Base 'Enigma' iterative exponential, tetrational and pentational

[sheldonison]

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Let \( \nu \) 'enigma' (Greek αίνιγμα, aínigma) be the base at which the tetrational has a parabolic fixed point, as shown above. This is analogous to \( \eta = e^{1/e} \), the base at which the exponential has a parabolic fixed point.

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Are these all 'coincidencies' or is there some good reason for the minute differences between these values? Any reason for it seems rather obscure, it's like a puzzle, a mystery to figure out. That is why I call that number, the 'enigma constant'.

Perhaps one can use base \( \nu \), instead of base \( \eta \) or \( e \) to evaluate non-integer iterations of exponential. There's no real fixed point of base \( \nu \) exponential so the half-iterate and such should be real-analytic on the entire domain


\( f_1(x) = \nu^x, \ f_2(x) = \exp_{\nu}^{\circ 1/2}(x), \ g(x) = \textrm{sexp}_\nu(x) \)


\( h(x) = \textrm{pent}_\nu(x) \)

Cherrina, Would you use the sexp(z) for the tetra-euler base, or the pent(z) for the tetra-euler base? b~1.6353245. In the pentation.gpp code, the genpen function uses the lower repelling fixed point (-1.6409), not the upper parabolic fixed point which is ~3.0885323. The parabolic fixed point would generate a similar but different pentation function for this base. Also, presumably one might be able to generate another pentation function going to infinity, from the parabolic fixed point. I haven't generated numerical approximations for either of those functions.
- Sheldon