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<h1 style="font-family:'Arial Black', sans-serif; text-align: center;">An Algebraic Structure of <br> 3n+1 Conjecture</h1>
$$ \text{Jimgo ,October 2021} $$
<section>
<h2>Abstract</h2>
<p>This paper studies the ternary operation of the 3n+1 Conjecture, proposes a new type of algebraic structure, and finds an equivalent conjecture </p>
</section>
<!-- =================================第一章=================================================== -->
<section>
<h2>1. Introduction</h2>
<p>3n+1 Conjecture is also known as the Collatz conjecture. The conjecture can be summarized as follows. Take any positive integer \(n\), do the following
operation:</p>
$$
f(n) =
\begin{cases}
n/2, & \text{ if } n \text{ is even} \\
(3n+1)/2, & \text{ if } n \text{ is odd}
\end{cases}
$$
<p>After servel times, it will always eventually goes into the infinite loop of \([2,1,2,1,\ldots]\)</p>
</section>
<!-- =================================第二章=================================================== -->
<section>
<h2>2 Ternary Calculation</h2>
<h3>2.1 Ternary</h3>
<p>For any positive integer represented by a decimal, which has its unique ternary representation, e.g. \(15_{(10)}
→ 120_{(3)}\) .</p>
<table class="example" id="exp1">
<caption><b>Example 1</b> Ternary process of 15</caption>
<tbody>
<tr>
<th>Decimal</th>
<th>Ternary</th>
</tr>
<tr>
<th><tt>15</tt></th>
<td><tt>120</tt></td>
</tr>
<tr>
<th><tt>23</tt></th>
<td><tt>212</tt></td>
</tr>
<tr>
<th><tt>35</tt></th>
<td><tt>1022</tt></td>
</tr>
<tr>
<th><tt>53</tt></th>
<td><tt>1222</tt></td>
</tr>
<tr>
<th><tt>80</tt></th>
<td><tt>2222</tt></td>
</tr>
<tr>
<th><tt>40</tt></th>
<td><tt>1111</tt></td>
</tr>
<tr>
<th><tt>20</tt></th>
<td><tt>202</tt></td>
</tr>
<tr>
<th><tt>10</tt></th>
<td><tt>101</tt></td>
</tr>
<tr>
<th><tt>5</tt></th>
<td><tt>12</tt></td>
</tr>
<tr>
<th><tt>8</tt></th>
<td><tt>22</tt></td>
</tr>
<tr>
<th><tt>4</tt></th>
<td><tt>11</tt></td>
</tr>
<tr>
<th><tt>2</tt></th>
<td><tt>2</tt></td>
</tr>
<tr>
<th><tt>1</tt></th>
<td><tt>1</tt></td>
</tr>
</tbody>
</table>
<h3>2.2 While \(n\) is Even</h3>
<p>A number is divided by 2 in ternary,each digit changes in the following three cases：</p>
$$ 0 → 0 ; 1 → 0 ;2 → 1 $$
<table class="example" id="exp2">
<caption><b>Example 2</b> Divide by 2 without borrow</caption>
<tbody>
<tr>
<td>Decimal</td>
<td>Ternary</td>
<td>Borrow</td>
</tr>
<tr>
<td><tt>20</tt></td>
<td><tt>202</tt></td>
<td><tt>0</tt></td>
</tr>
<tr>
<td><tt></tt></td>
<td><tt><b>1</b>02</tt></td>
<td><tt>0</tt></td>
</tr>
<tr>
<td><tt></tt></td>
<td><tt>1<b>0</b>2</tt></td>
<td><tt>0</tt></td>
</tr>
<tr>
<td><tt>10</tt></td>
<td><tt>10<b>1</b></tt></td>
<td><tt>0</tt></td>
</tr>
</tbody>
</table>
<p>While digit is 1,it need borrow till one digit is 1:</p>
$$ 0 → 1 ; 1 → 2 ;2 → 2 $$
<table class="example" id="exp3">
<caption><b>Example 3</b> Divide by 2 with borrow</caption>
<tbody>
<tr>
<td>Decimal</td>
<td>Ternary</td>
<td>Borrow</td>
</tr>
<tr>
<td><tt>34</tt></td>
<td><tt>1021</tt></td>
<td><tt>0</tt></td>
</tr>
<tr>
<td></td>
<td><tt><b>0</b>021</tt></td>
<td><tt>1</tt></td>
</tr>
<tr>
<td></td>
<td><tt>0<b>1</b>21</tt></td>
<td><tt>1</tt></td>
</tr>
<tr>
<td></td>
<td><tt>01<b>2</b>1</tt></td>
<td><tt>1</tt></td>
</tr>
<tr>
<td><tt>17</tt></td>
<td><tt>012<b>2</b></tt></td>
<td><tt>0</tt></td>
</tr>
</tbody>
</table>
<div class="pageBreakBefore"></div>
<h3>2.3 While \(n\) is Odd</h3>
<p>First step is \( 3 × n \). In ternary, shit left one digit and pad zero</p>
<p>E.g. \( 5 × 3 = 15\) in decimal, \( 12 ≪ 1 = 120\) in ternary. </p>
<p>Second step is \( (3n+1)/2 \). Becaurse \(n\) is odd, \(3n\) is also odd, it must be in borrowing at the last digit.
