Taylor series of upx function

[Kouznetsov]

..
I would like to know if a Taylor series development exists for the ultra exponential function. In my mind (correct me if I'm wrong), upx function as defined as the following:
- upx(1) = e
- upx(x+1) = exp ( upx(x) )
..

Dear Zagreus. If you are still interested...
I have expressed the holomorphic function which satisfies the equation you wrote through the Cauchi integral. It allows the straigihtforward differentiation. I claim that such a function is unique, if we insist that
\( \mathrm{upx}(z^*)= \mathrm{upx}(z)^* \)
\( \mathrm{upx'}(x)>0 \forall x>-2 \)
upx is holonorphic in the complex plane except real values smaller or equal 2.
I have evaluated several coefficients in the Tailor expansion:
\( \mathrm{upx}(z)=\sum_{n=0}^\infty c_n z^n
~\forall z\in \mathbb{C}: |z|<2 \)
Here is the table of the coefficients \( c_0, c_1,c_1,.. \):
1.
, 1.091767351258322138
, 0.271483212901696469
, 0.212453248176258214
, 0.069540376139988952
, 0.044291952090474256
, 0.014736742096390039
, 0.008668781817225539
, 0.002796479398385586
, 0.001610631290584341
, 0.000489927231484419
, 0.000288181071154065
, 0.000080094612538551
, 0.000050291141793809
, 0.000012183790344901
, 0.000008665533667382
, 0.000001687782319318
, 0.000001493253248573
, 0.000000198760764204
, 0.000000260867356004
, 0.000000014709954143
, 0.000000046834497327
,-0.000000001549241666
, 0.000000008741510781
,-0.000000001125787310
, 0.000000001707959267
,-0.000000000377858315
, 0.000000000349577877
,-0.000000000105377012
, 0.000000000074590971
,-0.000000000027175982
, 0.000000000016460766
,-0.000000000006741873
, 0.000000000003725329
,-0.000000000001639087
, 0.000000000000858364
,-0.000000000000394374
, 0.000000000000200252
,-0.000000000000094420
, 0.000000000000047121

The partial sum gives the polunomial approximation;
while modulus of argument is of order of unity or smaller, it gives of order of 14 correct decimal digits. I plot the partial sum
\( F(z)=\sum_{n=0}^{25} c_n z^n \) below:
   
In the upper right corner, in the complex \( z \) plane, the function \( f= F(z) \) is shown with lines of constant real and lines of constant imaginary part.
Levels \( \Re(f)=-2,-1,0,1,2,3,4 \) are shown with thick black curves.
Levels \( \Re(f)=-1.8,-1.6,-1.4,-1.2,-0.8,-0.6,-0.4,-0.2 \) are shown with thin red curves.
Levels \( \Re(f)=1.8,1.6,1.4,1.2,0.8,0.6,0.4,0.2 \) are shown with thin thin blue curves.
Levels \( \Im(f)=-2,-1 \) are shown with thick red curves.
Levels \( \Im(f)=1,2 \) are shown with thick blue curves.
Levels \( \Im(f)=\pm \pi,\pm 3\pi \) are shown with thick pink curves.
Levels \( \Im(f)=\pm 1.8, \pm 1.6, \pm 1.4, \pm 1.2, \pm 0.8, \pm 0.6, \pm 0.4,\pm 0.2 \) are shown with thin green curves.

In the upper left corner, I plot \( f=\log(F(z)) \) in the same notations. Singularities of log, id est, zeros of functions \( F \), appear as small dence loops.
You may interpret the left graphic as contour plot of constant \( |F(z)| \) and
constant arg\( (F(z)) \). Due to the second of your equations, in the central part, the left upper plot is just displacement od the upper right plot. If you print the pics at the transparencies you can see that the central parts overlap well. For those who have no overhead projector, I show the overlap of the two pics above in the botton left picture. For comparison, at the bottom right I show the central part of plot of tetration from my paper, which is expected to appear in Mathematics of Computation. Tetration is shown with lines of constant modulus and lines of constant phase. (If you displace the x-axis for unity, they become lines of constant real and constant imaginaty part)
You can copypast the coefficients imto your program and plot similar pics by yourself. Enjoy!

P.S. It is amaising how the polynomial tries to reproduce the behavior of lines \( \Im(f)=\pm 1 \) in vicinity of \( z=2 \), a little bit outside of the range of convergence of the series. If we take more terms, this detail disappears, although the lines incide the circle become more perfect.