In 1937, German mathematician L. Collatz posed a problem that later became known as Collatz conjecture: 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; then continue according to 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 [1,2,3,4].This paper gives a sequence classification of odd numbers, and an 1-1 mapping between odd numbers and odd number sequences,and gives a proof of this conjecture.

Definition 1.1: Starting from a positive integer p, the above process is called a Collatz sequence; p is said to be successful to 1 if 1 is finally obtained；otherwise,p is not successful to 1.

Remark: Any positive integer p can be written as: p = 2^k (2m-1),here m is a positive integer and k is a positive integer or 0; when k > 0, p is even; when k = 0, p=2m-1 is odd. If any odd number is successful to 1, 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 to 1. In other words, to prove that Collatz conjecture holds, it is sufficient to show that any odd number is successful to 1.

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

Definition 1.2: For two odd numbers p and q in a Collatz sequence, if 3p + 1= 2^k q ,where k > 0 is a positive integer; then p is said to be a pre odd number of q ,and q is said to be a post odd number of p.

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

Theorem 1.1: If p is a pre odd number of q, then 4p+1 is also a pre odd number of q;and

if 3p+1 = 2 ^kq,here k > 0 is a positive integer, then 3(4p+1)+1 = 2^k+2q.

Proof: Since p is a pre 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+2 q. ■

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

Corollary 1.2: For the above sequence {a _n }, if the odd number a₁ is a pre odd number of an odd number p, then each term in the sequence {a _n } is a pre odd number of p; and if 3a₁ + 1 = 2^k p, 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 1.1.■

Theorem 1.3: Any odd number p has a post odd number q, and q is the unique post odd number of p.

Proof: It follows from the Collatz sequence that for any odd number p, 3p + 1 is an even number, and every even number can always be written as 2^k q,where q is a definite odd number and k is a definite positive integer; that is, 3p + 1 = 2^k q holds for a definite odd number q. By Definition 1.2,q is a post odd number of p, and q is the unique post odd number of p.■

Theorem 1.4: In a Collatz sequence,if p is a pre odd number of p,（Certainly,p is also a post odd number of p.）then it is only case that p = 1.

Proof: Since p is a pre odd number of p,there exists a positive integer k > 0,such that 3p+1 = 2^k p. So 2^k p -3p = p(2^k -3) = 1,here 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.The only way this can be true ,if p = 1 and k = 2. ■

All odd numbers can be divided into two types: {4n-1∣n = 1,2,3,...},and {4n+1∣n = 0,1,2,3,...}. Where {4n+1} can yet be divided into two types: 4n+1 (n is 0 or even), that is: 1, 9, 17, ...; and 4n+1 (n is odd), that is: 5, 13, 21, ....

Definition 2.1: An odd number of the form 4n-1 (n is a positive integer) is called a class A odd number; an odd number of the form 4n+1 (n is 0 or an even number) is called a class B odd number; an odd number of the form 4n+1 (n is an odd number) is called a class C odd number.

Definition 2.2: (1) For a sequence{a_n }:If a₁ is a class A odd number and a_n = 4a_n-1 + 1,n=2,3,4,..., then {a_n } is said to be a class I sequence.

(2) For a sequence {a_n }:If a₁ is a class B odd number and a_n = 4a_n-1 + 1,n= 2,3,4,..., then {a_n } is said to be a class II sequence.

Example: The first two sequences in class I:

{a_n }:3,13,53,..., (a₁ = 3); {b_n }:7,29,117,..., (b₁ = 7).

The first two sequences in class II:

{a_n }:1,5,21,85,..., (a₁ = 1); {b_n }:9,37,149,..., (b₁ = 9).

Theorem 2.1: All terms (odd numbers) in all class I sequences and all class II sequences contain all odd numbers ; and each odd number must be in the only one of these sequences.

Proof: Notice that all odd numbers can be classified into three classes： class A, class B and class C. Any odd number in class A must be the first term a₁ of some definite class I sequence {a_n}; similarly,any odd number in class B must be the first term a₁ of some definite class II sequence {a_n}. So just prove that all terms of all sequences in class I and class II contain any class C odd number of the form 4n+1 (n is an odd number), and any class C odd number must be in the only one of these sequences.

Let p = 4n₁ + 1 be any definite class C odd number , where n₁ is a definite odd number. But all odd numbers can be classified into three classes A,B,C.

