On the Collatz Conjecture

The Collatz conjecture (or Syracuse conjecture) states: all Syracuse sequences converge to 1. We present a Syracuse sequence, and we prove that the conjecture is true, first by using the fact that all convergent integer sequences are eventually constant. We then prove wrong 2 hypotheses: the case where the sequence tends to infinity, and the case where the sequence has no limit and is eventually periodic. We conclude by elimination, afterward


Introduction
Let's consider natural numbers, which set is ℕ.We define a sequence of numbers built this way: given any non-zero natural number c, the next term is / 2 if c is even, (3 + 1)/2 if c is odd; repeatedly, we can build as many terms as we want.As a result, we have a sequence of odd and even numbers, with c as the initial term.We call it a Syracuse sequence.Equivalently, one can introduce another sequence of odd numbers only, with c as the initial term if c is odd.If c is even, the initial term is obtained after dividing c by 2 enough times to obtain an odd number, say r times and the initial term is  ′ =  2  ,  non-zero natural number.If we consider a Syracuse sequence with c' as the initial term, the next term is  ′′ = 3 ′ +1 2  ; after c'' we have  ′′′ = 3 ′′ +1 2  , . . ., the exponents s and t being the necessary times we divide by 2 to obtain odd numbers c'' and c''', respectively.Generally, any Syracuse sequence (  ) ∈ℕ has its 2 consecutive odd terms linked by the following expression: 3  + 1 2   ;   non − zero natural number. (1) It's tempting to seek a general term of the sequence using (1).But we quickly notice in (1) that the exponent   is not predetermined, and its value depends on 3  + 1, more specifically   .
The approach developped in this paper is new, and a deep study of syracuse sequences will be necessary so that, the tools proposed match the needs for the resolution of the conjecture.
Since each   can be known only between 2 consecutive odd terms; and based upon the particular property of a Syracuse sequence that, any odd term multiple of 3 cannot have a previous term, first we will establish all possible expressions of any previous term and its next one.Then, we will use the concept of onion-factorization, to test the consistency of the expression of the initial term, when we extend the sequence infinitely.
TERMINOLOGY -In the following, we deal with only the odd terms of the Syracuse sequence.And we choose to refer to the odd terms by "terms" rather than "odd terms"; -We mean by previous terms in a Syracuse sequence, all the terms with the same rank n behind the term with rank n+1; by next term, we mean the term with rank n+1 after the term with rank n. n is an arbitrary natural number; -We mean by same form-previous terms, all previous terms with the same algebraic form 6m+1, 6m+3 or 6m+5, m natural number; -"Initial term" refers to the first term of a Syracuse sequence, while "initial previous term" refers to the first term of a sequence of previous terms.

The General Expression of Previous Terms
(  ) ∈ℕ is a Syracuse sequence.By using the relation (1), we can write By choosing any term  +1 , expression (4) helps find its previous one in the Syracuse sequence.Since   is an integer, 3 has to divide the numerator 2    +1 − 1.But, if any term has a next one in the Syracuse sequence, do all terms necessarily have previous ones?We quickly notice that if 2    +1 is a multiple of 3, that is,  +1 multiple of 3, then 3 cannot divide 2    +1 − 1, since it is a subtraction between a multiple of 3 and a non-multiple of 3.This leads us to a property verified by all Syracuse sequences: Property.Given any random term   ,  ≥ 1, in a Syracuse sequence, it doesn't exist a previous term  −1 if and only if   is a multiple of 3.
(5) Now, let's select some examples to see how the relation (4) works.Let 7; 7 is not a multiple of 3, and by trying which   suits, we find But, since we have tried at random, is there any other   that can work too?How many?
We find for   = 4: If we continue, we figure out different values of  −1 for   = 6,8,10, ..., all being previous terms of   = 7 in the Syracuse sequence.
In comparison, to determine the next term to 7, we have and   = 2 is the only choice here; 11 is then the unique term that follows 7 in the Syracuse sequence.
Let any  +1 in the Syracuse sequence with rank i+1.We note ( , ) ∈ℕ the monotonically increasing sequence of its previous terms: i is there to signify that behind  +1 in the Syracuse sequence, all the previous terms have the same rank i, and p is to put order among them.Let  ,0 the initial previous term, and  , the (p+1)th; we have the following expressions: b fixed; and for  ≥ 1,   >  > 0.
Since  +1 has previous terms, it can't be a multiple of 3.

VALUE OF b
b is the exponent to determine the initial previous term of a monotonically increasing sequence.It is then the minimum positive integer needed to obtain  ,0 .Then we can try in the expression of  ,0 the lower possible value of b: 1, then add 1 if the test fails ( ,0 is not a natural number), and try the new value of b, till the test is valid ( ,0 is a natural number).
We notice that such a condition on s is the same if we replace 6a+5 by 6a+1, even if in addition we choose the algebraic form 6m+5 or 6m+3 for the two previous terms.In fact, (11) in this case becomes 9( 2 −  1 ) = 2 × 4  1 (4  − 1)(6 + 1), and we still have the condition mentioned above on s.Also, nothing changes about s no matter the value of  1 .Finally, whatever the two previous terms are, it only requires them to have the same algebraic form, for the condition on s to be applied.

CONDITION ON s
The number s is such that ∑ is a multiple of 3.
-For  = 3, n non-zero natural number, let's prove that is always a multiple of 3.
n=1,  1 = 21 is a multiple of 3. Let's suppose   is a multiple of 3, and let's prove it at order n+1.
We have which is a multiple of 3. In conclusion,   is a multiple of 3.
Finally, s=3n, n non-zero natural number, is the only possibility that validates the condition.This means, between two consecutive same form-previous terms, the gap is s = 3.

Case 1: a=3k
The initial previous term is the second previous term is and the third Since s=3 between two consecutive same form-previous terms, we have in general: ,   ,   are general terms of monotonically increasing sequences (  ) ∈ℕ , (  ) ∈ℕ , (  ) ∈ℕ , respectively.Notice that the three sequences are sub-sequences of ( , ) ∈ℕ .

Theorem
Let's consider  0 the initial term of a Syracuse sequence (  ) ∈ℕ .We define  0 as a previous term.Then it exists an onion-factorization in ℚ of the initial term such that, the more we extend the sequence by calculating new terms, the more onion factors we can append.Each of the appended onion factors is a value of k, found in the expression of any previous term and its next one.

Proof:
The general relations imply that, for any term 6 + 1, 6 + 3, 6 + 5 in a Syracuse sequence,  is always equal to  + ,  non-zero and  natural numbers.
Since  0 is a previous term, the initial term is either   ,   or   , ∈ ℕ.The 2 nd term is naturally a next term, then is either 6 + 1 or 6 + 5,  ∈ ℕ, but is also a previous term to the third one.Except the initial term, every term in the Syracuse sequence is both a previous and a next term.
∀  ≥ 1,   is not expressed the same whether we consider it as a previous term or a next term.In fact, for any 6 + 1or 6 + 5 as a next term, is either 3 , 3 + 1 or 3 + 2; but as a previous term,  is expressed differently, and its expression tells straight on its parity.For instance we have for  odd: 6(2 + 1) + 1 and 6(2 + 1) + 5.