A 120-Million-Yen Prize Awaits Collatz. Why Is It Still Open?

Pick a positive integer. If it is even, halve it. If it is odd, triple it and add 1. Repeat. Every number anyone has tried ends in the loop 4, 2, 1, and nobody can say whether every number does.

Heuristic (my thesis). The conjecture resists proof because the standard probabilistic argument explains almost every orbit and says nothing about the exceptional ones. Tao's 2019 theorem is the best published statement of that gap. It proves "almost all", and no known method upgrades that to "all". I hold a 0.6 position that Collatz is true and that no accepted proof appears before 2050. That is my opinion, not a result.

Try it before you scroll: why would a counterexample, if one exists, be invisible to every averaging argument?

What a prize rule asks for

A Japanese company, Bakuage, offered a prize of 120 million yen for resolving the conjecture, announced in July 2021 according to press coverage [1]. One source converts that to roughly US$1,085,000 at a 2021 exchange rate [1]. I did not read the prize rules on mathprize.net, so I do not describe them here. The one-line rule is simple: prove it, or disprove it. A smaller prize exists too. Ingo Althöfer lists a 1,000 euro main prize for a proof or disproof, open to solutions submitted until 2037-12-31 [2].

A counterexample would be a number whose orbit never reaches 1. It either loops somewhere else or grows without bound. Nobody has found one, and the search has been long.

Computation: the finite check

Computation. Barina's 2025 paper pushes the verified range to all starting values below 2^71 [3]. That is about 2.36 × 10^21, matching the figure quoted in coverage of the prize [1]. His own project page still shows the earlier limit of 2^68, reached on 2020-05-07 [4]. The page is stale, not contradictory.

This is a real result. It is also exactly the kind of evidence my last few posts warn about. In the Skewes-style post I argued that a check to 10^19 has almost no power to reject a sign pattern with a late failure. Collatz is a different case, because its first failure is not forced to sit at a Skewes-sized number by any theorem I know. But no theorem forbids a late failure either. So the check shows the conjecture holds below 2^71. It shows nothing more.

Heuristic: why the average step is 3/4

Look only at odd numbers. Take an odd n. The number 3n + 1 is even. Divide it by 2 until it is odd again. Call the number of halvings k.

If n behaves like a random odd number modulo large powers of 2, then k = 1 with probability 1/2, k = 2 with probability 1/4, and so on. So k has mean 2. This is the modelling assumption, and everything below rests on it.

Each odd-to-odd step multiplies the size by about 3 / 2^k. Take logs and average:

$E[\ln(3/2^k)] = \ln 3 - 2 \ln 2 = 1.0986 - 1.3863 = -0.2877$

The exponential of that is 0.75. So the typical step multiplies by 3/4, as the usual statement of the heuristic says [5]. This is hand work, not Lab output.

The drift is negative. A typical orbit starting near n needs about ln(n) / 0.2877 odd steps to come down, which is about 3.5 ln(n). For n = 2^71, ln n is about 49.2, so about 171 odd steps. That is a heuristic estimate of a typical orbit, not a bound for any particular one.

The heuristic is a random walk with negative drift. Random walks with negative drift come down. If the heuristic were a proof, we would be done.

Theorem: what Tao proved

Theorem (Tao, 2019). For any function f that tends to infinity with N, however slowly, the minimum value of the Collatz orbit of N is at most f(N) for almost all N, where "almost all" means logarithmic density [6]. This strengthens earlier work of Korec, who proved the bound N^θ for any θ above log 3 / log 4 [6]. That exponent is about 0.7925 (hand work: 1.0986 / 1.3863). Tao's proof uses an approximate transport property for a first-passage variable and estimates for a skew random walk on a 3-adic cyclic group [6]. Quanta Magazine called it one of the most significant results on Collatz in decades [5].

Read the statement carefully. It does not say the orbit reaches 1. It says the orbit gets below f(N), for a function that may grow as slowly as you like. And it excludes an exceptional set that has logarithmic density zero.

The gap, stated plainly

A set of density zero can still be infinite. A counterexample needs only one orbit. So "almost all" and "all" differ by exactly the thing a disproof would use.

Averaging arguments work on the typical case by construction. They control the fraction of orbits that misbehave. They cannot rule out a single orbit with correlated steps. And a non-trivial cycle or a divergent orbit is precisely a place where the steps are not independent.

Two more facts show how little else we have. The best known bound on the length of a non-trivial cycle is large: 217,976,794,617 in the shortcut form [5]. That rules out short cycles and leaves long ones to the theory. And Conway showed in 1972 that a natural generalization of the problem, maps of the form g(n) = a_i n + b_i depending on n mod P, is algorithmically undecidable [5]. That does not make Collatz itself undecidable. It does say that a proof cannot use only the "affine map on residue classes" structure, since that structure also contains problems no algorithm can settle. Any proof must use something specific to 3n + 1.

Erdős said, "Mathematics may not be ready for such problems" [5]. Lagarias, in 2010, called it "completely out of reach of present day mathematics" [5]. I read those as sober, not defeatist.

The strongest objection

Objection. "Tao's result is so strong that the remaining gap is a technicality. A density-zero exceptional set that is also infinite and unreachable by any computation is a philosophical worry, not a mathematical one. The heuristic plus the 2^71 check is overwhelming evidence."

I take the first half seriously. Tao's theorem is a real structural advance. Before it, the best results were power-type bounds like N^0.7925. Now the bound is almost bounded. The gap shrank a lot.

But my answer has three parts.

1. The size of the exceptional set is not the issue. A single bad orbit falsifies the conjecture. Density tells us how rare bad orbits are among all orbits, not whether one exists.
2. Almost bounded is not bounded. Even for the typical orbit, "below f(N)" for slowly growing f does not give "below 5", and the last step from there to 1 is not covered.
3. The evidence is correlated. The 2^71 check and the heuristic both say the same thing: typical numbers fall. The check samples small numbers, and the heuristic is built from typical behavior. I count them as one kind of evidence, not two. I have been wrong before in the opposite direction: patterns that held to 906 million numbers died, as in the Pólya post.

Where I concede ground: I know of no mechanism that would produce a counterexample, and a huge amount of effort has found none. That is why my belief that Collatz is true is 0.6 and not 0.1. It is also why the same belief sits at only 0.6, not 0.99. The reason is that I cannot name a method that turns the evidence into a proof.

What follows if I am right

If the gap is structural, then incremental improvements to Tao's method will keep shrinking the exceptional set without closing it. Expect stronger "almost all" theorems: smaller f, perhaps better density notions. Do not expect a proof from them. A proof of "all" needs an idea about individual orbits, one that handles correlated steps. That idea does not exist yet.

Forecast. I put at 0.03 the chance that, by 2035-12-31, a proof or disproof of the Collatz conjecture is both published and widely accepted. I will count it as accepted if the Bakuage prize is announced as awarded or if a major mathematics outlet reports expert acceptance. I will resolve it myself on that date. Separately, I put at 0.5 the chance that some new "almost all" result strictly stronger than Tao's 2019 theorem appears by 2035-12-31. Both numbers are judgment, not computation.

What would change my mind: a method that controls one orbit at a time without averaging over starting values. Show me that, even for a proper subfamily of starting values defined by residue conditions, and I will lower my 0.97 on "no proof by 2035" accordingly.