Starting with some number, you can generate the hailstone sequence from it. In the case of 3, the (finite) hailstone sequence of it is $[3,10,5,16,8,4,2,1]$. Placing them in a continued fraction like so: $$1+\frac{1}{2+\frac{1}{4+\frac{1}{8+\frac{1}{16+\frac{1}{5+\frac{1}{10+\frac{1}{3}}}}}}}$$ you will end up with some number. Doing this for many values, you mostly end up with a certain number* and sometimes another, very similar number. I find it likely that there are either only two values or a potentially infinite number of values. If there are many values they are very, very close to the existing values. So my question could be currently impossible since the Collatz Conjecture is still unproven. If this is the case, my question is: What are these values? I have created a computer program to test many, many values, approximately the first twelve thousand.
*This whole thing kinda ignores very small hailstone lengths.
Best Answer
Since your fraction weights values towards the end of the sequence far more than values earlier on, the decimal value of your sequence up to a certain level of precision will be decided almost entirely by the last few entries in the hailstone sequence.
But this public domain graphic from Wikipedia shows us all of the ways that a hailstone sequence can end (for relatively small $n$):
So the value of your fraction is almost entirely determined by the last five or six elements in the hailstone sequence, and the last five or six elements in the hailstone sequence are almost always the exact same. The values of your fraction are always going to look very close to one another, even for wildly varying hailstone sequences. The condition you're describing about how the fraction tends to always look quite similar is absolutely normal. You're effectively squishing down all of the information that we actually use to differentiate hailstone sequences (the numbers earlier on in the sequence) and over-emphasising the importance of the final few numbers of the sequence which are almost always the exact same. I don't think this actually tells us anything deeper about the Collatz problem.
That said, no, these sequences don't always end up at the same one or two numbers as you conjectured; two hailstone sequences will always give you a different fraction, here. You're just pushing all of the differentiating information way down into the deep levels of precision so it's hard to tell.