Peeping a bit into the Szekeres-Seminar somewhere on the forum mentioned by Andrew, and rearranging the elementary Schröder functions found by Schröder himself, I can add some more elementary superfunctions:
1. \( f(x)=2(x+x^2) \), \( F(x)=\frac{e^{2^x}-1}{2} \).
Let us check:
\( F(x+1)=\frac{\left(e^{2^x}\right)^2 -1 }{2} \) and on the other hand
\( 2(F(x)+F(x)^2)=e^{2^x}-1 + \frac{(e^{2^x}-1)^2}{2}=\frac{\left(e^{2^x}\right)^2 -1 }{2} \)
So its indeed a superfunction.
Edit: It is regular at fixed point \( 0 \): \( \lim_{x\to-\infty} \frac{e^{2^{x}}-1}{2} = 0 \)
2. \( f(x)=4(x+x^2) \), \( F(x)=\sinh\left(2^x\right)^2 \)
Let us check:
\( F(x+1)=\sinh\left(22^x\right)^2=4\sinh(2^x)^2\cosh(2^x)^2=4\sinh(2^x)^2(1 + \sinh(2^x)^2)=4(F(x)+F(x)^2) \)
It is again regular at 0.
1. \( f(x)=2(x+x^2) \), \( F(x)=\frac{e^{2^x}-1}{2} \).
Let us check:
\( F(x+1)=\frac{\left(e^{2^x}\right)^2 -1 }{2} \) and on the other hand
\( 2(F(x)+F(x)^2)=e^{2^x}-1 + \frac{(e^{2^x}-1)^2}{2}=\frac{\left(e^{2^x}\right)^2 -1 }{2} \)
So its indeed a superfunction.
Edit: It is regular at fixed point \( 0 \): \( \lim_{x\to-\infty} \frac{e^{2^{x}}-1}{2} = 0 \)
2. \( f(x)=4(x+x^2) \), \( F(x)=\sinh\left(2^x\right)^2 \)
Let us check:
\( F(x+1)=\sinh\left(22^x\right)^2=4\sinh(2^x)^2\cosh(2^x)^2=4\sinh(2^x)^2(1 + \sinh(2^x)^2)=4(F(x)+F(x)^2) \)
It is again regular at 0.