A Solution of The Collatz Conjecture Problem

1. Introduction about The Collatz Conjecture

The Collatz Conjecture is a famous math conjecture, named after mathematician LOTHAR COLLATZ, who introduced the idea in 1937. It is also known as the 3x + 1 conjecture, the Ulam conjecture [1] etc. Many mathematicians have tried to prove it true or false and have expanded it to more digits scale. But until today, it has not yet been proved.

The Collatz Conjecture concerns sequences of positive integers in which each term is obtained from the previous one as follows: if the previous integer is even, the next integer is the previous integer divided by 2, till to odd. If the previous integer is odd, the next term is the previous integer multiply 3 and plus 1. The conjecture is that these sequences always reach 1, no matter which positive integer is chosen to start the sequence [1].

Here is an example for a typical integer x = 27, takes up to 111 steps, increasing or decreasing step by step, climbing as high as 9232 before descending to 1 [1].

27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1.

If the conjecture is false, there should exists some starting number which gives rise to a sequence that does not contain 1. Such a sequence would either enter a repeating cycle that excludes 1, or increase without bound [1]. No such sequence has been found by human and computer after verified a lot of numbers can reach to 1. It is very difficult to prove these two cases exist or not.

This paper will try to prove the conjecture true from a special view. Because any even can become odd through $÷2^k$ operation, this paper will research only odd characters in the conjecture sequence. The equivalence conjecture become: with random starting odd x, do $(\times 3+1)÷2^k$ operation repeatedly, it always converges to 1. The above sequence can be written as following, in which numbers on arrows are k in $÷2^k$ in each step:

$$\begin{array}{l}27\stackrel{1}{\to }41\stackrel{2}{\to }31\stackrel{1}{\to }47\stackrel{1}{\to }71\stackrel{1}{\to }107\stackrel{1}{\to }161\stackrel{2}{\to }121\stackrel{2}{\to }91\stackrel{1}{\to }137\stackrel{2}{\to }103\stackrel{1}{\to }\\ 155\stackrel{1}{\to }233\stackrel{2}{\to }175\stackrel{1}{\to }263\stackrel{1}{\to }395\stackrel{1}{\to }593\stackrel{2}{\to }445\stackrel{3}{\to }167\stackrel{1}{\to }251\stackrel{1}{\to }377\stackrel{2}{\to }\\ 283\stackrel{1}{\to }425\stackrel{2}{\to }319\stackrel{1}{\to }479\stackrel{1}{\to }719\stackrel{1}{\to }1079\stackrel{1}{\to }1619\stackrel{1}{\to }2429\stackrel{3}{\to }911\stackrel{1}{\to }1367\stackrel{1}{\to }\\ 2051\stackrel{1}{\to }3077\stackrel{4}{\to }577\stackrel{2}{\to }433\stackrel{2}{\to }325\stackrel{4}{\to }61\stackrel{3}{\to }23\stackrel{1}{\to }35\stackrel{1}{\to }53\stackrel{5}{\to }5\stackrel{4}{\to }1\end{array}$$

2. Build Equation For The Conjecture

If odd x do n times $(\times 3+1)÷2^k$ calculation build odd y, we can get:

$$y=\frac{3^{n}x+3^{n-1}+3^{n-2}\times 2^{p1}+3^{n-3}\times 2^{p_1+p_2}\mathrm{...}+3\times 2^{p_{{}_1}+p_2^{}+\mathrm{...}+p_{n-2}{}_{}}{}^{}+2^{p_1+p_2+\mathrm{...}+p_{n-1}}}{2^{p_1+p_2+\mathrm{...}+p_n}}$$

In which p₁…p_n is k in $÷2^k$ operation in each step.

For example: $(7\times 3+1)÷2=11$, $(11\times 3+1)÷2=17$, then $17=\frac{3^2\times 7+3+2}{2^2}$

Suppose odd x can converge to 1 through $(\times 3+1)÷2^k$ calculation, then y=1, get:

$$3^{n}x+3^{n-1}+3^{n-2}\times 2^{p_1}+3^{n-3}\times 2^{p_1+p_2}\mathrm{...}+3\times 2^{p_{{}_1}+p_2^{}+\mathrm{...}+p_{n-2}{}_{}}{}^{}+2^{p_1+p_2+\mathrm{...}+p_{n-1}}-2^{p_1+p_2+\mathrm{...}+p_n}=0$$

(1)

We know $(1\times 3+1)÷2^2=1$, and can do any times this kind of operation. That is to say, 1 do random n steps $(\times 3+1)÷2^2$ operation can converge to 1, have:

$3^n+3^{n-1}+3^{n-2}\times 2^2+3^{n-3}\times 2^4\mathrm{...}+3\times 2^{2n-4}{}^{}+2^{2n-2}-2^{2n}=0$

Below we use this model to prove and search for solution of Formula (1) for any odd x converging to 1.

3. Solution to the Any Odd Converging To 1 Equation

First with odd x do reform:

$$x=a_m\times 3^m+a_{m-1}\times 3^{m-1}+\mathrm{...}+a_1\times 3+a_0,$$

a_m…a₀=0, 1 or 2. Then:

$$3^{n}x=3^n\times (a_m\times 3^m+a_{m-1}\times 3^{m-1}+\mathrm{...}+a_1\times 3+a_0)$$

If a_m>1 or a_m=1 but

$$(a_{m-1}\times 3^{n+m-1}+\mathrm{...}+a_1\times 3^{n+1}+a_0\times 3^n)>(3^{n+m-1}+3^{n+m-2}{}^{}\times 2^2\mathrm{...}+3^n\times 2^{2(m-1)}),$$

Make

$$x=3^{m+1}-3^m+a_{m-1}\times 3^{m-1}+\mathrm{...}+a_1\times 3+a_0$$

or:

$$x=3^{m+1}-2\times 3^m+a_{m-1}\times 3^{m-1}+\mathrm{...}+a_1\times 3+a_0$$

Build identical equation:

$$3^{n+m}+3^{n+m-1}+3^{n+m-2}\times 2^2+3^{n+m-3}\times 2^4\mathrm{...}+3^{n-1}\times 2^{2m}\mathrm{...}+3\times 2^{2(n+m)-4}{}^{}+2^{2(n+m)-2}-2^{2(n+m)}=0$$

(2)

If x can converge to 1, Formula (1) and Formula (2) should be equivalence. Below we try to reform Formula (2) to form of Formula (1), if success it proves that equation for Formula (1) has solution.

First let:

$$(3^{n+m-1}+3^{n+m-2}{}^{}\times 2^2\mathrm{...}+3^n\times 2^{2(m-1)})-(a_{m-1}\times 3^{n+m-1}+\mathrm{...}+a_1\times 3^{n+1}+a_0\times 3^n)=t_n\times 3^n,$$

because x is odd, this is odd minus even, t_n should be odd.

Because the max value of x-3^m is $2\times 3^{m-1}+2\times 3^{m-2}+\mathrm{...}+2\times 3+2$, min value is $-3^{m-1}+1$, then t_n has a range:

From $(3^{m-1}+3^{m-2}{}^{}\times 2^2\mathrm{...}+2^{2(m-1)})-(2\times 3^{m-1}+2\times 3^{m-2}+\mathrm{...}+2\times 3+2)$ to $(3^{m-1}+3^{m-2}{}^{}\times 2^2\mathrm{...}+2^{2(m-1)})-(-3^{m-1}+1)$.

Change t_n to binary form and let:

$$t_n\times (2+1)\times 3^{n-1}+3^{n-1}\times 2^{2m}-3^{n-1}=t_{n-1}\times 3^{n-1},$$

this is just with 3ⁿ part multiply (2+1) become 3^n-1 part, and plus corresponding part in Formula (2), minus corresponding part in Formula (1). From now on, t_n-1 become even. Continue:

$$t_{n-1}\times (2+1)\times 3^{n-2}+3^{n-2}\times 2^{2m+2}-3^{n-2}\times 2^{p_1}=t_{n-2}\times 3^{n-2},$$

and let 2^p1 be equal to the max value of even part(or the lowest bit of odd part).

Watch Formula (1) and Formula (2), in general, if do not consider 2^p1+… part (because we consider 2^p1+… as max value of even part of t_i-2) in Formula (1), corresponding parts in Formula (2) are bigger than corresponding part in Formula (1). Hence after a few times of $t_{i-1}\times (2+1)$, value of t_i-2 is mainly determined by corresponding part in Formula (2). And, after $t_{i-1}\times (2+1)$, odd part should add 1 or 2 bits, if add 1 bit, $+2^{2m+2}$ should operate in MSB bit; if add 2 bits, $+2^{2m+2}$ should operate in MSB-1 bit. Both cases odd part add 2 bits after $+2^{2m+2}$ operation, if MSB bit of t_i-2 is 2^k, k should be odd.

