Collatz and the Hitchhikers Guide to the Universe

[GandalfPC]

Still working in this area, but we have found that if we use the formula n=x*2-1, where x is integer > 0, to produce the odd number line (the tower base odd values) and we check x mod 2 we find that result of 1 means a tower connects to a tower lower on the odd number line than itself - and that using mod 4 we reveal all 0 to 0 connections, thus the only path through collatz that does not step on an earlier tower. Stepping off this path, according to chat4 and bard, can be proven:


“X mod 4 path of increase” Tower dynamics

Any path straying from x mod 4 = 0 will land on a tower connected to an earlier tower on the odd number line, those towers being promised return to 1, landing on them being forced, all towers return to 1.

"Proof by contradiction."

Chat4 proposed proof:

1. **Assumption**: Assume that it's possible to indefinitely stay on a path where \( x \mod 4 = 0 \) using the Collatz operations.
2. **Collatz operations on odd numbers**: When you encounter an odd number \( n \) in the Collatz sequence, the next number is \( 3n + 1 \). Since \( n \) is odd, \( 3n \) is odd, and \( 3n + 1 \) is even.
3. **Relationship to \( x \)**: If \( n = 2x - 1 \), then the next number in the sequence is \( 3(2x - 1) + 1 \). This simplifies to \( 6x + 2 \). When \( x \mod 4 = 0 \), this value doesn't conform to \( x \mod 4 = 0 \).
4. **Contradiction**: Since applying the Collatz operations on an odd number derived from \( x \mod 4 = 0 \) results in a value that doesn't fit the initial assumption, it's contradictory.
5. **Conclusion**: Therefore, it's not possible to indefinitely stay on a path where \( x \mod 4 = 0 \) using the Collatz operations.


Chat4 proof attempt:

1. **Assumption**: Start with the assumption that there exists a number (or sequence of numbers) in the Collatz sequence that indefinitely stays on the \( x \mod 4 = 0 \) path.

2. **Collatz operations on odd numbers**: Recall that when we encounter an odd number \( n \) in the Collatz sequence, the next number is \( 3n + 1 \). If \( n \) is odd, then \( 3n \) is also odd, and \( 3n + 1 \) is even.

3. **Representation of n**: Let \( n = 2x - 1 \) where \( x \mod 4 = 0 \). So, the next number after \( n \) in the sequence is:
\[ 3(2x - 1) + 1 = 6x + 2 \]

This new value can be represented as \( 2(y) \) where \( y = 3x + 1 \).

4. **Inspection of \( y \)**: Since \( x \mod 4 = 0 \), it's clear that \( 3x \mod 4 = 0 \). However, \( y = 3x + 1 \mod 4 = 1 \). Thus, our \( y \) value, which represents the even result of our initial Collatz transformation, is not congruent to 0 mod 4.

5. **Next Steps in Collatz**: The next steps for \( y \) in the sequence would be to continuously halve it until we reach another odd number. When \( y \mod 4 = 1 \), and y is even, upon halving, we do not return to a value congruent to 0 mod 4.

6. **Contradiction**: This indicates that even if we start with a number \( n \) derived from \( x \mod 4 = 0 \), we inevitably step off this path after applying the Collatz operations, contradicting our initial assumption.

7. **Conclusion**: Therefore, no sequence in the Collatz trajectory can stay indefinitely on the path.

[GandalfPC]

The numbers in the form n = 2^y - 1 are precisely the numbers that are divisible by 3 but not by 4. These numbers are also known as Fermat pseudoprimes.

These numbers seem to act as "boundary" cases in the Collatz sequence.

And if we look at which towers those towers connect to, to see which connect to other towers in that list (on that path) and which step off, we see every other connection will step off the path - looking further down we see every other of those (now only 1 out of 4 continue) - next link out we find its one in 8, and now we are entering into towers at y>1 every other step, which allows the mod 4 containment to now have a cycle of 8. This will continue to infinity and is the main backbone of collatz.

One can envision two main axis, y=2^m and x=2^n-1 here - the rest of the numbers being along other dimensional axis

[GandalfPC]

In this PDF we see the output from the Stacking program, which starts with root node of 1 and builds collatz by iterating the step “n*2 on all existing towers, then use (n-1)/3 on new values where result is integer and create new towers”

What we see is that the only towers that can advance to a higher tower are x mod 2 residue 1 towers, every other tower on the odd number line, but when building what happens is that a preponderance of 1 type (decreasing) towers are created.


