A  Proof of Collatz Conjecture

HUANG Qingxue

doi:10.20944/preprints202305.2161.v2

Submitted:

21 September 2023

Posted:

22 September 2023

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Abstract

In 1937,German mathematician L.Collatz proposed the following conjeture:for any positive integer,if it is even,divide it by 2,if it is odd,multiply it by 3 and add 1 to get an even number.Continuing with the above rule, the final result will be 1.This paper gives a proof of this conjecture.

Keywords:

number theory

;

Collatz conjecture

Subject:

Computer Science and Mathematics - Algebra and Number Theory

§1. Introduction

In 1937, German mathematician L. Collatz proposed the following conjecture that for a definite positive integer p, if p is even, divide it by 2,if p is odd, multiply it by 3 and add 1 to get an even number.Continuing with the above rule, the final result will be 1.

Collatz conjecture has been studied by many people and has long been regarded as an unsolved problem.See [1,2,3]. This paper gives a proof of this conjecture.

§2. Preliminaries

Definition 2.1: Starting from a positive integer p, the process with the rule in introduction is called a Collatz sequence.p is said to be successful if 1 is finally obtained.Otherwise,p is not successful.

Remark: Any positive integer p can be written as that p = 2^k(2m-1),where m is a positive integer and k is a positive integer or 0. When k > 0, p is even,and when k = 0, p=2m-1 is odd. If any odd number is successful, then since the even number p = 2^k(2m-1) is divided by 2 k times to get the odd number 2m-1, the even number p = 2^k(2m-1) is also successful. In other words, to prove that Collatz conjecture holds, it is sufficient to show that any odd number is successful.

In this paper, the following discussion focuses on the Collatz sequence for odd numbers, and the even numbers in the Collatz sequence are omitted.

Example: For a positive integer p = 44, its Collatz sequence is that 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1，4，2,1，.... By removing all even numbers, it becomes that 11, 17, 13, 5, 1,1....We will omit the duplicate odd number 1 later on.It becomes that 11, 17, 13, 5, 1. That is, the odd number 11 is successful. This implies that odd 17, odd 13, odd 5, odd 1 are all successful. This also implies that even 2^k , even 2^k × 5, even 2 ^k × 13, even 2 ^k × 17, even 2 ^k × 11 are all successful,where k = 1, 2, 3, ....

Definition 2.2: A sequence obtained by omitting all even numbers in a Collatz sequence starting with an odd number p is called a Collatz odd sequence.p is said to be successful if 1 is finally obtained.Otherwise,p is not successful.

Remark: Obviously,whether an odd number p is successful,and Definition 2.1 and Definition 2.2 are equivalent.

Definition 2.3: For two odd numbers p and q in a Collatz odd sequence, if 3p + 1= 2^k q ,where k > 0 is a positive integer, then p is said to be a front odd number of q ,and q is said to be a back odd number of p.

In the previous example , because 11 × 3 + 1 = 17 × 2, where k = 1, so 11 is a front odd number of 17, and 17 is a back odd number of 11. Similarly, because 17 × 3 + 1 = 13 × 2 ² , where k = 2, so 17 is a front odd number of 13, 13 is a back odd number of 17, etc.

Theorem 2.1: If p is a front odd number of q,then 4p+1 is also a front odd number of q. Further if 3p+1 = 2 ^kq,where k > 0 is a positive integer, then 3(4p+1)+1 = 2^k+2q.

Proof: Since p is a front odd number of q, there exists a positive integer k > 0,such that 3p+1 = 2^kq. So 3(4p+1)+1 = 12p+4 = 4(3p+1) = 4 × 2 ^kq= 2^k+2q. ■

Consider the sequence {a_n}:a₁ ,a₂ ,..., a_n ,...,where a₁ is an odd number, and a_n = 4a_n-1 + 1,n = 2,3,4,....

Corollary 2.2: For the above sequence {a _n}, if the odd number a₁ is a front odd number of the odd number p, then each term in the sequence {a_n} is a front odd number of p.Further if 3a₁ + 1 = 2^kp, then 3a₂ + 1 = 2^k+2 p,... ,3a_n + 1 = 2^k+2(n-1)p, where n = 2,3,4,....

Proof: This is a direct corollary of Theorem 2.1.■

Theorem 2.3: For any odd number p,there is a back odd number q, and q is the unique back odd number of p.

Proof: According to the rule of generating a Collatz odd sequence,it is immediate.■

Theorem 2.4: In a Collatz odd sequence,if p is a front odd number of p,（Of course,,p is also a back odd number of p.）then it is only case that p = 1.

Proof: Since p is a front odd number of p,there exists a positive integer k > 0,such that 3p+1 = 2^kp. So 2^kp -3p = p(2^k -3) = 1,where p is odd.

Case 1: k = 1. Then p(2-3) = -p = 1, which cannot be true.

Case 2: k > 1. Then 2^k -3 is a positive integer,and p(2^k -3) = 1.This equation can be true ,only if p = 1 and k = 2. ■

§3. Bijections between Odd Subsets and Sequence Sets

This paragraph gives two classification methods of the odd number set ,and gives a classification of sequence sets of the odd number set,and gives a bijection between odd subsets and sequence sets.

The odd number set can be divided into three subsets {4n-1∣n = 1,2,3,...}, {4n+1∣n is 0 or even}= {8n+1∣n = 0,1,2,3,...},{4n+1∣n is odd}.They are disjoint with each other.

Definition 3.1:Write A ={4n-1∣n = 1,2,3,...},B ={4n+1∣n is 0 or even}= {8n+1∣n = 0,1,2,3,...},C ={4n+1∣n is odd}.

Definition 3.2: (1)Write S₁ = {{a_n}∣a₁∈ A,and a_n = 4a_n-1+1,n = 2,3,4,...},where {a_n} is a sequence with the first term a₁ ∈ A,and a_n = 4a_n-1+1,n = 2,3,4,....

(2)Write S₂ = {{a_n}∣a₁∈ B,and a_n = 4a_n-1+1,n = 2,3,4,...},where {a_n} is a sequence with the first term a₁∈ B,and a_n = 4a_n-1+1,n=2,3,4,....

Examples: The first two sequences in S₁ are the following {a_n} and {b_n},where {a_n }:3,13,53,..., (a₁ = 3); {b_n }:7,29,117,..., (b₁ = 7).

The first two sequences in S₂ are the following {a_n} and {b_n},where {a_n }:1,5,21,85,..., (a₁ = 1); {b_n }:9,37,149,..., (b₁ = 9).

Theorem 3.1: All terms (odd numbers) in all sequences of S₁ and S₂ contain all odd numbers. And each odd number must be in the only one of these sequences.

Proof: Note that all odd numbers can be divided into three subsets A,B,C. Any odd number in A can be the first term a₁ of some definite sequence {a_n}in S₁.Similarly,any odd number in B can be the first term a₁ of some definite sequence {a_n}in S₂. So it is enough to prove that all terms of all sequences in S₁ and S₂ contain any odd number with the form 4n+1 (n is odd) in C, and that any odd number in C must be in the only one of these sequences.

Let p = 4n₁ + 1 ∈ C ,that is any definite odd number in C, where n₁ is a definite odd number. But all odd numbers can be divided into three subsets A,B,C.

Case 1: If n₁ ∈ A, then n₁ is the first term a₁ of some definite sequence {a_n}in S₁.By the construction of the sequence {a_n} in S₁, p = 4n₁ + 1 is the second term a₂ of this definite sequence {a_n} in S₁.Similarly, if n₁ ∈ B, then p = 4n₁ + 1 is the second term a₂ of some definite sequence {a_n} in S₂.

