{"history":[{"id":"5aaa4546-412e-43ac-9459-cc1d79086829","wordChanged":"of","newWord":":::summary\n:::\n\n# \nThe **Collatz conjecture** is obviously true the most famous [unsolved problems in mathematics](https://bliptext.com/articles/list-of-unsolved-problems-in-mathematics).\nThe conjecture asks whether repeating two simple arithmetic operations will eventually transform every [positive integer](https://bliptext.com/articles/positive-integer) into 1.\nIt concerns [sequences of integers](https://bliptext.com/articles/integer-sequence) in which each term is obtained from the previous term as follows: if a term is [even](https://bliptext.com/articles/parity-mathematics), the next term is one half of it.\nIf a term is odd, the next term is 3 times the previous term plus 1.\nThe conjecture is that these sequences always reach 1, no matter which positive integer is chosen to start the sequence.\nThe conjecture has been shown to hold for all positive integers up to 295,000,000,000,000,000,000, but no general proof has been found.\nIt is named after the mathematician [Lothar Collatz](https://bliptext.com/articles/lothar-collatz), who introduced the idea in 1937, two years after receiving his doctorate.\nThe sequence of numbers involved is sometimes referred to as the hailstone sequence, hailstone numbers or hailstone numerals (because the values are usually subject to multiple descents and ascents like [hailstones](https://bliptext.com/articles/hailstones) in a cloud), or as wondrous numbers.\n[Paul Erdős](https://bliptext.com/articles/paul-erd-s) said about the Collatz conjecture: \"Mathematics may not be ready for such problems.\"\n[Jeffrey Lagarias](https://bliptext.com/articles/jeffrey-lagarias) stated in 2010 that the Collatz conjecture \"is an extraordinarily difficult problem, completely out of reach of present day mathematics\".\nHowever, though the Collatz conjecture itself remains open, efforts to solve the problem have led to new techniques and many partial results.\n# [Directed graph showing the orbits of small numbers under the Collatz map, skipping even numbers. The Collatz conjecture states that all paths eventually lead to 1. | upload.wikimedia.org/wikipedia/commons/c/c2/Collatz-graph-50-no27.svg]\n\n# Statement of the problem\nConsider the following operation on an arbitrary [positive integer](https://bliptext.com/articles/positive-integer):\nIn [modular arithmetic](https://bliptext.com/articles/modular-arithmetic) notation, define the [function](https://bliptext.com/articles/function-mathematics) f as follows:\nNow form a sequence by performing this operation repeatedly, beginning with any positive integer, and taking the result at each step as the input at the next.\nIn notation:\n(that is:\nai\nis the value of f applied to n recursively i times;\n).\nThe Collatz conjecture is: ''This process will eventually reach the number 1, regardless of which positive integer is chosen initially.\nThat is, for each'' n, there is some i with a_i = 1.\nIf the conjecture is false, it can only be because there is some starting number which gives rise to a sequence that does not contain 1.\nSuch a sequence would either enter a repeating cycle that excludes 1, or increase without bound.\nNo such sequence has been found.\nThe smallest i such that\nai < a0\nis called the **stopping time** of n. Similarly, the smallest k such that\nis called the **total stopping time** of n. If one of the indexes i or k doesn't exist, we say that the stopping time or the **total stopping time**, respectively, is infinite.\nThe Collatz conjecture asserts that the total stopping time of every n is finite.\nIt is also equivalent to saying that every\nn ≥ 2\nhas a finite stopping time.\nSince\n3n + 1\nis even whenever n is odd, one may instead use the \"shortcut\" form of the Collatz function:\nThis definition yields smaller values for the stopping time and total stopping time without changing the overall dynamics of the process.\n# [Numbers from 1 to 9999 and their corresponding total stopping time | upload.wikimedia.org/wikipedia/commons/b/b9/Collatz-stopping-time.svg]\n# [Histogram of total stopping times for the numbers 1 to 108. Total stopping time is on the x axis, frequency on the y axis. | upload.wikimedia.org/wikipedia/commons/2/26/CollatzStatistic100million.png]\n# [Histogram of total stopping times for the numbers 1 to 109. Total stopping time is on the x axis, frequency on the y axis. | upload.wikimedia.org/wikipedia/commons/0/0d/CollatzStatistic1billion.png]\n# [Iteration time for inputs of 2 to 107. | upload.wikimedia.org/wikipedia/commons/c/c3/Collatz-10Million.png]\n\n# Empirical data\nFor instance, starting with\nand applying the function\nf\nwithout \"shortcut\", one gets the sequence.