2026-08-29 · article Collatz conjecturecyclescontinued fractions

What the sign of 1/3 decides — the sign fixes the side, and the sign of the asymmetry is fixed by 3³ < 2⁵

Leanmachine-checked (Lean 4 + mathlib, standard axioms only, no sorryAx, no native_decide; theorem names given) paperproved, not yet machine-checked computationchecked on this machine, within the range stated; not a claim made to the outside knowna restatement, a known theorem, or a check of the literature

Write the map as n → (3n + ε)/2ᵛ (ε = ±1, and v is how many times 2 divides in).

The first appearance is inside an expectation. From Terras's theorem, E[2⁻ᵛ] = 1/3, so—

3n+1 is a submartingale (the expectation rises); 3n−1 is a supermartingale (it falls). This is the one place where, as the criticality article puts it, "the sign differs".

The second appearance is in the cycle equation. Write the odd terms as n₁ → n₂ → … → n_a → n₁, let b be the total number of halvings, and multiply the relations around the loop—

The ε/3 of the drift and the ε/(3nᵢ) of (★) are the same thing.
One is a statement about expectations and the other an identity between integers, but the ε that appears is the identical ε.

The third appearance is inside the lower bound (the Eliahou-type bound of §03). Taking logarithms in (★), b·ln 2 − a·ln 3 = Σ log(1 + ε/(3nᵢ)). The sign of the right-hand side is the sign of ε, and bounding each term by 1/(3n_min) gives n_min ≤ a / (3·ln 2·‖a log₂3‖). The 3 in the denominator of the width of Eliahou's seat is the 3 of the drift 1/3.

(★) holds, as fractions, for every known cycle. computation

※ Agreement confirmed with exact rational arithmetic (not floating point).

The right-hand side of (★) is greater than 1 if ε = +1 and less than 1 if ε = −1. That alone gives the conclusion.

That is what the sign decides. The same claim has a public precedent — Helms, in a comment on Tao's blog (2019), derived from the same identity in the form 2^S = Π(3 + 1/a_k) that "the cycle solutions are not symmetric around 0". known Checking it against the known cycles—

Take logarithms in (★). Setting d = b·ln2 − a·ln3 gives d = Σ ln(1 + ε/(3nᵢ)). Feed in ln(1+u) ≤ u and |ln(1−u)| ≤ u/(1−u) and out comes an exact upper bound on the smallest element n_min of a cycle.

The size of the smallest element a cycle may have is decided purely by how well b/a approximates log₂3 from its side.
The better the approximation (the smaller |d|), the larger the element it permits.

If this bound is below the verified range, we can say immediately that no cycle with that a exists. That is the content of Eliahou-type lower bounds. known

For each side, take in order the values of a where the approximation sets a new record, and line up the corresponding bounds. computation

Ranks 1 to 3 are nearly even (1.00, 1.02, 1.12). From rank 4 the lower side suddenly gains, and by rank 9 the gap is a factor of 110,000. What produces that gap is the continued fraction of log₂3.

The convergent just before a large term (5 and 23) is an unusually good approximation. And those fall on the lower side. The gap at ranks 4 to 9 is that bias.

The three cycles of 3n−1 sit at ranks 1, 2 and 3. Exactly where the two sides are evenly matched.

Every one of them is right up against the bound. 5 against 5.99, 17 against 35.87. The cycles that exist sit just inside the Diophantine limit of what is permitted. And ranks 2 and 3 on the 3n+1 side (a=3, a=5) have bounds of 5.9 and 32.0 — almost the same amount of room as ranks 2 and 3 on the 3n−1 side (5.99 and 35.87). And yet 3n+1 has no cycle. The counts of candidate integers at ranks 1 to 3 are 1 against 1, 2 against 2, 11 against 12; at rank 2 both sides have the same two candidate integers {1, 5}, and only one side is taken.

The sign of 1/3 decides which side of log₂3 a cycle approaches from. That is exact, and it follows immediately from (★).

However, at the places where cycles actually exist, the two sides have almost the same amount of room. The cycles of 3n−1 are at ranks 1, 2 and 3, where the ratios are 1.00, 1.02 and 1.12. There is a seat of the same size on the 3n+1 side, and nobody is sitting in it.

So the sign decides the "side" but not the size of the seat.

It is true that from rank 4 onward the lower side gains a large advantage, but the cycles that exist are not there, so that gap does not explain the presence or absence of cycles. Moreover, at the current depth of verification the gap is levelled out.

The gap at ranks 4 to 9 is a local wobble of the continued fraction. Seen at large, the two sides have seats of the same size.

This question — why a seat of the same size is filled on only one side — is out of reach of the tools on the 3-adic side. LeanShiori702.syracuse_sign_conjugation: sign conjugation of the Syracuse map, Syr₊₁(−m) = −Syr₋₁(m). Every quantity built from the 3-adic limit measure is invariant under the sign, so no quantity on that side can tell 3n+1 from 3n−1.

There is a different object from the "size of the seat" of §06: the binary-side model that tracks whether the orbit is above or below its starting point. With a = log₂3 and V_k the cumulative number of halvings, set e_k := V_k − ⌊k·a⌋ (the same e as in the criticality article); one step changes it by

v is on the number-theoretic (binary) side and follows the Terras distribution P(v=j) = 2⁻ʲ; g is on the floor-function side and is the Sturmian word of slope log₂3. The model takes the two to be independentknown (Terras 1976 and Sturmian words are known ingredients; the independence is measured). Writing the ±1 word with D (descent: v = 1 and g = 2), F (flat: v = g) and U (ascent: v > g)—

A descent requires two conditions to hold at once; an ascent needs only one. So P(D) = (a−1)/2, P(U) = (3−a)/4, and

LeanCollatz1139.no_three_consecutive_descents, eK_three_step_lower, asymmetry_negative, three_cA_lt_five (only v_i ≥ 1 is assumed; standard axioms). The premises of the model — that v follows the Terras distribution and is independent of g — lie outside Lean.

Testing the model: measured over 8.45 million steps for each of 12 multipliers q ∈ {1, 3, …, 25}, the sign is negative for every q (magnitude −0.058 to −0.087). In windows of length 8 the measurements agree exactly with the set of 111 allowed and 145 forbidden words, and the longest run of descents is 2 for every q. The difference q between 3n+1 and 3n−1 cannot enter this model in principle — neither a nor Terras involves q. At Q = 5 the sign flips (measured +0.3656, paper +0.3707). computation

The entropy of the ±1 word, log₂3 − 1 = 0.585 bit per step, and the drift 2 − log₂3 = 0.415 are complementary: their sum is 1 bit. The seats for a descent are absent from the start at the positions with g = 1 (density 0.415), and that deficit is the drift. — "The seats for a cycle are the same size on both sides" (§06) and "in the ±1 word the seats for a descent are narrow from the start" (here) are separate statements about separate objects, and both hold.

Where the open items that have moved now stand is in What remains. Only what is open at present is placed here.

Sources and reproduction

There is no new mathematics in this article. (★), the Eliahou-type bound and Sturmian words are all known. What may be new is the restatement that "the sign of the asymmetry" and "the forbidden word" come from one inequality, and the number 2^5/3; for both, nothing more than "not found in the literature searched" is claimed.