Collatz Streetmap

Introduction

Collatz sequences are sequences of non-negative integer numbers with a simple construction rule: even elements a halved, and odd elements are multiplied by 3 and then incremented by 1. Since decades it is unknown whether the final cyle 4 - 2 - 1 is always reached for any start value. This problem is the Collatz conjecture, for which the english Wikipedia states:

It is also known as the 3n + 1 conjecture, the Ulam conjecture (after Stanisław Ulam), Kakutani's problem (after Shizuo Kakutani), the Thwaites conjecture (after Sir Bryan Thwaites), Hasse's algorithm (after Helmut Hasse), or the Syracuse problem; the sequence of numbers involved is referred to as the hailstone sequence or hailstone numbers (because the values are usually subject to multiple descents and ascents like hailstones in a cloud), or as wondrous numbers.

When we speak of numbers in this article, we normally mean natural integer numbers > 0. The zero is sometimes mentioned explicetely.

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

Collatz graph

When all Collatz sequences are read backwards, they form the Collatz graph starting with 1, 2, 4, 8 ... . At each node n > 4 in the graph, the path from the root (4) can be continued

• always to n * 2, and
• sometimes also to (n - 1) / 3.

When n ≡ 0 mod 3, the path will continue with duplications only, since these maintain the divisibility by 3.

The conjecture claims that the graphs contains all numbers, and that - except for the leading cycle 1 - 2 - 4 - 1 - 2 - 4 ... - it has the form of a tree without cylces.

Straightforward visualizations of the Collatz graph show now obvious structure. The sequences for the first dozen of start values seem to be rather harmless, but the sequence for 27 suddenly has 112 elements.

Read on in 3x+1 Problem.

This article proposes

• a rather regular structure for short portions of the Collatz graph, and
• a corresponding algorithm which
  • combines these portions and
  • uses them to walk the graph in a systematic, predictable way
• such that the reader could finally be convinced that the algorithm enumerates all numbers.

Steps

In order to walk back and forth in the Collatz graph, we will write

a step b

to denote a move from node (element) a to b. The following table lists such steps:

Name	Mnemonic	Direction	Operation	Condition	Remark
h	halve	-> root	b = a / 2	a ≡ 0 mod 2	next in sequence
t	triple	-> root	b = 3 * a + 1	true	next in sequence
m	multiply	root ->	b = a * 2	true	inverse of h
d	divide	root ->	b = (a - 1) / 3	a ≡ 1 mod 3	inverse of t

Steps may be combined, for example

a dm b : b = ((a - 1) / 3) * 2 

A starting number and a sequence of step names defines a unique, directed path in the Collatz graph.

Trivial paths

There are two types of paths whose descriptions are very simple:

(n = 2k) hhhh ... h 8 h 4 h 2 h 1  - powers of 2
(n ≡ 0 mod 3) mmm ... m (n * 2k) ... - multiples of 3

Collatz streets

Motivation: Patterns in sequences with same length

A closer look at the Collatz sequences shows a lot of pairs of adjacent start values which have the same sequence length, for example (from OEIS A070165):

142/104: [142 h  71 t 214 h 107 t 322 h 161 t 484 h  242 h 121 t 364 | 182, 91, ... 10, 5, 16, 8, 4, 2, 1]
143/104: [143 t 430 h 215 t 646 h 323 t 970 h 485 t 1456 h 728 h 364 | 182, 91, ... 10, 5, 16, 8, 4, 2, 1]
           +1  *6+4    +1  *6+4    +1  *6+4    +1   *6+4  *6+2    +0    +0 ...

The third line tells how the second line can be computed from the first. Walking from right to left, the step pattern is:

d m m d m d m d m 
m m d m d m d m d

or, in linearized form:

d/m m/m m/d d/m m/d d/m m/d d/m m/d ... 

Informally, a street is a parallel arrangement of 2 paths stemming from 2 sequences which have a common tail. A street starts with an element ≡ 4 mod 6 (364 in the example, before the bar), it proceeds to the left with a d/m and a m/m pair of steps, and then it extends to the left as long as a characteristical, alternating sequence of pairs of steps m/d - d/m - m/d - d/m ... can be continued. In the example, the street can be continued with 4 additional pairs of steps:

  q t | 62 h  31 t  94 h  47 t 142 h ...
126 h | 63 t 190 h  95 t 286 h 143 t ...
        +1  *6+4    +1  *6+4    +1  

The construction stops since there is no number q such that q * 3 + 1 = 62.

