Another new person! :)

[JmsNxn]

Hi Syzithryx, welcome on the forum. For me zeration is the increment. But yes I remember fiery discussions on this board with Gianfranco, there are surely other opinions about zeration. About the topic of fractional rank of hyperoperations I don't remember much said here. Do you have an approach to that?

So, define two functions: f(a,b) = (a+b)/2, and g(a,b) = ²√(ab). Then make an iterative sequence: c₀ = a+b, d₀ = ab; c_i = f(c_(i-1), d_(i-1)); d_i = g(c_(i-1), d_(i-1)). Then c_i and d_i should get closer together with each iteration and converge on a single value, then defined to be the 1.5ation of a and b; but unfortunately I don't know enough math to actually *prove* that or find any kind of closed expression for what that actually is.

Welcome to the forum!

This approach was constructed on here a long time ago! The problem with it, is it doesn't satisfy goodstein's equation. So it makes very very smooth solutions, but it doesn't actually satisfy the recursion necessary.

So, if you were to solve this iterative formula for \(0 \le s \le 1\) for \(0 = +\) and \(1 = \cdot\), this will not be extendable analytically to \(s = 1+\delta\) as \(\delta \to 0\).  Which is to say, if you extend this definition to \(0 \le s \le 2\) and you ask that:

\[
a [s] (a [s+1] b) = a [s+1] b+1\\
\]

Then the solution is not analytic at \(s=1\), and essentially makes a piece wise solution.

The graphs are somewhere on this forum, you should be able to find them. I forget who posted it it was so long ago. This essentially becomes a very pretty looking interpolation of addition and multiplication. But unfortunately, we have many many many of these. But we've yet to find one that is analytic and satisfies Goodstein's equation.

I'm excited to see what you bring to the table with fractional hyper-operators. It's a personal problem for me, that I've been stuck on for 10 years, lmao.