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!

Answer 1

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.

Answer 2

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).