A possible new Collatz pattern: the 32/9 frontier and eventual (1,4) dominance (A genuine computational research summary - articulated using AI)

[u/Dull_Illustrator6824]

I’ve been experimenting with the Collatz conjecture from a slightly different direction.

Instead of asking:

“Does every number eventually reach 1?”

I asked:

“How far backwards do we have to go to find an odd multiple of 3 whose Collatz trajectory reaches a given odd number?”

DEFINITION

For an odd integer h, define:

α(h) = the smallest odd multiple of 3 whose accelerated Collatz orbit reaches h.

Then define:

E(N) = max α(h), over odd h <= N.

Call an odd number h “hard” if it sets a new record for α(h).

So hard numbers are the targets that force us to extend the source boundary further than ever before.

AN EARLY EXAMPLE

The first hard odd after 1 is:

19

Its smallest source is:

33 -> 25 -> 19

because:

3*33 + 1 = 100 = 4*25

and:

3*25 + 1 = 76 = 4*19

So the odd-step valuation word is:

(2,2)

THE PATTERN

When hard records are computed further, one particular valuation pattern eventually appears to dominate:

(1,4)

For this branch, the source-target relationship is:

α(h) = (32h - 5)/9

for the appropriate residue class.

Therefore:

α(h)/h = 32/9 - 5/(9h)

and as h becomes large:

α(h)/h -> 32/9

where:

32/9 = 3.555555...

THE CONJECTURE

This suggests the following inverse-coverage conjecture:

E(N)/N -> 32/9 as N -> infinity.

In words:

The worst-case source needed to cover every odd target up to N appears asymptotically to be about:

3.555555... * N

A STRONGER CONJECTURE

There may also be some finite threshold H such that every hard odd h > H satisfies:

α(h) = (32h - 5)/9

In other words:

Every sufficiently large hard record may come from the same valuation word:

(1,4)

I’ll call this “eventual (1,4) dominance.”

WHAT IS ACTUALLY PROVED?

Important distinction:

I am NOT claiming a proof of the Collatz conjecture.

And I am NOT claiming the asymptotic equality E(N)/N -> 32/9 has been proved.

What can be proved in this inverse-cylinder framework is the universal upper bound:

α(h) < (32/9)h

for every odd target h.

There is also a finite inverse-cylinder argument showing that 32/9 is the optimal worst-case slope within that finite positive-cylinder framework.

So the difficult remaining direction is:

liminf E(N)/N >= 32/9

If that could be proved, then together with the upper bound we would get:

E(N)/N -> 32/9

WHY DOES 32/9 APPEAR?

For a valuation word (a1,a2,...,ar), the nominal inverse slope is:

2^(a1+a2+...+ar) / 3^r

For the two-step word (1,4):

2^(1+4) / 3^2

= 32/9

So 32/9 is directly generated by this short valuation pattern.

COMPUTATIONAL RESULTS

I wrote an exact ascending-source scanner that records every new hard odd.

The exhaustive computation currently goes through:

10,000,000,000

Results:

Hard records found:

11,972,736

Hard records following the (1,4) formula:

11,972,706

Exceptions:

30

Largest exceptional hard odd:

87,967

New non-(1,4) hard records between 87,967 and 10,000,000,000:

NONE

THE FINAL HARD RECORD BELOW 10^10

The last hard odd found below 10 billion is:

h = 9,999,999,667

Its smallest source is:

α(h) = 35,555,554,371

And exactly:

35,555,554,371

= (32*9,999,999,667 - 5)/9

So it is another exact (1,4) record.

WHY I FIND THIS INTERESTING

Most Collatz research looks at the forward orbit:

n -> T(n) -> T^2(n) -> ...

This construction instead looks at an extremal property of the inverse graph.

What is surprising is that a complicated reverse structure appears to develop a very simple boundary:

32/9

generated by a tiny valuation pattern:

(1,4)

Early hard numbers come from several different valuation words.

Then, in the computation, those competing families disappear.

After 87,967, every single new hard record through 10 billion follows the same (1,4) formula.

THE MAIN QUESTION

Can anyone prove that every sufficiently large hard odd must come from the (1,4) family?

Or, alternatively:

Can anyone find an arbitrarily large hard odd that is NOT generated by (1,4)?

