Let $x$ be a rational $2$-adic integer (i.e., an eventually repeating sequence of zeros and ones). Define the strictly increasing sequence $n_k$ of non-negative integers by the equation
$$x = -\sum_{k=0}^\infty \dfrac{2^{n_k}}{3^{k+1}}$$
This can be done inductively by defining $2^{n_0}$ to be the largest power of two that divides $x$. Then, if
$$x_m = -\sum_{k=0}^m \dfrac{2^{n_k}}{3^{k+1}}$$
we define $n_{m+1}$ to be the place of the first digit such that $x_m$ and $x$ differ in their $2$-adic expansions. This works because $-3^{-k-1}$ is always odd (first digit is one). Now, let $d_k = n_{k+1} - n_k$ be the sequence of differences.
Question: Is the sequence $d_k$ eventually periodic?
Or, equivalently, define the $2$-adic integer $y$ to have the digit $1$ in each place $n_k$ and $0$ elsewhere (i.e., $y = \sum_{k=0}^\infty 2^{n_k}$). Is $y$ rational?
Example: Take $x = -5$. Then, $x = \cdots 111011_2$, and we have
$$x = -\dfrac{1}{3} - \dfrac{2}{3^2} - \dfrac{2^3}{3^3} - \dfrac{2^4}{3^4} - \dfrac{2^6}{3^5} - \dfrac{2^7}{3^6} - \cdots$$
so that $y = \cdots 011011011011_2$, i.e., $y = -3/7$.
Motivation: If $x$ is odd, multiply by three and add one.
Update: Consider different series expansions of the same number, e.g.,
$$-\dfrac{1}{3} - \dfrac{2}{9} - \sum_{k=0}^\infty \dfrac{2^{2k+5}}{3^{k+3}} = 3 = -\sum_{k=0}^\infty \dfrac{2^{2k}}{3^k}$$
While the second series doesn't have the exact same form (the first power of $3$ is $0$ instead of $1$), they are similar in their tails. Might it be possible to show that, if there exists an expansion, such as on the right, that is rational -- in the sense of above -- then the series on the left (the one of interest) is also rational -- in the sense of above?