08/22/2017, 02:03 PM
(08/21/2017, 08:05 PM)JmsNxn Wrote: Hey Sheldon. Bo proved this somewhere on here. It's actually pretty straight forward to prove that ANY analytic hyper-operator necessarily satisfies \( f(-k) = 1-k \) for \( 0 \le k \le n-1 \) for \( n \) the rank of the hyper-operator. (exponentiation is rank 1, tetration is rank 2, pentation is rank 3, hexation is rank 4, so on and so forth.)
Yeah, that makes sense. I hadn't seen Henryk's proof. I'm still working on the finishing touches for analytic Hexation, with a theta mapping.
- Sheldon
