1. $G_{64}$ is a Graham Number:


  1. $TREE(3)$ is a particular value of a special sequence $TREE(k)$


  1. $D^{5}(99)$ is a output of program loader.c


  1. $Rayo(googol)$ is a Rayo number


How to prove that $1.<2.<3.<4.$ ?

First of all i know that there were several related question but in fact nobody has given a proper proof of any of these inequalities. That's why i wrote this question.

Let's take a look at all cases.


It is said that sequence that generates Graham numbers grows slower than sequence $TREE(k)$ But here are my questions:

  • Where are the proofs, sources of this assertion?
  • Even if some function $f$ grows faster than $g$ it doesn't prove that $f(n)>g(n)$ for certain value $n$.


Here is my idea to prove this:

First make a code that describes (not calculate) $TREE(3)$ Start it and see how much time it takes before you see a message (for example "Hello World") on the screen. The same with the code that describes $loader.c$. Compare these numbers. The number that has greater time is greater. Here are another problems.

  • The code from the site i gave you in the link doesn't work on codeblocks.
  • Is my reasoning even correct? If not( wich is very likely) then how to do this?


There is only one thing to know: an amount of symbols expressing $D^{5}(99)$ in language of First-Order-Set-Theory . If that number is smaller than $googol$ then proof follows.

  • How we can show that?


  • $\begingroup$ Related: math.stackexchange.com/questions/1950116/… $\endgroup$ – Maximilian Janisch Nov 24 '19 at 17:00
  • 1
    $\begingroup$ Also related: How big is TREE(3)? $\endgroup$ – Maximilian Janisch Nov 24 '19 at 17:04
  • $\begingroup$ And Wikipedia $\endgroup$ – Maximilian Janisch Nov 24 '19 at 17:06
  • $\begingroup$ TREE(3) is so much larger than Graham's number that we do not nedd a formal proof. The level of Graham's number is just $f_{\omega+1}$, whereas TREE(3) FAR exceeds level $f_{\Gamma_0}$. Maybe this site sites.google.com/site/largenumbers is a good start. TREE(3) is so large that , as far as I know, no upper bound is known. $\endgroup$ – Peter Nov 24 '19 at 18:08
  • $\begingroup$ @Peter - Although there may exist $N$ such that $\text{TREE}(n)\ge f_{\Gamma_0}(n)$ for all $n\ge N$, that says nothing about $\text{TREE}(3)$, and would not imply that $\text{TREE}(n)\ge f_{\Gamma_0}(n)$ for all $n\ge 3$. I think the OP would ask for a proof that $N=3$ suffices in this case. $\endgroup$ – r.e.s. Nov 27 '19 at 20:06

In the article https://cs.nyu.edu/pipermail/fom/2006-March/010260.html

Friedman showed that $$\text{TREE(3)} >\text{tree}(n(4)) + n(4)\tag{1}$$

where $n()$ is Friedman's block-subsequence function and $\text{tree}(n)$ is defined as the length of a longest sequence of unlabelled trees $T_1,T_2,\ldots$, such that, for all $i$, $T_i$ has at most $n+i$ vertices, and for all $i,j$ with $i<j,$ $T_i$ is not homeomorphically embeddable into $T_j.$

(A sketch of his proof is appended to the present answer.)

Furthermore, at https://mathoverflow.net/a/95588/20307 there is a proof that

$$\text{tree}(n)\geq H_{\vartheta (\Omega^{\omega}, 0)}( n) - n\tag{2}$$

where $H_{\alpha}$ is an accelerated version of the Hardy hierarchy; hence, as pointed out there,

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

We note that the RHS is clearly larger than $G_{64},$ because $H_{\omega^{\omega+2}}(3)>f_{\omega+2}(3)>G_{64},$ where $f_\alpha$ is the usual fast-growing hierarchy.

The following is an attempt to briefly elucidate the proof that $$\text{TREE}(3) > \text{tree}(q) + q,$$ where $q>n(4)$, based on the tree sequence and symbol-encodings used by Friedman in the article cited above. I'll use balanced parentheses of types $(\,),[\,],\{\,\}$ to denote vertices labelled with $1,2,3$ respectively.

T_1   {}
T_2   [[]]
T_3   [()()]
T_4   [((()))]
T_5   ([][][][])
T_6   ([][][](()))
T_7   ([][](()()()))
T_8   ([][](()(())))
T_9   ([][](((((()))))))
T_10  ([][]((((())))))
T_11  ([][](((()))))
T_12  ([][]((())))
T_13  ([][](()))
T_14  ([][]())
T_15  (A(B(((([]))))))
T_20  (C(D(E(((([])))))))
T_25  (F(G(H(I(((([]))))))))
T_30  (J(K(L(M(N(((([])))))))))
T_q   (X(Y(...(Z((((([])))))...))))  where q = 10 + 5p
T_q+1 (((...(())...)))  with q+1 ()s
T_q+tree(q)  ()

Here, each letter A,B,C,... denotes one of the following symbol-codes for a 4-symbol alphabet {1,2,3,4}:

(((())))   <- codes the symbol 1
((()()))   <- codes the symbol 2
(()()())   <- codes the symbol 3
((())())   <- codes the symbol 4

Now, by Friedman's results on the function $n()$, there exists a $p$-long sequence of words $x_1,...,x_p$ on alphabet $\{1,2,3,4\}$ such that $|x_i| = i+1$ and for all $i<j$, $x_i$ is not a subsequence of $x_j$, where $p = {n(4)-1\over 2}$.

So tree-embeddings are avoided by choosing the symbol-codes $A,B,C,...$ such that $AB$ encodes $x_1$, $CDE$ encodes $x_2$, $FGHI$ encodes $x_3$, etc.

Hence the sequence of trees $T_1,...,T_q$ (where $q = 10+5p>n(4)$) is such that $|T_i| \le i$ and for all $i<j$, $T_i$ is not homeomorphically embeddable in $T_j.$

Then the sequence continues $T_{q+1},...,T_{q+\text{tree}(q)}$ for another $\text{tree}(q)$ trees after $T_q$, these trees being constructed with $(\,)$-vertices only. QED


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