@AkivaWeinberger Say, I found an approximation to $G(13)$, the amount of steps it takes for $13$ to reach $0$ in the Goodstein sequence.
$$G(13)\approx g_{2\uparrow\uparrow\left(2\uparrow \uparrow\left[2\uparrow\uparrow\left(2^{ 2^{2^{70}+70}+2^{70}+70}\right)\right]\right]}$$
Where
$$2\uparrow\uparrow n=\underbrace{2^{2^{2^{\dots}}}}_n$$
And $g_n$ is the $n$th term of Graham's sequence with $g_{64}$ being Graham's number.
See [here](https://math.stackexchange.com/a/2345650/272831) for an explanation.