Rank-Wise Approximations of Hyper-Operations

[Catullus]

This implicit solution exists uniquely across iterations. But if you ask for a TETRATION solution, it's not enough to declare uniqueness. Even while moving your base value, it's not enough.

There are infinite TETRATION solutions to these equations. There are infinite \(\text{Tet}(s)\) which satisfy \(\text{Tet}(0) = 1\). All you have to do is find countable \(z\) in \(\exp^{\circ s}(z)\) which orbits eventually hit \(1\). Honestly this is an artifact of the exponential function; and home to the iteration of transcendental functions.

This formula converges as your describing. But there's little to no general uniqueness. There are countably infinite solutions, and just because an algorithm evaluates to something, that doesn't qualify as a uniqueness condition. I can design a wrench for your algorithm which makes everything converge different. But it doesn't dissway the general idea.

I said tetration has to be unique.
You are saying using different fixed points to iterate off of would produce different tetrations.
For bases not in the Shell-Thron region, maybe you could analytically continue the hyper-operations there.