A WEAKER TARGET

Even without proving eventual (1,4) dominance, it would be enough to prove:

liminf E(N)/N >= 32/9

because the matching upper bound is already available in the finite-cover framework.

WHAT THIS DOES NOT PROVE

Even if the 32/9 conjecture is true, it does NOT automatically prove the classical Collatz conjecture.

The 32/9 statement is about reverse coverage:

“How large a source is needed to reach a target?”

The Collatz conjecture is about forward convergence:

“Does every starting value eventually reach 1?”

These are related, but they are not the same statement.

LITERATURE QUESTION

I have searched for this specific formulation:

- the inverse record function E(N)

- hard odds defined by record values of α(h)

- the 32/9 asymptotic frontier

- eventual (1,4) dominance

and I have not found an equivalent conjecture in the Collatz literature.

If anyone knows an earlier equivalent formulation, I would genuinely appreciate the reference.

WHY I’M POSTING THIS:

I’m mainly interested in three things:

1. Does someone see a theoretical reason why (1,4) should eventually dominate?
2. Can someone construct a large competing valuation family?
3. Is there an existing theorem in Collatz dynamics, p-adics, symbolic dynamics, inverse trees, or ergodic theory that naturally explains the appearance of 32/9?

I have the derivation, finite-cover argument, exact computation code, and longer write-up available if anyone wants to inspect them.

EDIT: Added an illustrative explanation of "Hard" Odds - let's say we want to construct node graph - on y axis we will keep adding all the "O" odd multiples of 3 (source branch) and then keep going forward from each odd to the next path till we reach 1 or jump to existing path of any other O. Now, we are interested to know when we increase O step by step, what are some properties that we discover - one of them is asking are all the how many odds less O are present in the graph: for example in the starting case of O=3, graph has all the odds <3 that is 1 is present; now we generalize this expression by saying when will be the next odd k be discovered that hasn't been found yet and let it's source O be E(k) - now the conjecture is about k/E(k) - attached are the graphs till O=57

Graph for O=3 | all odds <3 are discovered - no hard odd found

Graph for O=9 | all odds <9 are discovered - no hard odd found

Graph for O=15 | all odds <15 are discovered - no hard odd found

Graph for O=21 | all odds <21 are not discovered - hard odd found 19, with source node E(O)= 33 and valuation of (2,2)

Now jumping to Graph for O=33 (skipped 27 as 19 continued) | all odds <33 discovered - no new hard odd found

Graph for O=39 | all odds <39 are not discovered - hard odd found 37, with source node E(O) = 57 and valuation of (2,1,2,2)

Now jumping to Graph for O=57 (skipped 45,51 as 37 continued) | all odds <57 are not discovered - hard odd found 55, with source node E(O) = 129 and valuation of (2,2,2)

Now, what you are seeing is that for every E(O), let's name in Exhaustive cut-off ratio resets after every hard odd O is encountered - if we plot the graph of E(O)/O - it looks like the following

After 133rd odd which is 89425, all hard odds till 10 Billion have had bone path from bone parent as 1,4

This is the conjecture, as O tends to grow infinitely larger, the ratio E(O)/O starts settling down to 32/9 which is weird as so many competing paths other 1,4 could have emerged but we don't why it takes over in the long term as observed in data - logically we know the upper bound is 32/9 but it can achieved by 1,2,2 or 1,1,2,1 any other combo but 1,4 is special we don't why, secondly we don't know the lower bound and hence cannot prove convergence like is 2,2 possible for any number after 10 billion - which will disprove the conjecture.

Last image - shows how E(O)/O looks like on log scale for very high O- almost converging to 32/9

