Collatz Streetmap

Introduction

When all Collatz sequences are read backwards, they form a graph starting with 1, 2 ..., hopefully without cycles (except for 1,2,4,1,2,4 ...). At each node n in the graph, the path starting at the root (4) and with the last node n can in principle be continued to 2 new nodes by a

• "m"-step: n * 2 (which is always possible), or a
• "d"-step: (n - 1) / 3 (which is possible only if n ≡ 1 mod 3).

When n mod 3 = 0, the path will continue with m-steps only, since the duplication maintains the divisibility by 3.

References

• Jeffry C. Lagarias, Ed.: The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010, ISBN 978-8218-4940-8. MBK78
• OEIS A07165: File of first 10K Collatz sequences, ascending start values, with lengths
• Gottfried Helms: The Collatz-Problem. A view into some 3x+1-trees and a new fractal graphic representation. Univ. Kassel.
• Klaus Brennecke: Collatzfolgen und Schachbrett, on Wikibooks

Motivation: Patterns in sequences with same length

When Collatz sequences are investigated, there are a lot of pairs of adjacent start values with the same sequence length, and with a characteristical neighbourhood of every other value, for example (from OEIS A070165):

142/104: [142 m  71 d 214 m 107 d 322 m 161 d 484 m  242 m 121 d | 364, 182, 91, ... 10, 5, 16, 8, 4, 2, 1]
           +1  *6+4    +1  *6+4    +1  *6+4    +1   *6+4  *6+2      =    =  ...
143/104: [143 d 430 m 215 d 646 m 323 d 970 m 485 d 1456 m 728 m | 364, 182, 91, ... 10, 5, 16, 8, 4, 2, 1]

Collatz roads

We define a "road" (with 2 parallel "lanes") as a sequence of pairs of elements (in 2 Collatz sequences with adjacent start values, read from right to left). A road is constructed by taking some n (364 in the example, the last common element of the 2 sequences) with n ≡ 4 mod 6, and by applying the steps

d m m d m d m d ... 
m m d m d m d m ...

in alternating sequence, until one of the elements in the pairs becomes divisible by 3. The construction yields one road with an upper lane (left elements of the pairs) and a lower lane (right elements), for example:

364: (121,728), (242,1456), (484,485), (161,970), (322,323), (107,646), (214,215), (71,430), (142,143) ...

The elements which are ≡ 4 mod 6 are emphasized. The construction continues until 63 ≡ 0 mod 3:

... (47,286), (94,95), (31,190), (62,63).

Roads table R

For better handling (e.g. in Excel), the roads for all starting values 4, 10, 16, 22 ... 4+6*n are listed as rows of a table. The columns in the rows are numbered r₀, r₁, r₂ ...

r0	r1	r2	r3	r4	r5	r6	r7	r8	r9	r10	r11	r12	r13	r14	r15	...
start	len	dr0	mr0	mr2	mr3	mr4	dr5	dr6	mr7	mr8	dr9	dr10	mr11	mr12	dr13	...
Δ6		Δ2	Δ12	Δ4	Δ24	Δ8	Δ8	3Δ8	3Δ48	3Δ16	3Δ16	9Δ16	9Δ96	9Δ32	9Δ32	...
4	5	1	8	2	16	4	5	1	10	2	3
10	3	3	20	6	40	12	13
16	3	5	32	10	64	20	21
22	4	7	44	14	88	28	29	9	58
28	3	9	56	18	112	36	37
34	3	11	68	22	136	44	45
40	9	13	80	26	160	52	53	17	106	34	35	11	70	22	23	7	46	14	15
46	3	15	92	30	184	60	61
52	3	17	104	34	208	68	69
58	5	19	116	38	232	76	77	25	154	50	51
64	3	21	128	42	256	84	85
70	3	23	140	46	280	92	93
76	4	25	152	50	304	100	101	33	202

There is a more elaborated example for elements <= 143248. The zoom factor of the web browser may be reduced (with Ctrl "-", to 25 % for example) such that the structure of the lengths of roads can be seen.

Relation to the Collatz graph

The Collatz graph is the union of all possible Collatz sequences, read backwards. Apart from the initial cycle 1-2-4-1 it should be a tree, and it should contain all positive numbers.

Each road contains 2 short paths in the Collatz graph, the root is left of the starting value of the road, and both proceed "upwards" in the tree. In contrast to the usual "chaotic" appearance of the Collatz graph, there is an amount of obvious structure in the road construction.

