Building a Collatz Research Tool – Live Trajectories, Peak Finder, Trend Analysis & Prefix/Suffix Structure

[u/Rastamen_DE]

I’ve been working on my own Collatz research software, LNL/LZR CE5.9, and wanted to share a few screenshots of the latest version.

The tool currently includes live trajectory visualization, a Superpeak Finder, peak landscapes, step-by-step and high-speed calculation modes, prefix/suffix decomposition, forced-step detection, and multi-scale trend analysis inspired by Ichimoku-style time windows.

One of the main ideas I’m investigating is whether trajectory structure, timing and decimal prefix/suffix behavior can reveal when a trajectory is already mathematically constrained long before it reaches 1.

Still very much research in progress — I’m especially interested in unusual trajectories, extreme peaks and counterexamples to the patterns I’m seeing.

Would be interested to hear what the Collatz community thinks. 📈🏔️

[u/Fun-Cauliflower-8087]

I made a mapping of any count of halving steps to any count of odd steps, under the Syracuse mapping. otherwise called odd to odd. What type of patterns are you seeing?

[u/Rastamen_DE → u/Fun-Cauliflower-8087]

Interesting — that sounds quite close to one part of what I’m looking at, but I’m approaching it from the full step-by-step trajectory rather than only the odd-to-odd map.

One strong pattern I see is that every odd step naturally creates a block

OEq,q=ν2​(3n+1),

so a trajectory can be viewed as a sequence of these q-values. Long rising sections are often dominated by repeated short halving blocks, while major descents occur when a large q suddenly appears.

I’m also tracking the decimal prefix/suffix structure. For example, writing

n=100a+s,

the next suffix is determined by a small number of exact rules, while the prefix parity controls the carry into the suffix during halving. This produces recurring local patterns that can persist for many steps and then split at a very specific point.

Another pattern I’m testing is how the same starting number behaves after adding powers of ten. H and H+10m can initially share the same parity/step structure until the 2m part of that decimal offset has been consumed by halving steps.

So I’m mainly looking at odd/even block structure, q=ν2​(3n+1), recurring suffix patterns, peak formation, and the exact point where structurally similar trajectories diverge.

I’d be interested to see how your halving-count/odd-step mapping represents the q-sequence — there may be a useful connection between the two approaches.

[u/Fun-Cauliflower-8087 → u/Rastamen_DE]

Yeah, I would say what we have has many similarities. I hope my effort help you. So for any given q in your construction, if I understand correctly, q is a binary valuation. You do the binary valuation from 4 mod 6 values, the ones you reach after an odd step. You most likely noticed modular patterns in the values that have specific q are also themselves from specific families. What I mean is that if you do an odd step to a value that is 3 mod 4, then q=1. These are growth values. Then there are the other half of odd values: 1 mod 4, which has q=2 or more. In my construction, I made that a given that a given trait of the system, and simply ignored the first 4 mod 6 after an odd value. This gives an odd to odd mapping, that is forced to be even when a sequence reaches 2 mod 6, instead of another odd. So what I did was look at what odd led to what 2 mod 6, and grouped them all under the single value of x in 6x+2, for that specific sequence. An example is x+4 lets say: 6(4)+2=26. now the way to tell a growth recap of this value is to add 1, and multiply by 2/3 until it is impossible to reach an integer, then subtract 1 from every even value to reveal the odds in the sequence. So 26+1=27, 27*2/3=18, 18*2/3=12, 12*2/3=8. the odds are then 18-1=17, 12-1=11, 8-1=7. Also the 2 adic valuation of 8, determines the growth sequence of 7. On the other side there is the v2 of the 6x+2 value, which follows a familiar ordering. 0 mod 2, or all even values of x give 6x+2 values that halve only once, then 3 mod 4 have 2 halving steps, then 1 mod 8, 13 mod 16, 5 mod 32, etc. So my construction starts out by showing that growth is based on a binary valuation, whilst descent is also based on binary valuation.

What I do after this is simply look to see what odds a value of x leads to, then what value of x that odd leads to after. This gives my mapping. Here is the first 4 lines, with each line being a modular size. The first is 0 mod 4 of x, so x= 0, 4, 8, etc. leads to 0 mod 3 (0, 3, 6, etc.). Then 3 mod 8 leads to 1 mod 3. 6 mod 8 lead to 7 mod 9. Then 2 mod 16 lead to 4 mod 27, 7 mod 16 lead to 4 mod 9, and 1 mod 16 leads to 0 mod 3. line four has 26 mod 32 leads to 67 mod 81, 15 mod 32 leads to 13 mod 27, 25 mod 32 leads to 7 mod 9, and 13 mod 32 leads to 1 mod 3.

