Collatz Trees: A Structural Framework for Understanding the 3x+1 Problem

1. Decomposing All Natural Numbers into Geometric Sequences

1.1. Background and Objective

We express $\mathbb{N}$ as a collection of rays parameterized by odd cores and powers of two, providing a structural stage for Collatz dynamics.

1.2. Definitions and Goals

Let

$$S=\{(2a+1)·2^b\mid a,b\in {\mathbb{Z}}_{\ge 0}\}.$$

We show $S=\mathbb{N}$ and the representation is unique.

1.3. Prime Factorization and Classification

Every $n\in \mathbb{N}$ decomposes uniquely as

$$n=2^b·k,b\in {\mathbb{Z}}_{\ge 0},k\mathrm{odd}.$$

1.4. Exhaustion of Odd Numbers

Any odd k is $k=2a+1$ with $a\ge 0$, giving $1,3,5,7,\dots$.

1.5. Exhaustion of Even Parts

For each odd k, the ray $k,2k,4k,\dots$ exhausts the even multiples of k.

1.6. Construction of S and Uniqueness

By the above, every $n=(2a+1)2^b$ with $a,b\ge 0$. If

$$(2a+1)2^b=(2a^{\prime }+1)2^{b^{\prime }},$$

then $(2a+1)/(2a^{\prime }+1)=2^{b^{\prime }-b}$, forcing $a=a^{\prime }$ and $b=b^{\prime }$ since the left side is odd rational and the right is a power of two. Hence $S=\mathbb{N}$ bijectively.

1.7. Remarks from the Collatz Perspective

For odd k, $3k+1$ is even and belongs to some ray $(2a^{\prime }+1)2^{b^{\prime }}$. This exhibits inter-branch connections. However, the assertion that every number lies on a finite forward path to 1 (global convergence) is a separate issue (coverage) made precise by Theorem 4; it is not implied by the mere classification $S=\mathbb{N}$.

Takeaway of Chapter 1.

We obtain a clean, bijective indexing of $\mathbb{N}$ by odd core and 2-adic height, furnishing a coordinate system on which later structural/affine arguments are staged.

2. The Structure of the Collatz Tree

2.1. Definition (Branches and Trunk)

Define the trunk $T_0=\{1·2^b:b\ge 0\}$ and for each odd $k\ge 3$ the branch $B_k=\{k·2^b:b\ge 0\}$. These rays partition $\mathbb{N}$ disjointly.

Figure 1. Trunk and branches (schematic; reverse orientation when embedded into the inverse graph: edges point to preimages). (finite cutoff; visualization, not a proof)

Figure 1. Trunk and branches (schematic; reverse orientation when embedded into the inverse graph: edges point to preimages). (finite cutoff; visualization, not a proof)

2.2. Branch–Branch Links via $3k+1$

Given odd k, $3k+1$ is even and decomposes as $(2a^{\prime }+1)2^{b^{\prime }}$, indicating where the branch from k can merge into another branch/trunk in forward dynamics. This shows linkage patterns but does not by itself prove global coverage of the tree by reverse generation.

Figure 2. Branch connections (schematic; reverse orientation: edges point to preimages). (finite cutoff; visualization, not a proof)

Figure 2. Branch connections (schematic; reverse orientation: edges point to preimages). (finite cutoff; visualization, not a proof)

2.3. Forward vs. Reverse Orientation

Let the standard forward map be

$$f\left(n\right)=\left\{\begin{array}{cc}n/2,\hfill & n\mathrm{even},\hfill \\ 3n+1,\hfill & n\mathrm{odd}.\hfill \end{array}\right.$$

The forward graph (edges $n\to f\left(n\right)$) is a functional digraph (outdegree 1). We do not call it a DAG because it contains the trivial $1\leftrightarrow 2\leftrightarrow 4$ 3-cycle; nontrivial finite cycles are excluded later (sec:affine).

The reverse (preimage) graph rooted at 1, with edges to preimages under f, is a true DAG: levels increase with each application of a reverse step.

2.4. Tree Language

When drawing a reverse BFS tree rooted at 1, each node is assigned a unique parent by construction (though a number may have up to two preimages as graph children). Connectivity of every node to 1 in the forward sense is equivalent to coverage of the reverse tree, which is equivalent to the Collatz convergence; see Theorem 4.

