(1) Any positive number D in Base-X Number System can be expressed as the product of a pure even-x number and an odd-x number:

$D=X_1^k\left(Q\ast X_1+r\right)r=1,2,3,\cdots ,X_4,X_3,X_2$ (1)

Q is either odd-x or even-x number.

For any odd-x number D, k = 0, D = Q * X₁ + r.

If Q = 0 and k = 0, D is a single digit odd-x number, D = r.

If Q = 0 and r = 1, D is a pure even-x number, D = $X_1^k$.

(2) multiple D by X did not affect k and r in property (1)

$\begin{array}{c}D\ast X=X_1^k\left(Q\ast X_1+r\right)X\\ =X_1^k\left(Q\ast X_1+r\right)\left(X_1+1\right)\\ =X_1^k\left[\left(Q\ast X_1+r\right)X_1+Q\ast X_1+r\right]\\ =X_1^k\left[\left(Q\ast X_1+r+Q\right)X_1+r\right]\\ =X_1^k\left(Q_1\ast X_1+r\right)\end{array}$ (2)

(3) from property (2) can be derived that multiple D by Xⁿ did not affect k and r

$D\ast X^n=X_1^k\left(Q_n\ast X_1+r\right)$ (3)

if D is a single digit r, the remainder of r*Xⁿ divided by X₁ is r itself.

(4) any odd-x D can be converted to even-x number by:

$\begin{array}{l}D\ast X^n+r^{\prime }{r}^{\prime }\text{is}{X}_1\text{'s}\text{compliment}\text{of}r\\ =Q_n\ast X_1+r+r^{\prime }\\ =Q_n\ast X_1+X_1\\ =\left(Q_n+1\right)X_1\end{array}$ (4)

(5) for any integer D, adding up all the digits of D until to get a single digit r, if the result r = X₁, D is even-x number, otherwise D is odd-x number, r is the remainder of D divided by X₁.

Any positive integer D can be expressed as:

$D={\displaystyle \underset{i=0}{\overset{n}{\sum }}{d}_i{X}^i}$ $d=1,2,3,\cdots ,X_3,X_2,X_1$

from property (3), each item d_iXⁱ divided by X₁, the remainder should be d_i, so the remainder R of D divided by X₁ should be:

$R={\displaystyle \underset{i=0}{\overset{n}{\sum }}{d}_i}$ $d=1,2,3,\cdots ,X_3,X_2,X_1$

R is a new number with less digits than D, repeating above processing, a single digit could be obtained.

Take Hexadecimal for example (X₁ = F):

for hex number $2E35F9BD67$

$2+E+3+5+F+9+B+D+6+7=55$

$5+5=A$

so it is an odd-16 number, and the remainder of divided by F is A.

$2E35F9BD67=314AA3FD3\ast F+A$

for hex number $7E35F9BD67$

$7+E+3+5+F+9+B+D+6+7=5A$

$5+A=F$

so it is an even-16 number.

$7E35F9BD67=869FF9529\ast F$

For any odd-x positive integer D

$D+r^{\prime }=Q\ast X_1+r+r^{\prime }=\left(Q+1\right)X_1$

divided by X₁ result is Q + 1. The Base-X Conjecture iteration is converging due to D > Q + 1, so the Conjecture is true.

The Base-X Conjecture iteration is mostly diverging, so the Conjecture could be false. This will not be discussed in this paper.

$\begin{array}{l}D\ast X+r^{\prime }\\ =\left(Q\ast X_1+r\right)X_1+r+r^{\prime }\\ =\left(Q\ast X_1+r\right)X_1+X_1\\ =\left(Q\ast X_1+r+1\right)X_1\end{array}$

(a) for a single digit number Q = 0

$r\ast X+r^{\prime }=r\left(X_1+1\right)+r^{\prime }=r\ast X_1+\left(r+r^{\prime }\right)=r\ast X_1+X_1=\left(r+1\right)\ast X_1$

the result r is increased by 1 after one odd-step and one even-step. If the result (r + 1) = X₁, take another even-step that will get 1, otherwise repeat the iteration until the result (r + 1) = X₁. Base-X conjecture is true for a single digit number.

(b) if (r + 1) = X₁ and (Q + 1) = $X_1^m$, Base-X conjecture is true.

$D=\left(X_1^{2n}-1\right)/X$, ( $n=1,2,3,\cdots$) meets this condition.

