I will make this simple, but hopefully it wont be too vague.
Let $x,n,q\in\mathbb{N}$.
I have recurrence relation $$x_{\alpha+1} = 2 (g(x_\alpha))$$
where $$g(n)=n - \frac{n\cdot q}{4}$$
and
well actually I know for a fact that $g(x_\alpha)$ is less than or equal to $x_\alpha$ always. So when $g(x_\alpha)$ reaches $2^p$, iterating with the factor $2$ for $2(g(x_\alpha))$ will mean that $p$ increases in $2^p$, thus giving a sequence of $\{2^p, 2^{p+1}, 2^{p+2},...\}$. Sorry for being vague on what $q$ is, I could not write it properly but $q$ perturbates the numerator, but is never larger than $g(n)$.
The problem is the factor $2$, but I hope to get some insight into how it can converge towards $2^p$. The $g$-function is behaving chaotically, but i know for a fact that $g(x_\alpha)$ is smaller than $x_{\alpha}$. I can proof that.
[Edit: and also $n\cdot q$ is always less than $n$.]
I want to proof that the orbit of $x$ reaches a power of two number that doubles on each iteration. How does one do that?
Is it called fixedpoint or periodic if $x_{\alpha+1} \rightarrow$ $2^p$? where $p$ increase?
I wanted to make a simplification because Im not formal maths writer.
If I shall be a bit more precise using my own words:
The factor $2$ is outside the parenthesis is a fixed number. But the expression inside is reducing if the factor is left out. Only reason its jumping up and down is because the $2$ left-shifts the binary expansion. but subtraction is happening and some borrow and carry techniques should make the dynamical system cancel itself out such that pow$2$ remains.
The question is simply how to proof the orbit reaches $2^p$ where $p$ can be any number at some point in time?