For example:

$$3+2^2=7,7\times (2+1)+2^4-1=9\times 2^2,9\times 2^2\times (2+1)+2^6-2^2=21\times 2^3$$

Continue:

$$t_{n-2}\times (2+1)\times 3^{n-3}+3^{n-3}\times 2^{2m+4}-3^{n-3}\times 2^{p_1+p_2}=t_{n-3}\times 3^{n-3},$$

let 2^p1+p2 be equal to the max value of even part. Because LSB bit sequence number of odd part of t_i increases continuously, this can be finished easily.

Watch t_i(i<n and decreases step by step), during iteration, the count of succession 1 in the highest part should be unchanged or increased. Why? This is because of characters of odd multiply 3 and $+2^{2m}$ operation. If t_i-1 is with binary form 10…, obviously, count of succession 1 in highest part of t_i-2 is unchanged or increased. Similar to binary form 110…. Suppose t_i-1 is with binary form 1…1(k>2 bits of 1)0…, do $\times (2+1)$, head part should become 101…1(k-2 bits of 1)01…, do $+2^{2m}$, become 1…1(k bits of 1)01…, if tail part carry 1 to head part after doing $\times (2+1)$, head part become 1…1(k+1 bits of 1)0….

Do this iteration continuously, count of succession 1 in the highest part of odd part of t_i is unchanged or increased, LSB bit sequence number is also increased. Hence, finally, t_i can become form of 11…, just $2^k\times (2^j-1)$ form (k+j=odd). Stop here, do not do $\times (2+1)$ again, odd x already converge to 1. Do $-2^{2(n+m)}$ operation, it should operate in MSB+1 bit, because MSB bit sequence number of $+2^{2k}$ is forever equal to MSB+1 bit sequence number of the previous item. Hence minus result can be equal to $-2^{p_1+p_2+\mathrm{...}+p_n}$, thus get solution of Formula (1).

Below give a specific example, x=7.

We know, with 7 do $(\times 3+1)÷2^k$, have:

$$7\stackrel{1}{\to }11\stackrel{1}{\to }17\stackrel{2}{\to }13\stackrel{3}{\to }5\stackrel{4}{\to }1$$

Suppose:

$$3^n\times 7+3^{n-1}+3^{n-2}\times 2^{p_1}+3^{n-3}\times 2^{p_1+p_2}\mathrm{...}+3\times 2^{p_{{}_1}+p_2^{}+\mathrm{...}+p_{n-2}{}_{}}{}^{}+2^{p_1+p_2+\mathrm{...}+p_{n-1}}-2^{p_1+p_2+\mathrm{...}+p_n}=0$$

$$3^n\times 7=3^n\times (2\times 3+1)=3^n\times (3^2-3+1)=3^{n+2}-3^{n+1}+3^n$$

Build:

$$3^{n+2}+3^{n+1}+3^n\times 2^2+3^{n-1}\times 2^4\mathrm{...}+3\times 2^{2n}{}^{}+2^{2n+2}-2^{2n+4}=0$$

$$3^{n+1}+3^n\times 2^2+3^{n+1}-3^n=(2^3+1)\times 3^n$$

$${ }^{*}(2+1) \mathrm{and}+2^4: (2^3+1)\times (2+1)\times 3^{n-1}+2^4\times 3^{n-1}=(2^5+2^3+2^{}+1)\times 3^{n-1}$$

$$-3^{\mathrm{n}-1}: (2^5+2^3+2^{}+1)\times 3^{n-1}-3^{n-1}=(2^5+2^3+2^{})\times 3^{n-1}$$

$${ }^{*}(2+1) \mathrm{and}+2^6: (2^5+2^3+2)\times (2+1)\times 3^{n-2}+2^6\times 3^{n-2}=(2^7+2^5+2^4+2^3+2^2+2)\times 3^{n-2},$$

Let p₁=1, and delete item 2:

$$(2^7+2^5+2^4+2^3+2^2+2-2)\times 3^{n-2}=(2^7+2^5+2^4+2^3+2^2)\times 3^{n-2}$$

$${ }^{*}(2+1) \mathrm{and}+2^8: (2^7+2^5+2^4+2^3+2^2)\times (2+1)\times 3^{n-3}+2^8\times 3^{n-3}=(2^9+2^8+2^5+2^4+2^2)\times 3^{n-3}$$

Let p₁+p₂=2, and delete item 2²:

$$(2^9+2^8+2^5+2^4+2^2-2^2)\times 3^{n-3}=(2^9+2^8+2^5+2^4)\times 3^{n-3}$$

$${ }^{*}(2+1) \mathrm{and}+2^{10}: (2^9+2^8+2^5+2^4)\times (2+1)\times 3^{n-4}+2^{10}\times 3^{n-4}=(2^{11}+2^{10}+2^8+2^7+2^4)\times 3^{n-4}$$

Let p₁+p₂+p₃=4, and delete item 2⁴:

$$(2^{11}+2^{10}+2^8+2^7+2^4-2^4)\times 3^{n-4}=(2^{11}+2^{10}+2^8+2^7)\times 3^{n-4}$$

$${ }^{*}(2+1) \mathrm{and}+2^{12}: (2^{11}+2^{10}+2^8+2^7)\times (2+1)\times 3^{n-5}+2^{12}\times 3^{n-5}=(2^{13}+2^{12}+2^{11}+2^7)\times 3^{n-5}$$

Let p₁+p₂+p₃+p₄=7, and delete item 2⁷:

$$(2^{13}+2^{12}+2^{11}+2^7-2^7)\times 3^{n-5}=(2^{13}+2^{12}+2^{11})\times 3^{n-5}$$

Now become 111…, the highest bit is 2¹³, iteration finished, steps n=5. And

$$2^{13}+2^{12}+2^{11}-2^{(2\times 5+4)}=-2^{11}=-2^{p_1+\mathrm{...}+p_5}.$$

This way, we get a solution for Formula (1), in which the value of n and p_i is exactly same with the result got from calculating directly.

4. Convergence Regularity Of Collatz Conjecture

4.1. Equivalent of (*3+1)/2^k and (*3+2^m-1)/2^k operation

If we calculate directly with odd through $(\times 3+1)÷2^k$ operation, the odd sequence built (called Sequence (1)) has no obvious converge regularity, elements in the sequence vary sometimes big, sometimes small. But if we do operation as introduced in above section, convergence regularity of the odd sequence built (called Sequence (2)) is more obvious.

If add two corresponding elements in each step in these two odd sequences, should be exactly 2^k(k is different with different elements). Such as

$7+9=16,11+21=32,17+47=64$…

in above example.

In general, first element in Sequence (2) is:

$a=(3^{m-1}+3^{m-2}\times 2^2\mathrm{...}+2^{2(m-1)})-(a_{m-1}\times 3^{m-1}+\mathrm{...}+a_1\times 3+a_0)$and first element in Sequence (1) is:

$x=3^m+a_{m-1}\times 3^{m-1}+\mathrm{...}+a_1\times 3+a_0$,

then

$x+a=3^m+3^{m-1}+3^{m-2}\times 2^2\mathrm{...}+2^{2(m-1)}=2^{2m}$,

is just the same form with Formula (2), and 2m should be the MSB+1 bit sequence number of x or a(along with the increase of a in Sequence (2), 2m should be the MSB+1 bit sequence number of a, because each corresponding part in Formula (2) is bigger than which in Formula (1), in fact we can select first a>x manually).

Below prove next elements also satisfy above regularity.

Suppose a in Sequence (2) and x in Sequence (1) satisfy above regularity, and:

$a=2^m+a_{m-1}\times 2^{m-1}+\mathrm{...}+a_1\times 2+1$,

$x=2^{m+1}-a$,

then

$3a+2^{m+1}-1=3\times 2^m+3\times a_{m-1}\times 2^{m-1}+\mathrm{...}+3\times a_1\times 2+3+2^{m+1}-1$,

$3x+1=3\times 2^{m+1}-3\times 2^m-3\times a_{m-1}\times 2^{m-1}-\mathrm{...}-3\times a_1\times 2-3+1$,

$(3x+1)+(3a+2^{m+1}-1)=4\times 2^{m+1}=2^k$

This states that the lowest bit of odd part of (3x+1) and (3a+2^m+1-1) is equal, and add these two odd parts should be 2ⁱ(i<k).

Above regularity states that the original odd sequence has no obvious regularity is because it is only the partial part, not the whole part.

