Solve the 3x+1 Problem

1. Introduction

The $3x+1$ problem is one of the unsolved problems in mathematics. It is also known as the Collatz conjecture, $3x+1$ mapping, Ulam conjecture, Kakutani’s problem, Thwaites conjecture, Hasse’s algorithm, or Syracuse problem [1]. Paul Erdos (1913-1996) commented on the intractability of problem $3n+1$ [2]: "Mathematics is not ready for those problems yet".

The $2x+1$ problem is that, take any positive integer x, If x is even, divide x by 2. If x is odd, multiply x by 3 and add 1. Repeat this process continuously. The conjecture states that no matter which number you start with, you will always reach 1 eventually.

2. Terminology and notations

We will use the notations as in [4]. we describe a Collatz function as

$$T\left(n\right)=\left\{\begin{array}{cc}3n+1,& \mathrm{if}n\mathrm{is}\mathrm{odd}\mathrm{number},\\ \frac{n}{2}& \mathrm{if}n\mathrm{is}\mathrm{even}\mathrm{number}.\end{array}\right.$$

(1)

Let N denote the set of positive integers. For $n\in N$, and $k=0,1,2,3,\cdots$, $T^0\left(n\right)$ and $T^{k+1}\left(n\right)$ denote n and $T(T^k\left(n\right)$, respectively. We will say p is a periodic point with period m if $p\in N$ and $p=T^m\left(p\right)$ and $p\ne T^k\left(p\right)$ for $1\le k<m$.

The $3x+1$ problem concerns the behavior of the iterates of the Collatz function, for any integer n, there must exist an integer r, so that

$$T^r\left(n\right)=1.$$

3. Proof of the Collatz Conjecture

3.1. The modified Sarkovskii ordering

We remove the last row number to the first column, get the modified Sarkovskii ordering as

$\begin{array}{ccccccccccc}{2}^0,& 3,& 5,& 7,& 9,& 11,& 13,& 15,& 17,& 19,& \cdots \\ 2,& 2·3,& 2·5,& 2·7,& 2·9,& 2·11,& 2·13,& 2·15,& 2·17,& 2·19,& \cdots \\ 2^2,& 2^2·3,& 2^2·5,& 2^2·7,& 2^2·9,& 2^2·11,& 2^2·13,& 2^2·15,& 2^2·17,& 2^2·19,& \cdots \\ 2^3,& 2^3·3,& 2^3·5,& 2^3·7,& 2^3·9,& 2^3·11,& 2^3·13,& 2^3·15,& 2^3·17,& 2^3·19,& \cdots \\ 2^4,& 2^4·3,& 2^4·5,& 2^4·7,& 2^4·9,& 2^4·11,& 2^4·13,& 2^4·15,& 2^4·17,& 2^4·19,& \cdots \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \end{array}$

In the first row, its are odd number from left to right, that are $1,3,5,7,9,11,13,\cdots$, from the second row, each number is multiplying each number in its previous row by 2, and so on.

3.2. Collatz graph

We can use the Collatz function $T\left(x\right)$, obtain that a algebraic formula of $\frac{1}{2^4},\frac{3}{2^7},\frac{3^2}{2^9},\cdots ,\frac{3^m}{2^r}$. If we draw a line segment of arrow between two digits, those are the original value and its value of Collatz function. When we repeat to iteration of Collatz function, we get a graph, which be called as Collatz graph. Here r is the number of perpendicular segments, m is the oblique segments. We can get a unique algebra formula about $3^m$ in numerator and $2^r$ in denominator, as

$$T^{m+r}\left(n\right)=T(m,r,n)=1,$$

For example,

$$T^{16}\left(7\right)=T(5,11,7)=\frac{1}{2^4}+\frac{3}{2^7}+\frac{3^2}{2^9}+\frac{3^3}{2^{10}}+\frac{3^4}{2^{11}}+\frac{3^5}{2^{11}}·7=1$$

ftbpFU2.911in1.951in0ptThe composition of the Collatz function $T(5,11,7)$tu333.eps

$$T^{21}\left(36\right)=T(6,15,36)=\frac{1}{2^4}+\frac{3}{2^7}+\frac{3^2}{2^9}+\frac{3^3}{2^{10}}+\frac{3^4}{2^{11}}+\frac{3^5}{2^{13}}+\frac{3^6}{2^{15}}·36=1$$

ftbpFU2.911in1.951in0ptThe composition of the Collatz function T(6,15,36)tu512.eps