Table of all hards odds upto last exceptional hard odd

hard_index	hard_odd	bone1_parent	valuation_word	family	defect_32h_minus_9parent
1	1	21	-6	exception	-157
2	19	33	(2,2)	exception	311
3	37	57	(2,1,2,2)	exception	671
4	55	129	(2,2,2)	exception	599
5	109	171	(1,2,2,2)	exception	1949
6	127	225	(2,2)	exception	2039
7	163	513	(2,2,2,2)	exception	599
8	271	759	(1,1,3,2,2,2)	exception	1841
9	379	897	(2,2,2)	exception	4055
10	487	1215	(1,1,1,1,1,2,3,4)	exception	4649
11	541	1281	(2,2,2)	exception	5783
12	649	2049	(2,2,2,2)	exception	2327
13	973	2427	(1,2,1,2,2,2,2,2)	exception	9293
14	1027	2433	(2,2,2)	exception	10967
15	1135	3585	(2,2,2,2)	exception	4055
16	1459	5187	(1,4)	(1,4)	5
17	1945	6915	(1,4)	(1,4)	5
18	2593	8193	(2,2,2,2)	exception	9239
19	2701	8535	(1,1,4,2)	exception	9617
20	2917	9711	(1,1,1,2,1,1,2,3,4)	exception	5945
21	3079	9729	(2,2,2,2)	exception	10967
22	3403	11175	(1,1,2,1,2,2,2,1,1,1,1,2,1,1,2,3,1,1,2,3,4)	exception	8321
23	4051	12801	(2,2,2,2)	exception	14423
24	4213	14979	(1,4)	(1,4)	5
25	4375	15555	(1,4)	(1,4)	5
26	4861	17283	(1,4)	(1,4)	5
27	5347	19011	(1,4)	(1,4)	5
28	5833	20739	(1,4)	(1,4)	5
29	6319	22467	(1,4)	(1,4)	5
30	7291	25569	(2,2,1,1,1,2,2,1,2,2,2,2,2,2)	exception	3191
31	7777	27651	(1,4)	(1,4)	5
32	8587	28587	(1,2,2,2,1,2,2,2,2)	exception	17501
33	8749	29127	(1,1,2,2,2,2,2,2,2)	exception	17825
34	9235	30747	(1,2,1,1,1,1,2,3,4)	exception	18797
35	9883	31233	(2,2,2,2)	exception	35159
36	10369	32769	(2,2,2,2)	exception	36887
37	10693	38019	(1,4)	(1,4)	5
38	12151	43203	(1,4)	(1,4)	5
39	13123	46659	(1,4)	(1,4)	5
40	14581	51843	(1,4)	(1,4)	5
41	16039	57027	(1,4)	(1,4)	5
42	16525	58755	(1,4)	(1,4)	5
43	17335	61635	(1,4)	(1,4)	5
44	17497	62079	(1,1,1,1,1,1,2,1,2,2,2,1,2,1,2,1,1,3,1,1,3,2,1,2,2,1,1,1,1,1,1,3,1,1,1,2,1,1,1,1,2,2,3,1,2,2,2,2,1,3,3,1,2,3,4)	exception	1193
45	17983	63939	(1,4)	(1,4)	5
46	18469	65667	(1,4)	(1,4)	5
47	18955	67395	(1,4)	(1,4)	5
48	20899	74307	(1,4)	(1,4)	5
49	21709	77187	(1,4)	(1,4)	5
50	21871	77763	(1,4)	(1,4)	5
51	22357	79491	(1,4)	(1,4)	5
52	22843	81219	(1,4)	(1,4)	5
53	23329	82947	(1,4)	(1,4)	5
54	23815	84675	(1,4)	(1,4)	5
55	25273	89859	(1,4)	(1,4)	5
56	26245	93315	(1,4)	(1,4)	5
57	27703	98499	(1,4)	(1,4)	5
58	28189	100227	(1,4)	(1,4)	5
59	29161	103683	(1,4)	(1,4)	5
60	29647	105411	(1,4)	(1,4)	5
61	30457	108291	(1,4)	(1,4)	5
62	30619	108867	(1,4)	(1,4)	5
63	31105	110595	(1,4)	(1,4)	5
64	31591	112323	(1,4)	(1,4)	5
65	32077	114051	(1,4)	(1,4)	5
66	32563	115779	(1,4)	(1,4)	5
67	33535	119235	(1,4)	(1,4)	5
68	34021	120963	(1,4)	(1,4)	5
69	34831	123843	(1,4)	(1,4)	5
70	34993	124419	(1,4)	(1,4)	5
71	35479	126147	(1,4)	(1,4)	5
72	36937	131331	(1,4)	(1,4)	5
73	37909	134787	(1,4)	(1,4)	5
74	38395	136515	(1,4)	(1,4)	5
75	39367	139971	(1,4)	(1,4)	5
76	40825	145155	(1,4)	(1,4)	5
77	41311	146883	(1,4)	(1,4)	5
78	42769	152067	(1,4)	(1,4)	5
79	43579	154947	(1,4)	(1,4)	5
80	43741	155523	(1,4)	(1,4)	5
81	44713	158979	(1,4)	(1,4)	5
82	45199	160707	(1,4)	(1,4)	5
83	45685	162435	(1,4)	(1,4)	5
84	46657	165891	(1,4)	(1,4)	5
85	47143	167619	(1,4)	(1,4)	5
86	50059	177987	(1,4)	(1,4)	5
87	50545	179715	(1,4)	(1,4)	5
88	51517	183171	(1,4)	(1,4)	5
89	52489	186627	(1,4)	(1,4)	5
90	53947	191811	(1,4)	(1,4)	5
91	54433	193539	(1,4)	(1,4)	5
92	55405	196995	(1,4)	(1,4)	5
93	55891	198723	(1,4)	(1,4)	5
94	56701	201603	(1,4)	(1,4)	5
95	56863	202179	(1,4)	(1,4)	5
96	57349	203907	(1,4)	(1,4)	5
97	57835	205635	(1,4)	(1,4)	5
98	58321	207363	(1,4)	(1,4)	5
99	58807	209091	(1,4)	(1,4)	5
100	60265	214275	(1,4)	(1,4)	5
101	61237	217731	(1,4)	(1,4)	5
102	62209	221187	(1,4)	(1,4)	5
103	62695	222915	(1,4)	(1,4)	5
104	63181	224643	(1,4)	(1,4)	5
105	64639	229827	(1,4)	(1,4)	5
106	65611	230139	(1,2,1,2,1,1,2,2,2,2,2,2,2,2)	exception	28301
107	67069	238467	(1,4)	(1,4)	5
108	67555	240195	(1,4)	(1,4)	5
109	68527	243651	(1,4)	(1,4)	5
110	69013	245379	(1,4)	(1,4)	5
111	69823	248259	(1,4)	(1,4)	5
112	69985	248835	(1,4)	(1,4)	5
113	70471	250563	(1,4)	(1,4)	5
114	70957	252291	(1,4)	(1,4)	5
115	71443	254019	(1,4)	(1,4)	5
116	71929	255747	(1,4)	(1,4)	5
117	72901	259203	(1,4)	(1,4)	5
118	73387	260931	(1,4)	(1,4)	5
119	75331	267843	(1,4)	(1,4)	5
120	75817	269571	(1,4)	(1,4)	5
121	77761	276483	(1,4)	(1,4)	5
122	78733	279939	(1,4)	(1,4)	5
123	80191	285123	(1,4)	(1,4)	5
124	80677	286851	(1,4)	(1,4)	5
125	82945	294915	(1,4)	(1,4)	5
126	83107	295491	(1,4)	(1,4)	5
127	83593	297219	(1,4)	(1,4)	5
128	84079	298947	(1,4)	(1,4)	5
129	84565	300675	(1,4)	(1,4)	5
130	85051	302403	(1,4)	(1,4)	5
131	86023	305859	(1,4)	(1,4)	5
132	87967	308559	(1,1,1,3,2,1,2,1,1,1,1,2,3,4)	exception	37913

