Okay, so this definitely requires more detail, but I see the solution. I made a mistake earlier. We cannot treat \(\theta\) alone. We actually have to consider \(\boldsymbol{\theta} = (\theta_1,\theta_2,\theta_3)\), for each occurrence of tetration in the Goodstein equation. If we force these \(\theta\)'s into one, we get the trivial solution \(\theta(s) = 1-s\). It's a little trickier than I originally thought. But I have it now. I'm just polishing off a proof of concept paper; that is no where near perfect rigor. But describes a good project outline.
This means, we're solving:
\[
1 \oplus_{s-1,\theta_1,F(1\oplus_{s,\theta_2,F(y)} y)} \left(1\oplus_{s,\theta_2,F(y)} y\right) = 1\oplus_{s,\theta_3,F(y+1)} y+1\\
\]
Where, \(F(y) = y^{1/y}\). And then we are relating \(\theta_1 = \theta_2 = \theta_3\), when we make the appropriate change of variables in each case. Where \(\theta_1\) is a function in \(s-1, 1\oplus_{s,\theta_2,F(y)} y\); \(\theta_2\) is a function in \(s,y\); and \(\theta_3\) is a function in \(s,y+1\). But after changing the variables they are all equivalent.
I forgot to pay close attention to the Jacobian above, once you pay attention to the Jacobian this forms a more natural solution; and actually takes care of all the trouble. It ensures we can extend indefinitely in any direction as well--upto branch cuts. This then creates a function \(\theta(s,y)\), such that:
\[
1<s> y = 1 \oplus_{s,\theta(s,y), F(y)} y = \exp_{F(y)}^{\circ s + \theta(s,y)}\left(\log_{F(y)}^{\circ s+\theta(s,y)}(1) + y\right)\\
\]
And \(\theta(s,y)\) has some very weird functional equations... It is not 1-periodic, that was a mistake. It just satisfies \(0 = \theta(0,y) =\theta(1,y) = \theta(2,y)\). This had me erroneously assume 1-periodic; it's close to it, but not exactly. Which explains how this will get us in between exponentiation and tetration (the value \(\theta\) will start to explode near here). This is very difficult to talk about though, so I'm going to go silent mode until I have a working project out line--then I'll start rigorously proving the claims in the outline, which aren't very hard (I know how to prove all the claims I make, and I'm making sure they can be shown rigorously).
This also works for \(\alpha <s> y\), for \(y \in \mathbb{C}/\{0,1\}\), and up to branch cuts which move for \(\alpha,y\). This becomes a branch cut nightmare. I'm going to try and program in a protocol for \(\alpha,s,y\) for \(\alpha ,y > 1\), \(0 \le s \le 2\). This should be fairly easy to program, it'll be slow, but all we really need is a Newton root finder algorithm. I'm hesitating though, because I think there should be a more efficient manner of calculating \(\theta\), it shouldn't be this hard, I'm just missing something...
EDIT:
It's also simpler to realize:
\[
1<s> y = \exp_{F(y)}^{\circ s}\left(\log_{F(y)}^{\circ s}(1) + y + \varphi(s,y)\right)\\
\]
Which is a slightly less convoluted functional equation. We can call:
\[
1<s>_\varphi y = \exp_{F(y)}^{\circ s}\left(\log_{F(y)}^{\circ s}(1) + y + \varphi\right)
\]
For any \(\varphi \in \mathbb{C}\). We are then solving:
\[
1<s-1>_{\varphi_1} 1 <s>_{\varphi_2} y = 1 <s>_{\varphi_3} y+1\\
\]
We choose the appropriate path, using the variable change, which allows us \(\varphi(s,y)\). This gives us a super easy proof for existence, that works for all \(0 \le \Re(s) \le 2\); but the construction is definitely not discovered. So just existence is feasible at the moment.
We have a surface \(\boldsymbol{\Phi} \subset \mathbb{C}^3\), such that, for all \(\boldsymbol{\varphi} \in \boldsymbol{\Phi}\) we get:
\[
1<s-1>_{\varphi_1} 1 <s>_{\varphi_2} y = 1 <s>_{\varphi_3} y+1\\
\]
Now, we choose a path in \(s\) and in \(y\), which reduces us to \(\varphi(s,y)\), which is our solution. We are literally tracing a path on a surface, Mphlee. That's literally! All we are doing!
