User blog comment:Nedherman1/New Notation Idea/@comment-35470197-20180713222712/@comment-35470197-20180714215242

I see. Then \(f_{\textrm{b}; \textrm{a}}(n) = \underbrace{f_{\textrm{b}}(f_{\textrm{b}}(f_{\textrm{b}} \cdots (f_{\textrm{b}}(n)) \cdots))}_{f_{\textrm{a}(n) \textrm{ times}}\), right?

Then I analyse it. \begin{eqnarray*} f_n(n) = & f_{\textrm{a} + 1}(n) & = f_{\omega}(n) \\ f_{\omega}(n) \leq & f_{\textrm{a} + x}(n) & \leq f_{\omega}^x(n) \\ f_{\omega + 1}^n(100) < & f_{\textrm{a}}(n) & < f_{\omega + 3}(n) \\ f_{\omega + 2}(n) < & f_{\textrm{b}}(n) & < f_{\omega + 5}(n) \\ & \vdots & \\ f_{\omega + 26}(n) < & f_{\textrm{z}}(n) & < f_{\omega + 53}(n) \\ f_{\omega + 2}(n) < & f_{\textrm{a}; \textrm{a}}(n) & < f_{\omega + 5}(n) \\ f_{\omega + 2}(n) < & f_{\textrm{a}; \textrm{b}}(n) & < f_{\omega + 7}(n) \\ f_{\omega + 3}(n) < & f_{\textrm{a}; \textrm{c}}(n) & < f_{\omega + 9}(n) \\ & \vdots & \\ f_{\omega + 26}(n) < & f_{\textrm{a}; \textrm{z}}(n) & < f_{\omega + 55}(n) \\ f_{\omega + 2}(n) < & f_{\textrm{b}; \textrm{a}}(n) & < f_{\omega + 7}(n) \\ f_{\omega + 3}(n) < & f_{\textrm{b}; \textrm{b}}(n) & < f_{\omega + 7}(n) \\ f_{\omega + 4}(n) < & f_{\textrm{b}; \textrm{c}}(n) & < f_{\omega + 9}(n) \\ & \vdots & \\ f_{\omega + 26}(n) < & f_{\textrm{b}; \textrm{z}}(n) & < f_{\omega + 55}(n) \\ & \vdots & \\ f_{\omega + 26}(n) < & f_{\textrm{z}; \textrm{a}}(n) & < f_{\omega + 28}(n) \\ & \vdots & \\ f_{\omega + 26}(n) < & f_{\textrm{z}; \textrm{z}}(n) & < f_{\omega + 57}(n) \\ \end{eqnarray*} So an upper bound of the growth rate of your funcions is \(\omega + 57\).