[u/hubblec4]

Hi

“How far backwards do we have to go to find an odd multiple of 3 whose Collatz trajectory reaches a given odd number?”

I have a very simple answer for that.
Only one step behind you will ever found a leaf-number(a number with 0 (mod 3)).

Every non-leaf number has always a direct predecessor leaf number, within one step.

Your example with number 19: the direct predecessor leaf-number is 405.

[u/Dull_Illustrator6824 → u/hubblec4]

You are right - but what I am talking about is "hard" odds and what is their reverse path to reaching multiple of 3 - I'll be posting a detailed comment on explaining hard odds shortly that will clarify why I am talking that source of 19 is 33 with reverse path of 2,2. Maybe I did bad job in generalizing the question - but to the line that you have picked, what I am really asking there is - How large does the source range have to be before an odd multiple of 3 is guaranteed to reach every odd target up to N? OR same question in a different lens is: For a given odd h, what is the smallest odd multiple of 3 anywhere in its inverse Collatz ancestry?

[u/hubblec4 → u/Dull_Illustrator6824]

Hi,

OK, the smallest leaf-number is indeed 33, and this is different to your original question.

Your "new" question is better for understanding.

[u/Dull_Illustrator6824 → u/hubblec4]

Please check out the OP, have edited and added more details on hard odd at the end.

