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Define $$a_n=3^{a_{n-1}}$$ for $n\ge 2$ with $a_1=3$. I observed that $$a_{11}=\ldots 2464195387,$$ $$a_{12}=\ldots 2464195387,$$ $$a_{13}=\ldots 2464195387$$ and this seems to hold as we continue increasing the index of $a$. I.e. the last decimal digits seem to be the same for all $a_n$ with $n\ge 11$.

How is this even possible? Can someone explain this? I'm familiar with Euler's theorem in number theory.

You can check the results with e.g. https://www.wolframalpha.com/input?i=3%5E3%5E3%5E3%5E3%5E3%5E3%5E3%5E3%5E3%5E3%5E3

Bill Dubuque
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Nomas2
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    Because $,3^a \equiv a\pmod{!10^{10}},$ for $,a = 2464195387.,$ Similar matters are discussed in prior questions. – Bill Dubuque Nov 11 '23 at 22:36
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    @BillDubuque How did you find this identity? – Nomas2 Nov 11 '23 at 22:41
  • Here is a related question with some explanations: Do the last digits of exponential towers really converge to a fixed sequence?. – Minus One-Twelfth Nov 11 '23 at 22:52
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    $a_{n+1} \equiv a_n!\iff 3^{a_n} \equiv a_n.,$ If so then by $\rm\color{#c00}{induction}$ $,a_{n+k} \equiv a_n,$ for all $k\ge 0,,$ i.e. $,a_{n+k} = 3^{\large \color{#c00}{a_{n+k-1}}}! \equiv 3^{\large\color{#c00}{a_n}}! \equiv a_n,,$ i.e. fixed points stay fixed upon iteration $,f(a_n) \equiv a_n \Rightarrow f^k(a_n) \equiv a_n\ \ $ – Bill Dubuque Nov 11 '23 at 23:01
  • @BillDubuque I meant how did you find $3^{2464195387}\equiv 2464195387\mod 10^{10}$? – Nomas2 Nov 11 '23 at 23:16
  • As I explained in my prior comment, that follows immediately from your equations, i.e. from the fact that $,a_{12},$ and $,a_{11},$ have the same last $10$ digits, i.e. $,a_{12}\equiv a_{11}\pmod{!10^{10}}\ \ $ – Bill Dubuque Nov 11 '23 at 23:21
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    @BillDubuque I know, but I would like to know how "my results" can be proved without WolframAlpha. – Nomas2 Nov 11 '23 at 23:22
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    Here's why: $,a\equiv b\pmod{!10^{k-1}}\Rightarrow 3^a\equiv 3^b\pmod{!10^{k}},$ for $k>2,,$ by ${\rm lcm}(\phi(2^k),\phi(5^k)) = 10^{k-1}.,$ So for $,a,b =, a_{k+1},,a_k$ this yields that if $,a_{k+1},$ and $,a_k,$ have the same last $k!-!1$ digits then $,a_{k+2},$ and $,a_{k+1},$ have the same last $,k,$ digits, so we get one more agreeing digit for each increment of $,k.\ \ $ – Bill Dubuque Nov 12 '23 at 00:54
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    $$\begin{align} {\rm e.g.}\ \ \ 3!\uparrow\uparrow!4 \equiv 3!\uparrow\uparrow!3 &\equiv\ \ 87! \pmod{!10^2}\[.3em] \Rightarrow\ \ \ 3^{3\uparrow\uparrow 4}\ \equiv\ 3^{3\uparrow\uparrow3}\ &\equiv\ \ 3^{87}!!!\pmod{!10^3}\[.3em] {\rm i.e.}\ \ \ 3!\uparrow\uparrow!5 \equiv 3!\uparrow\uparrow!4 &\equiv 387 !!\pmod{!10^3}\ \end{align}\qquad\qquad$$ – Bill Dubuque Nov 12 '23 at 02:21

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