regular tetration at b=e^(-e)

[bo198214]

Like an asymptotic series?

yes, yes. The series you obtain is asymptotic at the fixed point and has 0 convergence radius.

2. With help of the above formula there is a (dog-slowly but converging) limit formula.

What's that?

Perhaps I was too fast here generalzing from the case of multiplier 1. But I give this case for reference:
In the same way as above you can obtain the regular Abel function for \( h(z)=e^{z}-1 \):

\( \alpha(z)=\underbrace{\frac{1}{3}\log(z) - \frac{2}{z}}_{F(z)} + p(z) \), where \( p(z) \) is analytic in the left sector where \( h^{\circ -n}(z)\to 0 \) and has the asymptotic expansion at the fixed point 0.

Now we know that \( \alpha(h^{\circ -n}(z))=\alpha(z)-n \) and hence
\( \alpha(z)=n+\alpha(h^{\circ -n}(z))=n+F(h^{\circ -n}(z)) + p(h^{\circ -n}(z)) \). The rightmost summand converges to 0. So we can write:
\( \alpha(z)=\lim_{n\to\infty} n+F(h^{\circ -n}(z)) \).

But in your case we have two alternating sectors, I guess it needs some more thinking.