Explorations of the Collatz Conjecture - Moravian College

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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 . 30
Therefore, v and v ′ are block or string prefixes if and only if: ∣ ∑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 +v′ 1 ) v i+1 + . 2 k i=0 □ 4.2.3 Switching 0’s and 1’s Since we discovered earlier that corresponding blocks and strings are just permutations of each other, we began to look at what would happen if we took a parity sequence and moved the ones around to form a new permutation of the original. If we take a single 1 and have it switch positions with an adjacent 0, we have the following result. Lemma 5. Given a parity sequence v = 〈v 1 , v 2 , ..., v k−1 , v k 〉 such that v a = 1 and v a+1 = 0 for some a < k, if we form a new parity sequence v ′ by switching v a and v a+1 , then T v ′(x) − T v (x) = ∆ where ∆ = 3v a+2+v a+3 +...+v k 2 k−a+1 . Proof. Given v as stated above, we will use v ′ = 〈v 1 , v 2 , ...v a−1 , v a+1 , v a , v a+2 , ..., v k−1 , v k 〉 and we will let n be the number of ones in v and v ′ . From Equation 3.1 and Theorem 6 we know that when v is applied to a number x ∑ T v (x) = 3n k−1 2 x + k i=0 31 3 (v k+v k−1 +...+v i+2 ) 2 k−i v i+1 .
Page 1 and 2: Explorations of the Collatz Conject
Page 3 and 4: Dedication This project is dedicate
Page 5 and 6: Preface It all began as a research
Page 7 and 8: 6 Future Work 45 A Branch Count Pro
Page 9 and 10: search. Most of my work, however is
Page 11 and 12: strings to be pairs of parity seque
Page 13 and 14: Figure 2.1: 3x + 1 Tree integer. On
Page 15 and 16: Table 2.1: Total Stopping Times σ
Page 17 and 18: as stems. s i = 0 11...1 01 or s
Page 19 and 20: Now, if we look at the case where x
Page 21 and 22: 3.1 Applying Garner’s Stems In th
Page 23 and 24: Then T 0 (T i+1 (m + 1)) = 3T i+1 (
Page 25 and 26: Chapter 4 Blocks and Strings In the
Page 27 and 28: is an integer if and only if P k (y
Page 29 and 30: Definition 4. A string is a pair of
Page 31: 4.2 Block and String Structure Sinc
Page 35 and 36: Then, v ′ = 〈v 1 , v 2 , ...v a
Page 37 and 38: We can use Lemma 7, Equation 3.1, T
Page 39 and 40: Since, 2 k ∆ = 2 k − 3 n , 3 k
Page 41 and 42: 2 = 2 k − 2 r 1 1 = 2 k−1 − 2
Page 43 and 44: Chapter 5 Long Runs of Consecutive
Page 45 and 46: can be appended to the sequence in
Page 47 and 48: Chapter 6 Future Work Throughout th
Page 49 and 50: Appendix A Branch Count Procedure b
Page 51: Bibliography [1] David Applegate an

Explorations of the Collatz Conjecture - Moravian College

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