Coloring F(x) = F(exp(x))

[tommy1729]

Consider a 1-periodic real-analytic function of the real-analytic slog for re > 0.
That satisfies :
F(x) = F(exp(x))

However this functional equation can not hold for all nonreal x because exp is a chaotic map.

So we color the complex plane according to how many times the equation holds, more precisely :

Black x
F(x) =\= F(exp(x))
And F(x) =\= F(exp^[2](x))
Purple x
F(x) =\= F(exp(x))
And F(x) = F(exp^[2](x))
Blue x
F(x) = F(exp(x))
And F(x) =/= F(exp^[2](x))
Red x
F(x) = F(exp(x)) = F(exp^[2](x))

Tommy's 4 color conjecture

F is completely determined by its coloring and F(1).

Not to be confused with the 4 color theorem

Regards

Tommy1729