Case 2: If n₁ ∈ C, then there exists a definite odd number n₂ , such that n₁ = 4n₂ + 1. If n₂ ∈ A (or ∈ B), then according to Case 1, n₁ = 4n₂ + 1 is the second term a₂ of some definite sequence {a_n}in S₁ (or S₂), and p = 4n₁ + 1 is the third term a₃ of this definite sequence {a_n}in S₁ (or S₂).

Case 3: If n ₂ ∈ C, then we are back to the beginning of case 2. Continuing, since p = 4n₁ + 1 is a definite odd number in C, and the odd number 1 ∈ B and 3 ∈ A ,Case 2 cannot occur indefinitely.So, after finitely many case 2, we can always get some n_k , so that n_k-1 = 4n_k + 1, n₁ > n₂ > ... > n_k, n₁,n₂,...,n_k-1∈ C and n_k ∈ A (or ∈ B), where n_k is the first term a₁ of some definite sequence {a_n}in S₁ (or S₂) ,n_k-1 is the 2nd term a₂ of the sequence ,...,n₁ is the kth term a_{k ,}and p = 4n₁ + 1 is the k+1st term a_k+1 of the definite sequence {a_n}in S₁ (or S₂). ■

Note that all odd numbers can again be divided into three subsets : {6n-3| n = 1,2,3,...}, {6n-1| n = 1,2,3,...}, {6n+1| n = 0,1,2,3,...}.

Definition 3.3: Write D = {6n-1| n = 1,2,3,...},E = {6n+1| n = 0,1,2,3,...},and F ={6n-3| n = 1,2,3,...}. Since 6n-3 is divisible by 3,an odd number in F is also called a triple odd number.

Lemma 3.2: The odd number 4n-1 in A is a front odd number of the odd number 6n-1 in D, or 6n-1 is the back odd number of 4n-1,where n = 1,2,3,....

Proof: Since 3(4n-1) + 1 = 2(6n-1),the result holds. ■

Note that B ={4n+1∣n is 0 or even}= {8n+1∣n = 0,1,2,3,...}.

Lemma 3.3: The odd number 8n+1 in B is a front odd number of the odd number 6n+1 in E, or 6n+1 is the back odd number of 8n+1,where n = 0,1,2,3,....

Proof: Since 3(8n+1) + 1 = 2²(6n+1),the result holds. ■

Lemma 3.4: The triple odd number 6n-3 in F cannot be a back odd number of any odd number, where n = 1,2,3,....

Proof: Suppose that 6n-3 is a back odd number of some odd number p= 2m-1.Then 3(2m-1)+1 = 2^k(6n-3) holds for some positive integer k. At this point, the right side 2^k(6n-3) of the equation is divisible by 3, while the left side 3(2m-1)+1 of the equation is not divisible by 3. Contradiction. ■

Theorem 3.5:There exists a bijection ƒ₁: ƒ₁（D）= S₁，such that for any 6k-1 ∈ D,ƒ₁(6k-1) = {a_n} = {a_n∣a₁ = 4k-1,a_n = 4a_n-1 +1 ,n = 2,3,4,...}∈ S₁ ,k = 1,2,3,..., where all front numbers of 6k-1 are exactly all terms of the sequence {a_n},and 3a_n + 1 = 2^2n-1(6k-1) (n = 1,2,3,...);conversely, all terms of the sequence {a_n} have the unique back number 6k-1.

Proof: By Lemma 3.2,for a definite positive integer k,the first term 4k-1 of the sequence {a_n} is a front odd number of the odd number 6k-1.Then by Corollary 2.2, each term of {a_n} is a front odd number of 6k-1,and 3a_n + 1 = 2^2n-1(6k-1) (n = 1,2,3,...).We prove that any front odd numbers of 6k-1 is contained in this sequence {a_n}.

If an odd number p is not in {a_n},then by theorem 3.1,p is a term of some other sequence {b_n}.We prove that p cannot be a front odd number of 6k-1.

Case 1:{b_n}∈ S₁.Then the first term of {b_n} is b₁,and b₁ = 4s-1,where s is a definite positive integer with s ≠ k. As above, each term of {b_n} is a front odd number of 6s-1.So,p is also a front odd number of 6s-1.By Theorem 2.3,6s-1 is the unique back odd number of p.Since s ≠ k,6k-1 cannot be a back odd number of p.In other words,p cannot be a front odd number of 6k-1.

Case 2:{b_n}∈ S₂.Then b₁ = 8s+1,where s is 0 or a definite positive integer.Then by Lemma 3.3,b₁ = 8s+1 is a front odd number of 6s+1.Then by Corollary 2.2, each term of {b_n} is a front odd number of 6s+1.So,p is also a front odd number of 6s+1.Since D ∩ E = Ø,6k-1 ∈ D and 6s+1 ∈ E,6k-1 ≠ 6s+1.By Theorem 2.3, 6k-1 cannot be a back odd number of p.In other words,p cannot be a front odd number of 6k-1.

Therefore, any front odd numbers of 6k-1 is contained in this sequence {a_n}.So,for any odd number 6k-1 in D,ƒ₁(6k-1) = {a_n∣a₁ = 4k-1,a_n = 4a_n-1 +1 ,n=2,3,4,...},where k = 1,2,3,....

Conversely, by Theorem 2.3,for any sequence {a_n} = {a_n∣a₁ = 4k-1,a_n = 4a_n-1 +1,n = 2,3,4,...} in S₁,all terms of the sequence {a_n} have the unique back odd number 6k-1.So,ƒ₁^-1({a_n∣a₁ = 4k-1,a_n = 4a_n-1 +1 ,n=2,3,4,...}) = 6k-1,where k = 1,2,3,.... To sum up,we obtain that ƒ₁(D）= S₁ is a bijection.■

Examples:Take k = 1,ƒ₁(5) = {a_n}:3,13,53,213,.... Each term of {a_n} is a front odd number of 5 ∈ D.Take k = 2,ƒ₁(11) = {a_n}:7,29,117,469,.... Each term of {a_n} is a front odd number of 11 ∈ D.etc.

Theorem 3.6:There exists a bijection ƒ₂: ƒ₂（E）= S₂，such that for any 6k+1 ∈ E,ƒ₂(6k+1) = {a_n} = {a_n∣a₁ = 8k+1,a_n = 4a_n-1 +1 ,n=2,3,4,...}∈ S₂ ,k = 1,2,3,..., where all front numbers of 6k+1 are exactly all terms of the sequence {a_n},and 3a_n + 1 = 2²ⁿ(6k+1) (n = 1,2,3,...);conversely, all terms of the sequence {a_n} have the unique back number 6k+1.

Proof: Following the method proved in Theorem 3.5,the result holds.■

Examples: Take k = 0, ƒ₂(1) = {a_n}:1,5,21,85,....Each term of {a_n} is a front odd number of 1 ∈ E.Take k = 1, ƒ₂(7) = {a_n}:9,37,149,597,....Each term of {a_n} is a front odd number of 7 ∈ E.etc.

Corollary 3.7: There exists a bijection ƒ: ƒ（D ∪ E）= S₁ ∪ S₂ ，where if an odd number p ∈ D,then define ƒ(p) = ƒ₁(p),and if p ∈ E,then define ƒ(p) = ƒ₂(p).

Proof: Since D ∩ E = Ø,and S₁ ∩ S₂ = Ø,by Theorem 3.5 and Theorem 3.6, it is immediate.■

4. Successful Odd Sequence Set H

This paragraph gives such an odd sequence set H that an odd number p is successful,if and only if p is in a sequence of H.

