For a natural number $n$, let $c_b(n)$ denote the number of digit transitions in the representation of $n$ in base $b$. By a digit transition, we mean a pair of successive, unequal digits: for instance, the decimal number 114633366 has 4 transitions, given by 14, 46, 63 and 36, hence $c_{10}(114633366) = 4$.

Are there any known results surrounding the quantity $c_2(n) + c_3(n)$ (or, more generally, sums over different bases)? More specifically, I am trying to prove or disprove the following claim:

``For any $C > 0$, there exists a natural number $n$ such that, for any natural $k$ with $n \leq k  \leq 2n$, $c_2(k) + c_3(k) > C$.''

In other words, I want to know whether it is possible to always find, between any number and its double, a number with binary and ternary representations both having at most a fixed number of transitions. 

As a remark, I believe the numbers with few digit transitions in base $b$ can equivalently be described as the numbers expressible as $\sum e_ib^i$ where $i \geq 0$, $e_i \in \{-1, 0, 1\}$, and only few of the $e_i$ are non-zero.

As another remark, for a fixed number $k$ of transitions, when varying the number of digits $n$, the number of numbers with $k$ or fewer transitions is polynomial in $n$ (choose the places where the transitions happen, and the digits). The number of numbers, however, is exponential in $n$, so heuristically, one would expect the sets of numbers with few transitions in base 2 and base 3 respectively to have less and less opportunity to intersect as $n$ gets large; however, I am lacking intuition in the area, so I don't know if there is a way to make this rigorous.

My main field is not number theory, so I would be grateful for any advice of where to look for relevant results, no matter how obvious you think it is!

Update: I have checked the minimal value of $c_2(n) + c_3(n)$ between successive powers of 2. Between $2^i$ and $2^{i + 1}$, it is below 6 for $i \leq 22$. It then stabilises at 7 until $i = 30$ inclusive (except for $i = 25$ where it drops back to 6). It jumps to 9 for $i = 31$.