The Collatz Conjecture: Current Research and Approaches

March 29, 2025

The Collatz conjecture asks whether a tiny arithmetic recipe always takes a positive whole number to 1. Divide an even number by 2; multiply an odd number by 3 and add 1; then repeat. Starting with 6 gives 6 → 3 → 10 → 5 → 16 → 8 → 4 → 2 → 1. The calculation is elementary. Proving that every positive integer eventually reaches the same destination remains an open problem.

This guide explains the rule, checks concrete examples, and separates the different kinds of progress made on the problem. Computer verification, statements about most starting numbers, and restrictions on possible cycles answer different questions. Understanding their scope is essential for reading a research announcement or evaluating a proposed proof.

Try the 3n + 1 rule

Collatz sequence explorer

Choose a positive whole number and follow its exact sequence, one step at a time.

If the number is even, divide it by 2. If it is odd, multiply it by 3 and add 1. This explorer stops at the first 1.

JavaScript and a browser with BigInt support are needed to use this explorer. The controls stay disabled until it is ready.

Sequence results

Steps performed
Total stopping time: Steps to reach 1 for the first time
Largest value observed: Includes the starting number
First value below the start

Inspect the exact trajectory


Step	Exact value

All sequence values use exact integer arithmetic. A step limit is only a computing limit; reaching it does not show that a sequence never reaches 1. Testing examples does not prove the Collatz conjecture for every positive integer. This tool sends no data and stores nothing.

What is the Collatz conjecture?

Choose a positive integer n. If it is even, replace it with n/2. If it is odd, replace it with 3n + 1. Applying this operation repeatedly produces an orbit, also called a trajectory or Collatz sequence. The conjecture says that each positive starting integer eventually has 1 somewhere in its orbit. It is also called the 3n + 1 problem or the Syracuse problem.

C(n) = n/2 when n is even
C(n) = 3n + 1 when n is odd

The word “eventually” matters. It does not promise a decrease at every step, a common number of steps, or a maximum height shared by all sequences. Each starting value is allowed its own finite journey. Once 1 appears, continuing the rule produces 1 → 4 → 2 → 1 repeatedly. For calculations in this guide, we stop at the first 1 unless we explicitly discuss the continuing cycle.

The domain also matters. Zero stays at zero, and negative integers have other behavior. These observations do not disprove the conjecture because its claim concerns positive integers. Replacing the multiplier 3 with another number produces a different problem, even if the resulting calculation looks similar.

Three conventions that count different steps

Research papers often combine operations to remove predictable intermediate values. We will use C for the ordinary map above. A common shortcut map, T, divides an even input by 2 but sends an odd input directly to (3n + 1)/2. This is valid because 3n + 1 is even whenever n is odd. One odd T-step therefore combines two C-steps.

A second shortcut keeps only odd values. For an odd positive integer n, first calculate 3n + 1, then divide by 2 as many times as possible. Call the resulting odd number S(n). The notation v₂(m) means the number of factors of 2 in a positive integer m, so S(n) = (3n + 1)/2^{v₂(3n + 1)}. For instance, 3 × 13 + 1 = 40 = 2³ × 5, giving S(13) = 5.

All three formulations ask the same reachability question, but they have different clocks. Their terminal behavior is C: 1 → 4 → 2 → 1; T: 1 → 2 → 1; and S: 1 → 1. A claim about “cycle length” or “111 steps” is incomplete until the author specifies which map is being counted.

Worked examples: height, descent, and stopping time

Start with 7 and apply the ordinary rule carefully. The next value is 22 because 7 is odd. Halving 22 gives 11, which is odd again, so the sequence climbs to 34. It eventually reaches 52 before its first sustained-looking descent. The complete stopped trajectory is:

7 → 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1

There are seventeen displayed values and sixteen arrows. Thus the first arrival at 1 takes sixteen ordinary steps. The first value below the starting value 7 is 5, reached after eleven steps. Its peak before the first 1 is 52, attained at step five when the starting value is labeled step zero.

Every point is an exact integer from the worked sequence. Connecting segments help show the order; the map itself is defined only at the discrete steps.

