We define γ[α] = sup (γ[α[1]], γ[α[2]], γ[α[3]], ...) (for some limit ordinal α)

When we get this:

- Tape is erased
- Other ones aren't counted in limsup, exepted for those that are written with the state S
- State is set to S.

For γ[α+n], we can refer to all n states, and to the first state of the transfinite part.

I'm not sure if it works for λ and ζ.

## Values and bounds

We now that sup(γ[1], γ[2], γ[3], ...) = γ itself.

Further, for γ[ω+1] we can take the following ruleset:

S _ 1 l ω S 1 1 r halt ω 1 _ r 0

This'll work for γω steps.

## Why this doesn't work

There are finite machines for γ[n], but when we get to γ[ω], we can prove there are \(2^{\aleph_0}\) machines.

Therfore, according to LittlePeng9, this machine can clock every countable ordinal and γ[ω] is therefore undefined, and so are the higher γ[α].