u/MarcusOrlyius

Here is the correct image.

When taking sequences congruent to 666 using a 3 symbol dyadic dome representation, we produce the image shown here.

Is this the solution to Collapse conjecture?

Following on from my post here, I though it might be of interest to point out a consequence from that post.

Let ∘ be a string append operator such that "10" ∘ "456" = "10456". In base 10 you get:

3 = "" ∘ "3",
13 = "1" ∘ "3",
53 = "5" ∘ "3",
213 = "21" ∘ "3",
...

In base 2 you get:

3 = "11" ∘ "",
13 = "11" ∘ "01",
53 = "11" ∘ "0101",
213 = "11" ∘ "010101",
...

In base 4 you get:

3 = "3" ∘ "",
13 = "3" ∘ "1",
53 = "3" ∘ "11",
213 = "3" ∘ "111",
...

Base 4 is the "natural base" for the 3x+1 system.

Many people in this sub don't seem to grasp the basic structure of the Collatz tree. So, here's a refresher on the basics.

A branch in the Collatz tree is of the form B(x) = {x * 2^n | n in N} where x is an odd natural number. This is due to the division by 2 step.

If x ≡ 1 (mod 6) then x * 2^(2n+1) ≡ 2 (mod 6) and x * 2^(2n+2) ≡ 4 (mod 6).
If x ≡ 5 (mod 6) then x * 2^(2n+1) ≡ 4 (mod 6) and x * 2^(2n+2) ≡ 2 (mod 6).
If x ≡ 3 (mod 6) then x * 2^(n+1) ≡ 0 (mod 6).

If x * 2^n ≡ 4 (mod 6) then (x * 2^n - 1) / 3 ≡ 1 (mod 2).

So, if x ≡ 1 or 5 (mod 6) then B(x) has infinitely many child branches, if x ≡ 3 (mod 6) then B(x) has no child branches and is a leaf in the tree.

What these branches look like are shown below:

1,
2,
4, 1, 2, 4, 8, 16, ...
8,
16, 5, 10, 20, 40, 80, ...
32,
64, 21, 42, 84, 168, 336, ...
128,
256, 85, 170, 340, 680, 1360, ...
512,
...

5,
10, 3, 6, 12, 24, 48, ...
20,
40, 13, 26, 52, 104, 208, ...
80,
160, 53, 106, 212, 424, 848, ...
320,
640, 213, 426, 852, 1704, 3408, ...
1280,
...

3,
6,
12,
24,
48,
...

For the exact same values modulo 6 we get the following:

1,
2,
4, 1, 2, 4, 2, 4, ...
2,
4, 5, 4, 2, 4, 2, ...
2,
4, 3, 0, 0, 0, 0, ...
2,
4, 1, 2, 4, 2, 4, ...
2,
...

5,
4, 3, 0, 0, 0, 0, ...
2,
4, 1, 2, 4, 2, 4, ...
2,
4, 5, 4, 2, 4, 2, ...
2,
4, 3, 0, 0, 0, 0, ...
2,
...

3,
0,
0,
0,
0,
...

In the above, you can see that if x ≡ 1 (mod 6) then the order of the child branches of B(x) is 1,5,3,... which repeats indefinitely for all child branches of B(x). So, determining the residue class of the first child, determines the residue classes for all children.

We can also see that we have 3 main types of branch, 2 of which have children. For each of those branches that have children, there are 3 possible variations for the residue class of the first child branch. That gives a total of 7 different types of branch with unique ordering.

Also, look at the child branches of B(1). Let x_n = (x * 2^(2n+2) - 1) / 3 such that B(x_n) is the nth child branch of B(x). Then x_(n+1) = 4 * x_n + 1.

This is the basic structure of the Collatz tree and every "researcher" should by familiar with it.

A branch in the Collatz tree is of the form B(x) = {x * 2^n | n in N} where x is an odd natural number.

If x ≡ 1 (mod 6) then x * 2^(2n+1) ≡ 2 (mod 6) and x * 2^(2n+2) ≡ 4 (mod 6).
If x ≡ 5 (mod 6) then x * 2^(2n+1) ≡ 4 (mod 6) and x * 2^(2n+2) ≡ 2 (mod 6).
If x ≡ 3 (mod 6) then x * 2^(n+1) ≡ 0 (mod 6).

If x * 2^n ≡ 4 (mod 6) then (x * 2^n - 1) / 3 ≡ 1 (mod 2).

So, if x ≡ 1 or 5 (mod 6) it has infinitely many child branches, if x ≡ 3 (mod 6) it has no child branches.

Now, here is funny fact. 3/4 of all branches in the Collatz tree are the first child of some parent branch as defined above, despite the fact that every branch that has a first child has infinitely many children.

For example: B(3) is the first child branch of B(5) but B(13), B(53), B(213),... are also child branches of B(5). Yet, B(3) is an element of a set that contains 3/4 of all branches while B(13), B(53), B(213),... are elements of a set that contains only 1/4 of all branches.

From an previous post of mine we have the following. For all n ∈ N, m ∈ {0, 1, 2} and k ∈ {1, 5},

x = (3n + m) * 2^(((13-k)^)/4) + (k + 1) / 2

and B(x) is the first child branch of a parent branch, B(y), such that y ≡ k (mod 6).

This can also be expressed in matrix from.

We define a basis vector v_n in Z^3 that partitions the natural numbers into three residue classes modulo 3. For all n in N:

v_n = | 3n+0 |
      | 3n+1 |
      | 3n+2 |

This vector represents the pre-image space.

The backward Collatz map for odd numbers is determined by the residue of a parent y (mod 6). Specifically, for an odd parent y, the children x are generated by x = (2^p * y - 1) / 3, where p is the smallest integer such that x ≡ 1 (mod 2).

From the modular arithmetic of y ≡ (mod 6), we derive the scaling and translation constants. We define two vectors in R^2 to represent the two primary branching behaviours (k=1 and k=5):

the scaling vector, s, represents the dyadic shifts 2^p

s = | 2^3 | = | 8 |
    | 2^2 |   | 4 |

the translation vector, t, represents the additive constants required to satisfy the inverse mapping

t =  | 1 |
     | 3 |

To map the basis v_n into the state-space of the Collatz tree, we apply an affine transformation. We utilise the Kronecker product, ⊗, to distribute these transformations across all modular slots.

Let 1_3 = {1 1 1}^T be the all-ones vector. The root tensor, X(n) is defined as:

X(n) = s ⊗ v_n^T + t ⊗ 1_3^T

Expanding this expression:

X(n) = | 8 | (3n 3n+1 3n+2) + | 1 | (1 1 1)
       | 4 |                  | 3 |

Performing the matrix addition, we obtain the explicit state-space generator:

X(n) = | 8(3n) + 1, 8(3n+1) + 1, 8(3n+2) + 1 |
       | 4(3n) + 3, 4(3n+1) + 3, 4(3n+2) + 3 |

Simplifying the entries:

X(n) = | 24n + 1, 24n + 9, 24n + 17 |
       | 12n + 3, 12n + 7, 12n + 11 |

Any trajectory can be expressed as a composition of these affine maps. If T_i is the transformation corresponding to a specific row and column choice, a trajectory is a sequence x_(i+1) = T_i(x_i). The Collatz conjecture then becomes a question of whether the composition of these matrices always converges to the fixed point (1, 0, 0) in the coordinate space.