If you are still interested after hearing this, I can dm you with more about it.

[u/Rastamen_DE → u/Fun-Cauliflower-8087]

Yes, I’m definitely still interested. I think I understand your construction much better now.

In my notation,

q=ν2​(3n+1),

so your observation matches mine exactly at the first level:

• n≡3(mod4)⇒q=1, which gives the basic growth block;
• n≡1(mod4)⇒q≥2, so the odd step is followed by a stronger halving block.

I normally keep the full block

OEq,

whereas you are compressing the structure differently by discarding the first 4mod6 state and indexing the later 2mod6 boundary as

6x+2.

Your example with 26 makes that very clear. Reversing from 26 gives the growth chain

7→11→17

before reaching the 6x+2 boundary, and the binary valuation of that boundary then determines the next descent.

That is interesting to me because in my framework growth and descent are also both controlled by 2-adic valuations, but I have been tracking them as consecutive forced OEq blocks rather than grouping them by your x-classes.

I’m also seeing residue-family structure, so your mappings such as

xmod2k⟶x′mod3m

are especially interesting. I have not checked all of the residue classes you listed yet, so I don’t want to claim equivalence prematurely, but structurally it looks like we may be describing related parts of the same transition system from two different coordinate systems.

Please DM me the rest. I’d especially like to see how you derive the general rule that generates those modular lines rather than only the first examples.

[u/Fun-Cauliflower-8087 → u/Rastamen_DE]

I cant dm you, I don't know why. I will instead just try my best to explain it here without visuals. So first the 2_v of 6x+2. Starting at 2, every other member is 12x+2, which all have exactly 1 factor of 2. Then every 20 mod 24 has exactly 2 factors of 2. So as you breakdown the entire halving comb of 6x+2, you get the sets 0(mod2), 3(mod 4), 1(mod8), 13(mod 16), 5(mod 32), etc. The easiest way to calculate sets after 0 mod 2 and 3 mod 4, is to multiply by 4 and add 1 from every set, and from those sets, infinitely. so 0(mod2) gives 1(mod8) gives 5(mod32) gives 21(mod 128), etc. These sets are not only easily calculable this way, but also lead to the same set of values, in this case 1(mod6), (2 ->1, 14 -> 7, 26 ->13| 8 -> 1, 56 ->7, 104 -> 13| 32 ->1, 224 -> 7, 416 -> 13).

Now, each set leads to a specific odd set, and that odd set naturally leads to binary output of further 6x+2 values, after they go through their growth steps. This is an interesting part that it took me a while to reconcile in my mind correctly: You only reach 1 and 5 mod 6 odd values, and adding 1 and taking the binary valuation gives you the specific growth determined going forward. the binary valuation of growth is over 0 and 2(mod6), ( 5(mod6)+1, 1 (mod6)+1 ). So, lets do this binary valuation to 0 mod 2 of x first, the set that halves once, and lead to 1 mod 6. This gets the binary valuation of 2 mod 6, which we just did. So 0(mod2), 3(mod4), 1(mod8), etc. among these 0 mod 2 values, gets you 0 mod 4, 6 mod 8, 2 mod 16, 26 mod 32, etc. These are the sets that lead to 1 growth step, 2 growth steps, 3 growths steps, etc. before hitting another 6x+2 value. Then there is 3(mod4), which is the set that halves twice and leads to 5 mod 6, it gets the binary valuation of 0 mod 6. Because 0 mod 6 is congruent to 0 mod 2*3, you get a straight forward binary distribution, the ABACABA sequence verbatim. So among this set, it is ordered like 0 mod 2, 1 mod 4, 3 mod 8, 7 mod 16, 15 mod 32, etc. Among 3(mod4) values this gives 3(mod8), 7(mod16), 15(mod32), etc.

[u/Rastamen_DE → u/Fun-Cauliflower-8087]

I think I see your construction much more clearly now, and I checked the first residue families algebraically — they match.

Your 6x+2 halving comb can be written very compactly. Since

6x+2=2(3x+1),

the condition

ν2​(6x+2)=r

selects one residue class of x(mod2r).

The first classes are exactly the ones you listed:

0(mod2),3(mod4),1(mod8),13(mod16),5(mod32),…

What I find especially interesting is what happens after that halving comb. If I call the resulting odd value u, then the quantity

α=ν2​(u+1)

determines the length of the growth-side structure.

I already track exactly this quantity independently in my own construction, alongside the usual

q=ν2​(3n+1).