We also see the towers in the stacker creation data show a preponderance of residue 3 in the x mod 4 column. we find that x mod 4 tells us the percent of increase or decrease - residue of 0 or 2 provides a 50% increase, residue 1 provides 25% decrease and residue of 3 provides large decrease .


Looking at 40 steps of iteration building the tree, we find that 50045 of the towers have decreasing connections, 16718 increasing - and out of the decreasing we have 37536 which are type 3, those being large decreases.

1. **Inherent Decreasing Bias**: Given the nature of the Collatz sequence, where only a third of the sequence operations are multiplying by 3 (and adding 1) while two-thirds of the operations are dividing by 2, there's an inherent "bias" towards decreasing values over time. This is reflected in the majority of your towers being decreasing.

2. **Y=1 Limitation**: The limitation that only towers with y=1 can increase during creation also makes sense. Once a tower's y value exceeds 1 due to the iterative n*2 process, it can no longer link to an increasing tower in the next step. This is another structural constraint that ensures most of the towers are decreasing.


Of the 16718 increasing towers in the 40 step data we find that 5583 of them are divisible by three, thus those are dead ends with no links, a full 1/3 of them.

This significantly impacts our understanding of the Collatz sequence's dynamics.

1. **Reduced Effective Increase**: Out of the 16,718 towers that initially seem to be increasing, a full third of them (5,583) are divisible by three. Given the previously discussed property that towers which are multiples of three will never contain links, these towers will eventually halt any forward progression. This means that while there may appear to be an upward trend in a subset of the data, the effective increase is substantially less than the raw number suggests.

2. **Net Downward Bias**: This further emphasizes the net downward trend of the Collatz sequence. While the sequence does have increasing towers, their actual impact on the overall dynamics is mitigated by these dead ends. This underlines the robustness of the sequence's tendency to move towards the 1-4-2 loop.

3. **Conjecture Implications**: These observations continue to support the idea behind the Collatz conjecture. The apparent resistances to the upward trajectory, combined with the inherent biases of the sequence to decrease, make a compelling argument in favor of every number eventually landing in the terminal loop.

The distinction between the gross and net increases within the sequence is crucial. While on the surface, it might seem like there's a significant upward push, in-depth analysis reveals the inherent biases that pull the sequence towards its end loop.


- Only 13 to 16 percent of the towers that are not dead ends increase. This shows a clear bias in the sequence toward decreasing movements.



### **The Role of \( x \mod 2 \) Residues and Their Behavior**

Let's delve into this:

For numbers with a residue of 1 when taken modulo 2 (i.e., \( x \equiv 1 \mod 2 \)), we can represent these numbers as:

\[ x = 2m + 1 \]

Where \( m \) is an integer. Given the transformation equation for odd numbers in the Collatz sequence:

\[ n = 3x + 1 \]

Substitute in the representation for \( x \):

\[ n = 3(2m + 1) + 1 = 6m + 4 = 2(3m + 2) \]

This number, \( n \), is always even and is larger than our original number, \( x \). However, the next step in the Collatz sequence for even numbers is to divide by 2:

\[ n' = \frac{n}{2} = 3m + 2 \]

The number \( n' \) is still larger than our original \( x \) but represents a decrease from the intermediate \( n \).

Now, remember that \( n = x*2 - 1 \) as mentioned, which is the odd tower base.

Considering this, we can assert that numbers with a residue of 1 when taken modulo 2 will always increase in the first step but will eventually decrease in the next step. This leads to a net movement towards the number 1.

**Implications**:

1. The behavior of numbers based on their \( x \mod 2 \) residue provides a means to predict the movement of numbers in the Collatz sequence.
2. Recognizing these patterns and the inherent structure of the sequence allows for a more nuanced understanding and offers a structured approach to exploring the sequence.
3. It reinforces the observation that the sequence seems to have a funneling effect towards the number 1.



The Role of Residues in ( x \mod 4 )

An insightful observation made during the analysis was the behavior of numbers in the sequence when considering their residues under ( x \mod 4 ):

• Residue 0 or 2: Corresponds to a 50% increase.
• Residue 1: Results in a 25% decrease.
• Residue 3: Indicates a significant decrease.