Follow the rule in 2.1, last digit will add 1 and translate into 2 .
</p>
<p>Consequently, \((3n+1)/2 \) calculation in ternary, just needing divide by 2 at each digit, then padding 2 at last.</p>
<p>E.g. \( (3×15+1)/2 = 23 \) in decimal, \( 120 → 21 → 212\) in ternary.</p>
<h3>2.4 Summary</h3>
<p>Each digit of 3n+1 conjecture, in ternary should be the rules as follows:</p>
$$
\begin{cases}
0 → 0, & \text{ no borrow} \\
1 → 0, & \text{ no borrow} \\
2 → 1, & \text{ no borrow} \\
0 → 1, & \text{ need borrow} \\
1 → 2, & \text{ need borrow} \\
2 → 2, & \text{ need borrow} \\
0 → 2, & \text{ need borrow and at the last digit} \\
\end{cases}
$$
</section>
<!-- =================================第三章=================================================== -->
<section>
<h2>3 Visualization </h2>
<p>For more visualization, the following will use</p>
<tt>▁</tt> for value ０；<br>
<tt>▄</tt> for value １；<br>
<tt>█</tt> for value ２；<br>
<tt style="color: #e33;">□</tt> for borrow ０ <br>
<tt style="color: #e33;">■</tt> for borrow １ <br>
<tt style="color: #e33;">☆</tt> for borrow ０ <a class="ft" href="#f1">at the end<sup>①</sup></a>；<br>
<tt style="color: #e33;">★</tt> for borrow １ <a class="ft" href="#f1">at the end<sup>②</sup></a>；<br>
<table class="example" id="exp4">
<caption>
<b>Example Break4</b> Visualization of ( 3 × 22459 + 1 ) / 2 = 33689
<br>
\(1010210211_{(3)} → 1201012202_{(3)}\)
</caption>
<tbody class="track">
<tr>
<td><tt> </tt><tt>▄▁▄▁█▄▁█▄▄</tt><tt>１０１０２１０２１１</tt></td>
</tr>
<tr>
<td><tt>■</tt><tt>▁▁▄▁█▄▁█▄▄</tt><tt>００１０２１０２１１</tt></td>
</tr>
<tr>
<td><tt> ■</tt><tt>▁▄▄▁█▄▁█▄▄</tt><tt>０１１０２１０２１１</tt></td>
</tr>
<tr>
<td><tt> □</tt><tt>▁▄█▁█▄▁█▄▄</tt><tt>０１２０２１０２１１</tt></td>
</tr>
<tr>
<td><tt> □</tt><tt>▁▄█▁█▄▁█▄▄</tt><tt>０１２０２１０２１１</tt></td>
</tr>
<tr>
<td><tt> □</tt><tt>▁▄█▁▄▄▁█▄▄</tt><tt>０１２０１１０２１１</tt></td>
</tr>
<tr>
<td><tt> ■</tt><tt>▁▄█▁▄▁▁█▄▄</tt><tt>０１２０１００２１１</tt></td>
</tr>
<tr>
<td><tt> ■</tt><tt>▁▄█▁▄▁▄█▄▄</tt><tt>０１２０１０１２１１</tt></td>
</tr>
<tr>
<td><tt> ■</tt><tt>▁▄█▁▄▁▄█▄▄</tt><tt>０１２０１０１２１１</tt></td>
</tr>
<tr>
<td><tt> □</tt><tt>▁▄█▁▄▁▄██▄</tt><tt>０１２０１０１２２１</tt></td>
</tr>
<tr>
<td><tt> ■</tt><tt>▁▄█▁▄▁▄██▁</tt><tt>０１２０１０１２２０</tt></td>
</tr>
<tr>
<td><tt> ★</tt><tt>▁▄█▁▄▁▄██▁█</tt><tt>０１２０１０１２２０２</tt></td>
</tr>
</tbody>
</table>
<p>Obviously, calculation of current digit/borrow, just depend on previous digit's borrow & current digit's value.