4. Numerical example

We propose the following procedure,

$$T^{20}\left(18\right)=T(6,14,18)=\frac{1}{2^4}+\frac{3}{2^7}+\frac{3^2}{2^9}+\frac{3^3}{2^{10}}+\frac{3^4}{2^{11}}+\frac{3^5}{2^{13}}+\frac{3^6}{2^{14}}·18=1$$

$$T^{15}\left(23\right)=T(4,11,23)=\frac{1}{2^4}+\frac{3}{2^9}+\frac{3^2}{2^{10}}+\frac{3^3}{2^{11}}+\frac{3^4}{2^{11}}·23=1$$

$$T^{17}\left(15\right)=T(5,12,15)=\frac{1}{2^4}+\frac{3}{2^9}+\frac{3^2}{2^{10}}+\frac{3^3}{2^{11}}+\frac{3^4}{2^{12}}+\frac{3^5}{2^{12}}·15=1$$

$$T^{12}\left(17\right)=T(3,9,17)=\frac{1}{2^4}+\frac{3}{2^7}+\frac{3^2}{2^9}+\frac{3^3}{2^9}·17=1$$

$$T^{16}\left(397\right)=T(5,11,397)=\frac{1}{2^4}+\frac{3}{2^7}+\frac{3^2}{2^9}+\frac{3^3}{2^{10}}+\frac{3^4}{2^{11}}+\frac{3^5}{2^{17}}+\frac{3^6}{2^{20}}+\frac{3^7}{2^{20}}·397=1$$

$$T^{19}\left(61\right)=T(5,14,61)=\frac{1}{2^4}+\frac{3}{2^9}+\frac{3^2}{2^{10}}+\frac{3^3}{2^{11}}+\frac{3^4}{2^{14}}+\frac{3^5}{2^{14}}·61=1$$

We observe the three properties,

Note 1 In the algebraic formula, in numerators, it is $1,3,3^2,3^3,\cdots ,3^m$. there is a lack. But in denominator, there are many lacks,$2,2^2,2^3,\cdots ,2^r.$

Note 2 For positive integers $i,j,k,l$, and $l_k,l_{k-1},\cdots ,l_1$, if $i>j$, then there is a recurrence relation

$$T^i\left(n\right)=\frac{3^k}{2^l}{T}^j\left(n\right)+\frac{3^{k-1}}{2^{l_k}}+\cdots +\frac{3^2}{2^{l_3}}+\frac{3}{2^{l_2}}+\frac{1}{2^{l_1}}$$

where $k+l=i-j$, and $l\ge l_k\ge l_{k-1}\ge \cdots \ge l_1.$

Example 1 There are

$$T^3\left(97\right)=\frac{3}{2^2}·97+\frac{1}{2^2}=73$$

$$T^{18}\left(97\right)=\frac{3^7}{2^{11}}·97+\frac{3^6}{2^{11}}+\frac{3^5}{2^9}+\frac{3^4}{2^7}+\frac{3^3}{2^6}+\frac{3^2}{2^5}+\frac{3}{2^2}+\frac{1}{2}=107$$

We can get the currence formula about the Collatz function T(x)

$$T^{26}\left(97\right)=\frac{3^3}{2^5}·T^{18}\left(97\right)+\frac{3^2}{2^5}+\frac{3}{2^4}+\frac{1}{2^2}=91,$$

namely,

$$T^{26}\left(97\right)=\frac{3^{10}}{2^{16}}·97+\frac{3^9}{2^{16}}+\frac{3^8}{2^{14}}+\frac{3^7}{2^{12}}+\frac{3^6}{2^{11}}+\frac{3^5}{2^{10}}+\frac{3^4}{2^7}+\frac{3^3}{2^6}+\frac{3^2}{2^5}+\frac{3}{2^4}+\frac{1}{2^2}=91$$

$$T^{31}\left(97\right)=\frac{3^2}{2^3}·91+\frac{3}{2^3}+\frac{1}{2^2}=103$$

$$T^{40}\left(97\right)=\frac{3^4}{2^5}·103+\frac{3^3}{2^5}+\frac{3^2}{2^4}+\frac{3}{2^3}+\frac{1}{2}=263$$

$$T^{53}\left(97\right)=\frac{3^5}{2^8}·263+\frac{3^4}{2^8}+\frac{3^3}{2^7}+\frac{3^2}{2^6}+\frac{3}{2^4}+\frac{1}{2}=251$$

$$T^{53}\left(97\right)=\frac{3^9}{2^{13}}·103+\frac{3^8}{2^{13}}+\frac{3^7}{2^{12}}+\frac{3^6}{2^{11}}+\frac{3^5}{2^9}+\frac{3^4}{2^8}+\frac{3^3}{2^7}+\frac{3^2}{2^6}+\frac{3}{2^4}+\frac{1}{2}=251=T^{53}\left(97\right)$$