Stopping time is not total stopping time

The stopping time is the first positive step at which the orbit falls strictly below its starting value. The total stopping time is the number of steps needed to reach 1 for the first time. These definitions use C throughout the following table. A starting value of 1 has total stopping time zero, but no stopping time in the first sense because positive integer values cannot fall below 1.

Exact examples, stopping at the first 1
Start	First step below start	Steps to first 1	Peak before first 1
1	Never	0	1
2	1	1	2
3	6	7	16
6	1	8	16
7	11	16	52
9	3	19	52
19	6	20	88
27	96	111	9,232
97	3	118	9,232

The trajectory starting at 27 illustrates the danger of judging an orbit early. It first falls below 27 at step 96, when it reaches 23, and later reaches 1 at step 111. Its maximum is 9,232 at step 77. Yet 97 falls below its own starting value after only three steps, through 292 and 146 to 73, while taking longer overall to reach 1. A quick initial descent does not determine the remaining journey.

Recording the peak only through the first 1 is deliberate. If the sequence continues, a start of 1 or 2 eventually visits 4. Without that stopping convention, the first two peak entries would be different. Exact definitions prevent a perfectly correct computation from being compared with the wrong statistic.

Why is the Collatz conjecture so hard to prove?

One natural plan is strong induction: assume all positive integers smaller than n reach 1, then show that n reaches a smaller integer. This works immediately when n is even. For an odd n greater than 1, however, the next value 3n + 1 is larger than n, so the induction hypothesis cannot be applied to that value. The challenge is to establish an eventual descent for every n, regardless of the intervening increases.

In fact, proving that every n greater than 1 eventually drops below n would settle the conjecture. Once that property is available, strong induction completes the argument. Conversely, if a number reaches 1, it certainly drops below its start unless it started at 1. This is a useful reformulation because it identifies the exact missing step rather than declaring induction an unsuitable method.

An infinite family that really can be handled

Suppose n = 4k + 1 with k at least 1. Three ordinary steps give 4k + 1 → 12k + 4 → 6k + 2 → 3k + 1. The final value is smaller than the starting value because 3k + 1 is less than 4k + 1. Therefore, if a least positive counterexample existed, it could be neither even nor congruent to 1 modulo 4. It would have to be congruent to 3 modulo 4.

This calculation eliminates infinitely many candidates for the least counterexample. It does not prove that every remaining candidate falls into the eliminated classes in a helpful way. Nor does it independently prove convergence for every number of the form 4k + 1: the smaller value reached still needs to be known to converge. The distinction between reducing a case and finishing it is central to valid arguments.

There are two ways a counterexample could behave

A bounded orbit of positive integers can visit only finitely many values. It must eventually repeat one. Because the rule is deterministic, repeating a value repeats the entire future trajectory, creating an eventual cycle. If that cycle contains 1, the conjecture holds for that starting point; a genuinely different positive cycle would disprove it.

An orbit with no repeated value must eventually leave every finite set. For any chosen bound B, there are only finitely many positive integers at most B, and each can appear at most once. After their last appearances, the orbit stays above B. Consequently, a nonrepeating orbit tends to infinity, although individual steps can still decrease. Excluding other positive cycles and excluding such divergent orbits are separate requirements of a complete proof.

Current research: what has actually been established?

Computational verification reaches a huge finite range

Barina’s published 2025 paper reports verification for all positive starting integers below 2⁷¹. The project log dates that milestone to January 15, 2025. At the October 5, 2026 check, its live page reported a larger completed bound: n < 2075 × 2⁶⁰, or n < 2,392,312,122,059,207,475,200. This is a dated report of the project’s contiguous verified range, not a claim that no other numerical experiment has tested larger individual starts. See Sources 1 and 2.

The distinction between a contiguous range and a selected large example is important. Verifying one enormous integer does not verify its smaller neighbors. Conversely, range verification need not trace each start independently to 1. Previously established cases and mathematically justified reductions can remove duplicated work. A rigorous shortcut is still a check; an unsupported assumption about an untested trajectory is not.

