2026-08-29 · article Collatz conjecturecriticalitylarge deviationscontinued fractions

Only 3n+1 is exactly critical — criticality in full, the overshoot distribution, the boundary of exact counting, the lower bound on cycles

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

The usual answer to "why 3n+1?" is "because 3/4 < 1, so it shrinks." That is the logarithmic side. Here the answer comes from the other side.

Put Collatz back inside a wider family. For a prime p, with gcd(q,p)=1 and p ∤ r,

T(n) = (q·n + r) / p^(v_p(q n + r))    (n coprime to p)

p=2, q=3, r=1 is Collatz. Work out the distribution of v = v_p(qn+r) in this family and — it depends on neither q nor r, only on p.

Among the p−1 residues coprime to p, exactly one makes qn+r ≡ 0 (mod p). So P(v≥1) = 1/(p−1), and each further division has probability 1/p. Collecting this:

(Confirmed by measurement for p = 2, 3, 5, 7.) The expected ratio is q·E[p⁻ᵛ], so the multiplier that makes it 1 is—

q_crit(p) = (p² − 1) / (p² − p − 1)

3 at p = 2. Then it drops sharply, and for p ≥ 3 it always sits between 1 and 2. So the only integer multipliers available are q=1 (subcritical) and q=2 (supercritical), and neither can sit on criticality.

Theorem. q_crit(p) is an integer only for p = 2, and then q = 3.

Proof. (p²−p−1) | (p²−1) ⟺ (p²−p−1) | p (the difference is p). For p ≥ 3, p²−p−1 ≥ 5 > p, so it cannot divide. For p = 2, p²−p−1 = 1 and 1 | 2. ∎

LeanShiori.qcrit_eq_three, Shiori.qcrit_not_int_of_three_le (the ledger in the Lean verification bundle).

The answer to "why 3n+1?" lies in discreteness itself. A critical multiplier exists for every p, but it is realised as an integer only at p = 2. 3n+1 is the single point where the continuous critical curve meets the integer lattice.

Back to p = 2. The distribution of v is not an assumption but a theorem — by Terras (1976), looked at through residues of n modulo powers of 2, P(v = j) = 2⁻ʲ holds exactly. The mean is 2.

Now compute the expected "ratio" per step. An odd n moves to (3n+1)/2ᵛ, so the ratio is roughly 3/2ᵛ.

The computation gives E[2⁻ᵛ] = Σ 2⁻ʲ·2⁻ʲ = Σ 4⁻ʲ = 1/3. So—

E[ratio] = q/3 equals 1 only for q = 3. And at the same time the log expectation log(q/4) is negative precisely when q < 4. 3 sits exactly on criticality in the ratio and on the safe side in the logarithm — it is alone at that point.

Include the +1 exactly and the picture gets more uncomfortable still.

E[n_k+1 | n_k] = (3n_k+1)·(1/3) = n_k + 1/3

The expectation keeps rising by exactly 1/3 at every step. And yet the log expectation keeps falling by 0.288 at every step. The mean grows without bound while the typical orbit shrinks by a factor of 0.75 each time. The contradiction is filled in by rare enormous overshoots.

Measurement: 10 billion odd numbers, 34.9 billion steps in total. computation

How high does an orbit climb before it first drops below its starting point n? Call that ratio R. In the famous example n = 27 rises to 9232, so R = 342.

Take logarithms and this is the maximum of a random walk with negative drift. One step changes by X = log3 − v·log2. In such a case the tail of the maximum behaves as P(max > t) ≈ C·e^(−θ*t), where θ* is the nontrivial root of E[e^(θX)] = 1. Working it out—

E[e^(θX)] = 3^θ / (2^(1+θ) − 1) = 1   i.e.   3^θ = 2^(1+θ) − 1
Put θ = 1 and you get 3 = 4 − 1. It is exactly a root.

Written for the general family, E[e^(θX)] = q^θ · E[p^(−θv)], and the condition at θ = 1 is q · E[p^(−v)] = 1 — multiplicative criticality itself.

