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One of the most notorious problems in elementary mathematics that remains unsolved is the Collatz conjecture, concerning the function {f_0: {\bf N} \rightarrow {\bf N}} defined by setting {f_0(n) := 3n+1} when {n} is odd, and {f_0(n) := n/2} when {n} is even. (Here, {{\bf N}} is understood to be the positive natural numbers {\{1,2,3,\ldots\}}.)

Conjecture 1 (Collatz conjecture) For any given natural number {n}, the orbit {n, f_0(n), f^2_0(n), f^3_0(n), \ldots} passes through {1} (i.e. {f^k_0(n)=1} for some {k}).

Open questions with this level of notoriety can lead to what Richard Lipton calls “mathematical diseases” (and what I termed an unhealthy amount of obsession on a single famous problem). (See also this xkcd comic regarding the Collatz conjecture.) As such, most practicing mathematicians tend to spend the majority of their time on more productive research areas that are only just beyond the range of current techniques. Nevertheless, it can still be diverting to spend a day or two each year on these sorts of questions, before returning to other matters; so I recently had a go at the problem. Needless to say, I didn’t solve the problem, but I have a better appreciation of why the conjecture is (a) plausible, and (b) unlikely be proven by current technology, and I thought I would share what I had found out here on this blog.

Let me begin with some very well known facts. If {n} is odd, then {f_0(n) = 3n+1} is even, and so {f_0^2(n) = \frac{3n+1}{2}}. Because of this, one could replace {f_0} by the function {f_1: {\bf N} \rightarrow {\bf N}}, defined by {f_1(n) = \frac{3n+1}{2}} when {n} is odd, and {f_1(n)=n/2} when {n} is even, and obtain an equivalent conjecture. Now we see that if one chooses {n} “at random”, in the sense that it is odd with probability {1/2} and even with probability {1/2}, then {f_1} increases {n} by a factor of roughly {3/2} half the time, and decreases it by a factor of {1/2} half the time. Furthermore, if {n} is uniformly distributed modulo {4}, one easily verifies that {f_1(n)} is uniformly distributed modulo {2}, and so {f_1^2(n)} should be roughly {3/2} times as large as {f_1(n)} half the time, and roughly {1/2} times as large as {f_1(n)} the other half of the time. Continuing this at a heuristic level, we expect generically that {f_1^{k+1}(n) \approx \frac{3}{2} f_1^k(n)} half the time, and {f_1^{k+1}(n) \approx \frac{1}{2} f_1^k(n)} the other half of the time. The logarithm {\log f_1^k(n)} of this orbit can then be modeled heuristically by a random walk with steps {\log \frac{3}{2}} and {\log \frac{1}{2}} occuring with equal probability. The expectation

\displaystyle  \frac{1}{2} \log \frac{3}{2} + \frac{1}{2} \log \frac{1}{2} = \frac{1}{2} \log \frac{3}{4}

is negative, and so (by the classic gambler’s ruin) we expect the orbit to decrease over the long term. This can be viewed as heuristic justification of the Collatz conjecture, at least in the “average case” scenario in which {n} is chosen uniform at random (e.g. in some large interval {\{1,\ldots,N\}}). (It also suggests that if one modifies the problem, e.g. by replacing {3n+1} to {5n+1}, then one can obtain orbits that tend to increase over time, and indeed numerically for this variant one sees orbits that appear to escape to infinity.) Unfortunately, one can only rigorously keep the orbit uniformly distributed modulo {2} for time about {O(\log N)} or so; after that, the system is too complicated for naive methods to control at anything other than a heuristic level.

Remark 1 One can obtain a rigorous analogue of the above arguments by extending {f_1} from the integers {{\bf Z}} to the {2}-adics {{\bf Z}_2}. This compact abelian group comes with a Haar probability measure, and one can verify that this measure is invariant with respect to {f_1}; with a bit more effort one can verify that it is ergodic. This suggests the introduction of ergodic theory methods. For instance, using the pointwise ergodic theorem, we see that if {n} is a random {2}-adic integer, then almost surely the orbit {n, f_1(n), f_1^2(n), \ldots} will be even half the time and odd half the time asymptotically, thus supporting the above heuristics. Unfortunately, this does not directly tell us much about the dynamics on {{\bf Z}}, as this is a measure zero subset of {{\bf Z}_2}. More generally, unless a dynamical system is somehow “polynomial”, “nilpotent”, or “unipotent” in nature, the current state of ergodic theory is usually only able to say something meaningful about generic orbits, but not about all orbits. For instance, the very simple system {x \rightarrow 10x} on the unit circle {{\bf R}/{\bf Z}} is well understood from ergodic theory (in particular, almost all orbits will be uniformly distributed), but the orbit of a specific point, e.g. {\pi\hbox{ mod } 1}, is still nearly impossible to understand (this particular problem being equivalent to the notorious unsolved question of whether the digits of {\pi} are uniformly distributed).

The above heuristic argument only suggests decreasing orbits for almost all {n} (though even this remains unproven, the state of the art is that the number of {n} in {\{1,\ldots,N\}} that eventually go to {1} is {\gg N^{0.84}}, a result of Krasikov and Lagarias). It leaves open the possibility of some very rare exceptional {n} for which the orbit goes to infinity, or gets trapped in a periodic loop. Since the only loop that {1} lies in is {1,4,2} (for {f_0}) or {1,2} (for {f_1}), we thus may isolate a weaker consequence of the Collatz conjecture:

Conjecture 2 (Weak Collatz conjecture) Suppose that {n} is a natural number such that {f^k_0(n)=n} for some {k \geq 1}. Then {n} is equal to {1}, {2}, or {4}.

Of course, we may replace {f_0} with {f_1} (and delete “{4}“) and obtain an equivalent conjecture.

This weaker version of the Collatz conjecture is also unproven. However, it was observed by Bohm and Sontacchi that this weak conjecture is equivalent to a divisibility problem involving powers of {2} and {3}:

Conjecture 3 (Reformulated weak Collatz conjecture) There does not exist {k \geq 1} and integers

\displaystyle 0 = a_1 < a_2 < \ldots < a_{k+1}

such that {2^{a_{k+1}}-3^k} is a positive integer that is a proper divisor of

\displaystyle  3^{k-1} 2^{a_1} + 3^{k-2} 2^{a_2} + \ldots + 2^{a_k},

i.e.

\displaystyle  (2^{a_{k+1}} - 3^k) n = 3^{k-1} 2^{a_1} + 3^{k-2} 2^{a_2} + \ldots + 2^{a_k} \ \ \ \ \ (1)

for some natural number {n > 1}.

Proposition 4 Conjecture 2 and Conjecture 3 are equivalent.

Proof: To see this, it is convenient to reformulate Conjecture 2 slightly. Define an equivalence relation {\sim} on {{\bf N}} by declaring {a \sim b} if {a/b = 2^m} for some integer {m}, thus giving rise to the quotient space {{\bf N}/\sim} of equivalence classes {[n]} (which can be placed, if one wishes, in one-to-one correspondence with the odd natural numbers). We can then define a function {f_2: {\bf N}/\sim \rightarrow {\bf N}/\sim} by declaring

\displaystyle  f_2( [n] ) := [3n + 2^a] \ \ \ \ \ (2)

for any {n \in {\bf N}}, where {2^a} is the largest power of {2} that divides {n}. It is easy to see that {f_2} is well-defined (it is essentially the Syracuse function, after identifying {{\bf N}/\sim} with the odd natural numbers), and that periodic orbits of {f_2} correspond to periodic orbits of {f_1} or {f_0}. Thus, Conjecture 2 is equivalent to the conjecture that {[1]} is the only periodic orbit of {f_2}.