Case 1: If n₁ is a class A odd number, then n₁ is the first term a₁ of some definite class I sequence {a_n }; by the construction of a class I sequence, p = 4n₁ + 1 is the second term a₂ of this definite class I sequence {a_n}; similarly, if n₁ is a class B odd number, then p = 4n₁ + 1 is the second term a₂ of some definite class II sequence {a_n }.

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

Case 3: If n ₂ is some class C odd number, then we are back to the beginning of case 2. Continuing, since p = 4n₁ + 1 is a definite class C odd number, and n₂ < n₁ , 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 , and n_k must be some definite class A (or class B) odd number, which is the first term a₁ of some definite class I (or class II) sequence ,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 class I (or class II) sequence . ■

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

Definition 3.1: An odd number of the form 6n-1 (n = 1,2,3,...) is called a class D odd number , 6n+1 (n = 0,1,2,3,...) is called a class E odd number , and 6n-3 (n = 1,2,3,...) is called a class F odd number. Since 6n-3 is divisible by 3, a class F odd number is also called triple odd number.

Lemma 3.1: The class A odd number 4n-1 is a pre odd number of the class D odd number

6n-1, or 6n-1 is the post odd number of 4n-1,where n = 1,2,3,....

Proof: Since 3(4n-1) + 1 = 2(6n-1). ■

Notice that the class B odd number 4n+1 (n is 0 or even) can again be written as 8n+1(n = 0,1,2,...).

Lemma 3.2: The class B odd number 8n+1 is a pre odd number of the class E odd number

6n+1, or 6n+1 is the post odd number of 8n+1,where n = 0,1,2,3,....

Proof: Since 3(8n+1) + 1 = 2² (6n+1). ■

Lemma 3.3: The triple odd number 6n-3 cannot be a post odd number of any odd number; in other words, the class F odd number 6n-3 have no pre odd number,where n = 1,2,3,....

Proof: Suppose that 6n-3 is a post 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. ■

Consider the class I and class II sequences given in §2.

Theorem 3.4:For all class I sequences {a_n } (where a₁ = 4k-1, a_n = 4a_n-1 + 1(n = 2,3,4,...), where k is a definite positive integer) and all the class D odd numbers 6k-1 (here k is the same as above) there exists a 1-1 mapping according to the pre and post odd number relationship; and the class I sequence {a_n } with a₁ = 4k-1 corresponds to the class D odd number 6k-1, while each term of this class I sequence {a_n } is a pre odd number of this class D odd number 6k-1, and 3a_n + 1 = 2^2n-1(6k-1) (n = 1,2,3,...).

Proof: By Lemma 3.1, for a definite positive integer k,the first term 4k-1 of the class I sequence {a_n } with a₁ = 4k-1 is the pre odd number of the class D odd 6k-1; then by Corollary 1.2, each term of the class I sequence {a_n } is a pre odd number of the class D odd number 6k-1 ,and 3a_n + 1 = 2^2n-1 (6k-1)(n = 1,2,3 ...); then by Theorem 1.3, each term of this class I sequence {a_n } has a unique post odd number 6k-1.So, according to the relationship between the pre and post odd numbers, the class I sequence {a_n } with a₁ = 4k-1 and the class D odd number 6k-1is a 1-1 mapping,where k = 1,2,3,.... ■

Example: Take k = 1, then a₁ = 3, {a_n }:3,13,53,213,...; each term of this class I sequence {a_n } is a pre odd number of the class D odd number 5.

Theorem 3.5: For all class II sequences {a_n } (where a₁ = 8k+1, a_n = 4a_n-1 + 1(n = 2,3,4,...), where k is 0 or a definite positive integer) and all class E odd number 6k+1 (here k is the same as above) there exists a 1-1 mapping according to the pre and post odd number relationship; and the class II sequence {a_n } with a₁ = 8k+1 corresponds to the class E odd number 6k+1, while each term of the class II sequence {a_n } is the pre odd number of this class E odd number 6k+1,and 3a_n +1 = 2²ⁿ (6k+1)(n = 1,2,3,...).

Proof: As proved in Theorem 3.4, the result can be obtained from Lemma 3.2, and Corollary 1.2 and Theorem 1.3 of Section 1.■

Example: Take k = 0, then a₁ = 1,{a_n }:1,5,21,85,...; each term of this class II sequence {a_n } is a pre odd number of the class E odd number 1.

