Hey, Mphlee.
I haven't devoured this post fully yet. I'll definitely comeback to it; but you have me intrigued by "upper rank iterations respect lower rank iterations". This is absolutely true for the bounded analytic hyperoperators. And I think it's not as strict of an idea as you are suggesting. Any good iteration should have some level of this happening. That absolutely this happens; and probably in a good general sense.
For example, let:
\[
\begin{align}
\alpha \uparrow^0 z &= \alpha \cdot z\\
\alpha \uparrow^{n+1} z &= \frac{d^z}{dw^z}\Big{|}_{w=0} \sum_{k=0}^\infty \alpha \uparrow^{n,\,\circ k} (1)\frac{w^k}{k!}\\
\alpha \uparrow^{n,\,\circ y} z &= \frac{d^{z}}{dw^z}\Big{|}_{w=0} \sum_{k=0}^\infty \alpha \uparrow^{n,\circ y} k \frac{w^k}{k!}\\
\alpha \uparrow^{n,\circ y} \left(\alpha \uparrow^{n+1} z\right) &= \alpha \uparrow^{n+1}( z+y)\\
\end{align}
\]
And this holds for all \(\Re(y) > 0\) & \(\Re(z) > 0\)--where the trouble is getting this to work for \(s \Big{|}_{\mathbb{N}} = n\) and \(\Re(s) > 0\). But the moment we have fractional hyper operators/semi-operators/bounded analytic semi-operators... your equation definitely holds up. It's a built in feature from a generation ago
Also note that for \(s \Big{|}_{\mathbb{N}}\)--each of these expressions are convergent integral transforms. And upon which we can even define things like:
\[
H(w,u) = \sum_{y=0}^\infty \sum_{k=0}^\infty \alpha \uparrow^{n,\circ y}(k)\frac{u^yw^k}{y!k!}\\
\]
The action \(\frac{d^y}{du^y}\) pulls out the iteration of the rank at hand (finite iterations of tetration), the action \(\frac{d^z}{dw^z}\) pulls out the iteration of the rank previous (it gives the tetration value, as iterations of exponentiation).
Edit:
This is very incorrect notation, but I believe the notation can be descriptive and informative. When doing the bounded analytic approach:
\[
\alpha\uparrow^{(s_0,k_0),(s_1,k_1),...,(s_n,k_n)} z = \frac{d^{s_0}}{dw_0}...\frac{d^{s_n}}{dw_n^{s_n}}\Big{|}_{w_0,...,w_n=0} \sum_{j_1,...,j_n=0}^\infty \alpha \uparrow^{(j_0,k_0),(j_1,k_1),...,(j_n,k_n)} z \frac{w_0^{j_0}\cdots w_n^{j_n}}{j_0! \cdots j_n!}\\
\]
Where:
\[
\alpha \uparrow^{(s_n,k_n)} z = f(z)\\
\]
Where,
\[
g(z) = \alpha \uparrow^{k_n} z\,\,\text{for}\,\,k_n \in \mathbb{N}\\
\]
And:
\[
f(z) = g^{\circ s_n}(z)\\
\]
Whereby, the chaining is:
\[
\alpha \uparrow^{(s_n,k_n),(s_{n+1},k_{n+1})} z = \alpha \uparrow^{k_n,\circ s_n} \left(\alpha \uparrow^{k_{n+1}, \circ s_{n+1}} z\right)\\
\]
THIS ALWAYS CONVERGES AS I'VE WRIT IT!
To describe the chaining is very difficult (I.e: \((s_n,k_n),(s_{n+1},k_{n+1})\)). We actually have a good amount of cancellation. I do not know the rules, or how to do it. I've just seen the cancellation in effect. So although the notation I just posited is garbage. It'll look something like this. Where the "depth of iteration" is really just, how many Differintegrals we need.
And when nesting these things we are following your "Upper ranks are affected by lower ranks" rule.
