Growth rate of longest sequence of strings where no string is a subsequence of a later one We define $STR(n)$ to be the longest sequence of strings with $n$ symbols such that the $k$th string has at most k symbols, the symbols of the string are taken from an alphabet consisting of $n$ characters, and no string is a subsequence of a later one.
For example, $STR(1)=2$, because the longest sequence is "A", "". $STR(2)=4$, because the longest sequence is "A", "BB", "B", "". I'm not sure how big $STR(3)$ is (although it is at least 11 due to the sequence "A","BB","BC","CB","B","CCCCC","CCCC","CCC","CC","C","").
We do know, however, that $STR(n)$ is always finite, for the same reason that $TREE(n)$ and $SSCG(n)$ (see well-quasi-order).
My question, what is the growth rate of $STR(n)$? It is slower than $TREE(n)+1$, since strings can embed in trees.
(In particular, if we could find $n$ such that $STR(n)>TREE(3)$, that would be great.)
 A: I suppose it's a good idea to turn my comment into an answer.
The function $STR$ is basically the function $F$ defined by Friedman in this paper (more precisely, it's easy to show $STR(k)=F(k-1)+1$). Friedman pinpoints the growth rate of this function quite precisely in theorem 5.19 of that paper. He uses the following variant of fast-growing hierarchy: $$H_1(x)=2x+1$$ $$H_{\beta+1}(x)=H_\beta^{x+1}(x)$$ $$H_\lambda(x)=H_{\lambda(x)}(x)$$
(where $\lambda(x)$ denotes the $x$-th element of a standard fundamental sequence of $\lambda$ when $\lambda<\varepsilon_0$).
Theorem: The function $F$ eventually dominates all functions $H_\beta$ for $\beta<\omega^\omega$ and is itself eventually dominated by $H_{\omega^\omega+1}$.
Hence, in a very precise sense, $F$ has growth rate of approximately $H_{\omega^\omega}$.
On the other hand the function $TREE$ is growing very much faster. An often-quoted lower bound is $f_{\Gamma_0}$ in the fast-growing hierarchy, but I don't think an exact bound like above has been proven.
This way or another, regarding your last question - I haven't bothered to check the details, but I imagine that even with $TREE(3)$ we can construct sequences of trees of length exceeding things like $STR^{STR(5)}(5)$, so we are not going to easily find $n$ such that $STR(n)>TREE(3)$.
