OEIS/Collatz Story

Abstract

Small, finite trees with two branches are constructed with the operations defined by Collatz for his 3x+1 problem, These trees are connected to form bigger graphs in an iterative process. It is shown that this process finally builds a single graph which is a tree except for one cycle at the root. This graph is then embedded into the Collatz graph, and it is thereby shown that the latter is also a tree except for the cycle 4-2-1.

Introduction

Collatz sequences (also called trajectories) are sequences of integer numbers > 0. For any start value > 0 the elements of the sequence are constructed with two simple rules:

1. Even numbers are halved.
2. Odd numbers 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 all start values. 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.

Simple visualizations of Collatz sequences show no obvious structure. The sequences for the first dozen of start values are rather short, but the sequence for 27 suddenly has 112 elements.

Da sieht man den Wald vor lauter Bämen nicht.
German proverb: You cannot see the wood for the trees.

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
• Manfred Trümper: The Collatz Problem in the Light of an Infinite Free Semigroup. Chinese Journal of Mathematics, Vol. 2014, Article ID 756917, 21 p.

Collatz Graph

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

• always to m * 2, and
• to (m - 1) / 3 if m ≡ 1 mod 3.

The Collatz 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 cycles). We will not consider the leading cycle, and we start the graph with node 4, the root. Moreover, another trivial type of path starts when m ≡ 0 mod 3. We call such a path a sprout, and it contains duplications only. Sprout must be added to the graph for any node divisible by 3, therefore we will not consider them for the moment.

Graph Operations

Following Trümper, we use abbreviations for the elementary operations which transform a node (element, number) in the Collatz graph according to the following table (T1):

Name	Mnemonic	Distance to root	Mapping	Condition
d	down	-1	m ↦ m / 2	m ≡ 0 mod 2
u	up	-1	m ↦ 3 * m + 1	(m ≡ 1 mod 2)
s := ud	spike	-2	m ↦ (3 * m + 1) / 2)	m ≡ 1 mod 2
δ	divide	+1	m ↦ (m - 1) / 3	m ≡ 1 mod 3
µ	multiply	+1	m ↦ m * 2	(none)
σ := δµ	squeeze	+2	m ↦ ((m - 1) / 3) * 2	m ≡ 1 mod 3

We will mainly be interested in the reverse mappings (denoted with greek letters) which move away from the root of the graph.

3-by-2 Replacement

The σ operation, applied to numbers of the form 6 * m - 2, has an interesting property:

(6 * (3 * n) - 2) σ = 4 * 3 * n - 2 =  6 * (2 * n) - 2

In other words, as long as m contains a factor 3, the σ operation maintains the form 6 * x - 2, and it replaces the factor 3 by 2 (it "squeezes" a 3 into a 2). In the opposite direction, the s operation replaces a factor 2 in m by 3.

Motivation: Patterns in sequences with the 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 d  71 u 214 d 107 u 322 d 161 u 484 d  242 d 121 u 364 ] 182, 91, ... 4, 2, 1
143/104: 143 u 430 d 215 u 646 d 323 u 970 d 485 u 1456 d 728 d 364 ] 182, 91, ... 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 could be computed from the first. Walking from right to left, the step pattern is:

δ µ µ δ µ δ µ δ µ
µ µ δ µ δ µ δ µ δ

The alternating pattern of operations can be continued to the left with 4 additional pairs of steps:

 q? u [ 62 d  31 u  94 d  47 u 142 d ...
126 d [ 63 u 190 d  95 u 286 d 143 u ...
        +1  *6+4    +1  *6+4    +1

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

Segments

These patterns lead us to the construction of special subsets of paths in the Collatz graph which we call segments. They lead away from the root, and they always start with a node m ≡ -2 mod 6. Then they split and follow two subpaths in a prescribed sequence of operations. The segment construction process is stopped when the next node in one of the two subpaths becomes divisible by 3, resp. when a δ operation is no more possible.

Segment Directory Construction

We list the segments as rows of an infinite array C[i,j], the so-called segment directory.

Informally, and in the two examples above, we consider the terms betweeen the square brackets. For the moment, we only take those which are which are ≡ 4 mod 6 (for "compressed" segments, below there are also "detailed" segments where we take all). We start at the right and with the lower line, and we interleave the terms ≡ 4 mod 6 of the two lines to get a segment.

