Interesting value for W, h involving phi,Omega?

[Ivars]

Omega constant =0.5671432904097838729999686622 is defined by :

Omega*(e^Omega)=1 and
ln(Omega)=-Omega

So selfroot of Omega:

Omega^(1/Omega) = e^ln(Omega^(1/Omega) = e^((1/Omega)*(ln(Omega))=e^((1/Omega)*(-Omega)) =e^(-1)=1/e=0,367879441

And:
Omega^Omega = e^ln(Omega^Omega) = e^(Omega*ln (Omega) = e^(-Omega^2)=0,724950783
(Omega^(-Omega))=(1/Omega)^(Omega)=e^ln(Omega^(-Omega))=e^-Omega*ln(Omega)=e^(Omega^2)=1,379403986

Infinite tetration of selfroot of Omega:

h(Omega^(1/Omega))=h(1/e) = -W(-ln(1/e)/(ln(1/e))= W(1)/1=Omega=0,56714329=-ln(Omega),

Square superroot of (Omega^1/Omega) :

ssrt(Omega^(1/Omega) = ln(1/e)/W(ln(1/e))= -1/W(-1)= -1/(-0.318131505204764 + 1.337235701430689*I) = 0.16837688705553+0.707755195958823*I.

W(-1) = 0.318131505204764 + 1.337235701430689*I is one of two conjugate values. Jaydfox :"ith iteration of natural exponentiation of base e has two primary fixed points at 0.318131505204764 +- 1.337235701430689*I." in
Imaginary iterates of ....
Also:
h((1/Omega)^Omega))= h( e^(Omega^2)) = -W(-Omega^2)/(Omega)^2 = Omega/Omega^2= 1/Omega=-1/ln(Omega)

Generally, h( e^(Omega^n)= 1/(Omega^(n-1)) ; n>1

h((Omega^2)^(1/Omega^2))= Omega^2=0,321651512
h(Omega^3)^(1/Omega^3))= Omega^3=0,182422497 etc .Generally:

h(Omega^n)^(1/Omega^n))= Omega^n if n>=1.

We can compare this to h(i^(1/i)) = h((1/i)^i))= - i = i^3=1/i.