"E[ratio] = 1" and "tail exponent θ* = 1" are not two discoveries but one and the same claim. So the theorem of §01 can be restated: the only integer map whose tail exponent is exactly 1 is the Collatz map. The tail exponent 1 itself is already in Lagarias–Weiss (1992) as ρ_RW = 1known.

Hence P(R > λ) ≍ C/λ. The exponent is exactly 1. On a log₂ scale the count should halve exactly with every step. A least-squares fit to the measurement (10 billion odd numbers) gives θ* = 1.0028 — off by 0.28 %.

E[R] = ∞. But only logarithmically, only just. Meanwhile E[log R] is finite (measured 0.760).

So the question "how high does it overshoot on average?" has no answer. Think in expectations and it diverges; think in logarithms and it converges. And since the exponent is exactly 1, both intuitions are equally right.

The same 1 arrives by a completely different road

There is another way to write the computation. With a = log₂3 and Vk = Σvᵢ,

set e_k := V_k − ⌊k·a⌋  and then  the orbit stays at or above its starting point exactly while e_k ≤ 0, and the height there is {k·a} − e_k (in log₂).

e is integer-valued and moves as e → e + v − g at each step (g = ⌊(k+1)a⌋−⌊ka⌋ ∈ {1,2}). The drift is E[v] − a = +0.415, positive. So the problem becomes: how far below zero can e dive before it escapes upward?

The probability of diving to depth d falls as ρᵈ, where ρ is the root of E[ρ^(v−a)] = 1. Rearranged, ρ^(a−1)·(2−ρ) = 1. Put ρ = 1/2 in—

2^(1−a) · (3/2) = (2/3) · (3/2) = 1  exactly a root. ρ = 1/2.

ρ = 1/2 means "halved for every unit of log₂", that is, exponent exactly 1. A different road, and the same 1 at the end of it. This walk in e is the same object as the ±1-word model of the article on the sign.

And not just the exponent — the whole distribution agrees

The advantage of the e formulation is that it lets you compute not only the tail exponent but the distribution itself, exactly. The state is only the integer e and the time k, so a finite convolution suffices.

Across 7 orders of magnitude, the agreement is to 5 significant figures. That is a far stronger claim than "the exponent matches". It means the actual behaviour of the integers is explained by a binary random model. computation

The Collatz conjecture splits into two independent parts: (1) no orbit diverges, (2) there is no nontrivial cycle.

The probabilistic argument above works on (1). The negative drift log(3/4) = −0.288 is comfortably far from zero, so divergence requires an infinitely sustained large deviation in which "the mean of v stays below log₂3 = 1.585 forever". Computing that rate exactly gives I = 0.05498, so the probability of surviving k steps is 0.9465ᵏ.

On (2), however, it does nothing at all. Measure 3n−1 by the same method—

The statistics are identical. And yet one has cycles and the other does not.

Probabilistic arguments cannot tell 3n+1 from 3n−1. So about cycles they can say nothing in principle. Cycles are a problem of Diophantine approximation, not of statistics.

This silence closes the line of proof itself. The condition for the expectation of ns to equal ns is 3s = 21+s − 1 — the same equation that determined θ*, with roots s = 0 and s = 1. So "θ* = 1" is a restatement of "n itself is asymptotically harmonic". If n_k really were a martingale, a positive martingale converges almost surely, being integer-valued it is eventually constant, and it enters a cycle — the conclusion of the conjecture itself (on the non-divergence side). In reality E[nk+1|nk] = nk + 1/3 is a submartingale, and the convergence theorem does not apply. For 3n−1 the sign reverses and formally it is a supermartingale — a positive supermartingale converges almost surely and should fall into a fixed point, yet 3n−1 has cycles of length 2 and 7, and 5, 7, 5, 7, … does not converge.

The very existence of the cycles is a proof that this process is not a supermartingale. The expectation E[2⁻ᵛ] = 1/3 is an average over initial values, not a conditional expectation along an orbit — starting from 5, v is 1 and 2 in turn, every time, with nothing left to average. A probabilistic argument that concludes Collatz would, by the same computation, conclude something false about 3n−1.paper

What the sign separating 3n+1 from 3n−1 decides, and what it does not, is in What the sign of 1/3 decides.

