The Cycle Equation: Small Cases

This page works through the small cases of the cycle equation by hand. It does not rule out cycles in general — that is open — and the cases here are already covered by the literature (Steiner 1977, Eliahou 1993, Simons and de Weger 2005, Hercher 2023). It is a worked example of how the equation behaves.

The Cycle Equation

A Collatz cycle of total length $K$ with $S$ odd steps and $E = K - S$ even steps satisfies (Böhm and Sontacchi, 1978):

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

where $C_\text{total}$ is the sum of affine corrections from each odd step (see Affine Orbit Structure). 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

Any pair with $2^E > 3^S$ is a candidate. The pairs where the gap $|2^E - 3^S|$ is smallest relative to $2^E$ come from the convergents of $\log_2 3$, so they make natural test cases, but they are not the only candidates. Eliahou (1993) showed that for any nontrivial cycle, $E/S$ must be very close to $\log_2 3$ — close, not necessarily a convergent.

Negative gaps give no positive cycles

Lemma. If $3^S > 2^E$, there is no positive integer cycle with $S$ odd and $E$ even steps.

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:

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

This is elementary and well known. It is also exactly where the negative integers come in: $-1$, $-5$ and $-17$ sit on cycles of this kind.

The first few convergents

$E$	$S$	$K$	$E/S$	vs $\log_2 3$	Gap $2^E - 3^S$	What happens
1	1	2	1.000	below	$-1$	Negative gap
2	1	3	2.000	above	$1$	Trivial cycle (4→2→1)
3	2	5	1.500	below	$-1$	Negative gap
8	5	13	1.600	above	$13$	No cycle (checked below)
19	12	31	1.583	below	$-7153$	Negative gap
65	41	106	1.585	above	$\approx 4.2 \times 10^{17}$	Too large to enumerate by hand
84	53	137	1.585	below	$\approx -4.0 \times 10^{22}$	Negative gap
485	306	791	1.585	above	$\approx 1.0 \times 10^{143}$	Too large to enumerate

Gap = 13

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 $13 \mid C$.

Check. No Collatz cycle has $S = 5$ odd and $E = 8$ even steps. 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 $\{1, \ldots, 12\}$, and zero is missing.

The Trivial Cycle

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

• Parity word $(1, 0, 0)$: $C = 1/4$, $n = \tfrac14 \cdot 4 / 1 = 1$. The cycle $1 \to 4 \to 2 \to 1$. ✓
• 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

• No Loops? — the tour chapter, with the small cases and the convergent navigator
• Cycles as a Divisibility Question — the cycle problem restated term by term
• Affine Orbit Structure — the affine maps underlying the cycle equation
• abc Conjecture — the size of the gap $2^E - 3^S$