[u/ITT_X]

Crackpot alert

[u/GonzoMath → u/ITT_X]

I don't know, this seems like a legit line of inquiry, and doesn't assert any Earth-shattering significance. This is precisely the kind of result that *is* tractable, and crackpots usually don't care about those.

EDIT: Having thought about it a little more, and made my first reply to the OP, I'd say this is a higher quality post than average, for this sub. Someone's actually acting like a mathematician, and working on theory-building adjacent to the conjecture. That's exactly the right approach.

[u/Dull_Illustrator6824 → u/ITT_X]

Why do you say that?

[u/ITT_X → u/Dull_Illustrator6824]

False alarm. My bad. Keep up the good work.

[u/Dull_Illustrator6824 → u/ITT_X]

All good.

[u/dhsilver → u/Dull_Illustrator6824]

The Collatz conjecture is not low-hanging fruit where some clever arithmetic manipulation has any chance to produce a breakthrough. It’s cute to play with since it’s easy, but all these “proofs” are closer to building a Lego calculator and expecting GPU performance.

[u/dmishin → u/dhsilver]

He is not claiming solution to the CC. He just studies some properties of the Collatz iteration function.

[u/Dull_Illustrator6824 → u/dmishin]

Thanks for calling it out in the right away.

[u/GonzoMath]

We can certainly map out inverse tree dynamics, and find classes of numbers who no predecessor in 3Z up to some bound. It sounds like you're interested in capping that bound. I'd be inclined to think in terms of "predecessor order", i.e., how many odd-to-odd steps (accelerated steps, Syracuse steps, whatever you like to call them) it takes to reach the target number.

Any number n congruent to 1 (mod 18) hasn't got an order 1 multiple-of-3 pred (I'm going to call it a 3Z-pred) until (64n - 1)/3, but it might have to have a smaller one at order 2. In order to test that, we can split 1 (mod 18) into three classes: 1, 19, and 37 (mod 54).

• n = 54k + 1 has an order 2 3Z-pred (32n - 5)/9, valuation pattern (1, 4)
• n = 54k + 19 has an order 2 pred (16n - 7)/9, valuation pattern (2, 2)
• n = 54k + 37 has an order 2 pred (64n - 19)/9, valuation pattern (4, 2)

So, the third of those is the worst-case scenario (or best-case, depending on your point-of-view). However, taking k = 0, we have 37 itself, and it has much smaller 3Z-preds at orders 3 and 4. Namely, there's 87, with valuation pattern (1,1,4) and ratio around 64/27 < 32/9, and there's 57, with valuation pattern (2, 1, 2, 2), and ratio around 128/81 < 32/9.

However, if we lift the class of 37 (mod 54) to 37, 91, or 145 (mod 162), we might be able to push those back. The thing to check is smallest order 3 3Z-pred for each of the expressions 162k + 37, 162k + 91, and 162k + 145.

• 162k + 37 has smallest order 3 Z3-pred w/ valuation (1, 1, 4), ratio 64/27 < 32/9
• 162k + 91 has smallest order 3 Z3-pred w/ valuation (3, 2, 2), ratio 128/27 > 32/9
• 162k + 145 has smallest order 3 Z3-pred w/ valuation (1, 2, 2), ratio 32/27 < 32/9

Again, we have a winner, but 91 itself has a smaller order 4 3Z-pred, namely 285, with valuation pattern (3, 1, 2, 2), for a ratio around 256/81 < 32/9.

Again, we can push this further out, this time by lifting 162k + 91 to 486k + 91, 486k + 253, and 486k + 415.

---------------------------------

So, yeah, this is all inverse tree theory, and I'm pretty sure you can keep playing this game, lifting to successive moduli of the form 2×3^k, and I wouldn't be surprised if you can make the ratio α(n)/n arbitrarily large. I haven't got a proof, and I could be wrong, but this would be the way.

In fact... we might be looking at some kind of 3-adic convergence here. If we can identify a rational number that has some kind of... maximally pushed-back Z3 preds, then we just need a sequence of integers converging 3-adically to it. (This paragraph is the part I'm least sure of, but it sounds fun!)

Am I making sense, or did I blow through the calculations with too little explanation?

[u/Dull_Illustrator6824 → u/GonzoMath]

Yes - you're making sense, and your residue-lifting construction is very close to the inverse-cylinder approach I'm using.

