Friedman, in http://www.math.osu.edu/~friedman.8/pdf/EnormousInt112201.pdf, shows that TREE(3) is much larger than n(4), itself bounded below by $A^{A(187195)}(3)$ (where $A$ is the Ackerman function and exponentiation denotes iteration). But actually, using the fastgrowing hierarchy, $n(p)$ is smaller than $f_{\omega^{\omega^\omega}}(p)$ (shown by Friedman in http://www.math.osu.edu/~friedman.8/pdf/finiteseq10_8_98.pdf), while it seems that TREE grows faster than $f_{\Gamma_0}$ (${\Gamma_0}$ being the FefermanSchütte ordinal). So it could well be that in fact TREE(3) is larger than, say, n(n(4)), or even any number expressible by iterations of n. What is known on this question?

I believe I can state with some confidence that TREE(3) is larger than $f_{\vartheta (\Omega^{\omega}, 0)} (n(4))$, given a natural definition of $f$ up to $\vartheta (\Omega^{\omega}, 0)$. I can state with certainty that TREE(3) is larger than $H_{\vartheta (\Omega^{\omega}, 0)} (n(4))$, where H is a certain version of the Hardy hierarchy. To obtain this result, I will first define a version of TREE(n) for unlabeled trees: Let tree(n) be the length of the longest sequence of unlabeled rooted trees $T_1, T_2, \ldots, T_m$ such that $T_i$ has less than or equal to $n+i$ vertices and for no $i, j$ with $i < j$ do we have $T_i$ homeomorphically embeddable into $T_j$. (Note the term "embeddable" rather than "subtree"; the terms are different, and I believe using "subtree" would lead to infinite sequences.) In order to obtain a long sequence of trees, we will define a wellorder on unlabeled rooted trees. This definition will be by induction on the sum of the heights of the two trees being compared. Define an immediate subtree of a rooted tree $T$ to be a full subtree starting at one of its children. Given two rooted trees $S, T$, we define $S = T$ if the two trees are identical. We define $S \leq T$ if $S = T$ or $S < T$. Given two rooted trees $S, T$, we define $ < $ as follows. Say $S < T$ if $S \leq T_i$, where $T_i$ is an immediate subtree of $T$. Similarly, say $T < S$ if $T \leq S_i$, where $S_i$ is an immediate subtree of $S$. Otherwise, compare the number of children of $S$ and $T$. If $S$ has more children than $T$, then $S > T$, and vice versa. Otherwise, suppose $S$ and $T$ both have $n$ children. Let $S_1, S_2, \ldots, S_n$ and $T_1, T_2, \ldots T_n$ be the immediate subtrees of $S$ and $T$ respectively, ordered from smallest to largest. Compare $S_1$ to $T_1$, then $S_2$ to $T_2$, etc., until we get a pair of unequal trees $S_i$ and $T_i$. If $S_i > T_i$ then $S > T$, and vice versa. Of course, of all pairs of immediate subtrees are equal, then $S$ and $T$ will be equal. This gives a linear order on unlabeled rooted trees, and one can prove that this is a wellorder. Further, this wellordering has order type $\vartheta(\Omega^\omega,0)$. This definition is a modification of a wellordering of ordered rooted trees due to Levitz, and expounded on in papers by Jervell. From this wellordering we can define fundamental sequences for ordinals up to $\vartheta (\Omega^{\omega}, 0)$. Simply put, given an ordinal $\alpha$, let $\alpha[n]$ be the largest ordinal less than $\alpha$ corresponding to a tree of $n$ vertices or less. From this, we can define our version of the Hardy hierarchy: $H_0(n) = n$ $H_{\alpha + 1}( n) = H_{\alpha}( n+1)$ For $\alpha$ a limit ordinal, $H_{\alpha}( n) = H_{\alpha[n+1]}( n+1)$ Note the $n+1$'s in the last line  this differs from the usual values of $n$. Of course, this will only make the functions larger. $H_{\alpha}( n)$ for $\alpha < \vartheta (\Omega^{\omega}, 0)$ is the final index $m$ in the sequence of trees $T_n, T_{n+1}, \ldots, T_m$ where $T_n$ corresponds to $\alpha$ and $T_i$ is the largest tree with at most $i$ vertices that is smaller than $T_{i1}$, and $T_m$ is the tree with one vertex. Thus $H_{\vartheta (\Omega^{\omega}, 0)}( n)$ will be the final index $m$ in the sequence of trees $T_{n+1}, T_{n+2}, \ldots, T_m$ where $T_{n+1}$ is arbitrary. Thus tree(n) $\geq H_{\vartheta (\Omega^{\omega}, 0)}( n)  n$. So where does TREE(3) come in? Harvey Friedman himself explains in a post to the Foundations of Mathematics message boards: http://www.cs.nyu.edu/pipermail/fom/2006March/010260.html In the post he explains why a proof of the theorem "TREE(3) exists" in the theory $ACA_0 + Pi^1_2  BI$ must have more than 2^^1000 symbols. He does this by showing that TREE(3) must be very large  specifically, he constructs a sequence of more than $n(4)$ rooted trees labeled from {1,2,3} such that $T_i$ has at most $i$ vertices, for no $i, j$ with $i < j$ do we have $T_i$ homeomorphically embeddable into $T_j$, and each tree contains either a 2 label or a 3 label. We can obviously continue this with tree($n(4)$) trees with all labels 1. Thus we have TREE(3) $\geq$ tree$(n(4)) + n(4) \geq H_{\vartheta (\Omega^{\omega}, 0)}(n(4))$ In fact, we can do somewhat better than this; we can replace the $n(4)$ above by $F(4)$, where $F(4)$ is defined as the length of the longest sequence of sequences $x_1, x_2, \ldots x_n$ from {1,2,3,4} such that $x_i$ has length $i+1$ and for no $i,j$ with $i < j$ do we have $x_i$ a subsequence of $x_j$. I can prove much better bounds for $F(4)$ than Friedman's lower bound for $n(4)$; specifically, $F(4) > f_{\omega^2 + \omega + 1}f_{\omega^2 + \omega + 1}f_{\omega^2 + \omega}f_{\omega^2 + 1}f_{\omega^2 + 1}f_{\omega^2}f_{ \omega + 1}f_{ \omega + 1}f_{\omega}(30)$ But such specificity is perhaps unwarranted given how far from TREE(3) it may be. 


