Hi Daniel.
I encourage you to make your question more precise and to always define the terms of discourse. Defining the terms in mathematics is most of the time like solving 70% of the problem.
The existence of an hypothetical successor function for ranks, i.e. "adding one" to the rank parameter is a not only very non-trivial matter but also, I claim, the key to solving the whole non-integer ranks problem. This is not only a superficial guess: I've centered all my research around the problem of correctly define such a successor function and now I know it yields promising results.
As I said, it is non-trivial. It seems you approached this successor function the naive way. Let me rephrase what I think is evident from your definition.
Definition (Naive rank successor): let \(\uparrow^n:\mathbb N\times \mathbb N\to\mathbb N\) be the natural Goodstein hyperoperation sequences (Knuth-reparametrized at \(2=0\). Define the naive rank successor function to be the function over the natural numbers
\[\Sigma_{\rm naive}:\mathbb N\to\mathbb N\]
satisfying the functional equation
\[(*)\,\,\,\,\,\,\,\forall a,b,n\in\mathbb N.\, \, \Sigma_{\rm naive}(a\uparrow^n b)=a \uparrow^{n+1} b\]
As bo198214 suspects, this is not a good definition. Remember, a function is a binary relation that is right-univocal (functional) and left-total: here left-total means that it is defined for every argument; right-univocal means that we cannot have \(f(a)\neq f(b)\) if \(a=b\)... quite a reasonable requirement, and the very definition of what a function is.
I'll prove that the set of functions satisfying eq. \((*)\) is empty. In other words, the function \(\Sigma_{\rm naive}\) cannot exists.
To do that I'll shot that the naive definition contradicts the assumed right-univocal property of \(\Sigma_{\rm naive}\), a property assumed when we assume it is a function.
Proposition 1: the function \(\Sigma_{\rm naive}\) doesn't exist.
Proof: assume that the successor function exists. Then, by definition
\[(*)\,\,\,\,\,\,\, \forall a,b,n\in\mathbb N.\, \, \Sigma_{\rm naive}(a\uparrow^n b)=a \uparrow^{n+1} b\]
but also, if \(x=y\) then \(\Sigma_{\rm naive}(x)=\Sigma_{\rm naive}(y)\). Take \(x=2\uparrow^0 3\) and \(y=6\uparrow^1 1\). Clearly
\[x=y\]
because \(x=2\cdot 3=6\) and \(y=6^1=6\). By functionality we should deduce
\[\begin{align}
\Sigma_{\rm naive}(x)&=\Sigma_{\rm naive}(y)\\
\Sigma_{\rm naive}(2\uparrow^0 3)&=\Sigma_{\rm naive}(6\uparrow^1 1)
\end{align}\]
but by the functional equation defining the successor we must deduce
\[\begin{align}
2\uparrow^1 3&=6\uparrow^2 1\\
2^3&={}^1 6\\
8&=6
\end{align}\]
This concludes the proof. \(\Sigma_{\rm naive}\) is not a function, or if it is it can't satisfy eq \((*)\).
Note that we can find a much more elegant counterexample if we consider instead of the Knuth's family that is centered at \(0=\rm multiplication\), the original one centered at \(0=successor\). We can speed-run the proof
\[\begin{align}
0&=0\\
0+0&=1\cdot 0\\
0[1]0&=1[2]0\\
\Sigma_{\rm naive}(0[1]0)&=\Sigma_{\rm naive}(1[2]0)\\
0[2]0&=1[3]0\\
0\cdot 0&=1^0\\
\Sigma_{\rm naive}(0)=0&=1
\end{align}\]
We can prove that non-commutativity of the higher hos is an even greater obstruction to functionality/existence of the successor function.
Proposition 2: if \(\Sigma_{\rm naive}\) exists, then every \(\uparrow^n\) is commutative.
Proof: proceed by induction. The base of the induction is for \(\uparrow^0=\times\). Multiplication is commutative.
Then, assume \(\uparrow^n\) is commutative:
\[\begin{align}
\forall a,b.\, a \uparrow^n b&=b\uparrow^n a\\
\forall a,b.\, \Sigma_{\rm naive}(a \uparrow^n b)&=\Sigma_{\rm naive}(b\uparrow^n a)\\
\forall a,b.\, a \uparrow^{n+1} b&=b\uparrow^{n+1} a\\
\end{align}\]
using induction we deduce the proposition.
Bonus proof: thinking about the successor being defined for every argument we also obtain a great proof that such an object cannot do what it promises. In fact, if it was a function, we could trivially evaluate its value at each natural number
\[\Sigma_{\rm naive}(n)=\Sigma_{\rm naive}(1\cdot n)=1^n=1\]
I encourage you to make your question more precise and to always define the terms of discourse. Defining the terms in mathematics is most of the time like solving 70% of the problem.
