OEIS/3x+1 Problem
Abstract
The graph defined by Collatz for his 3x + 1 problem is overlaid with a set of small, finite trees with two branches. It is shown that this set of trees not only contains all numbers, but that it defines a permutation of the numbers. Furthermore, rules are defined how some subsets of these trees can be attached to other subsets. It is shown that no cycles can arise, and that finally the process leads to only one set of interconnected subtrees. Therefore, the Collatz graph is a tree which contains all numbers.
Introduction
Collatz sequences (also called trajectories) are sequences of integer numbers > 0. For some start value > 0 the elements of a particular sequence are constructed with two simple rules:
- Even numbers are halved.
- 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 do not show any 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 A070165: 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 n > 4 in the graph, the path from the root (4) can be continued
- always to n * 2, and
- to (n - 1) / 3 if n ≡ 1 mod 3.
The Collatz conjecture claims that the Collatz graph
- 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. Furthermore we observe that a path can only be continued with duplications once it reaches a node n ≡ 0 mod 3. We omit these trivial continuations.
Graph Operations
Following Trümper, we use abbreviations for the elementary operations which map a node (element, number) n in the Collatz graph to the neighbouring node as shown in the following table (T1):
| Name | Mnemonic | Distance to root | Mapping | Condition |
|---|---|---|---|---|
| d | "down" | -1 | n ↦ n / 2 | n ≡ 0 mod 2 |
| u | "up" | -1 | n ↦ 3 * n + 1 | (none) |
| s := ud | "spike" | -2 | n ↦ (3 * n + 1) / 2) | n ≡ 1 mod 2 |
| δ | "divide" | +1 | n ↦ (n - 1) / 3 | n ≡ 1 mod 3 |
| µ | "multiply" | +1 | n ↦ n * 2 | (none) |
| σ := δµ | "squeeze" | +2 | n ↦ ((n - 1) / 3) * 2 | n ≡ 1 mod 3 |
The operations will be noted as infix operators, with the source node as left operand and the target node as right operand, for example 10 δµ 6. In the following, we will mainly be interested in the reverse mappings (denoted by greek letters) which move away from the root 4 of the graph.
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):
143/104: 143 u 430 d 215 u 646 d 323 u 970 d 485 u 1456 d 728 d 364 ] d 182 ... 4 d 2 d 1 142/104: 142 d 71 u 214 d 107 u 322 d 161 u 484 d 242 d 121 u 364 ] d 182 ... 4 d 2 d 1
Beginning at some node of the form 6*i - 2 (364 in the example), these sequences join and follow the same path down to the root 4. The two differing parts of the sequences show a regular pattern, where nodes of the form 6*i - 2 alternate with other nodes, and where the operations u and d also alternate. This pattern of operations can be continued to the left with additional d and u operations:
126 d [ 63 u 190 d 95 u 286 d 143 u ... n? u [ 62 d 31 u 94 d 47 u 142 d ...
Finally the pattern stops because there is no integer n such that n * 3 + 1 = 62.
Example (E1) shows the sequences above, read from right to left, with the inverse operations:
[ 364 µ 728 µ 1456 δ 485 µ 970 δ 323 µ 646 δ 215 µ 430 δ 143 µ 286 δ 95 µ 190 δ 63 ] [ 364 δ 121 µ 242 µ 484 δ 161 µ 322 δ 107 µ 214 δ 71 µ 142 δ 47 µ 94 δ 31 µ 62 ]
Segments
These patterns lead us to the construction of special subtrees in the Collatz graph which we call segments. This construction is the main result of this article, and as we will see, it exhibits a nested structure reminding to a ruler. Though the structure is still complicated, it is - in contrast to the Collatz graph - very regular and "predictible" in all it's properties.
Starting at some node of the form 6*i - 2 (the left side, LS), the construction creates two subpaths (the upper branch and the lower branch) which to the right by a prescribed sequence of operations.
- The nodes of the form 6*i - 2 play a special role because both a δ and a µ operation can be applied to them. We could also have used 6*i + 4 instead, but the formulas seem easier with 6*i - 2, therefore we always use this form. Later we will consider nodes where 6*i - 2 is transformed to j, where j also may have the form j = 6*k - 2, etc.
Informally, a segment is constructed as follows:
- Step 1: Start with some node 6*i - 2 as the left side (LS).
- Step 2: Apply δ to the LS to get the first node in the lower branch.
- Step 3: Apply µ to the LS to get the first node in the upper branch.
- Steps 4, 5: Apply µ to both branches.
- In a loop, apply opposite operations to both branches, starting with µ (step 6) on the lower and δ (step 7) on the upper branch, continuing with δ (step 8) on the lower and µ (step 9) on the upper branch, and so on, as long as the next δ operation is possible.