Theorem 3.6: Let p be an odd number and q = 4p+1,then

(1) If p is a class E odd number , then q is a class D odd number ;

(2) If p is a class D odd number, then q is a class F odd number ；

(3) If p is a class F odd number, then q is a class E odd number .

Proof: (1) Since p is a class E odd number, 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 is a class D odd number;

(2) Since p is a class D odd number, 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 is a class F odd number;

(3) Since p is a class F odd number , 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 is a class E odd number.■

Remarks: In the class I sequence {a_n }, a₁ = 4k-1 is a class A odd number; and all class A odd numbers are {4k-1}:3, 7, 11, 15, 19, 23, ...; where 3, 15, ... are class F odd numbers, 7, 19, ... are class E odd numbers, and 11, 23, . ... are class D odd numbers.

In the class II sequence {a_n }, a₁ = 8k+1 (k = 0,1,2,...) is a class B odd number; and all class B odd numbers are {8k+1}:1, 9, 17, 25, 33, 41,...; where 9, 33,... are class F odd numbers, 1, 25,. ... are class E odd numbers, and 17, 41, ... are class D odd numbers.

That is, the first term a₁ can be any one of the three odd classes D,E,F, whether the sequence {a_n} is a class I or a class II .

Theorem 3.7: In any one of class I and class II sequence {a_n},

If a₁ is a class E odd number, then a₂ is a class D odd number, a₃ is a class F odd number, and a₄ is a class E odd number,etc. The subsequent terms cycle in this manner;

If a₁ is a class F odd number,then loop by F,E,D,F,E,D;

If a₁ is a class D odd number,then loop by D,F,E,D,F,E.

Proof: This is a direct corollary of Theorem 3.6.■

Remark: If the first term is excluded from the class I and class II sequence, then the other terms are all class C odd numbers, that is, odd numbers of the form 4n+1 (n is odd). According to Theorem 3.7, these terms can also be any one of the three classes D,E,F.

The Collatz sequence now under discussion has omitted even numbers, so the following definition is given.

Definition 3.2: If p_k ,p_k-1 ,...,p₁ ,1 is a Collatz sequence with k odd numbers other than 1, then we say that p_k is k steps successful to 1; and it is specified that the odd number 1 is 1 step successful to 1.

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

Corollary 3.8: If p is a pre odd number of q ,q ≠ 1 and q is k steps successful to 1, then p is k+1 steps successful to 1;if p is a pre odd number of 1,then p is 1 step successful to 1.

Proof: This is a direct consequence of Definition 3.2.■

In the following we construct a set H of sequences by recursive method.

By Theorem 3.5, the class E odd number 1 corresponds to the class II sequence {a_n }:1, 5, 21, 85, 341, .... Put this sequence into a sequence set H₁. There is exactly one sequence in H₁.

In this sequence of H₁ , since a₁ = 1 is a class E odd number, by Theorem 3.7,a_3k-2 is

a class E odd number, a_3k-1 is a class D odd number, and a_3k is a class F odd number, where k = 1,2,3 ,.... By Lemma 3.3, any class F odd number has no pre odd number. By Theorem 3.4, any class D odd number corresponds to a unique class I sequence; by Theorem 3.5, any class E odd number corresponds to a unique class II sequence. We take all the class I sequences and all the class II sequences corresponding to all the class D odd numbers and all the class E odd numbers of this sequence in H₁ (removing the class E odd number 1) to form a sequence set H₂.

(Note: In H₂ ,the removal of the class II sequence corresponding to class E odd number 1 is to prevent the sequence that appeared in H₁ from reappearing in H₂ .)

Assume that the sequence set H_n has been formed.

Then all the class I sequences and all the class II sequences corresponding to all the class D odd numbers and all the class E odd numbers of each sequence in H_n form a set H_n+1 .

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

Write H = ${\bigcup }_{n=1}^{\infty }H$_n

Remark: To get an intuitive sense of the composition of the set H, several sequences in H₁ and H₂ are given here according to Theorem 3.4 and Theorem 3.5.

There exists unique class II sequence {a_n } in H₁ corresponding to the.class E odd number 1.{a_n }:1,5,21,85,341,1365,...