This means, we're solving:
\[
1 \oplus_{s-1,\theta_1,F(1\oplus_{s,\theta_2,F(y)} y)} \left(1\oplus_{s,\theta_2,F(y)} y\right) = 1\oplus_{s,\theta_3,F(y+1)} y+1\\
\]
Where, \(F(y) = y^{1/y}\). And then we are relating \(\theta_1 = \theta_2 = \theta_3\), when we make the appropriate change of variables in each case. Where \(\theta_1\) is a function in \(s-1, 1\oplus_{s,\theta_2,F(y)} y\); \(\theta_2\) is a function in \(s,y\); and \(\theta_3\) is a function in \(s,y+1\). But after changing the variables they are all equivalent.
I forgot to pay close attention to the Jacobian above, once you pay attention to the Jacobian this forms a more natural solution; and actually takes care of all the trouble. It ensures we can extend indefinitely in any direction as well--upto branch cuts. This then creates a function \(\theta(s,y)\), such that:
\[
1<s> y = 1 \oplus_{s,\theta(s,y), F(y)} y = \exp_{F(y)}^{\circ s + \theta(s,y)}\left(\log_{F(y)}^{\circ s+\theta(s,y)}(1) + y\right)\\
\]
And \(\theta(s,y)\) has some very weird functional equations... It is not 1-periodic, that was a mistake. It just satisfies \(0 = \theta(0,y) =\theta(1,y) = \theta(2,y)\). This had me erroneously assume 1-periodic; it's close to it, but not exactly. Which explains how this will get us in between exponentiation and tetration (the value \(\theta\) will start to explode near here). This is very difficult to talk about though, so I'm going to go silent mode until I have a working project out line--then I'll start rigorously proving the claims in the outline, which aren't very hard (I know how to prove all the claims I make, and I'm making sure they can be shown rigorously).
This also works for \(\alpha <s> y\), for \(y \in \mathbb{C}/\{0,1\}\), and up to branch cuts which move for \(\alpha,y\). This becomes a branch cut nightmare. I'm going to try and program in a protocol for \(\alpha,s,y\) for \(\alpha ,y > 1\), \(0 \le s \le 2\). This should be fairly easy to program, it'll be slow, but all we really need is a Newton root finder algorithm. I'm hesitating though, because I think there should be a more efficient manner of calculating \(\theta\), it shouldn't be this hard, I'm just missing something...
EDIT:
It's also simpler to realize:
\[
1<s> y = \exp_{F(y)}^{\circ s}\left(\log_{F(y)}^{\circ s}(1) + y + \varphi(s,y)\right)\\
\]
Which is a slightly less convoluted functional equation. We can call:
\[
1<s>_\varphi y = \exp_{F(y)}^{\circ s}\left(\log_{F(y)}^{\circ s}(1) + y + \varphi\right)
\]
For any \(\varphi \in \mathbb{C}\). We are then solving:
\[
1<s-1>_{\varphi_1} 1 <s>_{\varphi_2} y = 1 <s>_{\varphi_3} y+1\\
\]
We choose the appropriate path, using the variable change, which allows us \(\varphi(s,y)\). This gives us a super easy proof for existence, that works for all \(0 \le \Re(s) \le 2\); but the construction is definitely not discovered. So just existence is feasible at the moment.
We have a surface \(\boldsymbol{\Phi} \subset \mathbb{C}^3\), such that, for all \(\boldsymbol{\varphi} \in \boldsymbol{\Phi}\) we get:
\[
1<s-1>_{\varphi_1} 1 <s>_{\varphi_2} y = 1 <s>_{\varphi_3} y+1\\
\]
Now, we choose a path in \(s\) and in \(y\), which reduces us to \(\varphi(s,y)\), which is our solution. We are literally tracing a path on a surface, Mphlee. That's literally! All we are doing!