Daedalus

John, in actuality, N goes on to infinity and will never cease. However, I'm hoping to show how all odd numbers are accounted for as N approaches infinity. I've already proven that odd numbers will never produce another odd number that contains factors of three.

From the odd rule we know (even = 3 x odd + 1) such that (odd = (even - 1) / 3). Given that all even numbers are odd numbers that contain factors of 2 or (odd x 2^m) we can show that (odd = (((6 x n - 3) x 2^m) - 1) / 3) will never produce integer / natural number results because after we divide the 3 we get: ((2 x n - 1) x 2^m) - 1/3.

This proves that odd numbers of the form (6 x n - 3) for n >= 1 can only start a hailstone sequence and will never occur in a sequence resulting from previous iterations. So, I've proposed another conjecture that all hailstone sequences start with an odd number that contains a factor of 3.