4.2. Weight Function and Its Monotonically Decrease Character

Research into odd multiplying 3, any odd can be written in binary form 1…1, both the highest and lowest bit is 1, after $\times 3$, although total bit number increases, first sub-step is to shift bit 1 to the middle of the result, second sub-step may make carry to higher bit due to 1+1 in the middle of the result(1-bits in the middle of odd also satisfy this regularity). Both substeps are beneficial to our final goal, because we need many 1 bits in final result. $+2^{2k}$ operation ensure succession 1 bits in the highest part, -1 operation reduce count of isolated 1 bits in the lowest part. Hence 0-bits in the odd part in t_i should shift right or bit-count reduce in each step, and its weight in total t_i should reduce step by step till to 0, when the odd part converges to 1...1. Build a simple weight model:

$w_i=\frac{{\mathrm{value} \mathrm{of} \mathrm{all} 0 \mathrm{bits} \mathrm{in} \mathrm{odd} \mathrm{part} \mathrm{in} \mathrm{t}}_{\mathrm{i}}}{2^{2\mathrm{k}}}$ Definition (1)

Which 2^2k is corresponding addition part in t_i in the step. Simply we can use w_i represent the weight of value of all 0 bits in odd part in t_i. We can also think $2^{2k}$ as the sum of t_i and its corresponding part in original sequence, the conclusion we final got is same. Specially, with any odd a, which highest bit is 2^m, define w_i for this odd:

$w_{[a]}=\frac{\mathrm{value} \mathrm{of} \mathrm{all} 0 \mathrm{bits} \mathrm{in} \mathrm{odd} \mathrm{a}}{2^{\mathrm{m}+1}}$ Definition (2)

Although the denominator may be bigger than which in Definition (1), the regularity is same.

Observe w_i, it should reduce step by step, and model value can and must converge to 0, because there is no possibility to exist a convergence value, which its corresponding odd part in t_i is not 1...1, and its model value can remain unchanged in next steps through multiplying 3 operation and other two operations.

t_i sequence in above example is: 9,42,188,816,3456,14336

odd part sequence is: 9,21,47,51,27,7

w_i sequence is(according to Definition (1)):

(2+4)/4=1.5,(4+16)/16=1.25,64/64=1,(64+128)/256=0.75,512/1024=0.5,0/4096=0

Through above introduction we know, with odd we do $(\times 3+1)÷2^k$ operation in the Collatz Conjecture, on the contrast, with odd we do $(\times 3+2^m-1)÷2^k$ in above iteration method. We can easily prove that odd 1…10a(a is in binary base) is equivalent to odd 10a in second method, count of succession 1 bits in the head part only represent the iteration steps roughly.

In fact, only one case 0 bits in t_i do not shift right or bit-count reduce when t_i has not converged. This is:

101->1011.

This case w_i do not change, both are 1/4, according to Definition (2). But next step 1011->11, t_i converges, hence this case is not worth worrying about.

Below we prove it strictly.

Suppose with odd a do $(\times 3+1)÷2^k$ operation, and use x represent iteration steps. We can reform w_i as following(according to Definition (1)), the numerator part is exactly equal to 0 bits in t_i:

$w(x)=\frac{3^{x}a+3^{x-1}+3^{x-2}\times 2^{p_1}+3^{x-3}\times 2^{p_1+p_2}\mathrm{...}+3\times 2^{p_{{}_1}+p_2^{}+\mathrm{...}+p_{x-2}{}_{}}{}^{}+2^{p_1+p_2+\mathrm{...}+p_{x-1}}-2^{p_1+p_2+\mathrm{...}+p_x}}{2^{2k_x}}$

Obviously w(x) is continuous derivable when a is in odd domain definition and x is in positive integer domain definition, and is bounded(>=0).

Now we try to take the derivative of w(x).

Here the derivation definition of the numerator and denominator is: (y(x+1)-y(x))/(x+1-x).

Then the derivation of the numerator is:

$2\times (3^{x}a+3^{x-1}+3^{x-2}\times 2^{p_1}+\mathrm{...}+3\times 2^{p_{{}_1}+p_2^{}+\mathrm{...}+p_{x-2}{}_{}}{}^{}+2^{p_1+p_2+\mathrm{...}+p_{x-1}})+2^{p_1+p_2+\mathrm{...}+p_x}+2^{p_1+p_2+\mathrm{...}+p_x}-2^{p_1+p_2+\mathrm{...}+p_{x+1}}$

The derivation of the denominator is:

$2^{2k_x+2}-2^{2k_x}=3\times 2^{2k_x}$

Then

$\begin{array}{l}{w}^{\text{'}}(x)=\frac{2\times 2^{2k_x}\times 2^{p_1+p_2+\mathrm{...}+p_x}+3\times 2^{2k_x}\times 2^{p_1+p_2+\mathrm{...}+p_x}-2^{2k_x}\times 2^{p_1+p_2+\mathrm{...}+p_{x+1}}-(3^{x}a+3^{x-1}+\mathrm{...}+2^{p_1+p_2+\mathrm{...}+p_{x-1}})\times 2^{2k_x}{}_{}}{2^{4k_x}}\\ =\frac{(5-2^{p_{x+1}})\times 2^{2k_x}\times 2^{p_1+p_2+\mathrm{...}+p_x}-b\times 2^{p_1+p_2+\mathrm{...}+p_x}\times 2^{2k_x}}{2^{4k_x}}=\frac{(5-2^{p_{x+1}}-b)\times 2^{p_1+p_2+\mathrm{...}+p_x}}{2^{2k_x}}\end{array}$

Which b is the odd after odd a doing x steps $(\times 3+1)÷2^k$ operation. that is:

$3^{x}a+3^{x-1}+3^{x-2}\times 2^{p_1}+\mathrm{...}+3\times 2^{p_{{}_1}+p_2^{}+\mathrm{...}+p_{x-2}{}_{}}{}^{}+2^{p_1+p_2+\mathrm{...}+p_{x-1}}=b\times 2^{p_1+p_2+\mathrm{...}+p_x}$

Observe w′(x), we know when b>3, w′(x)<0, w(x) monotonically decreases. Only when b=1(this case $2^{p_{x+1}}$should be equal to 4), or when b=3, $2^{p_{x+1}}=2$, w′(x)=0. Second case of b=3 is the exception case introduced above, the corresponding odd part of t_i is with form ‘101’, is not worth worrying about. First case is convergence case.

Totally, this kind of iteration calculation has these cases after doing as following:

Case 1: odd tail part decreases one bit, head part does not increase one bit, this case tail part should insert one bit of 1 and with zero or more 0 changing to 1, totally 1 bits weight should increase in tail part.

Case 2: odd tail part decreases one bit, head part increases one bit, if corresponding odd in  sequence changes bigger, is just because tail part carries one bit of 1 to head part; if corresponding odd changes smaller, is just we need.

Case 3: odd tail part decreases two bits, head part does not increase one bit, tail part 0 bits should shift right.

Case 4: odd tail part decreases two bits, head part increases one bit.

Case 5: odd tail part decreases three or more bits, head part increases zero or one bit.

All these cases, w(x) function are decreased step by step except some cases introduced above.

4.3. Case of Weight Function tending to 0 but not equal to 0

Does it exist some odds which its w_i tends to 0 but is not equal to 0 forever? In fact, it exists some odds which 0-bits distributions are similar and w_i decreases if they exist in same sequence. Such as, 10001 and 110001(+2⁵) or 11000011(*4-1), 10001 and 1100001(insert 0). Because the $(\times 3+2^m-1)÷2^k$ operation limits the varying of the highest part of odd, these odds could not be possible to appear in the same sequence, also could not repeatedly appear.

For example:

10001->101001->1011101->11001011->11011->111, could not produce similar 0-bits distribution.

Below prove it from another view.

Suppose odd a is in $(\times 3+1)÷2^k$ operation sequence, its corresponding odd in $(\times 3+2^m-1)÷2^k$ operation sequence is b, which highest bit is 2^m, then according to Definition (2), $w_{[b]}=\frac{\mathrm{a}-1}{2^{\mathrm{m}+1}}$.

Next Step, b become odd c, then $w_{[c]}=\frac{3\mathrm{a}+{1-2}^{\mathrm{p}}}{2^{\mathrm{m}+1}\times 4}$, where 2^p is the lowest bit of odd part.

$\frac{w_{[c]}}{w_{[b]}}=\frac{3\mathrm{a}+{1-2}^{\mathrm{p}}}{4\times (a-1)}=\frac{3}{4}-\frac{2^p-4}{4\times (a-1)}<\frac{3}{4}+\frac{1}{2\times (a-1)}$,

When a is big enough, for example a>=2¹⁰+1, $\frac{w_{[c]}}{w_{[b]}}<0.751$.

This means when odd in $(\times 3+1)÷2^k$ operation sequence is big enough, next step, w_i is smaller than which multiply 0.751 in current step.

In above example, for first odd, $w_{[10001]}=\frac{7}{16}$, for other odds, $w_{[110001]}=\frac{7}{32}$, $w_{[11000011]}=\frac{15}{64}$, $w_{[1100001]}=\frac{15}{64}$, w_i for all other odds is equal to or bigger than w_i *0.5 for first odd.

Any odds have this same regularity. Because when the tail part of the odds remain unchanged or insert 0(any tail position), the numerator part is same or bigger than 2 times of original, and the denominator become same or 2 times of original, when the head part(successive 1 part) of the odds add one 1, the denominator become 2 times again, then the final value should be bigger than 0.5 times than original.