Continuing the example above:

[ 62 d  31 u  94 d  47 u 142 d  71 u 214 d 107 u 322 d 161 u 484 d  242 d 121 u 364 ]
[ 63 u 190 d  95 u 286 d 143 u 430 d 215 u 646 d 323 u 970 d 485 u 1456 d 728 d 364 ]

Left-to-right reversed, only terms of the form 6*m+4, rows switched and without operations:

364  1456     970     644     430     286     190
364       484     322     214     142      94

The final, linearized example segment in row 61 of the directory looks like so:

61	364	1456	484	970	322	646	214	430	142	286	94	190

The following table (T2) tells how the columns in one row i of the array C must be constructed if the condition is fulfilled:

Column	Operation	Formula	Condition	Sequence
1		6 * i - 2		4, 10, 16, 22, 28, ...
2	C[i,1] µµ	24 * (i - 1) / 1 + 16		16, 40, 64, 88, 112, ...
3	C[i,1] δµµ	24 * (i - 1) / 3 + 4	i ≡ 1 mod 3	4, 28, 52, 76, 100, ...
4	C[i,2] σ	48 * (i - 1) / 3 + 10	i ≡ 1 mod 3	10, 58, 106, 134, ...
5	C[i,3] σ	48 * (i - 7) / 9 + 34	i ≡ 7 mod 9	34, 82, 130, 178, ...
6	C[i,2] σσ	96 * (i - 7) / 9 + 70	i ≡ 7 mod 9	70, 166, 262, 358, ...
7	C[i,3] σσ	96 * (i - 7) / 27 + 22	i ≡ 7 mod 27	22, 118, 214, 310, ...
8	C[i,2] σσσ	192 * (i - 7) / 27 + 46	i ≡ 7 mod 27	46, 238, 430, 622, ...
9	C[i,3] σσσ	192 * (i - 61) / 81 + 142	i ≡ 61 mod 81	142, 334, ...
...	...	...	...	...
...	C[i,2] σj-1	3 * 2j+2 * (i - m) / 3k + x	i ≡ m mod 3k	x, ...

The first column(s) C[i,1] will be denoted as the left side of the segment (or of the whole directory), while the columns C[i,j], j > 1 are the right part.

In general, the rows of T2 must be filled as follows:

• j and k are increased in every second row.
• The residues m of 3^k in the condition column are are the indexes 1, 7, 61, 547, 4921 ... (OEIS A066443) of the variable length segments with left sides (4), 40, 364, 3280, 29524 (OEIS A191681). They increase (jump) after 4 rows.
• The additive constants x in the formula are the values appearing first in columns 2-4 (in segment 1), 5-8 (in segment 7), 9-12 (in segment 61) etc.

The first few lines of the segment directory are the following:

	1	2	3	4	5	6	7	8	9	10	11	...	2*j	2*j+1
i	6*i‑2	µµ	δµµ	µµσ	δµµσ	µµσσ	δµµσσ	µµσ3	δµµσ3	µµσ4	δµµσ4	...	µµσj-1	δµµσj-1
1	4	16	4	10
2	10	40
3	16	64
4	22	88	28	58
5	28	112
6	34	136
7	40	160	52	106	34	70	22	46

There is a more elaborated segment directory with 5000 rows.

Properties of the Segment Directory

We make a number of claims for the segment directory C:

• (C1) All nodes in the segment directory are of the form 6 * n - 2.

This follows from the formula for columns C[i,1..3], and for any higher column numbers from the 3-by-2 replacement property of the σ operation.

• (C2) All segments have a finite length.

At some point the σ operations will have replaced all factors 3 by 2.

• (C3) All nodes in the right part of a segment have the form 6 * (3ⁿ * 2^m * f) - 2 with the same "3-2-free" factor f.

This follows from the operations for columns C[i,1..3], and from the fact that the σ operation maintains this property.

• (C4) All nodes in the right part of a particular segment are
  • different among themselves, and
  • different from the left side of that segment (except for the first segment for the root 4).

For C[i,1..2] we see that the values modulo 24 are different. For the remaining columns, we see that the exponents of the factors 2 and 3 are different. They are shifted by the σ operations, but they alternate, for example (in the segment with left part 40):

160 = 6 * (33 * 20 * 1) - 2
 52 = 6 * (32 * 20 * 1) - 2
106 = 6 * (32 * 21 * 1) - 2
 34 = 6 * (31 * 21 * 1) - 2
 70 = 6 * (31 * 22 * 1) - 2
 22 = 6 * (30 * 22 * 1) - 2
 46 = 6 * (30 * 23 * 1) - 2

• (C5) There is no cycle in a segment (except for the first segment for the root 4).

