The $3x+1$ problem asks if Collatz function reaches $1$ after a finite amount of iterations for any initial positive integer. If the Collatz function reaches $2^b$ it will drop down to $1$. So the conjecture can be asked accordingly.
Doing a slight modification to the Collatz function we get, $$f(x)=3x+2^y$$ Isolate the $3x$ part on the right side like this $$f(x)-2^y=3x$$
This latter equation shows that we can add a power of $2$ to the left side to reach a multiple of $3$. But when iterating $3x+2^y$, it will never reach a multiple of $3$ nor a power of $2$ since the result is always odd and will never be even. So how do we prove that the result will never be a multiple of three when we add a power of two to the result?
Im trying to understand specifically the $3x+2^y$-map. A computer scientist like to use the floor function when real numbers is not used, the $3x$ expression can also be written as:
Let $x \in\mathbb{N} $, $f:\mathbb{N}\rightarrow \mathbb{N}$:
$$ f(x) = \lfloor\frac{x}{2}\rfloor+\lfloor\frac{x}{2}\rfloor+\lfloor\frac{x}{2}\rfloor $$
Define $f^n$ as the $n$'th iterate of $f$.
Let $x\in\mathbb{N_1}$ and $y\in\mathbb{N_1}$
Question: How do we prove that for all positive initial integers $x$ and initial constant positive integer $y$, the orbit of $f^n(x)=3x+2^y$ will never reach $2^b$ nor a multiple of three where $b\in\mathbb{N}$ ?
Since $f(a) = 3\left\lfloor\frac{a}{2}\right\rfloor$ is divisible by $3$ but not $2^b$, $f(a) \neq 2^b$ for any $a,b \in \mathbb{N}$.