Theorem 4.1: Let p be an odd number and q = 4p+1.Then

(1) if p ∈ E, then q ∈ D;

(2) if p ∈ D, then q ∈ F；

(3) if p ∈ F, then q ∈ E.

Proof: (1) Since p ∈ E,let p = 6k+1, where k is 0 or a definite positive integer. Then q = 4p+1 = 4(6k+1)+1 = 24k+5 = 6(4k+1)-1, so q ∈ D.

Since p ∈ D,let p = 6k-1,where k is a definite positive integer.Then q = 4p+1 = 4(6k-1)+1 = 24k-3 = 6(4k)-3, so q ∈ F.

(3) Since p ∈ F,let p = 6k-3, where k is a definite positive integer. Then q = 4p+1 = 4(6k-3)+1 = 24k-11= 6(4k-2)+1, so q ∈ E.■

Remarks: For a sequence {a_n} ∈ S₁, a₁ = 4k-1 ∈ A. And all odd numbers in A are {4k-1}:3, 7, 11, 15, 19, 23, ...; where 3, 15, ... ∈ F,and 7, 19, ... ∈ E, and 11, 23, . ...∈ D.

For a sequence {a_n}∈ S₂, a₁ = 8k+1 ∈ B. And all odd numbers in B are {8k+1}:1, 9, 17, 25, 33, 41,...; where 9, 33,...∈ F,and 1, 25,...∈ E, and 17, 41, ...∈ D.

That is, the first term a₁ can be any one in three odd subsets D,E,F, whether the sequence {a_n} ∈ S₁ or {a_n}∈ S₂ .

Theorem 4.2: Let {a_n} ∈ S₁ or {a_n}∈ S₂, if a₁ ∈ E, then a_3k-2 ∈ E,a_3k-1 ∈ D,a_3k ∈ F,where k = 1,2,3,...;

(2) if a₁ ∈ F, then a_3k-2 ∈ F,a_3k-1 ∈ E,a_3k ∈ D,where k = 1,2,3,...;

(3) if a₁ ∈ D, then a_3k-2 ∈ D,a_3k-1 ∈ F,a_3k ∈ E,where k = 1,2,3,....

Proof: This is a direct corollary of Theorem 4.1.■

Remark: According to Theorem 4.2, any term of a sequence {a_n} in S₁ or S₂ can belong to any one in three odd subsets D,E,F.

Collatz sequences now under discussion are Collatz odd sequences, so the following definition is given.

Definition 4.1: If p_k ,p_k-1 ,...,p₁ ,1 is a Collatz odd sequence with k odd numbers other than 1, then we say that p_k is k steps successful; and it is specified that the odd number 1 is 0 step successful.

Example: 11, 17, 13, 5, 1 is a Collatz odd sequence given in §2. Then we have: odd 5 is 1 step successful, 13 is 2 steps successful, 17 is 3 steps successful, and 11 is 4 steps successful.Specially,1 is 0 step successful.

Corollary 4.3: If p is a front odd number of q ,p ≠ 1 and q is k steps successful, then p is k+1 steps successful. Specially,p = 1 is 0 step successful.

Proof: This is a direct consequence of Definition 4.1.■

Note that in the following,ƒ₁ ,ƒ₂ and ƒ are the three bijections given in Theorem 3.5, Theorem 3.6 and Corollary 3.7,where ƒ（D） = ƒ₁（D） = S₁,and ƒ（E）= ƒ₂（E）= S₂.For simplicity,if R is a sequence set,a sequence {a_n} ∈ R,and a_i∈ {a_n} is a term of {a_n},then write a_i ∈ R. If a_i ∈ R and a_i ∈ D, then write a_i ∈ R ∩ D,etc.

In the following, by a recursive method, we construct such a sequence set H that an odd number p is successful, if and only if p is in a sequence of H.

By Theorem 3.6 and Corollary 3.7,since 1 ∈ E, ƒ(1) = ƒ₂(1) = {a_n}:1,5,21,85,341,...∈ S₂.

Write H₁ ={ƒ(1)} = {ƒ₂(1)}={{a_n}:1,5,21,85,341,...},where H₁ is a sequence set,and there is exactly one sequence in H₁.

In this sequence of H₁ , since a₁ = 1 ∈ E, by Theorem 4.2,a_3k-2 ∈ E,a_3k-1 ∈ D,a_3k ∈ F,where k = 1,2,3,....By Lemma 3.4, a_3k (∈ F) has no front odd number,where k = 1,2,3,.... By Theorem 3.5,for a definite a_3k-1 (∈ D),there is the unique odd sequence {b_n}∈ S₁ such that ƒ₁(a_3k-1) = {b_n},where k = 1,2,3,...,and each term of {b_n} is a front odd number of a_3k-1. By Theorem 3.6,for a definite a_3k-2 (∈ E),there is the unique odd sequence {c_n}∈ S₂ such that ƒ₂(a_3k-2) = {c_n},where k = 1,2,3,...,and each term of {c_n} is a front odd number of a_3k-2.

Write H₂ = {ƒ₁(p)| p ∈ H₁ ∩ D} ∪ {ƒ₂(p)| p ∈ H₁ ∩ E and p ≠ 1 }.Note that (H₁ ∩ D)∪ (H₁ ∩ E) = H₁ ∩ (D ∪ E）.By Corollary 3.7,H₂ = {ƒ(p)| p ∈ H₁ ∩ (D ∪ E）and p ≠ 1}.(Note that the removal of ƒ(1) = ƒ₂(1) in H₂ is to prevent the sequence that appeared in H₁ from reappearing in H₂.)

Write H₃ = {ƒ₁(p)| p ∈ H₂ ∩ D} ∪ {ƒ₂(p)| p ∈ H₂ ∩ E } = {ƒ(p)| p ∈ H₂ ∩ (D ∪ E)}.

Assume that the sequence set H_n has been formed.

Write H_n+1 ={ƒ(p)| p ∈ H_n ∩ (D ∪ E)}.

Now the sequence sets H₁ ,H₂ ,H₃ ,...,H_n ,... have been constructed.

Write H =

⋃_{n = 1}^{\infty} H

_n.

Remark: To get an intuitive sense of the composition of the set H, several sequences in H₁,H₂ and H₃ are given here according to Theorem 3.5 and Theorem 3.6.

The unique sequence in H₁ is ƒ(1) = ƒ₂(1) = {a_n}:1,5,21,85,341,1365,....

Note that H₂ = {ƒ₁(p)| p ∈ H₁ ∩ D} ∪ {ƒ₂(p)| p ∈ H₁ ∩ E and p ≠ 1 }.In the sequence {a_n} of H₁, a₃ = 21 ,a₆ = 1365,... (∈ F) have no front odd number.Since p =5,341∈ H₁ ∩ D,by Theorem 3.5,ƒ₁(5) = {b_n}:3,13,53,...∈ H₂,ƒ₁(341) = {b_n}:227,909,3637,...∈ H₂.And since p = 85 ∈ H₁ ∩ E and p = 85 ≠ 1,by Theorem 3.6,ƒ₂(85)= {b_n}:113,453,1813,...∈ H₂. And so on. And see H₃.H₃ = {ƒ₁(p)| p ∈ H₂ ∩ D} ∪ {ƒ₂(p)| p ∈ H₂ ∩ E }.Because ƒ₁(5) = {b_n}:3,13,53,...∈ H₂, where 3 (∈ F) has no front odd number,13 ∈ H₂ ∩ E,and 53 ∈ H₂ ∩ D,so,by Theorem 3.6, ƒ₂(13) = {c_n}: 17,69,277,...∈ H₃,and by Theorem 3.5,ƒ₁(53) = {c_n}: 35,141,565,...∈ H₃.Because ƒ₁(341) = {b_n}:227,909,3637,...∈ H₂, where 909 (∈ F) has no front odd number, 227 ∈ H₂ ∩ D,and 3637 ∈ H₂ ∩ E,so,by Theorem 3.5,ƒ₁(227) = {c_n}: 151,605,2401,...∈ H₃, and by Theorem 3.6,ƒ₂(3637) = {c_n}: 4849,19397,77589,...∈ H₃.Because ƒ₂(85)= {b_n}:113,453,1813,...∈ H₂,where 453 (∈ F) has no front odd number,113 ∈ H₂ ∩ D,and 1813 ∈ H₂ ∩ E, so,by Theorem 3.5, ƒ₁(113) = {c_n}: 75,301,1205,...∈ H₃,and by Theorem 3.6,ƒ₂(1813) = {c_n}: 2417,9669,38677,... ∈ H₃.And so on.