The construction is such that increasing instances of σ operations are accumulated in each branch.
Segment directory
The resulting set of small trees is listed in a so-called segment directory (the link shows several thousand subtrees arranged by increasing left side, together with some additional information).
- The segment directory shows the segments in rows of a table. The steps are mapped to columns of the table. In order to maintain the subtree structure with two branches, each row contains a pair of columns numbered 3, 5, 7, ... for the upper branch and 2, 4, 6, ... for the lower branch.
- Nodes in the branches of a segment which have the form 6*i - 2 are shown in bold face. If i also has the form 6*k - 2, the node has a yellow background.
Details of a segment
The following table (T1) shows various properties of the steps or columns in the segment construction:
| Column j | Branch | Operation | Form of i | Formula | First elements | Covered nodes | Remaining nodes |
|---|---|---|---|---|---|---|---|
| 1 | (LS) | 6*i - 2 | 4, 10, 16, 22 | 1, 2, 4, 5 mod 6 | |||
| 2 | lower | δ | 2*i - 1 | 1, 3, 5, 7 | 1, 3, 5 mod 6 | 2, 4 mod 6 | |
| 3 | upper | µ | 12*i - 4 | 8, 20, 32, 44 | 8 mod 12 | 2, 4, 10 mod 12 | |
| 4 | lower | σ | 12*(i - 1)/1 + 2 | 2, 14, 16, 38 | 2 mod 12 | 4, 10 mod 12 | |
| 5 | upper | µ2 | 24*(i - 1)/1 + 16 | 16, 40, 64, 88 | 16 mod 24 | 4, 10, 22 mod 24 | |
| 6 | lower | σµ | i = 3*k + 1 | 24*(i - 1)/3 + 4 | 4, 28, 52, 76 | 4 mod 24 | 10, 22, 34, 46 mod 48 |
| 7 | upper | µ2δ | i = 3*k + 1 | 24*(i - 1)/3 + 5 | 5, 29, 53, 77 | (5 mod 24) | |
| 8 | lower | σµδ | i = 3*k + 1 | 24*(i - 1)/3 + 17 | 17, 41, 65, 89 | (17 mod 24) | |
| 9 | upper | µ2σ | i = 3*k + 1 | 48*(i - 1)/3 + 10 | 10, 58, 106, 154 | 10 mod 48 | 22, 34, 46 mod 48 |
| 10 | lower | σµσ | i = 9*k + 7 | 48*(i - 7)/9 + 34 | 34, 82, 130, 178 | 34 mod 48 | 22, 46, 70, 94 mod 96 |
| 11 | upper | µ2σδ | i = 9*k + 7 | 48*(i - 7)/9 + 35 | 35, 83, 131, 179 | (35 mod 48) | |
| 12 | lower | σµσδ | i = 9*k + 7 | 48*(i - 7)/9 + 11 | 11, 59, 107, 155 | (11 mod 38) | |
| 13 | upper | µ2σ2 | i = 9*k + 7 | 96*(i - 7)/9 + 70 | 70, 166, 262, 358 | 70 mod 96 | 22, 46, 94 mod 96 |
| 14 | lower | σµσ2 | i = 27*k + 7 | 96*(i - 7)/27 + 22 | 22, 118, 214, 310 | 22 mod 96 | 46, 94, 142, 190 mod 192 |
| 15 | upper | µ2σ2δ | i = 27*k + 7 | 96*(i - 7)/27 + 23 | 23, 119, 215, 311 | (23 mod 96) | |
| 16 | lower | σµσ2δ | i = 27*k + 7 | 96*(i - 7)/27 + 71 | 71, 167, 263, 359 | (71 mod 96) | |
| 17 | upper | µ2σ3 | i = 27*k + 7 | 192*(i - 7)/27 + 46 | 46, 238, 430, 622 | 46 mod 192 | 94, 142, 190 mod 192 |
| 18 | lower | σµσ3 | i = 81*k + 61 | 192*(i - 61)/81 + 142 | 142, 334, 526, 718 | 142 mod 192 | 94, 190, 286, 382 mod 384 |
| 19 | upper | µ2σ3δ | i = 81*k + 61 | 192*(i - 61)/81 + 143 | 143, 225, 527, 719 | (143 mod 192) | |
| j | i = qj*k + rj | sj*(i - rj)/qj + tj | tj, ... | tj mod sj |
The general formula (in the last row of T1) for a column j >= 5 uses the following parameters:
- qj = 3^(floor((j - 2) / 4),
- rj = A066443(floor((j - 2) / 8)), where A066443 is the OEIS sequence a(n) defined by a(0) = 1; a(n) = 9 * a(n-1) - 2 for n > 0; the terms are the indexes 1, 7, 61, 547, 4921 ... of the variable length segments with left sides 4, 40, 364, 3280, 29524 (OEIS A191681),
- sj = 3*2^floor((j + 7) / 4),
- tj = A309523(j) for j >= 3, where A309523 is an OEIS sequence computable by a linear recurrence with signature (1, -1, 1, 0, 0, 0, 0, 4, -4, 4, -4).
