User blog comment:Mh314159/Alpha numbers (and beyond)/@comment-35470197-20191007022858/@comment-35470197-20191015160616

I denote by \(f_{\alpha}(x)\) the function corresponding to an ordinal \(\alpha\) with respect to FGH. Here is my analysis. \begin{eqnarray*} f\{n\}(x) & = & f_{2n}(x) \\ f\{0,1\}(x) & = & f\{f\{1\}(x)\}(x) = f\{f_2(x)\}(x) = (f_{\omega} \circ 2f_2)(x) \\ f\{1,1\}(x) & \sim & (f_{2+2} \circ f\{0,1\})(x) = (f_4 \circ f_{\omega} \circ 2f_2)(x) \\ f\{2,1\}(x) & \sim & (f_{4+2} \circ f\{1,1\})(x) = (f_6 \circ f_4 \circ f_{\omega} \circ 2f_2)(x) \\ f\{n,1\}(x) & \sim & (f_{2n+2} \circ f_{\omega})(x) \\ f\{0,2\}(x) & = & f\{f\{2,1\}(x),1\}(x) \sim (f_{2 f_{\omega}} \circ f_{\omega})(x) \sim (f_{2 \omega} \circ f_{\omega})(x) \\ f\{1,2\}(x) & \sim & (f_{\omega+2} \circ f\{0,2\})(x) \sim (f_{\omega+2} \circ f_{\omega}^2)(x) \\ f\{2,2\}(x) & \sim & (f_{\omega+2} \circ f\{1,2\})(x) \sim (f_{\omega+2}^2 \circ f_{\omega}^2)(x) \\ f\{n,2\}(x) & \sim & \left\{\begin{array}{ll} (f_{\omega+2}^n \circ f_{\omega}^2)(x) & (n \ll f_{\omega}^2(x)) \\ (f_{n+3} \circ f_{\omega}^2)(x) & (n \gg f_{\omega}^2(x)) \right. \\ f\{0,3\}(x) & = & f\{f\{3,2\}(x),2\}(x) \sim (f_{(f_{\omega+2}^3 \circ f_{\omega}^2)(x)+3} \circ f_{\omega}^2)(x) \sim f_{\omega+2}^3(x) \\ f\{1,3\}(x) & \sim & (f_{\omega+4} \circ f\{0,3\})(x) \sim (f_{\omega+4} \circ f_{\omega+2}^3)(x) \\ f\{2,3\}(x) & \sim & (f_{\omega+4} \circ f\{1,3\})(x) \sim (f_{\omega+4}^2 \circ f_{\omega+2}^3)(x) \\ f\{n,3\}(x) & \sim & \left\{\begin{array}{ll} (f_{\omega+4}^n \circ f_{\omega+2}^3)(x) & (n \ll f_{\omega+2}^3(x)) \\ (f_{n+6} \circ f_{\omega+2}^3)(x) & (n \gg f_{\omega+2}^3(x)) \right. \\ f\{0,4\}(x) & = & f\{f\{4,3\}(x),3\}(x) \sim (f_{(f_{\omega+4}^4 \circ f_{\omega+2}^3)(x)+6} \circ f_{\omega+2}^3)(x) \sim f_{\omega+4}^4(x) f\{1,4\}(x) & \sim & (f_{\omega+6} \circ f\{0,4\})(x) \sim (f_{\omega+6} \circ f_{\omega+4}^4)(x) \\ f\{2,4\}(x) & \sim & (f_{\omega+6} \circ f\{1,4\})(x) \sim (f_{\omega+6}^2 \circ f_{\omega+4}^4)(x) \\ f\{n,4\}(x) & \sim & \left\{\begin{array}{ll} (f_{\omega+6}^n \circ f_{\omega+4}^4)(x) & (n \ll f_{\omega+4}^4(x)) \\ (f_{n+9} \circ f_{\omega+4}^4)(x) & (n \gg f_{\omega+4}^4(x)) \right. \\ f\{a_0,a_1\}(x) & \sim & \left\{\begin{array}{ll} (f_{\omega+(2a_1-2)}^{a_0} \circ f_{\omega+(2a_1-4)}^{a_1})(x) & (a_1 \ll f_{\omega+(2a_1-4)}^{a_1}(x)) \\ (f_{a_0+(3a_1-3)} \circ f_{\omega+(2a_1-4)}^{a_1})(x) & (a_1 \gg f_{\omega+(2a_1-4)}^{a_1}(x)) \right. \\ f\{1,0,1\}(x) & = & f\{1,f\{1,1\}(x)\}(x) \sim (f_{\omega+f_{\omega}(x)}^{1} \circ f_{\omega+f_{\omega}(x)}^{f_{\omega}(x)}(x) \\ & \sim & (f_{\omega \times 2} \circ f_{\omega})(x) \sim f_{\omega \times 2}(x) \\ f\{2,0,1\}(x) & \sim & (f_4 \circ f\{1,0,1\})(x) \sim (f_{\omega \times 2} \\ f\{n,0,1\}(x) & \sim & (f_{2n} \circ f_{\omega \times 2})(x) \\ f\{0,1,1\}(x) & = & f\{f\{1,0,1\},0,1\}(x) \sim (f_{2f_{\omega \times 2}(x)} \circ f_{\omega \times 2})(x) \\ & \sim & (f_{2 \omega} \circ f_{\omega \times 2})(x) \sim f_{\omega \times 2}(x) \\ f\{1,1,1\}(x) & = & (f_{\omega \times 2 + 2} \circ f_{\omega+2})(x) \sim f_{\omega \times 2 + 2}(x) \end{eqnarray*} It is really hard for me to continue this analysis, because the rule is quite different from FGH, but I guess that \(f\{\underbrace{n,\ldots,n}_n\}(x)\) roughly corresponds to \(\omega \times n\), which is bounded by \(\omega^2\). Since I have not fully verified the estimation, the guess might be wrong.