User blog comment:P進大好きbot/Elementary Large Number/@comment-30754445-20181118000005/@comment-35470197-20181118132714

> Pink function

Thank you for the definition. It is exactly the composite of the usual Goedel correspondence and the conversion of trees into \(\varepsilon_0\) used in the proof of \(\textrm{PTO}(\textrm{PA}) = \varepsilon_0\).

> Well, I didn't expect your explanatoin to include 2-adic integers and Godel representations! Are these things really needed to get the point across? The table of ordinals is useful, though.

I think that such an explanation might be helpful for googologists who know either one of them. Especially, Emlightened would understand what I did, because she also use a similar method related to Goedel correspondence. (Her method is the original Goedel correspondence, and my method is a variant of Goedel correspondence which does not need the enumeration function \(p_n\) of prime numbers.)

> Cool!

Thank you :)

> 33,554,433

I guess that you correctly understood the computation, but just miscalculated (or mistyped) a little.

The positive integer corresponding to \(\omega^{\omega+1}\) can be computed in the following way: \begin{eqnarray*} 1 & \mapsto & 1+2 = 3 \\ \omega & \mapsto & 1+2^{0 + \frac{3 + 1}{2}} = 5 \in [2^4,2^3) \\ \omega+1 & \mapsto & 5+2^3 = 13 \\ \omega^{\omega+1} & \mapsto & 1 + 2^{\frac{13 + 1}{2}} = 129 \end{eqnarray*}

On the other hand, the ordinal corresponding to \(33554433\) can be computed in the following way: \begin{eqnarray*} 1 & \mapsto & 0 \\ 3 & = & 1 + 2 = 1 + 2^{0 + \frac{1 + 1}{2}}\mapsto 0 + \omega^0 = 1 \\ 7 & = & 3 + 2^2 = 3 + 2^{1 + \frac{1 + 1}{2}} \mapsto 1 + \omega^0 = 2 \\ 17 & = & 1 + 2^4 = 1 + 2^{0 + \frac{7 + 1}{2}} \mapsto 0 + \omega^2 = \omega^2 \\ 49 & = & 17 + 2^5 = 17 + 2^{4 + \frac{1 + 1}{2}} \mapsto \omega^2 + \omega^0 = \omega^2 + 1 \\ 33554433 & = & 1 + 2^25 = 1 + 2^{0 + \frac{49 + 1}{2}} \mapsto 0 + \omega^{\omega^2 + 1} = \omega^{\omega^2 + 1} \end{eqnarray*}

Sorry if I miscalculated.