Essentially, I've noticed that tetrations of $\omega=\frac{\pi i}{\ln(2)}$ seem to converge on a cycle of three fixed points. Specifically, if $a_0$=1, and $a_{n+1}=\omega^{a_n}$, then we find
$$n\equiv 0\mod 3\implies a_n\approx 1$$ $$n\equiv 1\mod 3\implies a_n\approx \omega$$ $$n\equiv 2\mod 3\implies a_n\approx 0$$
for sufficently large $n$. Thus,
$$\lim_{k\to\infty}\frac{1}{k}\sum_{n=0}^{k-1} a_n\approx\frac{\omega+1}{3}$$
This is a sloppy approximation, however, according to my numerical approximations up to $k=1000$ with $512$-bit precision, the real and imaginary parts are only off by about $1.6\%$ and $0.059\%$, respectively.
Regardless, I am curious as to how I would go about deducing the exact values of the 3-cycle, and consequently, the average.
I'm stuck somewhere between knowing if this is trivially done algebraically or beyond my abilities. Hence, any insight would be greatly appreciated.
Hmm, the sequence $$\small {[1,\omega,\omega^\omega,...] = \\ [1 , 4.532360142\,î , 0.0006829063+0.0004341759\,î , 1.0003486+0.001729456\,î , \ldots]}$$ converges to a set of 3-periodic points (read columnwise then rowwise):
The convergence is rather fast:(showing the rowwise differences = differences between $a_{n+3}$ and $a_n$)
and the rowwise sums $a_{3n}+a_{3n+1}+a_{3n+2}$:
update In the end, your sum should converge to the sum of your three values $(a_1 +a_2+a_3)/3$ and just for visualization I've printed the partial averages when $n$ is increased for the sum.
This converges obviously, and to what? See
The value that you propose is $\large {{\pi\; î \over \ln 2 }+1 \over 3} \approx \frac13 + 1.51078671394 \; î $ and is a bit off.