Based on above analysis, use property (4) of Base-X number system with n = 1 to convert odd-x number to even-x.

In modular arithmetic notation, define the function f as follows:

$f\left(n\right)=\{\begin{array}{ll}n/X_1\hfill & \text{if}n\equiv 0\left(\mathrm{mod}{X}_1\right)\hfill \\ X\ast n+r^{\prime }\hfill & \text{if}n\equiv r\left(\mathrm{mod}{X}_1\right)\hfill \end{array}$

for Collatz Conjecture, it should be one of Base-X conjecture - Base-3 (Ternary). For Base-3 (Ternary), X₁ = 2, and there is only one remainder 1 if odd-3 (odd) number divided by 2, 2’s compliment of 1 is still 1, here comes 3n + 1. The Base-X conjecture for Base-3 (Ternary) is as follows:

$f\left(n\right)=\{\begin{array}{ll}n/2\hfill & \text{if}n\equiv 0\left(\mathrm{mod}2\right)\hfill \\ 3\ast n+1\hfill & \text{if}n\equiv 1\left(\mathrm{mod}2\right)\hfill \end{array}$

It is exactly the same as Collatz Conjecture.

Till now, the Base-X Conjecture is neither proved nor disproved, but a preliminary verification can be done by computer. There are a total of 28 bases from Base-3 (Ternary) to Base-30 (Tricenary) up to 10 digits have been verified, the verification results are shown in Table 1 .

Note: - not verified.

The table showed loops found for different digits number, and the steps for the loops found. There are a total of 28 bases verified, 16 bases are false due to more than 1 loop found. 12 bases could be true including Collatz Conjecture (Base-3). There is no diverging found, and no more loops found for more than 3 digits to maximum checked digits. Now we will try to prove the Base-3 (Collatz Conjecture) for it is the simple one, only dealing with a single digit 1.

It is already stated that “Collatz Conjecture, it is just one case of Base-X conjecture for Base-3 (Ternary)” in 3.4, so simple properties of Base-X number system could be used to figure it out. For any positive integer D in Base-3 can be expressed as following based on property (3):

$D=2^{k\left(0\right)}\left(Q\ast 2+1\right)k\left(0\right)=0,1,2,3,\cdots$ (6)

Now we can apply Collatz Conjecture iteration to the odd part and collect the divided by 2 and put it to the even part. Any odd number to another odd number, one odd-step and one or more even-step (simply called one step together) needed. For example:

1) Number 3:

$3\ast 3+1=10\to 5$,

one odd step and one even step, collect one divided by 2 → 2^k(0)+1.

2) Number 9:

$3\ast 9+1=28\to 14\to 7$,

one odd step and two even steps, collect two divided by 2 → 2^k(0)+2.

3) Number 13:

$3\ast 13+1=40\to 20\to 10\to 5$,

one odd step and three even steps, collect three divided by 2 → 2^k(0)+3.

4) Number 37:

$3\ast 37+1=112\to 56\to 28\to 14\to 7$,

one odd step and four even steps, collect four divided by 2 → 2^k(0)+4.

Generally for Equation [6] , just apply 3n + 1 and divided by 2 to the odd part (Q * 2 + 1) and pass divided by 2 to 2^k portion we got:

$\begin{array}{l}{2}^{k\left(0\right)}\left[3\left(Q\ast 2+1\right)+1\right]\\ =3\ast \left[2^{k\left(0\right)}\left(Q\ast 2+1\right)\right]+2^{k\left(0\right)}\\ =3\ast D+2^{k\left(0\right)}\text{collect}0\\ =D_1+Y_1{D}_1=3^1\ast D,Y_1=2^{k\left(0\right)},D_1>Y_1\end{array}$ (7)

in another way,

$\begin{array}{l}{2}^{k\left(0\right)}\left[3\left(Q\ast 2+1\right)+1\right]\\ =2^{k\left(0\right)}\left[3\ast Q\ast 2+4\right]\\ =2^{k\left(0\right)+1}\left[3\ast Q+2\right]\\ =2^{k\left(0\right)+1}\ast 2^x\ast \left(Q_1\ast 2+1\right)\\ =2^{k\left(0\right)+\left(1+x\right)}\ast \left(Q_1\ast 2+1\right)\text{collect}1+x,x=0,1,2,3,\cdots \\ =2^{k\left(1\right)}\ast \left(Q_1*2+1\right)k\left(1\right)=k\left(0\right)+\left(1+x\right)\end{array}$ (8)

after one 3n + 1 and one or more divided by 2 we got a new equation from Equation [6] :

