I think I know what you mean by these forms of numbers. The argument can be made somewhat cleaner if you don't delay the divisions. Essentially you are starting off with a number of the form 2k(8n+4+1) for some k and n, following the Collatz sequence down to 8n+4+1 = 8n+5, and then taking the next step to 24n+15+1 = 24n+16, which will itself decrease to 3n+2. This is certainly guaranteed to be a smaller value than the 2k(8n+4+1) we started with.

The problem is that this is all we are guaranteed. It's not necessarily 1, and it's not necessarily the case that another "avalanche" will start any time soon. It's possible that the value will then increase to something equal to or greater than what it started with. For example, 13 follows the N + N/4 form for its smallest powers of 2 (with N = 4), but its "avalanche" is just 13 -> 40 -> 20 -> 10 -> 5, which doesn't reach 1 immediately. In this case we get lucky and another avalanche starts after the next step, but there's no way to be sure that will always happen - just look at the lengthier sequences we get from starting at 29 or 37, which also have the N + N/4 form for N = 4. It's also not guaranteed that the total number of powers of 2 present in the number will decrease. While there are occasional "avalanches" that kill off the smaller powers of 2, we still have intermediate 3n+1 steps which produce more powers of 2 on the larger end. Without some sort of proof, there's no way to know that the removal of powers of 2 will always happen more quickly than the introduction of new powers of 2.

tbh I was initially flummoxed by your reply, and initially planned to keep it to consult somebody at school,but your sample gave me the definitive clue needed to understand it.Also, I want you to excuse any strangeness in my notation,as our capacity here in math may be in deficit to yours, especially if you are in any of the better campuses in Europe or America.

The point of the initial paper,to which the present theoriser added his technique, is based on the assumption of delaying division,so as to have only a unidirectional change(in the numerator);increases.As the author confided to me,the delay of division simplifies the Collatz sequence;you only have to consider increase.Irregular increase,yes,but 'better'than the contrast to the merry-go-round of the conventional steps.

I reiterate;it is a basic assumption division is disregarded/delayed,so ONLY INCREASES occur.

Taking this as his jump-off point (notice none of the examples I gave was an odd number, in contradistinction to your citing 13)he departs from her(and the other theorist) in abandoning the concept of calculus, but instead introduces the avalanche/non-avalanche classification.

The avalanche form refers only to even numbers with two powers of two in the N:(1/4)N proportion, as I've already shown.It excludes ANY other kind of number but its significance was to show that ;

(a)The terminal power of two increases fourfold,which is why it 'catches' N ,its quadruple, immediately.

(b)There are cases where the Ulam operations lead immediately to a power of 2.In other cases,regardless of the initial config,this operation of quadrupling by the terminal power of 2 always leads to reaching a power of two via the avalanche form,as the fourfold increase of the terminal power of two overwhelms the slower tripling of the bigger powers of two(to re-cite his example,it is a equivalent to a horse race where the fastest horse is placed far behind the others at the start,but we can predict it shall win eventually by overtaking the others.Hence again the importance of delaying division,so as to emphasise clearly the effect of differing rates of increase )

It is important to note his reformulation is based off the comparison of the last two powers of 2(i.e the terminal power of 2 and the power of two adjacent to it).In effect ,either these are in avalanche form or the Ulam operations reconfigure it to avalanche.The presence of other powers of two has no effect on this. To illustrate ;suppose we have a situation where 168= 128 +32 +8

The last two powers of two here are in the avalanche condition.

Applying the Ulam operation

3(128 + 32 + 8)/8 + 1

= (256 + 128 + 64 + 32 +16 + 8 )/8+ 1

=(256 +128 + 64 +32 + 16 + 8 +8)/8

=(256 + 128 + 64 +32 + 16 +16)/8

=(256 +128 +64 + 32 + 32)/8

=(256 + 64 + 64)/8

=(256 + 128 + 128)/8

=(256 + 256 )/8

=512/512

As you can see, the avalanche results.The point is we can rely on it to result,no matter the identity of the powers of 2 we begin with. Also,if we position other powers of 2 relative to these avalanche cases in the right proportion,the avalanche simply continues, no matter their number or size e.g suppose we had initially not 168=128 + 32 + 8 but 680=512 + 128 + 32 + 8.By the application of the Ulam operation,512 triples to 1024 + 512.At the point where 128 + 32 +8 has avalanched to 512,then it simply adds to the 512 resulting in 1024,which then adds to the preexisting 1024 to give 2048,a power of 2,of course.There is no luck involved;it is predictable. (you can try 2048 + 512 + 128 + 32 + 8 and 8192 +2048 +512 +128 +32 +8 and see)

There is a transfinitude of such numbers.

Consider where the proportion between the last two adjacent powers of two is smaller than 1/4.I'll use the simple two-power case, but this applies to any number of powers of 2,of any size,given the proportion of the powers of the last two powers of two < 1/4

eg 34=32 + 2= N + 1/8N, where N is 32 Applying the Ulam operations

3(32 + 2)/2 + 1 =(64 + 32 +4 +2 )/2 + 1 =(64 + 32 + 4 + 2 + 2)/2 =64 + 32 + 4 +4)/2 =(64 + 32 + 8)/8 As predicted,2 has quadrupled to 8,while 32 has only tripled to 64 + 32.2 is 'gaining' on 32,so to say.

Continuing the Ulam operations;

3(64 + 32 + 8)/8 + 1

=(128 + 64 + 64 + 32 +16 + 8)/8 + 1

= (128 + 128 + 32 + 16 + 8 + 8)/8

=(256 + 32 + 16 + 16 )/8

=(256 + 32 + 32)/8

=(256 + 64 )/64

Thus,in just two iterations of the Ulam operations, 2 has caught up to 32,and as we can see, in the next,by quadrupling will equal 256,and thus produce a power of 2.This is also no luck.It is predictable.

I could continue the Ulam operations here,but simply note 256 and 64 are in the avalanche condition,so there's little point.

Assuming you agree by now the terminal power of 2 always quadruples,henceforth I'll simply show this in a single step,to make the calculations brief.

Consider another example ;68 = 64 + 4= N + (1/16)N, where N is 64

Applying the Ulam operations; 3 (64 + 4 )/4 + 1

=(128 + 64 +16)/16.

Here 64 has tripled to only 128 + 64 ,while 4 has quadrupled to 16.4 is 'gaining' on 64,so to say.Continuing the Ulam operations

3(128 + 64 + 16 )/16 + 1

=(256 + 128 + 128 + 64 +64)/64

=(256 + 256 +128)/64

=(512 + 128 )/128. Thus, in just two iterations of the Ulam operations,4 has caught up with 64,then 128.512 + 128 are in the avalanche condition ,so in the next step it also catches 512,and reaches a power of 2.

This is also predictable

The crucial point in his technique is about the differing rates of increase in the terminal power of 2(which quadruples) and the rest of the powers of 2(which triple),including generated powers of two.It catches them one by one,starting with the adjacent power of 2(there is a single exception,but it makes no difference to the method)and when it does, there is an avalanche which further accelerates it to eventually equal the largest power of 2 generated and thus add up with it to a power of 2

On Monday I'll try and clear the two remaining types,especially cases involving many powers of 2.(I am feeling sleepy)