User blog comment:Deedlit11/Ordinal Notations IV: Up to a weakly inaccessible cardinal/@comment-24509095-20140507175906/@comment-5529393-20140508124758

King2218, if you remove $$\varphi$$ from the closure functions, then you actually get $$\psi_{\Omega_{I+1}}(0) = I\omega$$, since you can only use addition on I. I guess you mean if you replace $$\varphi$$ by $$\alpha \mapsto \omega^\alpha$$.

What Ikosarakt1 is suggesting is actually exactly how the above ordinal notations work! $$\psi_\alpha(\beta)$$ is defined as the smallest ordinal less than $$\alpha$$ and closed under the allowed operations, so $$\psi_1$$ is actually $$\psi_{\Omega_2}$$ under this notation. ($$\psi_{\Omega_2}$$ generates ordinals between $$\Omega_1$$ and $$\Omega_2$$, and $$\psi_\Omega_2(\Omega_2)$$ is the first fixed point of $$\psi_\Omega_2(\alpha)$$.

King, I'm not sure why you say this is not economical, unless you mean that you have to type in an extra $$\Omega$$, but I think it is worth it if the notation makes more sense conceptually.