TITLE: Ackermann function and $f_\omega$
QUESTION [7 upvotes]: The Wikipedia page of Ackermann function states that Ackermann function is "roughly comparable" to $f_\omega$ in fast-growing hierarchy. 
Is there some standard way to make the "roughly comparable" to more precise form? I would like to understand the claim better.

REPLY [3 votes]: Generally, when one compares functions to the fast growing hierarchy, we say that a function $g(n)$ grows at $f_\alpha(n)$ if $g(n)$ grows faster than $f_\beta(n)$ for every $\beta<\alpha$ and $g(n)$ grows slower than $f_\alpha^k(n)$ for a fixed $k$.  For functions/notations that come with multiple arguments, we diagonalize, for example,
$$g(n)=\chi(n,n,n,\dots,n)$$
Basically, replace every argument of the function with $n$ and call that $g(n)$.
Formally, we may define
$$g(n)\in f_\alpha(n)\iff\exists k\in\mathbb N\forall\beta<\alpha\exists0<N\forall n>N\text{ s.t. }g(n)>f_\beta(n)\land g(n)<f_\alpha^k(n)$$
Particularly here, all one needs to show is that
$$f_k(n)\ll A(n,n)\ll f_\omega(n)$$
Under this definition, you will find that every eventually monotonically increasing function can find a place on the fast growing hierarchy, and that there are simple properties, for example:
$$g(n)\in f_\alpha(n)\implies\begin{cases}g^k(n)\in f_\alpha(n)\\g^n(n)\in f_{\alpha+1}(n)\end{cases}$$
So repeatedly nesting Ackermann functions a fixed amount of times does not take you further down the fast growing hierarchy, instead, you need to nest the function a non-constant (and increasing) amount of times.