No Loops?

There are two ways the conjecture could fail: an orbit could grow forever, or it could go in circles. This chapter is about the circles. Nobody has ruled them out. What we can do is see why a loop is so hard to build, and what is actually known.

A Collatz cycle would be an orbit that returns to its starting value: $n \to \cdots \to n$. If such a cycle has $S$ odd steps and $E$ even steps (total $K = S + E$), then the starting value must satisfy the cycle equation (Böhm and Sontacchi, 1978):

$$n = \frac{C \cdot 2^E}{2^E - 3^S}$$

where $C$ is a constant determined by the specific pattern of odd and even steps (the "parity word"). The denominator $g = 2^E - 3^S$ is the gap — and it must divide $C \cdot 2^E$ for $n$ to be a positive integer.

Hunt for cycles

For a small pair $(S, E)$ the hunt can be done exhaustively: list every parity word, compute its constant $C$, and see whether the gap divides $C \cdot 2^E$.

• (S=5, E=8) — the first interesting case. 35 parity words that start at an odd value (91 if every starting point of the loop is counted), gap = 13, and 13 divides $C \cdot 2^E$ for none of them. No cycle of this shape.
• (S=1, E=2) — the trivial case. One word, gap = 1, and it gives the cycle 4 → 2 → 1.
• (S=2, E=3) — here $3^2 = 9 > 2^3 = 8$, so the gap is negative and the equation forces $n < 0$ (it lands on $-5$ and $-7$, which loop among the negative integers). No positive cycle of this shape.
• (S=41, E=65) — too large to list by brute force. My own meet-in-the-middle search found no cycle of this shape either, though the known results below already rule out cycles this short.

Why E/S has to sit just above log₂ 3

Along a cycle the values are multiplied by 3 a total of $S$ times and divided by 2 a total of $E$ times, with the $+1$ terms making up the difference. Coming back to $n$ means the two balance exactly: $2^E / 3^S$ equals the product of $1 + \frac{1}{3 n_i}$ over the odd members $n_i$ of the cycle. Every factor is a little above 1, so a cycle of positive numbers needs $2^E$ bigger than $3^S$. And since every number below $2^{68}$ is known to reach 1, every member of a nontrivial cycle is larger than that, every factor is within $2^{-69}$ of 1, and $E/S$ can exceed $\log_2 3 \approx 1.58496$ only by less than $2^{-69}$.

Because $\log_2 3$ is irrational, $2^E \neq 3^S$ for any positive integers, so the gap is never zero. But it can be relatively small, and the pairs where it is smallest come from the convergents of $\log_2 3$ — its best rational approximations:

$$1/1,\ 2/1,\ 3/2,\ 8/5,\ 19/12,\ 65/41,\ 84/53,\ 485/306,\ \ldots$$

log₂3 ≈ 1.5849625007

log₂3

1/1

2/1

3/2

8/5

19/12

65/41

84/53

485/306

1054/665

24727/15601

Zoom: 0

1/1K=2↑Negative gap: no positive cycle

2/1K=3↓The trivial cycle 1 → 4 → 2 → 1

3/2K=5↑Negative gap: no positive cycle

8/5K=13↓Enumerated: no cycle of this shape

19/12K=31↑Negative gap: no positive cycle

65/41K=106↓Meet-in-the-middle search: none found

84/53K=137↑Negative gap: no positive cycle

485/306K=791↓Too short for a cycle (Eliahou 1993)

1054/665K=1719↑Negative gap: no positive cycle

24727/15601K=40328↓Too short for a cycle (Eliahou 1993)

Convergent 8/5

E (even steps)	8
S (odd steps)	5
K = E + S	13
E/S	1.60000000
|log₂3 − E/S|	1.504e-2
Type	Descending (2ᴱ > 3ˢ)
Status	Enumerated: no cycle of this shape

Blue dots have $3^S > 2^E$ (negative gap, no positive cycle possible). Red dots have $2^E > 3^S$. Zoom in to see how the convergents cluster around $\log_2 3$ without ever reaching it.

The convergents are the most dangerous candidates, but they are not the only ones. Any pair whose ratio $E/S$ falls in that narrow window above $\log_2 3$ is a candidate in principle, and ruling out every one of them is exactly the open problem.

Counting words against the gap

There is a tempting heuristic. For a pair with $E/S$ close to $\log_2 3$, the number of parity words is roughly $\binom{E}{S} \approx 2^{0.95 E}$, while the gap is often close to $2^E$ in size. If the constants $C$ were spread out like random numbers, the expected number of words with $g \mid C \cdot 2^E$ would be about the number of words divided by the gap. Far out that is tiny: about one in a million at $(S, E) = (306, 485)$. For the small cases above it is not: about 2.7 at $(5, 8)$ and 0.6 at $(41, 65)$, and neither has a cycle. Added up over every $E$ it comes to roughly one for each value of $S$, and the trivial cycle run $S$ times is one such solution. Summed over all $S$ the naive count grows without bound, so by itself it says nothing about whether nontrivial cycles exist.

That is a heuristic, not a proof. The constants $C$ are not random, the gap can be far smaller than $2^E$ when $E/S$ is near a convergent, and an expected count below one does not show the actual count is zero.

What is actually known

Ruling out nontrivial cycles is open. The strongest results use the cycle equation together with lower bounds on $|2^E - 3^S|$ from Baker's theory of linear forms in logarithms, and computer verification of small numbers:

• Steiner (1977) — there is no nontrivial cycle made of a single run of odd steps followed by a single run of even steps (a "1-cycle").
• Eliahou (1993) — since every number below a verified bound reaches 1, the ratio $E/S$ of any nontrivial cycle must lie extremely close to $\log_2 3$, and the cycle must be enormously long; the bound grows as verification extends.
• Simons and de Weger (2005) — no nontrivial $m$-cycles for $m \leq 68$ (an $m$-cycle has $m$ separate climbing runs).
• Hercher (2023) — extended this to $m \leq 91$.

The small cases and the convergent list on this page are the first few cases of the problem these authors attacked.

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