3. Trunk–Branch Indexing of the Natural Numbers

Definition 1

(Odd core, 2-adic valuation). For $n\in \mathbb{N}$, write uniquely $n=\mathrm{odd}\left(n\right)·2^{{\nu }_2\left(n\right)}$ where $\mathrm{odd}\left(n\right)$ is odd and ${\nu }_2\left(n\right)\in {\mathbb{Z}}_{\ge 0}$ is the exponent of 2 in n.

Definition 2

(Trunk and branches). The trunk is $T_0=\{1·2^b:b=0,1,2,\dots \}=1,2,4,8,\dots$. For any odd $k\ge 3$, $B_k=\{k·2^b:b=0,1,2,\dots \}.$ Then $\left\{T_0\right\}\cup \{B_k:k\mathrm{odd}\ge 3\}$ is a disjoint partition of $\mathbb{N}$.

Definition 3

(Indices). Order the odd numbers as $1,3,5,7,\dots$. Assign the branch index $\mathrm{br}\left(\mathrm{odd}\right)=(\mathrm{odd}-1)/2\in {\mathbb{Z}}_{\ge 0}$, so that $\mathrm{br}\left(1\right)=0$ and $\mathrm{br}\left(3\right)=1$, $\mathrm{br}\left(5\right)=2$, etc. Define the height $\mathrm{ht}\left(n\right)={\nu }_2\left(n\right)$. Set

$$\mathrm{Idx}:\mathbb{N}\to {\mathbb{Z}}_{\ge 0}\times {\mathbb{Z}}_{\ge 0},\mathrm{Idx}\left(n\right)=\left(\mathrm{br}\left(\mathrm{odd}\right(n\left)\right),\mathrm{ht}\left(n\right)\right),{\mathrm{Idx}}^{-1}(i,b)=(2i+1)·2^b.$$

Theorem 1

(Complete classification). The map $n\mapsto \mathrm{Idx}\left(n\right)$ is a bijection from $\mathbb{N}$ onto ${\mathbb{Z}}_{\ge 0}\times {\mathbb{Z}}_{\ge 0}$.

Proof. 

Uniqueness of $\mathrm{odd}\left(n\right)$ and ${\nu }_2\left(n\right)$ is immediate; disjointness/exhaustiveness of rays follows.    □

Figure 3. Trunk–branch indexing (schematic; reverse orientation in the inverse graph). (finite cutoff; visualization, not a proof)

Figure 3. Trunk–branch indexing (schematic; reverse orientation in the inverse graph). (finite cutoff; visualization, not a proof)

Figure 4. Indexed reverse tree (schematic; edges point to preimages). (finite cutoff; visualization, not a proof)

Figure 4. Indexed reverse tree (schematic; edges point to preimages). (finite cutoff; visualization, not a proof)

In this table we use a 1-based display index $I\left(k\right)=\mathrm{br}\left(k\right)+1=(k+1)/2$.

Table 1. Trunk–Branch Indexing (sample)

Odd k	Index	Next Index (rule)	parity-based trend	transition factor
1	1	–	–	–
3	2	3	increase	$3/2$
5	3	1	decrease	$1/3$
7	4	6	increase	$3/2$
9	5	4	decrease	$4/5$
11	6	9	increase	$3/2$
13	7	3	decrease	$3/7$
15	8	12	increase	$3/2$
17	9	7	decrease	$7/9$
19	10	15	increase	$3/2$

Table 1. Trunk–Branch Indexing (sample)

Odd k	Index	Next Index (rule)	parity-based trend	transition factor
1	1	–	–	–
3	2	3	increase	$3/2$
5	3	1	decrease	$1/3$
7	4	6	increase	$3/2$
9	5	4	decrease	$4/5$
11	6	9	increase	$3/2$
13	7	3	decrease	$3/7$
15	8	12	increase	$3/2$
17	9	7	decrease	$7/9$
19	10	15	increase	$3/2$

4. Affine Word Method: Absence of Nontrivial Finite Cycles

We encode even/odd steps as affine maps and use finite-word compositions $F_W\left(x\right)=a_{W}x+c_W$ to prove absence of nontrivial finite cycles.

