There is a result posted on here about how the "eta constants" converge to 2 and the "euler constants" converge to 4. The n'th eta constant is the sup of the x'th n'th hyperoperator root of x. And the "euler constants" are the actual values x_n such that x_n'th n'th hyperoperator root of x_n = n'th eta constant. I think from this we can show that anything less than 2 does not grow unbounded, (just observe the meaning of the n'th hyperoperator root). Any base less than the n'th eta constant has a bounded hyperoperator for N>n-1.
We get something similar with the bounded hyperoperators as what you're talking about. It's a little messier but essentially there exists \( \mathbb{Z}^n \) n-2'th hyperoperators. They satisfy the following really weird recursion
\( \alpha \uparrow^n_{a_1a_2...a_n} (\alpha \uparrow^{n+1}_{a_1a_2...a_n b} z) = \alpha \uparrow^{n+1}_{a_1a_2...a_n b} (z+1) \)
for ANY arbitrary b, and \( a_1,...,a_n \in \mathbb{Z} \) fixed. The first n numbers have to equal for this recursion to work though, but the last number can be anything. Essentially, everytime we go up a hyperoperator, we get another \( \mathbb{Z} \) amount of iterates.
I haven't actually proved this, but it's really straightforward.
Firstly, there exists \( \mathbb{Z} \) exponentiations. namely \( f(z) = \,e^{2 \pi i a_1 z}\alpha^z \) for all integer a_1. Then we fix an a_1 and there exists a kind of fixed point that is unique (its uniqueness and what qualifies it is a little hard to explain, which is the weakest point of my argument). It's geometrically attracting and hence can be iterated. Now the iterate \( g(z,\xi) \) is conjugate to SOME exponential \( e^{2 \pi i a_2 z} \lambda^z \) for \( \lambda \) the multiplier of the above fixed point. This generates \( \mathbb{Z} \times \mathbb{Z} \) tetration functions g(z,1). (they interpolate the natural power towers, send to complex values and are bounded on SOME half plane of the complex plane that includes \( \mathbb{R}^+ \), and satisfy the recursion \( f(g(z)) = g(z+1) \). This allows us to find another FIXED point and it's unique because of Schwarz' lemma. Rinse and repeat, and we get \( \mathbb{Z} \times \mathbb{Z}\times \mathbb{Z} \) pentations. Then \( \mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}\times\mathbb{Z} \) hexations, so on and so forth.
This is all really roughly argued, more intuition than rigor. But I call it the "branches of hyper operators."
We get something similar with the bounded hyperoperators as what you're talking about. It's a little messier but essentially there exists \( \mathbb{Z}^n \) n-2'th hyperoperators. They satisfy the following really weird recursion
\( \alpha \uparrow^n_{a_1a_2...a_n} (\alpha \uparrow^{n+1}_{a_1a_2...a_n b} z) = \alpha \uparrow^{n+1}_{a_1a_2...a_n b} (z+1) \)
for ANY arbitrary b, and \( a_1,...,a_n \in \mathbb{Z} \) fixed. The first n numbers have to equal for this recursion to work though, but the last number can be anything. Essentially, everytime we go up a hyperoperator, we get another \( \mathbb{Z} \) amount of iterates.
I haven't actually proved this, but it's really straightforward.
Firstly, there exists \( \mathbb{Z} \) exponentiations. namely \( f(z) = \,e^{2 \pi i a_1 z}\alpha^z \) for all integer a_1. Then we fix an a_1 and there exists a kind of fixed point that is unique (its uniqueness and what qualifies it is a little hard to explain, which is the weakest point of my argument). It's geometrically attracting and hence can be iterated. Now the iterate \( g(z,\xi) \) is conjugate to SOME exponential \( e^{2 \pi i a_2 z} \lambda^z \) for \( \lambda \) the multiplier of the above fixed point. This generates \( \mathbb{Z} \times \mathbb{Z} \) tetration functions g(z,1). (they interpolate the natural power towers, send to complex values and are bounded on SOME half plane of the complex plane that includes \( \mathbb{R}^+ \), and satisfy the recursion \( f(g(z)) = g(z+1) \). This allows us to find another FIXED point and it's unique because of Schwarz' lemma. Rinse and repeat, and we get \( \mathbb{Z} \times \mathbb{Z}\times \mathbb{Z} \) pentations. Then \( \mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}\times\mathbb{Z} \) hexations, so on and so forth.
This is all really roughly argued, more intuition than rigor. But I call it the "branches of hyper operators."
