User blog comment:MilkyWay90/My ordinal function/@comment-35470197-20180630144403

For any strictly increasing function \(f = f(n)\), one has \begin{eqnarray*} f \uparrow^2 0 < & F(0,f) & = f \uparrow^2 1 \\ f \uparrow^2 1 < & F(1,f) & = f \uparrow^2 2 \\ f \uparrow^2 2 < & F(2,f) & = (f \uparrow^2 2) \uparrow^2 2 < f \uparrow^2 2^2 \\ f \uparrow^2 2^2 < & F(3,f) & < F(2,f \uparrow^2 2^2) < (f \uparrow^2 2^2) \upparrow^2 2^2 < f \uparrow^2 2^4 \\ & \vdots & \\ a \uparrow^2 2^{2^{x-2}} < & F(x,a) & < a \uparrow^2 2^{2^{x-1}}. \end{eqnarray*} For an ordinal number \(\alpha\) with a fundamental sequence, one has \begin{eqnarray*} f_{\alpha + 1} < & f_{\alpha} \uparrow^2 x & < f_{\alpha + \omega}. \end{eqnarray*} and hence \begin{eqnarray*} f_{\alpha + \omega}(n) < & F(n,f_{\alpha}(n)) & < f_{\alpha + \omega + 1}(n) \end{eqnarray*} for sufficiently large \(n\). When \(f = a\) in your definition, one has \(a \sim f_{\omega}\), and hence roughly \begin{eqnarray*} f_{\omega + \omega * \beta} < & a_{\beta} & < f_{\omega + (\omega + 1) * \beta + 1}. \end{eqnarray*}