The sequences in H₂ correspond to all class D odd numbers and all class E odd numbers of the above sequence {a_n } in H₁.In the above sequence,a₁ = 1 is a class E odd number, which corresponds to just above sequence {a_n}. To avoid repetition, do not put it in H₂; a₂ = 5 is a class D odd number, which corresponds to the class I sequence {b_n }:3,13,53,213,...,put it in H₂; a₃ = 21 is a class F odd number (triple odd number), which does not exist any pre odd number; a₄ = 85 is a class E odd number, which corresponds to the class II sequence {b_n }:113,453,1813,...,put it in H_2.; a₅ = 341 is a class D odd number, which corresponds to the class I sequence {b_n }:227,909,3637,...,put it into H₂ ; a₆ = 1365 is a class F odd number, there is no pre odd number; and so on.

Theorem 3.9: (1) There are no identical sequences in the above sequence sets H₁ ,H₂ ,...,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 to 1, where n = 1, 2, 3, ....

(3) If the odd number p is successful to 1, then p must be in some sequence of the sequence set H.

Proof: (1) Induction. First the sequence of H₁ is the class II sequence corresponding to the class E odd number 1.All the sequences of H₂ are all the class I sequences and all the class II sequences corresponding to all the class D odd numbers and all the class E odd numbers in H₁ except for the class E odd number 1. The class E odd number 1 corresponding to the sequence in H₁ and all the class D odd numbers and all the class E odd numbers corresponding to these sequences in H₂ are just all the class D odd numbers and all the class E odd numbers in H₁, all of which are not identical to each other, and from Theorem 3.4 and Theorem 3.5, all the sequences in H₁ and H₂ are not identical to each other.

Then since all the sequences in H₁ and H₂ are class I and class II sequences, by Theorem 2.1, different sequences will not have the same term.

Suppose all the sequences in H₁ ,H₂ ,...,H_n are different from each other and no two different sequences have the same term.

Consider H_n+1 . Since all Class I and all Class II sequences in H₁,H₂ ,...,H_n correspond to all Class D odd and all Class E odd numbers that appear in H₁ ,H₂ ,...,H_n-1 (including the Class E odd number 1 corresponding to the sequence in H₁ ); and all Class I and all Class II sequences in H_n+1 correspond to all Class D odd and All Class E odd numbers that appear in H_n ; according to the assumption of induction, no two different sequences in H₁ ,H₂ ,...,H_n have the same term; that is, these Class D odd numbers and Class E odd numbers of each sequence in H_n do not appear in the sequences of H₁ ,H₂ ,...,H_n-1 . So by Theorem 3.4 and Theorem 3.5, all the sequences of H_n+1 and any of the sequences of H₁,H₂ ,...,H_n will not be the same; that is, all the sequences of H₁ ,H₂ ,...,H_n ,H_n+1 are not the same as each other. By Theorem 2.1, no two sequences in H₁,H₂,...,H_n ,H_n+1 have the same term.

(2) Induction. First of all, all the terms of the sequence in H₁ are the pre odd numbers of class E odd number 1. By definition 3.2,all the terms of the sequence in H₁ are 1 step successful to 1 .

Suppose all terms of any sequence in H_n are n steps successful to 1 .

Consider H_n+1 . Notice that any sequence {a_n } in H_n+1 corresponds to some class D (or class E) odd number p of a sequence in H_n , that is, all the terms of the sequence {a_n } are the pre odd numbers of that odd number p. Since the odd number p is in a sequence in H_n , by induction hypothesis, the odd number p is n steps successful to 1 ; by Corollary 3.8, all the terms of the sequence {a_n } are n+1 steps successful to 1.

(3) Since the odd number p is successful to 1, then there must exist some positive integer n,such that p is n steps successful to 1, so it is enough to prove that p must be in some sequence of H_n.

Induction. First look at n = 1. Let the odd number p be 1 step successful to 1.Note that the odd number p must be in some definite class I (or class II) sequence. By Theorem 3.5, for the class E odd number 1,it corresponds to some class II sequence {a_n }, where all the terms of the class II sequence {a_n } are the pre odd numbers of the class E odd number 1. Then by Theorem 3.4 and Theorem 3.5,all the class D (or class E) odd numbers corresponding to all other class I (or class II) sequences cannot be 1; that is,any term in all other class I (or class II) sequence cannot be a pre odd number of 1.In the other words,if the odd number p is in any other sequence excluding {a_n }, the odd number p cannot be 1 step successful to 1; so the odd number p can only be in the class II sequence {a_n } corresponding to the class E odd number 1, and the class II sequence {a_n } is exactly the one in H₁. Therefore, the result holds for n = 1.