Definition of the three-way map T.

Define $T:{\mathbb{N}}_{\ge 1}\to {\mathbb{N}}_{\ge 1}$ by

$$T\left(n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{3}{2}}n,\hfill & n\mathrm{even},\hfill \\ {\displaystyle \frac{n-1}{2}},\hfill & n\mathrm{odd}\mathrm{and}{\nu }_2(3n+1)=1,\hfill \\ {\displaystyle \frac{3n+1}{4}},\hfill & n\mathrm{odd}\mathrm{and}{\nu }_2(3n+1)\ge 2.\hfill \end{array}\right.$$

Elementary steps.

$$E\left(x\right)={\displaystyle \frac{3}{2}}x,O_1\left(x\right)={\displaystyle \frac{x-1}{2}},O_2\left(x\right)={\displaystyle \frac{3x+1}{4}}.$$

(For odd n, choose $O_1$ if ${\nu }_2(3n+1)=1$, and $O_2$ if ${\nu }_2(3n+1)\ge 2$.)

Words and composition.

For any finite word W over $\{E,O_1,O_2\}$, the composition is affine:

$$F_W\left(x\right)=a_{W}x+c_W.$$

Writing $m=\#E$, $o_1=\#O_1$, $o_2=\#O_2$, we have

$$a_W={\left({\displaystyle \frac{3}{2}}\right)}^m{\left({\displaystyle \frac{1}{2}}\right)}^{o_1}{\left({\displaystyle \frac{3}{4}}\right)}^{o_2}={\displaystyle \frac{3^{m+o_2}}{2^{m+o_1+2o_2}}},$$

and the denominator of $c_W$ divides $2^{o_1+2o_2}$ (each $O_1$ contributes $-1/2$, each $O_2$ contributes $+1/4$; E does not increase the power-of-two denominator).

Lemma 1

(No $a_W=1$ for nonempty words). If W is nonempty then $3^m=2^{m+o_1+2o_2}$ cannot hold.

Lemma 2

(Odd numerator for $1-a_W$).

$$1-a_W={\displaystyle \frac{2^{m+o_1+2o_2}-3^{m+o_2}}{2^{m+o_1+2o_2}}},$$

so the numerator is odd (even minus odd).

Lemma 3

(If $m\ge 1$, a periodic solution cannot be integral). Since $\mathrm{den}\left(c_W\right)\mid 2^{o_1+2o_2}$, we have $c_W·2^{m+o_1+2o_2}=\left(\mathrm{integer}\right)·2^m$. By Lemma 2, $1-a_W=\left(\mathrm{odd}\right)/2^{m+o_1+2o_2}$. Hence

$$x={\displaystyle \frac{c_W}{1-a_W}}={\displaystyle \frac{\left(\mathrm{integer}\right)·2^m}{\mathrm{odd}}},$$

and the odd denominator cannot cancel $2^m$. Thus x is not an integer.

Lemma 4

(If $m=0$, contraction; the only integer fixed point in ${\mathbb{N}}_{\ge 1}$ is 1). When $m=0$, $a_W=2^{-(o_1+2o_2)}<1$, so any integer periodic point must be a fixed point. Solving $x=O_1\left(x\right)$ gives $x=-1$ (an integer but outside our domain ${\mathbb{N}}_{\ge 1}$), and solving $x=O_2\left(x\right)$ gives $x=1$. Hence, within ${\mathbb{N}}_{\ge 1}$ the only fixed point is 1.

Theorem 2

(Loop-freeness for T). Under the three-way rule above, the map T admits no finite cycle other than the trivial 1-cycle.

Proof. 

By Lemma 1, consider $x=c_W/(1-a_W)$. If $m\ge 1$, Lemma 3 rules out integer $x>1$. If $m=0$, Lemma 4 leaves only $x=1$ as a fixed point.    □

Interpretation. 

If an even step E appears at least once, a factor $2^m$ remains in the numerator of $x=c_W/(1-a_W)$ and cannot be canceled by the odd denominator, so no integer fixed point arises. If E never appears, the composition is a contraction and the only integer fixed point is 1. This algebraic loop-elimination aligns with the inverse-generation intuition.

