imho a core issue

[tommy1729]

... or in the case of tommysexp, \( \text{tommysexp}(z)=\lim_{n\to\infty} \log^{[n]}\text{superfunction2sinh}(z+n+k) \). I focused on cheta, the upper superexponential for base eta=exp(1/e), which is used for the base change sexp function, but it turns out, that iterated logarithms for cheta for n+2 logarithms behaves similar to iterated logarithms for tommysexp for n logarithms. I have some vacation time, so I will try and post the results later.
- Sheldon

So Sheldon , did you find time to compute the results ?
Guess im a bit impatient , but i expected them sooner since you mentioned vacation time.

and i guess im not alone, considering this has about 100 views for only 2 posts.

i understand that the computations might not be so simple as they might appear at first sight as you or someone else assumed... ( not a guess but the actual numbers up to some roundoff )

i also want to point out again that my method of computation is somewhat different and might even be distinct at some places.

tommy1729