Collatz conjecture

The length of a non-trivial cycle is known to be at least 17087915.[13] In fact, Eliahou (1993) proved that the period p of any non-trivial cycle is of the form

${\displaystyle p=301994a+17087915b+85137581c}$

where a, b and c are non-negative integers, b ≥ 1 and ac = 0. This result is based on the continued fraction expansion of ln 3/ln 2.

A similar reasoning that accounts for the recent verification of the conjecture up to 5×10¹⁸ leads to the improved lower bound 10439860591 (or 17026679261 without the "shortcut"). This lower bound is consistent with the above result, since ${\displaystyle 10439860591=17087915\times 33+85137581\times 116}$.

A k-cycle is a cycle that can be partitioned into 2k contiguous subsequences: k increasing sequences of odd numbers alternating with k decreasing sequences of even numbers. For instance, if the cycle consists of a single increasing sequence of odd numbers followed by a decreasing sequence of even numbers, it is called a 1-cycle.[14]

Steiner (1977) proved that there is no 1-cycle other than the trivial (1;2).[15] Simons (2004) used Steiner's method to prove that there is no 2-cycle.[16] Simons & de Weger (2005) extended this proof up to 68-cycles: there is no k-cycle up to k = 68.[14] Beyond 68, this method gives upper bounds for the elements in such a cycle: for example, if there is a 75-cycle, then at least one element of the cycle is less than 2385×2⁵⁰.[14] Therefore, as exhaustive computer searches continue, larger cycles may be ruled out. To state the argument more intuitively: we need not look for cycles that have at most 68 trajectories, where each trajectory consists of consecutive ups followed by consecutive downs.

As of 2017, the conjecture has been checked by computer for all starting values up to 87×2⁶⁰.[17] All initial values tested so far eventually end in the repeating cycle (4;2;1), which has only three terms. From this lower bound on the starting value, a lower bound can also be obtained for the number of terms a repeating cycle other than (4;2;1) must have.[18] When this relationship was established in 1981, the formula gave a lower bound of 35400 terms.[18]

This computer evidence is not a proof that the conjecture is true. As shown in the cases of the Pólya conjecture, the Mertens conjecture, and Skewes' number, sometimes a conjecture's only counterexamples are found when using very large numbers.

If one considers only the odd numbers in the sequence generated by the Collatz process, then each odd number is on average 3/4 of the previous one.[19] (More precisely, the geometric mean of the ratios of outcomes is 3/4.) This yields a heuristic argument that every Hailstone sequence should decrease in the long run, although this is not evidence against other cycles, only against divergence. The argument is not a proof because it assumes that Hailstone sequences are assembled from uncorrelated probabilistic events. (It does rigorously establish that the 2-adic extension of the Collatz process has two division steps for every multiplication step for almost all 2-adic starting values.)

And even if the probabilistic reasoning were rigorous, this would still imply only that the conjecture is almost surely true for any given integer, which does not necessarily imply that it is true for all integers.

Terence Tao (2019) proved using probability that almost all Collatz orbits are bounded by any function that diverges into infinity. Responding to this work, Quanta Magazine wrote that Tao "came away with one of the most significant results on the Collatz conjecture in decades."[20][21]

In a computer-aided proof, Krasikov and Lagarias showed that the number of integers in the interval [1,x] that eventually reach 1 is at least equal to x^0.84 for all sufficiently large x.[22]

The first 21 levels of the Collatz graph generated in bottom-up fashion. The graph includes all numbers with an orbit length of 21 or less.

There is another approach to prove the conjecture, which considers the bottom-up method of growing the so-called Collatz graph. The Collatz graph is a graph defined by the inverse relation

${\displaystyle R(n)={\begin{cases}\{2n\}&{\text{if }}n\equiv 0,1,2,3,5\\[4px]\left\{2n,{\frac {n-1}{3}}\right\}&{\text{if }}n\equiv 4\end{cases}}{\pmod {6}}.}$

So, instead of proving that all positive integers eventually lead to 1, we can try to prove that 1 leads backwards to all positive integers. For any integer n, n ≡ 1 (mod 2) if and only if 3n + 1 ≡ 4 (mod 6). Equivalently, n − 1/3 ≡ 1 (mod 2) if and only if n ≡ 4 (mod 6). Conjecturally, this inverse relation forms a tree except for the 1–2–4 loop (the inverse of the 4–2–1 loop of the unaltered function f defined in the Statement of the problem section of this article).

