3
$\begingroup$

$\newcommand\tetra[2]{{^{#1}{#2}}}$In a recent discussion on the Tetration Forum (see https://math.eretrandre.org/tetrationforum/showthread.php?tid=1703&page=2), it has been pointed out how my results on the constancy of the "congruence speed" of the integer tetration $\tetra b a$ (a peculiar property of hyper-$4$ described in The congruence speed formula), would imply that $\lim_{b \rightarrow \infty} \tetra b a$ is always an element of the set $\mathbb{Q}_p$.

At the time, I did not fully realize the power of this breakthrough, since my original goal was only to find a function $V(a) : \mathbb{Z}^+-\{M\} \rightarrow \mathbb{Z}^+$, where $M = \{a : a \not\equiv 0 \pmod {10} \land a \neq 1\}$, but after a little search, I have seen that this is a general open problem, explicitly solved only for a few cases (since 1953, for $a:=2$). Furthermore, I have not managed to find any proof that even $\lim_{b \rightarrow \infty} \tetra b 6$ is a rational $p$-adic number (or rather an irrational one, disproving the aforementioned claim).

Now, my first concern is if I am wrong and $\lim_{b \rightarrow \infty} \tetra b 6$ has already been proven to be an element of $\mathbb{Q}_p$ (or not), while my general request here is to know more about the implications of Equation 16 on page 15 of Ripà and Onnis - Number of stable digits of any integer tetration on the above mentioned general open problem (and related topics).

Improve this question
$\endgroup$
11
  • $\begingroup$ For those wishing to work on this without knowledge of tetration: Does the sequence $$6,\quad 6^6,\quad,6^{6^6},\cdots$$converge in the $p$-adic integers? That is: is this sequnce eventually constant mod $p$ and mod $p^2$ and mod $p^3$ and so on? Here $p$ is a prime, and perhaps the answer varies with $p$. $\endgroup$
    – Gerald Edgar
    Commented Feb 25, 2023 at 14:30
  • 2
    $\begingroup$ @Marco ChatGPT is stupid at math. Ignore anything it says. $\endgroup$
    – KConrad
    Commented Feb 25, 2023 at 15:21
  • 1
    $\begingroup$ If $a \equiv 1\bmod p$ for $p \not= 2$ and $a \equiv 1\bmod 4$ for $p = 2$, we have $|a^m -a^n|_p = |a-1|_p|m-n|_p$ for $m, n$ in $\mathbf Z$. So $|a^a -a|_p = |a-1|_p|a-1|_p = |a-1|_p^2$, $|a^{a^a}- a^a|_p = |a-1|_p|a^a- a|_p = |a-1|_p^3$, and in general $|^{b}a-\, ^{b-1}a|_p = |a-1|_p^{b}$. So $|^{b}a -\, ^{b-1}a|_p \to 0$ as $b \to \infty$. In the $p$-adics, a sequence $\{x_n\}$ is Cauchy if and only if it is "consecutively Cauchy" (meaning $|x_n - x_{n-1}|_p \to 0$. So $\{^b{a}\}_{b \geq 0}$ has a limit for $a \in 1 + p\mathbf Z_p$ if $p\not=2 $ and for $a\in 1 + 4\mathbf Z_2$ if $p = 2$. $\endgroup$
    – KConrad
    Commented Feb 25, 2023 at 15:24
  • 2
    $\begingroup$ So is the question really about convergence, or whether the limit is a rational number? $\endgroup$
    – R. van Dobben de Bruyn
    Commented Feb 25, 2023 at 15:50
  • 2
    $\begingroup$ I still have no idea what the actual question is. Please add more details to make it a self-contained precise mathematical question (or maybe ask a separate one, given that the question whether the limit exists in $\mathbf Q_p$ has been answered below). I do not see the link between speed of convergence and whether the limit is in $\mathbf Q \cap \mathbf Z_p$. Or do you mean something else by 'rational'? $\endgroup$
    – R. van Dobben de Bruyn
    Commented Feb 25, 2023 at 16:50

1 Answer 1

Reset to default
11
$\begingroup$

A sequence given by $x_1=a$, $x_{n+1}=a^{x_n}$, where $a$ is a positive integer, is eventually constant modulo every positive integer $T$. This is widely known.

A short proof. We induct in $T$, so assume that $T>1$ and the claim is proven for smaller numbers. Denote $T=T_1T_2$, where $T_1$ has prime divisors which divide $a$, and $T_2$ is coprime with $a$. Modulo $T_1$, obviously $x_n$ become equal to 0. Modulo $T_2$, the value of $x_n$ is determined by the remainder of $x_{n-1}$ modulo $\varphi(T_2)<T_2$, thus by induction hypothesis it is eventually constant.

I hope that this is what was asked.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Ah, I was just writing out a similar argument, but yours is even slicker. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Feb 25, 2023 at 15:48
  • $\begingroup$ Also for $a=-1$, that being a unique case for no positive integers. $\endgroup$
    – Oscar Lanzi
    Commented May 16 at 12:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .