I have been fascinated by the Collatz problem since I first heard about it in high school.

Take any natural number $n$. If $n$ is even, divide it by $2$ to get $n / 2$, if $n$ is odd multiply it by $3$ and add $1$ to obtain $3n + 1$. Repeat the process indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach $1$. [...]

Paul Erdős said about the Collatz conjecture: "Mathematics is not yet ready for such problems." He offered $500 USD for its solution.

How important do you consider the answer to this question to be? Why?

Would you speculate on what might have possessed Paul Erdős to make such an offer?

EDIT: Is there any reason to think that a proof of the Collatz Conjecture would be complex (like the FLT) rather than simple (like PRIMES is in P)? And can this characterization of FLT vs. PRIMES is in P be made more specific than a bit-length comparison?

Rodrigo de Azevedo
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Dan Brumleve
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    This should probably be community wiki. – Qiaochu Yuan Aug 18 '10 at 07:51
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    Also, Erdős had a habit of offering cash prizes for solutions to his favorite problems; this was by no means limited to the Collatz problem. – Qiaochu Yuan Aug 18 '10 at 07:55
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    Still, is the problem as profound as his quote suggests? – Dan Brumleve Aug 18 '10 at 08:00
  • $500 is a fairly large prize by Erdos' standards. See http://math.ucsd.edu/~fan/ep.pdf and http://www.math.niu.edu/~rusin/known-math/93_back/prizes.erd . (These are the only two lists I could find with dollar figures on them; neither actually includes the Collatz problem.) – Michael Lugo Aug 18 '10 at 18:07
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    http://xkcd.com/710/ – mmyers Aug 18 '10 at 21:04
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    @Michael, I believe neither list includes Collatz because (Wikipedia to the contrary notwithstanding) Erdos never offered money for its solution. He offered money for solutions to his own problems, not for solutions to problems posed by others. – Gerry Myerson Apr 26 '11 at 13:15

8 Answers8


Most of the answers so far have been along the general lines of 'Why hard problems are important', rather than 'Why the Collatz conjecture is important'; I will try to address the latter.

I think the basic question being touched on is:

In what ways does the prime factorization of $a$ affect the prime factorization of $a+1$?

Of course, one can always multiply out the prime factorization, add one, and then factor again, but this throws away the information of the prime factorization of $a$. Note that this question is also meaningful in other UFDs, like $\mathbb{C}[x]$.

It seems very hard to come up with answers to this question that don't fall under the heading of 'immediate', such as distinct primes in each factorization. This seems to be in part because a small change in the prime factorization for $a$ (multiplication by a prime, say) can have a huge change in the prime factorization for $a+1$ (totally distinct prime support perhaps). Therefore, it is tempting to regard the act of adding 1 as an essentially-random shuffling of the prime factorization.

The most striking thing about the Collatz conjecture is that it seems to be making a deep statement about a subtle relation between the prime factorizations of $a$ and $a+1$. Note that the Collatz iteration consists of three steps; two of which are 'small' in terms of the prime factorization, and the other of which is adding one:

  • multiplying by 3 has a small effect on the factorization.
  • adding 1 has a (possibly) huge effect on the factorization.
  • factoring out a power of 2 has a small effect on the factorization (in that it doesn't change the other prime powers in the factorization).

So, the Collatz conjecture seems to say that there is some sort of abstract quantity like 'energy' which cannot be arbitrarily increased by adding 1. That is, no matter where you start, and no matter where this weird prime-shuffling action of adding 1 takes you, eventually the act of pulling out 2s takes enough energy out of the system so that you reach 1. I think it is for reasons like this that mathematicians suspect that a solution of the Collatz conjecture will open new horizons and develop new and important techniques in number theory.

