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?