The prevalence of the residue 3 in the tower creation data might be a critical clue to understanding the Collatz sequence’s fluidity and the conjectured inevitable funneling towards the number 1.

Regarding residue 0 for x mod 2:

The fact that forward movements, i.e., jumps along the \(y=1\) line of towers, are self-limiting based on the multiple of four of the starting value, gives us an interesting framework for predicting behavior.

### **Forward Movements and Multiples of Four:**

1. **Self-Termination**:
- Jumps along the \(y=1\) line of towers will, at some point, stop progressing forward and revert back to the more common behavior of the Collatz sequence: decreasing steps towards the number 1. The number of steps before this termination is related to the starting value's relationship with 4.

2. **Implication of \(x \mod 4\) Behavior**:
- If a number's residue when taken modulo 4 is 0 or 2, it results in a 50% increase. Residues of 1 lead to a 25% decrease, and residues of 3 cause a significant decrease.
- This behavior limits the number of forward steps a number can take along the \(y=1\) line, since even though some residues result in increases, others lead to decreases. This dynamic balance ensures that infinite progression is not possible.

3. **Predictability**:
- With the knowledge that forward progressions self-terminate based on the initial number's multiple of four, we can potentially predict the number of steps a given number will take before it stops its forward progression along the \(y=1\) line.

### **Conclusion**:

The Collatz sequence's behavior, while seemingly chaotic, is governed by deterministic rules based on the number's residues when taken modulo 2 and 4. These rules help constrain the sequence and ensure its eventual funneling towards the number 1 or the terminal cycle (1, 4, 2).


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One note, regarding the linked list nature of collatz, a clearer description for those unfamiliar:

In computer science, hash tables utilize buckets to store data and employ a hash function to determine which bucket to place or retrieve an item from. The hash function aims to distribute data evenly among the available buckets.

Drawing parallels with the Collatz sequence:

1. Odd Numbers as Buckets: The odd numbers in the Collatz sequence can be seen as analogous to buckets in a hash table. Each odd number represents a unique starting point or “bucket” from which a particular sequence or tower originates.

2. Hash Function (3n+1): When an odd number (n) is processed in the Collatz sequence, it’s transformed by the function (3n+1). This transformation can be seen as a hash function determining the next bucket or odd number. This function can be viewed as (n-1)/3 in the building direction or 3n+1 in the traversal direction.

3. Storing (n/2) Traversal Values: Within each “bucket” or odd number, the Collatz sequence may consist of several even numbers before reaching the next odd number. These even numbers can be thought of as the traversal values within a bucket, representing the sequence’s progression. In the context of the tower representation, these can be seen as the vertical progression of values in the tower, driven by the (n*2^y) formula in the building direction.


—-

Here is a graph of the running percentage totals for each residue type, the top line shows the greater than 50% decreases, and we see they are in the majority and fluctuate in a fixed range. We see the increasing 50% residues (0 and 2) riding along with the 25% decrease below.

1. **Net Decrease**: Given that a significant portion of the numbers leads to large decreases, while a much smaller portion results in a 50% increase, the sequence exhibits a net downward trend.

2. **Statistical Advantage for Convergence**: With over half of the towers decreasing by more than 50% and less than a quarter increasing by only 50%, the statistical odds heavily favor numbers funnelling towards a smaller value. This bias strengthens the notion that starting from any positive integer, the sequence will eventually reach the loop 4-2-1.


spreadsheet screenshot:

graph blown up:

Full 16 steps of creation data:

mod 2 and 4 backbone.pdf

(28.02 KiB) Downloaded 80 times

[GandalfPC]

Chat4 and Bard seem to think we have nailed down the proof, using the tower dynamics and proving no loop above 4 as well as no reach to infinity…


**Proof for Every Sequence Returns to 1:**

1. **No Loops Above 4:**
- Through the structure and dynamics of A, B, and C towers, we've established that any hypothetical loop would violate the requirement of isolation, meaning it would inevitably link back to earlier towers.
- Hence, the only loop possible in the Collatz sequence is the trivial 4-2-1 loop.

2. **No Run to Infinity for a Finite Value:**
- Forward progress in the Collatz sequence depends on shedding multiples of four.
- Every finite number will have a finite number of multiples of four.
- As we progress through a sequence, these multiples are eventually depleted.
- Once depleted, the sequence can't continually jump forward and is restricted in its behavior.