Now suppose that Conjecture 2 failed, thus there exists {[n] \neq [1]} such that {f_2^k([n])=[n]} for some {k \geq 1}. Without loss of generality we may take {n} to be odd, then {n>1}. It is easy to see that {[1]} is the only fixed point of {f_2}, and so {k>1}. An easy induction using (2) shows that

\displaystyle  f_2^k([n]) = [3^k n + 3^{k-1} 2^{a_1} + 3^{k-2} 2^{a_2} + \ldots + 2^{a_k}]

where, for each {1 \leq i \leq k}, {2^{a_i}} is the largest power of {2} that divides

\displaystyle  n_i := 3^{i-1} n + 3^{i-2} 2^{a_1} + \ldots + 2^{a_{i-1}}. \ \ \ \ \ (3)

In particular, as {n_1 = n} is odd, {a_1=0}. Using the recursion

\displaystyle  n_{i+1} = 3n_i + 2^{a_i}, \ \ \ \ \ (4)

we see from induction that {2^{a_i+1}} divides {n_{i+1}}, and thus {a_{i+1}>a_i}:

\displaystyle  0 = a_1 < a_2 < \ldots < a_k.

Since {f_2^k([n]) = [n]}, we have

\displaystyle  2^{a_{k+1}} n = 3^k n + 3^{k-1} 2^{a_1} + 3^{k-2} 2^{a_2} + \ldots + 2^{a_k} = 3 n_k + 2^{a_k}

for some integer {a_{k+1}}. Since {3 n_k + 2^{a_k}} is divisible by {2^{a_k+1}}, and {n} is odd, we conclude {a_{k+1} > a_k}; if we rearrange the above equation as (1), then we obtain a counterexample to Conjecture 3.

Conversely, suppose that Conjecture 3 failed. Then we have {k \geq 1}, integers

\displaystyle 0 = a_1 < a_2 < \ldots < a_{k+1}

and a natural number {n > 1} such that (1) holds. As {a_1=0}, we see that the right-hand side of (1) is odd, so {n} is odd also. If we then introduce the natural numbers {n_i} by the formula (3), then an easy induction using (4) shows that

\displaystyle  (2^{a_{k+1}} - 3^k) n_i = 3^{k-1} 2^{a_i} + 3^{k-2} 2^{a_{i+1}} + \ldots + 2^{a_{i+k-1}} \ \ \ \ \ (5)

with the periodic convention {a_{k+j} := a_j + a_{k+1}} for {j>1}. As the {a_i} are increasing in {i} (even for {i \geq k+1}), we see that {2^{a_i}} is the largest power of {2} that divides the right-hand side of (5); as {2^{a_{k+1}}-3^k} is odd, we conclude that {2^{a_i}} is also the largest power of {2} that divides {n_i}. We conclude that

\displaystyle f_2([n_i]) = [3n_i + 2^{a_i}] = [n_{i+1}]

and thus {[n]} is a periodic orbit of {f_2}. Since {n} is an odd number larger than {1}, this contradicts Conjecture 3. \Box

Call a counterexample a tuple {(k,a_1,\ldots,a_{k+1})} that contradicts Conjecture 3, i.e. an integer {k \geq 1} and an increasing set of integers

\displaystyle 0 = a_1 < a_2 < \ldots < a_{k+1}

such that (1) holds for some {n \geq 1}. We record a simple bound on such counterexamples, due to Terras and to Garner :

Lemma 5 (Exponent bounds) Let {N \geq 1}, and suppose that the Collatz conjecture is true for all {n < N}. Let {(k,a_1,\ldots,a_{k+1})} be a counterexample. Then

\displaystyle  \frac{\log 3}{\log 2} k < a_{k+1} < \frac{\log(3+\frac{1}{N})}{\log 2} k.