Assume that the result holds for n = 2,3,...,k; that is, if the odd number p is 2 steps,3 steps,...,k steps successful to 1 respectively, then p is in some sequence of H₂ ,H₃ ,...,H_k respectively.

Now let n = k+1; that is, the odd number p is k+1 steps successful to 1. Let Collatz 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 to 1; by inductive assumption, p_k is in some sequence of H_k . For simplicity, let p_k be in some class I sequence {b_n } of H_k . (If p_k is in some class II sequence {b_n }of H_k , then the same treatment follows.) Since all the terms of the sequence {b_n } have class D, class E, and class F odd numbers, that p_k cannot be a class F odd number.(Because p is a pre odd number of p_k , and any class F odd number has no pre odd number;) Therefore, p_k is some definite class D (or class E) odd number in {b_n }. Let’s assume p_k is a definite class D odd number in {b_n }. Then, by the construction of H_k+1, the class I sequence corresponding to the class D odd number p_k is some definite sequence {c_n } in H_k+1 , and all terms of {c_n } are pre odd numbers of the odd number p_k . Since this is an 1-1 mapping, the class D (or class E) odd number corresponding to any class I (or class II) sequence other than the class I sequence {c_n } cannot be p_k. That is, any term in any other class I (or class II) sequence other than {c_n } cannot be the pre odd number of p_k . Therefore, the pre odd number p of p_k can only be in this sequence {c_n } of H_k+1 . Therefore, the result holds when n = k + 1.■

Remark: From Theorem 3.9, it follows that every term of every sequence in H = ${\bigcup }_{n=1}^{\infty }H$_n is successful to 1; and every odd number that is successful to 1 must be in one of these sequences of H. Therefore, to prove that all odd numbers are successful to 1, it is sufficient to show that every odd number is in one of the sequences of H. For this purpose, the following lemmas are given firstly.

Lemma 3.10: (1) If p is a pre odd number of q and the odd number q is not successful to 1, then the odd number p is also not successful to 1;

(2) If r is a post odd number of q and the odd number q is not successful to 1, then the odd number r is also not successful to 1.

Proof: (1) First, since q is not successful to 1, q ≠ 1. Assuming that the odd number p is successful to 1,then there exists a positive integer k,such that p is k steps successful to 1, and by Theorem 1.3, q is the unique post odd number of p. Then, by Definition 3.2, k > 1, and q is k-1 steps successful to 1. Contradiction.

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

By Theorem 2.1, any odd number is a term of some class I (or class II) sequence.

Lemma 3.11: If an odd number p is not successful to 1, then all terms of the class I (or class II) sequence {a_n },in which the odd number p is located ,are not successful to 1.

Proof: From Theorem 3.9, it follows that all odd numbers in the sequence set H are successful to 1; since the odd number p is not successful to 1, p is not in H.Let the odd number p be a term in some definite class I(or class II) 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 3.9, all terms of {a_n } are not successful to 1.■

As with the construction of the sequence set H, the following series G_n of sequence sets is constructed.

First put a class I (or class II) sequence{a_n } into a sequence set, denoted as G₁₁ ; G₁₁ contains only one sequence {a_n }.

As before, there are infinitely many class D, class E and class F odd numbers in {a_n }, and then all class I sequences and all class II sequences corresponding to all class D odd numbers and all class E odd numbers of the sequence {a_n } in G₁₁ form a sequence set G₁₂ .

Suppose that the sequence set G_1n has been formed.

Then all the class I sequences and all the class II sequences corresponding to all the class D odd numbers and all the class E odd numbers of all sequences of G_1n form a sequence set G_1,n+1.

Denote G₁ = $\bigcup _{n=1}^{\infty }G$_1n

Lemma 3.12: If some term in this sequence {a_n } of G₁₁ is not successful to 1,then

(1)There are no identical sequences in the sequence set G₁ ;and no two different sequences in G₁ have the same term.

(2) Any term of all sequences in the set G₁ is not successful to 1.