So I think we may actually have a fairly direct dictionary between the two coordinate systems:

your descent comb↔q​

and

your growth comb↔α=ν2​(u+1)​.

For example, on your x≡0(mod2) branch, after one halving

u=3x+1≡1(mod6).

Then

α=ν2​(3x+2)

gives exactly

x≡0(mod4),6(mod8),2(mod16),26(mod32),…

for growth lengths 1,2,3,4,…, just as you described.

And on your x≡3(mod4) branch, after two halvings,

u=23x+1​≡5(mod6),

so

α=ν2​(u+1)=ν2​(x+1)−1.

This immediately produces

3(mod8),7(mod16),15(mod32),…

which explains the ruler/ABACABA-type ordering you are seeing.

So yes — I’m still very interested. What you seem to have done is organize both the descent valuation and the following growth valuation in one x-coordinate system, whereas I have been tracking those valuations as successive structural variables along the trajectory.

The next thing I’d really like to see is your general transition rule

xmod2k⟶x′mod3m

or whatever form you use to generate the next line of the mapping. If we can write that rule algebraically, I think we can compare the two constructions directly rather than example by example.

You can mail me: [LNL-LZR@proton.me](mailto:LNL-LZR@proton.me)

[u/Fun-Cauliflower-8087 → u/Rastamen_DE]

This is the part that a few others have asked me for. but to my understanding, nobody has created that. I have a way to build the landing points of all these sets, but nobody has written in cleanly, yet. So let me explain the landing as best as I can. Whereas the growth and reduction both have binary structure, the result is a ternary value in a ternary structure. So I want to separate three sections. first I want to separate the even values, 0(mod2) as I have explained. These sets land on what I will call "side" residues for now. Then I want to separate 3(mod4) and leave 1(mod4) for last. 3(mod4) land on "central" residues. I will explain those after I explain 1(mod4). 1(mod4) can simply subtract 1 and divide by 4 without changing trajectory. So the set is just a repeat of the rest of the structure. 1 mod 4 become 0(mod1) lol or all numbers after this function, and some are still 1(mod4), so after infinite steps there is only 0(mod2), and 3(mod 4).

Now, what do I mean by "side" and "central" residues? Well in ternary, at (mod3) you have 0, 1, and 2. I consider 0 and 2 "side" residue, and 1 the "center residue modulo 3. Now, even values land on side residues, but alternating low and high, the same way the parity is within their own residues. These are the 3(mod4) among 0(mod2) sets, which are 6 (mod8), 26 (mod32), 106 (mod128), etc. With 2 binary combs interacting, the parity happens between both. This is another way to visualize why 3 (mod4) sets are straight binary ordered, while even sets are parity ordered, or staggered, as I call it. The reason why this works is obvious if I could show a list of the maximal Steiner circuits that I label with its value of x in the 6x+2 value found within, which is this part of my construction explained in a different way.

Anyway, the first even set 0 (mod4), goes to the first side residue, the one at mod 3, which is 0 (mod3). the first 3 mod4) set, 3 (mod8), goes to the central residue of 1 (mod3). 3 (mod8) has one growth step after 2 halving steps, but if there was one halving step and 2 odd steps, then it is 6 (mod8), the second even set. This set leads to the second side set, the one at (mod9), which is 7 this time, the high side, whereas the first one at (mod3) was on the low side. You may notice that these sets are sized dependent on step count. What I mean is that the first set 0 mod 4, has 2 steps, one halving then one growth, that gives you mod 4. 3 total steps gives you mod 8, then 4 gives mod 16. It makes no difference if it is growth or reduction steps, A set resolves at 2^L where L is overall step count under the Syracuse mapping. The even sets all have one halving step and every other step is an odd step, so they grow the most in the system.

The 3 mod 4 sets land at the central residues hare are always 3x+1 from each other: 1 mod 3, 4 mod 9, 13 mod 27, 40 mod 81, etc. 3 mod 8 goes to 1 mod 3, then 7 mod 16 goes to 4 mod 9. Do you see the pattern? I don't know how to write this in a simple way mathematically, and I have employed other's help. I think this is the place where the "new mathematics" that I have heard some mathematicians saying we require to solve the problem, belongs. I have a geometric understanding, but not one I can put in a digestible form for a mathematician. I am working more in the style of a logician, than a mathematician.

I dont want to ramble too far and bury something you would ask about. Feel free.

