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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?

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