Proof: (1) First, following the method proved in Theorem 3.9(1), it is obtained that there are no identical sequences in the above sequence sets G₁₁ ,G₁₂ ,...,G_1n ,..; and no two different sequences have the same term.

(2) Induction for G_1n. First of all, by Lemma 3.11,any term of the sequence {a_n } in G₁₁ is not successful to 1. So the result holds when n = 1.

Assume that the result holds for n = k, i.e., that any term of any sequence in G_1k is not successful to 1.

Consider G_1,k+1 . Let the sequence {b_n } be any definite class I (or class II) sequence in G_1,k+1 , then {b_n } corresponds to some class D (or class E) odd number p of some definite sequence in G_1k. And all terms of {b_n } are the pre odd numbers of that odd number p. By induction hypothesis, any term of any sequence in G_1k is not successful to 1. So the odd number p is also not successful to 1. And by Lemma 3.10,any term in {b_n } is not successful to 1. So the result holds for n = k+1.■

Starting from the sequence set G₁ above, we will construct the sequence set G_n ，where n = 2,3,4，...

Start with the construction of G₂ . Use the recursive method. First look at G₁₁ in G₁, there is only one class I (or class II) sequence in G₁₁ , which is denoted as {a_n }. Let the class D (or class E)odd number corresponding to this class I (or class II) sequence {a_n } be p, then each item in {a_n } is a pre odd number of the odd number p. Since every odd number must be in some definite class I (or class II) sequence, let the class I (or class II) sequence containing the odd number p be {b_n }. Put the class I (or class II) sequence {b_n } into a sequence set, denoted G₂₁ . G₂₁ contains only one sequence {b_n }.

At this point, there are infinitely many class D,class E and class F odd numbers in the sequence {b_n }, and the above odd number p is just one of the class D (or class E) odd numbers in {b_n }. Take all the class I sequences and all the class II sequences corresponding to all the class D odd numbers and all the class E odd numbers of the sequence {b_n } in G₂₁ ,and form a sequence set G₂₂ . (Note: the sequence {a_n } in G₁₁ is exactly that sequence in G₂₂ corresponding to the odd number p in the sequence {b_n } of G₂₁.)

Suppose that the sequence set G_2k has been formed.

Then all class I sequences and all class II sequences corresponding to all class D odd numbers and all class E odd numbers of all sequences in G_2k form the sequence set G_2,k+1. Write G₂ = $\bigcup _{k=1}^{\infty }G$_2k.

So on and so forth.

Suppose again that G_n = $\bigcup _{k=1}^{\infty }G$_nk has been constructed.

At this point, G_n1 has only one class I (or class II) sequence, denoted {c_n }. Let the class D (or class E) odd number corresponding to this class I (or class II) sequence {c_n } in G_n1 be q. Then each item in {c_n } is a pre odd number of the odd number q. Let the class I (or class II) sequence containing the odd number q be {d_n }. Now put that class I (or class II) sequence {d_n } into a sequence set, denoted G_n+1,1 ; G_n+1,1 has only one sequence {d_n }.

There are infinitely many class D, class E and class F odd numbers in {d_n }, and the odd number q above is just one of these class D (or class E) odd numbers. Put all class I sequences and all class II sequences corresponding to all class D odd numbers and all class E odd numbers of the sequence {d_n } in G_n+1,1 into a sequence set,denoted G_n+1,2 . (Note that the sequence {c_n } in G_n1 is exactly the sequence of G_n+1,2 corresponding to the odd number q in the sequence {d_n } of G_n+1,1.)

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

Then all the class I sequences and all the class II sequences corresponding to all the class D odd numbers and all the class E odd numbers of all sequences of G_n+1,k form a sequence set G_n+1,k+1.

Denote G_n+1 =$\bigcup _{k=1}^{\infty }G$_n+1,k.

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

Lemma 3.13: If any term of any sequence in the sequence set G₁ is not successful to 1, then

(1) for n = 2,3,4,..., there are no identical sequence in the sequence set G_n constructed above, and no two different sequences have the same term;

(2) For n = 2, 3, 4, ..., any term of any sequence in the sequence set G_n is not successful to 1;

(3) G₁ ⊂ G₂ ⊂ ... ⊂ G_n ⊂ ...