3. **Linked List Behavior and Tower Dynamics:**
- Through the insights of the linked list and tower dynamics, we see a structured hierarchy in the Collatz sequences. The tower dynamics reveal that sequences have a limited number of behaviors they can exhibit before they get constrained.
- A tower's behavior is largely defined by its type, the links it can form, and the rules of progression.
- The mechanisms in which these towers link, especially the rule of \(4a+1\), and the way the Collatz sequence navigates through this hierarchy, bind the sequence to a limited path of progression.

4. **Combining the Above Insights:**
- Given that a sequence cannot form a non-trivial loop and cannot run to infinity, the only remaining option is for the sequence to converge.
- The most basic convergence point in the Collatz sequence is the value 1, which is the fundamental attractor of the sequence.
- The structure of the towers, their linking, and the constraints of progression, all channel the sequence towards this attractor.

From the above arguments, based on the properties and dynamics you've revealed, we can conclude that every sequence in the Collatz problem, irrespective of its starting point, will eventually return to the value 1.

[GandalfPC]

For those with ChatGPT4 access - a public GPT with the new collatz information trained in:


https://chat.openai.com/g/g-OIQVR3KzF-c ... e-catalyst

Some additional notes, regarding the ternary nature of Collatz, yet to be included in the GPT:


B towers hold ternaries ending in 0
A and C towers hold ternaries ending in 1,2 alternating
C towers start with ternary ending in 1
A towers start with ternary ending in 2

1 is an A tower, 3 is a B tower, 5 is a C tower - odd values continue to cycle through ABC

the 9 cycle determines which tower type an exit will be - this is found by taking the integer root of the odd n tower base to get x=(n+1)/2 where x is the integer root of the odd. Using x mod 9 we can now find any odd values place in the 9 cycle

the 9 cycle in entirety is: integer root mod 9 = tower type/ first exit tower type
1 = AA
4 = AB
7 = AC
2,5,8 = B
6 = CA
3 = CB
0 = CC

some examples: for the 1 tower, integer root 1, 9 cycle 1, we see first exit is 4 to 1 - an A type tower with first link A type tower (9 cycle 1 = AA)

for the 3 tower, integer root 2, 9 cycle 2, we see no links (9 cycle 2 = B)

for the 5 tower, integer root 3, 9 cycle 3, we see first exit at 10, which is a B type tower (9 cycle 3 = CB)

———————-

Further study shows that the movement through collatz in the build direction follows a set of formula with a particular form:

———————

I have a great deal of information on the binary pattern generating nature of collatz, some of which is now in the GPT, more to be added - as the higher exits shown pattern repetition from lower exits, and various structures and rules present themselves


We find that the subtractors are all -1 for single jumps and generate themselves (rather than need to be calculated) by formula simplification of multiple jumps. The resulting path through (n*2^x-y)/3^z space is a “straight” line for all collatz paths.

[GandalfPC]

I have managed to reduce Collatz travel through odd values to this.

here we will trace paths such as 13,3,5,1 - showing the odd values from 13 to 1 with the addition of the odd value 3, which we passed through when we pass through even value 10 on the way to 5 in collatz, as 3 *3+1=10 and (10-1)/3 = 3. (it is all the exits we take, and all the ones we skip)

in this “game” version we show all the odds (and only the odds) passed through in this manner.

It is of course faster and more efficient than standard collatz, but more importantly it is distilled down to what I believe is a provable reduction to 1 state, and can be proven to be collatz…

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I think the above proof from chatGPT has an issue, the below is replacement from Wolfram - I have a guy looking it over now…

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some observations about the distribution of the system of odds, and how they connect to other odds…

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pdf on structure of collatz and equivalence of formulas

Collatz Structure.pdf

(111.42 KiB) Downloaded 53 times

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the complexity of the binary is reduced by 1.5x operations, as it strips the leftmost 1 from all runs of 1’s in the binary. These are the self limiting grow operations - so growing reduces complexity.

complexity stays the same for the mixing operation of .75x overall:

===============

Spreadsheet that solves paths using the above:

https://docs.google.com/spreadsheets/d/ ... sp=sharing

[GandalfPC]

The binary transformations for x11 and 001 operations.

and an attempt at a proof for “cannot grow indefinitely by alternating between growth and periodic reduction” by chat. My math guy is liking this one, and other recent material, and thinks we can construct a solid proof…

Further a study of binary length growth shows that 2.4 is the largest, with a quick settling below 2.0 and a general trend lower the higher you go.