No finite bound exhausts the positive integers. If every number below B has been verified, infinitely many starts remain at or above B. The result rules out small counterexamples and supplies valuable information for theoretical arguments, but another million-fold increase in B would leave the same logical gap between a finite statement and a universal one.

A 2026 algorithmic development

Vigleik Angeltveit’s February 11, 2026 arXiv preprint proposes a more efficient way to verify a finite range. It groups starting values by shared low binary digits and builds a recursive search, discarding cases through justified descent and path-merging tests. The paper reports that the fraction requiring explicit checking decreases as the target range grows. This is an algorithmic advance toward larger finite checks, rather than a new universal convergence proof or a claim that its projected larger ranges have already been verified. See Source 14.

Classical results about descent

Terras and, independently, Everett established that almost every starting integer eventually falls below its initial value, in the sense of natural density. Korec later obtained descent below n^θ for each fixed θ greater than log(3)/log(4), approximately 0.79248, again for a natural-density-one set. The strict inequality on θ matters: writing the threshold itself as an established exponent would strengthen the claim. Sources 3 and 4 identify the original work.

Natural density one means that the fraction of successful starts among 1 through X tends to 1 as X grows. It does not mean that the exceptional set is empty or finite. For comparison, the perfect squares form an infinite set whose fraction among the first X integers tends to zero. This example is about the meaning of density, not a suggestion that squares are Collatz counterexamples; powers and squares can be tested or analyzed on their own merits.

Tao’s theorem, with its quantifiers intact

Let C_min(n) be the smallest value anywhere in the orbit starting at n. Tao’s theorem states that, for every prescribed function f(n) tending to infinity as n tends to infinity, C_min(n) < f(n) for a set of n of logarithmic density one. The work first appeared in 2019 and was published in 2022. The arXiv version checked for this guide is v7, dated July 16, 2026. See Source 5, Theorem 1.3.

Logarithmic density weights an integer n by 1/n. For a set A, examine the sum of 1/n over members of A up to X, divided by the sum of 1/n over all positive integers up to X. Density one means that ratio approaches 1. Equal multiplicative ranges receive comparable weight: the intervals from X to 2X and from 2X to 4X contribute approximately the same total weight when X is large.

A threshold that tends to infinity can grow very slowly, but it cannot be replaced by the constant 1. Merely being unbounded is also insufficient: a function that repeatedly returns to 1 does not tend to infinity. The theorem supplies neither a stated percentage for an arbitrary finite interval nor a conclusion that almost all trajectories eventually reach 1.

Tao’s approach studies distributions of the places where orbits first cross size thresholds, rather than predicting each orbit individually. Its technical machinery includes estimates for a random walk on finite groups built from powers of 3. These 3-adic calculations should not be confused with the separate 2-adic extension discussed below. The author’s explanation in Source 13 describes how controlling distributions across scales makes a descent argument possible.

A more recent natural-density direction

Manuel Inselmann’s 2024 arXiv preprint studies trajectory estimates for the shortcut map T. Its Corollary 1.4 states that, for every fixed positive ε, a natural-density-one set of starts m reaches a value at most m^ε at a specified time proportional to log m. The cited record has no journal reference at this check, so this guide identifies it as a preprint. Its threshold and its notion of density should both be retained when comparing it with Tao’s theorem. See Source 6.

For example, choosing a very small positive exponent still produces a threshold that eventually grows without bound. One cannot set ε equal to zero after using a statement proved separately for each ε greater than zero. A parameter that may be arbitrarily small is not automatically allowed to take its limiting value. This logical issue arises throughout analysis, well beyond the Collatz problem.

Counting numbers that really reach 1

A different result by Krasikov and Lagarias, published in 2003, gives a lower bound of at least X^0.84 convergent starts below X for sufficiently large X. Their computer-assisted argument uses difference inequalities. It concerns how many integers reach 1, rather than establishing a universal restriction on how quickly an individual divergent orbit could grow. See Source 7.

Dividing X^0.84 by X gives X^−0.16, which tends to zero. Thus that lower bound guarantees many convergent integers without guaranteeing a positive limiting proportion. There is no contradiction with much larger observed proportions: a proven lower bound can be far below the true quantity. It is a floor on what has been established, rather than an estimate of what experiments must display.

