Explorations of the Collatz Conjecture - Moravian College

2 = 2 k − 2 r 1
1 = 2 k−1 − 2 r 1−1
Which implies that k = 2 and r 1
= 1 which is the case where our two parity
sequences are 〈10〉 and 〈01〉.
Case 2: Suppose that the second in the pair starts with 1. That makes r 1 = a 1 −1
and δ 1 = −1, which gives us
(−1) ( 2 2a1−2 − 2 ) a 1−1
= 2 k − 3
The left side of this equation will always be even, whereas the right side will
always be odd. This is a contradiction.
Therefore, a block or string prefix will only have a single one if it is of length
two and of the form specified above.
□
We can continue our argument above by writing similar proofs that will show
us all of the blocks that contain two ones, all of the blocks that contain three ones,
and so on. As we increase the number of ones that the sequence has, the number
of cases that the proof has grows rapidly, to the point that it is unreasonable to
continue using this method much beyond the case of a block or string prefix with
three ones.
From our block and string data in Tables 4.1 and 4.2, we have found values for
the minimum, maximum and average value of our n k
ratio. We believe that these
values will approximate the actual values for all blocks and strings. Therefore, we
have the following conjecture.
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