An intuitive explanation for a lack of cycles in the Collatz conjecture

2 points

3 years ago

As you traverse from one number (n0) to another (n1), n1 is either n0/2 or (3 n0 + 1). For a cycle to exist, the product of all of the increases must equal the decreases — you need to wind up exactly where you started.

For example, the trivial cycle 1 => 4 => 2 => 1 has one increase of (3 + 1) = 4, and two decreases of 2, also equaling 4.

For the alternative rule where odd numbers go up by 3n + 5, there are some cycles, like 19 => 62 => 31 => 98 => 49 => 152 => 76 => 38 => 19. The increases here are approximately 3.26, 3.16, and 3.10, the product of which is exactly 32, or 2^5.

As n gets larger, (3n + 1) approaches 3n. n=10,001 to 30,004 is an increase of 3.0001, for example. So as n gets larger, the product of the increases is near a power of 3, and that has to exactly equal a power of 2.

For small cycles with just a few odd numbers, this number is quite close to a power of 3. 3.0001^4, for example, is 81.01, a tiny bit larger than 81. Clearly, we need a big cycle with a lot of odd numbers so that the little error gets magnified. However, as the power of 3 gets larger, the next nearest power of 2 remains far away, as explained here. https://mathoverflow.net/a/116960

So, as you try ever larger values of n, you need ever larger cycle lengths to get the increases to produce a power of 2, but that power of 2 gets ever farther away. This is why all the cycles you can find for various rules tend to be of smaller n.

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