New Method for Solving the Collatz Conjecture 4 with High-Prize Money

    Please read "Introduction to a New Method in Chapter 1 (article 1)" before reading Chapter 3 (article 4).

   Chapter 1 also posts links to each chapter.

   When translating from Japanese to English, subtle nuances of the original articles may not be fully conveyed. If you would like to view the original Japanese articles, please see Chapter 1 of Japanese Article 1.

   Summary Video of the New Method in English

Chapter 3: Classification of Odd Numbers (article 4)   

   This article (the fourth in the series) is independent of the other chapters and covers content that will be familiar to readers who are already working on solving the Collatz conjecture. It can be used as a review of the knowledge required to solve the conjecture. For readers who have just started working on the solution, this is essential knowledge that they must acquire.

   In order to explain the descriptions in Chapter 4 and beyond, it is necessary to explain the odd number classification according to the Collatz rule. The equation that represents this odd number classification is often seen on the net and in papers, and although the appearance may differ, they are similar equations (or there should be equations that look the same).

3.1 Classification of Odd Numbers (Nc and Ns Types) 

   An operation that triples an odd number and adds 1 is referred to as a 3C+1 transformation.

   An odd number after the 3C+1 transformation will become even.
The operation of repeatedly dividing an even number by 2 until an odd number is reached is referred to as a 3S+1 transformation.

   The steps only track odd numbers. For example, the sequence $${13 \to 5 \to 1}$$ is 2 steps (Chapter 1).

   An odd number $${No}$$ becomes even after the 3C+1 transformation. An even number can be expressed as a product of an odd number $${m_{o}}$$​ and a power of 2, $${2^{n}}$$, where $${n}$$ is a positive integer.

　　　　　$${(3.1)          3No+1=m_{o} \cdot 2^n }$$

   Congruence is not taught in high school mathematics (in Japan), but the following will be sufficient for this article.

   If $${m_{o}}$$​ leaves a remainder of 2 when divided by 3, define  $${m_{c}}$$​ as follows: $${m_{c}\equiv 2 (mod 3)}$$ or simply: $${m_{c}\equiv ２}$$

   Similarly, If $${m_{o}}$$​ leaves a remainder of 1 when divided by 3, define $${m_{s}}$$​ as follows: $${m_{s}\equiv 1 (mod 3)}$$ or simply: $${m_{s}\equiv 1}$$

   Additionally:
        $${1\cdot 1\equiv 1 　2\cdot 2\equiv 1 　2\cdot 1 \equiv 2 　1\cdot 2\equiv 2 }$$　
   For a positive integer $${p}$$, $${2^{2p} \equiv 1}$$、$${2^{2p-1} \equiv 2}$$

   Since $${3No+1 \equiv 1}$$ in (3.1), the right-hand side can only be a combination of $${1\cdot 1}$$ or $${2\cdot 2}$$. $${m_{o}}$$​ cannot be a multiple of 3.

   Classify $${No}$$ into $${Nc}$$ and $${Ns}$$ according to $${m_{c}}$$ and $${m_{s}}$$​, respectively.

　　　　　$${(3.2)          3Nc+1=m_{c} \cdot 2^{2p-1} }$$
　　　　　$${(3.3)          3Ns+1=m_{s} \cdot 2^{2p}}$$

   Furthermore, let $${m_{c}(\equiv 2)}$$ and $${m_{s}(\equiv 1)}$$ be represented by non-negative integers $${t_{c}}$$​ and $${t_{s}}$$ as follows:

　　　　　$${(3.4)          m_{c}=6t_{c}+5}$$
　　　　　$${(3.5)          m_{s}=6t_{s}+1}$$

   Substituting (3.4) and (3.5) into (3.2) and (3.3), respectively,

　　　　　$${(3.6)          3Nc+1=(6t_{c}+5) \cdot 2^{2p-1} }$$
　　　　　$${(3.7)          3Ns+1=(6t_{s}+1) \cdot 2^{2p}}$$

   Solving for $${Nc}$$ and $${Ns}$$,

　　　　　$${(3.8)          Nc(t_{c},p)=t_{c}2^{2p}+\cfrac{5 \cdot 2^{2p}-2}{6}}$$

　　　　　$${(3.9)          Ns(t_{s},p)=t_{s}2^{2p+1}+\cfrac{2^{2p}-1}{3}}$$

   The odd number $${No}$$ is classified into two types of odd numbers,
type $${Nc}$$ and type $${Ns}$$, parameterized by (t,p) (where $${t_{c}}$$ and $${t_{s}}$$​ are non-negative integers and can be simply written as $${t}$$).

3.2 Characteristics of Nc and Ns Types

   When $${p}$$ is set to values from 1 to 4, (3.8) and (3.9) are expressed as follows:

　　　　$${(3.10)          Nc=4t_c+3,\ 16t_c+13, \ 64t_c+53, \ 256t_c+213, \cdots}$$
　　　　$${(3.11)          Ns=8t_s+1,\ 32t_s+5, \ 128t_s+21, \ 512t_s+85, \cdots}$$

   The elements on the right-hand side are sometimes referred to as type $${4t+3}$$, type $${16t+13}$$, type $${8t+1}$$, and so on, depending on the case.

   From (3.6) and (3.7), (3.10) and (3.11) are as follows:

• $${Nc}$$ is an odd number sequence that reaches $${6t_{c}+5}$$ in one step.

• $${Ns}$$ is an odd number sequence that reaches $${6t_{s}+1}$$ in one step.