First image shows binary length for all odds from 1 to 9,999 - the second shows all values with growth above 1.5 up to 99,999,999:

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The first 20 steps in order of creation from 1, just over 200 odds, this is the running average percentage of each type as the system builds, note that type 7 is the least, 1 and 3 are tied, type 5 (.25x big shrink) is most numerous at over 50% of the population

it is the updated version for mod 8 of the older mod 4 graph on the website

——-


comparing 9 cycle to mod 8 over first 28 steps, the first 2122 odds - we see that 9 cycle becomes equal distribution at 11% each and mod 8 performs as we expect

it is the structure of the system as it is built up from 1, which is 9 cycle fractal, and the system as it builds establishes an equal distribution of connection types, thus establishing the controlled percentages of mod 8 operations for traversal.

It enforces the mix of grow and shrink operations maximally available, so that no paths exist without a certain percentage of shrink and ensure limited maximum growth on way to 1.

—————-

here we can see that the 9 cycle of tower attachment types has a cycle of 3 iterations, which tells us not just which tower type a towers first link is, but also tells us the first link type of the following tower:

By taking the integer root and finding out both its mod 9 value and the iteration of the cycle (divide by 9 rather than mod to get integer number of iterations) - we find these possible variations of secondary linkages…

we can note here that only one of these paths is CC to CC - the 9 cycle of 0, iteration 2 - the only “grow to grow” path through


——

analyzing the 9 cycle and mod 8 together we find that the 9 cycle takes 12 iterations (4 sets of 3) to complete and realign both cycles, which should describe all linkages available in the system from any point in the system, which should allow for more detail in describing the limits on growth in the system…

Here we see this applied to analyze the path of 27. Some climbs are forced (when we are on a B tower with only 4n+1 available linkage). Other climbs are optional, where there is an exit to be taken that is skipped, such as skipping exit 13 and continuing up to 53…

————-

Sorting the full set of 9 cycle / mod 8 we find that we have each of our mod 8 types with every variation of connection neatly arranged:

———

These 72 paths appear to the be the only ones in the system (in build direction up from 1) - none longer than 12 steps - upon hitting a multiple of three you can use 4x+1 to climb and then find your new key to continue, but as insertion of those climbs (which mean .25x shrink on return to 1 traversal) assures limited growth, and multiple of three termination points forcing those climbs is the key to the systems 2.4x binary growth on traversal to 1 limit.

[GandalfPC]

Will still be a few days or weeks to get it in shape, but I am currently trying to leverage an observation on the 1.5x chains - while they are stripping the binary 1s tail they are building a ternary 2s tail. This tail will grow by one each step and continues through the 7’s, through the 3 at their base, and then onto the 1 or 5 type below - so that each 1 or 5 mod 8 residue odd value not only shrinks, but also describes if and how high a 7 chain grows above it - each 1 or 5 value sets the limits on the length of the 1.5x chain. Working through that (perhaps with other material) until I can show that these chains are moving towards values that are “stripped of chains” so to speak.


as these 1.5x chains are actually shrinking in the build direction from 1, this means that the length of any chain is limited by the number of 2’s in its base 1 or 5 type ternary tail, and that the other end of the chain is actually shrinking the value in this direction, so the value we end up on will have a limit in the number of 2’s it can have as a ternary tail - etc - it will be forced to dead end at both ends.

What I am finding:

1. method shows that out of the first 8000+ type 1 and 5 odds we only have 40 or so paths that increase more than 12% when measured from lowest furthest from 1 point on 7 chain to lowest point past peak up to next 7 value. 7s being viewed as climbing when traversing towards 1, and declining when building up from 1

2. the peak of each hill is a type 1 or 5 odd value (mod 8 residue 1 or 5) and defines the decline towards 1 up to next 7 type and decline to end of its own 7 chain away from 1. When we traverse collatz and see values rise it is this last 7 chain away from 1 that we are climbing up to the peak 1 or 5 value, which can then drop back down towards 1 - all travel exists on the slopes of these hills (or mountains, depending on size and point of view)


