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Instead of base 10 or base e, let the base of our logarithm be a positive integer B. Take log base B of TREE(3), B times, or stop if the answer is negative before B iterations. Say that TREE(3) does not "survive" if the answer is negative for B or fewer iterations. Now, if B is Graham's Number, will TREE(3) survive? What if B is Bowers array [3,3,3,3,3]?

Richard Peterson
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    So basically you're asking whether $$ \underbrace{\log \log \log \ldots \log}{g{64}} \left( \text{TREE}(3) \right)<0 $$, right? – Matti P. Nov 10 '23 at 12:49
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    The smallest number that survives $g_{64}$ many applications of the logarithm is a power tower of height $g_{64}$. In the grand scheme of huge numbers, this isn't that much bigger than $g_{64}$. It's for instance much much smaller than $g_{65}$, which is much smaller than $\mathrm{TREE}(3)$. See also this. – Izaak van Dongen Nov 10 '23 at 13:02
  • Matti P., when the base of the logarithm is the same as the number of iterations, which is B. B might be Graham's number. I suggested a larger number also. – Richard Peterson Nov 10 '23 at 17:29

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Let $G$ be Graham's number and let the Bower's array be $b = [3,3,3,3,3].$ Now, according to here we have $G\lt b<f_{\omega^3}(3),$ and according to here we have $f_{n\uparrow^n n}(n)\lt f_{\omega+1}(n)$; furthermore, here proves $f_{\epsilon_0}(G)\ll f_{\Gamma_0}(G)\lt TREE(3)$. Also, to illustrate just how conservative these bounds are, we note that

$b\uparrow^b b \\ = b\uparrow^{b-1} (b\uparrow^{b}(b-1))\\ \gg b\uparrow\uparrow (b\uparrow^{b}(b-1)) $

hence

$b\uparrow\uparrow (b\uparrow^{b}(b-1))\\ \lt b\uparrow^b b\\ \lt f_{b\uparrow^b b}(b)\\ \lt f_{\omega+1}(b)\\ <f_{\omega+1}f_{\omega^3}(3)\\ \lt f_{\omega^3+1}(3)\\ \lt f_{\epsilon_0}(G)\\ \ll TREE(3).$

Therefore, if $B$ is either $G$ or $b$, then $$TREE(3)\gg B\uparrow\uparrow (B\uparrow^{B}(B-1))$$

yielding

$$\underbrace{\log_B\log_B\dots\log_B}_{B\text{ applications of $\log_B$}}\ TREE(3)\gg B\uparrow\uparrow (B\uparrow^{B}(B-1)-B)$$

i.e., $TREE(3)$ survives this by a margin of inconceivable magnitude: it would survive even if $\log_B$ were applied vastly more than a further $B\uparrow^{B}(B-1)-B$ times!

NB: I'm adding the following to show just how understated are the above results ...

Your questions are special cases (for $B=G$ and $B=[3,3,3,3,3]$) of this:

How big is $B$, if ${\log_B}^B TREE(3)< 0$?

Applying the operation $B\uparrow $ repeatedly to both sides of the inequality, this can be rewritten as:

How big is $B$, if $B\uparrow\uparrow(B-1) >TREE(3)$?

More generally, when $F$ is some given increasing function $F$:

How big is $n$, if $F(n)>TREE(3)$ ?

Defining $n^*:=\min\{n: F(n)>TREE(3)\},$ a recent answer elsewhere shows that $$n^*>\underbrace{f_{\epsilon_0}(f_{\epsilon_0}(\dots f_{\epsilon_0}(G)\dots ))}_{G-1 \text{ applications of } f_{\epsilon_0}}\tag{1}$$

if $F(n)\le f_{\epsilon_0}(n)$ for all $n>1$, where $(f_\alpha)$ is a specific fast-growing hierarchy.

Therefore, because $B\uparrow\uparrow(B-1)\le f_{\epsilon_0}(B)$ for all $B>1$, that same bound in (1) applies to your question as well; i.e., $TREE(3)$ will "survive" if
$$B\lt \underbrace{f_{\epsilon_0}(f_{\epsilon_0}(\dots f_{\epsilon_0}(G)\dots ))}_{G-1 \text{ applications of } f_{\epsilon_0}}$$

which is certainly the case for both $B=G<f_{\omega+1}(64)$ and $B=[3,3,3,3,3]<f_{\omega^3}(3).$

An irony here is that although this result makes the previous one look puny, it also happens that the same methods described at the cited link easily give much much stronger bounds than this one!

r.e.s.
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