The existence of an hypothetical successor function for ranks, i.e. "adding one" to the rank parameter is a not only very non-trivial matter but also, I claim, the key to solving the whole non-integer ranks problem. This is not only a superficial guess: I've centered all my research around the problem of correctly define such a successor function and now I know it yields promising results.
As I said, it is non-trivial. It seems you approached this successor function the naive way. Let me rephrase what I think is evident from your definition.
Definition (Naive rank successor): let \(\uparrow^n:\mathbb N\times \mathbb N\to\mathbb N\) be the natural Goodstein hyperoperation sequences (Knuth-reparametrized at \(2=0\). Define the naive rank successor function to be the function over the natural numbers
\[\Sigma_{\rm naive}:\mathbb N\to\mathbb N\]
satisfying the functional equation
\[(*)\,\,\,\,\,\,\,\forall a,b,n\in\mathbb N.\, \, \Sigma_{\rm naive}(a\uparrow^n b)=a \uparrow^{n+1} b\]
As bo198214 suspects, this is not a good definition. Remember, a function is a binary relation that is right-univocal (functional) and left-total: here left-total means that it is defined for every argument; right-univocal means that we cannot have \(f(a)\neq f(b)\) if \(a=b\)... quite a reasonable requirement, and the very definition of what a function is.
I'll prove that the set of functions satisfying eq. \((*)\) is empty. In other words, the function \(\Sigma_{\rm naive}\) cannot exists.
To do that I'll shot that the naive definition contradicts the assumed right-univocal property of \(\Sigma_{\rm naive}\), a property assumed when we assume it is a function.
Proposition 1: the function \(\Sigma_{\rm naive}\) doesn't exist.
Proof: assume that the successor function exists. Then, by definition
\[(*)\,\,\,\,\,\,\, \forall a,b,n\in\mathbb N.\, \, \Sigma_{\rm naive}(a\uparrow^n b)=a \uparrow^{n+1} b\]
but also, if \(x=y\) then \(\Sigma_{\rm naive}(x)=\Sigma_{\rm naive}(y)\). Take \(x=2\uparrow^0 3\) and \(y=6\uparrow^1 1\). Clearly
\[x=y\]
because \(x=2\cdot 3=6\) and \(y=6^1=6\). By functionality we should deduce
\[\begin{align}
\Sigma_{\rm naive}(x)&=\Sigma_{\rm naive}(y)\\
\Sigma_{\rm naive}(2\uparrow^0 3)&=\Sigma_{\rm naive}(6\uparrow^1 1)
\end{align}\]
but by the functional equation defining the successor we must deduce
\[\begin{align}
2\uparrow^1 3&=6\uparrow^2 1\\
2^3&={}^1 6\\
8&=6
\end{align}\]
This concludes the proof. \(\Sigma_{\rm naive}\) is not a function, or if it is it can't satisfy eq \((*)\).
Note that we can find a much more elegant counterexample if we consider instead of the Knuth's family that is centered at \(0=\rm multiplication\), the original one centered at \(0=successor\). We can speed-run the proof
\[\begin{align}
0&=0\\
0+0&=1\cdot 0\\
0[1]0&=1[2]0\\
\Sigma_{\rm naive}(0[1]0)&=\Sigma_{\rm naive}(1[2]0)\\
0[2]0&=1[3]0\\
0\cdot 0&=1^0\\
\Sigma_{\rm naive}(0)=0&=1
\end{align}\]
We can prove that non-commutativity of the higher hos is an even greater obstruction to functionality/existence of the successor function.
Proposition 2: if \(\Sigma_{\rm naive}\) exists, then every \(\uparrow^n\) is commutative.
Proof: proceed by induction. The base of the induction is for \(\uparrow^0=\times\). Multiplication is commutative.
Then, assume \(\uparrow^n\) is commutative:
\[\begin{align}
\forall a,b.\, a \uparrow^n b&=b\uparrow^n a\\
\forall a,b.\, \Sigma_{\rm naive}(a \uparrow^n b)&=\Sigma_{\rm naive}(b\uparrow^n a)\\
\forall a,b.\, a \uparrow^{n+1} b&=b\uparrow^{n+1} a\\
\end{align}\]
using induction we deduce the proposition.
Bonus proof: thinking about the successor being defined for every argument we also obtain a great proof that such an object cannot do what it promises. In fact, if it was a function, we could trivially evaluate its value at each natural number
\[\Sigma_{\rm naive}(n)=\Sigma_{\rm naive}(1\cdot n)=1^n=1\]
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)