   Let $${\bm{cs(m_o)}}$$ denote an odd number sequence that reaches the odd number $${m_o}$$ in one step. Then, (3.10) and (3.11) become the following:

$${(3.12)          cs(6t_c+5)=\{4t_c+3,\ 16t_c+13, \ 64t_c+53, \ 256t_c+213, \cdots\}}$$
$${(3.13)          cs(6t_s+1)=\{8t_s+1,\ 32t_s+5, \ 128t_s+21, \ 512t_s+85, \cdots\}}$$

   For a simple example of (3.12) and (3.13), set $${t_{c}=t_{s}=0}$$:

　　　$${cs(5)=\{3,\ 13, \ 53, \ 213, \cdots\}}$$
　　　$${cs(1)=\{1,\ 5, \ 21, \ 85, \cdots\}}$$

   (Familiar) odd number sequences that reach 1 or 5 in one step have appeared. There are an infinite number of elements.

   It can be seen that there are odd number sequences with an infinite number of elements that can display all odd numbers except multiples of 3
with $${6t+1}$$ and $${6t+5}$$.

   From another perspective, the odd number sequences on the right-hand side of (3.12) are odd number sequences with the same $${t_{c}}$$​, even though they have different $${p}$$. They all reach the odd number $${6t_c+5}$$ in one step, which depends only on $${t_{c}}$$.

   Regardless of the value of$${p}$$, if the value of $${t_{c}}$$​ is the same, they all reach the same value, $${6t_c+5}$$, in one step. The same applies to $${t_{s}}$$​.

   The odd number sequences on the right-hand sides of (3.12) and (3.13) can represent all positive odd numbers.
   Only the odd numbers of the type $${4t+3}$$ become larger after the 3S+1 transformation, becoming $${6t+5}$$. All other odd numbers become smaller after the 3S+1 transformation.

   Since all odd numbers are either of the type $${4t+3}$$ or $${4t+1}$$, half of the odd numbers will become larger after the 3S+1 transformation, and the other half will become smaller.

   It can be shown that there is only 1 that returns to the original odd number in one step:
               In (3.1), let $${m_{o}=No}$$. Then,
              $${3No+1=No \cdot 2^n }$$
              $${(2^n-3)No=1}$$
               For the left-hand side to be equal to 1,
              $${2^n-3=1}$$ and$${No=1}$$
               Therefore, only $${n=2}$$ and $${No=1}$$ satisfy the conditions.

   Let's take a closer look at (3.12) and (3.13).
   These two equations represent all the number sequences (positive odd number sequences) that follow the Collatz rule (triple an odd number and add 1, and add 2 to an even number) in the two equations (3.12) and (3.13) (even numbers can be ignored).

   As mentioned above, the elements on the right-hand sides of the two equations (3.12) and (3.13) together represent all odd numbers. All odd numbers correspond to one of these elements.

   Substitute various values for $${t}$$ into the right-hand sides of the two equations and find the odd numbers on the left-hand sides that can be reached in one step. At this point, all multiples of 3 are eliminated.

   The numbers reached in one step are odd numbers of the type $${6t+5}$$ or $${6t+1}$$. Since they are odd, they must be one of the elements on the right-hand sides of equations (3.12) and (3.13) (returning again). However, they are different from the original $${t}$$.

   Repeating this process creates an odd tree (tree diagram). This will be referred to as the number sequence of the (general) Collatz space.  The word "general" may be included after Chapter 4 to distinguish it from another space.

   Let's take $${17}$$ as the initial value.
Since $${17=8\cdot 2+1}$$, it is of the type $${8t+1}$$ with $${t=2}$$.
This moves to $${6t+1}$$ in one step, and the value becomes $${13}$$.

   However, the odd numbers that move to $${13}$$ are all odd numbers on the right-hand side of equation (3.13) with the same $${t}$$, and are $${17, 69, 277,\cdots}$$ when $${t=2}$$ is substituted. These all move to the same $${13}$$.。

   Next, since $${13=16\cdot 0+13}$$, it belongs to the type $${16t+13}$$ with $${t=0}$$. The right-hand side of equation (3.12) with $${t=0}$$ is $${3,13,53,\cdots}$$ and these all move to $${6\cdot 0+5=5}$$. This is repeated.

   This is how the tree of all odd numbers in the Collatz space is created. However, whether or not it eventually reaches 1 is a separate issue.

   Starting from a particular odd number, the Collatz process may end up far away or return to the original number. The Collatz conjecture states that all positive odd numbers will eventually reach 1, without getting stuck in a loop or going to infinity. (Since all even numbers become odd numbers under the Collatz rule, if all positive odd numbers reach 1, then all positive integers will reach 1.)

   Here is a summary of the classification and properties of odd numbers.
For readers who have started working on the Collatz conjecture, please feel free to leave a comment in the comments section if you have any additional properties.

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Variables, Definitions, and Functions up to Chapter 3:

$${N}$$, $${No}$$,
total number of steps, step number, CS-vibration, C-transformation,
CS-plot, CS-linear-equations, 3C+1 transformation, 3S+1 transformation,
(general) Collatz space,
type $${6t+5}$$, $${6t+1}$$, $${4t+3}$$, $${8t+1}$$, etc.
$${c(No)}$$, $${z_{s}}$$, $${f_{c}(x)}$$, $${f_{s}(x)}$$, distance$${D}$$,
$${Nc}$$, $${Ns}$$, $${m_o}$$, $${m_{c}}$$, $${m_{s}}$$, $${t_{c}}$$, $${t_{s}}$$, $${t}$$, $${cs(m_o)}$$

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