In above example, obviously, first odd could not become other odds in within 3 steps(case of huge odds is same). But 0.751*0.751*0.751=0.423564751<0.5, it is contradictious.

If steps increase, it is also not possible to become other odds, because if steps increase, count of 1 in head part should also increase, this consumes many steps, there are no enough steps left to finish the need deformation.

We know, normally if only think about varying of head part, it needs 2 or 3 steps periodically to finish adding one 1 to head part, if tail part carries one bit of 1 to head part, it decreases 1 step. And tail part is not possible to carry 1 bit two times to head part when head part add two 1 successively, because each time head part add one 1 or tail part carry 1 bit to head part, highest part of tail part produces two more 0 bits, it could not produce carrying bit successively. This is to say, normally in long odd sequence, each time head part add one 1, it at least need about 2 more steps(we ignore odds needing only 1 step to add one 1 to head part in first step here, and we also ignore odds with form 10111(many many 1)..., because although this kind of odds need 2 steps to finish adding one 1 to head part successively during some steps, it decreases count of successive 1 in tail part after each step, this is not good for changing to similar 0-bits distribution).

We know loop odd sequence and divergence odd sequence both are long sequence which has much more than 4 elements(3 steps). Suppose any huge start odd a(its corresponding odd in $(\times 3+1)÷2^k$ sequence is bigger than 2¹⁰+1), a add x bits of 1 in head part and become huge odd b with similar 0-bits distribution of a, it at least need y steps to finish. Then w_[b] should be bigger than 0.5^x times of W_[a] from calculation directly, and should be smaller than 0.751^y times of W_[a] through iteration calculation character introduced in above. This is:

${0.751}^y>{0.5}^x$

$y\times \ln (0.751)>x\times \ln (0.5)$

$y<2.4207\times x$

But, no matter whether the deformation is finished or not, only to finish adding enough bits of 1 to head part, it need at least more than 2x steps(about 2.5x steps), there is no enough steps to do tail deformation. So far, the needing steps from these two angles may be contradictious.

5. The Complement Weight Function Of W_[a]

5.1. Convergence Regularity of Complement Weight Function

To avoid proving weight function W_[a] converging to 0(it is not easily to prove strictly the numerator part equal to 0 finally), we build its complement weight function. Build:

$w_{c[a]}=\frac{\mathrm{a}}{2^{\mathrm{m}+1}}$,

the highest bit of a is 2^m.

Through the proof and introduction above, we know W_c[a] monotonically increases except when corresponding odd b_i in $(\times 3+1)÷2^k$ sequence of a_i is 1 or 3, and these except cases are not worth worrying about. And we also know the convergence state of W_c[a] is $\frac{2^{\mathrm{k}}-1}{2^{\mathrm{k}}}$.

How much does W_c[a] increase in each step? Suppose odd a₀, a₁, a₂ are three elements in order in $(\times 3+2^m-1)÷2^k$ sequence, a₀ is equal to a, then

$w_{c[a_0]}=\frac{\mathrm{a}}{2^{\mathrm{m}+1}}$, $w_{c[a_1]}=\frac{3\mathrm{a}+2^{\mathrm{m}+1}-1}{2^{\mathrm{m}+3}}$, $w_{c[a_2]}=\frac{3^2\mathrm{a}+3\times 2^{\mathrm{m}+1}-3+2^{m+3}-2^p}{2^{\mathrm{m}+5}}$,

where 2^p is 2^k in first step $(\times 3+2^m-1)÷2^k$ operation.

$w_{c[a_1]}-w_{c[a_0]}=\frac{3\mathrm{a}+2^{\mathrm{m}+1}-1-4a}{2^{\mathrm{m}+3}}=\frac{2^{m+1}-a-1}{2^{m+3}}$, $w_{c[a_{{}_2}]}-w_{c[a_1]}=\frac{3^2\mathrm{a}+3\times 2^{\mathrm{m}+1}-3+2^{m+3}-2^p-12a-4\times 2^{m+1}+4}{2^{\mathrm{m}+5}}=\frac{3\times 2^{m+1}-3a-2^p+1}{2^{m+5}}$,

$\frac{w_{c[a_{{}_2}]}-w_{c[a_1]}}{w_{c[a_1]}-w_{c[a_0]}}=\frac{3\times 2^{m+1}-3a-2^p+1}{2^{m+5}}\times \frac{2^{\mathrm{m}+3}}{2^{m+1}-a-1}=\frac{3}{4}+\frac{4-2^p}{4\times (2^{m+1}-a-1)}$

Observe this formula, when 2^p is equal to 2 or 4, $\frac{w_{c[a_{{}_2}]}-w_{c[a_1]}}{w_{c[a_1]}-w_{c[a_0]}}$is $\ge \frac{3}{4}$, suppose this ratio is $\frac{3}{4}$, then

$w_{c[a_n]}=\frac{\mathrm{a}}{2^{\mathrm{m}+1}}+\frac{2^{m+1}-a-1}{2^{m+3}}\times (1+\frac{3}{4}+{(\frac{3}{4})}^2+{(\frac{3}{4})}^3+\mathrm{...}{(\frac{3}{4})}^{n-1})$,

When n->∞, $w_{c[a_n]}=\frac{\mathrm{a}}{2^{\mathrm{m}+1}}+\frac{2^{m+1}-a-1}{2^{m+3}}\times 4=\frac{2^{m+1}-1}{2^{\mathrm{m}+1}}$, this is a convergence state, and we know, in actual case, it needs a limit number n steps to reach to(or bigger than) $\frac{2^{m+1}-1}{2^{\mathrm{m}+1}}$, because the ratio is $\ge \frac{3}{4}$.

When 2^p is bigger than 4, $\frac{w_{c[a_{{}_2}]}-w_{c[a_1]}}{w_{c[a_1]}-w_{c[a_0]}}$is $<\frac{3}{4}$, but still $>\frac{1}{2}$, W_c[a] also increases, this time, there is not any other limit, it can increase till to its convergence state.

Of course, W_c[a] can converge in $\frac{2^{\mathrm{k}}-1}{2^{\mathrm{k}}}$(k is any positive integer), not only $\frac{2^{\mathrm{m}+1}-1}{2^{\mathrm{m}+1}}$. This increases the convergence chance of W_c[a].

5.2. Complement Weight Function Has Many Chances to Converge

Is it possible that W_c[a] increases continuously but never equal to $\frac{2^{\mathrm{k}}-1}{2^{\mathrm{k}}}$? For example, 87/128, 177/256, 357/512, 717/1024…(here does not consider ratio temporarily).

It is not possible. Observe the varying of fraction in lowest terms of W_c[a], the denominator part is equal, smaller, or 2 times of previous(because the numerator part at least can be divided by 2 in each step) in each step, when is equal, the numerator part should increase, it is possible to converge, when is 2 times of previous, the total value also increase, when is smaller, the total value should not only bigger than the value of front W_c[a] with same denominator part(if exist), but also bigger than all W_c[a] follow it. And in long sequence, usually appear the smaller case, it has many chances to appear $\frac{2^{\mathrm{k}}-1}{2^{\mathrm{k}}}$, especially when the front element is already close to its convergence state. For example, suppose 177/256 is in sequence, if some following element with same denominator part 256 appear after many steps, its value should be bigger than all the elements between 177/256 and itself, it is much possible to equal to 255/256.

Continuously observe W_c[a], even in the 2 times case, elements are closer to convergence state by themselves. Suppose the denominator part of fraction in lowest terms of $w_{c[a_1]}=\frac{3\mathrm{a}+2^{\mathrm{m}+1}-1}{2^{\mathrm{m}+3}}$ is 2^m+2,

$\frac{2^{m+2}-1}{2^{m+2}}-\frac{(3\mathrm{a}+2^{\mathrm{m}+1}-1)÷2}{2^{\mathrm{m}+2}}=\frac{3\times 2^m-\frac{3}{2}a-\frac{1}{2}}{2^{m+2}}$

$\frac{2^{m+1}-1}{2^{m+1}}-\frac{\mathrm{a}}{2^{\mathrm{m}+1}}=\frac{2^{m+1}-a-1}{2^{m+1}}$

$\frac{3\times 2^m-\frac{3}{2}a-\frac{1}{2}}{2^{m+2}}-\frac{2^{m+1}-a-1}{2^{m+1}}=\frac{3\times 2^m-\frac{3}{2}a-\frac{1}{2}-2^{m+2}+2a+2}{2^{m+2}}=\frac{\frac{1}{2}a-2^m+\frac{3}{2}}{2^{m+2}}=\frac{a-2^{m+1}+3}{2^{m+3}}$

We know 2^m<a<2^m+1-1, if a is not equal to 11…101, which is very close to its convergence state 11…1, the above formula is <0. Thus prove the above conclusion.