That means, after first digit's calculation in first step,it can calculate synchronously first and second step in second step , an so on.</p>
<div class="pageBreakBefore"></div>
<table class="example" id="exp5">
<caption style="width: 18em;"><b>Example 5</b> Synchronous calculation of \(210_{(3)}\)</caption>
<tbody class="track">
<tr>
<th>0</th>
<td><tt> </tt><tt>█▄▁</tt></td>
</tr>
<tr>
<th>1</th>
<td><tt>□</tt><tt>▄▄▁</tt></td>
</tr>
<tr>
<th>2</th>
<td><tt>■■</tt><tt>▁▁▁</tt></td>
</tr>
<tr>
<th>3</th>
<td><tt>□■■</tt><tt>▁▄▄</tt></td>
</tr>
<tr>
<th>4</th>
<td><tt>□■□★</tt><tt>▁▁██</tt></td>
</tr>
<tr>
<th>5</th>
<td><tt>□□■□</tt><tt>▁▁█▄</tt></td>
</tr>
<tr>
<th>6</th>
<td><tt>□□□□☆</tt><tt>▁▁▄█</tt></td>
</tr>
<tr>
<th>7</th>
<td><tt>□□■□☆</tt><tt>▁▁▁▄</tt></td>
</tr>
<tr>
<th>8</th>
<td><tt>□□□□☆</tt><tt>▁▁▁█</tt></td>
</tr>
</tbody>
</table>
<p>For more convenience, 7 cases in <a class="ft" href="#s2_4">2.4</a> are represented by numbers \([0-6]\),
of which \(6\) only for current digit is the last digit.</p>
<table class="example" >
<tbody class="track">
<tr>
<td><tt>□□□■■■ </tt><tt>▁▄█▁▄█ </tt><tt>０１２３４５６</tt></td>
</tr>
</tbody>
</table>
<p>The Example 5 is transformed as follows:</p>
<table class="example" id="exp5_2">
<caption style="width: 18em;"><b>Example 5’</b> Synchronous calculation of \(210_{(3)}\)</caption>
<tbody class="track">
<tr>
<th>0</th>
<td>210</td>
</tr>
<tr>
<th>1</th>
<td>110</td>
</tr>
<tr>
<th>2</th>
<td>330</td>
</tr>
<tr>
<th>3</th>
<td>044</td>
</tr>
<tr>
<th>4</th>
<td>0322</td>
</tr>
<tr>
<th>5</th>
<td>0051</td>
</tr>
<tr>
<th>6</th>
<td>0012</td>
</tr>
<tr>
<th>7</th>
<td>0031</td>
</tr>
<tr>
<th>8</th>
<td>0002</td>
</tr>
</tbody>
</table>
</section>
<!-- =================================第四章=================================================== -->
<section>
<h2>4. Equivalent Model</h2>
<h3>4.1 Base \(H\) System<a class="ft" href="#f2"><sup>③</sup></a></h3>
<p>In the third step of <a class="ft" href="exp5">Example 5</a>, The last digit's borrow has been involved in the operation;
we could get numbers \(44_{(H)}\) at this step,
by erasing the zero that the first parameter no longer affects the calculation in subsequent steps.</p>
<p> Here defind operations in steps 1 to 3 \(210_{(3)} → 44_{(H)}\) as \(H\) conversion,
and \(44_{(H)}\) is the \(H\) number of \(210_{(3)}\).</p>
<p>Here are two obvious introductions:</p>
<p><em>\(I\)</em> :<q>For any integer \(n\) , \(n \leq 3^k\), can translate into \(H\) base number after \(k\) step calculation.</q></p>
<p><em>\(II\)</em>:<q>the \(H\) number of any integer \(n\) is unique</q></p>
<p>Thus, the proof of the 3n+1 conjecture can be obtained by the proof of the \(H\) base system.