Research approaches: probability, arithmetic, and dynamics

Why a downward-drift heuristic is persuasive

Consider an odd-to-odd step S(n). Its size is approximately multiplied by 3/2^a, where a counts the factors of 2 in 3n + 1. Among odd residue classes, half give a = 1, a quarter give a = 2, an eighth give a = 3, and so on. These proportions can be checked by solving congruences modulo successive powers of 2: multiplication by the odd number 3 permutes the residue classes.

Under a model that treats successive exponents as independent samples from those proportions, the mean exponent is 2. The mean logarithmic size change is therefore log 3 − 2 log 2 = log(3/4), a negative number. This suggests a tendency to shrink over many odd-to-odd steps. The calculation concerns an average of logarithms, or geometric growth, rather than an assertion that each actual step multiplies its input by exactly 3/4. Lagarias discusses this classical heuristic in Source 8.

Here is a small independent illustration of that distinction. A process that alternately multiplies by 3 and by 1/4 shrinks by a factor of 3/4 over each pair. Yet its arithmetic mean multiplier is (3 + 1/4)/2 = 1.625, greater than 1. For repeated products, adding logarithms captures the cumulative effect; averaging the raw multipliers answers a different question.

The missing ingredient in the Collatz heuristic is a justification for applying a random model to every deterministic trajectory for as long as necessary. Once a start is fixed, every later parity and division exponent is fixed too. A correct distribution across starting residues at one stage does not by itself guarantee independence along one infinite orbit. Exceptional trajectories are exactly where a universal conjecture can fail.

What cycle equations can tell us

Suppose a hypothetical odd-only cycle contains r odd values x₀, …, x_r−1. Let a_i count the divisions by 2 after multiplying x_i by 3 and adding 1, and let A be the sum of these exponents. Every transition satisfies 2^aᵢx_i+1 = 3x_i + 1, with indices wrapping around at the end.

Divide each transition by x_i and multiply the equations. The ratios of successive x-values cancel around the closed loop, leaving the exact identity below. This short derivation explains why powers of 2 and 3 become central to cycle research.

2^A = ∏_i=0^r−1 (3 + 1/x_i)

If m is the smallest odd value in the cycle, each factor exceeds 3 and is at most 3 + 1/m. Taking base-two logarithms gives r log₂3 < A ≤ r log₂(3 + 1/m). Dividing by r shows that A/r lies in an interval of width log₂(1 + 1/(3m)), which becomes small as m grows. Hence the integers A and r must produce a close rational approximation A/r to log₂3. Additional arithmetic conditions are needed to turn a candidate ratio into an actual cycle.

Bounds excluding short cycles are therefore meaningful, but their units must be clear. The odd-only length here is r, whereas the ordinary cycle length is r + A. Some specialized results count monotone blocks instead. Quoting an enormous length without specifying the map or the counted objects can turn a valid theorem into a misleading statement. This guide avoids presenting an unverified numerical cycle-length record; Source 9 explains the classical arithmetic framework.

What 2-adic dynamics adds, and what it does not

The 2-adic integers organize numbers by congruences modulo 2, 4, 8, and higher powers of 2. Two values are close in the 2-adic sense when their difference is divisible by a large power of 2. For example, 1 and 1 + 2²⁰ agree modulo 2²⁰, even though they are far apart on the ordinary number line. The notion of closeness is different, so conclusions about size cannot be transferred casually between the two settings.

Odd 2-adic integers have multiplicative inverses; 2 does not have an inverse inside the 2-adic integers. Indeed, reducing a proposed equation 2x = 1 modulo 2 would give 0 = 1. The shortcut map T nevertheless extends because its even branch divides an even input by 2, while its odd branch divides the even quantity 3n + 1 by 2. No inverse of 2 within the integer ring is required.

Bernstein and Lagarias describe a conjugacy between this extended shortcut map and a binary shift. Parity itineraries provide a way to encode points, and the dynamics has strong mixing properties for the natural 2-adic measure. These are results about T and its larger domain, not an automatic theorem about the ordinary map C on positive integers. See Source 10.

