Solve the 3x+1 Problem by the Multiplication and Division of Binary Numbers

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 [4,7]. 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. 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

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

2.1. The Modified Sarkovskii Ordering and Integer Lattice

We remove the last row number to the first column, get an integer lattice[6] of the modified Sarkovskii ordering as

$\begin{array}{ccccccccccc}1,& 3,& 5,& 7,& 9,& 11,& 13,& 15,& 17,& 19,& \cdots \\ 2,& 2·3,& 2·5,& 2·7,& 2·9,& 2·11,& 2·13,& 2·15,& 2·17,& 2·19,& \cdots \\ 2^2,& 2^2·3,& 2^2·5,& 2^2·7,& 2^2·9,& 2^2·11,& 2^2·13,& 2^2·15,& 2^2·17,& 2^2·19,& \cdots \\ 2^3,& 2^3·3,& 2^3·5,& 2^3·7,& 2^3·9,& 2^3·11,& 2^3·13,& 2^3·15,& 2^3·17,& 2^3·19,& \cdots \\ 2^4,& 2^4·3,& 2^4·5,& 2^4·7,& 2^4·9,& 2^4·11,& 2^4·13,& 2^4·15,& 2^4·17,& 2^4·19,& \cdots \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \end{array}$

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.

2.2. The Algebraic Formula and Collatz Graph

If we draw a line segment of arrow between two digits in the lattice of integer in the modified Sarkovskii ordering, those are the original value x, and its value of Collatz function $T\left(x\right)$, and connect $T\left(x\right)$ to $T^2\left(x\right)$, and so on $T^2\left(x\right)$ to $T^3\left(x\right),\cdots$, thus we get a graph, which can be called as Collatz graph. For different integer m, n, besides 1, 4, 2, there is not other common vertices in their Collatz graphs. Using the Collatz function $T\left(x\right)$, We obtain an algebraic formula of $\frac{1}{2^4},\frac{3}{2^7},\frac{3^2}{2^9},\cdots ,\frac{3^m}{2^r}·x$. Here r is the number of perpendicular segments, m is the oblique segments in the Collatz graph,

$$T^{m+r}\left(n\right)=T(m,r,n)=\frac{1}{2^4}+\frac{3}{2^7}+\frac{3^2}{2^9}+\cdots +\frac{3^m}{2^r}·x=1,$$

For example, $n=7$, $7\to 22\to 11\to 34\to 17\to 52\to 26\to 13\to 40\to 20\to 10\to 5\to 16\to 8\to 4\to 2\to 1\to \cdots$, the algebraic formula is

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

and the Collatz graph is Figure 1.

And $n=36$, the algebraic formula is

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

and the Collatz graph is Figure 2.

3. Numerical Example

We propose the following algebraic formulas,

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

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

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

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

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

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

4. Convert the Integer Number from Decimal to Binary

Be inspired by the above, we use binary to describe the Collatz function (1) an the follows. We denote binary number which is a string of 0s and 1s, $n={(1\cdots \times )}_2,\times$ is 1 or 0, $3={\left(11\right)}_2$,

$$T\left(n\right)=T\left({(1\cdots \times )}_2\right)=\left\{\begin{array}{cc}{\left(11\right)}_2·{(1\cdots 1)}_2+1,& \mathrm{if}n\mathrm{is}\mathrm{odd}\mathrm{number},\\ \frac{{(1\cdots 0)}_2}{{\left(10\right)}_2},& \mathrm{if}n\mathrm{is}\mathrm{even}\mathrm{number}.\end{array}\right.$$

(2)

$$T\left(n\right)=T\left({(1\cdots \times )}_2\right)=\left\{\begin{array}{cc}{(1\times \times \times \times 0\cdots 0)}_2,& \mathrm{if}n\mathrm{is}\mathrm{odd}\mathrm{number},\\ {(1\times \times \times 1)}_2,& \mathrm{if}n\mathrm{is}\mathrm{even}\mathrm{number}.\end{array}\right.$$

(3)

Namely, when n is odd number, we multiple it with ${\left(11\right)}_2$ and add 1 to the end of the binary number, we give an example, $T\left(97\right)=T\left(1100001\right)$ in Figure 3 in the follows. When n is even number, the division is equal to delete zeros at the end in binary number. We give the iteration of the Collatz function for 7, 97 in binary as the following two tables.

Example 1.

For positive integer 1, we manipulate the iteration of the Collatz function in becimal and binary numbers,

Example 2.

For 7=(111)${ }_2,$ we manipulate the iteration of the Collatz function in becimal and binary numbers,

Example 3.

For 97=(100100)${ }_2,$ we manipulate the iteration of the Collatz function in binary as the following table,

We can rewrite the Collatz conjecture in binary as the follows, it becomes an easy problem.

Fact 4

For any positive integer, under the Collatz function, the sequence of integer number in binary eventually must reach the integer 1.

Proof. 

For an binary number of odd integer, when we add 1 to the end, and add the shifted binary number one place to the left, finally, the result number must with at least one zero at the end, we delete this zeros. Thus repeat this process as long as we can, eventually we must reach 1. □

Remark 5

We can say that $3x+1$ problem is an converse proposition of period three implies chaos [4] and an example.