Finite, arbitrary length of the segments
First we have to show that the segment construction process always stops. For this purpose, we observe that the σ operation, when applied to left sides of the form 6*(3*k) - 2 has the interesting property that it maintains the general form 6*i - 2:
- 6*(3*k) - 2 σ (2*(3*k) - 1) * 2 = 12*k - 2 = 6*(2*k) - 2
That means that
- σ replaces one factor 3 by a factor 2
The mnemonic is that σ "squeezes" a 3 into a 2. In the opposite direction, the s ("spike") operation replaces one factor 2 by a factor 3.
The branch prolongation in the segment construction process must stop when the successive σ operations have exhausted all factors of 3 in the left side. Therefore, all segments have a finite length.
It is also clear that there is no limit on the length of a segment, since we only need to take a segment which has a factor of 3 with a sufficiently high power in its left side. The σ operations will then stretch out to the corresponding length.
Segments do not contain cycles
The modulo conditions of the columns (column "Covered nodes" in T1) ensure that all nodes in a segment are different. Therefore the two branches remain disjoint, and they cannot lead to a cycle inside of the segment. (The question whether cyles could occur when differnt segments are combined is still open).
Coverage of the right part
First we state again that we are not interested in even nodes of the form 3*2*k, because only µ operations can be applied on them. We disregards such nodes.
We see that column 2 contains all odd numbers (i ≡ 1, 3, 5 mod 6). Columns 7, 8, 11, 12, 15, 16, 19, 20 ... also contain subsets of odd numbers, but we disregard these columns, since their terms are duplicated in column 2.
We define the right part of a segment or of the whole segment directory to consist of the terms in columns 2-6, and all higher numbered columns which do not contain odd numbers (that are columns 9, 10, 13, 14, 17, 18 and so on).
We are now interested how the numbers > 0 are distributed over that right part. Corresponding remarks are contained in the last two columns of table T1. We want to show that the nodes in the right part are a permutation of the numbers > 0.
The left sides are in arithmetic progression, and a specific column is computed by a fixed combination of δ and µ operations. Therefore, a column is also in a simple, arithmetic progression. The modulo conditions are all different. The initial terms tj from OEIS A309523 are also disjoint.
- The terms of OEIS A308709 are all disjoint, since that sequence is the set {2^k} union {2*2^k} for k > 0, and OEIS A309523 is the superset {6*A308709 - 2} union {6*A308709 - 1}.
Together that shows that the terms in the right part are all disjoint.
So far we did not yet cover numbers of the form i ≡ 2, 4 mod 6 (or i ≡ 2, 4, 8, 10 mod 12; 0 mod 6 was excluded above). We proceed by looking into each column in the right part and by observing the modulo conditions of the terms contained in that column.
Column 3 contains numbers of the form i ≡ 8 mod 12, such that 2, 4, 10 mod 12 remain to be covered. Column 4, if it exists, contains numbers of the form i ≡ 2 mod 12. This leaves us with i ≡ 4, 10 mod 12.
As indicated in table T1 above, the modulo conditions in the right part become more and more refined. Since the segment directory contains arbitrarily long segments (respectively arbitrarily many columns), the modulo conditions of any height are finally reached.
Furthermore, since the modulo conditions lead to disjoint subsets of the numbers, it is ensured that any number can occur only once in the right part. In total, the nodes in the right part are a permutation of the numbers which are not divisible by 6.
Following a Collatz trajectory
Since we are sure that all numbers occur once in the right part of our segment directory, we can pick any of them and start a Collatz trajectory, hopefully proceeding down to the root (4). We try to trace the trajectory in the segment directory.
Covering the Collatz graph by the right part
The fact that all numbers occur at exactly one place in the segment directory is a good indication
- that the whole Collatz graph can be covered by the right part of the segment directory,
- that there are no cycles in the graph, and
- that it really has the form of a tree.
Any number n, be it a start value or a node in a trajectory, can easily and uniquely be located in some column in the right part just be determining the rest modulo 6, 12, 24, 48 ... 6*2^k and looking it up in T1 resp. OEIS A309523. Once the column is known, the segment directory row index i (and thereby the left side 6*i - 2) can also easily be determined via table T1.