What are sharp lower and upper bounds of the fast growing hierarachy? With fast growing hierarchy, I mean the Wainer hierarchy, which starts with
$$f_0(n)=n+1$$
 $$f_1(n)=2n$$
 $$f_2(n)=2^n n$$
A lower bound for $f_m(n)$ is $2 \uparrow^{m-1} n$.
If $f(n,m):=2 \uparrow^{m-1}n$ and $g(n,m):=f_m(n)$, is it true that
 $$f(n,m) < g(n,m) < f(n+1,m)$$ for all $m,n\ge 2$ ?
If not, what is the least k such that $$f(n,m) < g(m,n) < f(n+k,m)$$
 for all $m,n\ge 2$ ?
The next inequality I found is $f_{\omega+m}(n)>[n,n,m,2]$ , where
 the square brackets denote bowers linear array notation. Is there
 a similar inequality chain as above ?
 A: The answer to your first question is no.  For the second, there is no k that satisfies the inequality chain.
For example, take m = 2:
$$
f(n,2) = 2^n n = 2^{n + \log n} \approx g(n + \log n, 2)
$$
For m = 3, we have $f(n,3) > 2 \uparrow 2 \uparrow 2 \ldots \uparrow n$, and $n \approx 2 \uparrow \uparrow \log_* n$, so 
$$f(n,3) \approx 2 \uparrow \uparrow (n + \log_* n) \approx g(n + \log_* n, 3)$$
and of course $\log_* n$ will eventually exceed any constant k.
More generally, for fixed m let h(n,m) be the number-theoretic inverse of f(n,m). Then
$$
f(n,m) > 2 \uparrow^{m-1} 2 \uparrow^{m-1} \ldots n \text{ with n up arrows } \approx 2 \uparrow^m (n + h(n,m)) = g(n + h(n,m), m)$$
and h(n,m), while extremely slow growing, will still eventually consant k, unfortunately.  So the best we can do is something like
$$
f(n,m) < g(n,m) < f(2n,m)
$$
which is true for all $m,n \ge 2$, and is reasonably provable.  With additional work you could replace $2n$ with $n + \lceil \log n \rceil$.
A: For your chain inequality, one has $$f_{\omega+m}(n) > n \rightarrow n \rightarrow (n-1) \rightarrow (m+1)$$
You can use your inequality between FGH and Bowers' array notation in combination with Birds proof to prove this. 