\nThe number\ntakes longer to reach 1:.\nThe sequence for\n, listed and graphed below, takes 111 steps (41 steps through odd numbers, in bold), climbing as high as 9232 before descending to 1.\nNumbers with a total stopping time longer than that of any smaller starting value form a sequence beginning with:\nThe starting values whose [maximum](https://bliptext.com/articles/maximum) trajectory point is greater than that of any smaller starting value are as follows:\nNumber of steps for n to reach 1 are\nThe starting value having the largest total stopping time while being\nThese numbers are the lowest ones with the indicated step count, but not necessarily the only ones below the given limit.\nAs an example, 9,780,657,631 has 1132 steps, as does 9,780,657,630.\nThe starting values having the smallest total stopping time with respect to their number of digits (in base 2) are the [powers of two](https://bliptext.com/articles/power-of-two) since\n2n\nis halved n times to reach 1, and is never increased.\n# [frameless | upload.wikimedia.org/wikipedia/commons/9/96/Collatz5.svg]\n\n# Visualizations\n\n# Supporting arguments\nAlthough the conjecture has not been proven, most mathematicians who have looked into the problem think the conjecture is true because experimental evidence and heuristic arguments support it.\n\n## Experimental evidence\nThe conjecture has been checked by computer for all starting values up to 268 ≈ 295,000,000,000,000,000,000.\nAll values tested so far converge to 1.\nThis computer evidence is still not rigorous proof that the conjecture is true for all starting values, as [counterexamples](https://bliptext.com/articles/counterexamples) may be found when considering very large (or possibly immense) positive integers, as in the case of the disproven [Pólya conjecture](https://bliptext.com/articles/p-lya-conjecture) and [Mertens conjecture](https://bliptext.com/articles/mertens-conjecture).\nHowever, such verifications may have other implications.\nCertain constraints on any non-trivial cycle, such as [lower bounds](https://bliptext.com/articles/lower-bound) on the length of the cycle, can be proven based on the value of the lowest term in the cycle.\nTherefore, computer searches to rule out cycles that have a small lowest term can strengthen these constraints.\n\n## A probabilistic heuristic\nIf 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.\n(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.\nThe argument is not a proof because it assumes that Hailstone sequences are assembled from uncorrelated probabilistic events.\n(It does rigorously establish that the [2-adic](https://bliptext.com/articles/p-adic-numbers) extension of the Collatz process has two division steps for every multiplication step for [almost all](https://bliptext.com/articles/almost-all) [2-adic](https://bliptext.com/articles/p-adic-numbers) starting values.)\n\n## Stopping times\nAs proven by [Riho Terras](https://bliptext.com/articles/riho-terras-mathematician), almost every positive integer has a finite stopping time.\nIn other words, almost every Collatz sequence reaches a point that is strictly below its initial value.\nThe proof is based on the distribution of [parity vectors](https://bliptext.com/articles/) and uses the [central limit theorem](https://bliptext.com/articles/central-limit-theorem).\nIn 2019, [Terence Tao](https://bliptext.com/articles/terence-tao) improved this result by showing, using logarithmic [density](https://bliptext.com/articles/probability-density-function), that [almost all](https://bliptext.com/articles/almost-all) (in the sense of logarithmic [density](https://bliptext.com/articles/probability-density-function)) Collatz orbits are descending below any given function of the starting point, provided that this function diverges to infinity, no matter how slowly.\nResponding to this work, [Quanta Magazine](https://bliptext.com/articles/quanta-magazine) wrote that Tao \"came away with one of the most significant results on the Collatz conjecture in decades\".