Below give an example of start number 27 in $(\times 3+1)÷2^k$ odd sequence to verify, some decimals are written in the form which is easily to be judged equal to, bigger or smaller than 0.75.

Odds in $(\times 3+2^m-1)÷2^k$ sequence are:

37,87,97,209,441,917,1887,1927,1957,3959,3993,8037,16151,16209,32505,65141,130479,130627,65369,130821,261767,261861,523863,523969,1048097,2096433,4193225,8386989,16774787,8387697,16775849,33552381,67105787,16776639,16776783,16776891,4194243,2097129,4194269,8388555,1048571,262143

W_c[a] sequence:

37/64,87/128,97/128,209/256,441/512,917/1024,1887/2048,1927/2048,1957/2048,3959/4096,3993/4096,8037/8192,16151/16384,16209/16384,32505/32768,65141/65536,130479/(65536*2),130627/(65536*2),65369/65536,130821/(65536*2),261767/(65536*4),261861/(65536*4),523863/(65536*8),523969/(65536*8),1048097/(65536*16),2096433/(65536*32),4193225/(65536*64),8386989/(65536*128),16774787/(65536*256),8387697/(65536*128),16775849/(65536*256),33552381/(65536*512),67105787/(65536*1024),16776639/(65536*256),16776783/(65536*256),16776891/(65536*256),4194243/(65536*64),2097129/(65536*32),4194269/(65536*64),8388555/(65536*128),1048571/(65536*16),262143/262144

$w_{c[a_{i+1}]}-w_{c[a_i]}$ sequence:

13/128,10/128,15/256,23/512,35/1024,53/2048,40/2048,30/2048,45/4096,34/4096,51/8192,77/16384,58/16384,87/32768,131/65536,197/(65536*2),148/(65536*2),111/(65536*2),83/(65536*2),125/(65536*4),94/(65536*4),141/(65536*8),106/(65536*8),159/(65536*16),239/(65536*32),359/(65536*64),539/(65536*128),809/(65536*256),607/(65536*256),455/(65536*256),683/(65536*512),1025/(65536*1024),769/(65536*1024),144/(65536*256),108/(65536*256),81/(65536*256),15/(65536*64),11/(65536*64),17/(65536*128),13/(65536*128),1/(65536*16)

$\frac{w_{c[a_{{}_{i+2}}]}-w_{c[a_{i+1}]}}{w_{c[a_{i+1}]}-w_{c[a_{{}_i}]}}$ sequence:

10/13≈0.77,0.75,0.77,0.76,0.76,0.755,0.75,0.75,0.76,0.75,0.755,0.753,0.75,0.753,0.752,0.751,0.75,0.748,0.753,0.752,0.75,0.752,0.75,0.752,0.751,0.751,0.750,0.750,0.749,0.751,0.750,0.750,0.749,0.75,0.75,0.741,0.73,0.77,0.76,0.62

5.3. An Equivalent Description of Collatz Conjecture

Through above we know$w_{c[a_1]}=\frac{3\mathrm{a}+2^{\mathrm{m}+1}-1}{2^{\mathrm{m}+3}}$, it can be written in following forms:

$w_{c[a_1]}=\frac{3\mathrm{a}+2^{\mathrm{m}+1}-1}{2^{\mathrm{m}+3}}=\frac{4\mathrm{a}+2^{\mathrm{m}+1}-a-1}{2^{\mathrm{m}+3}}=\frac{4\mathrm{a}+b-1}{2^{\mathrm{m}+3}}$

$w_{c[a_1]}=\frac{2\mathrm{a}+[\frac{b-1}{2}]}{2^{\mathrm{m}+2}}$,

b-1<>0 mod 4, or

$w_{c[a_1]}=\frac{\mathrm{a}+[\frac{b-1}{4}]}{2^{\mathrm{m}+1}}$,

b-1≡0 mod 4, in which b is the corresponding odd of a in $(\times 3+1)÷2^k$ sequence, b-1 reflects the 0-bits in the tail part of a.

Then Collatz Conjecture can be described as: With any odd a in range of 2^k to 2^k+1-1, set its initial goal set is 2^j+1-1(j<=k), its tail part is b, do operation: try to do (b-1) divided by 4, if can not, shift left one bit of a, plus the result of shifting right one bit of b(the 0-bits in the tail part of a), and add 2^k+2-1 to goals set of a, this operation makes the 0-bits in the tail part of a shift right or count reduce; if can, a plus the result of (b-1) divided by 4, this operation not only makes the 0-bits in the tail part of a shift right or count reduce, but also reduces the odds count about 1/4 to its goal 2^k+1-1, furthermore, if the last result is even, it can reduce a fraction of using 2^k+1 as denominator, this makes it can reach its previous goal 2^j+1-1(j<=k) possibly. Do these operations repeatedly, it have unlimited chances to reach one of its goal set.

Through above we know, if $(\times 3+1)÷2^k$ sequence have only /2 and(or) /4 cases, the sequence can never converge, /2 case makes goal of a in $(\times 3+2^m-1)÷2^k$ sequence larger, /4 case needs ∞ steps. But it is not possible in long sequence, this is determined by the regularity of tail binary bits of odd doing $(\times 3+1)÷2^k$operation. Odds of form with *10…01(many 0), both its initial value and result can do (-1)/4, Odds of form with *11…11(many 1), both its initial value and result can do (-1)/2, these two cases must become other forms after several steps. Odds with other forms, themselves and their following steps can appear alternately /2, /4, /2^k cases.

6. (*3+2^m-1)/2^k Odd Tree and Its Regularity

6.1. (*3+2^m-1)/2^k Odd Tree and Its Characters

Characters of 2^k are very regular, if we set odds of $(\times 3+1)÷2^k$ between 4^p+4^p-1+…+1 and 4^p+1+4^p+…+1 as one layer, call 2^k are the properties of these odds after doing $(\times 3+1)÷2^k$ operation, we can find each layer count of 2^2p+1, 2^2p,…2², 2 are 1, 1 2 4, 1 2 4 8 16…, their positions have equal interval space, 2^2p+1 is in the middle between 4^p and 4^p+1, 2^2p is in the middle of left part…, first position and step length of odds of different 2^k property are different after doing $(\times 3+1)÷2^k$ operation in different layers. In brief, characters of 2^k are very regular, we do not introduce in detail. Here we still put focus on $(\times 3+2^m-1)÷2^k$ odds. See following tree:

…

L6: 129(321.1) 131(81.3) 133(327.1) 135(165.2) 137(333.1) 139(21.5) 141(339.1) 143(171.2) 145(345.1) 147(87.3) 149(351.1) 151(177.2) 153(357.1) 155(45.4) 157(363.1) 159(183.2) 161(369.1) 163(93.3) 165(375.1) 167(189.2) 169(381.1) 171(3.8) 173(387.1) 175(195.2) 177(393.1) 179(99.3) 181(399.1) 183(201.2) 185(405.1) 187(51.4) 189(411.1) 191(207.2) 193(417.1) 195(105.3) 197(423.1) 199(213.2) 201(429.1) 203(27.5) 205(435.1) 207(219.2) 209(441.1) 211(111.3) 213(447.1) 215(225.2) 217(453.1) 219(57.4) 221(459.1) 223(231.2) 225(465.1) 227(117.3) 229(471.1) 231(237.2) 233(477.1) 235(15.6) 237(483.1) 239(243.2) 241(489.1) 243(123.3) 245(495.1) 247(249.2) 249(501.1) 251(63.4) 253(507.1) 255

L5: 65(161.1) 67(41.3) 69(167.1) 71(85.2) 73(173.1) 75(11.5) 77(179.1) 79(91.2) 81(185.1) 83(47.3) 85(191.1) 87(97.2) 89(197.1) 91(25.4) 93(203.1) 95(103.2) 97(209.1) 99(53.3) 101(215.1) 103(109.2) 105(221.1) 107(7.6) 109(227.1) 111(115.2) 113(233.1) 115(59.3) 117(239.1) 119(121.2) 121(245.1) 123(31.4) 125(251.1) 127

L4: 33(81.1) 35(21.3) 37(87.1) 39(45.2) 41(93.1) 43(3.6) 45(99.1) 47(51.2) 49(105.1) 51(27.3) 53(111.1) 55(57.2) 57(117.1) 59(15.4) 61(123.1) 63

L3: 17(41.1) 19(11.3) 21(47.1) 23(25.2) 25(53.1) 27(7.4) 29(59.1) 31

L2: 9(21.1) 11(3.4) 13(27.1) 15

L1: 5(11.1) 7

L0: 3

In above tree, a.b in () means result is a*2^b after front odd doing $(\times 3+2^m-1)÷2^k$ operation. m_th layer has 2^m elements, the last element is the convergence state. Characters of 2^k are also very regular, for example, upward from a specific layer, positions of 2 are 1+2i(i>=0), upward from another specific layer, positions of 2² are 4+4i, positions of 2³ are 2+8i, positions of 2⁴ are 14+16i…, this can be easily proved strictly. For example, odds of position 2+8i in m layer are 2^m+1-1+(2+8i)*2, (0=<i<=[(2^m-1-1)/4]).

