The attempt to make my own ordinals collapse function

Note: The first several parts may be the same as existing one. But just hold on, it will soon get crazy.

Up to Omega

definition

I use ρ(Greek letter rho) for my OCF.

(x and y are ordinals)

$C_0(\alpha)=\{0, x : x\leq \rho (\beta) \;\; (\beta < \alpha) \}$

$C_n(\alpha)=\{ x+y, xy : x,y \in C_{n-1}(\alpha) \}$

$C(\alpha)=\bigcup_{n<\omega} C_n(\alpha)$

$\rho(\alpha)=min \{ \gamma| \gamma \notin C(\alpha) \}$

I don't know if it works, but I mean that one step in rho is "the smallest number that can't be acieved by addition and multiplication from current step."

analysis

$\rho(0)=1$, because the first step is zero and you can't get one from zero.

$\rho(1)=\omega$

$\rho(2)=\omega^{\omega}$

$\rho(3)=\omega^{\omega^2}$

$\rho(n)=\omega^{\omega^n}$

$\rho(\omega)=\omega^{\omega^{\omega}}$

$\rho(\omega+1)=\omega^{\omega^{\omega+1}}$

$\rho(\omega 2)=\omega^{\omega^{\omega 2}}$

$\rho(\omega^2)=\omega^{\omega^{\omega^2}}$

$\rho(\omega^{\omega})=\omega^{\omega^{\omega^{\omega}}}$

The limit is $\epsilon_0$.

Stil calm

$\Omega$ means a fixed point and keep going on.

$\rho(\Omega)=\epsilon_0$

$\rho(\Omega+1)=\epsilon_0^{\omega}$

$\rho(\Omega+\alpha)=\epsilon_0^{\omega^{\alpha}}$

$\rho(\Omega 2)=\epsilon_1$

$\rho(\Omega 3)=\epsilon_2$

$\rho(\Omega \omega)=\epsilon_{\omega}$

$\rho(\Omega^2)=\zeta_0$

$\rho(\Omega^2+\Omega)=\epsilon_{\zeta_0+1}$

$\rho(\Omega^2 2)=\zeta_1$

$\rho(\Omega^\Omega)=\Gamma_0$

$\rho(\Omega^{\Omega+\alpha})=\phi(1,\alpha,0)$

$\rho(\Omega^{\Omega 2})=\phi(2,0,0)$

$\rho(\Omega^{\Omega^2})=\phi(1,0,0,0)$

$\rho(\Omega^{\Omega^{\Omega}})=SVO$ (Is it correct?)