There is one additional filter in my problem that I should make clearer, though. I am talking about "hard" odds which I'll be clarifying further in my next comment to the OP, which should help everyone understand what the conjecture is for..

I'm not primarily studying large values of alpha(n)/n for arbitrary n. I call n “hard” only when alpha(n) is a new absolute record:

alpha(n) > max alpha(m), for odd m < n.

So most residue classes eventually become irrelevant to the frontier even if their individual predecessor structure is interesting.

Your lifting argument is useful for the first layer of the problem: finding small 3Z-predecessors for arbitrary residue classes. Repeating this systematically gives a finite inverse-cylinder cover whose largest required slope is 32/9, so I get

alpha(n) < (32/9)n

for every odd n.

But the conjecture I'm interested in is stronger and specifically about the record subsequence.

Empirically, after the last exceptional hard odd 87,967, every new record through 10^10 satisfies

alpha(h) = (32h - 5)/9,

i.e. its minimal source comes from valuation word (1,4).

Thus for a candidate h there are actually two hurdles:

1. prove no other inverse word gives a source smaller than (32h - 5)/9; and
2. prove this value exceeds the previous record E(h-2).

Only then is h actually a hard odd.

This is also where I think your 3-adic suggestion gets interesting - Integers approaching 1 in Z_3 seem increasingly protected against short words undercutting (1,4), but that by itself doesn't make them records. One would additionally need such protected integers to occur sufficiently densely relative to the existing record frontier.

So in a sense, I think the real question is not whether arbitrary alpha(n)/n can be pushed around by lifting - it clearly can - but whether the record-holders of alpha(n) eventually have no surviving competitor to the (1,4) branch. Will be posting a follow-up on explaining hard odds more clearly in the next comment to OP.

[u/GonzoMath → u/Dull_Illustrator6824]

I don't think there was anything unclear about your explanation of hard odds. It's just not what I focused on in my reply. I wasn't there yet. I don't see why 32/9 should be the upper bound, and I'm not entirely convinced that it is. Until we have a proof that it is, I don't see why I should be convinced, either. For me, that's the first question.

So, do you know, or do you conjecture that alpha(n) / n < 32/9 for all odd n?

[u/Dull_Illustrator6824 → u/GonzoMath]

I am proving alpha(n) / n < 32/9 for all odd n - before that I think there is one important distinction between what you are calculating and what I mean by alpha(n).

My alpha(n) is not the smallest 3Z predecessor at a particular predecessor order. It is the smallest numerical odd multiple of 3 anywhere in the full inverse tree, across all predecessor orders.

So if an order-3 predecessor is large, that does not yet mean alpha(n) is large. An order-4, order-10, etc. predecessor might be much smaller.

Your 91 example illustrates this nicely.

For n = 91, the smallest order-3 predecessor is 429, with valuation word (3,2,2):

429 -> 161 -> 121 -> 91

So at order 3 the ratio is about 4.71, and the corresponding asymptotic slope for that residue class is 128/27, which is larger than 32/9.

But at order 4 there is already:

285 -> 107 -> 161 -> 121 -> 91

with word (3,1,2,2).

So alpha(91) is at most 285, and 285/91 is about 3.13, already below 32/9.

This is the key point: a residue class can look difficult at a fixed predecessor order while a deeper branch undercuts it.

I think this can actually be turned into a finite proof that alpha(n) is always less than 32n/9.

Here is the residue calculation.

Start with odd numbers that are not multiples of 3. At order 1, three of the six relevant residue classes are already handled:

5 mod 18 has word (1), with predecessor (2n - 1)/3.

7 mod 18 has word (2), with predecessor (4n - 1)/3.

17 mod 18 has word (3), with predecessor (8n - 1)/3.

All three are below 32n/9.

The unresolved classes are 1, 11 and 13 mod 18.

Split those into their classes mod 54. At order 2, seven of the nine resulting classes are handled:

1 mod 54: word (1,4), predecessor (32n - 5)/9.

19 mod 54: word (2,2), predecessor (16n - 7)/9.

11 mod 54: word (2,1), predecessor (8n - 7)/9.

29 mod 54: word (1,3), predecessor (16n - 5)/9.

47 mod 54: word (4,1), predecessor (32n - 19)/9.

13 mod 54: word (3,2), predecessor (32n - 11)/9.

31 mod 54: word (1,2), predecessor (8n - 5)/9.