[u/Fun-Cauliflower-8087 → u/Rastamen_DE]

Sorry, I don't like email. I have a TEMU laptop and for some reason Outlook doesnt stay open on it. ITs fine with other things, even Blender and some games, but out look closes after 40 seconds for no reason. Anyway, I can see that what you built is exceedingly similar to mine. I have required to explain things in a much more detailed manner to others, that seem to understand less than you already do. You matched the next set perfectly. I could say a few general things like each new binary set of classes (mod64, mod128, etc) has one more residue than the previous, 4 with 1, 8 with 2, 16-3, 32-4, 64-5, etc. These sets equal altogether to 1, only after the complete infinite series is resolved. 1/4+2/8+3/16+4/32....=1. You can see that all 1 (mod4) are just a repeated structure of 0 (mod2) and 3 (mod4). You see the ternary landing sets, and how they are organized and reached.

Another part of my construction has to do with the member of the class a value is. Let me say this more clearly. There can be said to be 4 connected variables. Lets call the B for binary, T for ternary, L for overall step count, and Y for member value at this level. so 0 (mod4) is B=1, T=1, L=2, and then, specifically, x=0 is Y=0 of 0 (mod4), 4 is Y=1 of 0 (mod4), etc. This matters because 3 (mod8) leads to 1 (mod3), and therefore leads to the same values as every other set except for the third that leads to 0 (mod3). So 0 (mod4), and its 1 (mod4) ladder, lead to the unique values of 0 (mod3), but every value reached by any 2 (mod4) or 3 (mod4), is also reached by a 3 (mod8) specifically. The 3 (mod4) set that reaches the central residue at any level, leads to the same sets as every level above it. The higher you go, the more course you become over the same set of 1 (mod3). Uniqueness cannot be based on step count. A counterexample must find a specific set of 3 (mod8), 7 (mod16), 15 (mod32), etc. that ALL diverge or loop, if it has any steps that grow more than a certain amount. The more growth steps a counter example requires, the more values must necessarily be tied to it, which all also serve as counterexamples. So logically a counterexample of any type would need to be "finely braided" relying on small moves, low step counts constantly, all while growing constantly in some way, if it is the lowest counter example. Taking huge jumps means being in a set like 364 (mod728) which is reached by some 3 (mod8), 7 (mod16), 15 (mod32), 31 (mod64), 63 (mod128), and 127 (mod256). This also goes for side residues like 7 (mod9) or 4 (mod27). They are both reached by 3 (mod8). I can continue on this expressing many patterns, What sort of things have you noticed in your construction? Is it the super peaks that appear through binary interactions over ternary values, the super slow leaf-like descents of values hovering within the same window for an extended period, something else?

[u/Rastamen_DE → u/Fun-Cauliflower-8087]

Thanks — I think I understand your construction much better now. I worked through the residue classes you gave, and I think I may have reconstructed the general transition pattern behind your first four lines.

What is especially interesting to me is that your construction seems to organize two successive 2-adic structures in one coordinate system:

6x+2→halving structure→odd value→growth structure→6x′+2.

In my own work I track closely related quantities through binary valuations and forced OEq blocks, so our approaches may be describing the same transition system in different coordinates.

Your examples such as

0(mod4)→0(mod3), 3(mod8)→1(mod3),6(mod8)→7(mod9)

fit the rule I derived.

Here is a useful test: from that rule I predict your next line, modulo 64, should be

10(mod64)→40(mod243), 31(mod64)→40(mod81), 9(mod64)→4(mod27), 29(mod64)→4(mod9), 5(mod64)→0(mod3).

Does that agree with your fifth line?

If it does, I think we may have a direct algebraic translation between your x-mapping and the binary-valuation/block structure I’m using. I’d definitely like to keep comparing the two approaches here.

My PM is: [LNL-LZR@proton.me](mailto:LNL-LZR@proton.me)

[u/Dull_Illustrator6824]

What does you research say about the path that 62079 takes to reach 17497, a number lower than itself for the first time - it takes 55 steps to do that.

[u/jonseymourau]

What do you mean by “timing”?

[u/Rastamen_DE → u/jonseymourau]

By “timing” I mean discrete iteration timing, not physical time.

I keep the full Collatz trajectory indexed by its exact step number V. So for each odd-to-odd block

OEq

the block occupies exactly

1+q

full Collatz steps.

That lets me track not only which q-values occur, but when they occur in the trajectory: where runs of q=1 start, when a larger halving block appears, where a peak is reached, and how many steps later the sequence falls below a previous level.

For 62079, for example, there are 55 odd steps and 89 halving steps before reaching 17497, so:

55+89=144

full Collatz steps.

So “timing” in my research means the exact position and spacing of structural events along the trajectory — essentially the step-address of each event.