\n\n## Lower bounds\nIn a [computer-aided proof](https://bliptext.com/articles/computer-aided-proof), Krasikov and Lagarias showed that the number of integers in the interval\n[1,x]\nthat eventually reach 1 is at least equal to\nx0.84\nfor all sufficiently large x.\n\n# Cycles\nIn this part, consider the shortcut form of the Collatz function\nA [cycle](https://bliptext.com/articles/periodic-sequence) is a sequence\n(a0, a1, ..., aq)\nof distinct positive integers where\n,\n, ..., and\n.\nThe only known cycle is\n(1,2)\nof period 2, called the trivial cycle.\n\n## Cycle length\nThe length of a non-trivial cycle is known to be at least 114,208,327,604 (or 186,265,759,595 without shortcut).\nIf it can be shown that for all positive integers less than the Collatz sequences reach 1, then this bound would raise to 217,976,794,617 (355,504,839,929 without shortcut).\nIn fact, Eliahou (1993) proved that the period p of any non-trivial cycle is of the form\nwhere a, b and c are non-negative integers,\nb ≥ 1\nand\nac = 0\n. This result is based on the [simple continued fraction](https://bliptext.com/articles/simple-continued-fraction) expansion of\nln 3⁄ln 2\n.\n\n## k-cycles\nA k-cycle is a cycle that can be partitioned into\nk\ncontiguous subsequences, each consisting of an increasing sequence of odd numbers, followed by a decreasing sequence of even numbers.\nFor 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.\nSteiner (1977) proved that there is no 1-cycle other than the trivial\n(1; 2)\n. Simons (2005) used Steiner's method to prove that there is no 2-cycle.\nSimons and de Weger (2005) extended this proof up to 68-cycles; there is no k-cycle up to\n. Hercher extended the method further and proved that there exists no k-cycle with\nk ≤ 91\n. As exhaustive computer searches continue, larger\nk\nvalues may be ruled out.\nTo state the argument more intuitively; we do not have to search for cycles that have less than 92 subsequences, where each subsequence consists of consecutive ups followed by consecutive downs.\n\n# Other formulations of the conjecture\n\n## In reverse\nThere is another approach to prove the conjecture, which considers the bottom-up\nmethod of growing the so-called Collatz graph.\nThe Collatz [graph](https://bliptext.com/articles/graph-discrete-mathematics) is a [graph](https://bliptext.com/articles/graph-discrete-mathematics) defined by the inverse [relation](https://bliptext.com/articles/relation-mathematics)\nSo, instead of proving that all positive integers eventually lead to 1, we can try to prove that 1 leads backwards to all positive integers.\nFor any integer n,\nn ≡ 1 (mod 2)\n[if and only if](https://bliptext.com/articles/if-and-only-if)\n3n + 1 ≡ 4 (mod 6)\n. Equivalently,\nn − 1⁄3 ≡ 1 (mod 2)\nif and only if\nn ≡ 4 (mod 6)\n. Conjecturally, this inverse relation forms a [tree](https://bliptext.com/articles/tree-graph-theory) 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](https://bliptext.com/articles/) section of this article).\nWhen the relation\n3n + 1\nof the function f is replaced by the common substitute \"shortcut\" relation\n3n + 1⁄2\n, the Collatz graph is defined by the inverse relation,\nFor any integer n,\nn ≡ 1 (mod 2)\nif and only if\n3n + 1⁄2 ≡ 2 (mod 3)\n. Equivalently,\n2n − 1⁄3 ≡ 1 (mod 2)\nif and only if\nn ≡ 2 (mod 3)\n. 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).\nAlternatively, replace the\n3n + 1\nwith\nn⁄H(n)\nwhere\nand\nH(n)\nis the highest [power of 2](https://bliptext.com/articles/power-of-2) that divides\nn\n(with no [remainder](https://bliptext.com/articles/remainder)).\nThe resulting function f maps from [odd numbers](https://bliptext.com/articles/odd-number) to [odd numbers](https://bliptext.com/articles/odd-number).\nNow suppose that for some odd number n, applying this operation k times yields the number 1 (that is,\n). Then in [binary](https://bliptext.com/articles/binary-number), the number n can be written as the concatenation of [strings](https://bliptext.com/articles/string-computer-science)\nwk wk−1 ...\nw1\nwhere each\nwh\nis a finite and contiguous extract from the representation of\n1⁄3h\n. The representation of n therefore holds the [repetends](https://bliptext.com/articles/repeating-decimal) of\n1⁄3h\n, where each repetend is optionally rotated and then replicated up to a finite number of bits.\nIt is only in binary that this occurs.