OEIS/3x+1 Problem
Introduction
Collatz sequences are sequences of non-negative integer numbers with a simple construction rule:
- Even elements are 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.
Straightforward visualizations of the Collatz sequences no obvious structure. The sequences for the first dozen of start values are rather short, but the sequence for 27 suddenly has 112 elements.
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 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.
When m ≡ 0 mod 3, the path will continue with duplications only, since these maintain the divisibility by 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 cylces.
Following Trümper, we use abbreviations for the elementary operations which transform a node (element, number) in the Collatz graph:
| Name | Mnemonic | Distance to root | Mapping | Condition |
|---|---|---|---|---|
| d | down | -1 | m ↦ m / 2 | m ≡ 0 mod 2 |
| u | up | -1 | m ↦ 3 * m + 1 | (none) |
| s := ud | spike | -2 | m ↦ (3 * m + 1) / 2 | (none) |
| δ | divide | +1 | m ↦ (m - 1) / 3 | m ≡ 1 mod 3 |
| µ | multiply | +1 | m ↦ m * 2 | (none) |
| σ := δµ | step | +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.
Segment Construction
We will now construct special portions of paths in the Collatz graph which we call segments. They lead away from the root, and they always start with a node m ≡ 4 mod 6. Then they split into a northern and a southern subsegment by applying the following operations:
- northern: µ µ δ µ δ µ δ ...
- southern: δ µ µ δ µ δ µ ...
The two subsegements are built in parallel, and the process is stopped when one of the two new nodes becomes divisible by 3, resp. when a δ operation is not possible.
We will call these segments detailed, and a segment directory can easily be created by a Perl program.
- Claim 1: All segments have a finite length.
- If we look at the segment nodes which have the form 6*n - 2, we observe that the σ operations successively replace factors 3 in n by factors 2. This process stops when all factors 3 are exhausted.
We will later come back to more properties of detailed segments.
Compressed Segments
For the moment we will concentrate on the nodes m ≡ 4 mod 6 in the detailed segments (which are highlighted in the HTML directory). For each segment we define a row i in an array C[i,j] which is filled with all nodes from the corresponding detailed segment which are ≡ 4 mod 6, according to this table:
| Column | Operation | Formula | Condition | Sequence |
|---|---|---|---|---|
| C[i,1] | = 6*i-2 | |||
| C[i,2] | C[i,1] µµ | 24 * (i - 1) + 16 | 16, 40, 64, 88, 112, ... | |
| C[i,3] | C[i,1] δµµ | 24 * (i - 1) / 3 + 4 | i ≡ 1 mod 3 | 4, 28, 52, 76, 100, ... |
| C[i,4] | C[i,2] σ | 48 * (i - 1) / 3 + 10 | i ≡ 1 mod 3 | 10, 58, 106, 134, ... |
| C[i,5] | C[i,3] σ | 48 * (i - 7) / 9 + 34 | i ≡ 7 mod 9 | 34, 82, 130, 178, ... |
| C[i,6] | C[i,2] σσ | 96 * (i - 7) / 9 + 70 | i ≡ 7 mod 9 | 70, 166, 262, 358, ... |
| C[i,7] | C[i,3] σσ | 96 * (i - 7) / 27 + 22 | i ≡ 7 mod 27 | 22, 118, 214, 310, ... |
| C[i,8] | C[i,2] σσσ | 192 * (i - 7) / 27 + 46 | i ≡ 7 mod 27 | 46, 238, 430, 622, ... |
| C[i,9] | C[i,3] σσσ | 192 * (i - 61) / 81 + 142 | i ≡ 61 mod 81 | 142, 334, ... |
| ... | ... | ... | ... | ... |
The segment directories are obviously 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. At the starting values 4, 40, 364, 3280, 29524 (OEIS A191681), the segment lengths have high values 4, 8, 12, 16, 20 which did not occur before. Those starting values are (9n+1 - 1) / 2, or 4 * Sum(9i, i=0..n).
- Claim 2: All nodes in a compressed segment (except for the segment starting with 4) are different.
- This should be provable from the formula for the fixed length portion, and for the variable length portions by the 3-by-2 replacement mentioned above.
- Therefore, the compressed segments themselves do not contain a cycle.
Coverage of Compressed Segments
- Claim 3: All numbers of the form 6*n - 2 occur exactly once in the right side C[i,j], j > 1. There is a one-to-one mapping between the left side C[i,1] and the right side.
In the first few columns of C we find the following sequences:
C[i,2] = 24*(i - 1) + 16 : 16, 40, 64, 88, 112, ... C[i,3] = 24*(i - 1)/3 + 4 if i ≡ 1 mod 3 : 4, 28, 52, 76, 100, ... C[i,4] = 48*(i - 1)/3 + 10 if i ≡ 1 mod 3 : 10, 58, 106, 134, ... C[i,5] = 48*(i - 7)/9 + 34 if i ≡ 7 mod 9 : 34, 82, 130, 178, ...
We thereby cover the highlighted numbers of the form 6*p - 2:
x ≡ 4, 10, 16, 22, 28, 34, 40, 46 mod 48
We missed the numbers:
x ≡ 22, 46, 70, 94 mod 96
For these, we continue our observations:
C[i,6] = 96*(i - 7)/9 + 70 if i ≡ 7 mod 9 : 70, 166, 262, 358, ... C[i,7] = 96*(i - 7)/27 + 22 if i ≡ 7 mod 27 : 22, 118, 214, 310, ...
This leaves us with
x ≡ 46, 94, 142, 190 mod 192
Then with
C[i,8] = 192*(i - 7)/27 + 46 if i ≡ 7 mod 27 : 46, 238, 430, 622, ... C[i,9] = 192*(i - 61)/81 + 142 if i ≡ 61 mod 81 : 142, 334, ...
we only miss
x ≡ 94, 190, 286, 382 mod 384
"and so on": We can always exclude the first and the third element missed so far by looking in the next column of segments with sufficient length.
It seems plausible that there is a proof that all numbers of the form 6*p - 2 are finally contained in the right side of C, and the modular conditions make it rather obvious that these numbers are all different.
Compressed Segment Tree Construction
We now claim that we can construct a tree from the compressed segments by:
- Starting at the segment for 4,
- at every node > 4 we attach the unique segment with that starting element.
- We repeat this process for all nodes collected in the tree so far.
We cannot enter a cycle, since that would imply a node which is reached by two edges, but such a node would occur more than once in the right side of the segment directory. We saw above that the segments themselves cannot have cycles.
Collatz Tree
Obviously we can replace all compressed segments by the corresponding detailed segments. The claim is that this will lead to the Collatz tree which reaches all numbers.
In the directory of detailed segments we see that we can quickly get reach all odd numbers:
C[i,1] δ = ((6*i-2) - 1) / 3 = 2*i - 1 : 1, 3, 5, 7
We remain concerned with the numbers
x ≡ 0,2 mod 6 resp. x ≡ 0, 2, 6, 8, 12, 14, 18, 20 mod 24
The following table lists the operations which are necessary to reach most of these by starting at C[i,1]:
| Operation | Expression | Result | Condition | Coverage |
|---|---|---|---|---|
| µ | (6*i-2)*2 | 12*i - 4 | 8, 20 mod 24 | |
| δµ | ((6*i-2-1)/3)*2 | 4*i - 2 | 2, 6, 10, 14, 18, 22 mod 24 | |
| δµµ | (6*i-2-1)/3*2*2 | 8*i - 4 | i ≡ 2 mod 3 | 12 mod 24 |
The multiples of 12 can be reached from the multiples of 3 (contained in the odd numbers) by a µµ operation.