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

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

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

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.

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:

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:

The derivation of the denominator is:

Then

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

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.

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:

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.

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.

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

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].

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,

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

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:

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

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]).

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.

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: $\frac{c-2^{m+4-p_1-p_2}+1}{2^{5-p_1-p_2^{}}}$

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

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

Position of e in m-1 layer is: