Cauchy integral also for b< e^(1/e)?

[Kouznetsov]

why not apply a similar Cauchy-integral technique for bases \( b<e^{1/e} \)?
...

Because, in the case \( b>e^{1/e} \), the Cauchi integral seems to be the only robust way to evaluate the superexponential; but for \( b<e^{1/e} \), the direct expansion is much faster, giving the same 14 decimal digits.
If you mention, even for b=e, in the final implementation, I use the truncated series instead of the numerical approximation of the integrals. It is at least 2 orders of magnitude faster. There is difference, to get the figure within few seconds or tomorrow, ah?

Now I feel that I can evaluate any tetration with 14 digits, and I work making the evaluation fast; and the algorithm - simple and portable (the same pieces of code run with C++, Mathematica and Maple).
If you implement the integral formulas, we may compare the results with my expansion. I expect, of order of 14 digits will coincide.

In the recursion formula, I would specify the base, to avoid confusions:
\(
f(z)=\frac{1}{2\pi i}\oint \frac{f(w)}{w-z} \mathrm{d}w=\frac{1}{2\pi i}\int_{-iA}^{iA}\Big(-\frac{\log_b(f(w))}{w-1-z} + \frac{\exp_b(f(w))}{w+1-z}\Big) \mathrm{d}w + \frac{1}{2\pi i}\int_{-1}^1 \Big(-\frac{f(w)}{w+iA-z} + \frac{f(w)}{w-iA-z} \Big) \mathrm{d}w
\)
If you want your implementation to converge, in the last integral, \( f(w) \) should be replaced to \( \log_b(w+1) \) at the left hand tip of the interval of integration, and to the \( \exp_b(w-1) \) at the right hand tip, in order to avoid the evaluation of the function in a point close to the contour of integration.
Be careful with the order of updates of the estimates of the values of the function:
the speed of convergence (if at all) may depend on this order.