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.
Detailled Segment
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 detailled, and a segment directory can easily be created by a Perl program. We will later come back to more properties of detailled segments.
Compressed Segements
For the moment we will concentrate on the nodes x ≡ 4 mod 6 in the detailled 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] | ... |
|---|---|---|---|---|---|---|---|
| 6*i-2 | C[i,1] µµ | C[i,1] δµµ | C[i,2] δµ | C[i,2] δµ | C[i,3] δµ | C[i,3] δµ | ... |
The row for the compressed segment is filled as long as the corresponding node in the detailled segment is ≡ 4 mod 6.
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
- si+1 = si / 3 and
- si+2 = si+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.