The Collatz Conjecture

Pick a starting value and watch where the rule takes it. Then look at the second picture, which is every starting value up to the limit at once.

Worked example — The Collatz Conjecture

Total stopping time	111
Highest value reached	9,232
Highest value ÷ starting value	341.93
Biggest gap to a neighbour	101
Halvings per tripling	1.7073
The orbit with its +1s deleted	0.8341
Longest orbit up to 10,000	261
Starting values up to 10,000 peaking at 9,232	1,579

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

27 takes 111 steps. Its neighbour takes ten. 🖖

Put 26 in the starting value. The orbit is 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1. Ten steps, and it never rises above 40.

Now put 27. It takes 111 steps and climbs to 9,232, which is 341.93 times where it started.

Nothing about 27 announces this. It sits between 26 and 28, and 28 is home in 18 steps; the Biggest gap to a neighbour row works that out as 101. Drag the starting value along and the dots in the lower panel never settle into a curve.

The sharpest step below 10,000 is 225, and six pairs of neighbours manage it. The first is 6,170 and 6,171: one takes 36 steps, the other 261.

Three quarters per tripling, and it keeps being right 🖖

An odd number becomes 3n + 1, which is even, so a halving always follows. That value is divisible by 4 half the time and by 8 a quarter of the time, so on average two halvings follow. One tripling and two halvings multiply by ¾.

Run that to the end and the number of triplings needed to bring n down to 1 should be roughly ln n ÷ ln(4/3), which is 3.4761 ln n.

The results table measures it against every starting value in the scan. Up to 10,000 the mean number of triplings is 28.21 and ¾ per tripling predicts 28.54. Up to 1,000 it is 19.67 against 20.57. Up to 100,000, 35.65 against 36.54.

The phrase carrying the whole argument is on average. The halvings along one orbit are not independent coin tosses; they are settled the moment you choose n, and nobody has found a way to say so and keep the answer. Paul Erdős, quoted by Jeffrey Lagarias in 1985: “Mathematics is not yet ready for such problems.”

Change the 3 to a 5 and the same rule traps 27 forever 🖖

Press 5n + 1 and set the start to 13. The orbit runs 13, 66, 33, 166, 83, 416, 208, 104, 52, 26, and then it is back at 13. Ten numbers, and it will go round them for as long as you let it. Nothing there is conjectured; it is a list you can check by hand in a minute.

27 is caught too, in a different loop of ten: 27, 136, 68, 34, 17, 86, 43, 216, 108, 54.

The heuristic says why. A tripling and two halvings is ¾; a quintupling and two halvings is 5/4, and a rule that multiplies by 5/4 on average has no reason to come down. Leave 5n + 1 selected and look at the lower panel, which has gone nearly empty. Of the first 10,000 starting values, 256 reach 1. Another 470 fall into a loop. The remaining 9,274 climb past 2⁵³, where the tool's arithmetic stops being exact and it stops following them.

So a proof of the Collatz conjecture has to be a proof that knows the difference between 3 and 5. As of 2025 the rule has been checked for every starting value up to 2⁷¹, about 2.36 × 10²¹ of them, by David Bařina on European supercomputers. The next one is still open.

Problem solved in full

Why 27 needs 41 triplings when the arithmetic says twelve 8 steps
The ¾ rule says an orbit should need about ln n ÷ ln(4/3) triplings to get home. For n = 27 that is between eleven and twelve. The tool says 41. Find where the extra twenty-nine came from, then use the same bookkeeping on the largest starting value anyone has ever checked.

Open this problem in the tool
1. From an odd n the next value is 3n + 1, which is even, so at least one halving follows. It is divisible by 4 half the time, by 8 a quarter of the time, by 16 an eighth of the time, and the expected number of halvings is the sum of k·2^−k, which is exactly 2. One tripling followed by two halvings multiplies by ¾.
  
  $E [v_{2} (3 n + 1)] = k \geq 1 \sum k 2^{- k} = 2 ⟹ \frac{3}{2 ^{2}} = \frac{3}{4}$
2. Multiplying by ¾ each time, the orbit should arrive at 1 once (¾)^o has cut n down to nothing. Solving for o gives ln 27 ÷ ln(4/3) = 11.46. Call it twelve triplings, and with two halvings apiece, about thirty-five steps altogether.
  
  $(\frac{3}{4})^{o} = \frac{1}{n} ⟹ o = \frac{ln 27}{ln ( 4/3 )} = \frac{3 . 2958}{0 . 28768} = 11.46$
3. Now load 27 into the tool. Total stopping time reads 111, and Halvings per tripling shows the split it came from: 41 triplings and 70 halvings. The prediction is out by a factor of 3.58. So 27 is not a typical starting value.
  
