6.9.16

Trapped In Finite Fields




I think I know why the algorithm behind the Collatz conjecture always reaches 1 instead of reaching some infinite cycle for all examples tried so far and I think that it will always reach 1 first.

I think the Collatz Conjecture is bridging the gap between continuous and discrete Mathematics.

Unfortunately, what I think is irrelevant if I can't state it correctly or prove it. I will attempt to do so because I'm stubborn. I already attempted to do so in my first entry on the Collatz conjecture but looking at it now it looks like a lot of rambling. So I give it another, hopefully less rambly, try.

Any integer s is either even or odd. If is even, then it will be divided by 2 by the Collatz algorithm until it becomes odd and then it will peak as it gets multiplied by 3 and added to 1 and so on.

I conjecture that at some point during the peaking, the resulting odd integer, which I will call k, is such that 3*k+1 = (k*((n+1)/2) mod n) for some odd integer n.

At this point the trajectory of s under the Collatz map collides with or it gets trapped into a portion of the ((n+1)/2)th row of the exponentiation table mod n starting at 3*k+1.

The ((n+1)/2)th row of the exponentiation table modulo n for some odd integer n is the 2nd row of that same exponentiation table in reverse order.

In other words, the set of all consecutive powers of ((n+1)/2) mod n is the set of all consecutive powers of 2 in reverse order.

This means that the cardinality of the two sets is equal, and it is equal to the order of 2 mod n. Furthermore, the last few elements of the set of all consecutive powers of ((n+1)/2) mod n are powers of 2 mod n

If an element r of the set of all consecutive powers of ((n+1)/2) mod n is even then the next element in the set is r/2.

So if r is even, some x trajectory under the Collatz map will overlap with the portion of the set of all consecutive powers of ((n+1)/2) mod n  starting at r and if k is the first odd integer on this trajectory such that 3*k+1 = (k*((n+1)/2) mod n) then the two trajectories become identical starting at r, and they reach a stopping point 1 after hitting powers of 2.

The reason why it will always reach this particular stopping point is because even if s starts out as a multiple of 2 (which is the only time when the ((n+1)/2)th row of the exponentiation table modulo n will reach an infinite cycle as opposed to powers of 2), s will be divided by 2 by the Collatz algorithm until it reaches an odd integer, which exposes the trajectory for a lack of a better way to describe this.