Theorem 4.4: (1) There are no identical sequences in all sequences of H =

⋃_{n = 1}^{\infty} H

_n.And no two different sequences in them have the same term.

(2) All terms of any sequence in H_n are n steps successful, where n = 1, 2, 3, ...,except for the odd number 1 in H₁.

(3) If an odd number p is successful, then p must be in some sequence of the sequence set H.

Proof: (1) Induction. Because H₁ = {ƒ(1)} and H₂ = {ƒ(p)| p ∈ H₁ ∩ (D ∪ E）and p ≠ 1},so H₁ ∪ H₂ = {ƒ(p)| p ∈ H₁ ∩ (D ∪ E）}.

Because H₁ = {ƒ(1)} is a sequence set = {{a_n}:1,5,21,85,341,1365,...},and it has not the same term,so,there is no same odd number in H₁ ∩ (D ∪ E）. Since ƒ is a bijection,all sequences in H₁ ∪ H₂ are not identical to each other.Then since H₁ ∪ H₂ ⊂ S₁ ∪ S₂ ,by Theorem 3.1,no two different sequences in H₁ ∪ H₂ have the same term.

Consider H₃ = {ƒ(p)| p ∈ H₂ ∩ (D ∪ E)}.Then H₁∪H₂∪H₃ ={ƒ(p)| p ∈ H₁ ∩(D ∪ E）}∪{ƒ(p)| p ∈ H₂ ∩ (D ∪ E)} = {ƒ(p)| p ∈ (H₁∪H₂ )∩(D∪E)}.Since no two different sequences in H₁ ∪ H₂ have the same term,there is not the same odd number in (H₁∪H₂ )∩(D∪E).Since ƒ is a bijection,all sequences in H₁∪H₂∪H₃ are not identical to each other.By Theorem 3.1,no two different sequences in H₁∪H₂∪H₃ have the same term.

Suppose all sequences in

⋃_{k = 1}^{n} H

_k are different from each other, and no two different sequences have the same term.

Consider H_n+1 = {ƒ(p)| p ∈ H_n ∩ (D ∪ E)}.Then

⋃_{k = 1}^{n + 1} H

_k ={ƒ(p)| p ∈ H₁ ∩(D ∪ E）}∪{ƒ(p)| p ∈ H₂ ∩ (D ∪ E)} ∪...∪{ƒ(p)| p ∈ H_n ∩ (D ∪ E)} = {ƒ(p)| p ∈ (

⋃_{k = 1}^{n} H

_k)∩(D∪E)}.By the inductive assumption,no two different sequences in

⋃_{k = 1}^{n} H

_k have the same term.So,there is no same odd number in (

⋃_{k = 1}^{n} H

_k)∩(D∪E).Since ƒ is a bijection,all sequences in

⋃_{k = 1}^{n + 1} H

_k are not identical to each other.By Theorem 3.1,no two different sequences in

⋃_{k = 1}^{n + 1} H

_k have the same term.

(2) Induction. First of all, all terms of the sequence in H₁ are the front odd numbers of the odd number 1. By Definition 4.1,all terms of the sequence in H₁ are 1 step successful,except for the odd number 1.

Consider any sequence {a_n} in H₂.Then H₂ = {ƒ(p)| p ∈ H₁ ∩ (D ∪ E）and p ≠ 1}.So,there is a definite odd number q,such that q∈ H₁ ∩ (D ∪ E),q ≠ 1, and ƒ(q) = {a_n}.Note that all terms of {a_n} are all front numbers of q,and q ∈ H₁ ,q ≠ 1 is 1 step successful.By Corollary 4.3, all terms of {a_n} are 2 steps successful. Therefore,all terms of any sequence in H₂ are 2 steps successful.

Assume that all terms of any sequence in H_n are n steps successful,n ≧ 2.

Consider any sequence {b_n} in H_n+1.Then H_n+1 = {ƒ(p)| p ∈ H_n ∩ (D ∪ E）}.So,there is a definite odd number q,such that q∈ H_n ∩ (D ∪ E), and ƒ(q) = {b_n}.By the inductive assumption, q ∈ H_n is n steps successful.Note that all terms of {b_n} are all front numbers of q.By Corollary 4.3, all terms of {b_n} are n+1 steps successful.Therefore,all terms of any sequence in H_n+1 are n+1 steps successful.

(3) Firstly,the odd number p = 1 is 0 step successful,1 ∈ H₁.Now let p ≠ 1.Since the odd number p is successful, there must exist some positive integer n,such that p is n steps successful.It is enough to prove that p must be in some sequence of H_n.

Induction. Look at n = 1.Let p be 1 step successful.Then p ≠ 1.And let Collatz odd sequence of p be p,1.Then p is a front odd number of 1.Note that H₁ ={ƒ(1)} = {{a_n}:1,5,21,85,341,...},and ƒ is a bijection,so,all front odd numbers of 1 belong to H₁.So,p ∈ H₁.

Look at n = 2.Let p be 2 steps successful. And let Collatz odd sequence of p be p ,p₁ ,1.Then the odd number p₁ is 1 step successful,and p₁ ≠ 1.As above,p₁ ∈ H₁ and p₁ ≠ 1.Because p₁ has the front odd number p,so p₁ ∈ D ∪ E,and p₁ ∈ H₁ ∩ (D ∪ E）and p₁ ≠ 1. Note that H₂ = {ƒ(q)| q ∈ H₁ ∩ (D ∪ E） and q ≠ 1},then ƒ(p₁) ∈ H₂,where ƒ(p₁) is a sequence of H₂.And ƒ(p₁) contain all front odd numbers of p₁ ,and p is a front odd number of p₁.So,p ∈ƒ(p₁)∈ H₂.

Assume that the result holds for n = k,that is, if the odd number p is k steps successful, then p ∈ H_k.

Now let n = k+1,that is, the odd number p is k+1 steps successful.And let Collatz odd sequence of the odd number p be p ,p_k ,p_k-1 ,...,p₁ ,1. Then the odd number p_k is k steps successful. By inductive assumption, p_k ∈ H_k.Because p_k has the front odd number p,so p_k∈ D ∪ E,and p_k ∈ H_k ∩ (D ∪ E）. Note that H_k+1 = {ƒ(q)| q ∈ H_k ∩ (D ∪ E）},then ƒ(p_k) ∈ H_k+1, where ƒ(p_k) is a sequence of H_k+1.And ƒ(p_k) contain all front odd numbers of p_k ,and p is a front odd number of p_k.So,p ∈ƒ(p_k)∈ H_k+1.■

§5. Possibility of Not Successful Odd Sequence Set

Remark: From Theorem 4.4, it follows that every term of every sequence in H =

⋃_{n = 1}^{\infty} H

_n is successful, and every odd number that is successful must be in one of these sequences of H. Therefore, to prove that all odd numbers are successful, it is sufficient to show that every odd number is in one of these sequences of H.

