Peano arithmetic vs. fast-growing hierarchy with pathological fundamental sequences Fundamental sequence for a countable limit ordinal $\alpha$ is an increasing sequence $\{\alpha[i]\}$ of ordinals of length $\omega$ such that $\lim_{i\rightarrow\omega}\alpha[i]=\alpha$. There are many (continuum many, in fact) possible choices for fundamental sequence for any ordinal, some are quite natural, like $\omega^2[n]=\omega n$, and some are quite odd, like $\omega[n]=\Sigma(n)$. Most common definition of fundamental sequences below $\varepsilon_0$ is via Wainer hierarchy. Using these, it's known that in fast-growing hierarchy, $F_{\varepsilon_0}(n)$ is a total recursive function which outgrows all recursive functions which Peano axioms can prove total. A friend of mine posed a question, if this necessarily hold under different choices for fundamental sequences. For me, it seems like the answer would be no, because we can choose some very slow fundamental sequences for all ordinals, possible making it slower than $F_\alpha(n)$ for some $\alpha<\varepsilon_0$ in Wainer hierarchy, but my friend believes the answer to be yes.
To put it into a single question:

Is it true that for any choice of fundamental sequences for ordinals below $\varepsilon_0$ we have that, in fast-growing hierarchy, $F_{\varepsilon_0}(n)$ outgrows all functions provably total recursive in PA?

Does it make any difference if we replace it with Hardy hierarchy?
Thanks for your feedback.
 A: The answer is no. Choose a fundamental sequence for $\epsilon_0$ itself in the usual way, which I think is $\epsilon_0[n]=\omega^{\omega^{{\vdots}^\omega}}$, and then modify the earlier fundamental sequences for $\alpha=\epsilon_0[n]$ by making it start with $0,1,2,\ldots,n$, before resuming with the usual values. In particular, we have thereby ensured $\alpha[n]=n$ for $\alpha=\epsilon_0[n]$. It now follows, according to the rules of the fast-growing hierarchy, that $F_{\epsilon_0}(n)$, which by definition is $F_{\epsilon_0[n]}(n)$, is the same as $F_\alpha(n)$ where $\alpha=\epsilon_0[n]$, but since this is a limit ordinal, it is equal to $F_{\alpha[n]}(n)$, which is the same as $F_n(n)$, by construction. So with these modified fundamental sequences, the top function $F_{\epsilon_0}$ is basically the same as $F_\omega$. This seems completely to confirm your intuition that slowing the fundamental sequences down could make the diagonal function at the top very small.
If we dropped the requirement that the fundamental sequences must be strictly increasing, we could pad with $n$ many $0$'s instead, ensuring that $\alpha[n]=0$ for limit $\alpha=\epsilon_0[n]$, and get a more extreme situation $F_{\epsilon_0}(n)=F_{\epsilon_0[n]}(n)=F_{(\epsilon_0[n])[n]}(n)=F_0(n)$.