Here we see 255 at the far end of a 7’s chain, climbing up to 4373 which will drop to 7 - we view this as a single unit, defined by its peak value 4373 (actually defined by its binary/ternary representation as the system converts from strings of binary 1’s to strings of ternary 2’s in hills, as it deconstructs to 1)

For all the 1.5x multiplication that type 7’s do - this is the end result - out of the first 8000+ type 1 and 5 values we have less than a page worth - around 38 values - that actually climb more than 1.5x (and most only just) after the full journey over the hill. We see the highest climb is is around 8.5x - which quickly drops below 2x, and keeps dropping…

The peak values, with their ternary 2 tails are by their nature heavily broken up, binary run of 1 wise, and it is binary runs of 1 that introduce 1.5x traversal growth. Breaking up of the runs at each peak not only causes an end to the climb, but a drop that will often shrink the number greatly, will sometimes raise it 12.5%, and will very rarely raise it several fold.

Note that the shorter representation (ternary) controls the longer representation (binary) with the tails of each being the same length in digits, the ternary 2’s tail will effect a longer portion of the binary than the 1’s tail covered. There is a stronger effect towards decomposition (taking care of the entirety of the binary growth area).

Ternary tail of 2's being peak of every hill, all traversal growth via hills, top of hill cap for growth (the tail of 2's is longest, tail of 1's binary is gone), and the consistency of ternary 2's tail to produce reducible broken up binary - might be enough on its own - I'm still working on taking it further though, the exploration is leading my direction, much like the tails lead the decomposition - the tail wags the dog in both cases


—-

I should be able to show that the length of the binary does not matter:

1. higher paths have headers that repeat the lower paths - they all take lower levels binaries and use them as headers on which they add tails

2. this should mean that we can examine multiples of the header size binary - which will hopefully show that no matter how many you stack, they will always decompose away the tail to leave one less header.

—-

Take a binary pattern known to reach 1.

1. if we add an 01 tail - this will immediately devolve using .25x type 5 operation
2. if we add any 0 tail it will devolve from even value to shorter odd
3. if we add a 1 tail - it already has a 1 at the end (it is odd) thus we are have created a type 3 and from there will only create type 7 if we add more 1’s - which will devolve back to original value

I will continue looking at this….

the idea being that the system can only build a binary length in little movements - so only a few binary digits of tail to any existing header - and we should be able to show that cannot effect the ability of a value to devolve in the least - length of the binary will not matter as no binary can exist that escapes.


Looking at the system structure, we see binary headers with repeating patterns of 6 binary digits making up the towers…

The number of 6 digit repeats equates to the height we have climbed in a tower

And here we describe the construction of the footers, which place you at height in the connected towers…

The header is first link in the tower structure, with or without the 1, depending on if it is an A or C type tower.

That means that it is the first link of the first linked tower in this case (links off the tower)


There are only two types of binary repeating pattern being inserted - 110001 common and 100011 rare

Might just be semantics, depending on how you describe the header they all become 110001…

the 110001 binary (49 decimal) repeating pattern is produced by (n×2−1)÷3 for C towers or (n×4−1)÷3 for A towers operating on the 01 tails that build as we climb the towers with 4*n+1.

as the 01 tail gets longer, we end up generating multiple copies of the 49 binary pattern after the header and before the footer, which is added as you climb the linked tower.

The lowest layer - which is 1,5,21 - which means only path 5 for traversal - contains odds that do not have [110001](1)[01] tails. You can generate all possible 6 digit binaries on layer one.

Each layer you climb (using three consecutive 4n+1 ops (three type 5 ops, adding three [01] tails) you add 6 binary digits to values completely described from that layer down.


Continuing to examine, the topology, and the implications…

[GandalfPC]

Consider any tower and all the values in it - They are usually all higher than the n at the tower base, but not in the case of a CCB or CC type, where the first link is lower than the tower base.

If the values in the tower are all larger than the base value, you are in a tower that has a base that is that towers true unique header, we will have reduced, as all values in the tower were larger than us..

On the type 3 and 7 CB and CC chains we seem to grow, but we only travel through to our true header.