Proof: The first bound is immediate from the positivity of {2^{a_{k+1}}-3^k}. To prove the second bound, observe from the proof of Proposition 4 that the counterexample {(k,a_1,\ldots,a_{k+1})} will generate a counterexample to Conjecture 2, i.e. a non-trivial periodic orbit {n, f(n), \ldots, f^K(n) = n}. As the conjecture is true for all {n < N}, all terms in this orbit must be at least {N}. An inspection of the proof of Proposition 4 reveals that this orbit consists of {k} steps of the form {x \mapsto 3x+1}, and {a_{k+1}} steps of the form {x \mapsto x/2}. As all terms are at least {N}, the former steps can increase magnitude by a multiplicative factor of at most {3+\frac{1}{N}}. As the orbit returns to where it started, we conclude that

\displaystyle  1 \leq (3+\frac{1}{N})^k (\frac{1}{2})^{a_{k+1}}

whence the claim. \Box

The Collatz conjecture has already been verified for many values of {n} (up to at least {N = 5 \times 10^{18}}, according to this web site). Inserting this into the above lemma, one can get lower bounds on {k}. For instance, by methods such as this, it is known that any non-trivial periodic orbit has length at least {105{,}000}, as shown in Garner’s paper (and this bound, which uses the much smaller value {N = 2 \times 10^9} that was available in 1981, can surely be improved using the most recent computational bounds).

Now we can perform a heuristic count on the number of counterexamples. If we fix {k} and {a := a_{k+1}}, then {2^a > 3^k}, and from basic combinatorics we see that there are {\binom{a-1}{k-1}} different ways to choose the remaining integers

\displaystyle 0 = a_1 < a_2 < \ldots < a_{k+1}

to form a potential counterexample {(k,a_1,\ldots,a_{k+1})}. As a crude heuristic, one expects that for a “random” such choice of integers, the expression (1) has a probability {1/q} of holding for some integer {n}. (Note that {q} is not divisible by {2} or {3}, and so one does not expect the special structure of the right-hand side of (1) with respect to those moduli to be relevant. There will be some choices of {a_1,\ldots,a_k} where the right-hand side in (1) is too small to be divisible by {q}, but using the estimates in Lemma 5, one expects this to occur very infrequently.) Thus, the total expected number of solutions for this choice of {a, k} is

\displaystyle  \frac{1}{q} \binom{a-1}{k-1}.

The heuristic number of solutions overall is then expected to be

\displaystyle  \sum_{a,k} \frac{1}{q} \binom{a-1}{k-1}, \ \ \ \ \ (6)

where, in view of Lemma 5, one should restrict the double summation to the heuristic regime {a \approx \frac{\log 3}{\log 2} k}, with the approximation here accurate to many decimal places.

We need a lower bound on {q}. Here, we will use Baker’s theorem (as discussed in this previous post), which among other things gives the lower bound

\displaystyle  q = 2^a - 3^k \gg 2^a / a^C \ \ \ \ \ (7)

for some absolute constant {C}. Meanwhile, Stirling’s formula (as discussed in this previous post) combined with the approximation {k \approx \frac{\log 2}{\log 3} a} gives

\displaystyle  \binom{a-1}{k-1} \approx \exp(h(\frac{\log 2}{\log 3}))^a

where {h} is the entropy function

\displaystyle  h(x) := - x \log x - (1-x) \log (1-x).

A brief computation shows that

\displaystyle  \exp(h(\frac{\log 2}{\log 3})) \approx 1.9318 \ldots

and so (ignoring all subexponential terms)

\displaystyle  \frac{1}{q} \binom{a-1}{k-1} \approx (0.9659\ldots)^a

which makes the series (6) convergent. (Actually, one does not need the full strength of Lemma 5 here; anything that kept {k} well away from {a/2} would suffice. In particular, one does not need an enormous value of {N}; even {N=5} (say) would be more than sufficient to obtain the heuristic that there are finitely many counterexamples.) Heuristically applying the Borel-Cantelli lemma, we thus expect that there are only a finite number of counterexamples to the weak Collatz conjecture (and inserting a bound such as {k \geq 105{,}000}, one in fact expects it to be extremely likely that there are no counterexamples at all).