  $o = 41, e = 70, t = o + e = 111, \frac{41}{11 . 46} = 3.58$
4. Take the +1s out and the orbit becomes pure multiplication: 41 triplings and 70 halvings applied to 27. That product is 0.8341, and the tool prints it under The orbit with its +1s deleted. It would be exactly 1 if the +1s did nothing at all, and it lands below 1 because each +1 hands the halvings a little extra to remove. On a power of two, where no odd step ever happens, it is exactly 1.
  
  $L = n \cdot \frac{3 ^{o}}{2 ^{e}} = 27 \cdot \frac{3 ^{41}}{2 ^{70}} = 0.8341 < 1, 2^{k} \cdot \frac{3 ^{0}}{2 ^{k}} = 1$
5. Rearrange that. Writing L for the product, n·3^o ÷ 2^e = L, so taking logarithms base 2 gives e = o·log₂3 + log₂(n/L), and dividing by o gives the ratio the tool prints. For 27 that is 1.58496 + 0.12236 = 1.70732, which the tool rounds to 1.7073. For an orbit that reaches 1 the ratio can never fall below log₂3 = 1.5850, and the ones that come closest are those spending the most triplings for the size of their starting value. Below 100,000 nothing gets closer than 27 does.
  
  $\frac{e}{o} = lo g_{2} 3 + \frac{lo g _{2} ( n / L )}{o} = 1.58496 + \frac{lo g _{2} ( 27/0 . 8341 )}{41} = 1.58496 + 0.12236 = 1.70732$
6. That is what a long orbit is: one that runs short of halvings. Here are 27 and the three starting values that top the scan's three ranges, with 26 alongside for contrast. A typical starting value averages about two halvings per tripling; every record holder here runs between 1.70 and 1.74. The orbit is spending its triplings and not getting the divisions back.
  
  $n 2627871 6, 171 77, 031 t 10111178261350 o 2416596129 e 870113165221 e / o 4.0000 1.7073 1.7385 1.7188 1.7132$
7. Every starting value up to 2⁷¹ has been checked, about 2.36 × 10²¹ of them, by David Bařina on European supercomputers and published in 2025. Every one came home. How long should the journey be down there? ln(2⁷¹) is 49.213, so the heuristic gives 171 triplings and around 513 steps for a typical number of that size.
  
  $n = 2^{71} \approx 2.36 \times 1 0^{21} : o \approx \frac{71 ln 2}{ln ( 4/3 )} = \frac{49 . 213}{0 . 28768} = 171, t \approx 3 o = 513$
8. The numbers that make this problem interesting are not typical ones. Against the mean total stopping time in each scanned range — 59.60, 84.98 and 107.54 steps — the record holder runs 2.99, 3.07 and 3.25 times as long. Carry the last of those across and the slowest starting values near 2⁷¹ want something nearer 1,700 steps than 500. The ratio is still climbing at the largest range the tool scans, so 1,700 is a floor rather than an estimate, and every figure in this step is extrapolation from a heuristic. That is the only kind of statement anyone has about numbers that size.
  
  $\frac{178}{59 . 60} = 2.99, \frac{261}{84 . 98} = 3.07, \frac{350}{107 . 54} = 3.25 ⟹ 3.25 \cdot 513 \approx 1, 700$
Answer

41 triplings rather than twelve, because 27's orbit runs at 1.7073 halvings per tripling where a typical one runs at about 2. All that is left to bring 27 down is the gap between 1.7073 and log₂3 = 1.5850, and it is thin, so the descent takes 111 steps.

None of this proves anything. The ¾ comes from treating each 3n + 1 as though a coin decided how many times it is divisible by two, and the coin is a fiction: the halvings along one orbit are fixed the moment you pick n. Every number checked so far has come home anyway, all 2.36 × 10²¹ of them. Terence Tao came closest, proving that almost every starting value eventually drops below f(n) for any f that grows to infinity however slowly, with almost every meaning logarithmic density 1. Almost every is not every, and that gap is where the problem still sits.

$\frac{e}{o} = lo g_{2} 3 + \frac{lo g _{2} ( n / L )}{o} o \to \infty lo g_{2} 3 = 1.5850$

References (3)

The verification limit the page quotes, and where the 2.36 × 10²¹ comes from: David Bařina, "Improved verification limit for the convergence of the Collatz conjecture", The Journal of Supercomputing 81(1), 1–14, 2025. doi:10.1007/s11227-025-07337-0
The strongest partial result, which the answer quotes: Terence Tao, "Almost all orbits of the Collatz map attain almost bounded values", Forum of Mathematics, Pi 10, e12, 2022 (posted 2019). doi:10.1017/fmp.2022.8
Where the Erdős line comes from, and the standard survey of the problem: Jeffrey C. Lagarias, "The 3x + 1 problem and its generalizations", American Mathematical Monthly 92(1), 3–23, 1985.

By the 42Math team · Reviewed against published sources. Corrections welcome.