addendum 5: found a nice generalization that cover most of the cases that uses the "hypersquare" equation \(b+_{s+2}2=b+_{s+1}b\), and also found the general formula that uses the "four-rule" equation \(2+_{s+1}2=4\).
Setting \(T_{b,k}(s):=b+_s k\) just note that \(T_{b,k}(s+2)=h_{b,s+1}^{k-2}T_b(s+2)\)
\[\begin{align}T_{b,k}(s+2)&=h_{b,s+2}(k)\\
&=h_{b,s+1}^{k-2}h_{b,s+2}(2)&& 2\leq k\\
&=h_{b,s+1}^{k-2}h_{b,s+1}(b) && \textrm{"hypersquare" equation for} \,\, s\in\mathbb N\\
&=h_{b,s+1}^{k-2}h_{b,s}^{b-k}h_{b,s+1}(k)&& k\leq b \\
&= h_{b,s+1}^{k-2}h_{b,s}^{b-k}T_{b,k}(s+1)
\end{align}\]
This gives an extension of the recursion when \(2\leq k\leq b\). Just to be enlightening, let's convert it in box notation with \(b\) implicit
\[\begin{align}T_{b,k}(s+n+2)&=\prod_{i=0}^n h_{b,s+i+1}^{k-2}h_{b,s+i}^{b-k}\bullet T_{b,k}(s+1)\\
[s+n+2]k&=[s+n+1]^{k-2}[s+n]^{b-k}[s+n]^{k-2}[s+n-1]^{b-k}...[s+3]^{k-2}[s+2]^{b-k}[s+2]^{k-2}[s+1]^{b-k}[s+1]^{k-2}[s]^{b-k}[s+1]k \\
&=[s+n+1]^{k-2}[s+n]^{b-2}[s+n-1]^{b-2}...[s+3]^{b-2}[s+2]^{b-2}[s+1]^{b-2}[s]^{b-k}[s+1]k\\
&=h_{b,s+n+1}^{k-2}\circ \left(\prod_{i=1}^n h_{b,s+i}^{b-2}\bullet h_{b,s}^{b-k}T_{b,k}(s+1)\right)
\end{align}\]
To extend the exponent in the other case can perform a similar trick using the \(4\) rule for hos, i.e. 4 is a fixed point. Set \(E_k(s):=h_s(k)=2+_sk\)
\[\begin{align}E_k(s+2):&=h_{s+2}(k)\\
&=h_{s+1}^{k-3}h_{s+2}(3)&& 3\leq k\\
&=h_{s+1}^{k-3}h_{s+1}h_{s+2}(2)\\
&=h_{s+1}^{k-3}h_{s+1}(4) && \textrm{"four-rule" equation for} \,\, s\in\mathbb N\\
&=h_{s+1}^{k-3}h_{s}^{4-k}h_{s+1}(k)&& k\leq 4 \\
&= h_{s+1}^{k-3}h_s^{4-k}E_{k}(s+1)
\end{align}\]
In this case we are less lucky: we can only have \(k=3\) or \(k=4\).
Note: There should be some hidden structure underneath this that generates all these iterated compositions... but idk what is the ultimate source. I have a feeling that all the identities of this kind are important to improve our understanding of the behaviour of possible rank-extensions in a neighborhood of the positive integers (bigger than 1).
Wtf is going to happen if you apply your Ramanujan black magic or your Infinite iteration/limit-trick to these outer compositions? Genuine non-integer ranks?
Because, if this is the case, I have a bunch of iterated compositions (outer and inner) that I've derived for Goodstein's H.Os that we can work on.
Setting \(T_{b,k}(s):=b+_s k\) just note that \(T_{b,k}(s+2)=h_{b,s+1}^{k-2}T_b(s+2)\)
\[\begin{align}T_{b,k}(s+2)&=h_{b,s+2}(k)\\
&=h_{b,s+1}^{k-2}h_{b,s+2}(2)&& 2\leq k\\
&=h_{b,s+1}^{k-2}h_{b,s+1}(b) && \textrm{"hypersquare" equation for} \,\, s\in\mathbb N\\
&=h_{b,s+1}^{k-2}h_{b,s}^{b-k}h_{b,s+1}(k)&& k\leq b \\
&= h_{b,s+1}^{k-2}h_{b,s}^{b-k}T_{b,k}(s+1)
\end{align}\]
This gives an extension of the recursion when \(2\leq k\leq b\). Just to be enlightening, let's convert it in box notation with \(b\) implicit
\[\begin{align}T_{b,k}(s+n+2)&=\prod_{i=0}^n h_{b,s+i+1}^{k-2}h_{b,s+i}^{b-k}\bullet T_{b,k}(s+1)\\
[s+n+2]k&=[s+n+1]^{k-2}[s+n]^{b-k}[s+n]^{k-2}[s+n-1]^{b-k}...[s+3]^{k-2}[s+2]^{b-k}[s+2]^{k-2}[s+1]^{b-k}[s+1]^{k-2}[s]^{b-k}[s+1]k \\
&=[s+n+1]^{k-2}[s+n]^{b-2}[s+n-1]^{b-2}...[s+3]^{b-2}[s+2]^{b-2}[s+1]^{b-2}[s]^{b-k}[s+1]k\\
&=h_{b,s+n+1}^{k-2}\circ \left(\prod_{i=1}^n h_{b,s+i}^{b-2}\bullet h_{b,s}^{b-k}T_{b,k}(s+1)\right)
\end{align}\]
To extend the exponent in the other case can perform a similar trick using the \(4\) rule for hos, i.e. 4 is a fixed point. Set \(E_k(s):=h_s(k)=2+_sk\)
\[\begin{align}E_k(s+2):&=h_{s+2}(k)\\
&=h_{s+1}^{k-3}h_{s+2}(3)&& 3\leq k\\
&=h_{s+1}^{k-3}h_{s+1}h_{s+2}(2)\\
&=h_{s+1}^{k-3}h_{s+1}(4) && \textrm{"four-rule" equation for} \,\, s\in\mathbb N\\
&=h_{s+1}^{k-3}h_{s}^{4-k}h_{s+1}(k)&& k\leq 4 \\
&= h_{s+1}^{k-3}h_s^{4-k}E_{k}(s+1)
\end{align}\]
In this case we are less lucky: we can only have \(k=3\) or \(k=4\).
Note: There should be some hidden structure underneath this that generates all these iterated compositions... but idk what is the ultimate source. I have a feeling that all the identities of this kind are important to improve our understanding of the behaviour of possible rank-extensions in a neighborhood of the positive integers (bigger than 1).
Wtf is going to happen if you apply your Ramanujan black magic or your Infinite iteration/limit-trick to these outer compositions? Genuine non-integer ranks?
Because, if this is the case, I have a bunch of iterated compositions (outer and inner) that I've derived for Goodstein's H.Os that we can work on.
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)