There is a way to write TREE(3) via $F^a(n)$?

Question

I read about Graham number and TREE(3).
Graham number is: $f^{64}(4)$ where $f(n)=3\uparrow^n 3$

My question is: If there is a way to write TREE(3) via $f^a(b)$?
(and of course $f(n)$ can be different, but at the form of: $ x\uparrow^yz $)

Thank you!

To clarify : Do you mean the pure existence of such a number $a$ , or that a "reasonable-sized" $a$ does the job ? — Peter, Jun 24 '18 at 12:45
Concerning the question whether we can hit $TREE(3)$ EXACTLY , this is currently unknown and might never be known. But I agree to Arthur that chances are almost $0$ that a function of the kind you want does, unless you start with $TREE(3)$ which would however be "cheating" — Peter, Jun 24 '18 at 12:58
Not sure about whether [tag:hyperoperation] is an appropriate tag here. But number theory certainly isn't. — Jyrki Lahtonen, Jun 24 '18 at 17:11

score 2 · Accepted Answer · answered Jun 24 '18 at 12:42

2

No, $TREE(3)$ is in a completely different league. Graham's number has level $f_{\omega+1}$ in the fast-growing hierarchy, $TREE(3)$ is far beyond level $f_{\Gamma_0}$

So, the index $a$ in the function you want to arrive at $TREE(3)$ would be indistinguishable of $TREE(3)$ itself.

answered Jun 24 '18 at 12:42

Peter

86,576

1

Sorry for the question, but what is $f_{\Gamma_0}$? I try to find the answer here: https://en.wikipedia.org/wiki/Fast-growing_hierarchy but I didn't found.... Thank you!! – CS1 Jun 24 '18 at 13:00
Check for example Saibians site : https://sites.google.com/site/largenumbers/home – Peter Jun 24 '18 at 13:02
The next major milestone above $f_{\omega}$ is $f_{\omega^2}$ , which grows like a Conway chain of $n$ $n's$ , then it continues to $f_{\omega^\omega}$ and so on until we arrive at $f_{\epsilon_0}$ which is nothing compared to $f_{\Gamma_0}$ – Peter Jun 24 '18 at 13:04
You mean this: https://sites.google.com/site/largenumbers/home/4-2 ? – CS1 Jun 24 '18 at 13:05
Even "insane huge" would not emphasize enough the magnitude of $TREE(3)$ – Peter Jun 24 '18 at 13:06
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Yes, this is exactly the chapter you can read to get a feeling for the fast growing hierarchy. – Peter Jun 24 '18 at 13:07
And BTW, what is the symbol "#" means? like E100#^#100#(1+E100#^#100#3) from here: https://sites.google.com/site/largenumbers/home/appendix/a/numbers3 – CS1 Jun 24 '18 at 13:07
Let us continue this discussion in chat. – Peter Jun 24 '18 at 13:07
@CS1 See also: David Metzler's and Giroux Studio's videos on extremely large numbers. Also feel free to stop by in my chatroom if you wish to discuss this with me. – Simply Beautiful Art Aug 04 '18 at 21:37

score 2 · Answer 2 · answered Jun 24 '18 at 12:48

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The numbers of the form $x\uparrow^y z$ are quite sparse. So while it's definitely possible to write a number larger than TREE(3) that way (not physically in our universe, but mathematically possible), I sincerely doubt any of them are equal to TREE(3).

answered Jun 24 '18 at 12:48

Arthur

204,511

I understand. We can also find $a,b$ such that: $a+b>TREE(3)$, but we can't write them at our universe :-) But you answer my question because when you wrote: ...not physically in our universe... - This what I look for! – CS1 Jul 23 '18 at 11:37

There is a way to write TREE(3) via $F^a(n)$?

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