Question

Trying to prove the Collatz Conjecture... https://www.math.hmc.edu/funfacts/ffiles/10008.5.shtml

Answer 1

So far it looks like the numbers that go to 1 are all powers of 2, and trying to find more numbers that will easily go to 1, I set up this equation for when n is odd: \[\Large 2^{2k}=3n+1\] This can be rearranged to this: \[\Large \frac{4^k-1}{4-1}=n\] which is sort of the trick to why I chose the 2k so that it's a geometric series: \[\Large n=1+4+4^2+4^3+ \cdots + 4^k\] And since this number is always odd, I don't have to worry about it "breaking" the requirement that it's odd. Here's as far as I've got, now I'm jut playing around to find out more to try to prove this.

Answer 2

Looking at this in base 4 we can see that all of the numbers of these forms will go to 1

The first row represents powers of 4 and the second row represents 2 times powers of 4 which over all give all the powers of 2\[\Large 1, 10, 100, 1000, ... \\ \Large 2, 20, 200, 2000, ...\]
Then from the geometric series we can also write these: \[\Large 1, 11, 111, 1111,... \\ \Large 2, 22, 222, 2222,...\] Since the second row is just the first row multiplied by 2, we can easily see that in one step it will become the geometric series represented by the first row. Hmm...

Answer 3

Hmm! Looking at it now.

Answer 4

Interesting. Are we really trying to prove something that has never been proved yet? :O

Answer 5

Hahaha yes. I mean not that we are going to prove it exactly but we might learn something fascinating in trying to prove it.

Answer 6

Yeah..! Agreed. Go on Sir.. am following you :D :D

Answer 7

No that's it, that's all I figured out in trying to prove this unprovable thing like it doesn't lead to the answer hahaha

Answer 8

Haha! :D I am in a mood to work on something interesting. Wanna do some thermodynamics? :D

Answer 9

Ahahaha alright sure let's do that.