# Collatz Conjecture

The ultimate convergence to the 4,2,1 loop for any whole number when using the 3x+1 algorithm of the Collatz Conjecture appears to be a consequence of the nature of powers of two. For 2^N-1, half of all of the possible positive integer N results is an odd number that is divisible by 3, thus there is an infinitely large net that can “catch” the algorithm cycle and bring it down to the 4,2,1 loop.

This net idea can be further extended such that, the Collatz Conjecture can be proven to be true if the following is true:

There is a set of odd positive integers A_1 such that 3*A_1 + 1 = 2^M.

Furthermore, there are additional sets of odd positive integers A_(N) satisfying this: 3*A_(N+1) + 1 = A_(N)*2^M, where N and M are any positive integer. So, A_2 is the set of odd positive integers satisfying the relation 3*A_2 + 1 = A_1 * 2^M, and A_3 is the set of odd positive integers satisfying 3*A_3 + 1 = A_2 * 2^M, and so on. Note that, when the sets are refered to in the relations, that is meant to be interpreted as “at least one member of” the referenced set. Such that 3*A_2 + 1 = A_1 * 2^M is to be interpreted as 3*(at least one member of A_2) + 1 = (at least one member of A_1) * 2^M.

The Collatz Conjecture, therefore, is proven true if it is proven true that the sets A_(N) together include all odd positive integers.

So, for A_5, we already have the relation 3*A_5 + 1 must be equal to at least one member of A_4 * 2^M. But, we can tweak the definition a little. We know that all of A_4,A_3,A_2, and A_1 satisfy the requirements of the Collatz Conjecture, so we can define B_5 as having to satisfy the requirement that 3*B_5+1 must be equal to at least one member of (A_4,A_3,A_2, or A_1) x 2^M. Or more generally, we can redefine B_(N+1) as being a list of odd positive integers satisfying:

3*B_(N+1) + 1 = (at least one member of {B_N,…,B_1}) x 2^M, where N and M are any positive integer, and where 3*(a member of B_1)+1 = 2^M.

This definition makes B_(N) a list such that we can say that if we prove that B_(N) as N approaches infinity is equal to the set of all odd positive integers, then we have proven the Collatz Conjecture.

The trick to proving that B_N approaches the set of all odd positive integers as N approaches infinity is to prove that each iteration of N must increase the size of the list such that the number of members of B_N divided by the number of members of the set of all odd positive integers converges to unity with increasing N. This is trifling easy to show. Starting at B_1, which is a list with infinite members, we go to B_2 and find that each of those members of B_1, allows for at least one new member to B_2. When we get to B_3 then, it is easy to show (easy may not be the best descriptor here. All of the multiples of three have no further back-solutions, all non-multiples of three have infinite back-solutions, and none of the non-multiples of three ever has only back-solutions that are multiples of three) that we get a new set of members at least as numerous as was acquired in the first step which was at least as numerous as the set B_1. So we have learned that the length of B_N is greater then N*(length of B_1) for all N. We know that B_1 has infinite members, but we also know that the set of all odd positive integers is infinitely larger than B_1, such that the ratio of the length of B_1 to the goal set is equivalent to 1/infinity or zero. But, for B_N we now have N*(length of B_1)/(length of list of all odd positive integers), and as N approaches infinity, we arrive at unity.

This is not yet a complete proof, for though it proves that the Conjecture is true for the bulk of the target set, it fails to positively disprove that B_N does not have discontinuities. A discontinuous set that satisfies the size relation above would be, for instance, all odd positive integers except 10855. The size of that set divided by the size of the full set is still unity. So this, as yet, falls short of being a solid proof of the Collatz Conjecture.