5. Bridge to the Accelerated Collatz Map (Rigorous Version)

Let the standard accelerated Collatz map on odd integers be

$$A\left(n\right)={\displaystyle \frac{3n+1}{2^{{\nu }_2(3n+1)}}},n\in {\mathbb{N}}_{\mathrm{odd}}.$$

Recall the three-way affine map $T:{\mathbb{N}}_{\ge 1}\to {\mathbb{N}}_{\ge 1}$:

$$E\left(x\right)={\displaystyle \frac{3}{2}}x,O_1\left(x\right)={\displaystyle \frac{x-1}{2}},O_2\left(x\right)={\displaystyle \frac{3x+1}{4}},$$

where $O_1$ is used if ${\nu }_2(3x+1)=1$ and $O_2$ otherwise. For any finite word $W\in {\{E,O_1,O_2\}}^{*}$, the composition $F_W\left(x\right)=a_{W}x+c_W$ satisfies

$$a_W={\displaystyle \frac{3^{m+o_2}}{2^{m+o_1+2o_2}}},$$

where $m,o_1,o_2$ denote the counts of $E,O_1,O_2$ in W.

Lemma 5

(Coefficient identity for a cycle of A). Let $(x_0,\dots ,x_{L-1})$ be a cycle of A with $r_j={\nu }_2(3x_j+1)$ for each step. Setting $E={\sum }_{j=0}^{L-1}{r}_j$ and $e=L$, there exists an integer C such that

$$(2^E-3^e)x_0=C,D:=2^E-3^e\ne 0.$$

Moreover, $gcd(3,D)=1$.1

Lemma 6

(Coefficient matching on the T-side). Choose a word W with counts $\#E=m=e$ and $m+o_1+2o_2=E$. Then

$$a_W={\displaystyle \frac{3^{m+o_2}}{2^{m+o_1+2o_2}}}={\displaystyle \frac{3^e}{2^E}}.$$

Lemma 7

(Adjacent-swap residue control). Let ${\tilde{c}}_W:=c_W{2}^E$. Adjacent swaps of blocks $EO\leftrightarrow OE$ with $O\in \{O_1,O_2\}$ alter ${\tilde{c}}_W$ by

$${\tilde{c}}_W\mapsto {\tilde{c}}_W\pm u{3}^t\left(\mathrm{mod}D\right),$$

for some odd unit u modulo D and integer $t\ge 0$. Hence, by successive swaps, one can realize any residue class modulo D.

Lemma 8

(Simultaneous satisfiability of parity conditions). At each occurrence of $O_1$ or $O_2$, the requirement on ${\nu }_2(3x+1)$ reduces to a linear congruence

$${\alpha }_{j}x\equiv {\beta }_j\left(\mathrm{mod}{2}^{s_j}\right)\left({\alpha }_j\mathrm{odd}\right),$$

and adjacent swaps adjust ${\beta }_j$ by controlled powers of two. Therefore, the entire system admits a simultaneous solution by the Chinese Remainder Theorem.

Theorem 3

(Bridge Theorem (Equivalence of Cyclicity)). If the accelerated map A possesses a nontrivial finite cycle of length $L\ge 1$, then the three-way map T also admits a nontrivial finite cycle.

Proof. 

By Lemma 5, an A-cycle yields parameters $(e,E,D)$ with $D=2^E-3^e$ and $gcd(3,D)=1$. Pick W with counts as in Lemma 6 so that $a_W=3^e/2^E$. By Lemma 7, using adjacent swaps we can tune ${\tilde{c}}_W\equiv (2^E-3^e)x\left(\mathrm{mod}D\right)$ to match any prescribed residue class. Finally, Lemma 8 ensures the local parity constraints can be satisfied simultaneously. Thus there exists $x\in \mathbb{Z}$ such that

$$F_W\left(x\right)=a_{W}x+c_W=x,$$

so x is periodic for T with the same length L.    □

Corollary 1.

For the accelerated Collatz map on odd integers,

$$A\left(n\right)={\displaystyle \frac{3n+1}{2^{{\nu }_2(3n+1)}}},$$

there exists no nontrivial finite cycle.

