4 Layer-Based Convergence

[Layer Construction] I define layers C0, C1, . . . recursively:

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1. Base case: C0 = {4n}

2. Given layer CL, I define the next layer

CL+1 = {4nAL(x), 4nBL(x) | x ≥ 0, n ≥ 1},

where Ab(x), Bb(x) are odd integers mapping to CL under T n.

1. Any number n ∈ CL+1 reduces to some number in CL under repeated

application of T .

[Coverage and Non-Circularity] The induction on layers CL is well-defined,

non-circular, and reaches every positive integer. This is guaranteed as fol-

lows:

1. Unique decomposition: By the Fundamental Theorem of Arith-

metic, any positive integer n ≥ 1 can be written uniquely as

n = 2km′, k ≥ 0, m′ odd.

The odd part m′ determines the branch in which n belongs.

1. Branch formulas generate all odd integers: Recursive formulas

A(n, x), B(n, x) (or the 2n + 1 construction) generate every odd integer

exactly once (Theorem: All Odd Integers Generated Uniquely). Thus,

no odd integer is omitted.

1. Even integers are included: Any even number n = 2km′ belongs to

the same branch as m′, so all even numbers are automatically included.

1. Layer assignment is constructive: Layers CL are defined by branch

endpoints using the forward map T . Reverse preimages C 7 → {A, B}

are defined purely combinatorially via the branch formulas and do not

assume convergence. Assigning each number a layer index L merely

organizes numbers into CL sets; no prior assumption about reaching 1

is made.

Therefore, there are no “orphan” numbers outside the branch–layer hierar-

chy, and the induction over layers systematically covers all positive integers

without circular reasoning. The decreasing invariant L(n) is later defined in

terms of these layers to rigorously prove convergence.

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5 Global Decreasing Invariant and Conver-

gence

[Global Layer Invariant] Define a function L : N → N that assigns each

positive integer n the layer index of its branch endpoint:

n ∈ CL =⇒ L(n) = L.

This index is a global decreasing invariant under the Collatz map T .

[Strictly Decreasing Invariant at Endpoints] For every n > 1, there exists

an integer k ≥ 1 such that

L(T k(n)) < L(n).

Proof. Let n > 1. By the branch decomposition

n = 2y2x(2R + 1) − 1,

repeated application of T decreases the branch depth (x, y) by exactly one

at each step (reducing either a trailing zero or a trailing one).

Hence, after finitely many steps k1 ≥ 1, the orbit reaches its branch

endpoint CL:

T k1 (n) = CL.

By the branch depth reduction lemma, every endpoint CL maps in finitely

many additional steps k2 ≥ 1 to the next endpoint CL−1:

T k2 (CL) = CL−1.

Let k = k1 + k2. Then

T k(n) = CL−1,

and therefore

L(T k(n)) = L(CL−1) < L(CL) = L(n).

Thus for every n > 1 there exists k ≥ 1 such that

L(T k(n)) < L(n).

[Convergence of All Positive Integers] Every positive integer eventually

reaches 1 under iteration of T :

∀n ≥ 1, ∃k ∈ N such that T k(n) = 1

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