(01/05/2023, 08:53 PM)MphLee Wrote: I agree, in fact I'm working on Goodstein-like solutions on my quest for naturality.

But there can be a non-trivial zerations zoo of solutions. And this thread is all about that alternative exotic landscape. One day we will be able to appreciate better its taste and meaning I believe, eg. its relationship to tropical math.

Hello MphLee!

I'm interested in negative-rank hyperoperations, and I posted this on the googology fandom of japan. I tried to solve the equation for bi-infinite sequence \((\star_r: \mathbb{Z}_{\geq 2} \times \mathbb{Z}_{\geq 2} \to \mathbb{Z}_{\geq 2})_{r = -\infty}^{\infty}\) as follows:

\[

\forall m \forall r \forall n; m \star_{r+1} (n+1) = m \star_r (m \star_{r+1} n) \land m \star_{r+1} 2 = m \star_r m \land m \star_1 n = m+n.

\]

I proofed the existence of solutions and found one, \(\star_0 = \heartsuit\) such that

\[

m \heartsuit n =

\begin{cases}

n & \text{if}\; 2 \leq n \leq m-1 \\

n+2 & \text{if}\; n = m \\

n & \text{if}\; n = m+1 \\

n+1 & \text{if}\; n \geq m+2.

\end{cases}

\]

Then all negative elements are uniquely determined as

\[

m \star_{-1} n =

\left\{

\begin{array}{lll}

& & \text{if \(m = 2\) and} \\

n+1 & \leq 1 & n \leq 0 \\

n+1 & = 2 & n = 1 \\

n+2 & = 4 & n = 2 \\

n+2 & = 5 & n = 3 \\

n-1 & = 3 & n = 4 \\

n+1 & \geq 6 & n \geq 5 \\

& & \text{if \(m = 3\) and} \\

n+1 & \leq 1 & n \leq 0 \\

n+2 & = 3 & n = 1 \\

n+3 & = 5 & n = 2 \\

n-1 & = 2 & n = 3 \\

n+2 & = 6 & n = 4 \\

n-1 & = 4 & n = 5 \\

n+1 & \geq 7 & n \geq 6 \\

& & \text{if \(m \geq 4\) and} \\

n+1 & \leq 1 & n \leq 0 \\

m & = m & n = 1 \\

n+1 & \geq 3, \leq m-1 & 2 \leq n \leq m-2 \\

n+3 & = m+2 & n = m-1 \\

2 & = 2 & n = m \\

n+2 & = m+3 & n = m+1 \\

n-1 & = m+1 & n = m+2 \\

n+1 & \geq m+4 & n \geq m+3

\end{array}

\right.

\]

\[

m \star_{-2} n =

\left\{

\begin{array}{lll}

& & \text{if \(m = 2\) and} \\

n+1 & \leq 1 & n \leq 0 \\

n+1 & = 2 & n = 1 \\

n+2 & = 4 & n = 2 \\

n+3 & = 6 & n = 3 \\

n+1 & = 5 & n = 4 \\

n-2 & = 3 & n = 5 \\

n+1 & \geq 7 & n \geq 6 \\

& & \text{if \(m = 3\) and} \\

n+1 & \leq 1 & n \leq 0 \\

n+2 & = 3 & n = 1 \\

n+4 & = 6 & n = 2 \\

n+2 & = 5 & n = 3 \\

n+3 & = 7 & n = 4 \\

n-3 & = 2 & n = 5 \\

n-2 & = 4 & n = 6 \\

n+1 & \geq 8 & n \geq 7 \\

& & \text{if \(m = 4\) and} \\

n+1 & \leq 1 & n \leq 0 \\

n+3 & = 4 & n = 1 \\

n+5 & = 7 & n = 2 \\

n+3 & = 6 & n = 3 \\

n-1 & = 3 & n = 4 \\

n+3 & = 8 & n = 5 \\

n-4 & = 2 & n = 6 \\

n-2 & = 5 & n = 7 \\

n+1 & \geq 9 & n \geq 8 \\

& & \text{if \(m \geq 5\) and} \\

n+1 & \leq 1 & n \leq 0 \\

m & = m & n = 1 \\

m+3 & = m+3 & n = 2 \\

n+1 & \geq 4, \leq m-1 & 3 \leq n \leq m-2 \\

n+3 & = m+2 & n = m-1 \\

3 & = 3 & n = m \\

n+3 & = m+4 & n = m+1 \\

2 & = 2 & n = m+2 \\

n-2 & = m+1 & n = m+3 \\

n+1 & \geq m+5 & n \geq m+4

\end{array}

\right.

\]

...

If the domain and codomain are limited to \(\mathbb{Z}_{\geq 2}\), we can interpolate the hyperoperation sequence.