Road lengths

• The lengths r₁ of the roads seem to be finite. The highlighted pairs of numbers ≡ 4 mod 6 are decreasing to the end of the road.
• The lengths show a repeating pattern for the start values mod 54. The fixed lengths 3, 4, 5 can probably be explained from the road construction rule.

4 mod 54	10 mod 54	16 mod 54	22 mod 54	28 mod 54	34 mod 54	40 mod 54	46 mod 54	52 mod 54
5	3	3	4	3	3	n	3	3

• At the starting values 4, 40, 364, 3280, 29524 (OEIS A191681) the road lengths n_i have high values 5, 9, 13, 17, 21 which did not occur before. Those starting values are (9ⁿ⁺¹ - 1) / 2, or 4 * Sum(9ⁱ, i=0..n).
• The pattern of increasing and decreasing lengths is replicated when subsets of the rows (mod 9, 27, 81 ...) are regarded.

Coverage

The elements of the roads are strongly interconnected, and the array "obviously" shows all positive integers which are not multiples of 24:

r0 ≡ 4 mod 6	≡ 4,10,16,22 mod 24
r3 ≡ 8 mod 12	≡ 8,20 mod 24
r4 ≡ 2 mod 4	≡ 2,6,10,14,18,22 mod 24
r5 ≡ 16 mod 24	≡ 16 mod 24
r6 ≡ 4 mod 8	≡ 4,12,20 mod 24
r2 ≡ 1 mod 2	all odd numbers

All odd multiples of 3 are contained in column r₂. All multiples of 24 are obtained by 3 m-steps.

So if we can show that we reach all start values ≡ 4 mod 6, we have a proof that all positive integers are reached.

Starting with 4, it seems possible that a continuous expansion of all numbers ≡ 4 mod 6 into roads would finally yield all roads up to some start value. Experiments show that there are limits for the numbers involved. Roads above the clamp value are not necessary in order to obtain all roads below and including the start value:

start value	clamp value
4	4
40	76
364	2308
3280	143248

Subset table S

We may build derived table from the table of roads. We take columns r₀ and r₅ ff., and therein we keep the highlighted entries (those which are ≡ 4 mod 6) only, add 2 to them and divide them by 6. The resulting subset table starts as follows:

s0  s1   s2   s3   s4   s5   s6   s7   s8   ...
 n  len  
 1   3    3    1    2
 2   1    7
 3   1   11
 4   3   15    5   10
 5   1   19
 6   1   23
 7   7   27    9   18    6   12    4    8
 8   1   31
 9   1   35
10   3   39   13   26
11   1   43
12   1   47
13   3   51   17   34
14   1   55
15   1   59
16   5   63   21   42   14   28
...

This table can be described by simple rules: s₂ is always s₀ * 4 - 1. When s₂ ≡ 0 mod 3, the following columns s₃, s₄ ... are obtained by an alternating sequence of steps s_i+1 = s_i / 3 and s_i+2 = s_i+1 * 2, until all factors 3 in s₂ are replaced by factors 2.

This rule is probably provable from the construction rule for the roads.

Does S contain all positive integers?

The answer is yes. As above, we can regard the increments in successive columns:

ss ≡ 3 mod 4	half of the odd numbers
s3 ≡ 1 mod 4	other half of odd numbers
s4 ≡ 2 mod 8	≡ 2,10 mod 16
s5 ≡ 6 mod 8	≡ 6,14 mod 16
s6 ≡ 12 mod 16	≡ 12 mod 16
s7 ≡ 4 mod 16	≡ 4 mod 16
s4 ... s7	≡ 2,4,6, 10,12,14 mod 16

The numbers divisible by 8 are missing so far. Half of them would show up if the next column s₈ is considered, but multiples of 16 would be missing then, etc.

A simpler argument notes that S contains the rows for 27 * n for all n >= 1, and these rows also contain 8 * n because of the 3 -> 2 replacement process.

Can S be generated starting at 1?

We ask for an iterative process which starts with the row of S for index 1:

 1:    3    1    2

Then, all additional rows for the elements obtained so far are generated:

 2:    7
 3:   11

These rows are also expanded:

 7:   27    9   18    6   12    4    8
11:   43

Since we want to cover all indexes, we would first generate the rows for lower indexes. This process fills all rows up to 13 rather quickly. The next portion of completely filled rows up to ... is obtained when generating row ..., however.