Segment Lengths

Oviously the segment directory is very structured. The lengths of the compressed segments follow the pattern

4 2 2 4 2 2 L1 2 2 4 2 2 4 2 2 L2 2 2 4 2 2 ...

with two fixed lengths 2 and 4 and some variable lengths L₁, L₂ ... > 4. For the left parts 4, 40, 364, 3280, 29524 (OEIS A191681), the segment lengths have high values 4, 8, 12, 16, 20 which did not occur before. Those left parts are (9ⁿ⁺¹ - 1) / 2, or 4 * Sum(9ⁱ, i = 0..n).

Coverage of the Right Part

We now examine the modular conditions which result from the segment construction table (T2) in order to find out how the numbers of the form 6 * n - 2 are covered by the right part of the segment directory. The following table (T3) shows the result:

Columns j	Covered	Remaining
2-3	4, 16 mod 24	10, 22, 34, 46 mod 48
3-4	10, 34 mod 48	22, 46, 70, 94 mod 96
5-6	70, 22 mod 96	46, 94, 142, 190 mod 192
7-8	46, 142 mod 192	94, 190, 286, 382 mod 384
...	...	...

We can always exclude the first and the third element remaining so far by looking in the next two columns of segments with sufficient length.

• (C6) There is no limit on the length of a segment.

We only need to take a segment which, in its right part, has a factor of 3 with a sufficiently high power, and the σ operations will stretch out the segment accordingly.

Therefore we can continue the modulus table above indefinitely, which leads us to the claim:

• (C7) All numbers of the form 6 * n - 2 occur exactly once in the right part of the segment directory, and once as a left side. There is a bijective mapping between the left sides and the elements of the right parts.

The sequences defined by the columns in the right part all have different modulus conditions. Therefore they are all disjoint. The left sides are disjoint by construction.

Segment Tree

So far we possess the segment directory C which represents the root segment and an infinite set of small trees with disjoint nodes and two branches. We know that the segments represent trees, and that their right parts are all disjoint and different from the left side.

We now want to attach (or connect) the segments to other graphs until we get a single big graph which will later become the backbone of the Collatz graph. Ideally the attachment process should maintain the tree property of the graphs all the time.

The verb attach emphasizes the direction of the operation better than the verb connect.

Attachment Directory Construction

Parallel to the segment directory we maintain the attachment directory A which, for any source segment in C:

1. tells whether the tree corresponding to the segment was already attached to the graph represented by some other segment, and if so,
2. tells the target row and column numbers i, j in the segment directory C where the source segment was attached.

Initially all segments are unattached.

We operate on A as follows: Considering simultaneously a set of source rows i > 1 (i.e. omitting the root segment) in C - which fulfill some modularity condition (the source row set), and which are so far unattached, we attach their segments in parallel to the unique occurrences of their left sides in the right part of C (target row set and target column).

These operations on A involve infinite sets. They are similiar to the gedankenexperiment of Hilbert's hotel.

Attachment rules

The following table (T4) tells the computation rules for the target position, depending on the modularity condition of the source row index i. We identify and denote these attachment rules by the target column number (which is also the branch level, below).

Target Column	Source rows i	First source rows	Target rows	First target rows		Remaining rows
1	i = 1		1	1	=	i 2, 3, 4 ...	2/2
2	i ≡ 3 mod 4	3, 7, 11, 15 ...	30 * (i - 3) / 4 + 1	1, 2, 3, 4 ...	<	i ≡ 0, 1, 2 mod 4	3/4
3	i ≡ 1 mod 4 and i > 1	5, 9, 13 ...	31 * (i - 1) / 4 + 1	4, 7, 10 ...	<	i ≡ 0, 2, 4, 6 mod 8	2/4
4	i ≡ 2 mod 8	2, 10, 18, 26 ...	31 * (i - 2) / 8 + 1	1, 4, 7, 10 ...	<	i ≡ 0, 4, 6 mod 8	3/8
5	i ≡ 6 mod 8	6, 14, 22, 30 ...	32 * (i - 6) / 8 + 7	7, 16, 25, 34 ...	>	i ≡ 0, 4, 8, 12 mod 16	2/8
6	i ≡ 12 mod 16	12, 28, 44, 60 ...	32 * (i - 12) / 16 + 7	7, 16, 25, 34 ...	<	i ≡ 0, 4, 8 mod 16	3/16
7	i ≡ 4 mod 16	4, 20, 36, 52 ...	33 * (i - 4) / 16 + 7	7, 34, 61, 88 ...	>	i ≡ 0, 8, 16, 24 mod 32	2/16
8	i ≡ 8 mod 32	8, 40, 72, 104 ...	33 * (i - 8) / 32 + 7	7, 34, 61, 88 ...	<	i ≡ 0, 16, 24 mod 32	3/32
9	i ≡ 24 mod 32	24, 56, 88, 120 ...	34 * (i - 24) / 32 + 61	61, 142, 223, 304 ...	>	i ≡ 0, 16, 32, 48 mod 64	2/32
10	i ≡ 48 mod 64	48, 112, 176, 240 ...	34 * (i - 48) / 64 + 61	61, 142, 223, 304 ...	>	i ≡ 0, 16, 32 mod 64	3/64
11	i ≡ 16 mod 64	16, 80, 144, 208 ...	35 * (i - 16) / 64 + 61	61, 304, 547, 790 ...	>	i ≡ 0, 32, 64, 96 mod 128	2/64
...	...	...	...	...	...	...	...
...	i ≡ {3, 1} * 2j-2 mod 2j	{3, 1} * 2j-2 ...	3k * (i - {1, 3} * 2j-2) / 2j + m	m ...	>	... mod 2k+2	{3, 2}/2j

