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TREE(3) is an extremely large number that requires ordinal arithmetic to prove it is finite. For what value of n would $G(n)>TREE(3)$? The length of the Goodstein sequence $G(n)$ is how many numbers in the sequence beginning with n are traversed in reaching 0. Proving that $G(n)$ is finite for all n also requires ordinal arithmetic and I'm more familiar with the Goodstein sequence.

For $n\ge8$, an $\omega^{\omega+1}$ sequence of decreasing ordinals is required to describe the Goodstein sequence and $G(12)$ is greater than Graham's number. For $G(16)$; $n\ge16$ a sequence beginning with $\omega^{\omega^\omega}$ iteration is required, and numbers that large are difficult to imagine even in terms of Graham's number type iterations. $G(65536)$ requires $\omega\uparrow\uparrow 4$ iteration.

How large a value of n would be required so that $G(2\uparrow\uparrow n)>TREE(3)$? I chose the $2\uparrow\uparrow n$ notation on the assumption that using $G(n)>TREE(3)$ might require an extremely large value for $n$.

I found this How large is TREE(3)? post on mathoverflow.net, which should be useful in putting a lower bounds on the answer to this question, but I'm still trying to understand the post on mathoverflow. A partial answer with the lower bounds answer would also be appreciated!

I also found this related mathstack question giving a lower bound for Kruskal's weak tree function, which along with the wikipedia TREE(3) article may give a lower bounds for this answer.

EDIT: The Fast-growing hierarchy along with another Tree(n) mathstack question provides a possible answer. Here is the Goodstein sequence written exactly in terms of the fast growing hierarchy; hopefully my notation here is correct:

  • $G(4)=f_3(3)-2$
  • $G(8)=f_{\omega+1}(3)-2$
  • $G(16)=f_{\omega^\omega}(3)-2$
  • $G(2\uparrow\uparrow n) = f_{\omega\uparrow\uparrow(n-1)}(3)-2$

The mathoverflow Tree3 link gives this inequality:

TREE(3) $\geq H_{\vartheta (\Omega^{\omega}, 0)}(n(4))$

Based on the links, the Fast-growing hierachy can be written in term's of Hardy's H function but I don't understand the $\Omega^\omega$ notation term in the inequality. The implication seems to be that *more * layers of diagonalization are required and perhaps $TREE(3) > G(2\uparrow \uparrow n)$ for any value of n that could possibly be written down!

Sheldon L
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    It's worth trying the Googology Wiki. – Benjamin Wang Oct 29 '23 at 21:26
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    Thanks for the link Benjamin! On the one hand, it appears that this is a very difficult question, yet on the other hand, Friedman writes, "Also, numbers derived from Goodstein sequences or Paris/Harrington Ramsey theory, although bigger than n(4), are also completely UNNOTICEABLE in comparison to TREE[3]." So perhaps $G(2\uparrow\uparrow n)<TREE(3)$ for any value of n that we could hope to write down as a decimal integer .... – Sheldon L Oct 29 '23 at 22:00

1 Answers1

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How large a value of n would be required so that $G(2\uparrow\uparrow n)>TREE(3)$?

perhaps $TREE(3) > G(2\uparrow \uparrow n)$ for any value of n that could possibly be written down!

Yes, and this should not be surprising! Using the fast-growing hierarchy with fundamental sequences defined as in the reference below at Footnote(2), we have, for all $n\gt 2,$ $$G(2\uparrow\uparrow n) = f_{\omega\uparrow\uparrow(n-1)}(3)-2 \lt f_{\epsilon_0}(n)$$

whereas$^\dagger$

$$TREE(3) \gt f_{\epsilon_0+1}(k) $$ where $k$ is Graham's number. (Using $\epsilon_0+1$ here is an extremely conservative choice.)

If $n^*$ denotes the least $n$ s.t. $G(2\uparrow\uparrow n)>TREE(3),$ then $n^*$ must be greater than the least $n$ s.t. $$f_{\epsilon_0}(n)\gt f_{\epsilon_0+1}(k)=f_{\epsilon_0}^{k}(k), $$ i.e., $$n^* \gt f_{\epsilon_0}^{k-1}(k).$$

The decimal digits of Graham's number $k$ are already impossible to "write down", let alone doing so for the digits of $\underbrace{f_{\epsilon_0}(f_{\epsilon_0}(\dots f_{\epsilon_0}(k)\dots ))}_{k-1 \text{ applications of } f_{\epsilon_0}}.$

NB: The cited references prove (rigorously) that $TREE(3) > f_{\Gamma_0}(k).$ Thus $f_{\epsilon_0}(n^*)\gt f_{\Gamma_0}(k),$ from which we could derive much stronger results of the form $n^*>f_{\epsilon_0}^{-1}f_{\Gamma_0}(k);$ but nothing stronger than our very weak bound is needed to answer your question.

$^\dagger$ This extremely weak lower bound is proved by combining the following two facts that are established at the cited links:

(1) https://mathoverflow.net/a/95588/20307: $$TREE(3) \gt tree(n(4))\gt tree(5)$$

(2) https://recursion-theory.blogspot.com/2020/05/lower-bounds-for-tree4-and-tree5.html: $$tree(5) > f_{\Gamma_0}(\text{Graham's number}) \gt f_{\epsilon_0+1}(\text{Graham's number}) $$

r.e.s.
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  • Are there any differences in the definitions of the Hardy fast growing hierarchy for the goodstein function and for the Tree function that would make it difficult to prove: $$TREE(3) \ge H_{\vartheta (\Omega^{\omega}, 0)}(k) \gg f_{\epsilon_0+1}(k) $$ In the https://googology.fandom.com/wiki/TREE_sequence footnote 3 says that estimates for a lower bounds of TREE(3) cannot be made rigorous. – Sheldon L Nov 08 '23 at 20:58
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    @SheldonL I prefer to answer your question (the one I quoted) without addressing any controversy about the lower bound $H_{\vartheta (\Omega^{\omega}, 0)}(n(4)).$ Thus, I've edited my answer to avoid using that lower bound altogether. – r.e.s. Nov 09 '23 at 02:02
  • The equations are brilliant, and I especially like the approach in $G(2\uparrow\uparrow n)$ in terms of $f_{\epsilon_0(n-1)}-2$ and then showing that Tree(3) is at least $f_{\epsilon_0+1}$. Thanks! – Sheldon L Nov 10 '23 at 03:19