## More thoughts on the Collatz Conjecture

I made a quick housekeeping sweep of the blog recently, and wanted to revisit the Collatz Conjecture. I’m sure it’s been suggested before, but I believe there’s another method of proving the conjecture:

for all where

That is to say, we only need prove that for an n > 2, all integers between consecutive powers of 2 resolve according to the collatz conjecture. I’m not sure where to go from there, but for instance we can show that in general:

where r is the number of (divisible by 3) iterations such that

## Derivation:

Consider where n > 1

First step: must be odd, so multiply by 3 and add 1.

Second step: must be even, so divide by 2.

Third step: must be even, so divide by 2.

We’ve gone through 3 steps. If n = 2, we’re at 4 and we’re done. Let’s say n is very large, so that we can go through another 3 iterations and find a pattern.

Fourth step: is odd again, so multiply by 3 and add 1.

Fifth step: Even again, divide by 2.

Sixth step: Still even, divide by 2.

The next iteration will increment the power of 3, and give us a number divisible by 4, and so on.

More generally, we’ll have:

for k iterations where k is divisible by 3.

Letting we have

When we can simplify:

I’m not sure where we go from here. The next thing I want to do is a 3d plot of iterations to resolve against the position (from 0 to 1) between and for varying n.

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