The first rule applies to the root segment only and is not used. It should be obvious how the rows of this infinite table must be filled in general:

• j and m are increased in every second row.
• The residues of 2^k in the source row column are 3 * 2^k-2, 1 * 2^k-2 with the pattern 3 1 1 3 3 1 1 3 3 ... for the factor. (This are also the values to be subtracted from i in the target row colum.)
• The additive constants n in the target row column are the indexes 1, 7, 61, 547, 4921 ... (OEIS A066443: a(n) = a(n-1) * 9 - 2) of the variable length segments with left sides (4), 40, 364, 3280, 29524 (OEIS A191681) mentioned above. They increase after 4 rows.

As an example, we apply rule 7 to source row 4. (This example does not show the result of of the whole process, but a single step only.)

	1	2	3	4	5	6	7	8	9	10	11	...
1	4	16	4	10
2	10	40
3	16	64
4
5	28	112
6	34	136
7	40	160	52	106	34	70	22	46
							22	88	28	58

Properties of the Attachment Rules

For the attachment directory A we note respectively claim:

• (A1) The source rows met by the conditions in the rules are all disjoint.
• (A2) Therefore, a source row is chosen by the process exactly once.
• (A3) The construction is such that the target column always exists in the target rows.

Table (T4) is derived from (T2) which has similiar modularity conditions.

• (A4) The target column (or rule number) depends on the modularity condition for i alone, but not on the value of i.

This can be shown by the graph operations (δ / µ / σ) which are tied to the columns.

Moving up or down

There are three major groups of attachment rules (column New pos. in (T4)):

• Rules 2, 3 and 4 attach to a row with a lower index i.
• Rules 5 and 7 attach to higher, 6 and 8 to or lower indexes.
• Rules 9 and above all attach to higher indexes.

This can be seen from the powers of 2 and 3 in the source and target row columns. Starting at rule 9, we have 3^k > 2^k+2 for k >= 4.

With the single exception of the root segment 1, the rules obviously never attach a row to itself. The target row may have been hit by the same rule, and may already have been attached elsewhere. This is no problem, since the attachment process maintains the attachment state in A for the source row only.

• (A5) It does not matter whether the single attachment steps are performed with increasing source row number, or thought to happen with decreasing row number.

Order of Rule Application

• (A6) The resulting graphs do not depend on the order of application of the attachment rules.

The rules may well hit the same target rows, but they always do so in different target columns. It does not matter whether the target row is already attached.

Despite of (A6) we will apply the rules in a well-defined order, because only with this order we can show that the tree property of the subgraphs is always maintained.

Attachment Process

We will now use the rules of (T4) to reduce the set of unattached segments in C in an iterative process. Our goal is to show that all segments are attached - mostly indirectly - to the root segment. (Rule 1 would state that the root segment should be attached to itself.)

Branch Levels

In general, when dealing with the 3x+1 problem, it seems difficult to introduce a measure, that is a numerically ordered property of some object related to the Collatz graph. This would be desireable in order to conduct a proof by induction, infinite descent, leading a minimal element to a contradiction etc.

Here we use the branch level, that is the column index j of the unique position C[i, j] in a segment where a second segment should be attached.

