Collatz conjecture

As of 2020, the conjecture has been checked by computer for all starting values up to 2⁶⁸ ≈ 2.95×10²⁰. All initial values tested so far eventually end in the repeating cycle (4; 2; 1) of period 3.[13]

This computer evidence is not sufficient to prove that the conjecture is true for all starting values. As in the case of some disproved conjectures, like the Pólya conjecture, counterexamples might be found when considering very large numbers.

However, such verifications may have other implications. For example, one can derive additional constraints on the period and structural form of a non-trivial cycle.[14][15][16]

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.[17] (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.)

As proven by Riho Terras, almost every positive integer has a finite stopping time.[18] In other words, almost every Collatz sequence reaches a point that is strictly below its initial value. The proof is based on the distribution of parity vectors and uses the central limit theorem.

In 2019, Terence Tao improved this result by showing, using logarithmic density, that almost all Collatz orbits are descending below any given function of the starting point, provided that this function diverges to infinity, no matter how slowly. Responding to this work, Quanta Magazine wrote that Tao "came away with one of the most significant results on the Collatz conjecture in decades".[19][20]

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.[21]

The length of a non-trivial cycle is known to be at least 17087915. 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.[15]

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.[16] 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.

Steiner (1977) proved that there is no 1-cycle other than the trivial (1; 2).[22] Simons (2005) used Steiner's method to prove that there is no 2-cycle.[23] Simons & de Weger (2005) extended this proof up to 68-cycles: there is no k-cycle up to k = 68.[16] For each k beyond 68, this method gives an upper bound for the smallest term of a k-cycle: for example, if there is a 77-cycle, then at least one element of the cycle is less than 38137×2⁵⁰.[16] Along with the verification of the conjecture up to 2⁶⁸, this implies the nonexistence of a non-trivial k-cycle up to k = 77.[13] As exhaustive computer searches continue, larger k values may be ruled out. To state the argument more intuitively: we need not look for cycles that have at most 77 circuits, where each circuit consists of consecutive ups followed by consecutive downs.

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.[24] 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.[25] 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 0s (that is, repeatedly divide by 2 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][18]

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. For example, 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: for example, 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 four 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 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, that is, 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[27] 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 ${\displaystyle \mathbb {Z} _{2}}$ 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 ${\displaystyle \mathbb {Z} _{2}}$ 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.[28] 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 (with shortcut) can be viewed as the restriction to the integers of the smooth 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 iterations of this map on the real line lead to a dynamical system, further investigated by Chamberland.[29] He showed that the conjecture does not hold for positive real numbers since there are infinitely many fixed points, as well as orbits escaping monotonically to infinity. The function f has two attracting cycles of period 2, (1; 2) and (1.1925...; 2.1386...). Moreover, the set of unbounded orbits is conjectured to be of measure 0.

Letherman, Schleicher, and Wood extended the study to the complex plane, where most of the points have orbits that diverge to infinity (colored region on the illustration).[30] The boundary between the colored region and the black components, namely the Julia set of f, is a fractal pattern, sometimes 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.[31] 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.[14] 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.[32] For example, the only surviving residues mod 32 are 7, 15, 27, and 31.