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Let us define these two sequences as follows:

$a_0=1$, $b_0=c$

$a_{n+1}=x^{a_n}$, $b_{n+1}=x^{b_n}$

$b_{n+1}\ne b_n$ for any $n$.

$x,c\in\mathbb C$

Is it possible for $a_n$ to diverge but $b_n$ to converge under these conditions? For example, if $x=2$ and $c=i$, we have

$b_1=2^i=\operatorname{cis}(\ln2)$

$b_2=2^{\operatorname{cis}(\ln2)}=2^{\cos(\ln2)}\operatorname{cis}(\ln2\sin(\ln2))$

etc.

I haven't much clue as to whether it is the case that $b_n$ can converge when $a_n$ diverges, and I can hardly work out if $b_n$ converges with $x=2$ and $c=i$.

Simply Beautiful Art
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    I hope this and this will help you. –  Dec 29 '16 at 14:09
  • @Rohan I'm not so sure, both papers cover the topic of my sequence $a_n$, which I already know about, I'm more interested in $b_n$. – Simply Beautiful Art Dec 29 '16 at 16:06
  • Then the second paper may be useful. –  Dec 29 '16 at 16:09
  • Hmm, the clause $c \ne x^c$ intends to exclude the "trivial" case of $c$ being the fixpoint. But consider cases , where the exponentialtower to base $x$ has a *set of cyclic* fixpoints (which might also be accumulation points such that the height-increasing exponentialtower of $b_n$ "converges" to a cycle over such a set of points) then you might want to add an analogue clause/to extend the given clause. (I've not yet a true answer to your question so far) – Gottfried Helms Dec 30 '16 at 07:38
  • hmmm, for real values of x, the first requirement is that x>exp(1/e) so that the first power tower diverges. But all complex fixed points for bases>exp(1/e) are repelling. So doesn't that mean the 2nd power tower won't converge? And therefore there are no real valued solutions to the Op's problem? – Sheldon L Dec 30 '16 at 15:31
  • @SheldonL Hm, I'm not entirely sure. I'm also wondering about $x<e^{-e}$. – Simply Beautiful Art Dec 30 '16 at 15:32
  • @SimpleArt ouch!, but I think. $0<b<e^{-e}$ has repelling fixed points.... negative bases? complex bases? it might be easier to work with iterating $z \mapsto \exp(z)+k;;;;k=\ln(\ln(b))$ which is congruent to $y \mapsto b^y;;;;y=(z-k)/ln(b)$ – Sheldon L Dec 30 '16 at 16:12
  • @SheldonL Oddly, though I am asking such questions, such vocabulary is new to me. But it's ok, I get the gist of what you are saying. If you can think up a worthy answer, go for it and I'll ask clarifications later. And do complex fixed points behave like real ones? It seems quite hard to discern – Simply Beautiful Art Dec 30 '16 at 16:15
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    Even considering complex values of x, I believe the surprising result is that there are no solutions! I wrote a pari-gp loop that verified this for over 300,000 real and complex bases with attracting fixed points .... I think this is the case since I think the domain of the basin of attraction of the attracting fixed point can be extended to f'=0; but I will need to do some reading to brush up on complex dynamics, and iterated functions. – Sheldon L Dec 31 '16 at 04:04
  • @SheldonL: if you'll get something substantial - I'd like to see a workout also in our tetrationforum! (It seems to be an interesting facet in context with the D. Shell article) – Gottfried Helms Dec 31 '16 at 10:41
  • @SheldonL If you still wish to I would be interested in seeing your numerical results. – Simply Beautiful Art Dec 19 '19 at 00:42
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    One can work out that the only times that $x^{\ldots^c}$ can converge without hitting a fixed point is whenever $W(-\log(x))$ has absolute value at most $1$, where $W$ is the product log (and there are definitely converging $c$ when this absolute value is strictly less than $1$). I would maybe bet that there's an example of this happening where the absolute value is exactly $1$ and that where it's less than $1$, the tower converges starting at $1$. – Milo Brandt Dec 19 '19 at 01:37
  • @miloBrandt b=exp(1/e) has a neutral fixed point=e, and iterating b^^n converges towards e. – Sheldon L Dec 21 '19 at 00:23

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