$D_1+Y_1=2^{k\left(1\right)}\ast \left(Q_1\ast 2+1\right)k\left(1\right)>k\left(0\right)$ (9)

Compare to Equation [6] :

$D\to D_1+Y_1,2^{k\left(0\right)}\to 2^{k\left(1\right)}=2^{k\left(0\right)+\left(1+x\right)},Q\to Q_1$

Apply another 3n + 1 and divided by 2 to the odd part (Q₁ * 2 + 1) of Equation [9] we got:

$\begin{array}{l}{2}^{k\left(1\right)}\left[3\left(Q_1\ast 2+1\right)+1\right]\\ =3\ast \left[2^{k\left(1\right)}\left(Q_1\ast 2+1\right)\right]+2^{k\left(1\right)}\\ =3\ast D_1+3\ast Y_1+2^{k\left(1\right)}\\ =D_2+Y_2{D}_2=3^2\ast D,Y_2=3\ast Y_1+2^{k\left(1\right)}\end{array}$

in other way,

$\begin{array}{l}{2}^{k\left(1\right)}\left[3\left(Q_1\ast 2+1\right)+1\right]\\ =2^{k\left(1\right)}\left[3\ast Q_1\ast 2+4\right]\\ =2^{k\left(1\right)+1}\left[3\ast Q_1+2\right]\\ =2^{k\left(1\right)+\left(1+x\right)}\left(Q_2\ast 2+1\right)\text{collect}1+x,x=0,1,2,3,\cdots \\ =2^{k\left(2\right)}\left(Q_2\ast 2+1\right)k\left(2\right)>k\left(1\right)\end{array}$

same as Equation [9] we got

$D_2+Y_2=2^{k\left(2\right)}\left(Q_2\ast 2+1\right)$

repeat the 3n + 1 and divided by 2 on the odd part to step n we could get:

$D_n+Y_n=2^{k\left(n\right)}\left(Q_n\ast 2+1\right)$ (10)

The above process could be interpreted as it approaches a certain pure even-x number 2^m with continuing iteration, then we can get the following equation by combining Equation [5] and [10] :

$2^{m-1}<D_n+Y_n=2^{k\left(n\right)}\left(Q_n\ast 2+1\right)\le 2^m$ (11)

let Y₀ = 0, then

$Y_1=2^{k\left(0\right)}=3\ast Y_0+2^{k\left(0\right)}$

it can be figured out that Y is built up as:

$Y_0=0,Y_1=3\ast Y_0+2^{k\left(0\right)},Y_2=3\ast Y_1+2^{k\left(1\right)},\cdots ,Y_n=3\ast Y_{n-1}+2^{k\left(n-1\right)}$

$k\left(0\right)<k\left(1\right)<\cdots <k\left(n-1\right)<k\left(n\right)$

The minimum step of k is 1, and for any odd number k(0) = 0, the minimum k sequence should be:

$k=0,1,2,3,\cdots$

and the minimum Y sequence is as:

$Y=0,1,5,19,65,211,\cdots ,3^n-2^n$ [17] .

As described above, D and Y are both multiplied by 3, D is not changed, but Y is added 2^k each time, so that Y is getting bigger and bigger, at the beginning Y₁ < D₁, along with the iteration continuing, Y will exceed D and turn to Y > D from Y ≤ D at a certain step. Figure 3 shows the change in Base-3 (Ternary) - D is simply shifting left and Y is growing up along with left shifting.

Assuming Y_n−1 ≤ D_n−1 at step n − 1, and Y_n > D_n at step n, from Equation (11) we got:

$2^{m-1}<D_n+Y_n=D_n+3\ast Y_{n-1}+2^{k\left(n-1\right)}\le 2^m$ (12)

$2^{m-1}<D_n+3\ast Y_{n-1}+{\Delta }_n=2^m$ (13)

