User blog:Simply Beautiful Art/Bump Funk and its OCF

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

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

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

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

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

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

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 $$\psi_B$$.

$$C(\alpha)_0=\{0,1,\Omega\}$$

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

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

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

And so we have...

$$\psi_B(0)=\omega$$

$$\psi_B(1)=\omega^2$$

$$\psi_B(2)=\omega^\omega$$

$$\psi_B(3)=\varepsilon_0$$

$$\psi_B(4)=\varepsilon_\omega$$

$$\psi_B(3+\alpha)=\varepsilon_{\omega^\alpha}$$

$$\psi_B(\Omega)=\zeta_0$$

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

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

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

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

$$\psi_B(\Omega2)=\zeta_1$$

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

$$B(\psi(3),\Omega)=\varepsilon_{\Omega+1}$$

In general, we have

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

For $$\alpha\ge\beta$$. Likewise, the statement is reversed if $$\alpha\le\beta$$.

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...