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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.

Infinite Planes
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  • It's not clear exactly what you're asking - but continued fractions give a unique representation of every number. You could generalize this to infinite reversed sequences of Collatz numbers, but they're not going to be the same. They'll be close because every reverse sequence begins [1,2,4,8,16], which defines the number pretty closely. – Milo Brandt Jun 18 '19 at 00:59
  • I don't know that the conjecture you're making is very sound, but if you want to quickly perform a sanity check on it, have you tried writing a computer program that just tests this for you for many different hailstone sequences? – Jack Crawford Jun 18 '19 at 01:01
  • I have tried making a program do this, which showed that any other values that can occur are very, very close to the other, often seen, values. The precision of my program isn't the best for how close the values are. – Infinite Planes Jun 18 '19 at 01:05
  • How big was the sample size you tried? And how close were the values? What was the value they seemed to be close to? You are making each successive step of the hailstone sequence contribute less and less to the total value of the fraction (that's how this kind of repeated fraction works) so you should expect that they don't vary too wildly, even for radically varying hailstone sequences. – Jack Crawford Jun 18 '19 at 01:07
  • The value of this fraction is almost entirely determined by just the last few elements in the hailstone sequence. The only way the hailstone sequence can terminate is with $...4,2,1$ (some string of powers of $2$) so your fraction is always going to be just a little bit less than $1.5$, and almost any hailstone sequence's fractions will look identical since almost every hailstone sequence ends in the same few numbers.

    The only thing it tells us if the fraction varies a lot for two hailstone sequences is a rough indication of how many powers of $2$ there were at the end of the sequence.

     – Jack Crawford Jun 18 '19 at 01:18
  • When you say "end up with some number", you mean an integer, because all rational numbers (e.g. $22/7$) are... numbers. – Jean Marie Jun 18 '19 at 11:12

1 Answers1

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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 $n$):

Collatz directed graph

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.

Jack Crawford
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