This, of course, is far short of any rigorous proof of Conjecture 2. In order to make rigorous progress on this conjecture, it seems that one would need to somehow exploit the structural properties of numbers of the form

\displaystyle  3^{k-1} 2^{a_1} + 3^{k-2} 2^{a_2} + \ldots + 2^{a_k}. \ \ \ \ \ (8)

In some very special cases, this can be done. For instance, suppose that one had {a_{i+1}=a_i+1} with at most one exception (this is essentially what is called a {1}-cycle by Steiner). Then (8) simplifies via the geometric series formula to a combination of just a bounded number of powers of {2} and {3}, rather than an unbounded number. In that case, one can start using tools from transcendence theory such as Baker’s theorem to obtain good results; for instance, in the above-referenced paper of Steiner, it was shown that {1}-cycles cannot actually occur, and similar methods have been used to show that {m}-cycles (in which there are at most {m} exceptions to {a_{i+1}=a_i+1}) do not occur for any {m \leq 63}, as was shown by Simons and de Weger. However, for general increasing tuples of integers {a_1,\ldots,a_k}, there is no such representation by bounded numbers of powers, and it does not seem that methods from transcendence theory will be sufficient to control the expressions (8) to the extent that one can understand their divisibility properties by quantities such as {2^a-3^k}.

Amusingly, there is a slight connection to Littlewood-Offord theory in additive combinatorics – the study of the {2^n} random sums

\displaystyle  \pm v_1 \pm v_2 \pm \ldots \pm v_n

generated by some elements {v_1,\ldots,v_n} of an additive group {G}, or equivalently, the vertices of an {n}-dimensional parallelepiped inside {G}. Here, the relevant group is {{\bf Z}/q{\bf Z}}. The point is that if one fixes {k} and {a_{k+1}} (and hence {q}), and lets {a_1,\ldots,a_k} vary inside the simplex

\displaystyle  \Delta := \{ (a_1,\ldots,a_k) \in {\bf N}^k: 0 = a_1 < \ldots < a_k < a_{k+1}\}

then the set {S} of all sums of the form (8) (viewed as an element of {{\bf Z}/q{\bf Z}}) contains many large parallelepipeds. (Note, incidentally, that once one fixes {k}, all the sums of the form (8) are distinct; because given (8) and {k}, one can read off {2^{a_1}} as the largest power of {2} that divides (8), and then subtracting off {3^{k-1} 2^{a_1}} one can then read off {2^{a_2}}, and so forth.) This is because the simplex {\Delta} contains many large cubes. Indeed, if one picks a typical element {(a_1,\ldots,a_k)} of {\Delta}, then one expects (thanks to Lemma 5) that there there will be {\gg k} indices {1 \leq i_1 < \ldots < i_m \leq k} such that {a_{i_j+1} > a_{i_j}+1} for {j=1,\ldots,m}, which allows one to adjust each of the {a_{i_j}} independently by {1} if desired and still remain inside {\Delta}. This gives a cube in {\Delta} of dimension {\gg k}, which then induces a parallelepiped of the same dimension in {S}. A short computation shows that the generators of this parallelepiped consist of products of a power of {2} and a power of {3}, and in particular will be coprime to {q}.

If the weak Collatz conjecture is true, then the set {S} must avoid the residue class {0} in {{\bf Z}/q{\bf Z}}. Let us suppose temporarily that we did not know about Baker’s theorem (and the associated bound (7)), so that {q} could potentially be quite small. Then we would have a large parallelepiped inside a small cyclic group {{\bf Z}/q{\bf Z}} that did not cover all of {{\bf Z}/q{\bf Z}}, which would not be possible for {q} small enough. Indeed, an easy induction shows that a {d}-dimensional parallelepiped in {{\bf Z}/q{\bf Z}}, with all generators coprime to {q}, has cardinality at least {\min(q, d+1)}. This argument already shows the lower bound {q \gg k}. In other words, we have

Proposition 6 Suppose the weak Collatz conjecture is true. Then for any natural numbers {a, k} with {2^a > 3^k}, one has {2^a-3^k \gg k}.