Lemma 5.1: (1) If p is a front odd number of q,and the odd number q is not successful, then the odd number p is also not successful.

(2) If r is a back odd number of q,and the odd number q is not successful,then the odd number r is also not successful.

Proof: (1) First, since q is not successful, q ≠ 1. Assume that the odd number p is successful,then there exists a positive integer k,such that p is k steps successful. And by Theorem 2.3, q is the unique back odd number of p. Then, by Definition 4.1, k > 1,and q is k-1 steps successful. Contradiction.

Assume that the odd number r is successful,then there exists a positive integer m, such that r is m steps successful, so q is m+1 steps successful. Contradiction.■

By Theorem 3.1, any odd number is a term of only one sequence in S₁ ∪ S₂.

Lemma 5.2: If an odd number p is not successful,and p is a term of some sequence {a_n} in S₁ ∪ S₂,then all terms of {a_n} are not successful.

Proof: From Theorem 4.4, it follows that all odd numbers in the sequence set H are successful. Since the odd number p is not successful, p is not in H.And the odd number p is a term in some sequence {a_n}.Since H is the sequence set , that sequence {a_n} is also not in H. In the other words, all terms of {a_n} are not in H. By Theorem 4.4, all terms of {a_n} are not successful.■

Remarks:(1)In a Collatz odd sequence,if there is a situation like q₁,q₂,...,q_k,q₁,where q₁,q₂,...,q_k are different from each other,and the odd number q₁ recurring,then we call these k odd numbers to form a k-cycle.In fact,it is impossible to have such a k-cycle in a successful Collatz odd sequence.Because there must be a positive integer s for a successful odd number q₁,so that q₁ is s steps successful,so,there is a Collatz odd sequence q₁,p_s-1,p_s-2,...,p₁,1.

Assume q₁ is in a k-cycle q₁,q₂,...,q_k,q₁.Then q₁ has a Collatz odd sequence q₁,q₂,...,q_r,p_t,p_t-1,...,p₁,1,where q₂ = p_s-1,...,q_r = p_t+1,1 ≦ r ≦ k,r is a definite integer.Then q_r has 2 back odd numbers q_r+1 and p_t,where if r = k,then q_r+1 = q_k+1 = q₁,and if r = 1,then p_t = p_s-1.By Theorem 2.3,this is impossible.

(2)Assuming that p₁ is not successful,and p₁ is in a k-cycle p₁,p₂,...,p_k,p₁.Then by Theorem 2.3, Collatz odd sequence of these k odd numbers is the k-cycle,and any one of these k odd numbers can not appear in other j-cycle.And these k odd numbers are not successful.

(3)Let p be a front odd number of q.Then there exists a positive integer i,such that 3p+1 = 2ⁱq.If i = 1,then q = (3p+1)/2 > p.If i ≧ 2,then q = (3p+1)/2ⁱ < p.Note that any odd number is a term of some sequence in S₁ ∪ S₂.By Theorem 3.5,for any sequence {a_n} = {a_n∣a₁ = 4k-1,a_n = 4a_n-1 +1 ,n=2,3,4,...}∈ S₁ ,3a_n + 1 = 2^2n-1(6k-1) (n = 1,2,3,...),where any term of {a_n} is a front odd number of 6k-1.And when n =1,that is,i = 2n-1 = 1,6k-1 > a₁.When n ≧ 2,that is,i = 2n-1 ≧ 3,6k-1 < a_n.And by Theorem 3.6,for any sequence {a_n} = {a_n∣a₁ = 8k+1,a_n = 4a_n-1 +1 ,n=2,3,4,...}∈ S₂,3a_n + 1 = 2²ⁿ(6k+1) (n = 1,2,3,...),where any term of {a_n} is a front odd number of 6k+1.When n = 1,2,3,...,that is,i = 2n ≧ 2,6k+1 < a_n.

Lemma 5.3:(1)There are no 2-cycles in Collatz odd sequences.

(2)There are no 3-cycles in Collatz odd sequences.

Proof: (1)Suppose that there is a 2-cycle q₁,q₂,q₁ in Collatz odd sequences.Let q₁ < q₂.Because q₁ is a front odd number of q₂,so there is a positive integer i,such that 3q₁+1 = 2ⁱq₂.By Remark(3) above,i = 1,that is,3q₁+1 = 2q₂.And because q₂ is a front odd number of q₁,so there is a positive integer j,such that 3q₂+1 = 2^jq₁,i.e,(3q₂+1)/2^j = q₁.By Remark(3) above,since q₂ > q₁,j ≧ 2.But (3q₁+1)/2 = q₂.When j = 2,(3q₂+1)/2^j =(3((3q₁+1)/2)+1)/2² = (9q₁+5)/8 > q₁.When j ≧ 3,(3q₂+1)/2^j ≦ (3((3q₁+1)/2)+1)/2³ = (9q₁+5)/16 < q₁.That is to say, (3q₂+1)/2^j ≠ q₁.So,there are no 2-cycles in Collatz odd sequences.

(2)And suppose that there is a 3-cycle q₁,q₂,q₃,q₁ in Collatz odd sequences.Let q₁ < q₂ and q₁ < q₃.The following are divided into two cases.

Case 1：q₂ > q₃.Since q₁ < q₂,and q₁ is a front odd number of q₂,by Remark(3) above,3q₁+1 = 2q₂,i.e,(3q₁+1)/2 = q₂.Since q₂ > q₃,and q₂ is a front odd number of q₃,3q₂+1 = 2ⁱq₃,i.e,(3q₂+1)/2ⁱ = q₃,where i ≧ 2.Since q₃ > q₁,and q₃ is a front odd number of q₁,3q₃+1 = 2^jq₁,i.e,(3q₃+1)/2^j = q₁,where j ≧ 2.But (3q₃+1)/2^j ≦ (3q₃+1)/2² = (3( (3q₂+1)/2ⁱ)+1)/2² ≦(3( (3q₂+1)/2²)+1)/2² = (9q₂+7)/16 = (9((3q₁+1)/2)+7)/16 = (27q₁+23)/32 < q₁.(Note that because small odd numbers must be successful,and odd numbers in k-cycle are all not successful,so,we can set q₁ > 99.) That is to say,(3q₃+1)/2^j ≠ q₁.Contradiction.

Case 2：q₂ < q₃.Since q₁ < q₂,and q₁ is a front odd number of q₂,by Remark(3) above,3q₁+1 = 2q₂,i.e,(3q₁+1)/2 = q₂.Since q₂ < q₃,and q₂ is a front odd number of q₃,3q₂+1 = 2q₃,i.e,(3q₂+1)/2 = q₃.Since q₃ > q₁,and q₃ is a front odd number of q₁,3q₃+1 = 2ⁱq₁,i.e,(3q₃+1)/2ⁱ = q₁,where i ≧ 2.But (3q₃+1)/2ⁱ = (3((3q₂+1)/2)+1)/2ⁱ = (9q₂+5)/2ⁱ⁺¹ =(9((3q₁+1)/2)+5)/2ⁱ⁺¹ = (27q₁+19)/2ⁱ⁺²,

where i ≧ 2.And when i = 2,(3q₃+1)/2ⁱ = (27q₁+19)/2ⁱ⁺² = (27q₁+19)/16 > q₁.when i ≧ 3,(3q₃+1)/2ⁱ = (27q₁+19)/2ⁱ⁺² ≦ (27q₁+19)/32 < q₁,where set q₁ > 99. That is to say,(3q₃+1)/2ⁱ ≠ q₁.

