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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?

Abanob Ebrahim
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  • An upper bound is , if $G=G64$ denotes Graham's number, $3 \uparrow^{G_n+1} 3$ which is slightly larger than $G_{65}$ , hence the level is still $f_{\omega+1}$ , as Graham's number. In fact, this number is not "much larger than" $G$ in googology-standards. Maybe , someone else can prove that it is vastly smaller than $G_{65}$. – Peter Oct 17 '23 at 14:49
  • Hi @Peter I am not sure I understand why G64 ↑↑ G64 should be smaller than G65. Would you please explain in more detail why this would be the case? – Abanob Ebrahim Oct 18 '23 at 14:16
  • This is due to Saibian who explained that teration (taking $N\uparrow \uparrow N$) has virtually no effect , if $N$ is way beyond the tetrational level. This is of course not a proof , but he proved a theorem that might help to upper-bound this expression , but I am not an expert in such (usually difficult) manipulations. Maybe , it helps if you read his site about large numbers. Just google "large number site" – Peter Oct 18 '23 at 16:16

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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 $k\ge 6.$)

r.e.s.
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