This is just something from my notes I wanted to put out there for those who might be interested.
I will try to show that q=\frac{d}{\Delta(n)} and q > \frac{2d}{n} for cycle denominator d, 3n+1 stopping sequence starting number n, and system 3n+q in which the trajectory is a fully reduced cycle, q>1. I apologize in advance for all the inline equations.
If you generalize the cycle equation n = \frac{β(v)}{2^l-3^s} to include the difference between n and the final number in the sequence, and call this difference \Delta(n), you get n = \frac{β(v)+\Delta(n)2^l}{2^l-3^s}. Note that the cycle equation is the case where \Delta(n)=0. From here you can see that \Delta(n) < 2^l-3^s for all n < 2^l (we will only consider trajectories such that n < 2^l, i.e. the instance of a trajectory with the smallest n). However, \Delta(n) has another bound when you consider only stopping sequences (sequences such that the final number is the first number to be less than the starting number n). The bound is \Delta(n) < \frac{n}{2}, which comes from the basic geometric fact that if you have a number greater than n and divide it by two to get a number less than n, the amount you overshoot n will be less than half of n.
Now consider divisibility. If you get \frac{β(v)+\Delta(n)2^l}{2^l-3^s} to divide, you know that β(v) and \Delta(n)2^l sum to 0 \pmod{2^l-3^s}, so you can say β(v) \equiv -\Delta(n)2^l \pmod{2^l-3^s}. Now, to get the math to work for 3n+q, all we have to do is turn β(v) into q \cdot β(v). Conventionally q is the smallest multiplier such that the division works, otherwise you could multiply a working q by any factor to get an unreduced cycle.
Now, say \Delta(n) = 5. Then β(v) \equiv -5 \cdot 2^l \pmod{2^l-3^s}. Remember how q is just what you multiply β(v) by to get it to divide? Well in this case just multiply β(v) by \frac{2^l-3^s}{5} and it will divide. So if I’m not mistaken here, you can find q, or at least one valid q which seems to be the smallest and therefore most reduced, by taking the denominator of the cycle equation divided by \Delta(n). So we have q=\frac{d}{\Delta(n)}.
Now to put it all together with another example. Again we are only considering cycle and stopping sequences. Let’s call 2^l-3^s, the denominator, d for short. For n=11, the possible values for \Delta(x) are 0, 1, 2, 3, 4, and 5. This is calling back to \Delta(n) < \frac{n}{2}. Using our new formula q=\frac{d}{\Delta(n)} we can also translate this to say the possible values for q are \frac{d}{0}, \frac{d}{1}, \frac{d}{2}, \frac{d}{3}, \frac{d}{4} and \frac{d}{5}. To fix the divide by 0 we just have to understand that \Delta(n)=0 and \Delta(n)=d both describe cycle vectors, so the real q for \Delta(n)=0 is \frac{d}{d}=1. Aside from this case, the smallest possible q is \frac{d}{5}. And since \Delta(n)=5 was just from our example n=11, and in general the maximum \Delta(n) is no more than \frac{n}{2}, the bound is q > \frac{2d}{n}. An interesting side-note about this example is that the actual \Delta(n) for n=11 is 1, which is the only option besides 0 where d could theoretically divide. Given that d is never a multiple of 2 or 3, I have a hypothesis that a \Delta(n) which is a multiple of 2 or 3 forces β(v) to be coprime to d, but I haven’t looked much into this.
Edited to add: The cycle in 3n+5 with minimum element 19 has vector 11100 (d=5) and this vector in 3n+1 corresponds to n=23, \Delta(n)=3. Therefore we have a case where \Delta(n) is a multiple of 2 or 3. The cycle fraction doesn’t reduce at all, as suspected. This suggests that the relationship q=\frac{d}{\Delta(n)} does not apply to such cases, but rather q=d. This suggests I’m missing the more general relationship. Sorry for not polishing this out before posting. Since I brought in an example for an unreducible 3n+5 cycle, let’s look at a reducible one. The 3n+5 cycle with minimum element 187 has a vector whose n in 3n+1 is 26843583, \Delta(n)=1015513, and d=5077565. We can see that indeed \frac{5077565}{1015513}=5.
If you want you can plug in Ellison’s bound for d to get q > \frac{2 \cdot 1.8^l}{n} when l>27.
Just to clarify what this inequality represents, If you take some vector (aka trajectory) of some length l such that using 3n+1 rules the starging number n reaches a number less than itself, there is a system 3n+q where this vector is a cycle (it doesn’t have to be a stopping sequence for this to be true but it does for the inequality to hold). The inequality concerns these three variables under this circumstance.
So yeah, this is more of a sketch than a proof. I also just wanted to share the ideas I used along the way because I think they’re interesting.