Consider
$$n \rightarrow ... \rightarrow an+b$$
to be a sequence of natural numbers to which we apply $3n+1$ or $\frac{n}{2}$ operations.
This sequence is called a cycle, if and only if $an+b=2n$.
(one can define it as $an+b=\frac{n-1}{3}$ as well, but it is the same thing, just shifted to other place)
If we use operations above and rewrite the necessary condition of a cycle, we get its general form
$$ \frac{3^k}{2^l}n+\frac{3^{k-1}}{2^{l_1}}+\frac{3^{k-2}}{2^{l_2}}+...+\frac{1}{2^{l_k}}=2n, $$ where $l \geq l_1 \geq l_2 \geq ... \geq l_k \geq 0$ are natural numbers, including $k$.
Let's define $m=2 \cdot 2^l$ and $m_k=2 \cdot 2^{l_k}$, so we get the final form:
$$ \frac{3^k}{2^m}n+\frac{3^{k-1}}{2^{m_1}}+\frac{3^{k-2}}{2^{m_2}}+...+\frac{1}{2^{m_k}}=n $$
and $\frac{3^k}{2^m} \in (0; 1)$.
$\mathbf Question:$ Is there any progress on solving such exponential diophantine equations?
From the first look it feels like a solid approach to the cycle problem, but perhaps it's way more difficult complication than it seems.
By intuition, can you think of a reason why only $n=1;2$ would be the solutions to this problem? Why $n=3$ has no solution?
$\mathbf Edit:$ To make things more clear, $n \rightarrow ... \rightarrow an+b$ sequence precisely follows the Collatz Conjecture and its procedure (when and which operation to apply). Also the question is mainly aimed at the constructed diophantine equation and its behaviour.