Proof. 

By the Bridge Theorem Theorem 3, any nontrivial finite cycle of A would yield a nontrivial finite cycle of the three-way map T. But Theorem 2 (loop-freeness for T) excludes such cycles. Hence A has no nontrivial finite cycle.    □

Theorem 4

(Reduction to convergence). For the accelerated map A, the following are equivalent:

• Every positive integer reaches 1 in finitely many steps (global convergence).
• The inverse generation tree rooted at 1 covers all positive integers (reachability/coverage).

Proof. 

Each connected component of a functional digraph is a directed cycle with an in-tree. Since only the 1-cycle is permitted (Corollary 1), global convergence holds iff every node is in the basin of 1, i.e. iff the inverse tree covers $\mathbb{N}$.    □

6. Coverage of the Inverse Collatz Tree

The previous sections established the absence of nontrivial finite cycles. Hence global convergence reduces to the remaining structural pillar: coverage—whether the inverse Collatz tree rooted at 1 covers all natural numbers. We formalize and partially prove this pillar.

6.1. Inverse Graph Definition

For any $n\in {\mathbb{N}}_{\ge 1}$, define its reverse (preimage) set:

$$\mathcal{C}\left(n\right)=\left\{2n\right\}\cup \left\{\frac{n-1}{3}|(n-1)\equiv 0\left(\mathrm{mod}3\right),\frac{n-1}{3}\mathrm{odd}\right\}.$$

The inverse Collatz graph has root 1 and edges $n\to m$ for all $m\in \mathcal{C}\left(n\right)$. It is a directed acyclic graph (DAG) whose connectedness determines coverage.

Definition 4

(Coverage). The inverse Collatz tree is said to becoveringif for every $n\in \mathbb{N}$, there exists a finite sequence $(n_0,n_1,\dots ,n_L)$ with $n_0=1$ and $n_L=n$, where each $n_{j+1}\in \mathcal{C}\left(n_j\right)$.

By Theorem 4, this condition is equivalent to the global convergence of the Collatz map.

6.2. Branch-Level Analysis

For each odd core k and height b, $n=k·2^b$ belongs to a branch $B_k$. We classify the reverse-generation behavior along this branch.

Lemma 9

(Criterion for two-child branching). If $n\equiv 4\left(\mathrm{mod}6\right)$, then $\mathcal{C}\left(n\right)=\{2n,(n-1)/3\}$, so a new odd preimage branch emerges at this level. Otherwise $\mathcal{C}\left(n\right)=\left\{2n\right\}$.

Proof. 

Since n even, write $n=2m$. Then $(n-1)/3$ is integer iff $n\equiv 4\left(\mathrm{mod}6\right)$. When this holds, $(n-1)/3$ is necessarily odd. □    □

Corollary 2

(Infinite supply of branching levels). If the odd core k is not a multiple of 3, the sequence $k·2^b$ attains $4\left(\mathrm{mod}6\right)$ infinitely often as b varies. Hence $B_k$ contains infinitely many levels with two-child branching.

Proof. 

Modulo 6, we have $2^b\equiv 2,4,2,4,\dots$. If $k\equiv 1\left(\mathrm{mod}3\right)$, then $k·2^b\equiv 4\left(\mathrm{mod}6\right)$ for even b. If $k\equiv 2\left(\mathrm{mod}3\right)$, then $k·2^b\equiv 4\left(\mathrm{mod}6\right)$ for odd b. □    □

6.3. Local Descent of the Odd Core

Definition 5

(Odd core). For $n=k·2^{{\nu }_2\left(n\right)}$, the integer $k=\mathrm{odd}\left(n\right)$ is called theodd coreof n.

Proposition 1

(Descent direction at branching points). When $n\equiv 4\left(\mathrm{mod}6\right)$, its children are $2n$ and $(n-1)/3$. The odd core of the second child satisfies

$$\mathrm{odd}\left(\frac{n-1}{3}\right)\le \frac{n-1}{3}\le \frac{2}{3}n.$$

Hence at every branching level, one child is a strict odd-core descent.

Proof. 

