User blog comment:MilkyWay90/Yet another attempt at making a fast-growing function/@comment-35470197-20180709135900

It is easy to show \begin{eqnarray*} f_{\omega \uparrow \uparrow a}(n) < \textrm{RandamFunction}(n) \end{eqnarray*} for any \(a \in \mathbb{N}\), after noting that \(\varepsilon_0\) is the PTO of \(\textrm{PA}\). So the growth rate is at least \(\epsilon_0\).

However, you need to add precise computation process of \(RandomFunction(n)\) as long as you want a computable function. Such a Rayo-like definition yields an uncomputable function because there is no way to determine a formula with given bound of length which actually gives the greatest number.

For example, under the assumption of \(\textrm{Con}(\textrm{PA})\) on the meta theory of \(\textrm{PA}\), \(\textrm{Con}(\textrm{PA})\) is independent of \(\textrm{PA}\). Therefore the formula \(F(x)\) given as \begin{eqnarray*} (\textrm{Con}(\textrm{PA}) \to (x = 0)) \wedge (\neg(\textrm{Con}(\textrm{PA})) \to (x = 2)) \end{eqnarray*} is a definition of a natural number \(X\) in \(\textrm{PA}\). However, none of the relations \(X < 1\), \(X = 1\), or \(X > 1\) is provable or unprovable under \(\textrm{PA}\). Therefore by the completeness theorem, the value of \(X\) depends on the choice of a model of \(\textrm{PA}\). If you consider the greatest one among natural numbers given by abstract definitions in \(\textrm{PA}\), you need to avoid such a problem in order to compare all the natural numbers that you are considering.