Dmitrii Kouznetsov's Tetration Extension

[Kouznetsov]

No, that does not. The analytic tetrations should coincide. Does your soluiton show period
\( \frac{2\pi \rm i}{ \ln\big(\ln(\sqrt{2})\Big)+\ln(2)}\approx 17.1431 \!~\rm i \)?

Hm, not exactly, what I derived some posts earlier would give merely \( \frac{2\pi i}{\ln(\ln(\sqrt{2}))} \), yours is \( \frac{2\pi i}{\ln(\ln(2))} \).

Bo, namely for \( b=\sqrt{2} \) my expression can be simplified to yours.

Btw. there are infinitely many analytic functions that have no singularity on right halfplane: If \( F \) has no singularity there, also \( F(x+c\sin(2\pi x)) \), \( 0<c<\frac{1}{2\pi} \), is a superexponential with no singularities on the right halfplane and any 1-periodic function \( p \) with \( p(x)>-x \) in \( 0<x\le 1 \) would do.

I disagree. At some smooth contour \( u \) in your strip, the contour \( v=u+c \sin(2\pi u) \) enters the left halfplane and crosses the cut \( v<-2 \). You have no need to evaluate any tetration in order to see it. Corresponding contour \( F(v) \) is not continuous.

Fig.2 is ready at
http://en.citizendium.org/wiki/TetrationAsymptoticParameters00. Try to run it.