Suppose a cycle exists. Write the odd terms as n₁ → n₂ → … → nₐ → n₁, let b be the total number of halvings, and multiply all the relations together—

2ᵇ / 3ᵃ = ∏ (1 + 1/(3nᵢ))

Every nᵢ is at least the smallest element m, so b/a must be an extremely good rational approximation to log₂3 = 1.5849625…. Concretely, 0 < b/a − log₂3 ≤ log₂(1 + 1/(3m)).

So the problem becomes: "among the fractions falling in the interval (log₂3, log₂3 + ε], which has the smallest denominator?" — a classical Stern–Brocot tree question, solvable exactly. Put in the value from the era when the verification record was 2⁴⁰ and out comes b = 17,087,915, exactly the lower bound of Eliahou (1993)known.

The counting convention must be fixed. "Cycle length" has two conventions, and mixing them changes the order of magnitude.

The current verification record is 2⁷¹ (reported as Barina, 2025). Substituting it—

This lower bound has a staircase property. The places where the continued-fraction approximation improves are isolated, so as verification extends the bound stays put for a while and then jumps at a single point.

Whether the verification record is 2⁶⁸ or 2⁷¹, the bound is the same 186,265,759,595. The current bound holds until verification reaches 2^71.88; the moment it passes that, the bound jumps to 355,504,839,929 terms — the extension of verification needed is only 1.85×, and the gain in the bound is 1.91×. This is a matter of compute, not of mathematics.

By Terras's theorem, the v vector of the first k steps of an odd n is determined by n mod 2^(Σv), with each pattern corresponding to exactly one residue class. So—

"The number of odd n < 2⁷¹ that do not fall below their starting point during 44 steps" can be counted as an exact integer. The answer is 2,483,007,077,973,438,304. No randomness, no approximation.computation

But only up to 44 steps. ⌊70 / log₂3⌋ = 44. Beyond that, Σv exceeds 70 and n mod 2⁷¹ no longer determines the v vector.

Beyond 44 steps the exact computation of the probability can continue (the figure above goes to 1200 steps). The decay rate per step converges to the theoretical e⁻ᴵ = 0.946505. That the prefactor of this tail is asymptotic to a 1-periodic function (the stopping-time tail theorem) is in the entrance article. But the claim about a count — "how many of the n < 2⁷¹ survive" — stops at 44 steps.

This is where probabilistic methods stop on this problem. You can compute that "the expected number of survivors falls below 1". But that is not a proof of nonexistence. A measure-zero set is invisible to any analytic method. Tao's 2019 result stops at "almost every" because of the same wall.

known The wall of §06 has a name. "Almost every orbit attains an almost bounded value" (Tao 2019) is a claim about a positive proportion of orbits, within reach of tools that take averages. "There is no cycle" and "no orbit diverges" are claims about the absence of a zero-proportion defect. A counterexample, if it existed, could have density 0, and something of density 0 is not seen by any correlation, any moment, or any asymptotic formula. Knowing that the orbits are perfectly regular by every measure would not give the conjecture.

The Riemann hypothesis has the same shape. Λ ≥ 0 (Rodgers–Tao 2018) is the positive-proportion claim "the zeros are not equally spaced in local average", and it is proved. Λ ≤ 0 (the Riemann hypothesis) is "not one zero lies off the critical line", the absence of a zero-proportion defect, and it is open. The same tools proved the one and did not prove the other; only one side could be proved because only one side is a claim about the crowded part. The five problems lined up in this shape are in Problems remain only at the exact edge.

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

Most of this article is not new. The tail exponent 1 is Lagarias–Weiss; the cycle lower bound is Eliahou's method itself, and all that is new is the number substituted in. The three-line proof of uniqueness was not found in the literature, and nothing more than "not found in the literature searched" is claimed. If you know of a prior appearance, please say so on the board.