When the relation 3n + 1 of the function f is replaced by the common substitute "shortcut" relation 3n + 1/2, the Collatz graph is defined by the inverse relation,

${\displaystyle R(n)={\begin{cases}\{2n\}&{\text{if }}n\equiv 0,1\\[4px]\left\{2n,{\frac {2n-1}{3}}\right\}&{\text{if }}n\equiv 2\end{cases}}{\pmod {3}}.}$

For any integer n, n ≡ 1 (mod 2) if and only if 3n + 1/2 ≡ 2 (mod 3). Equivalently, 2n − 1/3 ≡ 1 (mod 2) if and only if n ≡ 2 (mod 3). Conjecturally, this inverse relation forms a tree except for a 1–2 loop (the inverse of the 1–2 loop of the function f(n) revised as indicated above).

Alternatively, replace the 3n + 1 with n′/H(n′) where n′ = 3n + 1 and H(n′) is the highest power of 2 that divides n′ (with no remainder). The resulting function f maps from odd numbers to odd numbers. Now suppose that for some odd number n, applying this operation k times yields the number 1 (that is, f^k(n) = 1). Then in binary, the number n can be written as the concatenation of strings w_k w_k−1 … w₁ where each w_h is a finite and contiguous extract from the representation of 1/3^h.[23] The representation of n therefore holds the repetends of 1/3^h, where each repetend is optionally rotated and then replicated up to a finite number of bits. It is only in binary that this occurs.[24] Conjecturally, every binary string s that ends with a '1' can be reached by a representation of this form (where we may add or delete leading '0's to s).

Repeated applications of the Collatz function can be represented as an abstract machine that handles strings of bits. The machine will perform the following three steps on any odd number until only one "1" remains:

1. Append 1 to the (right) end of the number in binary (giving 2n + 1);
2. Add this to the original number by binary addition (giving 2n + 1 + n = 3n + 1);
3. Remove all trailing "0"s (i.e. repeatedly divide by two until the result is odd).

The starting number 7 is written in base two as 111. The resulting Collatz sequence is:

           111
          1111
         10110
        10111
       100010
      100011
      110100
     11011
    101000
   1011
  10000

For this section, consider the Collatz function in the slightly modified form

${\displaystyle f(n)={\begin{cases}{\frac {n}{2}}&{\text{if }}n\equiv 0\\[4px]{\frac {3n+1}{2}}&{\text{if }}n\equiv 1\end{cases}}{\pmod {2}}.}$

This can be done because when n is odd, 3n + 1 is always even.

If P(…) is the parity of a number, that is P(2n) = 0 and P(2n + 1) = 1, then we can define the Collatz parity sequence (or parity vector) for a number n as p_i = P(a_i), where a₀ = n, and a_i+1 = f(a_i).

Which operation is performed, 3n + 1/2 or n/2, depends on the parity. The parity sequence is the same as the sequence of operations.

Using this form for f(n), it can be shown that the parity sequences for two numbers m and n will agree in the first k terms if and only if m and n are equivalent modulo 2^k. This implies that every number is uniquely identified by its parity sequence, and moreover that if there are multiple Hailstone cycles, then their corresponding parity cycles must be different.[3][25]

Applying the f function k times to the number n = 2^ka + b will give the result 3^ca + d, where d is the result of applying the f function k times to b, and c is how many increases were encountered during that sequence (e.g. for 2⁵a + 1 there are 3 increases as 1 iterates to 2, 1, 2, 1, and finally to 2 so the result is 3³a + 2; for 2²a + 1 there is only 1 increase as 1 rises to 2 and falls to 1 so the result is 3a + 1). When b is 2^k − 1 then there will be k rises and the result will be 2 × 3^ka − 1. The factor of 3 multiplying a is independent of the value of a; it depends only on the behavior of b. This allows one to predict that certain forms of numbers will always lead to a smaller number after a certain number of iterations, e.g. 4a + 1 becomes 3a + 1 after two applications of f and 16a + 3 becomes 9a + 2 after 4 applications of f. Whether those smaller numbers continue to 1, however, depends on the value of a.

For the Collatz function in the form

${\displaystyle f(n)={\begin{cases}{\frac {n}{2}}&{\text{if }}n\equiv 0\\[4px]{\frac {3n+1}{2}}&{\text{if }}n\equiv 1.\end{cases}}{\pmod {2}}}$

Hailstone sequences can be computed by the extremely simple 2-tag system with production rules

a → bc, b → a, c → aaa.

