Finding a function satisfies $\ln F(x+1)=a F(x)$

Question

I want to find a smooth function $F$ satisfies $\ln F(x+1)=a F(x),\ x\in[0,2]$ and $F(0)=1$
I didn't prove the existance of the function but I think it exists.
I can easily get $F(1)=e^a$ and $F(2)=e^{ae^a}$
I found it hard to evaluate it. So I only want to find the coefficients of Taylor series of $F$.
So I let $$F(x)=\sum_{k\geq0}c_kx^k\text{ (suppose it is also smooth in [-2,0])}$$ And I can get$$\ln F(x+1)\\ =\ln(1+\sum_{k\geq1}{\frac{c_k}{k!}(x+1)^k})\\ =\sum_{n\geq1}{\sum_{k\geq1}{(-1)^{n+1}(\sum_{m\geq0}\frac{c_k}{k!}C_k^mx^m)^n}}$$ I want to compare the coefficients of $x$ to get the value of $c_k$ but it looks like impossible for me.
Then, I had a thought: transfer the equation into $$F(x)=\exp(aF(x-1))\text{ or}\\\ln F(x)=aF(x-1)$$ But all of these contain $\ln F(...)$ or $\exp F(...)$
So, I have to evaluate formulas like $(a+bx+cx^2+...)^n$.
I tried using computer to find the coefficients but there are infinite terms of $c_0$. I can't treat them well.
On the other hand, I found it's increment speed is very fast, faster than $e^x$, $x!$, $x^{x^x}$ and many other functions. I don't think it is an easy question. What can I do?
Edit: I have no idea about $F(x),\ x\in[0,1)$, but I think the smoothness of $F$ can make it unique.

Answer 1

This problem is called "tetration". Discussion can be seen on the Wikipedia page on Tetration

A paper about it can be found at http://myweb.astate.edu/wpaulsen/tetration2.pdf