Convergent Elimination

The Cycle Equation

From the Affine Orbit Structure, a Collatz cycle of total length $K$ with $S$ odd steps and $E = K - S$ even steps satisfies:

$$n = \frac{C_\text{total} \cdot 2^E}{2^E - 3^S}$$

where $C_\text{total}$ is the sum of affine corrections from each odd step. For a cycle to exist:

1. $2^E - 3^S$ must be nonzero (it is, since $\log_2 3$ is irrational)
2. The result must be a positive integer
3. $n$ must actually follow the proposed parity pattern

The viable $(S, E)$ pairs are the convergents of $\log_2 3$ — the best rational approximations give the smallest gaps $|2^E - 3^S|$.

Ascending Convergent Elimination

Theorem. All convergents with $3^S > 2^E$ (ascending ratio) cannot produce positive integer cycles.

The affine constant $C_\text{total}$ is always positive: each odd step contributes $+1$, subsequently multiplied by positive factors ($3^j / 2^m$ for various $j, m$).

When $3^S > 2^E$, the denominator $2^E - 3^S < 0$, so:

$$n = \frac{C_\text{total} \cdot 2^E}{2^E - 3^S} = \frac{(\text{positive})}{(\text{negative})} < 0$$

No positive integer cycle exists.

This eliminates half of all convergents for free.

Convergent Status Table

$E$	$S$	$K$	$p/q$	vs $\log_2 3$	Gap	Status
1	1	2	1.000	below	$-1$	Eliminated (ascending)
2	1	3	2.000	above	$1$	✓ Trivial cycle (4→2→1)
3	2	5	1.500	below	$-1$	Eliminated (ascending)
8	5	13	1.600	above	$13$	Eliminated (divisibility)
19	12	31	1.583	below	$-7153$	Eliminated (ascending)
65	41	106	1.585	above	$\sim 10^{1.3}$	Open (too large to enumerate)
84	53	137	1.585	below	$\sim -10^{1.3}$	Eliminated (ascending)
485	306	791	1.585	above	$\sim 10^{0.6}$	Open

Gap = 13 Elimination

For $(S=5, E=8, K=13)$ with gap $= 2^8 - 3^5 = 256 - 243 = 13$:

A valid parity word is a circular binary string of length 13 with exactly 5 ones (odd positions) and no two consecutive ones (since $3n+1$ always produces an even number). There are exactly 91 such words.

For each word, the affine composition gives a specific constant $C$. The cycle equation requires:

$$13 \mid C \cdot 256$$

Since $\gcd(256, 13) = 1$, this reduces to $13 \mid C$.

Theorem. No 13-step Collatz cycle exists. Among all 91 valid parity words, the remainder $C \cdot 256 \bmod 13$ is distributed over $\{1, 2, \ldots, 12\}$ — zero never appears.

Distribution of $C \cdot 256 \bmod 13$:

Remainder	Count
1	7
2	6
3	9
4	7
5	6
6	10
7	7
8	6
9	10
10	6
11	8
12	9
0	0

The distribution is roughly uniform over $\{1, \ldots, 12\}$, but zero is structurally excluded.

The Trivial Cycle

The convergent $(S=1, E=2, K=3)$ with gap $= 1$ produces the known cycle:

• Parity word $(1, 0, 0)$: $C = 1$, $n = 1 \cdot 4 / 1 = 4$. The cycle $4 \to 2 \to 1 \to 4$. ✓
• Parity word $(0, 1, 0)$: $n = 2$. The cycle $2 \to 1 \to 4 \to 2$. ✓
• Parity word $(0, 0, 1)$: $n = 4$. Same cycle, different starting point.

Related

• Divisibility Obstruction — the conjecture that generalizes gap=13
• Affine Orbit Structure — the affine maps underlying the cycle equation
• abc Conjecture — stronger bounds on the gap