$3\times (2^{m+1}-1+(2+8i)\times 2)+2^{m+2}-1=2^{m+3}+2^{m+1}+48i+8$

Can be divided by 2³, result is odd if m+1>3. And because the highest bit of the result odd is 2^m, it must be in m-1 layer, downward one layer from m layer.

Through above, we can easily prove that if the property of an odd is 2¹, it moves upward one layer(and also moves forward some location), if the property of an odd is 2², it moves forward in the same layer, if the property of an odd is 2^k(k>2), it moves downward k-2 layers(and also moves forward some location).

In this tree, because element count of each layer is 2 times of which of the downward layer, we can transform all positions to one specific layer. M-1 layer transform to m layer do *2, m+1 layer transform to m layer do /2, etc. Then all transformed positions can not exceed 2^m!

Below we try to prove odds in any layer can converge. Normally, we suppose the research sequence is long huge(odds in $(\times 3+1)÷2^k$ sequence are huge) sequence.

6.2. Transform Position Sequence and Its Convergence

Suppose a is an odd in m-1 layer, its highest bit is 2^m.

Pos of a in m-1 layer is: $\frac{a-2^m+1}{2}$,

$3\times a+2^{m+1}-1=b\times 2^{p_1}$, b is in layer m-p₁+1

Pos of b in m-p₁+1 layer is: $\frac{b-2^{m-p_1+2}+1}{2}$,

Pos of b in m-1 layer is: $\frac{b-2^{m-p_1+2}+1}{2^{3-p_1}}$

$3^2\times a+3\times 2^{m+1}-3+2^{m+3}-2^{p_1}=c\times 2^{p_1+p_2}$, is in layer m+3-p₁-p₂

Pos of c in m+3-p₁-p₂ layer is: $\frac{c-2^{m+4-p_1-p_2}+1}{2}$

Pos of c in m-1 layer is: $\frac{c-2^{m+4-p_1-p_2}+1}{2^{5-p_1-p_2^{}}}$

$\frac{b-2^{m-p_1+2}+1}{2^{3-p_1}}-\frac{a-2^m+1}{2}=\frac{b+1-2^{2-p_1}\times a-2^{2-p_1}}{2^{3-p_1}}$

$\frac{c-2^{m+4-p_1-p_2}+1}{2^{5-p_1-p_2^{}}}-\frac{b-2^{m-p_1+2}+1}{2^{3-p_1}}=\frac{c+1-2^{2-p_2}\times b-2^{2-p_2}}{2^{5-p_1-p_2}}$, ratio p is:

$\begin{array}{l}p=\frac{c+1-2^{2-p_2}\times b-2^{2-p_2}}{2^{5-p_1-p_2}}\times \frac{2^{3-p_1}}{b+1-2^{2-p_1}\times a-2^{2-p_1}}=\frac{c\times 2^{p_2}+2^{p_2}-2^2\times b-2^2}{2^2}\times \frac{2^{p_1}}{b\times 2^{p_1}+2^{p_1}-2^2\times a-2^2}\\ =\frac{c\times 2^{p_1+p_2}+2^{p_1+p_2}-2^{2+p_1}\times b-2^{2+p_1}}{2^2}\times \frac{1}{3\times a+2^{m+1}-1+2^{p_1}-2^2\times a-2^2}\\ =\frac{3^2\times a+3\times 2^{m+1}-3+2^{m+3}-2^{p_1}+2^{p_1+p_2}-2^{2+p_1}\times b-2^{2+p_1}}{2^2}\times \frac{1}{2^{m+1}+2^{p_1}-a-5}\\ =\frac{3^2\times a+3\times 2^{m+1}-3+2^{m+3}-2^{p_1}+2^{p_1+p_2}-2^2\times (3\times a+2^{m+1}-1)-2^{2+p_1}}{2^2}\times \frac{1}{2^{m+1}+2^{p_1}-a-5}\\ =\frac{3\times 2^{m+1}-3\times a-5\times 2^{p_1}+2^{p_1+p_2}+1}{2^2}\times \frac{1}{2^{m+1}+2^{p_1}-a-5}\\ =\frac{3\times (2^{m+1}+2^{p_1}-a-5)+2^{p_1+p_2}-8\times 2^{p_1}+16}{2^2\times (2^{m+1}+2^{p_1}-a-5)}\\ =\frac{3}{4}+\frac{2^{p_1+p_2}-8\times 2^{p_1}+16}{2^2\times (2^{m+1}+2^{p_1}-a-5)}=\frac{3}{4}+\frac{2^{p_1}\times (2^{p_2}-8)+16}{2^2\times (2^{m+1}+2^{p_1}-a-5)}\end{array}$

Next try to prove the average value of position increment ratio p is >=3/4 in long valid sequence(or rebuilt sequence).

From formula above we know, in long huge sequence, 2^m+1-a is very big, the ratio is very close to 3/4. Only these cases ratio p<3/4: p₂=1, p₁>=2; p₂=2, p₁>=3. This is to say: cases of (forward, upward), (downward, upward), (downward, forward) ratio <3/4; cases of (upward, upward), (downward, downward), (upward, downward), (upward, forward) ratio >3/4; case of (forward, forward) ratio =3/4. cases of >3/4 is more than cases of <3/4. And most importantly, in long huge sequence, the general trend of the sequence is upward in the tree, cases of (upward, upward), (upward, forward), (forward, forward) should appear frequently, (upward, upward) should appear most times.

Because one step can only upward one layer, and one step can downward one more layers, we can consider some successive upward steps as one step to achieve reciprocity operation.

For example, if appear (4, 1) or (4, 2), normally it should upward 2 or more layers before(or after) to guarantee general upward trend. Suppose after (1,1,4,1) operation get odd e, then

$3^4\times a+3^3\times 2^{m+1}-3^3+3^2\times 2^{m+3}-3^2\times 2^{p_1}+3\times 2^{m+5}-3\times 2^{p{}_{\backslash 1}+p_2}+2^{m+7}-2^{p_1+p_2+p_3}=e\times 2^{p_1+p_2+p_3+p_4}$

Position of e in m-1 layer is:

$\begin{array}{l}\mathrm{pos}1=\frac{e-2^{m+8-p_1-p_2-p_3-p_4}+1}{2^{9-p_1-p_2^{}-p_3-p_4}}=\frac{e\times 2^{p_1+p_2+p_3+p_4}-2^{m+8}+2^{p_1+p_2+p_3+p_4}}{2^9}\\ =\frac{3^4\times a+3^3\times 2^{m+1}-3^3+3^2\times 2^{m+3}-3^2\times 2^{p_1}+3\times 2^{m+5}-3\times 2^{p{}_{\backslash 1}+p_2}+2^{m+7}-2^{p_1+p_2+p_3}-2^{m+8}+2^{p_1+p_2+p_3+p_4}}{2^9}\\ =\frac{3^4\times a+3^3\times 2^{m+1}-3^3+3^2\times 2^{m+3}-3^2\times 2+3\times 2^{m+5}-3\times 2^2+2^{m+7}-2^6-2^{m+8}+2^7}{2^9}\\ =\frac{3^4\times a+2^{m+3}+2^{m+5}+27\times 2^{m+1}+7}{2^9}\\ =\frac{81\times a+47\times 2^{m+1}+7}{2^9}\end{array}$

If use proportional sequence of ratio 3/4, consider two steps of (1,1) as one step, still use a and b-a as start position and start position increment to estimate, position is:

$\begin{array}{l}\mathrm{pos}2\approx \frac{a-2^m+1}{2}+(\frac{b+1-2^{2-p_1}\times a-2^{2-p_1}}{2^{3-p_1}})\times (1+\frac{3}{4}+{(\frac{3}{4})}^2)\\ =\frac{a-2^m+1}{2}+(\frac{b\times 2^{p_1}+2^{p_1}-2^2\times a-2^2}{2^3})\times (1+\frac{3}{4}+{(\frac{3}{4})}^2)\\ =\frac{a-2^m+1}{2}+(\frac{2^{m+1}-a-3}{2^3})\times (1+\frac{3}{4}+{(\frac{3}{4})}^2)\\ =\frac{a-2^m+1}{2}+(\frac{2^{m+1}-a-3}{2^3})\times \frac{37}{16}\\ =\frac{27\times a-2^{m+6}+2^6+37\times 2^{m+1}-37\times 3}{2^7}\\ =\frac{27\times a+5\times 2^{m+1}-47}{2^7}=\frac{108\times a+20\times 2^{m+1}-188}{2^9}\end{array}$pos1>pos2 because 2^m+1>a, it is thus clear that the average ratio is >3/4.