David Kipper
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Greg Muller
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    What delights me most about the Collatz conjecture is your observation about what the iteration does to the factorizations combined with an observation on the sizes of the numbers. Multiplication by 3 and adding 1 more than triples the number, while dividing by 2 only halves it. If you ended up doing a large number of iterations to compute the sequence, and each was equally likely, then you should expect to see exponential growth in the terms of the sequence over the long term. This is why I don't quite believe the conjecture myself, but love that a counterexample is elusive. – Barry Smith Apr 26 '11 at 19:43
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    @Barry Smith: But after a triple-and-add-1 step you're guaranteed to divide by 2 at least once, so you actually expect a ~19% decline rather than a ~34% increase. – Charles Apr 26 '11 at 20:36
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    Ah yes, true. So now I guess I believe it! Unfortunately, it has now become slightly less interesting to me (but not too much). – Barry Smith Apr 26 '11 at 20:58
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    Just to highlight the effect that the factorization of $a$ and $a+1$ can be radically different, consider the largest known prime number, a Mersenne prime, $a=2^{43,112,609}-1$. $a+1=2^{43,112,609}$. – Jackson Walters Feb 02 '12 at 17:25
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    What about the 5x+1-problem? We have the same factorization-observation by *5, +1 and /2 - but we have cycles and also (very likely) divergent trajectories. How does that fit with the above considerations concerning the 3x+1? – Gottfried Helms Feb 24 '13 at 18:21
  • I agree with Gottfried. – mick Aug 18 '13 at 00:20
  • Responding to this 3 years after the fact because I really like your connection to the primes, but then @GottfriedHelms came along... Could you please comment on this? – JorenHeit Aug 20 '16 at 11:04
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    @Charles I know it's 5 years later, but can you comment on where you got $19\%$ from? What I get is, you expect to mutiply by $3$ and then you expect to divide by two half the time, divide by 4 a quarter of the time, divide by 8 an eighth of the time, and so on, so the expected factor you mutiply by before getting to the next odd number is $3 \left( \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{8^2} + \cdots\right) = 3 \frac{1/4}{3/4} = 1$. – 6005 Sep 04 '16 at 19:32
  • @JorenHeit I don't know if this helps, but if you see the comment I just made, replacing $3$ with $5$ the loosely-calculated expected factor you multiply by will be above $1$, instead of $1$, so you might expect to keep going off to infinity. Either way, I don't think provides a serious challenge the value of Greg Muller's excellent answer. The question just then becomes why $3$ seems to not increase the energy too much, but $5$ does. The point that there seems to be some kind of "energy" that is not increased too much by adding $1$ still stands. – 6005 Sep 04 '16 at 19:34
  • @GottfriedHelms A years late response, but maybe you will be interested in my rough calculation in my comment to Charles in which while multiplying by 3 you expect to be roughly in the same place, on the other hand multiplying by 5 would be too much. – 6005 Sep 04 '16 at 19:36
  • Thanks to @6005 for the notification. I'll look at it another day... – Gottfried Helms Sep 05 '16 at 02:00
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    @6005: A number like "19%" is calculated by assuming 50% chance even (becomes $\frac{x}{2}$, at which point we assume random parity again) and 50% chance odd (becomes $3x+1$ which is necessarily even so then it becomes $\frac{3x+1}{2}$, at which point we assume random parity again). Combining an equal number of $k$ "even" steps and $k$ "odd" double-steps, we see that with $3k$ steps we expect a shrinkage factor of $(\frac{3}{4})^k$, which yields an average shrinkage factor of $\sqrt[3]\frac{3}{4} = 0.908$ per step, so really it's a ~9% decline per step. – Matt Jan 10 '17 at 02:05
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    On the topic of expected shrinkage/growth, a more interesting example in this regard is another iterative procedure (due to Conway I think) which considers numbers mod 3: 3x→2x, 3x+1→4x+1, 3x-1→4x-1. Note that this mapping is reversible — just consider the result mod 4. Now note that the expected change in magnitude is positive *in both directions*! You can even try it and see that indeed, if you start with a random number, it will slowly grow in the long run regardless of whether you use the forward mapping or the reversed mapping. Wrapping your head around that one is a good exercise! – Matt Jan 10 '17 at 02:16
  • @Matt Both comments of yours are very interesting. Thanks for the infos. – 6005 Jan 10 '17 at 21:59
  • It's funny how fast the sciences run, 6 years ago the largest Mersenne prime was $2^{43,112,609}-1$ (according to a previous comment, which is correct) and recently I read somewhere that the last Mersenne's prime has been computed to be $2^{74,207,281}-1$, which is an amazingly huge gap compared to the previous one for the intuition of a human being! Stunning! –  Mar 11 '17 at 17:52
  • @Matt: Thanks for that very interesting example! Others may be interested in the [two envelops problem](https://en.wikipedia.org/wiki/Two_envelopes_problem) as well, which is of a similar nature. However, in your example, is it already known that there are acyclic sequences? My rudimentary computer search finds only a 1-cycle (1) and a 2-cycle (2,3) and a 5-cycle (4,5,7,9,6) and a 12-cycle starting from 44, but is there a proof that say the sequence starting from 8 is acyclic? – user21820 Jun 09 '17 at 13:44
  • what about just add one, why the need to multiply by 3? seems like just adding one to make it even should work too, but in less steps. – luckyguy73 Aug 08 '19 at 21:33
  • the number 9 takes 19 steps (28,14,7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1) using 3n+1, but only 7 steps just using n+1. if the whole point is to get to 1 from any integer i am just asking why not n+1 instead of 3n+1 when the number is odd? – luckyguy73 Aug 08 '19 at 21:37