In this system, the positive integer n is represented by a string of n copies of a, and iteration of the tag operation halts on any word of length less than 2. (Adapted from De Mol.)

The Collatz conjecture equivalently states that this tag system, with an arbitrary finite string of a as the initial word, eventually halts (see Tag system#Example: Computation of Collatz sequences for a worked example).

An extension to the Collatz conjecture is to include all integers, not just positive integers. Leaving aside the cycle 0 → 0 which cannot be entered from outside, there are a total of 4 known cycles, which all nonzero integers seem to eventually fall into under iteration of f. These cycles are listed here, starting with the well-known cycle for positive n:

Odd values are listed in large bold. Each cycle is listed with its member of least absolute value (which is always odd) first.

The generalized Collatz conjecture is the assertion that every integer, under iteration by f, eventually falls into one of the four cycles above or the cycle 0 → 0. The 0 → 0 cycle is often regarded as "trivial" by the argument, as it is only included for the sake of completeness.

The Collatz map can be extended to (positive or negative) rational numbers which have odd denominators when written in lowest terms. The number is taken to be 'odd' or 'even' according to whether its numerator is odd or even. Then the formula for the map is exactly the same as when the domain is the integers: an 'even' such rational is divided by 2; an 'odd' such rational is multiplied by 3 and then 1 is added. A closely related fact is that the Collatz map extends to the ring of 2-adic integers, which contains the ring of rationals with odd denominators as a subring.

When using the "shortcut" definition of the Collatz map, it is known that any periodic parity sequence is generated by exactly one rational.[26] Conversely, it is conjectured that every rational with an odd denominator has an eventually cyclic parity sequence (Periodicity Conjecture [3]).

If a parity cycle has length n and includes odd numbers exactly m times at indices k₀ < … < k_m−1, then the unique rational which generates immediately and periodically this parity cycle is

${\displaystyle {\frac {3^{m-1}2^{k_{0}}+\cdots +3^{0}2^{k_{m-1}}}{2^{n}-3^{m}}}.}$

(1)

For example, the parity cycle (1 0 1 1 0 0 1) has length 7 and four odd terms at indices 0, 2, 3, and 6. It is repeatedly generated by the fraction

${\displaystyle {\frac {3^{3}2^{0}+3^{2}2^{2}+3^{1}2^{3}+3^{0}2^{6}}{2^{7}-3^{4}}}={\frac {151}{47}}}$

as the latter leads to the rational cycle

${\displaystyle {\frac {151}{47}}\rightarrow {\frac {250}{47}}\rightarrow {\frac {125}{47}}\rightarrow {\frac {211}{47}}\rightarrow {\frac {340}{47}}\rightarrow {\frac {170}{47}}\rightarrow {\frac {85}{47}}\rightarrow {\frac {151}{47}}}$.

Any cyclic permutation of (1 0 1 1 0 0 1) is associated to one of the above fractions. For instance, the cycle (0 1 1 0 0 1 1) is produced by the fraction

${\displaystyle {\frac {3^{3}2^{1}+3^{2}2^{2}+3^{1}2^{5}+3^{0}2^{6}}{2^{7}-3^{4}}}={\frac {250}{47}}}$.

For a one-to-one correspondence, a parity cycle should be irreducible, i.e., not partitionable into identical sub-cycles. As an illustration of this, the parity cycle (1 1 0 0 1 1 0 0) and its sub-cycle (1 1 0 0) are associated to the same fraction 5/7 when reduced to lowest terms.

In this context, assuming the validity of the Collatz conjecture implies that (1 0) and (0 1) are the only parity cycles generated by positive whole numbers (1 and 2, respectively).

If the odd denominator d of a rational is not a multiple of 3, then all the iterates have the same denominator and the sequence of numerators can be obtained by applying the "3n + d" generalization of the Collatz function

${\displaystyle T_{d}(x)={\begin{cases}{\frac {x}{2}}&{\text{if }}x\equiv 0{\pmod {2}},\\[4px]{\frac {3x+d}{2}}&{\text{if }}x\equiv 1{\pmod {2}}.\end{cases}}}$

The function

${\displaystyle T(x)={\begin{cases}{\frac {x}{2}}&{\text{if }}x\equiv 0{\pmod {2}}\\[4px]{\frac {3x+1}{2}}&{\text{if }}x\equiv 1{\pmod {2}}\end{cases}}}$

is well-defined on the ring ℤ₂ of 2-adic integers, where it is continuous and measure-preserving with respect to the 2-adic measure. Moreover, its dynamics is known to be ergodic.[3]

Define the parity vector function Q acting on ℤ₂ as

${\displaystyle Q(x)=\sum _{k=0}^{\infty }\left(T^{k}(x)\mod 2\right)2^{k}}$.