See Figure 4 [A, B]. To ensure from Y_n−1 ≤ D_n−1 to Y_n > D_n, 2^k(n−1) (difference between D_n and 3*Y_n−1) must take the possible maximum value. From Equation (12), the maximum value of 2^k(n−1) could be 2^m−1. But if 2^k(n−1) = 2^m−1, D_n + 3*Y_n−1 should be ≤2^m−1, and result in 3*Y_n−1 = 0 and D_n = 2^m−1. For Y_n−1 cannot be 0, so that 2^k(n−1) must be less than 2^m−1, and the maximum possible value should be 2^m−2. So we got:

$2^{m-2}\le D_n<2^{m-1}$

$2^{m-2}\le 3\ast Y_{n-1}<2^{m-1}$

$2^{m-1}<D_n+3\ast Y_{n-1}\le 2^m$

See Figure 4 [C, D]

From above analysis, we can get:

${\Delta }_n=2^{k\left(n-1\right)}=2^{m-2}$

so that

$3\ast Y_{n-1}+{\Delta }_n=3\ast Y_{n-1}+2^{k\left(n-1\right)}=Y_n$ (see Figure 4 [F, G, H])

Y_n satisfies Equation [5] , Collatz Conjecture should be TRUE for any positive integer.

Please note that, the Y_n based on condition “ $Y_{n-1}\le D_{n-1}$ to $Y_n>D_n$“ is not the first one satisfies Equation [5] , it is just used to prove such Y_n exists. The test results shown that the first Y satisfies Equation [5] should be 2 steps earlier. See Figure 5 (for number 7). At step 7, Y changed from less than D to greater than D, and the first Y satisfies Equation [5] happened at step 5.

From step 5, m became a straight line which means the odd part of Equation [6] keeps the same number. The minimum difference between m and k should be 4 (from number 5, 3 × 5 + 1 = 16 = 2⁴, will be explained later) at the first time that Y satisfies Equation [5] (step 5 in Figure 5 ). From step 6, k became a straight line, and m − k = 2, which means the odd part of Equation [6] goes into the 1 → 4 → 2 → 1 loop as Collatz Conjecture iteration.

All odd number are grouped by 4n + 1 sequence for Collatz Conjecture. All numbers in the same sequence will go to the same odd number on Collatz Conjecture iteration with one odd step and one or more even steps:

$3\ast \left(4n+1\right)+1=3\ast 4n+3+1=3\ast 4n+4=4\ast \left(3n+1\right)\to 3n+1$

Start from 1 we got:

S(1) = 1, 5, 21, 85, 341, ...

the nth number is: $S\left(1\right)\left[n\right]=\left(2^{2n}-1\right)/3$ $n=1,2,3,\cdots$

All numbers in S(1) are going to 1 on one odd-step and

2, 4, 6, 8, 10, ... even-step. (even sequence).

S(3) = 3, 13, 53, 213, 841, ...

the nth number is: $S\left(3\right)\left[n\right]=2^{2n-1}+S\left(1\right)\left[n\right]$ $n=1,2,3,\cdots$

All numbers in S(3) are going to 5 (SeqN) on one odd-step and

1, 3, 5, 7, 9, ... even-step. (odd sequence).

5 is 2nd number of S(1).

S(7) = 7, 29, 117, 469, 1877, ... - odd sequence, SeqN = 11.

the nth number is: $S\left(7\right)\left[n\right]=3\ast 2^{2n-1}+S\left(1\right)\left[n\right]$ $n=1,2,3,\cdots$

S(9) = 9, 37, 149, 597, 2389, ... - even sequence, SeqN = 7.

the nth number is: $S\left(9\right)\left[n\right]=2\ast 2^{2n}+S\left(1\right)\left[n\right]$ $n=1,2,3,\cdots$

S(11) = 11, 45, 181, 725, 2901, ... - odd sequence, SeqN = 17.

the nth number is: $S\left(11\right)\left[n\right]=5\ast 2^{2n-1}+S\left(1\right)\left[n\right]$ $n=1,2,3,\cdots$

13 is the 2nd number of S(3).

S(15) = 15, 61, 245, 981, 3925, ... - odd sequence, SeqN = 23.

the nth number is: $S\left(15\right)\left[n\right]=7\ast 2^{2n-1}+S\left(1\right)\left[n\right]$ $n=1,2,3,\cdots$

S(17) = 17, 69, 277, 1109, 4437, ... - even sequence, SeqN = 13.

the nth number is: $S\left(17\right)\left[n\right]=4\ast 2^{2n}+S\left(1\right)\left[n\right]$ $n=1,2,3,\cdots$

S(19) = 19, 77, 309, 1237, 4949, ... - odd sequence, SeqN = 29.

the nth number is: $S\left(19\right)\left[n\right]=9\ast 2^{2n-1}+S\left(1\right)\left[n\right]$ $n=1,2,3,\cdots$

21 is the 3rd number of S(1).