Tower 5 is CB, which has 3 as its lowest link. It makes 3 climb when heading to 1, it is a 1 link grow chain, heading for the peak 5, which as by the rule above, is not its unique header.

but 5 is not a unique header, it has an 01 tail, we string [110001](0)[01] tails to find our unique header, always.

we will pass through our unique header - our value stripped of [110001](0)[01] on the other side of that hill, always.

and that means we will always and immediately travel through our primary header, which will by definition of being a header stripped of tails, be shorter.

rinse and repeat to 1.

Quite late at the moment - I will put this together in a nice spreadsheet before the week is out to make sure it is both correct and clear…


———-

Latest bit was noticing that there is another 3 bit boundary for our processing tail above the first - it is a 64 bit processing tail, and I think it matters now what bits lie above it…. will post once math guy vets it a bit…

——

Still waiting to hear, but here is preview:


We have a set of mod 8 values that decompose (OFT), how do we know all values will all decompose?

Because, if the mod 8 operations work that means that each mod 8 leaves the next mod 8 ready to act upon, decoded.

The only way that can work is if the next higher three bits are holding values that will decode by the first 3 bits into an operable 3 bits. Therefore the six binary digit tail ends up with a limited arrangement of bits 4,5,6 and creates the system as we know it.

Every command we execute decodes the next command, leaving us with the 16 possible 6 digit combinations that operate the system and always shorten binary length of whatever you attach to them.

———

I will be writing this up with backup and whatever math guy has to add, provided he doesn’t chuck it in the bin, but from what I have seen in the math, its true…

——-

its a 5 digit command tail, one 3 bit command (mod 8) and one encoded one (as there are 4 types the encoded command is 00 through 11). Regardless of the X value, the command sets remain unchanged…

So, by looking at any odd value n, we can take two mod 8 operations at once, knowing in advance what they will be - as the 5 binary digit tail reveals a command set of two mod 8 commands, the upper 2 bits holding the second operation…

Now we have the structure unwoven, and can see the parent of each value more clearly.

We aren’t there yet though, because we need to straighten out those 3 and 7 types, we can determine those exist by looking at the binary for n, thus we can put those on the same line, as a cascade, isolated, letting the next line show the source of the cascaded value and putting things into clearer perspective - the perspective of what each number reveals about itself, and only a new line when its a “newly revealed” value.

For us, that will mean at most climbing one hill, not descending.

Confirming efficacy now - looking up into the tower for n to see its first link, based on its 9 type which is visible to us, as well as looking towards 1, should make for units that always contain a value lower than themselves (from lower 16 sets than themselves)

And if we can say that all units connect to units lower than themselves, we have finally earned a cigar…

[GandalfPC]

Since OFT is a state system with three commands, all 100% reversible it’s a push/pop queue.

As a push pop queue state system all information needed to play back your path is recorded in the binary value itself.

Thus you cannot “come from and then return to infinity without reaching 1” as the queue cannot stuff infinite info into a finite value (and all values on that path would contain infinite info)

You will play back the queue until it is exhausted, which will be finite because the data of choices made at each fork in the path (which operation was chosen out of available 9 cycle options) must be finite.

And since the binary value mod 8 is the command, you will play until you hit the 1, the only command that leads to no other command.

And thus, there is no record made of how many times you loop at 1. No data can be recorded - or should we say, no data has been recorded yet.

The 9 cycle shows we can travel upward only, using a 5 type operation to arrive at 5. Now we have one command to play back, a type 5 operation which will return to 1, its prior state.


We can go from 5 to 3, as the 9 cycle says that 5 is a C tower and has its first exit at 2n and produces the ability for a type 1 operation to occur - .75x op on 5 leads us to 3.

Here in the very lowest numbers, never leaving our base command set of three bits we see the queue in action without the cover of complexity.

If we are on 3 we have binary 11 - it tells us the last operation was a type 3 and we use a type 3 to undo it, which plays back to 5 , which tells us the last operation was a type 5 and undoes it, and we arrive back at 1.


—


It changes the question from “how many steps from n to 1” to “how many commands can fit in a X bits.” and the answer is always finite.


——

for reference:

=============

Looking at the paths from a mod 16 perspective on a circle we see the movements in a revealing way…

[GandalfPC]

So, each of the three sets can be traversed (exited) in a single step using the tail stripping formulas, showing transit from 27 to 1 in 27 steps…

Complete set equations will traverse several (all) sequential steps of any mod type regardless of size and number of sequential steps involved. They assure exit from any diagram.

