Explorations of the Collatz Conjecture - Moravian College

Then
T n (m + 1) = 3n
2 (m + 1) + 1 ∑n−2
( ) n−i 1 n 2 + 3 n−i−1
2
i=0
⎛
= 3 ⎜⎝ 1 ( ) n 3
m + 1 3 2 2 + 1 ∑n−2
( ) n−i 3
3 2
⎞⎟⎠ − 2 ∑n−2
( ) n−i 3
+ 1 ( ) n ( ) n 3 3
+ − 1
3 2 3 2 2
= 3 (T n (m)) − 2 3
= 3 (T n (m)) + 2.
i=1
∑n−2
( ) n−i 3
+ 4 ( ) n 3
− 1
2 3 2
i=1
i=1
Then,
T n (m + 1) = 3 (T n (m)) + 2 ≡ 1 mod 2.
Therefore, P n+1 (m + 1) = 〈1 11...1〉 which concludes our proof. □
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n 1’s
Theorem 2. Given an integer m > 0, if P i+3 (m) = 〈0 11...1 01〉, then P i+3 (m + 1) =
〈1 11...1 00〉 and T i+3 (m) = T i+3 (m + 1).
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i 1’s
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i 1’s
Proof. Suppose P i+3 (m) = 〈0 11...1 01〉. Then, T i+1 (m) ≡ 0 mod 2 and T i+2 (m) ≡
1 mod 2. So, we have
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i 1’s
T i+2 (m) = T i+1 (m)
2
T i+3 (m) = 3T i+1 (m)
4
≡ 1 mod 2
+ 1 2 .
Then from the previous lemma, P i+1 (m + 1) = 〈11...1〉 and, T i+1 (m + 1) =
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i+1 1’s
3T i+1 (m) + 2. Now, since T i+1 (m) ≡ 0 mod 2, T i+1 (m + 1) ≡ 0 mod 2. Therefore,
P i+2 (m + 1) = 〈1 11...1 0〉.
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i 1’s
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