Proof attempt
So u/IntelligentTwo2175 had a post that seemed really interesting to me. Basically numbers of the form 2^(n) -1 and 2^(n+1) -1 merge for n >= 0. I think we can make it a little stronger by changing it to
m * 2^(n) -1 and m * 2^(n+1) -1 for odd m. If m was even, we can just pull out the factors of 2 into the 2^(n)
So quick proof.
m * 2^(n) -1
is a steiner circuit, so let's jump ahead to
m * 3^(n) - 1
which we know must be even, except let's back up an additional even step, so we know we are at a multiple of 4 of some number x.
2 * (m * 3^(n) - 1 ) = 4 x
Now let's follow
m * 2^(n+1) -1 for only n odd steps
m * 2 * 3^(n) -1
We can easily see this is equal to 4x + 1.
And we know x and 4x + 1 numbers must merge after O, and OEE respectively.
Recursively, we can say m * 2^(n) must be connected on the tree for a specific m and all n.
Now could this be used in a proof of collatz? Let's try.
So all even numbers can divide by 2 until an odd number.
All odd numbers can be written as the even number 1 above it - 1.
By repeatedly alternating these two steps, we create a monotonically decreasing sequence. And thus, must go to 1.
Ex. Odd number 29, write it as the even number above - 1, so 30 -1. Pull out all factors of 2
15 * 2^(1) - 1.
Using the property above, we know this must connect to
15 * 2^(0) - 1
Once we are out of factors of 2, it becomes odd - 1 which must be even. So we can divide those factors of 2 via standard collatz rules until odd again.
15 - 1 = 14
14 / 2 = 7
Once again, write it as the even above - 1
7 = 8 - 1 = 2^(3) - 1
Which using the property goes
2^(3) - 1
2^(2) - 1
2^(1) - 1 = 1
So can anyone check my logic? It's way too late and this seems way too simple for what I have lost too many nights of sleep over.