Let's say I have this function $$f(x) = \sum_{i=0}^{x-1}f(i)$$ provided $f(0) = 0, f(1) = 1$ and $x \in \mathbb Z$. This function is evidently one-to-one on $[3, \infty) $. Is there an inverse to this and if yes, how to find it? Thanks in advance.
PS. If you see something wrong please write it in the comments. I will take action ASAP.
Use the recursion to compute the values of $f(x)$ for a few small values of $x$:
$f(2) = 1$, $f(3) = 2$, $f(4) = 4$, $f(5) = 8$, $f(6) = 16$, etc.
Do you see a pattern? (Think powers of $2$) This will give you a formula for $f(x)$ valid for $x \ge 2$.
As you said, the function is one-to-one on $[3,\infty)$, so it must have an inverse on $[3,\infty)$.
To invert this function, simply set $x = f(y)$ and solve for $y$.