S(23) = 23, 93, 373, 1493, 5973, ... - odd sequence, SeqN = 35.

the nth number is: $S\left(23\right)\left[n\right]=11\ast 2^{2n-1}+S\left(1\right)\left[n\right]$ $n=1,2,3,\cdots$

S(25) = 25, 101, 405, 1621, 6485, ... - even sequence, SeqN = 19.

the nth number is: $S\left(25\right)\left[n\right]=6\ast 2^{2n}+S\left(1\right)\left[n\right]$ $n=1,2,3,\cdots$

S(27) = 27, 109, 437, 1749, 6997, ... - odd sequence, SeqN = 41.

the nth number is: $S\left(3\right)\left[n\right]=13\ast 2^{2n-1}+S\left(1\right)\left[n\right]$ $n=1,2,3,\cdots$

29 is the 2nd number of S(7).

S(31) = 31, 125, 501, 2005, 8021, ... - odd sequence, SeqN = 47.

the nth number is: $S\left(31\right)\left[n\right]=15\ast 2^{2n-1}+S\left(1\right)\left[n\right]$ $n=1,2,3,\cdots$

S(33) = 33, 133, 533, 2133, 8533, ... - even sequence, SeqN = 25.

the nth number is: $S\left(33\right)\left[n\right]=8\ast 2^{2n}+S\left(1\right)\left[n\right]$ $n=1,2,3,\cdots$

S(35) = 35, 141, 565, 2261, 9045, ... - odd sequence, SeqN = 53.

the nth number is: $S\left(35\right)\left[n\right]=17\ast 2^{2n-1}+S\left(1\right)\left[n\right]$ $n=1,2,3,\cdots$

37 is the 2^nd number of S(9).

··· ···

For any odd number D = Q*2 + 1,

if the integer quotient of SeqN of S(x) divided by 3 is q, for a odd sequence, the nth number is:

$S\left(x\right)\left[n\right]=q\ast 2^{2n-1}+S\left(1\right)\left[n\right]$ $n=1,2,3,\cdots$

for an even sequence, the nth item is:

$S\left(x\right)\left[n\right]=q\ast 2^{2n}+S\left(1\right)\left[n\right]$ $n=1,2,3,\cdots$

If line up the sequences described above, and put the odd numbers in a chart, we can get a spiral as shown in Figure 6 Collatz Spiral. All numbers on the same radial line are in one sequence. An odd sequence will go to an odd number greater than the first sequence number. An even sequence will go to an odd number less than the first sequence number. S(1) is a special even sequence just loop on its first number 1.

From Figure 6 we can see that each time the number across S(1), the new sequences will be added. If breaks at that point and makes rings, the sequences can show the sequences more clearly as shown in Figure 7 . Expanding the ring by applying 4n + 1 on each number, 3 new sequences added between any adjacent two numbers.

From Collatz Ring, it can be found that “the corresponding relationship of an odd number and a 4n + 1 sequence can be only one-to-one correspondence”.

S(1) -> 1

S(3) -> 5

S(7) -> 11

S(9) -> 7

S(11) -> 17

S(15) -> 23

S(17) -> 13

S(19) -> 29

S(23) -> 35

S(25) -> 13

S(27) -> 41

S(31) -> 47

S(33) -> 25

S(35) -> 53

... ...

So that only odd numbers in S(1) can go to 1 on Collatz Conjecture iteration, and the minimum odd number turn to 1 is 5 (the 2nd number). To take S(1) as the trunk, there should be only one branch (4n + 1 sequence) connected to each number in the S(1). In the same way, there should be only one sub branch connected each number in the branches. For any branch, there should no loop exist. For any odd number on the tree, there should be only one route down to the trunk S(1) on Collatz Conjecture iteration, then go to 1. All of the odd numbers should be on the tree which means for any odd number, there is a route and only one route to the trunk S(1) then to 1 on Collatz Conjecture iteration. Figure 8 and Figure 9 show a Collatz tree with 21 numbers in the S(1) trunk. There is no branch connected to Group 0 number.