I was working on my answer for Golf a number bigger than TREE(3) and I realized I couldn't use The Hardy Hierarchy in the way wanted to. So I defined a slightly modified version: $$H'_0(n)=n+1$$ $$H'_\alpha(n)=H'_{\alpha[n]}(H'_{\alpha[n]}(n))$$ the goal was that this would be a balance between the simplicity (to program) of The Hardy Hierarchy and the recursion of the Fast-growing-/Ackermann- hierarchy.
note:
- The second rule assumes $(\alpha+1)[n]=\alpha$
- This is a Grzegorczyk hierarchy: the result depends on the choice of fundamental sequence, you may assume I'm using standard fundamental sequences.
- $H'_\alpha(n)>H_\alpha(n)$ because $\alpha[n]\ge 0 \Rightarrow H'_{\alpha[n]}(n)\ge n+1$
Trough some manual working out I found:
- $H'_k(n)=n+2^k$
- $H'_ω(n)>n+2^{n+1}$
- $H'_{ω+k}(n)>(2↑↑2^k)^{n+1}$
- $H'_{ω*k}(n)>2↑^k2^n$
- $H'_{ω^ω}(n)\approx n→n→...→n $ (n arrows) but that was mostly just a guess based on the trend of $H'_{ω^0}(n),H'_{ω^1}(n),H'_{ω^2}(n)$
This suggests that $H'_{ω*\alpha}(n)\approx f_\alpha(n)$
or $H'_\alpha(n)\approx f_\alpha(n)$ for $\alpha>ω^ω$
But this is mainly my (and Simply Beautiful Art's) suspicion could someone with more knowledge about ordinals please try to approximate $H'_\alpha$ better (more precisely, more rigidly,...) using an other ordinal hierarchy?
Assuming you are using standard fundamental sequences, I get
$$H'_\omega(n)=n+2^n>2\cdot n$$
From there we can prove with induction that
$$H'_{\omega\cdot\alpha}(n)>f_\alpha(n)\quad\forall\alpha>0$$
Base case:
$$\alpha=1\implies H'_\omega(n)=n+2^n>2\cdot n=f_1(n)$$
Successor ordinal induction step:
$$\text{assume that }H'_{\omega\cdot\alpha}(n)>f_\alpha(n)$$ \begin{align}H'_{\omega\cdot(\alpha+1)}(n)&=H'_{\omega\cdot\alpha+n}(H'_{\omega\cdot\alpha+n}(n))\\&>H'_{\omega\cdot\alpha}(H'_{\omega\cdot\alpha+n-1}(H'_{\omega\cdot\alpha+n-1}(n)))\\&>H'_{\omega\cdot\alpha}(H'_{\omega\cdot\alpha}(H'_{\omega\cdot\alpha+n-2}(H'_{\omega\cdot\alpha+n-2}(n))))\\&>\dots\tag{left up to the reader}\\&>(H'_{\omega\cdot\alpha})^n(n)\\&>f_\alpha^n(n)\\&=f_{\alpha+1}(n)\end{align}
Limit ordinal induction step:
$$\text{assume that }H'_{\omega\cdot\beta}(n)>f_\beta(n)\forall\beta<\alpha,~\alpha[n]<\alpha$$ $$H'_{\omega\cdot\alpha}(n)=H'_{\omega\cdot(\alpha[n])}(n)>f_{\alpha[n]}(n)=f_\alpha(n)$$
Where the same ordinal hierarchy is used on both sides.
I fail to find a good upper bound, but hopefully a lower bound is all you really wanted.