Explorations of the Collatz Conjecture - Moravian College

Now, if we look at the case where x = 0 then,
d = c − 3m
2 k .
Next, we substitute this for d in Equation 3.2 and obtain the following:
3 n
2 k x + c = 3m
2 k x + 3m
2 k + c − 3m
2 k .
Then, we simplify,
3 n
2 x + c = 3m
k 2 x + c k
3 n
2 x = 3m
k 2 x k
3 n = 3 m .
Therefore, n = m, and so s and s ′ must have an equal number of ones.
□
From our definition of corresponding stems and Theorem 1, we can form an
equation to test whether or not a pair of parity sequences is a set of corresponding
stems.
Corollary 1. Parity sequences 〈v 1 , v 2 , ..., v k 〉 and 〈v ′ 1 , v′ 2 , ..., v′ k
〉 are corresponding
stems if and only if v 1 v ′ 1 and
k∑ (
1 =
(2 i−1 v i
)
∣ 3 − v ′ ) ∣ i ∣∣∣∣∣
.
(v 1+v 2 +...+v i−1 +v ′ i ) 3 (v′ 1 +v′ 2 +...+v′ i−1 +v′ i )
i=1
Proof. Recall, v = 〈v 1 , v 2 , ..., v k 〉 and v ′ = 〈v ′ 1 , v′ 2 , ..., v′ k
〉 are corresponding stems
if and only if T v (x) = T ′ v(x + 1). So,
3 (v k+v k−1 +...+v 2 +v 1 ) ∑k−1
1 ( )
(x) + 3
(v k +v k−1 +...+v i+2 ) v
2 k 2 k−i i+1 =
i=0
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