The function Q is a 2-adic isometry.[27] Consequently, every infinite parity sequence occurs for exactly one 2-adic integer, so that almost all trajectories are acyclic in ${\displaystyle \mathbb {Z} _{2}}$.

An equivalent formulation of the Collatz conjecture is that

${\displaystyle Q\left(\mathbb {Z} ^{+}\right)\subset {\tfrac {1}{3}}\mathbb {Z} .}$

Cobweb plot of the orbit 10-5-8-4-2-1-2-1-2-1-etc. in the real extension of the Collatz map (optimized by replacing "3n + 1" with "3n + 1/2")

The Collatz map can be viewed as the restriction to the integers of the smooth real and complex map

${\displaystyle f(z)={\frac {1}{2}}z\cos ^{2}\left({\frac {\pi }{2}}z\right)+(3z+1)\sin ^{2}\left({\frac {\pi }{2}}z\right).}$

If the standard Collatz map defined above is optimized by replacing the relation 3n + 1 with the common substitute "shortcut" relation 3n + 1/2, it can be viewed as the restriction to the integers of the smooth real and complex map

${\displaystyle f(z)={\frac {1}{2}}z\cos ^{2}\left({\frac {\pi }{2}}z\right)+{\frac {3z+1}{2}}\sin ^{2}\left({\frac {\pi }{2}}z\right).}$

Collatz map fractal in a neighbourhood of the real line

The point of view of iteration on the real line was investigated by Chamberland (1996),[28]. He showed that the conjecture does not hold for real numbers since there are infinitely many fixed points, as well as orbits escaping monotonically to infinity. He also showed that there is, at least, another attracting cycle: 1.1925 → 2.1386.

On the complex plane, it was investigated by Letherman, Schleicher, and Wood (1999).[29] Most points of the plane diverge to infinity, as seen in blue on the illustration below. The boundary between diverging and non-diverging regions show a fractal pattern called the "Collatz fractal".

The section As a parity sequence above gives a way to speed up simulation of the sequence. To jump ahead k steps on each iteration (using the f function from that section), break up the current number into two parts, b (the k least significant bits, interpreted as an integer), and a (the rest of the bits as an integer). The result of jumping ahead k steps can be found as:

f^k(2^ka + b) = 3^c(b)a + d(b).

The c (or better 3^c) and d arrays are precalculated for all possible k-bit numbers b, where d(b) is the result of applying the f function k times to b, and c(b) is the number of odd numbers encountered on the way.[30] For example, if k = 5, one can jump ahead 5 steps on each iteration by separating out the 5 least significant bits of a number and using:

c(0...31) = {0,3,2,2,2,2,2,4,1,4,1,3,2,2,3,4,1,2,3,3,1,1,3,3,2,3,2,4,3,3,4,5}

d(0...31) = {0,2,1,1,2,2,2,20,1,26,1,10,4,4,13,40,2,5,17,17,2,2,20,20,8,22,8,71,26,26,80,242}.

This requires 2^k precomputation and storage to speed up the resulting calculation by a factor of k, a space–time tradeoff.

For the special purpose of searching for a counterexample to the Collatz conjecture, this precomputation leads to an even more important acceleration, used by Tomás Oliveira e Silva in his computational confirmations of the Collatz conjecture up to large values of n. If, for some given b and k, the inequality

f^k(2^ka + b) = 3^c(b)a + d(b) < 2^ka + b

holds for all a, then the first counterexample, if it exists, cannot be b modulo 2^k.[18] For instance, the first counterexample must be odd because f(2n) = n, smaller than 2n; and it must be 3 mod 4 because f²(4n + 1) = 3n + 1, smaller than 4n + 1. For each starting value a which is not a counterexample to the Collatz conjecture, there is a k for which such an inequality holds, so checking the Collatz conjecture for one starting value is as good as checking an entire congruence class. As k increases, the search only needs to check those residues b that are not eliminated by lower values of k. Only an exponentially small fraction of the residues survive.[31] For example, the only surviving residues mod 32 are 7, 15, 27, and 31.