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Lately, I've been trying to use ordinals as the argument in FGH and I came out with these:

• $f(\omega)=\omega+1$
• $f_1(\omega)=\omega2$
• $f_2(\omega)=\omega^2$
• $f_3(\omega)=\varepsilon_0$
• $f_4(\omega)=\zeta_0$

I don't know if the pattern continues. (EDIT: It does)

If the pattern continues like that, then,

$f_\omega(\omega)=\phi(\omega,0)$

where $\phi$ is the Veblen Function.

The first hierarchy ordinal is

$H_1 = f_{f_{f_{f_{...f_3(\omega)...}(\omega)}(\omega)}(\omega)}(\omega) = \{f_3(\omega),f_{f_3(\omega)}(\omega),f_{f_{f_3(\omega)}(\omega)}(\omega),...\}$

which has the property:

$H_1=f_{H_1}(\omega)$

Extending

At this point, we need to extend it to create larger ordinals.

• $f[0](n)=f(n)$
• $f[m+1](n)=f[m]_{f[m]_{f[m]_{f[m]_{f[m]_{f[m]_{...}(n)}(n)}(n)}(n)}(n)}(n)$ (n f[m](n)'s)

The second hierarchy ordinal is

$H_2=f[f[...f[1](\omega)...](\omega)](\omega)=\{f[1](\omega),f[f[1](\omega)](\omega),f[f[f[1](\omega)](\omega)](\omega),...\}$

which has the property:

$H_2=f[H_2](\omega)$

More

Wythagoras and googleaarex extended the notation to easily write $H_2$. (See comments below)

Let $\#$ represent rest of expression.

• $f\#[0]_m(n)=f\#_m(n)$
• $f[0][0]...[0][m](n)=\underbrace{f[0][0]...[f[0][0]...[...][m-1](n)][m-1](n)}_n$ where m>0.
• $f[m]\#(n)=\underbrace{f[m-1]\#_{f[m-1]\#_{f[m-1]\#_{f[m-1]\#_{...}(n)}(n)}(n)}(n)}_n$ where m>0

Then,

$H_2=f[0][1](\omega)$

There are lots of ordinals that you can create with this and, of course, I wish to extend it more.

Even More

Of course, it doesn't end there.

Let's define this:

• $f\#[l]^m(n)=f\underbrace{\#[l][l]...[l]}_m(n)$

Then, the third hierarchy ordinal is:

$H_3=f[\omega]^{f[\omega]^{f[\omega]^{f[\omega]^{f[\omega]^{...}(\omega)}(\omega)}(\omega)}(\omega)}(\omega)=\{f[\omega](\omega),f[\omega]^{f[\omega](\omega)}(\omega),f[\omega]^{f[\omega]^{f[\omega](\omega)}(\omega)}(\omega),...\}$

which has the property:

$H_3=f[\omega]^{H_3}(\omega)$

And, of course,

We can extend forever!

• $f\#[l]\uparrow^1m(n)=f\#[l]^m(n)$
• $f\#[l]\uparrow^m1(n)=f\#[l](n)$
• $f\#[k]\uparrow^lm(n)=f\#[k]\uparrow^{l-1}\{f\#[k]\uparrow^lm-1(n)\}(n)$ where l>1

Using these new rules we can form ordinals like

$f[\omega]\uparrow^3\omega[\omega2]\uparrow^23_{\omega3}(\omega)$

Also,

$H_3=f[\omega]\uparrow^2\omega(\omega)$

The fourth hierarchy ordinal is:

$H_4=f[\omega]\uparrow^{f[\omega]\uparrow^{f[\omega]\uparrow^{f[\omega]\uparrow^{...}\omega(\omega)}\omega(\omega)}\omega(\omega)}\omega(\omega)=\{f[\omega]\uparrow\omega(\omega),f[\omega]\uparrow^{f[\omega]\uparrow\omega(\omega)}\omega(\omega),...\}$

which has the property:

$H_4=f[\omega]\uparrow^{H_4}\omega(\omega)$

Next

$\uparrow\uparrow$ is now different from $\uparrow^2$.

New rules:

Let $\uparrow_n$ represent a group of n up-arrows.

• $f\#[k]\uparrow_lm(n)=\underbrace{f\#[k]\uparrow_{l-1}^{f\#[k]\uparrow_{l-1}^{...f\#[k]\uparrow_{l-1}m(n)...}m(n)}m(n)}_m$
• $f\#[j]\uparrow_l^km(n)=\underbrace{f\#[j]\uparrow_l^{k-1}\{f\#[j]\uparrow_l^{k-1}\{...f\#[j]\uparrow_l^{k-1}m(n)...\}(n)\}(n)}_m$

Then,

$H_4=f[\omega]\uparrow_2\omega(\omega)=f[\omega]\uparrow\uparrow\omega(\omega)$

And,

$H_5=f[\omega]\uparrow_{f[\omega]\uparrow_{f[\omega]\uparrow_{...}\omega(\omega)}\omega(\omega)}\omega(\omega)=\{f[\omega]\uparrow\omega(\omega),f[\omega]\uparrow^{f[\omega]\uparrow\omega(\omega)}\omega(\omega),...\}$

Ordinals

$f(\omega)=\omega+1$

$f_1(\omega)=\omega2$

$f_2(\omega)=\omega^2$

$f_3(\omega)=\varepsilon_0$

$f_4(\omega)=\zeta_0$

$f_\omega(\omega)=\varphi(\omega,0)$

$f_{f_3(\omega)}(\omega)=\varphi(\varphi(1,0),0)$

$f[1](\omega)=\varphi(1,0,0)$

$f[f[1](\omega)](\omega)=\varphi(\varphi(1,0,0),0,0)$

$f[0][1](\omega)=\varphi(1,0,0,0)$

$f[0][0][1](\omega)=\varphi(1,0,0,0,0)$

$f[\omega]\uparrow\omega(\omega)\approx\theta(\Omega^\omega)$ Close enough :P

$f[\omega]\uparrow^2\omega(\omega)\approx\theta(\Omega^\Omega)$

THIS IS A WORK IN PROGRESS FOREVER?