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 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.

It is convenient to use abbreviations for the elementary operations which transform a node (element, number) x into y:

Name	Mnemonic	Distance to 4	Neighbour location	Condition
x d y	down	-1	y = x / 2	x ≡ 0 mod 2
x u y	up	-1	y = 3 * x + 1	(none)
x δ y	divide	+1	y = (x - 1) / 3	x ≡ 1 mod 3
x µ y	multiply	+1	y = x * 2	(none)

When moving towards the root (4) of the graph, d/u operations are used, while δ/µ operations are used to move away from the root.

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 x ≡ 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. We will later come back to more properties of detailed segments.

Compressed Segments

For the moment we will concentrate on the nodes x ≡ 4 mod 6 in the detailed segments (which are highlighted in the directory). For each segment we define a row i in an array C[i,j] as follows:

C[i,1]	C[i,2]	C[i,3]	C[i,4]	C[i,5]	C[i,6]	C[i,7]	C[i,8]	...
6*i-2	C[i,1] µµ	C[i,1] δµµ	C[i,2] δµ	C[i,3] δµ	C[i,4] δµ	C[i,5] δµ	C[i,6] δµ	...

The row for the compressed segment is filled as long as the corresponding node in the detailed segment is ≡ 4 mod 6.

The segment directories are obviously very structured. The lengths of the compressed segments follow the pattern

4 2 2 4 2 2 m 2 2 4 2 2 4 2 2 n 2 2 4 2 2 ...

with two fixed lengths 2 and 4 and some variable lengths m, n ... > 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).

If we interprete the C[i,j] as 6*p - 2, we observe that factors 3 in p are successively replaced by factors 2 for increasing j >= 2.

Coverage of Compressed Segments

We now claim that

• (1) All numbers of the form 6*p - 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 at the next column of segements with sufficient length.

It seems plausible that there is a proof by induction 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.

Connectivity of Compressed Segments

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.