Evidence (probably not proof) that the Collatz conjecture is true

If you're not familiar with the Collatz conjecture, see here: http://en.wikipedia.org/wiki/Collatz_conjecture

The best evidence I can come up with that the conjecture is true has to do with the density of counterexamples in the integers.

Restricting the Collatz map to the odd integers produces an interesting result. Let G(x) be the set of odd numbers which pass to an odd number x without going through any other odd numbers. For instance, G(5) would include 3, since 3(3) + 1 = 10, and 10/2 = 5. G(5) would also include 13, since 3(13) + 1 = 40, 40/2 = 20, 20/2 = 10, and 10/2 = 5.

This set of numbers can be generated in this case by the formula G(5) = {(4^k * 2 * 5 - 1)/3 where k is a non-negative integer}. If we choose k = 0, we get 3. k = 1 produces 13, and k = 2 produces 53.

However, not every number has the same odd inverse generating function. For instance, if we look at the number 7, the numbers that map to it without passing through other odd integers include 9, 37 and 149. G(7) = {(4^k * 4 * 5 - 1)/3 where k is a non-negative integer}. k = 0 produces 9, k = 1 produces 37, and k = 2 produces 149.

The functions are similar, but not identical.

If we look at the number 3 in the same fashion, another interesting result occurs. There are no odd numbers which map to it.

It turns out that the general result for G(x) is determined by the value of x mod 6. If x = 1 mod 6, then G(x) = {(4^k * 4 * x - 1)/3}. If x = 3 mod 6, G(x) is empty. If x = 5 mod 6, then G(x) = {(4^k * 2 * x - 1)/3.

Another interesting result comes from looking at the mod 6 values of G(x). If G(x) is non-empty, then the values of G(x) cycle between 1, 3, and 5 mod 6.

Putting this all together, it implies that for all odd numbers which are not of the form 3 mod 6, the tree which is generated by looking at successive iterations of G has the same shape. What this means is that if we have a number x which is not in the set of values which converge to 1, then the tree generated by iterations of G on x is the same shape as the tree which converges to 1. Since anything that maps to x must also not map to 1, the density of numbers which don't map to 1 should be non-zero over the integers.

However, we can show that if we take a set of numbers A and apply the Collatz function to all of them, as long as there are roughly as many even and odd values, the average of f(A) will be about 3/4 the average of A. Similarly, after two iterations of f, the average of values will be about 9/16 what they were when they started. As long as we can assume that the results of the set after an iteration will be approximately equal numbers of even and odd (which has not been proven, but seems likely) we can show that the average of the set will decrease regularly unless the majority of the elements are less than 4. Thus, the set of numbers which don't converge to 1 must be small in the integers. Otherwise, it would follow that this average would stop decreasing well before all of the elements dropped to a sufficiently low level.

So, there's one argument that counterexamples, if they exist, must be dense. And another that says that if there are counterexamples, they must be non-dense. The only possibility then would be that either the arguments are flawed (totally possible, since they are not rigorous proofs) or there are no counterexamples.