If $(n-1)/3$ is odd, its odd core equals itself and is $<\frac{2}{3}n$. If even, further division by 2 only decreases it. □    □

Corollary 3

(Guaranteed descent opportunities). Every branch $B_k$ with $3\nmid k$ encounters infinitely many levels where a descent edge in $\mathrm{odd}(·)$ is available. Thus such branches can always locally reduce their odd core.

6.4. Constructive Reachability for $3\nmid k$

We now show that branches with odd cores not divisible by 3 are fully covered by the inverse tree.

Lemma 10

(Finite design reachability via CRT). For any finite sequence of inverse steps

$$E^{e_0}O{E}^{e_1}O\dots E^{e_{t-1}}O,$$

where $E\left(x\right)=2x$ and $O\left(x\right)=(x-1)/3$, there exists a starting point $n_0$ such that each O-step is valid, i.e. the operand of O satisfies $n\equiv 4\left(\mathrm{mod}6\right)$.

Proof. 

Each O-validity condition is a linear congruence $n\equiv 4\left(\mathrm{mod}6\right)$ shifted by powers of two through E. The resulting system of congruences is consistent mod $2^r$ and mod 3, hence solvable simultaneously by the Chinese Remainder Theorem. □    □

Proposition 2

(Finite descent construction). For every odd k with $3\nmid k$, there exists a finite inverse sequence of the above form leading from 1 to some $n=k·2^b$, whose forward image (under the usual Collatz iteration) returns to a smaller odd core $k^{\prime }<k$.

Proof. 

By y Corollary 2, the branch $B_k$ contains infinitely many levels $n\equiv 4\left(\mathrm{mod}6\right)$. At such a level, the inverse step O reduces the odd core by at least the factor $2/3$ ((Proposition 1). By chaining a finite number of these O-steps separated by powers of E, and invoking lem:crt to ensure legality of all O, we obtain a finite construction lowering the core. □    □

Theorem 5

(Coverage for $3\nmid k$). Every branch $B_k$ with $3\nmid k$ is entirely covered by the inverse tree, and all its nodes reach the root 1 in finitely many forward steps.

Proof. 

Repeatedly apply Proposition 2. The odd core strictly decreases at each finite stage and cannot descend infinitely without reaching 1. □    □

6.5. Branches with $3\mid k$

The remaining case is $k\equiv 0\left(\mathrm{mod}3\right)$, for which $k·2^b\equiv 0,3\left(\mathrm{mod}6\right)$ so no direct O-type edge occurs.

Theorem 6

(Sufficient condition for full coverage). Suppose that for every odd $k\equiv 0\left(\mathrm{mod}3\right)$, there exists a finite sequence of inverse operations of the form $E^{e_0}O{E}^{e_1}O\dots E^{e_{t-1}}O$ leading to a node whose odd core $k^{★}$ satisfies $3\nmid k^{★}$. Then the entire inverse Collatz tree is covering, and the Collatz Conjecture holds.

Proof. 

By assumption, each $3\mid k$ branch eventually transfers to a branch with $3\nmid k$. By Theorem 5, that branch reaches 1. Hence every integer is connected to 1. □    □

Interpretation. 

The remaining open step is to verify the existence of such finite transfer sequences for all $3\mid k$. This problem is purely combinatorial and can be attacked with the same adjacent-swap and Chinese Remainder techniques used in the Bridge Theorem.

Summary. 

The coverage pillar is thus partially proven:

• All branches $B_k$ with $3\nmid k$ are provably covered.
• Full coverage reduces to verifying finite transfer from $3\mid k$ branches to non-multiples of 3.

This completes the reduction of the Collatz Conjecture to a finite residue-class verification problem.

7. Related Work

The affine-composition viewpoint with coefficient $3^m/2^E$ is classical in studies of cycles and their lengths (in our three-way encoding, $a_W=3^{m+o_2}/2^{m+o_1+2o_2}$). Our novelty is to integrate (i) the trunk–branch indexing and inverse-tree structure, with (ii) a complete loop-elimination for the three-way map T, and (iii) a bridge transporting hypothetical cycles of A into T, thereby isolating cycle-freeness as a stand-alone pillar and reducing full convergence to coverage.