A Proof of Collatz Conjecture

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 = ${\bigcup }_{n=1}^{\infty }H$_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 = ${\bigcup }_{n=1}^{\infty }H$_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 ${\bigcup }_{k=1}^{n}H$_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 ${\bigcup }_{k=1}^{n+1}H$_k ={ƒ(p)| p ∈ H₁ ∩(D ∪ E）}∪{ƒ(p)| p ∈ H₂ ∩ (D ∪ E)} ∪...∪{ƒ(p)| p ∈ H_n ∩ (D ∪ E)} = {ƒ(p)| p ∈ (${\bigcup }_{k=1}^{n}H$_k)∩(D∪E)}.By the inductive assumption,no two different sequences in ${\bigcup }_{k=1}^{n}H$_k have the same term.So,there is no same odd number in (${\bigcup }_{k=1}^{n}H$_k)∩(D∪E).Since ƒ is a bijection,all sequences in ${\bigcup }_{k=1}^{n+1}H$_k are not identical to each other.By Theorem 3.1,no two different sequences in ${\bigcup }_{k=1}^{n+1}H$_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.■