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It is well-known that (in radix-$10$) Graham's number, $G$, can be expressed as a tetration with base $3$ and a very large hyperexponent $\tilde{b}$. Thus, we can write that $\exists! \hspace{1mm} \tilde{b} \in \mathbb{Z}^+ : G=g_{64}={^{\tilde{b}}3}$.

In recent years, by assuming the decimal numeral system as above, I proven a peculiar property of tetration involving every integer base that is not a multiple of $10$, and in this paper published in 2021 I named it the constancy of the congruence speed of tetration, while this other paper published in 2022 fully describes it.
In particular, if the tetration base is $3$, then ${^{b}3} \equiv {^{b+c}3} \pmod {{10}^{b-1}} \wedge {^{b}3} \not\equiv {^{b+c}3} \pmod {{10}^b}$ holds for all $b,c \in \mathbb{Z}^+$.

Now, I need to compactly state which is the exact number of stable digits of Graham's number, which corresponds to the value $\overline{b}:=\tilde{b}-1$ (since ${G} \equiv {^{\overline{b}}3} \pmod {{10}^{\tilde{b}-1}} \wedge {G} \not\equiv {^{\overline{b}}3} \pmod {{10}^\tilde{b}}$ holds for all $\mathbb{Z} \ni \overline{b}>\tilde{b}$).

Thus, all I need to get is a very compact definition of $\tilde{b}$ by knowing that $G={^{\tilde{b}}3}$.
Which is the most elegant/efficient way to achieve this goal?

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  • $\begingroup$ Could you clarify the question? $G= {}^b3$ is almost definitely the simplest definition you can hope to get. $\endgroup$ Commented Jan 3, 2024 at 14:43
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    $\begingroup$ @CommandMaster Yeah, $\tilde{b}$ such that $G={^{\tilde{b}}3}$ is certainly a short proposition, but my goal is to define $\tilde{b}$ (or $\overline{b}:=\tilde{b}-1$) more explicitly. I mean, tetration has two inverse operations (i.e., super-root and super-logarithm) and maybe we could simply state that $\tilde{b}=Slog_3(G)$, specifying also the definition of $G$ as a reference... Would it be clear enough in your opinion? $\endgroup$ Commented Jan 3, 2024 at 17:14

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Let me update the thread just to state that the exact number of stable decimal digits of Graham’s number is $\operatorname{slog}_3(G)-1$ (where "slog" is the integer super-logarithm and $G$ denotes Graham’s number, as usual).
The full proof is given in my paper Graham's number stable digits: An exact solution, Notes on Number Theory and Discrete Mathematics, 31(3), 2025, pp. 607–616 (DOI: 10.7546/nntdm.2025.31.3.607-616). It relies on the fact that $G$ is a (huge) base-$3$ tetration, and that base 3 has congruence speed $0$ at height $1$ and stabilizes to $1$ from height $2$ onwards (thus remaining unitary thereafter).

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    $\begingroup$ First sentence on p. 608 (2nd page) of your paper: In detail, assuming that each digit of Graham’s number occupies one Planck length cubed (i.e., a Planck volume), the observable universe is far too small to contain an ordinary digital representation of $G.$ A comparison such as this is so far off that it’s misleading. There are roughly ${10}^{185}$ Planck volumes in the observable universe, so you're talking about a number with ${10}^{185}$ digits. Compare that with my description of 2^^(2^^(2^^9)), (continued) $\endgroup$ Commented Oct 13 at 20:09
  • $\begingroup$ a number too small to usefully express using just $3$ up-arrows $\uparrow\uparrow\uparrow,$ in Comment #157 in Scott Aaronson's 28 June 2025 Shtetl-Optimized blog post BusyBeaver(6) is really quite large. $\endgroup$ Commented Oct 13 at 20:09
  • $\begingroup$ I cannot see anything misleading there: it is an introductory sentence where “far too small” means exactly that $G$ >> $10^{185}$. Even $G$ >> $g_{63}$. $\endgroup$ Commented Oct 15 at 4:21

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