Another way to maximize/bound the powers of 2 and 3 sum (\beta(v_h)) is to divide by the highest possible exponent of 2 at each step (this is how the high-cycle parity vector is build, 11011011010…). I think that’s what you did more or less.
So the recurrence relation would be \beta_{i+1}=3\beta_i+2^{\lfloor i\log_2(3)\rfloor}, with \beta_0=0 and at each step we take the highest power of 2 or 2^{\lfloor i\log_2(3)\rfloor}<3^i.
From there you have the same bound: \beta_{x}<x3^{x-1}
The issue is that this recurrence is based on the assumption (not proven even for integers) that the largest possible exponent of 2, to keep a_i>a_0, is 2^{\lfloor i\log_2(3)\rfloor}.
With your rationals at b=1, this is not the case. e.g. at step 8, you have x=5 and k= 8 whilst still having
\frac{539966123744915182447}{37313979917667420061}>\frac{519869004464865760111}{37313979917667420061}
So the parity vector is 11011010… and not the high-cycle 11011011…
Note that for integers, you also can have x=5 and k= 8 whilst still having 8>7 (but here the trajectory went to 5<7 before reaching 8).
So this is probably not related to the rational nature of the terms, but more to the assumption made on the maximum power of 2 that keeps a_i>a_0.
Anyway, if you assume integer terms (and nothing about the maximum power of 2 dividing each term) by using the matchexchange formula, you can trust the bound \beta_{x}<x3^{x-1}. From there i wonder if we can deduce that the maximum power of 2 to keep a_i>a_0 when dealing with integers, is effectively 2^{\lfloor i\log_2(3)\rfloor}. In which case I think you also prove k= \lceil x \log_2 3 \rceil