Alternate continuum sum formula?

[mike3]

Also, with base e we get a zero denominator since

This is not twice-iterated logarithm, this is square of logarithm.

Thanks for clearing that up. What about my other question, then, about whether there is an iterating formula for the function similar to the one from the original continuum sum formula?

By iterating formula, I mean this:

Original continuum sum formula:
\( \log_z\left(\frac{\mathrm{sexp}'_z(x)}{\mathrm{sexp}'_z(0) (\log z)^x}\right) = \sum_{k=0}^{x-1} \mathrm{sexp}_z(k) \)

Iterating formula:
\( \mathrm{sexp}_z(x) = \int_{-1}^{z} \mathrm{sexp}'_z(0) (\log z)^x \exp_z\left(\sum_{k=0}^{x-1} \mathrm{sexp}_z(k)\right) \).

So is there an analogue of the second formula with these alternate formulas, that doesn't require knowing in advance (or allows one to determine) what the function of x \( D_z \mathrm{sexp}_z(x) \) is? If so, what is it?