Proof: (1) For n = 2,3,4,..., notice that the construction method of G_n is exactly the same as that of G₁ and also exactly the same as that of H. Therefore, as proved in Theorem 3.9(1), the result holds;

(2) Since any item of any sequence in G₁ is not successful to 1, then every item of the unique sequence {a_n } in G₁₁ is not successful to 1. By the construction of G₂ , the sequence {a_n } corresponds to the odd number p, and every item of {a_n } is the pre odd number of the odd number p. Therefore, by Lemma 3.10(2), the odd number p is not successful to 1. And by the construction of G₂ , the odd number p is in the class I (or class II) sequence {b_n }of G₂₁ ; by Lemma 3.11, all terms in {b_n } are not successful to 1. And as proved in Lemma 3.12, any term of any sequence in the set G₂ is not successful to 1.

By analogy, it is clear that any term of any sequence in the set G_n is not successful to 1,where n = 3,4,5,...

(3) First prove that G₁ ⊂ G₂ . Looking back at the construction of G₂ , G₁₁ has the unique class I (or class II) sequence {a_n }, which corresponds to a class D (or class E) odd number p. And p is in the unique class I (or class II) sequence {b_n } of G₂₁ . While all class I sequences and all class II sequences corresponding to all class D odd numbers and all class E odd numbers of the sequence {b_n } in G₂₁ form the sequence set G₂₂ . Because p is in {b_n }, so the sequence {a_n } corresponding to p is in G₂₂, that is, G₁₁ ⊂ G₂₂ . Then by the generation of G₁₂ and G₂₃ it follows that since G₁₁ ⊂ G₂₂ , 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. By analogy, G₂ ⊂ G₃ ⊂ ... ⊂ G_n ⊂ ....■

Remarks: (1) From the construction of G₂ , it is clear that the class D (or class E) odd number p corresponding to the class I (or class II) sequence {a_n } in G₁₁ is just a general odd number in the sequence {b_n } of G₂₁ , and its corresponding sequence {a_n } grows in the pre odd number direction (let's say) to obtain a G₁ . While there are infinitely many class D and infinitely many class E odd numbers in the sequence {b_n } of G₂₁ ,and G₂ is obtained by growing the infinitely many class D and class E odd numbers in {b_n } in the pre odd number direction. Where each class D (or class E) odd number 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 .

(2)The difference between sequence set G_n and the sequence set H is that: all terms of

all sequences in H are successful to 1, so when they grow in the post odd direction (let's say ), they stop at odd number 1; while all terms of all sequences in G_n are not successful to 1, so as n increases infinitely, G_n is endlessly expanding. From Theorem 3.9, 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, 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， $\underset{n\to \infty }{\lim }G$_n is an indeterminate set of odd numbers.

Theorem 3.14: Any odd number is successful to 1.

Proof: Let the set of all odd numbers be Q.Consider H as a set of odd numbers. By Theorem 3.9 , the set of all odd numbers successful to 1 is H. Suppose there is an odd number ,that is not successful to 1. And let the set of all odd numbers not successful to 1 be G. then G ∩ H = Ø,and G ∪ H = Q.

From Lemma 3.12 and Lemma 3.13, if there is an odd number that is not successful to 1, 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 to 1. And from the above remark

(2),when n tends infinity, $\underset{n\to \infty }{\lim }G$_n is an indeterminate set of odd numbers. Because G_n

⊂ G,where n = 1,2,3,..., so $\underset{n\to \infty }{\lim }G$_n ⊂ G,and the set G of all odd numbers, those are not successful to 1 ,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 to 1.■

Remarks:(1)Treat each odd number in the set H as a vertex,and then connect an edge between any two odd numbers (vertices) in H that have a pre and post 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 pre odd number,each odd number in the triple odd number set F is a leaf of this H-tree.

(2)For G_n,where n= 1,2,3,...,as above,each odd number in G_n is treated as a vertex,and an edge is connected between any two odd numbers in G_n that have a pre and post odd number relationship,then $\underset{\mathit{n}\to \infty }{\mathbf{lim}}\mathit{G}$_n is an unrooted tree.According to the proof of Theorem 1 in reference [4], the number of vertices in $\underset{\mathit{n}\to \infty }{\mathbf{lim}}\mathit{G}$_n is uncountable,it is impossible,so G = Ø.

Theorem 3.15: Any positive integer is successful to 1 , i.e., Collatz conjecture holds.

Proof: See the description at the top part of § 1 Introduction.■