Where is $G64\uparrow \uparrow G64$ in the Fast-growing hierarchy?

Question

I am trying to find out how $G64\uparrow \uparrow G64$ can be represented by the Fast-growing hierarchy, but I do not know how this can be done.

Is there a way to simply convert between those two notations?

Answer 1

By the Knuth Arrow Theorem (KAT) we have, for any integers $x \ge 2, a \ge 2, b \ge 1, c \ge 1,$ $$(a \uparrow ^x b) \uparrow ^xc < a \uparrow ^x(b+c).$$

Thus, with $G=G_{64},$ we have

$G_{65}\\ =3\uparrow^G3\\ =3\uparrow^{G-1}(3\uparrow^{G-1}3)\\ \gt (3\uparrow^{G-1}3)\uparrow^{G-1}(3\uparrow^{G-1}3-3)\quad \text{by KAT}\\ \gg G\uparrow^{G-1}G\\ \gg G\uparrow^2 G $

Therefore, $G\uparrow^2 G$ is still at level $f_{\omega+1}$ in the FGH, because $G_{65}\lt f_{\omega+1}(65)$. (It can be shown by induction that $G_{k}\lt f_{\omega+1}(k)$ for all $k\ge 6.$)