Contradiction.

So,there are no 3-cycles in Collatz odd sequences.■

Remark:The same method can be used to prove that there are no 4-cycles,5-cycles,etc.But there are too many cases to consider,and it will not go on.Later,in Theorem 5.5,we will prove that the k-cycle does not exist.

Remark:Suppose that an odd number q₁ is not successful.Then a Collatz odd sequence q₁,q₂,...,q_n,...can be obtained,where q_i+1 is the unique back odd number of q_i,i = 1,2,3,....By Lemma 5.1,each odd number in the Collatz odd sequence is not successful.In this point,there exist two cases for the Collatz odd sequence q₁,q₂,...,q_n,...

Case 1:All odd numbers of the Collatz odd sequence q₁,q₂,...,q_n,... are different from each other.

Case 2:q₁,q₂,...,q_k,q_k+1,q_k+2 are different from each other,and q_k+3 is an odd number in q₁,q₂,...,q_k+2,such as,q_k+3 = q₃.Then q₃,q₄,...,q_k+2,q₃ is a k-cycle,k ≧ 4.(k ≧ 4 is because the non-existence of 2-cycle and 3-cycle has been proved in Lemma 5.3.)And the Collatz odd sequence q₁,q₂,...,q_n,... is changed to q₁,q₂,q₃,...,q_k,q_k+1,q_k+2,q₃,...,q_k,q_k+1,q_k+2,q₃,...,by Theorem 2.3,here this k-cycle keeps appearing repeatedly.For convenience,in the following, the Collatz odd sequence has been replaced with r₁,r₂,q₁,q₂,q₃,...,q_k,q₁,q₂,q₃,...,q_k,q₁,....

Now,As with the construction of the sequence set H,from not successful Collatz odd sequence q₁,q₂,...,q_n,..., the following series G_n of sequence sets is constructed.In the above Case 2,q₁,q₂,...,q_n,...has been replaced with q₁,q₂,q₃,...,q_k,q₁,q₂,q₃,...,q_k,q₁,....

The following construction is unified for Case 1 and Case 2.In the construction of G_n,for Case 2,when n = ik + m,let q_n = q_m,where i = 0,1,2,...,m = 1,2,...,k.

Firstly, we construct a sequence set G₁.

Note that each odd number in the Collatz odd sequence q₁,q₂,...,q_n,...is not successful, and q₂ is the unique back odd number of q₁ .

Write G₁₁ = {ƒ(q₂)} = {{a_n}},where {a_n} ∈ S₁ ∪ S₂,all terms of {a_n} are all front odd numbers of q₂,and the odd number q₁ is a term of {a_n}.

Write G₁₂ = {ƒ(p)| p ∈ G₁₁ ∩ (D ∪ E)}.

Write G₁₃ = {ƒ(p)| p ∈ G₁₂ ∩ (D ∪ E)}.

Assume that the sequence set G_1n has been formed.

Write G_1,n+1 = {ƒ(p)| p ∈ G_1n ∩ (D ∪ E)}.

Denote G₁ =

⋃_{n = 1}^{\infty} G

_1n.

And construct a sequence set G₂.

Note that q₃ is the unique back odd number of q₂.

Write G₂₁ = {ƒ(q₃)} = {{b_n}},where the odd number q₂ is a term of {b_n}.

Write G₂₂ = {ƒ(p)| p ∈ G₂₁ ∩ (D ∪ E)}.

Write G₂₃ = {ƒ(p)| p ∈ G₂₂ ∩ (D ∪ E)}.

Assume that the sequence set G_2n has been formed.

Write G_2,n+1 = {ƒ(p)| p ∈ G_2n ∩ (D ∪ E)}.

Denote G₂ =

⋃_{n = 1}^{\infty} G

_2n.

So on and so forth.

Suppose again that G_n =

⋃_{k = 1}^{\infty} G

_nk has been constructed.In the construction of G_n,for Case 2,when n = ik + m,let q_n = q_m,where i = 0,1,2,...,m = 1,2,...,k.

At this point, G_n1 has only one sequence, denoted {c_n},and q_n ∈ {c_n}.q_n+1 is the unique back odd number of q_n.

Write G_n+1,1 ={ƒ(q_n+2)}= {{d_n}}.where the odd number q_n+1 is a term of {d_n}.

Write G_n+1,2 = {ƒ(p)| p ∈ G_n+1,1 ∩ (D ∪ E)}.

Write G_n+1,3 = {ƒ(p)| p ∈ G_n+1，2 ∩ (D ∪ E)}.

Assume that the sequence set G_n+1,k has been formed.

Write G_n+1,k+1 = {ƒ(p)| p ∈ G_n+1,k ∩ (D ∪ E)}.

Denote G_n+1 =

⋃_{k = 1}^{\infty} G

_n+1,k.

Now G_n has been constructed recursively, where n = 1,2,3,....

Remark:For r₁ and r₂ mentioned in Case 2,because r₂ is a front odd number of q₁,and r₁ is a front odd number of r₂,and q₁ ∈ G₁₁,so,r₂ ∈ ƒ(q₁) ∈ G₁₂ and r₁∈ ƒ(r₂) ∈ G₁₃.In other words, r₁ and r₂ are all in G₁.

The following Theorem 5.4 is valid for both Case 1 and Case 2.

Theorem 5.4:(1) Any term of all sequences in the set G_n is not successful,where n = 1,2,3,....

G₁ ⊂ G₂ ⊂ ... ⊂ G_n ⊂ ....，where G_i is a proper subset of G_i+1,i = 1,2,3,....

Proof:(1)Induction for G_nk,where n = 1,2,3,.... Firstly,consider G_n1.Since q_n is not successful, and q_n ∈ {c_n},(see G_n1.)by Lemma 5.2,any term of the unique sequence {c_n} in G_n1 is not successful.The result holds for k = 1.

Consider G_n2 = {ƒ(p)| p ∈ G_n1 ∩ (D ∪ E)}.Let ƒ(q) = {b_n}(∈ S₁ ∪ S₂) be any sequence in G_n2.Then q ∈ G_n1 ∩ (D ∪ E),and all term of {b_n} are all front odd numbers of q.Since q ∈ G_n1,q is not successful.And by Lemma 5.1(1),all term of {b_n} are not successful.So the result holds when k = 2.

Assume that the result holds for k = i, that is, any term of any sequence in G_ni is not successful.

Consider G_n,i+1 = {ƒ(p)| p ∈ G_ni ∩ (D ∪ E)}.Let ƒ(q) = {d_n}(∈ S₁ ∪ S₂) be any sequence in G_n,i+1.Then q ∈ G_ni ∩ (D ∪ E),and all term of {d_n} are all front odd numbers of q.Since q ∈ G_ni,by inductive assumption,q is not successful.And by Lemma 5.1(1),all term of {d_n} are not successful.So the result holds when k = i+1.