The following only discusses the convenience of proving \(H\) base system.</p>
<h3>4.2 Mathematical Description</h3>
<p>Agree \(H^i_j\) : mark positions by \([0,1,2,…,i]\) on superscript，mark steps by \([0,1,2,…,j]\) on subscript,
thus \(H^i_j\)'s value only depend on \(H^{i-1}_{j-1}\) & \(H^i_{j-1}\).
According rules in <a class="ft" href="#s2_4">2.4</a>, we can get the table</p>
<!--
0 3 1 4 2 5 6 next
0 0 0 3 3 1 1 6
1 0 0 3 3 1 1 6
2 0 0 3 3 1 1 6
3 4 4 2 2 5 5 2
4 4 4 2 2 5 5 2
5 4 4 2 2 5 5 2
6 6
-->
<table class="example" id="Matrix">
<caption style="width: 18em;">
Table 2 Calculation of \(H\) base system<br>
<span style="color: #fa3;">\(H^{i-1}_{j-1}\)</span>
<span style="color: #3ae;">\(H^i_{j-1}\)</span>
<span style="color: #fff;">\(H^i_j\)</span>
</caption>
<tbody>
<tr>
<td></td>
<td>0</td>
<td>3</td>
<td>1</td>
<td>4</td>
<td>2</td>
<td>5</td>
<td>6</td>
</tr>
<tr>
<td>0</td>
<td>0</td>
<td>0</td>
<td>3</td>
<td>3</td>
<td>1</td>
<td>1</td>
<td>6</td>
</tr>
<tr>
<td>1</td>
<td>0</td>
<td>0</td>
<td>3</td>
<td>3</td>
<td>1</td>
<td>1</td>
<td>6</td>
</tr>
<tr>
<td>2</td>
<td>0</td>
<td>0</td>
<td>3</td>
<td>3</td>
<td>1</td>
<td>1</td>
<td>6</td>
</tr>
<tr>
<td>3</td>
<td>4</td>
<td>4</td>
<td>2</td>
<td>2</td>
<td>5</td>
<td>5</td>
<td>2</td>
</tr>
<tr>
<td>4</td>
<td>4</td>
<td>4</td>
<td>2</td>
<td>2</td>
<td>5</td>
<td>5</td>
<td>2</td>
</tr>
<tr>
<td>5</td>
<td>4</td>
<td>4</td>
<td>2</td>
<td>2</td>
<td>5</td>
<td>5</td>
<td>2</td>
</tr>
<tr>
<td>6</td>
<td>-</td>
<td>-</td>
<td>-</td>
<td>-</td>
<td>-</td>
<td>-</td>
<td>6</td>
</tr>
</tbody>
</table>
<p>For any series in \(H\) base system \(\mathbb{H}_j = H^{1}_{j}H^{2}_{j}…H^{i}_{j}, \ (H^{i}_{j}\ in\ 0,1,2,3,4,5) \), </p>
<p>Define function \(S(\mathbb{H}_j) = 0H^{1}_{j}H^{2}_{j}…H^{i}_{j}6 \) irrelevant to the steps, just add 0 at the begining and add 6 at the end.</p>
<p>Define function \(T(\mathbb{H}_j) = H^{1}_{j+1}H^{2}_{j+1}…H^{i}_{j+1} = \mathbb{H}_{j+1} \), and \(H^{i}_{j+1}\) mapping from the table above \(H^{i-1}_{j-1}H^{i}_{j-1} → H^{i}_{j+1} \).</p>
<p>Thus, define function calculated in each step:</p>
<p>\(ST(\mathbb{H}_j) = T(S(\mathbb{H}_j)) \)</p>
<p>The author has checked all 7-digit series in \(H\) base system by computer program, after calculation of finite steps by function \(ST\),
the final series must be 12 → 31 → 2 → 1 ... , and we can get the spanning tree.</p>
<pre class="BlackBoard">
1 - 3
\ /
2
|
31
|
12
/ \
5 51
| |
32 322
/ \ / \
4 41 44 441
</pre>
<p>It can find out the similarity by contrasting the spanning of origin conjecture,
which implications both struct are aim to the same maths problem.
