User blog comment:Mh314159/Natural number recursion - first 4 rule sets/@comment-35470197-20191019143654/@comment-35470197-20191020234550

Here is the result of my analysis. \begin{eqnarray*} A \langle n \rangle(x) & \approx & f_{2n}(x) \\ A \langle 0,1 \rangle(x) & \approx & f_{2A \langle 1 \rangle(x)}(x) \approx (f_{2 \omega} \circ f_2)(x) \\ A \langle n,1 \rangle(x) & \approx & f_{2 \omega + 2n}(x) \\ A \langle 0,2 \rangle(x) & \approx & f_{2 \omega + 2A \langle 2,1 \rangle(x)}(x) \approx (f_{2 \omega \times 2} \circ f_{2\omega + 4})(x) \\ A \langle n,2 \rangle(x) & \approx & f_{2 \omega \times 2 + 2n}(x) \\ A \langle n_0,n_1 \rangle(x) & \approx & f_{2 \omega \times n_1 + 2n_0}(x) \\ A \langle n,0,1 \rangle(x) & \approx & f_{2 \omega \times A \langle n,1 \rangle(x) + 2}(x) \approx (f_{2 \omega^2} \circ f_{2 \omega + 2n})(x) \\ A \langle 0,1,1 \rangle(x) & \approx & (f_{2 \omega^2} \circ f_{2 \omega + 2A \langle 1,0,1 \rangle(x)})(x) \\ & \approx & (f_{2 \omega^2} \circ f_{2 \omega \times 2} \circ f_{2 \omega^2} \circ f_{2 \omega + 2})(x) \approx f_{2 \omega^2}^2(x)\\ A \langle n,1,1 \rangle(x) & \approx & f_{2 \omega^2+2n}(x) \\ A \langle 0,2,1 \rangle(x) & \approx & f_{2 \omega^2+2A \langle 2,1,1 \rangle(x)} \approx (f_{2 \omega^2 + 2 \omega} \circ f_{2 \omega^2 + 4})(x) \\ A \langle n,2,1 \rangle(x) & \approx & f_{2 \omega^2 + 2 \omega + 2n}(x) \\ A \langle n_0,n_1,1 \rangle(x) & \approx & f_{2 \omega^2 + 2 \omega \times n_1 + 2n_0}(x) \\ A \langle n,0,2 \rangle(x) & \approx & f_{2 \omega^2 + 2 \omega \times A \langle n,2,1 \rangle(x) + 2n}(x) \\ & \approx & (f_{2 \omega^2 \times 2} circ f_{2 \omega^2 + 2 \omega \times 2 + 2n})(x) \\ A \langle 0,1,2 \rangle(x) & \approx & (f_{2 \omega^2 \times 2} circ f_{2 \omega^2 + 2 \omega \times 2 + 2A \langle 1,0,2 \rangle(x)})(x) \\ & \approx & (f_{2 \omega^2 \times 2} circ f_{2 \omega^2 + 2 \omega \times 3} \circ f_{2 \omega^2 \times 2} circ f_{2 \omega^2 + 2 \omega \times 2 + 2})(x) \\ & \approx & f_{2 \omega^2 \times 2}^2(x) \\ A \langle n,1,2 \rangle(x) & \approx & f_{2 \omega^2 \times 2 + 2n}(x) \\ A \langle 0,2,2 \rangle(x) & \approx & f_{2 \omega^2 \times 2 + 2A \langle 2,1,2 \rangle(x)}(x) \approx (f_{2 \omega^2 \times 2 + 2 \omega} \circ f_{2 \omega^2 \times 2 + 4})(x) \\ A \langle n,2,2 \rangle(x) & \approx & f_{2 \omega^2 \times 2 + 2 \omega + 2n}(x) \\ A \langle n_0,n_1+1,2 \rangle(x) & \approx & f_{2 \omega^2 \times 2 + 2 \omega \times n_1 + 2n_0}(x) \\ A \langle n,0,3 \rangle(x) & \approx & f_{2 \omega^2 \times 2 + 2 \omega \times A \langle n,3,2 \rangle(x) + 2n}(x) \\ & \approx & (f_{2 \omega^2 \times 3} \circ f_{2 \omega^2 \times 2 + 2 \omega \times 2 + 2n})(x) \\ A \langle 0,1,3 \rangle(x) & \approx & (f_{2 \omega^2 \times 3} \circ f_{2 \omega^2 \times 2 + 2 \omega \times 2 + 2A \langle 1,0,3 \rangle(x)})(x) \\ & \approx & (f_{2 \omega^2 \times 3} \circ f_{2 \omega^2 \times 2 + 2 \omega \times 3} \circ f_{2 \omega^2 \times 3} \circ f_{2 \omega^2 \times 2 + 2 \omega \times 2 + 2})(x) \\ & \approx & f_{2 \omega^2 \times 3}^2(x) \\ A \langle n,1,3 \rangle(x) & \approx & f_{2 \omega^2 \times 3 + 2n}(x) \\ A \langle 0,2,3 \rangle(x) & \approx & f_{2 \omega^2 \times 3 + 2A \langle 2,1,3 \rangle(x)}(x) \\ & \approx & (f_{2 \omega^2 \times 3 + 2 \omega} \circ f_{2 \omega^2 \times 3 + 4})(x) \\ A \langle n,2,3 \rangle(x) & \approx & f_{2 \omega^2 \times 3 + 2 \omega + 2n}(x) \\ A \langle n_0,n_1+1,3 \rangle(x) & \approx & f_{2 \omega^2 \times 3 + 2 \omega \times n_1 + 2n_0}(x) \\ A \langle n_0,n_1+1,n_2 \rangle(x) & \approx & f_{2 \omega^2 \times n_2 + 2 \omega \times n_1 + 2n_0}(x) \\ A \langle n_0,n_1,\ldots,n_k \rangle(x) & \approx & f_{2 \omega^k \times n_k + \cdots + 2 \omega \times n_1 + 2n_0}(x) \\ A \langle \underbrace{n,\ldots,n}_n \rangle(x) & \approx & f_{2 \omega^{n-1} \times n}(x) \\ A(x) & \approx & f_{2 \omega^{\omega}}^x(x) = f_{2 \omega^{\omega}+1}(x) \\ A_n(x) & \approx & f_{2 \omega^{\omega}+1+2n}(x) \\ A_x(x) & \approx & f_{2 \omega^{\omega}+1+2 \omega}(x) \end{eqnarray*} Therefore the limit corresponds to \(2 \omega^{\omega}+1+2 \omega\), which is \\(\omega^{\omega}+\omega\) as an ordinal, and is of level 9 with respect to my googological googical if I am correct.

The great improvement is not only that the level reached \(\omega^{\omega} + \omega\), but also the recursion is quite compatible with FGH, i.e. much easier to enjoy the analysis than before.

Also, the way to write definitions are much better than before. I note that there are several unnecessary descriptions. I guess that it is because I could not fully help you to understand the issues, and you actually did well. Anyway, writing unnecessary descriptions is much better than not writing necessary descriptions. If you want to improve the way to write definitions, it might be also good to observe others' stuffs. In addition, I guess that you can now analyse them up to at least \(\omega^2\) level. Analysing functions will improve the ability to create new functions, and hence it is good to start analysing others. Of course, I am looking forward to seeing your stuffs, too.