User blog comment:Nedherman1/New Notation Idea/@comment-35470197-20180713222712/@comment-35470197-20180715141149

It is better to define functions by using any functions which have already been defined.

For example, in the definition of \(f_{\textrm{a};\textrm{a}}\), you just used \(f_{\textrm{a}}\). If you use \(f_{\textrm{z}}\) instead, then \(f_{\textrm{a};\textrm{a}}\) will become bigger.

In the definition of \(f_{\textrm{a};\textrm{b}}\), you just used \(f_{\textrm{a}}\) and \(f_{\textrm{b}}\). If you use \(f_{\textrm{a};\textrm{a}}\) instead, then \(f_{\textrm{a};\textrm{b}}\) will become bigger. You can repeat this process to \(f_{\textrm{a};\textrm{z}}\).

Also you can use \(f_{\textrm{a};\textrm{z}}\) in the definition of \(f_{\textrm{b};\textrm{a}}\). Repeating this process to \(f_{\textrm{z};\textrm{z}}\), the growth rate will become \(\omega + 1405\).

The reason why such an extension will never go beyond \(\omega + \omega\) is because you did not diagonalise the construction. The \(\omega\) in \(\omega + 57\) in your growth rate is derived from that of the original FGH \(f_n(n)\) in the definition of \(f_{\textrm{a}}\). So you need some diagonalisation like the following: \begin{eqnarray*} f_{\textrm{a}}(n) & := & n+1 f_{\textrm{a}+0}(n) & := & f_{\textrm{a}}(n)(n) f_{\textrm{a}+(x+1)}(n) & := & f_{\textrm{a}+x}^n(n) f_{\textrm{b}}(n) & := & f_{\textrm{a}+n}(n) \textrm{(diagonalisation)} f_{\textrm{b}+0}(n) & := & f_{\textrm{b}}(n)(n) f_{\textrm{b}+(x+1)}(n) & := & f_{\textrm{b}+x}^n(n) f_{\textrm{c}}(n) & := & f_{\textrm{c}+n}(n) \textrm{(diagonalisation)} \end{eqnarray*} Then the growth rate of \(f_{\textrm{c}}\) becomes \(\omega + \omega\).

Roughly, diagonalisation is a technique to use \(n\) to change the parameter of functions.