Anyway, I drank too much now--I'm a little tipsy and I should probably shut up. Just trying to say the rules you seem to be describing; bounded analytic hyper operators follow them to a tee
I haven't devoured this post fully yet. I'll definitely comeback to it; but you have me intrigued by "upper rank iterations respect lower rank iterations". This is absolutely true for the bounded analytic hyperoperators. And I think it's not as strict of an idea as you are suggesting. Any good iteration should have some level of this happening. That absolutely this happens; and probably in a good general sense.
For example, let:
\[
\begin{align}
\alpha \uparrow^0 z &= \alpha \cdot z\\
\alpha \uparrow^{n+1} z &= \frac{d^z}{dw^z}\Big{|}_{w=0} \sum_{k=0}^\infty \alpha \uparrow^{n,\,\circ k} (1)\frac{w^k}{k!}\\
\alpha \uparrow^{n,\,\circ y} z &= \frac{d^{z}}{dw^z}\Big{|}_{w=0} \sum_{k=0}^\infty \alpha \uparrow^{n,\circ y} k \frac{w^k}{k!}\\
\alpha \uparrow^{n,\circ y} \left(\alpha \uparrow^{n+1} z\right) &= \alpha \uparrow^{n+1}( z+y)\\
\end{align}
\]
And this holds for all \(\Re(y) > 0\) & \(\Re(z) > 0\)--where the trouble is getting this to work for \(s \Big{|}_{\mathbb{N}} = n\) and \(\Re(s) > 0\). But the moment we have fractional hyper operators/semi-operators/bounded analytic semi-operators... your equation definitely holds up. It's a built in feature from a generation ago
Also note that for \(s \Big{|}_{\mathbb{N}}\)--each of these expressions are convergent integral transforms. And upon which we can even define things like:
\[
H(w,u) = \sum_{y=0}^\infty \sum_{k=0}^\infty \alpha \uparrow^{n,\circ y}(k)\frac{u^yw^k}{y!k!}\\
\]
The action \(\frac{d^y}{du^y}\) pulls out the iteration of the rank at hand (finite iterations of tetration), the action \(\frac{d^z}{dw^z}\) pulls out the iteration of the rank previous (it gives the tetration value, as iterations of exponentiation).
Edit:
This is very incorrect notation, but I believe the notation can be descriptive and informative. When doing the bounded analytic approach:
\[
\alpha\uparrow^{(s_0,k_0),(s_1,k_1),...,(s_n,k_n)} z = \frac{d^{s_0}}{dw_0}...\frac{d^{s_n}}{dw_n^{s_n}}\Big{|}_{w_0,...,w_n=0} \sum_{j_1,...,j_n=0}^\infty \alpha \uparrow^{(j_0,k_0),(j_1,k_1),...,(j_n,k_n)} z \frac{w_0^{j_0}\cdots w_n^{j_n}}{j_0! \cdots j_n!}\\
\]
Where:
\[
\alpha \uparrow^{(s_n,k_n)} z = f(z)\\
\]
Where,
\[
g(z) = \alpha \uparrow^{k_n} z\,\,\text{for}\,\,k_n \in \mathbb{N}\\
\]
And:
\[
f(z) = g^{\circ s_n}(z)\\
\]
Whereby, the chaining is:
\[
\alpha \uparrow^{(s_n,k_n),(s_{n+1},k_{n+1})} z = \alpha \uparrow^{k_n,\circ s_n} \left(\alpha \uparrow^{k_{n+1}, \circ s_{n+1}} z\right)\\
\]
THIS ALWAYS CONVERGES AS I'VE WRIT IT!
To describe the chaining is very difficult (I.e: \((s_n,k_n),(s_{n+1},k_{n+1})\)). We actually have a good amount of cancellation. I do not know the rules, or how to do it. I've just seen the cancellation in effect. So although the notation I just posited is garbage. It'll look something like this. Where the "depth of iteration" is really just, how many Differintegrals we need.
And when nesting these things we are following your "Upper ranks are affected by lower ranks" rule.
Anyway, I drank too much now--I'm a little tipsy and I should probably shut up. Just trying to say the rules you seem to be describing; bounded analytic hyper operators follow them to a tee