The Collatz conjecture is the simplest open problem in mathematics. You can explain it to all your non-mathematical friends, and even to small children who have just learned to divide by 2. It doesn't require understanding divisibility, just evenness.

The lack of connections between this conjecture and existing mathematical theories (as complained of in some other answers) is not an inadequacy of this conjecture, but of our theories.

This problem has led directly to theoretical work by Conway showing that very similar questions are formally undecidable, certainly a surprising result.

The problem also relates directly to chaotic cellular automata. If you look at a number in base 6, you will see that multiplying by 3 and dividing by 2 are the same operation (differing only by a factor of 6, i.e. the location of the decimal point), and the operation is local: each new digit only depends on two of the previous step's digits. Using a 7th state for cells that are not part of the number, a very simple cellular automaton is obtained where each cell only needs to look at one neighbor to compute its next value. (Wolfram Mathworld has some nonsense about a CA implementation being difficult due to carries, but there are no carries when you add 1, because after multiplying by 3 the last digit is either 0 (becomes a non-digit because number was even so we should divide by 6) or 3 (becomes 4), so there are never any carries.)

It is easy to prove that this CA is chaotic: If you change the interior digits in any way, the region of affected digits always grows linearly with time (by $\log_6 3$ digits per step). This prevents any engineering of the digit patterns, which are quickly randomized. If the final digit behaves randomly, then the conjecture is true. Clearly any progress on the Collatz conjecture would immediately have consequences for symbolic dynamics.

Emil Post's tag systems (which he created in 1920 expressly for studying the foundations of mathematics) have been studied for many decades, and they have been the foundation of the smallest universal Turing machines (as well as other universal systems) since 1961. In 2007, Liesbeth De Mol discovered that the Collatz problem can be encoded as the following tag system:

$\begin{eqnarray} \hspace{2cm} \alpha & \longrightarrow & c \, y \\ \hspace{2cm} c & \longrightarrow & \alpha \\ \hspace{2cm} y & \longrightarrow & \alpha \alpha \alpha \\ \end{eqnarray}$

In two passes, this tag system processes the word $\alpha^{n}$ into either $\alpha^{n/2}$ or $\alpha^{(3n+1)/2}$ depending on the parity of $n$. Larger tag systems are known to be universal, and any progress on the 3x+1 problem will be followed with close attention by this field.

In short the Collatz problem is simple enough that anyone can understand it, and yet relates not just to number theory (as described in other answers) but to issues of decidability, chaos, and the foundations of mathematics and of computation. That's about as good as it gets for a problem even a small child can understand.