(2) First prove that G₁ ⊂ G₂ . Look back at the construction of G₂.Note that G₁₁ = {ƒ(q₂)}

= {{a_n}}, where {a_n} ∈ S₁ ∪ S₂,and the odd number q₁ is a term of {a_n}.G₂₁ = {ƒ(q₃)} = {{b_n}},where {b_n} ∈ S₁ ∪ S₂,and the odd number q₂ is a term of {b_n}. G₂₂ = {ƒ(p)| p ∈ G₂₁ ∩ (D ∪ E)}.Since q₂∈ {b_n} ∈ G₂₁,and ƒ(q₂) = {a_n},(any term of {a_n} is a front odd numbers of q₂.)q₂ ∈ G₂₁ ∩ (D ∪ E).Thus ƒ(q₂) ∈ G₂₂,i.e, G₁₁ ⊂ G₂₂.Because G₁₂ = {ƒ(p)| p ∈ G₁₁ ∩ (D ∪ E)},and G₂₃ = {ƒ(p)| p ∈ G₂₂ ∩ (D ∪ E)}，and (G₁₁ ∩ (D ∪ E)) ⊂ (G₂₂ ∩ (D ∪ E)),so,G₁₂ ⊂ G₂₃.By analogy,it follows that G₁₃ ⊂ G₂₄ ,...,G_1n ⊂ G_2,n+1 , etc.

Since G₁ = G₁₁ ∪ G₁₂ ∪ ... ∪ G_1n ∪ ... G₂ = G₂₁ ∪ G₂₂ ∪ ... ∪ G_2n ∪ G_2,n+1 ∪ ... So G₁ ⊂ G₂.Obviously,G₁ is a proper subset of G₂. By analogy, G₂ ⊂ G₃ ⊂ ... ⊂ G_n ⊂ ....■

Theorem 5.5:Case 2 cannot exist,that is,there are no k-cycles in not successful Collatz odd sequence.

Proof: Firstly,for Case 2,by Theorem 5.4,G₁ ⊂ G₂ ⊂ ... ⊂ G_k+1,where G_i is a proper subset of G_i+1,i = 1,2,...,k.So,G₁ is a proper subset of G_k+1, On the other hand, consider the constructions of G_k+1 and G₁.Note that G_k+1,1 = {ƒ(q_k+2)}, where q_k+2 = q₂.So,G_k+1,1 = {ƒ(q₂)}.But G₁₁ = {ƒ(q₂)},so,G_k+1,1 = G₁₁.And because G_k+1,2 = {ƒ(p)| p ∈ G_k+1,1 ∩ (D ∪ E)} = {ƒ(p)| p ∈ G₁₁ ∩ (D ∪ E)},and G₁₂ = {ƒ(p)| p ∈ G₁₁ ∩ (D ∪ E)}, so,G_k+1,2 = G₁₂.By analogy,G_k+1,i = G_1i,where i = 3,4,5,....Thus,G₁ =

⋃_{n = 1}^{\infty} G

_1n =

⋃_{n = 1}^{\infty} G

_k+1,n = G_k+1. Contradiction. Case 2 cannot exist.■

Next, we just need to discuss Case 1.Consider G₁ first.

Theorem 5.6:If q₁,q₂,...,q_n,...is the Collatz odd sequence in the construction of G_n,where all odd numbers of the Collatz odd sequence are different from each other,then

(1)q₂

\notin

G_1i,where i = 1,2,3,....

(2)There are no identical sequences in the sequence set

⋃_{i = 1}^{\infty} G

_1i,and no two different sequences in

⋃_{i = 1}^{\infty} G

_1i have the same term.

Proof:(1)Firstly,G₁₁ = {ƒ(q₂)} = {{a_n}},where all terms of {a_n} are all front odd numbers of q₂.Because q₂ is not successful,so,q₂≠ 1, and by Theorem 2.4,q₂

\notin

G₁₁.Suppose that q₂ is a term of some sequence {d_n} in some G_1i,where i ≧ 2 is a definite integer.Since G_1i = {ƒ(p)| p ∈ G_1,i-1 ∩ (D ∪ E)},there exists a definite odd number s ∈ G_1,i-1 ∩ (D ∪ E),such that ƒ(s) = {d_n},and all terms of {d_n} are all front odd numbers of s.But q₃ is the unique back odd number of q₂ (∈{d_n}),so,q₃ = s ∈ G_1,i-1.By analogy,q₄ ∈ G_1,i-2,...,q_i ∈ G₁₂,q_i+1 ∈ G₁₁.But G₁₁ = {ƒ(q₂)} = {{a_n}},where all terms of {a_n} are all front odd numbers of q₂.So,q₂ is a back odd number of q_i+1(∈ {a_n}).Then q_i+1 has two back odd number q₂,q_i+2(≠ q₂),where i ≧ 2.By Theorem 2.3,this is impossible.

(2)Induction.First prove that the result holds for G₁₁ ∪ G₁₂.Note that G₁₁ = {ƒ(q₂)} = {{a_n}}= {ƒ(p)| p ∈ {q₂} ∩ (D ∪ E)},and G₁₂ = {ƒ(p)| p ∈ G₁₁ ∩ (D ∪ E)}.So,G₁₁ ∪ G₁₂ = {ƒ(p)| p ∈ ({q₂}∪ G₁₁)∩ (D ∪ E)}.By (1),q₂

\notin

G₁₁.And ({q₂}∪ G₁₁)∩ (D ∪ E) has not the same odd number.Because ƒ is a bijection,so,there are no identical sequences in the sequence set G₁₁ ∪ G₁₂.Then since G₁₁ ∪ G₁₂ ⊂ S₁ ∪ S₂ ,by Theorem 3.1,no two different sequences in G₁₁ ∪ G₁₂ have the same term.

Consider G₁₃ = {ƒ(p)| p ∈ G₁₂ ∩ (D ∪ E)}.Then

⋃_{i = 1}^{3} G

_1i = {ƒ(p)| p ∈ ({q₂}∪ G₁₁)∩ (D ∪ E)} ∪ {ƒ(p)| p ∈ G₁₂ ∩ (D ∪ E)} = {ƒ(p)| p ∈({q₂}∪ G₁₁∪ G₁₂ ) ∩ (D ∪ E)}.By (1),q₂

\notin

G₁₁ ∪ G₁₂.And because there is no same odd number in G₁₁∪ G₁₂,so, there is no same odd number in ({q₂}∪ G₁₁∪ G₁₂ ) ∩ (D ∪ E).Because ƒ is a bijection,so,there are no identical sequences in the sequence set

⋃_{i = 1}^{3} G

_1i .Then by Theorem 3.1,no two different sequences in

⋃_{i = 1}^{3} G

_1i have the same term.

Assume there are no identical sequences in the sequence set

⋃_{i = 1}^{n} G

_1i,and no two different sequences in

⋃_{i = 1}^{n} G

_1i have the same term,i,e.there is no same odd number in

⋃_{i = 1}^{n} G

_1i.

Note that

⋃_{i = 1}^{n + 1} G

_1i = {ƒ(p)| p ∈({q₂}∪(

⋃_{i = 1}^{n} G

_1i) ) ∩ (D ∪ E)}.By (1),q₂

\notin

⋃_{i = 1}^{n} G

_1i. By the inductive assumption, there is no same odd number in

⋃_{i = 1}^{n} G

_1i.So,there is no same odd number in ({q₂}∪(

⋃_{i = 1}^{n} G

_1i) ) ∩ (D ∪ E).Because ƒ is a bijection,so,there are no identical sequences in the sequence set

⋃_{i = 1}^{n + 1} G

_1i .Then by Theorem 3.1,no two different sequences in

⋃_{i = 1}^{n + 1} G

_1i have the same term.■

The same method can be used to prove the following Theorem 5.7.

Theorem 5.7:If q₁,q₂,...,q_n,...is the Collatz odd sequence in the construction of G_n,where all odd numbers of the Collatz odd sequence are different from each other,then for n = 2,3,4,...,

(1)q_n+1

\notin

G_ni,where i = 1,2,3,....

