Always One, So Far

Halve it if it is even; triple it and add one if it is odd; repeat. Every starting number below 271 has been checked by computer, every one reaches 1, and nobody can prove that all of them must. Run the rule on a number of any size, watch the last k binary digits of a number decide its first k moves, change the 3 to a 5 and watch the orbits fly off, and check every possible loop up to length 20 yourself. Then a ledger of what is proven, what was only computed, and what is merely believed.

if n is even: n → n / 2
if n is odd:  n → 3n + 1
repeat until you reach 1

Start at 6 and you get 6, 3, 10, 5, 16, 8, 4, 2, 1. Start at 7 and you wander for sixteen steps, as high as 52, and still come home to 1. Start at 27 and you go up to 9,232 and take 111 steps. The claim that every positive whole number ends at 1 is the Collatz conjecture. It was being passed around by the 1950s, it has been checked by computer on every starting number below 271, about 2.4 × 1021 of them, and it is not proven. This page lets you run it, then shows you three of the actual reasons it is hard, each one as something you can operate rather than take on trust.

The machine

Type any positive whole number, of any size. The page does exact arithmetic on it (no rounding, however many digits), and draws the whole journey on a logarithmic scale, so a number with a hundred digits and a number with two share one picture.

the number at each step   the coin-flip prediction

The coin-flip line. Pretend each number's parity is a fair coin. An odd step multiplies by about 3 and must be followed by a halving (3n+1 is always even), so it is worth ×3/2 over two steps; an even step is worth ×1/2. Half of each, on average, gives ×(3/2 · 1/2)1/2 = ×0.866 per odd-or-even decision, or about one decimal digit lost every 24 steps of the rule. The dashed line is that prediction. Try the random 100-digit number a few times: the model says about 2,400 steps, and the real journeys land around it. Then try 27, 871 and 8,400,511, which were chosen because they overshoot it badly (8,400,511 is the slowest start below ten million), and 2¹⁰⁰ − 1, which climbs for a hundred decisions in a row before it can fall. The average says the numbers must come down. The average is not a proof, because the numbers are not coins, and the next section shows you exactly where they are not.

The last k bits decide the first k moves

Write your number in binary. It turns out that its last k binary digits completely decide its first k odd-or-even decisions (counting the shortcut step, n → (3n+1)/2 for odd n, as one decision). Two numbers that end in the same k bits make the same k moves, however different they are otherwise. And every one of the 2k possible up-down patterns is made by exactly one ending. This was proved by Riho Terras in 1976. Here it is on your number and a stranger with the same ending:

▲ an odd step, n → (3n+1)/2   ▼ an even step, n → n/2. The stranger is your number plus a random multiple of 2k.

That bijection is why the coin-flip model is almost right: across all numbers, the up-down patterns are exactly as evenly spread as coin flips. It is also a weapon. Some endings force a number below where it started within k moves (the pattern has too many ▼ to recover), and for those endings the question is settled for every number that has them, provided every smaller number is already known to reach 1. Computer searches use exactly this to skip most numbers without running them. Here is how much is left after you sieve by the last k bits:

share of numbers not yet forced down, by k (log scale)

It goes to zero, which is a proof (Terras's, and C. J. Everett's in 1977) that almost every number eventually drops below where it started. But "almost every" is a statement about proportions. A set of exceptions can have proportion zero and still be infinite, and one exception is all a counterexample needs. And "drops below itself" is not "reaches 1": it only helps if every smaller number is already known to be fine, which is the induction nobody can close.