But it's difficult to strat with the spanning tree, another more intuitive model will be given below.</p>
<h3>4.3 Jigsaw Model</h3>
<p>Equally easy to find, just two numbers (0,1) of borrow of \(H^{i-1}_{j-1}\) and three numbers (0,1,2) of value of \(H^i_{j-1}\),
take (x,y) for (0,1) of borrows and (a,b,c) for (0,1,2) of values,
then only 7 kinds of jigsaw can impicate the calculation of \(H\) base system</p>
<pre class="BlackBoard">
x a x c y b x b y a y c x b
\ / \ / \ / \ / \ / \ / \ /
0 1 2 3 4 5 6
/ \ / \ / \ / \ / \ / \ / \
a x b x c x a y b y c y b x
</pre>
<p>Here restate the steps 3-8 in <a class="ft" href="exp5">Example 5</a> in jigsaw model to clarify the intuitiveness</p>
<pre class="BlackBoard">
0 4 4 6
/ \ / \ / \ / \
a x b y b y x b
/ \ / \ / \ / \
0 3 2 2 6
/ \ / \ / \ / \ /
a x a y c x c x b
/ \ / \ / \ / \ /
0 0 5 1 6
\ / \ / \ / \ /
x a x c y b x b
\ / \ / \ / \ /
0 1 2 6
/ \ / \ / \ /
a x b x c x b
/ \ / \ / \ /
0 3 1 6
\ / \ / \ /
x a y b x b
\ / \ / \ /
0 2 6
</pre>
<p>Step 1, arrange two pieces of jigsaw 4 into 44 in first lind, follow the function \(ST\),
get first series 0446 by add jigsaw 0 at the head and jigsaw 6 at the tail;</p>
<p>Step 2, arrange pieces 322 following top nodes are (x,b) (y,b) (y,x),
get second series 03226 by add jigsaw 0 at the head and jigsaw 6 at the tail;</p>
<p>Follow this rule, in the step 6, only one piece of jigsaw 2 in the middle while others are 0 & 6.</p>
<p>Notice: this papper require jigsaw 6 can only appear at the end though the top nodes of 3 & 6 are same in (x,b).</p>
<p>The conjecture of jigsaw model is:</p>
<p><q>There is a serie arranged by finite pieces of jigsaw 0,1,2,3,4,5 ,
and follow the rules to add jigsaw 0 at the head and jigsaw 6 at the tail in the following lines,
it finally get the circle of 2 → 1 → 3 in finite lines beside 0 & 6.</q></p>
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<section>
<h2>5 Postscript</h2>
<p>I have studied this problem as an amateur for many years. Until the end of 2019 <a class="ft" href="#ft3">Terence Tao</a> made out an almost proved
and I had a long vacation for COVID-19 in early 2020 ,I decided to write down my ideas and some were added during Oct 2021.
Maybe some mathematicians have done the same research before, or maybe I haven't consulted the corresponding literature;
and it is very likely that the ideas completely coincide,
and the mathematical language used in the literature is beyond the scope of my understanding.
This article only gives a model, and my personal mathematical ability is not enough to prove this world problem.
I hope this paper can bring some help to professors; If not, please point out the mistakes and omissions.</p>
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<p class="ft" id="f1">① ② Convenience for intermediate state, actually no need to be specially pointed out.</p>
<p class="ft" id="f2">③ Base H System can also be deduced by Binary with carry, they won't be covered here.</p>
<h2>References</h2>
<p class="ft"><a class="lk" href="http://www.matrix67.com/blog/archives/6756">1. Matrix67's blog</a></p>
<p class="ft" id="ft3">2. TERENCE TAO, ALMOST ALL ORBITS OF THE COLLATZ MAP ATTAIN ALMOST BOUNDED VALUES</p>
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