User blog comment:Nedherman1/Arrow Destruction (AD)/@comment-31966679-20180903190222/@comment-35470197-20180905010936

An effective upperbound is \(\omega + \omega\).

First of all, we have \begin{eqnarray*} [n] & \leq & f_{\omega+1}(n) \\ & & \\ [a;1] & \leq & f_{\omega+1}^{f_{\omega+1}(a)}(a) \\ & \leq & f_{\omega+2}^2(a) \\ & & \\ [a;b+1] & \leq & f_{\omega+1}^{f_{\omega+1}(b)}([f_{\omega+1}^{f_{\omega+2}^2(b)}(a);b]) \\ & \leq & f_{\omega+1}^{b f_{\omega+1}(b)} f_{\omega+2}^2 f_{\omega+1}^{b f_{\omega+2}^2(b)}(a) \\ & \leq & f_{\omega+2}^7(a+b+1) \\ & & \\ [a;b;1] & \leq & f_{\omega+1}^{f_{\omega+1}^{f_{\omega+2}^7(a+b)}(f_{\omega+2}^7(a+b))}(f_{\omega+2}^7(a+b)) \\ & \leq & f_{\omega+2}^9(a+b) & & \\ [a;b;c+1] & \leq & f_{\omega+1}^{f_{\omega+1}(c+1)}([a;f_{\omega+1}^{f_{\omega+1}^{f_{\omega+1}(c)}(c)}(b);c]) \\ & \leq & (f_{\omega+1}^{cf_{\omega+1}(c+1)} f_{\omega+2}^9 f_{\omega+1}^{c f_{\omega+2}^2(c)})(a+b+c+1) \\ & \leq & f_{\omega+2}^{14}(a+b+c+1) \end{eqnarray*} and such an estimation goes in a similar way. Therefore the growth rate of \([a]^n\) is bounded by \(f_{\omega+2}^{7n}(a)\).

The reason why it stops below \(\omega+3\) is because you just repeated a single function \([x]\). In order to increase the growth rate in FGH, you need to repeat the strongest function. For example, it is better to repeat \([a;b-1]) in order to define \([a;b]\).

Next, we have \begin{eqnarray*} [a]^{[a]} & \leq & f_{\omega+2}^{7 f_{\omega+1}(a)}(f_{\omega+1}(a)) \\ & \leq & f_{\omega+3}^2(a) \\ & & \\ [a] \downarrow & \leq & f_{\omega+2}^{7 f_{\omega+3}^2(a) f_{\omega+1}(a)}(f_{\omega+1}(a)) \\ & \leq & f_{\omega+3}^4(a) \\ & & \\ [a] \downarrow \downarrow & \leq & (f_{\omega+3}^4)^{f_{\omega+3}^4(a)}(a) \\ & \leq & f_{\omega+4}^2(a) \\ & & \\ [a] \downarrow_n & \leq & f_{\omega+n+2}(a) \end{eqnarray*} Similar for the multivariable version. Therefore the growth rate of your functions are bounded by \(\omega + \omega\). The reason why you succeeded in going beyond \(\omega + 3\) is because you repeated the strongest functions. Namely, you repeated \(a \downarrow_{n-1}\) in order to define \(a \downarrow_n\).