Change one number

If the coin-flip drift were the whole story, any rule with a downward drift would behave. It does not. Keep ×3 but subtract 1 instead of adding it (which is the same as running the real rule on the negative numbers: put in −n and every value comes out negated). The drift is identical. Then change the 3 to a 5 or a 7, which tips the drift upward.

the odd step

With 3n − 1 every orbit still comes down, as the drift promises, but it comes down into one of three loops, not one: 1 → 2, a loop of five through 5, and a loop of eighteen through 17. The drift cannot tell those apart from a world with a single loop. So whatever rules out a second loop for 3n + 1, if anything does, is not the average; it is something specific about +1 that nobody has found. With 5n + 1 the drift points up, and most starts climb past 1030 (where the page stops watching), while a few fall into small loops. Nobody has proved that 7 escapes under 5n + 1 forever. It is believed, for the same coin-flip reason, and unproven for the same reason as everything else here.

Every possible loop, up to length k

Here is a way to hunt for a second loop that does not involve running any starting number at all. Suppose some n returns to itself after k shortcut steps, j of them odd. Unwinding the rule gives one exact equation:

n = c / (2k − 3j)

where c is a whole number fixed by the up-down pattern (each odd step at position i contributes 2i, tripled by every odd step after it). So a loop of length k exists exactly when, for one of the 2k patterns, the division comes out whole. No searching of starting values, no ceiling, no "we stopped watching": for a given length it is a finite check that covers every whole number at once, positive and negative. Take the pattern ▲▼: k = 2, j = 1, c = 1, and n = 1 / (4 − 3) = 1, the loop 1 → 2 → 1. Now let your browser try every pattern:

Up to 16 takes a moment; 20 checks two million patterns and takes a few seconds.

The negative numbers have their three loops (the 3n − 1 loops above, negated), and the positive numbers have one. Published work goes enormously further, with the same equation plus a great deal of number theory. Simons and de Weger (2005) ruled out every other loop built from up to 68 climbs-then-falls, and Hercher (2023) pushed that to 91. Measured in plain length, Lagarias's 2010 overview already gives 10,439,860,591 shortcut steps as the least any other positive loop could have, and Hercher shows that once every number below 3 × 269 is verified, any other loop must contain more than 1.375 × 1011 odd numbers. Bařina's 2025 verification passed that threshold. (Putting those two together is our reading of the papers; we did not find the combined statement in print.) The same equation explains why that bound is so large: a loop needs 2k and 3j to be very close, which ties the length of any loop to how well fractions approximate log23, and those approximations are famously poor.

What is known, sorted by how it is known

Computed (true for every case checked, and only those)

  • Every starting number below 271 (about 2.4 × 1021) reaches 1. David Bařina, 2025, after a computation that has run for years; his project page says it is still running.
  • No other loop among the positive numbers shorter than 10,439,860,591 shortcut steps (Eliahou's method, quoted in Lagarias 2010), and none made of 91 or fewer climbs-then-falls (Hercher 2023).
  • On this page: every loop of length up to 20 in your browser, over all integers; up to 24 in the verifier.

Proven (for all numbers, forever)

  • The last k bits of n decide its first k moves, and every up-down pattern occurs exactly once (Terras 1976; Everett 1977). You checked it above for k up to 16.
  • Almost every number (proportion 1) eventually falls below its starting value (Terras 1976; Everett 1977).
  • Almost every orbit (in a logarithmic sense of "almost") falls below any bound that grows, however slowly: below log log log n, say (Terence Tao, 2019; published 2022). This is the strongest result of its kind and it still says nothing about reaching 1.
  • At least x0.84 of the numbers below x reach 1, for all large enough x (Krasikov and Lagarias 2003). Compare x itself: this is far from all of them.
  • Widen the family slightly and the question becomes undecidable: for maps that multiply by a fraction chosen by the remainder of n, there is no general procedure that can settle every question of this kind (John Conway 1972; Kurtz and Simon 2007). This says nothing about 3n + 1 itself; it says the tools cannot be completely general.

Believed (heuristic, not proven)

  • Every orbit reaches 1: the coin-flip drift you saw above, plus the enormous verification.
  • Under 5n + 1, most orbits, including the one from 7, grow without bound. Kontorovich and Lagarias call proving even one such divergence an open problem.
  • On the negative numbers, the three loops through −1, −5 and −17 are all there are (a conjecture recorded by Lagarias; we found no published computation bounding it).