If the sequence is (4,1,1,1), position of e in m-1 layer is:

$\begin{array}{l}\mathrm{pos}1=\frac{e-2^{m+8-p_1-p_2-p_3-p_4}+1}{2^{9-p_1-p_2^{}-p_3-p_4}}=\frac{e\times 2^{p_1+p_2+p_3+p_4}-2^{m+8}+2^{p_1+p_2+p_3+p_4}}{2^9}\\ =\frac{3^4\times a+3^3\times 2^{m+1}-3^3+3^2\times 2^{m+3}-3^2\times 2^{p_1}+3\times 2^{m+5}-3\times 2^{p{}_{\backslash 1}+p_2}+2^{m+7}-2^{p_1+p_2+p_3}-2^{m+8}+2^{p_1+p_2+p_3+p_4}}{2^9}\\ =\frac{3^4\times a+3^3\times 2^{m+1}-3^3+3^2\times 2^{m+3}-3^2\times 2^4+3\times 2^{m+5}-3\times 2^5+2^{m+7}-2^6-2^{m+8}+2^7}{2^9}\\ =\frac{3^4\times a+2^{m+3}+2^{m+5}+27\times 2^{m+1}-203}{2^9}\\ =\frac{81\times a+47\times 2^{m+1}-203}{2^9}\end{array}$

If use proportional sequence of ratio 3/4, consider two steps of (1,1) after (4) as one step, position is:

$\begin{array}{l}\mathrm{pos}2\approx \frac{a-2^m+1}{2}+(\frac{b+1-2^{2-p_1}\times a-2^{2-p_1}}{2^{3-p_1}})\times (1+\frac{3}{4}+{(\frac{3}{4})}^2)\\ =\frac{a-2^m+1}{2}+(\frac{b\times 2^{p_1}+2^{p_1}-2^2\times a-2^2}{2^3})\times (1+\frac{3}{4}+{(\frac{3}{4})}^2)\\ =\frac{a-2^m+1}{2}+(\frac{2^{m+1}-a+11}{2^3})\times (1+\frac{3}{4}+{(\frac{3}{4})}^2)\\ =\frac{a-2^m+1}{2}+(\frac{2^{m+1}-a+11}{2^3})\times \frac{37}{16}\\ =\frac{27\times a-2^{m+6}+2^6+37\times 2^{m+1}+37\times 11}{2^7}\\ =\frac{27\times a+5\times 2^{m+1}+471}{2^7}=\frac{108\times a+20\times 2^{m+1}+471\times 4}{2^9}\end{array}$pos1>pos2 if 2^m+1-a is very big, then the average ratio is >3/4.

Generally, from above we know the transform position increment from odd a to b is: $\begin{array}{l}\Delta =\frac{b-2^{m-p_1+2}+1}{2^{3-p_1}}-\frac{a-2^m+1}{2}=\frac{b+1-2^{2-p_1}\times a-2^{2-p_1}}{2^{3-p_1}}\\ =\frac{b\times 2^{{\mathrm{p}}_1}+2^{{\mathrm{p}}_1}-2^2\times a-2^2}{2^3}=\frac{3\times a+2^{m+1}-1+2^{{\mathrm{p}}_1}-2^2\times a-2^2}{2^3}\\ =\frac{2^{\mathrm{m}+1}-\mathrm{a}+2^{p_1}-5}{2^3}\end{array}$

Only when a=3(or 1…11 in binary form), p₁=2 or a=5(or 1…101 in binary form), p₁=1, the position increment is equal to 0, these two cases are convergence state or quasi convergence state. Other cases position increment is bigger than 0. The bigger p₁ is, the bigger position increment will get. Since a is random, we can get result: To any specific odd, upward step has the smallest transform position increment. Of course, to any specific a, p₁ is actually fixed, here suppose it can vary for purposes of comparison.

The position increment of two successive upward steps is

$\begin{array}{l}\mathrm{delp}1=\frac{c-2^{m+4-p_1-p_2}+1}{2^{5-p_1-p_2^{}}}-\frac{a-2^m+1}{2}=\frac{c\times 2^{p_1+p_2}+2^{p_1+p_2}-2^{m+4}}{2^5}-\frac{a-2^m+1}{2}\\ =\frac{3^{2}a+3\times 2^{m+1}-3+2^{m+3}-2+2^2-2^{m+4}}{2^5}-\frac{a-2^m+1}{2}\\ =\frac{7\times (2^{m+1}-a)-17}{2^5}\end{array}$

The position increment of downward 2 layers in one step is

$\mathrm{delp}2=\frac{2^{\mathrm{m}+1}-\mathrm{a}+2^{p_1}-5}{2^3}=\frac{2^{m+1}-a+11}{2^3}$

$\mathrm{delp}1-\mathrm{delp}2=\frac{7\times (2^{m+1}-a)-17}{2^5}-\frac{2^{m+1}-a+11}{2^3}=\frac{3\times (2^{m+1}-a)-61}{2^5}>0$,

When in long huge sequence.

$\frac{\mathrm{delp}1-\mathrm{delp}2}{\mathrm{delp}2}=\frac{3\times (2^{m+1}-a)-61}{2^5}÷\frac{2^{m+1}-a+11}{2^3}\approx \frac{3}{4}$

This means the position increment of merging (1,1) to one step in long huge sequence is about equivalent to one downward step. Merging (1,2), (1,2+), (2,2), (2,2+), (2,1), (2+,1), (2+,2) is similar. If needed, we can merge any successive steps to one step, because we have already transformed odd position to absolute position of same layer, nothing to do with layer sequence number again.

Hence, if appear position increment ratio<3/4 cases in long huge sequence, we can merge two or more behind successive steps to one step, finally get a new sequence which its average position increment ratio of transform position sequence is bigger than 3/4. If merge all behind steps the ratio is still <3/4(convergence sequence could appear this case), it reinforces convergence of the sequence, because the current transform position is not far away 2^m-1.

Note, during the merging procedure, make sure the ratio is a little bigger than 3/4, not very big, especially not bigger than 1. If the previous ratio is much bigger than 3/4, we can split some transform position value to next step, not influencing the final position value. Never suppose the transform position limit of long huge sequence is 2^m-1, because position increment is forever >0 before convergence.

So we can use proportional sequence of ratio 3/4 to estimate the rebuilt sequence.

After a do n times $(\times 3+2^m-1)÷2^k$ operation, pos in m-1 layer is:

$pos>=\frac{a-2^m+1}{2}+(\frac{b+1-2^{2-p_1}\times a-2^{2-p_1}}{2^{3-p_1}})\times (1+\frac{3}{4}+{(\frac{3}{4})}^2+{(\frac{3}{4})}^3+\mathrm{...}+{(\frac{3}{4})}^{n-1})$

When n->∞(although we merge some successive steps to one step, non convergence sequence still has ∞ steps),

$\begin{array}{l}pos>=\frac{a-2^m+1}{2}+(\frac{b+1-2^{2-p_1}\times a-2^{2-p_1}}{2^{3-p_1}})\times 4=\frac{a-2^m+1}{2}+\frac{b\times 2^{p_1}+2^{p_1}-2^2\times a-2^2}{2^{}}\\ =\frac{a-2^m+1}{2}+\frac{3\times a+2^{m+1}-1+2^{p_1}-2^2\times a-2^2}{2}=\frac{2^m+2^{p_1}-4}{2}\end{array}$

When first number property $2^{p_1}$>4(this is very easy to achieve in original long sequence), and when n->∞, the final transform position is >2^m-1, is contradictory. This means, the sequence should become small sequence(once one element become a small odd in our range, the sequence becomes), or converge before a limit steps, or overstep the boundary of the tree(it is not possible in real world).

6.3. Another Way to Build a New Sequence

Below we rebuild sequence using other method and give a much stronger proof. Look below. The common transform position formula is:

$s_0=\frac{a-2^m+1}{2}=2^{m-1}-\frac{2^{m+1}-a-1}{2}=2^{m-1}+\frac{1-b_0}{2}$

$s_1=\frac{3\times a+2^{m+1}-1-2^{m+2}+2^{p_1}}{2^3}=2^{m-1}-\frac{3^1\times (2^{m+1}-a)+1-2^{p_1}}{2^{2\times 1+1}}=2^{m-1}+\frac{(1-b_1)\times 2^{p_1}}{2^{2\times 1+1}}$

$s_2=\frac{3^{2}a+3\times 2^{m+1}-3+2^{m+3}-2^{p_1}-2^{m+4}+2^{p_1+p_2}}{2^5}=2^{m-1}-\frac{3^2\times (2^{m+1}-a)+3+2^{p_1}-2^{p_1+p_2}}{2^{2\times 2+1}}=2^{m-1}+\frac{(1-b_2)\times 2^{p_1+p_2}}{2^{2\times 2+1}}$

$s_3=2^{m-1}-\frac{3^3\times (2^{m+1}-a)+3^2+3\times 2^{p_1}+2^{p_1+p_2}-2^{p_1+p_2+p_3}}{2^{2\times 3+1}}=2^{m-1}+\frac{(1-b_3)\times 2^{p_1+p_2+p_3}}{2^{2\times 3+1}}$…

$\begin{array}{l}{s}_i=2^{m-1}+\frac{2^{p_1+p_2+\mathrm{...}{p}_i}-3^i\times (2^{m+1}-a)-3^{i-1}-3^{i-2}\times 2^{p_1}-{\mathrm{...2}}^{p_1+p_2+\mathrm{...}{p}_{i-1}}}{2^{2i+1}}\\ =2^{m-1}+\frac{(1-b_i)\times 2^{p_1+p_2+\mathrm{...}{p}_i}}{2^{2i+1}}\end{array}$which b_i is the corresponding odd in $(\times 3+1)÷2^k$ sequence.