$$\begin{array}{cc}251& x\\ 754& T\left(x\right)=3x+1\\ 377& T^2\left(x\right)=\frac{3}{2}x+\frac{1}{2}\\ 1132& T^3\left(x\right)=\frac{3^2}{2}x+\frac{3}{2}+1\\ 566& T^4\left(x\right)=\frac{3^2}{2^2}x+\frac{3}{2^2}+\frac{1}{2}\\ 283& T^5\left(x\right)=\frac{3^2}{2^3}x+\frac{3}{2^3}+\frac{1}{2^2}\\ 850& T^6\left(x\right)=\frac{3^3}{2^3}x+\frac{3^2}{2^3}+\frac{3}{2^2}+1\\ 425& T^7\left(x\right)=\frac{3^3}{2^4}x+\frac{3^2}{2^4}+\frac{3}{2^3}+\frac{1}{2}\\ ⋮& ⋮\\ 958& T^{64}\left(97\right)=\frac{3^5}{2^6}{T}^{53}\left(97\right)+\frac{3^4}{2^6}+\frac{3^3}{2^5}+\frac{3^2}{2^3}+\frac{3}{2^2}+1\\ 61& T^{100}\left(97\right)=\frac{3^7}{2^{15}}{T}^{77}\left(97\right)+\frac{3^6}{2^{15}}+\frac{3^5}{2^{14}}+\frac{3^4}{2^{13}}+\frac{3^3}{2^{12}}+\frac{3^2}{2^8}+\frac{3}{2^6}+\frac{1}{2^4}\\ 1& T^{119}\left(97\right)=T^{100}\left(97\right)·\frac{3^5}{2^{14}}+\frac{3^4}{2^{14}}+\frac{3^3}{2^{11}}+\frac{3^2}{2^{10}}+\frac{3}{2^9}+\frac{1}{2^4}\end{array}$$

Note 3 We observe that

$$3^5·61+3^4+3^3·2^3+3^2·2^4+3·2^5+2^{10}=2^{14}$$

$$3^7·397+3^6+3^5·2^3+3^4·2^9+3^3·2^{10}+3^2+3·2^{13}+2^{16}=2^{20}$$

Theorem 2 We can use the Collatz function $T\left(x\right)$, obtain that a series of $\frac{1}{2^4},\frac{3}{2^7},\frac{3^2}{2^9},\cdots ,\frac{3^m}{2^r}$, where r is the number of perpendicular segments, m is the oblique segments in the graph. We can get a unique algebra denotion about $3^m$ in numerator and $2^r$ in denominator, as

$$T(m,r,n)=1,$$

So, there is

$$T^{m+r}\left(n\right)=T(m,r,n)=\frac{3^m}{2^r}·n+\frac{3^{m-1}}{2^{r_{m-1}}}+\cdots +\frac{3^2}{2^{r_2}}+\frac{3}{2^{r_1}}+\frac{1}{2^4}=1$$

where $r\ge r_{m-1}\ge r_{m-2}$$\ge \cdots \ge r_1>4.$

and there is a recurrence relation

$$T^i\left(n\right)=\frac{3^k}{2^l}{T}^j\left(n\right)+\frac{3^{k-1}}{2^{l_k}}+\cdots +\frac{3^2}{2^{l_3}}+\frac{3}{2^{l_2}}+\frac{1}{2^{l_1}}$$

where $l\ge l_k\ge l_{m-2}$$\ge \cdots \ge l_1.$

Theorem 3 For positive integer, n, there must exist positive integer $m,r$ and $r_{m-1},\cdots ,r_1$, such that

$$2^r=3^m·n+3^{m-1}·2^{m-r_{m-1}}+\cdots +3^2·2^{m-r_2}+3·2^{m-r_1}+2^{m-4},$$

where $r\ge r_{m-1}\ge r_{m-2}$$\ge \cdots \ge r_1\ge 4$. This mean 3 multiply to the initial value n gradually with the digits of the abacus, must equal to $2^r.$

Exampe 4 For example, for the formula

$$T(6,14,18)=\frac{1}{2^4}+\frac{3}{2^7}+\frac{3^2}{2^9}+\frac{3^3}{2^{10}}+\frac{3^4}{2^{11}}+\frac{3^5}{2^{13}}+\frac{3^6}{2^{14}}·18=1,$$

namely,

$$2^{10}+3·2^7+3^2·2^5+3^3·2^4+3^4·2^3+3^5·2+3^6·18=2^{14},$$

PROOF. We calculate

$$\begin{array}{ccc}\hfill 3& =& 2+1\hfill \\ \hfill 3^2& =& 2^3+1\hfill \\ \hfill 3^3& =& 2^4+2^3+2+1\hfill \\ \hfill 3^4& =& 2^6+2^4+1\hfill \\ \hfill 3^5& =& 2^7+2^6+2^5+2^4+2+1\hfill \\ \hfill 3^6& =& 2^9+2^7+2^6+2^4+2^3+1\hfill \\ \hfill 18& =& 2^4+2\hfill \end{array}$$

and substitute them in the expression

$$\begin{array}{ccc}& & 3^6·18+3^5·2+3^4·2^3+3^3·2^4+3^2·2^5+3·2^7+2^{10}\hfill \\ & =& (2^9+2^7+2^6+2^4+2^3+1)·(2^4+2)+(2^7+2^6+2^5+2^4+2+1)·2\hfill \\ & & +(2^6+2^4+1)·2^3+(2^4+2^3+2+1)·2^4+(2^3+1)·2^5+(2+1)·2^7+2^{10},\hfill \end{array}$$

and get the value $2^{14}$.

Remark 5 We can say that $3x+1$ problem is the convert statement of period three implies chaos [3].

5. Conclusion

By modifying the Sarkovskii ordering, denote the composition of the Collatz function as a algebraic formula about the$\frac{3^m}{2^r}$, we give a bridge of algebraic formula with graphs. We completely solve the 3x + 1 problem.