I should perhaps have asked this question to Harvey Friedman himself... Anyway, a reasonable answer is given on the Talk page for Wikipedia article on Kruskal tree theorem : http://http://en.wikipedia.org/wiki/Talk:Kruskal%27s_tree_theorem#Correcting_TREE.282.29 , where one can find this quotation (from H. Friedman himself) : "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]." To precise my remarks (and answer Dylan Thurston), I meant that TREE(3) is bigger than expressions like $n^{n^{n(100)}(100)}(100)$, say, where again exponentiation means iteration, and the whole expression has no more than $n(4)$ symbols ; this would indeed be true if, for instance, as suggested, we had TREE(3)$>f_{\Gamma_0}(n(4))$. For reference, I should have added that TREE(3) is the incredibly (at first, or even second look) large answer to the question : "which is the length of the longest sequence $(T_2,T_3,T_4,\dots,T_n)$ of labeled trees such that $T_k$ has at most $k$ nodes labeled $a$ or $b$, and $T_i$ is not a subtree of $T_j$ for $i < j$ ?". 


(Following are two comments, posted this way because I ("r.e.s.") cannot post comments directly.) Comment on the answer by "Deedlit":
That's not quite right. His first tree $T_1$ uses label 3 (so this label cannot be used later at all), followed by more than $n(4)$ trees using labels 1,2  not, as you wrote, using labels 2,3. It's because of the way these latter {1,2}labelled trees are constructed, that they can nevertheless be followed by a long sequence of trees using only label 1. (I show the beginning of his sequence in my other comment below, using bracket expressions in which the brackettypes
Actually, we can do even better, although these may be relatively "small" adjustments: by playing with various ways to start a long embeddingfree sequence, one can improve upon the sequence constructed by Friedman (displayed below), but still using his method of coding $n()$ or $F()$type longest wordsequences via certain subtrees. For example, one can find sequences that demonstrate (in Deedlit's notation) TREE(3) $\ \geq \ $ tree$(N) + \\ N \ \geq \ H_{\vartheta (\Omega^{\omega}, 0)}(N)$ where $N \ = \ F_\omega F_\omega F_\omega F_{\omega+1} F_\omega F_\omega \ F(4)$ with $F_\alpha$ being a fastgrowing hierarchy that begins with $F_0 = F$, rather than beginning as usual with $F_0(x) = x+1$. (Friedman showed that $F$ eventually dominates every $f_{\lt \omega^\omega}$ in the usual fastgrowing hierarchy.) I've posted a very terse derivationsketch of this result. Comment on the answer by "Feldman Denis":
Rather than "is not a subtree of", that should be "is not homeomorphically embeddable into", which is a very much more stringent requirement. (There might not even exist a longest such sequence in the lessstringent case. A similar situation occurs for Friedman's $n()$ and $F()$ functions concerning wordsequences: these use "is not a subsequence of" rather than the lessstringent "is not a substring of", there being no longest wordsequence in the latter case.) With this correction, and by starting with $T_2$, the length of the resulting sequence will of course be TREE(3)  1. A convenient way of representing these trees is to use nested bracket expressions (wellformed in the usual way with pairs of matching brackets) involving only three brackettypes  say Because $X_1$ must be some single bracketpair which cannot then appear in any later nest in an embeddingfree sequence, it may be assumed that $X_1=\ $ To illustrate the use of bracket expressions, here is a representation of the initial tree sequence used by Friedman to prove the lower bound mentioned by the OP:
NB: It should be noted that the article linked by the OP does not treat Friedman's TREE function, but a rather different function TR. The confusion may be partly due to the fact that "TR" is also what Friedman called the TREE function before he changed it to the latter name in a followup article to the one mentioned in Deedlit's posting. 


Chris Bird has a series of articles that list functions with high growth rates, all the way to the BachmannHoward ordinal. They're all over my head, but they might be useful for getting a more accurate (although still extremely broad) approximation of TREE(3). 


What larger: TREE(4) or Busy Beaver(1000) or Dupertri = {3,tritri (1) 2} in BEAF? 