The distinction in the domain is decisive. Each single 2-adic point has measure zero, since it lies in residue classes of measure 2^−k for every k. A countable union of such points still has measure zero, so all ordinary positive integers together form a measure-zero subset. An “almost every 2-adic point” theorem may leave that entire subset untreated. Source 11 explicitly warns that ergodicity does not settle the integer problem.

Why extending the system can still help

A larger setting can expose structure that is hard to see in a list of integers. Parity encodings, inverse branches, and arithmetic restrictions can suggest precise lemmas to prove back in the original domain. The value of the extension comes from a successful bridge between settings. A result proved for a typical point in a larger space becomes relevant to every positive integer only after the bridge itself is justified.

The same discipline applies to related computational problems. Certain generalized Collatz-type systems can encode undecidable behavior, as discussed in Lagarias’s overview, Source 12. This does not establish that the original conjecture is undecidable or independent of a particular axiom system. Similar-looking rules may have substantially different mathematical properties.

How to run a useful Collatz experiment

A careful experiment begins with a specification. State the input range, the map being used, whether the initial value counts as step zero, and when the run stops. Record at least the number of steps, the largest intermediate value, and whether 1 was actually reached. An experiment should distinguish a completed trajectory from one that was interrupted after a chosen time or step limit.

Use exact integer arithmetic. A sequence can rise far above its starting value, so a numeric type that safely stores the input may still overflow during 3n + 1. Floating-point arithmetic brings a different risk: once integers exceed the format’s exact range, parity tests or additions may no longer represent the intended integer calculation. An arbitrary-precision integer type is a straightforward choice for a small educational implementation.

Reject inputs that are not positive integers.
Set the current value to the input, the step count to zero, and the peak to the input.
If the current value is 1, report a completed run.
Otherwise perform exactly one C-step, increase the count, and update the peak.
Repeat, while treating any imposed resource limit as an inconclusive result rather than a counterexample.

Check the implementation against short hand-worked trajectories before exploring large values. The starts 1, 2, 3, and 7 catch several common counting mistakes. Testing 27 checks that a long upward excursion is handled correctly. A second implementation using the odd-only map can provide a useful cross-check, provided the skipped divisions are restored when comparing ordinary step counts.

Memoization stores results for values already analyzed. For example, once the complete trajectory from 5 is known, a later run that reaches 5 can reuse its remaining step count. But if the goal is to measure a peak, storing only the remaining length is insufficient; the peak of the stored tail must also be considered. Optimizations should preserve each reported statistic, not merely the eventual destination.

Collatz makes a good programming exercise because input validation, loops, integer representation, and testing all matter. Students reviewing those foundations can use the AP Computer Science A study guide alongside this investigation. The experiment is enrichment in mathematical computing; it is not a claim that proving Collatz is an exam requirement.

How to assess a claimed Collatz proof

Begin by identifying the exact conclusion. A paper may prove descent for a restricted residue class, a finite verified range, or a statement holding outside a small exceptional set. Each can be worthwhile. Problems arise when the conclusion is silently strengthened to cover all starts, all times, or a threshold that the theorem does not allow.

Look next for the place where the proof handles a potentially exceptional orbit. Statements such as “on average there are enough divisions” need a theorem whose hypotheses apply to that orbit. A graph of many successful trajectories is evidence about those trajectories. A computer search with a timeout cannot certify that an unfinished one never reaches 1. A pattern seen in a fixed number of binary digits cannot be assumed to control arbitrarily many future steps.

Check that the argument has not assumed convergence in disguise. Defining a function as “the number of steps until 1” and then treating it as finite for every start already uses the desired conclusion. It is legitimate to define the total stopping time as infinity when an orbit never reaches 1, but every later algebraic manipulation must respect that possibility.

Finally, trace references to their actual statements and publication status. A recent revision date does not turn an old computational bound quoted inside a paper into the current record. An arXiv upload is a public research manuscript, not by itself a guarantee of correctness or a declaration of community acceptance. Clear definitions and a checkable chain of reasoning matter more than a dramatic title.

