User blog comment:P進大好きbot/What is the greatest ordinal notation now?/@comment-31966679-20180623154722/@comment-35470197-20180625041007

I see. So your definition can be written as \begin{eqnarray*} a_0(n) & = & lvl(n+1,n+1,n+1) \sim f_{\omega}(n)\\ a_{b+1}(n) & = & a_b^{n!}(n) \\ a_b(n) & = & a_{b[n]}^{n!}(n). \end{eqnarray*} I am sure that you did not use the factorial, but the equality \(F(x,a(n)) = a^{x!}(n)\) automatically holds by the recursive definition of the transformation \(F\). Therefore your hierarchy is roughly estimated as \(f_{\omega + b} < a_b < f(\omega + 2b + 1)\).