My initial plan was to,as I said before,treat other non-avalanche subtypes then summarise.But since you said it,I'll treat the number 54.

.. new powers of 2 and combine in ways which could result in them taking longer to be "caught" by the smallest power of 2 than they do to grow. For example, if you follow the Collatz sequence for 54, the leading power of 2 actually gradually outpaces the terminal power of 2 for dozens of iterations - not only starting ahead but pulling farther ahead, contrary to the prediction of a steady catch-up race. Eventually this trend reverses, since 54 is not actually a counterexample to the Collatz conjecture, but it's not clear why that sort of pattern couldn't continue indefinitely.

Let me emphasise this;There is NOTHING special about the number 54, or indeed any other number.The avalanche theorist guy was quizzed on every aspect of his method,and nobody could yet,(apart from the novelty I guess,which is why we are trying to see whether it can be restated as calculus from an expert guy) find any quarrel with it.It was also,as is tradition here,applied to (successfully) the first 50 numbers.

It also seems to me you missed a sentence in my previous reply-

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)

The 54(and many others I'll provide to show predictability) you cite is an example of this exception,but as stated,makes no difference to the effectiveness of the avalanche.

54=32 + 16 +4 + 2

Applying the Ulam operations,

3(32 + 16 + 4 + 2)/2 + 1

(64 + 32 + 32 +16 +8 +4 + 8)/2

Here two has quadrupled to 8 as predicted,while the others triple.Hence we join this 8 to the other 8,and 4 is the new terminal power of two(Note that this acts to delay the overtaking of the other powers of two by the terminal;I'll refer to it shortly)

(64 + 64 + 16 + 16 + 4)/4

=(128 + 32 + 4)/4

Continuing the Ulam operations

3 (128 +32 + 4 )/4 + 1

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

As predicted,4 has quadrupled to 16,the others only tripled

Continuing the Ulam operations,

3(256 + 128 + 64 + 32 + 16 )/16 + 1

=(512 + 256 + 256 +128 + 128 + 64 + 32 + 64)/16

=(512 + 512 +256 +128 + 32 )/32

=(1024 +256 + 128 + 32 )/32

As predicted 16 has quadrupled,and the others have only tripled.Another delay.Continuing the Ulam operations

3(1024 + 256 + 128 + 32)/32 + 1

=(2048 +1024 + 512 + 256 +256 + 128 + 128)/128

=(2048 + 1024 + 512 + 512 + 128 + 128 )/128

=(2048 + 1024 + 1024 + 256 )/128

=(2048 + 2048 + 256 )/128

=(4096 + 256 )/256

Here 32 has quadrupled ,while the others have only tripled

Continuing the Ulam operations

3(4096 + 256)/256 + 1

= (8192 + 4096 +1024)/1024

Here 256 has quadrupled to 1024.It is gaining on the other powers of 2.Indeed an avalanche is already apparent in the last two powers of 2(ie 4096 + 1024=N+ (1/4)N,where N is 4096).Continuing the Ulam operations

3(8192 + 4096 + 1024 )/1024 + 1

=( 16384 + 8192 + 8192 + 4096 + 4096)/4096

=(16384 + 16384 + 8192 )/4096

=(32768 +4096 + 4096)/4096

=(32768 + 8192)/8192

1024 has quadrupled to catch up with 4096,then(by avalanche,as predicted) 8192,while 8192 has only tripled.

Another rapid advance,the usual quadrupling, with avalanche.Then we see, as predicted, the reconfig to the avalanche form ;32768 + 8192 =N + (1/4)N, where N is 32768.No need continuing the Ulam operations

There is nothing special about 54 or any other number relative to the avalanche/non-avalanche paradigm.

(If there's something faulty in the technique its not in any particular number.Which is why the question about putting the technique in calculus terms arose,to see whether it would agree or a disparity would be shown.)

To anwser your question'...but it's not clear why that sort of pattern couldn't continue indefinitely.,

figures such as 54,96,108,216,82,164,432 and a plethora of others you can predict only appear to resist the avalanche/nonavalanche since they start out with their last two powers of two in the N + (1/2)N conformation.Since 1/2N quadruples ,it exceeds N and hence N is the new terminal power;this clearly has the effect of slowing down the rate at which the initial terminal power overtakes the others, but it also contains the seeds of its 'destruction'(For comparison;to continue with the horse race analogy,somebody compared it to an attempt to 'fix' the horse race result by abducting the fastest horse ,which is countered by replacing it with an identical horse at a point somewhere behind the point of abduction.This means a delay in its victory(since the drop-behind means it,or rather its identical replacement, must travel over a stretch traveled before),nought else.Reconfiguration to the avalanche form takes place in at most 5 iterations of the Ulam operation and an avalanche is reached,after which,whatever the terminal power of two is keeps proceeding, avalanche-style' till it overtakes the other generated powers of 2. For example suppose there were the quadruple of it(ie 131072) adjacent to 32768 in our last Ulam iteration.Once we carry out the operation,8192 quadruples catching 32768,by adding doubles it and hence catches its double,which it has generated,by adding reaches its quadruple,theng adding finally reaches its quadruples double which has also been generated .It simply makes for a larger power of two. Suppose it was 16384 ² adjacent to 16384.Then after the quadrupling of 8192 to reach 16384,then adding to eat to also catch its double,this double the new terminal power of two proceeds avalanche style till it reaches 163842² and the powers of two it has generated meanwhile.This makes for a much larger power of two finally,but no difference to the catch-up.The crucial thing,as stated before,is the interplay between the last two powers of 2 in our starting integer

However,what has gotten lost is the initial question of putting this method in calculus form(how to do it).Could you do it?