Collatz - restated without the halving.

Here's the way of reformulating Collatz chains that I have been using. This gives us a way to express the conditions for 1) the existence of a loop and 2) the conjecture being true.

Start from n, any odd number

Step 1.
Multiply by 3 and add 1 = 3n+1

 Step 2
Multiply by 3 = 9n+ 3
Factor as (2^a)(an odd number)
Add 2^a = 9n+3+2^a
 
Step 3...
Multiply by 3 = 27n + 9 + 3(2^a)
Factor as (2^b)(an odd number)...
Add 2^b = 27n + 9 + 3(2^a) + 2^b

Repeat until we reach 2^z(n) or 2^z

As the total increases, the cumulative total is

[3^y x n] + [3^(y-1)] + [2^a x 3^(y-2)] + [2^b x 3^(y-3)] …. + [2^c x 3]+ 2^z...

The power of 3 falls by one in each term, until we reach 3^0 in the final term
The power of 2 increases in each term but can rise by more than 1 at a time – so in the example above d>c>b>a.

If the Collatz conjecture is true, we will always reach an equation of the form

[3^y x n] + [3^(y-1)] + [2^a x 3^(y-2)] + [2^b x 3^(y-3)] …. + [2^c x 3]+ 2^d = 2^z

For there to be a 3n+1 loop, which would be a counterexample to the conjecture, we would need this equation to be true:

[3^y x n] + [3^(y-1)] + [2^a x 3^(y-2)] + [2^b x 3^(y-3)] …. + [2^c x 3]+ 2^d = 2^z(n)

We can adapt this method for 5n+1 chains by replacing all the 3s with 5s.