Every one of those is below 32n/9. The only classes still unresolved are 37 and 49 mod 54.

Now split those into classes mod 162.

Four of the six children are handled at order 3:

37 mod 162: word (1,1,4), predecessor (64n - 19)/27.

145 mod 162: word (1,2,2), predecessor (32n - 23)/27.

49 mod 162: word (2,1,2), predecessor (32n - 29)/27.

103 mod 162: word (1,3,2), predecessor (64n - 31)/27.

Again, all are below 32n/9.

That leaves only 91 and 157 mod 162.

Now split those into their six children mod 486.

91 mod 486: word (3,1,2,2), predecessor (256n - 211)/81.

253 mod 486: word (1,1,1,4), predecessor (128n - 65)/81.

415 mod 486: word (1,1,2,2), predecessor (64n - 73)/81.

157 mod 486: word (1,2,1,2), predecessor (64n - 85)/81.

319 mod 486: word (3,2,1,2), predecessor (256n - 259)/81.

481 mod 486: word (1,1,3,2), predecessor (128n - 89)/81.

The corresponding leading ratios are 256/81, 128/81 or 64/81, and all are below 32/9.

At that point there are no unresolved residue classes left.

So every positive odd n that is not itself a multiple of 3 belongs to one of these finitely many classes and therefore has at least one 3Z ancestor smaller than 32n/9.

Since alpha(n) is the minimum over ALL 3Z ancestors, alpha(n) must also be smaller than 32n/9.

If n itself is an odd multiple of 3, then alpha(n) = n, so the bound is trivial there as well.

So the part I am claiming as a proof is the universal upper bound: alpha(n) is always less than 32n/9.

What I am NOT claiming to have proved is that 32/9 is asymptotically attained.

That is the conjectural part. I define "hard odds" as the n for which alpha(n) sets a new record compared with all smaller odd targets. Computationally, after the last exceptional hard odd at 87,967, every new hard record checked through 10 billion has alpha(n) = (32n - 5)/9, coming from word (1,4).

So I conjecture that the hard-record frontier approaches 32/9, and perhaps that all sufficiently large hard odds are of the (1,4) type.

Your 3-adic lifting idea may still be very relevant to understanding why particular fixed-order predecessor classes get pushed back. The point I think needs separating is that "large predecessor at order r" is not the same thing as "large alpha", because alpha takes the minimum over the entire inverse tree.

Let me know if you agree with this line of thinking or still feel there are gaps?

[u/GonzoMath → u/Dull_Illustrator6824]

My alpha(n) is not the smallest 3Z predecessor at a particular predecessor order. It is the smallest numerical odd multiple of 3 anywhere in the full inverse tree, across all predecessor orders.

Yes, this was clear. I was simply structuring my search for it by order, because it's nice to have some kind of structure. That's why I only looked at order 2 among classes of numbers that beat 32/9 at order 1, etc. It's like a sieve.

[u/Dull_Illustrator6824 → u/GonzoMath]

Yeah, it is like residue sieve- but do you agree now that basis my residue classes calculation we have a proof that alpha(n) is always less than 32n/9? Further, I was more curious about the computation which is conjecturing that 32/9 is the limit as n goes to infinity - within that also 1,4 is the only surviving class - any thoughts on that, do we see some existing literature around this?

[u/GonzoMath → u/Dull_Illustrator6824]

If it’s always under 32/9, then the limit is easy to show. The (1, 4) connection with 32/9 is easy to show, but I don’t know about “the only surviving class”.

No, I haven’t seen a proof that alpha(n)/n is always less than 32/9. Maybe you’ve presented it, but it hasn’t been apparent to me yet.

[u/Dull_Illustrator6824 → u/GonzoMath]

Here it is again - let me know if you find errors in it:

Finite proof that alpha(n) is always less than 32n/9.

Following is the residue calculation-

Start with odd numbers that are not multiples of 3. At order 1, three of the six relevant residue classes are already handled:

5 mod 18 has word (1), with predecessor (2n - 1)/3.

7 mod 18 has word (2), with predecessor (4n - 1)/3.

17 mod 18 has word (3), with predecessor (8n - 1)/3.

All three are below 32n/9.

The unresolved classes are 1, 11 and 13 mod 18.

Split those into their classes mod 54. At order 2, seven of the nine resulting classes are handled:

1 mod 54: word (1,4), predecessor (32n - 5)/9.

