)$$
Thus,
$$a=\log(a)$$
$$e^a=a\implies-1=-ae^{-a}\implies a=-W_k(-1)\stackrel{k=0}\approx0.318-1.337i$$
is the Lambert W function.  Note that this is constant, and dependent only on whether or not the choice of $z$ converge and where $z$ is.  The position of $z$ will determine which branch of the Lambert W function it will converge to.  A couple notes:
If we $z$ is a perfect super power of $e$, then it diverges due to $\log(0)$. That is, $z\ne e^{e^{e^{\dots}}}$ and $z\ne0$.
I am pretty sure it converges everywhere else, with the exception of $z=-W(-1)$.
Not finished with the rest:
$\color{white}{Let \$z=re^{i\theta}\$.  Then we have
\$\$\log(z)=\log(r)+i\theta=r_1e^{i\theta_1}\$\$
We can see that \$r_1=\sqrt{\log^2(r)+\theta^2}\$ and \$\theta_1=\arg(\log(r)+i\theta)\$.  From this, we can discern that \$r\$ must be bounded, and likewise, we are left to look at $\theta$.  As \$r\to r'\$ be the limit of \$r\$, \$\theta\$ also approaches a limit}$