Street directory S

Though the graph usually has a "chaotic" appearance, the streets exhibit quite some amount of regular structure. This can be seen if we list the paired elements of the streets for all possible starting values 4, 10, 16, 22 ... 6n-2 for n = 1, 2 ... as rows of a table, in reversed direction (extending to the right). The elements ≡ 4 mod 6 are highlighted:

c1	c2	c3	c4	c5	c6	c7	c8	c9	c10	c11	c12	c13	c14	c15	...
4	1	8	2	16	4	5	1	10	2	3
10	3	20	6	40	12	13
16	5	32	10	64	20	21
22	7	44	14	88	28	29	9	58
28	9	56	18	112	36	37
34	11	68	22	136	44	45
40	13	80	26	160	52	53	17	106	34	35	11	70	22	23	7	46	14	15
46	15	92	30	184	60	61
52	17	104	34	208	68	69
58	19	116	38	232	76	77	25	154	50	51
64	21	128	42	256	84	85
70	23	140	46	280	92	93
76	25	152	50	304	100	101	33	202

There is a more elaborated example for elements <= 143248.

When this file is displayed in a browser, the zoom factor may be reduced (with Ctrl "-", to 25 % for example) such that the structure of the lengths of streets can be seen.

Street construction rules

The following table shows the rules for the construction of the first 9 columns S[n,1..9] of row n (n = 1, 2, 3 ...) in the street directory:

Column	Steps	Expression	Formula	Condition	Coverage
1			6n-2		4,10,16,22 mod 24
2	d	(6n-2-1)/3	2n-1		all odd numbers
3	m	(6n-2)*2	12n-4		8,20 mod 24
4	dm	((6n-2-1)/3)*2	4n-2		2,6,10,14,18,22 mod 24
5	mm	(6n-2)*2*2	24n-8		16 mod 24
6	dmm	((6n-2-1)/3)*2*2	8n-4		4,12,20 mod 24
7	mmd	((6n-2)*2*2-1)/3	8n-3		5,13,21 mod 24
8	dmmd	((6n-2-1)/3)*2*2-1)/3	(8n-5)/3	n ≡ 1 mod 3	(1,9,17,25 ...)
9	mmdm	((6n-2)*2*2-1)/3)*2	16n-6	n ≡ 1 mod 3	(10,58,106,154, ...)

The first 6 columns of the table cover the odd numbers and all numbers ≡ 2,4,6,8,10,12,16,18,20,22 mod 24.

It is not shown so far that all multiples of 24 are contained in the table.

Highlighted numbers

The numbers of the form 6p-2 were highlighted in the example above. They have the special property that, when p > 0 and p ≡ 0 mod 3, a dm-step yields a number of the same form, but with one factor 3 in p replaced by 2:

6(3q)-2 dm ((6(3q)-2)-1)/3*2 = (6q-1)*2 = 6(2q)-2

This implies that a dm-step decreases any number by about one third.

Street lengths > 7

• Columns 4(k+1)+1 result by dm-steps from columns 4k+1 for k=1,2,... (and so do columns 4(k+1)+2 result from columns 4k+2). Sequences of dm-steps decrease the numbers. Therefore the lengths of all streets are finite.
• Column 5 is 24n-8, and the lengths depend on the power of 3 contained in that n.
• At the starting values 4, 40, 364, 3280, 29524 (OEIS A191681) the street lengths 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).

Coverage

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

r1 ≡ 4 mod 6	≡ 4,10,16,22 mod 24
r2 ≡ 1 mod 2	all odd numbers
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

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 streets would finally yield all streets up to some start value. Experiments show that there are limits for the numbers involved. Streets above the clamp value are not necessary in order to obtain all streets 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 streets. 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 S 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 which are hopefully provable from the construction rule for the streets:

• 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.

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
s8 ≡ 8 mod 32	8, 40, 72, ...
s9 ≡ 24 mod 32	24, 56, 88, ...
s10 ≡ 48 mod 64	48, 112, 176, 240 ...
s11 ≡ 16 mod 64	16, 80, ...

This shows that the columns s₄ ... s₇ contain all numbers ≡ 2,4,6,10,12,14 mod 16, but those ≡ 0,8 mod 16 are missing so far. The ones ≡ 8 mod 16 show up in s₈ resp. s₉, half of the multiples of 16 are in s₁₀ resp. s₁₁ but ≡ 0,32 mod 64 are missing, etc.

Since s₂ contains arbitray high powers of 3, S has rows of arbitrary length, and for the missing multiples of powers of 2 the exponents can be driven above all limits.

Thus S contains all positive integers.

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 s₀ = 13 rather quickly, but the first 27 completely filled rows involve start numbers s₀ up to 1539, and for the first 4831 rows, start values up to 4076811 are involved.