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  • Thanks for pointing out the relation to symbolic dynamics. I've encountered the sequence just now (on Stewart's Calculus actually) and Sharkovski's Theorem came to my mind.. – Alp Uzman Sep 13 '15 at 20:29
  • Thank you for your answer! I came up with this isomorphism to a cellular automata idea myself last night! Yay for insomnia! It's lovely that for a want-to-be mathematician like myself, both statements of the algorithm are extremely clear and simple. I wonder if, when the CA is run, discernible structures are produced that a higher level meta-representation could re-produce. Perhaps it could be shown that there is a meta series, and it is endless, meaning the problem is undecidable… The tag system sounds rewarding – I'll have to look at that. – Benjohn Nov 26 '15 at 09:11
  • "The Collatz conjecture is the simplest open problem in mathematics." I am not sure such a strong statement is warranted. The [Union-closed sets conjecture](https://en.wikipedia.org/wiki/Union-closed_sets_conjecture) and the [Ramsey number $R(5,5)$](https://en.wikipedia.org/wiki/Ramsey%27s_theorem#Ramsey_numbers) come to mind as other possible candidates. – 6005 Sep 04 '16 at 19:50
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    @6005: Yes, those are also simple and the first one is also a wonderful problem. But try explaining them to a small child -- I think you'll quickly find that of the three, the 3x+1 problem is most easily understood. In my opinion the Ramsey number question, for example, is only a simple question once you have already understood that Ramsey numbers exist at all, a non-obvious fact that requires proof. – Matt Sep 06 '16 at 15:12
  • @Matt well, I don't think I agree. For Collatz you need to understand arithmetic and casework and even and odd numbers, which are nontrivial and besides the setup of the problem is pretty unnatural/confusing. For Ramsey number all you have to do is say, how many people in a room before five people must be mutual friends or strangers? – 6005 Sep 07 '16 at 08:54
  • @6005: Most people find even and odd numbers easy to understand, as well as casework (doing one thing if it's even and another if it's odd). But if you feel these are "nontrivial" (which is fine, it's a matter of taste), the Collatz problem doesn't need these things. Let $f(x)=x/2$, and $g(x)=(3x+1)/2$. Now, the conjecture is that for any positive integer $N$, there exists a sequence of applications of $f$ and $g$ that reduces $N$ to 1. No parity, no casework, just an existence question, built from just the 3x+1 operation and the halving operation. – Matt Aug 12 '19 at 23:23
  • Putting aside the small child discussion, though, your alternate formulation is really cool. Do you have a reference on mathSE or elsewhere that proves that your formulation is equivalent? – 6005 Aug 13 '19 at 03:48
  • @Matt (let's try this again) Are we still talking about explaining it to a small child? You've removed the casework at the cost of making the entire thing incomprehensible (at the child's level) :) OK, I'm being unfair, you weren't trying to explain it to a child when you said "let $f(x) = ", "sequence of applications", etc. But I actually think it would be hard to do so. I think you have to understand rational number arithmetic before you understand the conjecture, and I think that's not true for Ramsey numbers or Union-closed sets. – 6005 Aug 13 '19 at 03:54
  • @6005: You can show this formulation is equivalent by considering the number written in binary. If you ever choose the "wrong" operation (among $f$ and $g$), then you get a non-integer with a finite number of binary digits after the decimal point. This number of digits after the decimal point can never be reduced -- both $f$ and $g$ will increase it by one. (Multiplying by 3 doesn't change it, and adding 1.0 doesn't change it, but dividing by two lengthens it by one digit.) So if you make a "wrong" step, you will never get back to the land of integers, so 1 becomes unobtainable. (QED) – Matt Aug 13 '19 at 20:54
  • @Matt I see, yes, that works out nicely. I think I like this formulation better than the original. – 6005 Aug 14 '19 at 15:53

So many mathematicians, and famous ones among them, have tried various ways to attack this problem, and it is still as elusive as it was when first posed. So the importance of the problem is that genuinely new mathematical ideas will have to be created to solve it, and such ideas may be helpful in other domains where "truly important" problems are at stake. Note that Erdős himself has said something along the lines that "we don't have the mathematics yet to solve this problem".

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Christian Blatter
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    +1: Even though the problem might seem irrelevant, attempts to solve it might spring up new important branches in mathematics. For instance, consider Fermat's Last Theorem. – Aryabhata Aug 18 '10 at 13:39
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    How much description will such a _genuinely_ new idea require? – Dan Brumleve Sep 01 '10 at 06:44
  • According to the Wikipedia, Erdös said "Mathematics is not yet ready for such problems." – lhf Nov 16 '10 at 17:44
  • The Erdős quote referred to in this answer (and in the Nov16 comment) was already correct and highlighted in the original post... – Matt Apr 21 '11 at 06:34

I do not think this is a conceptually important problem. It is an example of a down-to-earth problem that can be checked numerically up to a large value and it has resisted a solution for many years. Not all such problems are automatically important (e.g., nonexistence of odd perfect numbers).