Practice questions with explained answers

Use the ordinary map C unless a question specifies otherwise. Try the arithmetic first, then explain what the calculation does and does not establish. That second step is where this topic becomes an exercise in mathematical reasoning rather than simply repeated multiplication and division.

1. Trace the start 12

Question: Write its sequence through the first 1. Find the total stopping time, the ordinary stopping time, and the peak.

Answer: The sequence is 12 → 6 → 3 → 10 → 5 → 16 → 8 → 4 → 2 → 1. There are nine arrows, so the total stopping time is nine. The first step already reaches 6, which is below 12, giving ordinary stopping time one. The peak is 16. The initial decrease does not prevent the sequence from later rising above its start.

2. Compare ordinary and odd-only steps

Question: The odd-only sequence from 7 is 7 → 11 → 17 → 13 → 5 → 1. Why does it represent sixteen ordinary steps rather than five?

Answer: Each odd-only transition includes one multiplication-and-addition step followed by all necessary halvings. The halving counts are 1, 1, 2, 3, and 4. Adding these eleven divisions to the five odd operations gives sixteen. The first compressed transition, for instance, represents 7 → 22 → 11.

3. Prove convergence for powers of 2

Question: Explain why every start of the form 2^k, where k is a nonnegative integer, reaches 1.

Answer: For k greater than zero, each step halves the value and reduces the exponent by one. After k steps the value is 2⁰ = 1. When k is zero, the starting value is already 1. This proves an infinite family without requiring a separate computation for every member, illustrating the difference between an argument and a finite list of examples. It also proves that total stopping times are unbounded: choosing larger k gives arbitrarily many required steps. That fact is elementary, even though convergence for all other starts remains unresolved.

4. Find the gap in repeated “almost all” descent

Question: Suppose almost every start eventually drops below itself. Why can we not simply repeat that claim until every orbit reaches 1?

Answer: The smaller value reached might belong to the exceptional set. A statement that applies to most inputs does not automatically apply to the outputs of a deterministic process. Those outputs could be distributed very unevenly. A valid repeated-descent proof needs to control the new starting values or establish descent for every relevant value.

5. Interpret an unfinished run

Question: A program tests a start for one million steps and never encounters 1. Has it disproved Collatz?

Answer: No. It has shown only that this run did not reach 1 within that limit, assuming its arithmetic and implementation are correct. It might reach 1 later. A verified repeated cycle excluding 1 would be a counterexample. A proof that the orbit diverges would also suffice, but a long rising segment alone is neither of those certificates.

6. Check a proposed constant-threshold conclusion

Question: If a theorem allows every fixed ε greater than zero in the threshold n^ε, may we substitute ε = 0 to obtain a threshold of 1?

Answer: No. Zero lies outside the stated parameter range. The sets of successful starts and the size from which an estimate becomes useful can depend on ε. Passing to a limit requires an additional argument; it is not supplied by the phrase “every positive ε.” This is the same kind of quantifier care needed when interpreting an almost-bounded-orbit theorem.

Frequently asked questions

Has the Collatz conjecture been solved?

No accepted proof of the full positive-integer conjecture is known at this guide’s October 5, 2026 research check. There are substantial partial results and extensive numerical verification. A claim that a particular range or family has been settled should be read with its range and assumptions intact.

Did Terence Tao prove that almost all numbers reach 1?

No. The theorem discussed here concerns reaching below a prescribed threshold that tends to infinity, with exceptional starts measured using logarithmic density. Reaching the fixed value 1 is a stronger conclusion. The distinction is mathematical, not merely a cautious choice of wording.

Why is 27 used so often as an example?

Its small starting value leads to a comparatively long and high excursion: 111 ordinary steps to first reach 1 and a peak of 9,232. That makes it a useful demonstration that initial size, peak height, and total stopping time are different quantities. It is an example, not a universal worst case.

Does every even number have an easy proof?

An even number immediately reduces to a smaller number. That proves convergence if the smaller number is already known to converge. Repeated halving eventually reaches an odd number, so the unresolved behavior is concentrated in what happens after odd starts. The simple even step helps organize the problem without solving it alone.

