Why does this cycle of 44 show up in the Collatz Conjecture

collatz conjecturedynamical systemsmodular arithmeticnumber theorypolar coordinates

Consider this function:

$$f\left(x\right)=\frac{x-b^{\left(\operatorname{floor}\left(\log_{b}x\right)\right)}}{b^{\left(\operatorname{floor}\left(\log_{b}x\right)\ +\ 1\right)}-b^{\left(\operatorname{floor}\left(\log_{b}x\right)\right)}}$$

Here is a StackExchange post with more information about the function. But basically, you could describe the function as:

A relative measure of how far away a number is from the smallest number with the same number of digits.

or

A relative measure of how far a number 'x' is from the biggest power of 'b' that is still less than or equal to 'x'

So anyways, if we set b=6, then apply f(x) to elements of Collatz orbits that have a length > 44, we see a cycle. It's easy to see this when you graph the points (i, f(x)) on a polar graph.

i = the index of the element in the sequence
x = the value at COLLATZ[i] where COLLATZ is an array created from the elements of the Collatz orbit starting at n and terminating at 1.

Here is the graph for n = 27. The points are labeled (i, COLLATZ[i])

graph showing cycle of 44 for n=27

It clearly cycles at 44.

To test this hypothesis further, I tried graphing n = 670617279, which has 949 steps. Here is the result.

graph showing cycle of 44 for n=670617279

I thought it may be simply because I was using polar coordinates, but I see similar symmetry when using cartesian coordinates.

I think I'm on the verge of understanding why, but I just wanted to ask the community for any insight. I find it odd that 44 shows up.

By the way, here is a medium article where I go over the use of this function in more detail as it applies to Collatz.

Best Answer

There are two types of steps involved in a Collatz sequence: One that multiplies a number by $\frac{1}{2}$, and one that approximately multiplies a number by 3. (The “approximately” part is because the “+1” in the step).

If we assume that a Collatz counterexample cycle exists, that means that after m halving steps and n tripling (plus 1) steps, we get back to where we started, i.e.,

$$(\frac{1}{2})^m \times 3^n \approx 1$$

Taking logarithms, we get:

$$n \log 3 - m \log 2 \approx 0$$

or equivalently:

$$\frac{n}{m} \approx \frac{\log 2}{\log 3}$$

The 10 smallest integer solutions to this equation, having $|n \log 3 - m \log 2| < 0.1$, are:

  • m=8, n=5 → m+n=13, error=0.052116
  • m=11, n=7 → m+n=18, error=0.065667
  • m=19, n=12 → m+n=31, error=0.013551
  • m=27, n=17 → m+n=44, error=0.038565
  • m=30, n=19 → m+n=49, error=0.079218
  • m=35, n=22 → m+n=57, error=0.090681
  • m=38, n=24 → m+n=62, error=0.027102
  • m=46, n=29 → m+n=75, error=0.025014
  • m=49, n=31 → m+n=80, error=0.092769
  • m=54, n=34 → m+n=88, error=0.077130

So it happens that one of these near-cycle possibilities has 44 steps, with 27 halvings and 17 triplings.

That still leaves the question: Why ask about 44-cycles as opposed to the smaller 13-, 18-, or 31-cycles? And that's because of your choice of the polar coordinate system, using integer values of $\theta$. It happens that 44 radians is 7.002817 turns, very close to an integer. So the cyclic behavior is most visible there.

By adjusting your coordinate system, you should see similar cyclic behavior for the other $m+n$ values I listed above.

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