Regular slog for base sqrt(2) - Using z=2

[andydude]

Two comments.

@Henryk
The infinitely iterated exponential with branch indecies is probably the best way:

\( H_k(x) = {}^{\infty}(x)_k = \frac{W_k(-\log(x))}{-\log(x)} = e^{-W_k(-\log(x))} \)

with the value substituted for x as in \( H_k(\sqrt{2}) \). Using the Lambert W function branches, this gives \( H_{-1}(\sqrt{2}) = 4 \) and \( H_{0}(\sqrt{2}) = 2 \) which are exactly those that you would expect.

@Jay
Why is it 18-periodic? I understand it would be periodic in the imaginary direction just like base-e, but why 18?

Andrew Robbins