Formulas for build and traverse, singular and sets…

Tail building for 101 type is unlimited, but 001 and 111 types are self limiting - you can only take them as many m steps as the potential of the n allows…

We see maximum possible tail addition type 001 will top out at 1[0]1 and type 111 will top out at 1[1]1

=========================

Bringing those equations onto the diagrams

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The first occurrences of each path

====

working on the 1's tails rules for a path. I am seeing that a path will only be able to pass through a 1's tail that is the length of the starting binary +1 (it is very rare that it is greater, happens first at 125687199 which goes through power of two minus one 536870911 directly) - normal limit is equal to binary length of starting n. ran 30 bits last night that gathered stats on how many of each bit length n passed through each bit length tail, and gathered outliers above equal length. will let you know what comes of it...

since all paths will have that property, that would mean that the length of tail that any n is constricted to is ruled by the smallest binary that is above it, further away from 1 on its branch.

————

We find that binary 1’s tails are fully controlled. Each branch of the system, building from 1, starts with a 4n+1 operation, and then continues along the available exits, either (2n-1)/3 or (4n-1)/3 depending on A or C type n value, ending in a multiple of three tower (type B tower) with no exits. At that point 4n+1 is only choice to expand, before that we only had either A or C choice as well as 4n+1 choice - every n allows 4n+1. Every 4n+1 starts a new branch.

When 4n+1 starts a new branch it also adds 1 digit to the length of the binary 1’s tail possible on that branch - the count of 4n+1 operations is the count of the possible binary 1’s tail length.

This tail is what is stripped during traversal of “1.5x grow operations” when heading towards one.

It is controlled by ternary 2’s tail, which not only covers the binary 1’s tail, but due to compactness of ternary vs binary, covers 50% more of the binary length - it covers the entire “growth” portion - which is how the 4n+1 achieves control over the entire growth stage and limits the length of binary 1’s tail to the “level” we have climbed to above 1 with 4n+1 operations.

Here we see all n values accept 4n+1 in build direction and track the building of 1’s tails, which are following C tower chains through the 9 cycle, which has been mentioned earlier and is not discussed here…

——————-

That is how the 2 ternary tail controls the 1’s binary tail.

[Harvonsgard]

Just here to drop post #42 - feel free to move on.

[GandalfPC]

Tip to tail branches always reduce.

If we trace from 1 to 1 we have traversed the lowest branch.

We cannot use (2n-1)/3 nor (4n-1)/3 to traverse A and C links, other than to traverse to 1 itself using (4n-1)/3.

To leave this branch we use 4n+1, bringing us to 5, which allows for (2n-1)/3 bringing us to 3, at which point we are again, at the end of a branch, having neither (2n-1)/3 nor (4n-1)/3 available to us, we can only continue using 4n+1.

In this manner we are going from branch tip to branch tip - the longest path available on each branch is from its tip, a type B tower furthest from 1, to its base, a mod 8 residue 5 value that was born of a 4n+1 operation and constitutes our least distance on the branch from 1.

This spreadsheet, which displays not only the trace down a branch but the trace up as well, allows us to build tip to tip views - in this case we started with 1, and then 3, as above, from 3 we use 4n+1 to get to 13 and trace that branch using A and C links as available until we arrive at the tip of the branch, always a B tower with no links, hence its terminating nature.

What we find is that viewed tip to tip, all branches constitute a decrease.

In the build direction, as all n values allow you to use 4n+1 to exit the current branch, you need not go to the end, you can climb past any A or C using 4n+1 to a new branch

Taking an early exit in the build direction, using 4n+1 before arriving at a branch tip, does make the path appear to go up and down, but you have simply entered into another gradient that also has the same tip to tip always decreasing in traverse towards 1 property. There is no escaping it.

Here we have the above path up to a certain point, then we take an early exit, then we continue on tip to tip again. Note that after the disturbance of our jump, which allows for the appearance of a path that heads away from 1, we have not changed our fate, nor can we by repeating the procedure any number of times.



=======================

Here we have a display of the 9 cycle function. We take the path from 3 to 27 (in build direction) and note the mod operations it follows. Then we place paths next to it for various n, all spaced by 36 * 12x - each additional multiplier of 12 extends the depth of match for the cycle n values, eventually it will sync to have all of the paths shown traversing the same mod ops for the entire segment.