(2)There are no identical sequences in the sequence set

⋃_{i = 1}^{\infty} G

_ni,and no two different sequences in

⋃_{i = 1}^{\infty} G

_ni have the same term.■

§6. Final Results

In order to make a comparison between the sequence sets H and G_n,, another construction method of H is introduced here.

Let the odd number q₁ be k steps successful. And let q₁,q₂,...,q_k,1 (here set q_k+1 = 1)be the Collatz odd sequence.

Next the following series T_i of sequence sets is constructed,where i = 1,2,...,k.

Firstly, we construct a sequence set T₁.

Note that each odd number in the Collatz odd sequence q₁,q₂,...,q_k,1 is successful,and q₂ is the unique back odd number of q₁ .

Write T₁₁ = {ƒ(q₂)} = {{a_n}},where {a_n} ∈ S₁ ∪ S₂,and the odd number q₁ is a term of {a_n}.

Write T₁₂ = {ƒ(p)| p ∈ T₁₁ ∩ (D ∪ E)}.

Assume that the sequence set T_1n has been formed.

Write T_1,n+1 = {ƒ(p)| p ∈ T_1n ∩ (D ∪ E)}.

Denote T₁ =

⋃_{n = 1}^{\infty} T

_1n.

Note that q₃ is the unique back odd number of q₂.

Write T₂₁ = {ƒ(q₃)} = {{b_n}},where {b_n} ∈ S₁ ∪ S₂,and the odd number q₂ is a term of {b_n}.

Write T₂₂ = {ƒ(p)| p ∈ T₂₁ ∩ (D ∪ E)}.

Assume that the sequence set T_2n has been formed.

Write T_2,n+1 = {ƒ(p)| p ∈ T_2n ∩ (D ∪ E)}.

Denote T₂ =

⋃_{n = 1}^{\infty} T

_2n.

So on and so forth.

Suppose again that T_i =

⋃_{n = 1}^{\infty} T

_in has been constructed,where i = 3,4,...,k-2.

At this point, T_i1 has only one sequence, denoted {c_n},and q_i ∈ {c_n}.q_i+1 is the unique back odd number of q_i.

Write T_i+1,1 = {ƒ(q_i+2)}= {{d_n}},where {d_n} ∈ S₁ ∪ S₂,and the odd number q_i+1 is a term of {d_n}.

Write T_i+1,2 = {ƒ(p)| p ∈ T_i+1,1 ∩ (D ∪ E)}.

Assume that the sequence set T_i+1,j has been formed.

Write T_i+1,j+1 = {ƒ(p)| p ∈ T_i+1,j ∩ (D ∪ E)}.

Denote T_i+1 =

⋃_{j = 1}^{\infty} T

_i+1,j.

And consider i = k-1.Then q_i+2 = q_k+1 = 1,T_i+1,1 = T_k1 = {ƒ(q_i+2)} = {ƒ(1)},where ƒ(1) is the odd sequence {e_n}:1,5,21,85,341,...,q_i+1 = q_k is a term of {e_n}.

Write T_k2 = {ƒ(p)| p ∈ T_k1 ∩ (D ∪ E),and p ≠ 1}.

Assume that the sequence set T_kj has been formed,where j ≧ 2.

Write T_k,j+1 = {ƒ(p)| p ∈ T_kj ∩ (D ∪ E)}.

Denote T_k =

⋃_{j = 1}^{\infty} T

_kj.

Remarks:(1)Note that T_k = H,that is,the sequence set H in §5 can also be constructed by the above method,and this process is the same as the generation process of G_n.

(2)See §5.Consider the constructions of G₁ and G₂.G₁₁ = {ƒ(q₂)},and G₂₁ = {ƒ(q₃)} = {{b_n}},where the odd number q₂ is just a general odd number in the sequence {b_n} of G₂₁，or in G₂₁ ∩ (D ∪ E).G₁₁ = {ƒ(q₂)} grows in the front odd number direction (let's say) to obtain a G₁. While there are infinitely many odd numbers in G₂₁ ∩ (D ∪ E),and G₂ is obtained by growing the infinitely many odd numbers of G₂₁ ∩ (D ∪ E) in the front odd number direction.Where each odd number in G₂₁ ∩ (D ∪ E) also generates a sequence set equivalent to G₁.Thus,G₂ is "infinitely many times" larger than G₁ .

Similarly, for n = 3,4,...,G_n is "infinitely many times" larger than G_n-1.It can also be obtained for i = 1,2,...,k-1,T_i+1 is "infinitely many times" larger than T_i.

(3)By the constructions of G_n and T_i,the difference between sequence set G_n and the sequence set T_i is that all terms of all sequences in T_i are successful,so when it grows in the back odd direction (let's say ),it stops at odd number 1 to obtain T_k = H; while all terms of all sequences in G_n are not successful, so as n increases infinitely, G_n is endlessly expanding at a great speed. From Theorem 4.4, we know that the sequences in H are not identical to each other,and no two sequences have the same term (odd number).So H can be regarded as a set of odd numbers.Similarly,from Theorem 5.7, G_n can also be regarded as a set of odd numbers. From the above, it is obtained that as a set of odd numbers, H is a definite set of odd numbers, and as n increases infinitely，

\lim_{n \to \infty} G

_n is an indeterminate set of odd numbers.

Theorem 6.1: Any odd number is successful.

Proof: Let the set of all odd numbers be Q.Consider H as a set of odd numbers. By Theorem 4.4,the set of all successful odd numbers is H.Suppose there is an odd number that is not successful.And let the set of all not successful odd numbers be G. then G ∩ H = Ø,and G ∪ H = Q.

From Theorem 5.4 and Theorem 5.7,suppose there is an odd number that is not successful,then the sequence set G_n ,which is regarded as the set of odd numbers, can be obtained. And it is known that any odd number of G_n is not successful. And from the above remark (3),

\lim_{n \to \infty} G

_n is an indeterminate set of odd numbers. Because G_n ⊂ G,where n = 1,2,3,...,so

\lim_{n \to \infty} G

_n ⊂ G,and the set G of all odd numbers, those are not successful ,is also an indeterminate set of odd numbers.And the sets of odd numbers H and Q are both definite sets of odd numbers. So the equation G ∪ H = Q does not hold, contradiction.Thus,G = Ø, H = Q,and any odd number is successful.■

Remark:Treat each odd number in the set H as a vertex.Then connect an edge between any two odd numbers (vertices) in H that have a front and back odd number relationship.In particular,connect an edge between odd number 1 and any other odd number in the odd number set H₁ except for odd number 1.At this point,H can be regarded as a tree with an odd root 1.We call it the H-tree.This H-tree contains all odd numbers.Because any triple odd number has no front odd number,each odd number in the triple odd number set F is a leaf of this H-tree.

Theorem 6.2: Any positive integer is successful,i.e.,Collatz conjecture holds.

Proof: See Remark at the top part of §2.■

Problem:For any odd number p,how to get a integer k with a formula,so that p is k-steps successful,or,p ∈ H_k in H.

Acknowledgment

Author is very grateful to the reviews for their comments on the original manuscript and some valuable suggestions,especially the suggestion to use the bijection to make the paper in its current form.

References

Richrd K. Guy : Unsolved Problem In Number Theory.The 3x+1 Problem.Springer-Verlag, New York.330-336,2007.
Jeffrey C.Lagarias: "The 3n+1 Problem and its generalization", Amer. Math. Monthly, 92:1 pp3-23.1985. [CrossRef]
Jerffrey C.Lagarias,The 3x+1 problem:An Annotated Bibliography, II(2000-2009). arXiv:math/0608208 V₆ [math NT]. [CrossRef]

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A Proof of Collatz Conjecture