Explorations of the Collatz Conjecture - Moravian College

as stems.
s i = 0 11...1 01 or s ′ i = 1 11...1 00 with i ≥ 0.
}{{}
i 1’s
}{{}
i 1’s
Garner conjectured that s i and s ′ i
were the only possible form for corresponding
stems. In addition, Garner conjectured that any consecutive integers must
contain stems. Since we have not found any evidence that this is not the case, we
formally state this conjecture below.
Conjecture 1. (Garner) Any pair of consecutive integers with equal total stopping
time contain stems of the form s i and s ′ i
in their parity vectors.
In order to explore the properties of consecutive integers, we must first consider
what happens to any integer when it goes through a given sequence of odd
and even pieces of our function regardless of the integer’s parity.
We define
T 0 (x) = x 2 and T 1(x) = 3x+1
2 . Given the partial parity vector v = 〈v 1, v 2 , ..., v k 〉,
the result of applying v to x, denoted as T v (x), is the result of the composition
T vk (T vk−1 (...(T v1 (x))...)). For example,
T 〈1,0,0〉 (13) = T 0 (T 0 (T 1 (13))) =
We generalize Garner’s stems as follows:
3(13)+1
2
2
2
= 5.
Definition 1. A pair of parity sequences s and s ′ of length k are corresponding
stems if for any integer x, T s (x) = T s ′(x + 1) and for any initial subsequences v
and v ′ of s and s ′ of equal length, |T v (x) − T v ′(x + 1)| 1 and T v (x) T v ′(x + 1).
Note that the second half of our definition ensures that our corresponding
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