Suppose one long huge sequence has n upward steps, k forward steps, l downward steps, finally upward h layers, then $\mathrm{h}=2\times (\mathrm{n}+\mathrm{k}+\mathrm{l})-\mathrm{n}-2\times {\mathrm{k}-(\mathrm{q}}_1+\mathrm{...}{q}_l)=n+2\times l-{(\mathrm{q}}_1+\mathrm{...}{q}_l)$, which q₁…q_l is k of 2^k in $(\times 3+2^m-1)÷2^k$ downward steps., n>h(normally n>>h in long huge sequence) and n>l. The final transform position is:

$\begin{array}{l}{s}_{n+k+l}=2^{m-1}+\frac{2^{p_1+p_2+\mathrm{...}{p}_{n+k+l}}-3^{n+k+l}\times (2^{m+1}-a)-3^{n+k+l-1}-3^{n+k+l-2}\times 2^{p_1}-{\mathrm{...2}}^{p_1+p_2+\mathrm{...}{p}_{n+k+l-1}}}{2^{2(n+k+l)+1}}\\ =2^{m-1}+\frac{2^{p_1+p_2+\mathrm{...}{p}_{n+k+l}}-3^{n+k+l-1}-3^{n+k+l-2}\times 2^{p_1}-{\mathrm{...2}}^{p_1+p_2+\mathrm{...}{p}_{n+k+l-1}}}{2^{2(n+k+l)+1}}-\frac{1}{2}\times {(\frac{3}{4})}^{n+k+l}\times (2^{m+1}-a)\\ =2^{m-1}+\frac{(1-b_{n+k+l})\times 2^{p_1+p_2+\mathrm{...}{p}_{n+k+l}}}{2^{2(n+k+l)+1}}\end{array}$

Now rebuild, do not change first step, we move number property 2^k of all the downward steps next to the first step(suppose we can move in order to compare, we can do it from math calculation angle), then forward steps and upward steps. From common transform position formula we know, since $2^{p_1+p_2+\mathrm{...}{p}_{n+k+l}}$ and $2^{2(n+k+l)+1}$are not changed, the new sequence has smaller final transform position because more previous the position of 2^k is, more great influence it produce, bigger the value of minus part is. And at least previous downward steps have ratio >3/4. Merge all upward steps to one step(If needed, we can even merge all upward steps and forward steps to one step, if we merge upward steps, forward steps and the last downward step before the forward steps to one step, the new sequence already has a increment ratio>3/4 only according to the ratio formula which indicates that (upward, downward), (forward, downward), (downward, downward) have a transform position increment ratio>3/4.). Use new sequence to estimate transformation position. Below prove the estimation position is smaller than the original.

Use proportional sequence of ratio 3/4 and new step count to estimate, then

$\begin{array}{l}s\approx \frac{a-2^m+1}{2}+(\frac{2^{m+1}-a+2^{p_1}-5}{2^3})\times (1+\frac{3}{4}+{(\frac{3}{4})}^2+\mathrm{...}{(\frac{3}{4})}^{k+l+1-1})\\ =2^{m-1}+\frac{2^{p_1}-4}{2}-\frac{2^{m+1}-a+2^{p_1}-5}{2}\times {(\frac{3}{4})}^{k+l}\end{array}$

$\begin{array}{l}s-s_{n+k+l}\\ \approx \frac{2^{p_1}-4}{2}-\frac{2^{m+1}-a+2^{p_1}-5}{2}\times {(\frac{3}{4})}^{k+l}+{(\frac{3}{4})}^{n+k+l}\times \frac{(2^{m+1}-a)}{2}-\frac{2^{p_1+p_2+\mathrm{...}{p}_{n+k+l}}-3^{n+k+l-1}-3^{n+k+l-2}\times 2^{p_1}-{\mathrm{...2}}^{p_1+p_2+\mathrm{...}{p}_{n+k+l-1}}}{2^{2(n+k+l)+1}}\end{array}$

$\frac{2^{m+1}-a+2^{p_1}-5}{2}\times {(\frac{3}{4})}^{k+l}$ is >>${(\frac{3}{4})}^{n+k+l}\times \frac{(2^{m+1}-a)}{2}$,is about ${(\frac{4}{3})}^n$ times, $\frac{2^{p_1+p_2+\mathrm{...}{p}_{n+k+l}}-3^{n+k+l-1}-3^{n+k+l-2}\times 2^{p_1}-{\mathrm{...2}}^{p_1+p_2+\mathrm{...}{p}_{n+k+l-1}}}{2^{2(n+k+l)+1}}$ is the tail part, its absolute value is <<${(\frac{3}{4})}^{n+k+l}\times \frac{(2^{m+1}-a)}{2}$, can be ignored in long huge sequence. $\begin{array}{l}\frac{2^{p_1+p_2+\mathrm{...}{p}_{n+k+l}}-3^{n+k+l}\times (2^{m+1}-a)-3^{n+k+l-1}-3^{n+k+l-2}\times 2^{p_1}-{\mathrm{...2}}^{p_1+p_2+\mathrm{...}{p}_{n+k+l-1}}}{2^{2(n+k+l)+1}}\\ =\frac{(1-b_{n+k+l})\times 2^{p_1+p_2+\mathrm{...}{p}_{n+k+l}}}{2^{2(n+k+l)+1}}\approx -\frac{b_{n+k+l}\times 2^{p_1+p_2+\mathrm{...}{p}_{n+k+l}}}{2^{2(n+k+l)+1}}\end{array}$

Because the sequence is long huge sequence, we can select a $b_{n+k+l}\ge 2^{m+1}-a$. Let $b_{n+k+l}\approx r\times (2^{m+1}-a)$, then $-\frac{b_{n+k+l}\times 2^{p_1+p_2+\mathrm{...}{p}_{n+k+l}}}{2^{2(n+k+l)+1}}\approx -r\times \frac{(2^{m+1}-a)}{2}\times \frac{1}{2^h}$

Hence ${(\frac{3}{4})}^{n+k+l}\times \frac{(2^{m+1}-a)}{2}$ is about $r\times \frac{(2^{m+1}-a)}{2}\times \frac{1}{2^h}$, its ${(\frac{4}{3})}^n$ times is

$r\times \frac{(2^{m+1}-a)}{2}\times \frac{1}{2^h}\times {(\frac{4}{3})}^n>>\frac{2^{p_1}-4}{2}$ when n(or n+k)>>h in long huge sequence and p₁ is not very big(for example =3), hence, $s-s_{n+k+l}<0$. If the long huge sequence is non-convergence, the count of the previous downward steps(ratio>3/4) of new sequence can be infinite. This means we can use proportional sequence of ratio 3/4 to estimate the convergence of original sequence, and its transformation position must >2^m-1 after infinite steps.

Still has one puzzle, the transformed positions of equivalence elements(add binary 1s in head) of elements in left half part in m-1 layer are all in right half part in m-1 layer, it is as if exist many loops. It is of course not correct, this is because, although they are equivalence, their functions are different. Other odds can change to them, and they can also converge. If some long sequence exist loops, the transformation position(to m-1 layer) can never reach to or bigger than 2^m-1, it is contradictory.

Maybe it is possible to use proportional sequence of ratio 3/4 to estimate the convergence steps for some long huge sequence(guarantee the average ratio is >3/4). For some odds in m-1 layer, if start odd can reach to or bigger than 2^m-1 in limit steps n using ratio 3/4, indicates that the convergence step count should be smaller than n multiply a number(because we merge some successive steps to one step to estimate, the suitable value of the number is difficult to get, but should not be very large); if can not reach to forever, indicates should use average ratio >3/4, but we don’t know suitable value of the ratio, we can do $(\times 3+2^m-1)÷2^k$ operation several steps until found a suitable odd(normally the number property $2^{p_1}$of the odd is bigger than 4) as start odd and do estimation again.

7. Conclusion

This way, we have proved that the Collatz Conjecture is true.