User blog comment:Mh314159/Natural number recursion - first 4 rule sets/@comment-35470197-20191019143654/@comment-35470197-20191024004509

The α-subrscript contributes as "diagonalisation + 1" to ordinals. Since \(A_n(x)\) corresponds to \(2 \omega^{\omega}+1+2n\), its diagonalisation corresponds to \(2 \omega^{\omega}+1+2 \omega\), and hence \(A_{\alpha}(x)\) correspond to \(2 \omega^{\omega}+1+2 \omega + 1\).

The α-superscript contributes as "+2" to ordinals. Since \(A_{\alpha}(x)\) corresponds to \(2 \omega^{\omega}+1+2 \omega + 1\), \(A^{\alpha}(x)\) correspond to \(2 \omega^{\omega}+1+2 \omega + 3\). It is \(\omega^{\omega} + \omega + 3\) as an ordinal, and hence is of level 9.