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:

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

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.