Explorations of the Collatz Conjecture - Moravian College

and
∣ ∑k−1
1 =
1 ( 3
(v ′
∣ 2 k−i k +v′ k−1 +...+v′ i+2 ) v ′ i+1 − ) 3(v k+v k−1 +...+v i+2 ) 3 (v′ k +t k−1+...+v ′ 2 +t 1) ∣∣∣∣∣∣
v i+1 + .
2 k
i=0
Proof. Recall, v = 〈v 1 , v 2 , ..., v k 〉 and v ′ = 〈v ′ 1 , v′ 2 , ..., v′ k
〉 are block prefixes if and
only if T v (x) + 1 = T v ′(x + 1). From Equation 3.1, v and v ′ are block prefixes if
and only if
⎛
⎜⎝ 3(v k+v k−1 +...+v 2 +v 1 )
2 k (x) +
3 (v′ k +v′ k−1 +...+v′ 2 +v′ 1 )
∑k−1
i=0
2 k (x + 1) +
1
2 k−i (
3
(v k +v k−1 +...+v i+2 ) v i+1
) ⎞ ⎟⎠ + 1 =
∑k−2
i=0
1 ( )
3
(v ′
2 k−i k +v′ k−1 +...+v′ i+2 ) v ′ i+1 .
Now, assume v and v ′ are block prefixes. Then, from Theorem 6, we know
that v and v ′ must have the same number of ones so we can simplify our equation
from above to the following:
1 +
∑k−1
i=0
So,
1 =
1 ( )
3
(v k +v k−1 +...+v i+2 ) 3 (v′ k +v′ k−1 +...+v′ 2 +v′ 1 ) ∑k−1
v
2 k−i i+1 = +
2 k
∑k−1
i=0
i=0
1 ( )
3
(v ′
2 k−i k +v′ k−1 +...+v′ i+2 ) v ′ i+1 .
1
2 k−i (
3
(v ′ k +v′ k−1 +...+v′ i+2 ) v ′ i+1 − 3(v k+v k−1 +...+v i+2 ) v i+1
)
+
3 (v′ k +v′ k−1 +...+v′ 2 +v′ 1 )
2 k .
On the other hand, if v and v ′ have an equal number of 1’s and the equality in
our corollary holds, then do the steps above backwards and we have our result.
Similarly, v and v ′ are string prefixes if and only if:
−1 =
∑k−1
i=0
1
2 k−i (
3
(v ′ k +v′ k−1 +...+v′ i+2 ) v ′ i+1 − 3(v k+v k−1 +...+v i+2 ) v i+1
)
+
3 (v′ k +v′ k−1 +...+v′ 2 +v′ 1 )
2 k .
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