An analogy with the significance of Fermat's last theorem is apt. Before the link was made between FLT and deep conjectures of elliptic curves, there was no over-arching significance to knowing whether or not FLT was true. (I think the link between FLT and the abc-conjecture was made at around the same time.) Yes, work on FLT was responsible for useful developments in algebraic number theory, but all the same it was not clear for a long time that settling the problem, or rather finding a counterexample, would have any other repercussions.

If tomorrow someone showed that the Collatz conjecture were a consequence of the abc-conjecture or some other recognizably important unsolved problem, then I would change my mind about its importance (because, as with the link between the modularity conjecture and FLT, a counterexample to Collatz would then have real implications elsewhere in mathematics). But as long as it stays isolated, having no implications to other problems, I don't think on its own it is a mathematically profound question. The same goes for odd perfect numbers: unless someone shows the existence of an odd perfect number has effects elsewhere that we do not expect (like a counterexample to FLT having a very unexpected implication for elliptic curves), I don't think the mainstream would consider odd perfect numbers to be important either.

On the pedagogical side, however, I will definitely grant that this is a nice problem to show students unfamiliar with advanced mathematics that there really are unsolved math problems. People don't necessarily realize this, e.g., they may think that everything can be solved by computers or something.

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  • With regards to my answer http://math.stackexchange.com/questions/2949/which-one-result-in-maths-has-surprised-you-the-most/2974#2974 can we also relate the Collatz Conjecture to PRIMES is in P for similar reasons? – Dan Brumleve Aug 24 '10 at 07:40
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    It is far more down-to-earth than the non-existence of odd perfect numbers. Any child who can multiply by 3 and divide by 2 can understand this problem and wonder about it. – Matt Apr 21 '11 at 06:40

I believe the $3x+1$ problem is considered a test case for ergodic theory, i.e., proving that certain probabilistic expectations true are true for orbits of a specific system. There are papers by Sinai, Lagarias and others giving probabilistic models, similar to the asymptotic predictions made in questions about distribution of prime numbers, where the predictions are reliable but proving them in any particular case (twin primes, $n^2+1$, Goldbach, ...) is a centuries-old open problem.

This is analogous to transcendence or irrationality proofs of specific numbers: proving that what is "true with probability 1" really holds in a particular instance is extremely hard and pushes the theory forward. Other than that there are no problems outside the $3x+1$ conjecture that use the same iteration, so it is a sink and not a source in the applications-of-theory graph.

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Paul Erdős offered cash prizes for the solution to problems according to his assessment of their difficulty and importance. I believe his judgement of the difficulty of this problem has been shown to be correct by the fact that it remains open.

I think this problem is very important in the sense that a large proportion of the people reading this response will have had a go at it, at one time or another. Thus its solution would be of interest to many.

Another reason for its importance is that, like Fermat's Last Theorem (Wiles's Theorem), it's easy to state and understand and thus has the potential to attract young people towards mathematics. I too learned about it in high school and could not resist its allure.

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Derek Jennings
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I can't add to the question: "what is the importance?", but there is one more aspect which was not yet mentioned. This is the relation of powers of 3 and powers of 2. A subproblem of the question of cycles in the Collatz leads to a critical inequality where the possibility of such cycles depends on the relative distance of perfect powers of 2 to perfect powers of 3. This can also be expressed in terms of approximation of log(3)/log(2) to rational numbers. Kurt Mahler had studied this in terms of his notion of z-numbers (but had only partial success); but it is related also to an unsolved detail in the Waring-problem where the rational approximation of powers of 3/2 to integers is the focus of a conjecture.

Hmm, I don't know whether it is meaningful to go more in detail here, I tend to think: no; the idea of the critical inequality was considered by Ray Steiner, and later by John Simons and Benne de Weger; I've linked to an article of the latter in the wikipedia-article on the collatz-problem. A discussion of mine of the approximation-problem is here (but should possibly get enriched with more context).

Gottfried Helms
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Aside from the mathematical answers provided by others, the computational verification of the Collatz problem is a good exercise for programmers. There are many optimization opportunities (e.g., time-space trade-off using lookup tables, parallelism), many pitfalls (e.g., integer type overflow), possibilities of exploiting various implementation tricks (e.g., count trailing zeros instructions available in modern hardware), etc. It is a simple task where you can practice many basic programming constructions (branching the program, do-while loops, recursion). And for these reasons, this is arguably the most common task you can find in many online or university courses (e.g., Harvard University's CS50 course).

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