Let r be a real number 0<r<1
We can numerically calculate c₍ᵣ₎=ʳa and c₍₋ᵣ₎=⁻ʳa with Sheldonison's knesser-gp.
No proof, but numerically:
^{^{-(r+1)}a}=(^{-r}a)^{^{r}a}})
and in general
}a)^{^{-(r+(n+1))}a}=(^{-(r+n)}a)^{^{r+n}a}})
This remains valid for -(r+n)<-2, where the iterated logarithms are complex numbers.
This gives a natural way to define °a for the Z branch. °a should be the value that makes:
\,^{(a^{(^0a))}=(a^{(^0a)})\,^{(log_a{^0a})})
then it also satisfies:
Example (for the main branch)
![[Image: FqooTc3.jpg?1]](http://i.imgur.com/FqooTc3.jpg?1)
For the Z branch
![[Image: zUWE8Xh.png?1]](http://i.imgur.com/zUWE8Xh.png?1)
We can numerically calculate c₍ᵣ₎=ʳa and c₍₋ᵣ₎=⁻ʳa with Sheldonison's knesser-gp.
No proof, but numerically:
and in general
This remains valid for -(r+n)<-2, where the iterated logarithms are complex numbers.
This gives a natural way to define °a for the Z branch. °a should be the value that makes:
then it also satisfies:
Example (for the main branch)
For the Z branch
I have the result, but I do not yet know how to get it.