This bound is very weak when compared against the unconditional bound (7). However, I know of no way to get a nontrivial separation property between powers of {2} and powers of {3} other than via transcendence theory methods. Thus, this result strongly suggests that any proof of the Collatz conjecture must either use existing results in transcendence theory, or else must contribute a new method to give non-trivial results in transcendence theory. (This already rules out a lot of possible approaches to solve the Collatz conjecture.)

By using more sophisticated tools in additive combinatorics, one can improve the above proposition (though it is still well short of the transcendence theory bound (7)):

Proposition 7 Suppose the weak Collatz conjecture is true. Then for any natural numbers {a, k} with {2^a > 3^k}, one has {2^a-3^k \gg (1+\epsilon)^k} for some absolute constant {\epsilon > 0}.

Proof: (Informal sketch only) Suppose not, then we can find {a, k} with {q := 2^a-3^k} of size {(1+o(1))^k = \exp(o(k))}. We form the set {S} as before, which contains parallelepipeds in {{\bf Z}/q{\bf Z}} of large dimension {d \gg k} that avoid {0}. We can count the number of times {0} occurs in one of these parallelepipeds by a standard Fourier-analytic computation involving Riesz products (see Chapter 7 of my book with Van Vu, or this recent preprint of Maples). Using this Fourier representation, the fact that this parallelepiped avoids {0} (and the fact that {q = \exp(o(k)) = \exp(o(d))}) forces the generators {v_1,\ldots,v_d} to be concentrated in a Bohr set, in that one can find a non-zero frequency {\xi \in {\bf Z}/q{\bf Z}} such that {(1-o(1))d} of the {d} generators lie in the set {\{ v: \xi v = o(q) \hbox{ mod } q \}}. However, one can choose the generators to essentially have the structure of a (generalised) geometric progression (up to scaling, it resembles something like {2^i 3^{\lfloor \alpha i \rfloor}} for {i} ranging over a generalised arithmetic progression, and {\alpha} a fixed irrational), and one can show that such progressions cannot be concentrated in Bohr sets (this is similar in spirit to the exponential sum estimates of Bourgain on approximate multiplicative subgroups of {{\bf Z}/q{\bf Z}}, though one can use more elementary methods here due to the very strong nature of the Bohr set concentration (being of the “{99\%} concentration” variety rather than the “{1\%} concentration”).). This furnishes the required contradiction. \Box

Thus we see that any proposed proof of the Collatz conjecture must either use transcendence theory, or introduce new techniques that are powerful enough to create exponential separation between powers of {2} and powers of {3}.

Unfortunately, once one uses the transcendence theory bound (7), the size {q} of the cyclic group {{\bf Z}/q{\bf Z}} becomes larger than the volume of any cube in {S}, and Littlewood-Offord techniques are no longer of much use (they can be used to show that {S} is highly equidistributed in {{\bf Z}/q{\bf Z}}, but this does not directly give any way to prevent {S} from containing {0}).

One possible toy model problem for the (weak) Collatz conjecture is a conjecture of Erdos asserting that for {n>8}, the base {3} representation of {2^n} contains at least one {2}. (See this paper of Lagarias for some work on this conjecture and on related problems.) To put it another way, the conjecture asserts that there are no integer solutions to

\displaystyle  2^n = 3^{a_1} + 3^{a_2} + \ldots + 3^{a_k}

with {n > 8} and {0 \leq a_1 < \ldots < a_k}. (When {n=8}, of course, one has {2^8 = 3^0 + 3^1 + 3^2 + 3^5}.) In this form we see a resemblance to Conjecture 3, but it looks like a simpler problem to attack (though one which is still a fair distance beyond what one can do with current technology). Note that one has a similar heuristic support for this conjecture as one does for Proposition 3; a number of magnitude {2^n} has about {n \frac{\log 2}{\log 3}} base {3} digits, so the heuristic probability that none of these digits are equal to {2} is {3^{-n\frac{\log 2}{\log 3}} = 2^{-n}}, which is absolutely summable.

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