Rule 1

This degenerate rule is inserted for completeness only. It puts the root segment in the "attached" state.

Rule 2

For this rule, all source rows indexes are contained in the target rows indexes.

We look more closely at the first of these chains of coincidences: row 3 is attached to the root, row 11 to row 3, 43 to 11 and so on. In the end, the trees corresponding to all rows of the form (4^k + 2) / 6, k >= 0 (OEIS A007583, with left sides 4, 16, 64, 256 ...) are stacked on the root segment (row 1). All involved segments are different, and because of the uniqueness of the attachment positions, we have built one tree above the root segment.

3  7 11 15 19 23 27 31 35 39 43 47 51 55 59 63 67 71 75 ... source rows
1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 ... target rows
^-----v                  
      ^-----------------------v  1  3 11 43  ...              
   ^--------------v                        
                  ^------- 2  7 27 107 ... (10 * 4k + 2) / 6 = OEIS A136412
                                           
         ^---------------- 4 15 59  ...    (22 * 4k + 2) / 6 = OEIS A199210     
            ^------------- 5 19 75  ...    (28 * 4k + 2) / 6 = OEIS A206373     
               ^---------- 6 23 91  ...    (34 * 4k + 2) / 6  
                                           
                     ^---- 8 31 123 ...    (46 * 4k + 2) / 6 

Likewise, all trees for rows of the form (10 * 4^k + 2) / 6 are stacked on segment 2. The general formula for the rows stacked on row 4 * i + 3 is ((6 * i - 2) * 4^k + 2) / 6, k >= 0.

As a preliminary result, we have all source rows 3, 7, 11, 15 ... attached somewhere, and we have built bigger trees above all remaining segments i ≡ 0, 1, 2 mod 4 (2, 4, 5, 6, 8, 9, 10, 12 ..., OEIS A004773).

Rule 3

5  9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77 80 ... source rows
4  7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 ... target rows
  ~~          ~~          ~~          ~~          ~~           already attached by rule 2

Source rows of the form 16 * k + 1 coincide with target rows for this rule, and for k = 4, 16, 64, ... 4^m, the length of the chains increases:

17->13=>10
33->25=>19
49->37=>28
65->49=>37=>28
81->61=>46
...

The target of every fourth source row was already attached by rule 2. After the application of rules 2 and 3, we have attached all odd source rows.

Rule 4

This is similiar to rule 3. Chains occurs for source rows 64 * k + 26, and the lengths of the chains increases for k = 6, 38, ... 32^m + 6.

Rule 5

This is the first rule which moves upwards.

No Cycles

• (A7) The attachment process does not create any new cycle (in addition to the one in the root segment).

Let a segment/tree t₁ with left side n₁ and right part R₁ be attached to node n₁ in the right part R₂ of the unique segment/tree t₂ which has the left side by n₂. t₁ and t₂ are disjoint trees by (C4), therefore the result of such a single attachment step is a tree again (u₂, still with left side n₂).

Proof for the Collatz Tree

• (P1) The remaining single tree is a subgraph of the Collatz graph.

The edges of the compressed tree carry combined operations µµ, δµµ and σ = δµ.

So far, numbers of the form x ≡ 0, 1, 2, 3, 5 mod 6 are missing from the compressed tree.

We insert intermediate nodes into the compressed tree by applying operations on the left parts of the segments as shown in the following table (T5):

Operation	Condition	Resulting Nodes	Remaining Nodes
δ		2 * i - 1	i ≡ 0, 2, 6, 8 mod 12
µ		12 * i - 4	i ≡ 0, 2, 6 mod 12
δµ	i ≡ 1, 2 mod 3	4 * i - 2	i ≡ 0, 12 mod 24
δµµ	i ≡ 2 mod 3	8 * i - 4	i ≡ 0 mod 24
δµµµ	i ≡ 2 mod 3	16 * i - 8	(none)

The first three rows in T5 care for the intermediate nodes at the beginning of the segment construction with columns 1, 2, 3. Rows 4 and 5 generate the sprouts (starting at multiples of 3) which are not contained in the segment directory.

We call such a construction a detailed segment (in contrast to the compressed segments described above).

A detailed segment directory can be created by the same Perl program. In that directory, the two subpaths of a segment are shown in two lines. Only the highlighted nodes are unique.

• (P2) The connectivity of the compressed tree remains unaffected by the insertions.
• (P3) With the insertions of (T5), the compressed tree covers the whole Collatz graph.
• (P4) The Collatz graph is a tree (except for the cycle 4-2-1.