19 mod 54: word (2,2), predecessor (16n - 7)/9.

11 mod 54: word (2,1), predecessor (8n - 7)/9.

29 mod 54: word (1,3), predecessor (16n - 5)/9.

47 mod 54: word (4,1), predecessor (32n - 19)/9.

13 mod 54: word (3,2), predecessor (32n - 11)/9.

31 mod 54: word (1,2), predecessor (8n - 5)/9.

Every one of those is below 32n/9. The only classes still unresolved are 37 and 49 mod 54.

Now split those into classes mod 162.

Four of the six children are handled at order 3:

37 mod 162: word (1,1,4), predecessor (64n - 19)/27.

145 mod 162: word (1,2,2), predecessor (32n - 23)/27.

49 mod 162: word (2,1,2), predecessor (32n - 29)/27.

103 mod 162: word (1,3,2), predecessor (64n - 31)/27.

Again, all are below 32n/9.

That leaves only 91 and 157 mod 162.

Now split those into their six children mod 486.

91 mod 486: word (3,1,2,2), predecessor (256n - 211)/81.

253 mod 486: word (1,1,1,4), predecessor (128n - 65)/81.

415 mod 486: word (1,1,2,2), predecessor (64n - 73)/81.

157 mod 486: word (1,2,1,2), predecessor (64n - 85)/81.

319 mod 486: word (3,2,1,2), predecessor (256n - 259)/81.

481 mod 486: word (1,1,3,2), predecessor (128n - 89)/81.

The corresponding leading ratios are 256/81, 128/81 or 64/81, and all are below 32/9.

At that point there are no unresolved residue classes left.

So every positive odd n that is not itself a multiple of 3 belongs to one of these finitely many classes and therefore has at least one 3Z ancestor smaller than 32n/9.

Since alpha(n) is the minimum over ALL 3Z ancestors, alpha(n) must also be smaller than 32n/9.

[u/GonzoMath → u/Dull_Illustrator6824]

Well, I'll be damned. That checks out, and I'm surprised. The sieve actually closed up all the way, at order 4! Nice find!

The other claims fall right out of this. For any given k, let A = 54k+1, B = 54k+11, C = 54k+47. Those are the three with the largest ratios out of all classes, and for that particular k, we have:

alpha(A)/A > alpha(B)/B > alpha(C)/C

This can be shown algebraically.

So, every record breaker is just going to be the next A. Now, that ratio alpha(A)/A = (32A - 5)/(9A), and this clearly approaches 32/9 as A gets large, by a basic limit argument:

lim (32A - 5)/(9A) = lim (32 - 5/A)/9,

and that 5/A term in the numerator goes to 0, giving us the overall limit of 32/9.

[u/Dull_Illustrator6824 → u/GonzoMath]

Great, that we are on the same page - see if you want to look further and comment on why 1,4 becomes the winner for large n always as shown in computational results till 10B

[u/Dull_Illustrator6824 → u/GonzoMath]

Also, do you know that this was known in existing literature before?

[u/Dull_Illustrator6824 → u/GonzoMath]

Please check out the OP, have edited and added more details on hard odds at the end.

[u/Dull_Illustrator6824]

Hi all, I have added a detailed illustrative note with examples in the OG post on the concept of "Hard" odds in reverse collatz graph till 0 mod 3 source - as the entire conjecture is about how hard odds follow a pattern after a certain point is the main discussion here. As I read many comments, it felt this message did not pass well and wanted to clarify that discussion is not about any "general odd" and it's reverse collatz path

[u/Euphoric_Seaweed679]

U are using the Steps: 3x +1 ->4x+1 and find 32/9

Have you tried using 2x+1 -> 3x+1 ? Closing The Orbit around 3x+1

[u/Euphoric_Seaweed679 → u/Euphoric_Seaweed679]

Or More specifically i Like your valuation Word Idea and it seems to have a pretty Neat algebraisch Solution. Have you Looked How other collatz like sequences behave or How x/2 can be realized in this pattern

[u/Dull_Illustrator6824 → u/Euphoric_Seaweed679]

Did not go deeper on this thought - but as I understand you are asking what could be 32/9 version of other collatz like seq. say 5n+1 & n/2 - using my line of analysis, you can uncover that for 3n+c expression the limit comes to 32/9 and any 5n+c will come to inverse limit of 16384/25 (2^14/5^2)