Could a different positive cycle exist?

It has not been ruled out in complete generality. Any candidate must satisfy stringent arithmetic restrictions and avoid all starting values already verified to reach 1. A cycle in a changed rule, in negative integers, or in a larger number system would not automatically be a counterexample to the original positive-integer conjecture.

Can a bounded trajectory avoid both 1 and a cycle?

No. A bounded positive-integer trajectory has only finitely many possible states, so eventually one repeats. Determinism then forces a repeating tail. This finite-state argument is exact and requires no probabilistic assumption. The open question is whether a positive cycle other than the familiar one can actually occur.

Would faster computers eventually prove the conjecture?

Faster computers can enlarge finite verification ranges and assist with rigorously formulated mathematical arguments. Exhaustive checking of successive starts alone never completes an infinite list in finite time. A computer-assisted proof would need a justified reduction that covers all remaining integers, rather than merely a very large successful search.

The central lesson

The Collatz conjecture is a compact introduction to the difference between calculating, modeling, and proving. Exact sequences show what happens in particular cases. Density theorems control large sets with carefully stated qualifications. Arithmetic and dynamical methods reveal structure while leaving demanding bridges to cross. The full problem asks for one argument that covers every positive integer, including any exceptional start that ordinary experiments or averages might miss.

Sources and further reading

Primary research and author-maintained sources were checked on October 5, 2026. The live verification figure is a dated snapshot. Worked sequences, graph coordinates, elementary deductions, and practice answers were calculated for this guide. Historical surveys are used for their stated scope, not as evidence of a current computational record.

David Barina, Convergence verification of the Collatz problem. Live project status and dated verification milestones.
David Barina, Improved verification limit for the convergence of the Collatz conjecture, The Journal of Supercomputing 81, article 810, published May 2, 2025. Published 2⁷¹ result and verification algorithms.
Riho Terras, A stopping time problem on the positive integers, Acta Arithmetica 30 (1976), 241–252. Classical stopping-time research; the related Everett result is also summarized in Tao’s introduction.
Ivan Korec, A density estimate for the 3x + 1 problem, Mathematica Slovaca 44 (1994), 85–89. Power-threshold density estimate.
Terence Tao, Almost all orbits of the Collatz map attain almost bounded values, Forum of Mathematics, Pi 10 (2022), e12; arXiv v7, July 16, 2026. Definition 1.2 and Theorem 1.3 state the density convention and main theorem.
Manuel Inselmann, An approximation of the Collatz map and a lower bound for its average total stopping time, 2024 arXiv preprint, v3. Corollary 1.4 gives the stated natural-density descent result for the shortcut map.
Ilia Krasikov and Jeffrey C. Lagarias, Bounds for the 3x + 1 problem using difference inequalities, Acta Arithmetica 109 (2003), 237–258. Lower bound for the number of starts reaching 1.
Jeffrey C. Lagarias, A heuristic argument, from The 3x + 1 Problem and its Generalizations. Classical probabilistic heuristic and its limitations.
Jeffrey C. Lagarias, Are there non-trivial cycles?, from the same survey. Cycle equations and rational-approximation methods; historical numerical bounds are not treated as current records.
Daniel J. Bernstein and Jeffrey C. Lagarias, The 3x + 1 conjugacy map. Shortcut-map conjugacy and 2-adic dynamics.
Jeffrey C. Lagarias, Connections of the problem to ergodic theory. The measure-zero limitation for ordinary integers.
Jeffrey C. Lagarias, The 3x + 1 Problem: An Overview, a 2010 survey made available on arXiv in 2021. Research approaches, generalized problems, and the limits of undecidability analogies.
Terence Tao, Almost all Collatz orbits attain almost bounded values, author’s research explanation, September 10, 2019. Motivation, first-passage distributions, and the approach to the theorem.
Vigleik Angeltveit, An improved algorithm for checking the Collatz conjecture for all n < 2^N, arXiv preprint, February 11, 2026. Recursive binary search and sieving for finite-range verification.