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Suppose $p$ is a prime number and $c$ is some constant value which is coprime to $p$. I found that $c,c^c,c^{c^c},c^{c^{c^c}},\ldots \bmod p$ have period $1$ or $2$.

In other words, it seems sequence,$(\ ^nc )_{\geq1}$, have period $1$ or $2$.

Can we prove this?

The code which I used is here: link

Gottfried Helms
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ueir
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    Are you specifically taking each $a_j$ to be between $0$ and $p$? If $p=3$ and $c=2$ and $a_0=1$, then the $a_n$ alternate back and forth between $2$ and $1$. – Greg Martin Dec 01 '19 at 09:20
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    The limit is simply the solution of the problem $c^x\mod (p)=x$ – Uday Khanna Dec 01 '19 at 09:23
  • I'm sorry. I specifically choose $a_0=c$ in the code. I change the statement of question now. – ueir Dec 01 '19 at 09:36
  • In the example Mindlack provides even with $a_{0}=c=3$ we don't have a solution. I see it as a problem of fixed points, you could even reinterpret it as a problem in Group theory by considering then as permutations if you don't want to apply number theory. – Uday Khanna Dec 01 '19 at 09:37
  • Maybe you could provide the numerical data you were dealing with. – Uday Khanna Dec 01 '19 at 09:40
  • May I suspect, that you actually work with $c \pmod p$, $c^c \pmod p$,$c^{c^c} \pmod p$ ,$ \cdots \pmod p$ instead? (Note that this formulation against yours gives different results, possibly according better to your proof -intent and I remember to have looked at this in the context of modular tetration sometime ago and might have seen the constant sequence for many examples) – Gottfried Helms Dec 01 '19 at 09:53

  • Gottfried Helms I'm sorry for confusing. Yes.

     – ueir Dec 01 '19 at 09:55
  • Ah, well, so possibly you better use the notation (known from discussion of tetration) $a_n(c,p)= ,^n c \pmod p \qquad \text{ for } n>0 $ and whether $a_{n+1}(c,p)=a_n(c,p) \qquad \text{for } n \gt n_0$ with some $n_0$ – Gottfried Helms Dec 01 '19 at 09:59
  • I check the sequence once again and it's not converge at all. The period is 1 or 2.I add code which I used. https://wandbox.org/permlink/VLWZrrutUiQaL7NV – ueir Dec 01 '19 at 10:03
  • ueir - if my previous comment is what really happened then you might also add the tag tetrationto your question. – Gottfried Helms Dec 01 '19 at 10:04
  • Thank you. I add a tag and change the notation. – ueir Dec 01 '19 at 10:15

2 Answers2

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That’s false. Consider $c=3$ and $p=7$. Note that $c^2=2$, $c^4=4$ mod $7$. So depending on the value of $a_0$, you can have two constant sequences. Worse: with $a_0=3$, the sequence is a cycle $3,6,1,3,6,1,\ldots$.

Aphelli
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  • I'm sorry. I specifically choose $a_0=c$ in the code. I change the statement of question now. – ueir Dec 01 '19 at 09:36
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update
If you actually wanted to prove that $ \,^nc \pmod p$ converge to a constant sequence then the following might help for a proof:

assume $p=7$, $c=5$
Then what you ask for is whether $\{5, 5^5 , 5^{5^5}, ...\} \pmod 7$ converge to a constant sequence.
This is a question of recursive application of "order of cyclic subgroup" :

  • we know $5^k \pmod 7= 5 ^ { k \pmod {\varphi(7)}}\pmod 7= 5 ^ { k \pmod 6}\pmod 7$
  • Now if $k$ is itself a power of $5$ then we ask furtherly for the residues of $k=5^j \pmod 6$.
  • we find that this is $5^{j \pmod 2} \pmod 6$
  • And one step higher this becomes a constant sequence.

Here it is surely meaningful to try a proof. (I think, this should be easy to derive one from that example)

old version (removed. You can see it in the "edit-history")

Gottfried Helms
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    This answer seems to show that the sequence is eventually constant. – pregunton Dec 01 '19 at 10:31
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    The key fact is $\phi(p),\phi(\phi(p)),\phi(\phi(\phi(p))),\ldots$ is strictly decreasing sequence until it becomes 1. Then, maybe there is some mistake in the code. Thank you. – ueir Dec 01 '19 at 10:35
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    @pregunton - a very nice answer you've found! Perhaps it could be a bit shaped (upps my comment crossed with that last one of ueir) – Gottfried Helms Dec 01 '19 at 10:36