Question

Collatz conjecture! @dan815

if you have an even number, divide by 2
if you have an odd number, multiply by 3 and then add 1.

Conjecture says, all numbers go to 1 eventually.

Answer 1

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Answer 2

1. 6
2. 5

Answer 3

So here's a fun observation, is take all the possible numbers and we can represent them by 3n, 3n+1, and 3n+2 and see where they map to to try to help us out a little. So depending on if n=2k or n=2k+1 we'll shuffle them between these classes, just like evens and odds except 3n, 3n+1, and 3n+2. Here we go:



So here I am showing how all the 3 classes map back onto the 3 classes after doing the rules.

First level I split them into even and odd so there are 6, then if the number is even I divide by 2. These all nicely map back to the 3 classes, but if they're odd they end up as 9n+?? class. However, we can see that 3n, 3n+1, and 3n+2 are all plugged into them which makes up all the numbers so we can simplify that they all end up mapping to 3n+2 which is a great simplification. A graph of the flow of these looks like this:

This directed graph shows that all 3n numbers eventually map to 3n+1 and 3n+2 numbers so we now only have to prove the Collatz Conjecture for numbers of this form. Luckily 3n+1 is a form we get from applying the odd rule once, and there's more I'd like to say, but I don't wanna puke too much out hahahaha.