Open

  • Whether every positive number reaches 1. Equivalently: is there no other loop, and no orbit that climbs forever? Either would be a counterexample, and both are unresolved.
  • Paul Erdős, as quoted by Lagarias in 1985: “Mathematics is not yet ready for such problems.” Prizes have included $500 from Erdős and, since 2021, 120 million yen from the Japanese company Bakuage.

The check, run in your browser on load

Every number on this page is computed live by collatz.mjs, which you can read. A separate program, verify-collatz.mjs, imports those same bytes and checks each function against a deliberately naive second implementation: the step counts and peaks of every start below 100,000 (and the slowest start below ten million, found independently), the Terras bijection for every k up to 20, the survivor counts against all 3,508 terms of OEIS A076227, the census of the four rules, and the loop search to length 24 by an independent method (the divisibility test done in modular arithmetic instead of exact fractions). It needs only Node: run it inside a copy of this site's source, or on its own in an empty folder, where it downloads the engine from this page and the OEIS terms from oeis.org. It prints 30 checks and takes about twenty seconds.

What this page does not do: it does not add to the evidence for the conjecture. Checking lengths up to 20 in a browser, or 24 in the verifier, is far below what is published. The checks are here so that every claim the page makes about what you are looking at is one you can reproduce.

Sources

  1. J. C. Lagarias, “The 3x+1 problem and its generalizations”, American Mathematical Monthly 92 (1985) 3–23. Web edition: the names, the history, the Erdős remark, the prizes, the negative-number loops.
  2. J. C. Lagarias, “The 3x+1 problem: an overview”, in The Ultimate Challenge: The 3x+1 Problem (AMS, 2010) 3–29. arXiv:2111.02635: the 10,439,860,591 loop-length bound and 2m − 1 climbing for m steps.
  3. J. C. Lagarias, annotated bibliographies, arXiv:math/0309224 and arXiv:math/0608208.
  4. R. Terras, “A stopping time problem on the positive integers”, Acta Arithmetica 30 (1976) 241–252; C. J. Everett, “Iteration of the number theoretic function f(2n) = n, f(2n+1) = 3n+2”, Advances in Mathematics 25 (1977) 42–45. The bijection as stated here is Theorem 2.1 of A. Kontorovich and J. C. Lagarias, arXiv:0910.1944, which also records 5n + 1 and the orbit of 7 as open.
  5. T. Tao, “Almost all orbits of the Collatz map attain almost bounded values”, Forum of Mathematics, Pi 10 (2022) e12. arXiv:1909.03562.
  6. I. Krasikov and J. C. Lagarias, “Bounds for the 3x+1 problem using difference inequalities”, Acta Arithmetica 109 (2003) 237–258. arXiv:math/0205002.
  7. J. H. Conway, “Unpredictable iterations”, Proc. 1972 Number Theory Conference, Boulder, 49–52; S. A. Kurtz and J. Simon, “The undecidability of the generalized Collatz problem”, TAMC 2007, LNCS 4484, 542–553.
  8. J. Simons and B. de Weger, “Theoretical and computational bounds for m-cycles of the 3n+1 problem”, Acta Arithmetica 117 (2005) 51–70; C. Hercher, “There are no Collatz m-cycles with m ≤ 91”, Journal of Integer Sequences 26 (2023) 23.3.5, arXiv:2201.00406.
  9. D. Bařina, “Improved verification limit for the convergence of the Collatz conjecture”, Journal of Supercomputing 81 (2025) 810, doi:10.1007/s11227-025-07337-0; the live project page.
  10. OEIS A076227, number of surviving Collatz residues mod 2n. E. Roosendaal, On the 3x+1 problem (the records; 8,400,511 is the slowest start below 107).
  11. Bakuage Co., the Collatz prize (announced 7 July 2021).