## FANDOM

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So I made this Bump Funk, and the first version was basically defined as follows:

(all ordinals in Cantor Normal Form)

$B(k,\delta)=k\forall k<\omega$

$B(\omega,\delta)=\delta$

$B(\alpha+\beta,\delta)=B(\alpha,\delta)+B(\beta,\delta)$

$B(\alpha\beta,\delta)=B(\alpha,\delta)B(\beta,\delta)$

$B(\alpha^\beta,\delta)=B(\alpha,\delta)^{B(\beta,\delta)}$

If your ordinal is not one of the above, then

$B(\alpha,\delta)[\eta]=B(\alpha[\eta],\delta)$

Where $\alpha[\eta]$ represents the $\eta$-th ordinal in the fundamental sequence of $\alpha$.

Let's look at an example:

$B(\varepsilon_0,\varepsilon_0)$

$=\sup\{B(1,\varepsilon_0),B(\omega,\varepsilon_0),B(\omega^\omega,\varepsilon_0),\dots\}$

$=\sup\{1,\varepsilon_0,\varepsilon_0^{\varepsilon_0},\dots\}$

$=\varepsilon_1$

Pretty basic right? Now we have my Bump Ordinal Collapsing Funk, BOCF or $\psi_B$.

$C(\alpha)_0=\{\Omega\}$

$C(\alpha)_{n+1}=C(\alpha)_n\cup\{\gamma+\delta,\gamma\delta,\gamma^\delta,B(\gamma,\delta),\psi_B(\eta)|\gamma,\delta,\eta\in C(\alpha)_n,\eta<\alpha\}$

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

$\psi_B(\alpha)=\min\{\beta|\beta\notin C(\alpha)\}$

And so we have...

$\psi_B(0)=0$

$\psi_B(1)=\omega$

$\psi_B(2)=\varepsilon_0$

$\psi_B(3)=\varepsilon_\omega$

$\psi_B(2+\alpha)=\varepsilon_{\omega^\alpha}$

$\psi_B(\Omega)=\zeta_0$

$\psi_B(\Omega+1)=\varepsilon_{\zeta_0\omega}$

$\psi_B(\Omega+2)=\varepsilon_{\zeta_0\omega^2}$

$\psi_B(\Omega+\alpha)=\varepsilon_{\zeta_0\omega^\alpha}$

$\psi_B(\Omega+\varepsilon_0)=\varepsilon_{\zeta_0\varepsilon_0}$

$\psi_B(\Omega2)=\zeta_1$

In general, for $\alpha\le\varepsilon_{\Omega+1}$ and $\alpha$ being a perfect multiple/power of $\Omega$, we have that this is equal to Madore's psi function.

We can then go further that Madore's psi function since

$B(\psi(2),\Omega)=\varepsilon_{\Omega+1}$

In general, we have

$B(\varepsilon_\alpha,\varepsilon_\beta)=\varepsilon_{\alpha+1+\beta}$

So now we can do neat little things like

$\psi_B(\varepsilon_{\Omega+\psi_B(\varepsilon_{\dots})})$

And furthermore, since $B(\Omega,\Omega)=\varepsilon_{\Omega2}$, we can even go further...