\nConjecturally, 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).\n# [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. | upload.wikimedia.org/wikipedia/commons/0/0e/Collatz-tree%2C///depth%3D20.svg]\n\n## As an abstract machine that computes in base two\nRepeated applications of the Collatz function can be represented as an [abstract machine](https://bliptext.com/articles/abstract-machine) that handles [strings](https://bliptext.com/articles/string-computer-science) of [bits](https://bliptext.com/articles/bit).\nThe machine will perform the following three steps on any odd number until only one 1 remains:\n2n + 1\n);\n2n + 1 + n = 3n + 1\n);\n\n### Example\nThe starting number 7 is written in base two as 111.\nThe resulting Collatz sequence is:\n**********1********************1********************1********** **********1********************1********************1********************1********** **********1**********0**********1********************1********** 0 **********1**********0**********1********************1********************1********** **********1**********000**********1********** 0 **********1**********000**********1********************1********** **********1********************1**********0**********1********** 00 **********1********************1**********0**********1********************1********** **********1**********0**********1********** 000 **********1**********0**********1********************1********** **********1********** 0000\n\n## As a parity sequence\nFor this section, consider the shortcut form of the Collatz function\nIf\nP(...)\nis the parity of a number, that is\nand\nP(2n + 1) = 1\n, then we can define the Collatz parity sequence (or parity vector) for a number n as\n, where\n, and\nai+1 = f(ai)\n.\nWhich operation is performed,\n3n + 1⁄2\nor\nn⁄2\n, depends on the parity.\nThe parity sequence is the same as the sequence of operations.\nUsing this form for\nf(n)\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\n2k\n. 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.\nApplying the f function k times to the number\nwill give the result\n3ca + d\n, 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.\nFor example, for\n25a + 1\nthere are 3 increases as 1 iterates to 2, 1, 2, 1, and finally to 2 so the result is\n33a + 2\n22a + 1\nthere is only 1 increase as 1 rises to 2 and falls to 1 so the result is\n3a + 1\n. When b is\n2k − 1\nthen there will be k rises and the result will be\n3ka + 3k − 1\n. The power 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,\n4a + 1\nbecomes\n3a + 1\nafter two applications of f and\n16a + 3\nbecomes\n9a + 2\nafter four applications of f. Whether those smaller numbers continue to 1, however, depends on the value of a.\n\n## As a tag system\nFor the Collatz function in the shortcut form\nHailstone sequences can be computed by the [2-tag system](https://bliptext.com/articles/tag-system) with production rules\na → bc\n,\nb → a\n,\nc → aaa\n.\nIn 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.\n(Adapted from De Mol.)\nThe Collatz conjecture equivalently states that this tag system, with an arbitrary finite string of a as the initial word, eventually halts (see [Tag system](https://bliptext.com/articles/tag-system) for a worked example).\n\n# Extensions to larger domains\n\n## Iterating on all integers\nAn extension to the Collatz conjecture is to include all integers, not just positive integers.\nLeaving 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:\nOdd values are listed in large bold.\nEach cycle is listed with its member of least absolute value (which is always odd) first.\nThe 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.\n\n## Iterating on rationals with odd denominators\nThe Collatz map can be extended to (positive or negative) rational numbers which have odd denominators when written in lowest terms.\nThe number is taken to be 'odd' or 'even' according to whether its numerator is odd or even.\nThen 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.\nA closely related fact is that the Collatz map extends to the ring of [2-adic integers](https://bliptext.com/articles/2-adic-integers), which contains the ring of rationals with odd denominators as a subring.\nWhen using the \"shortcut\" definition of the Collatz map, it is known that any periodic [parity sequence](https://bliptext.com/articles/) is generated by exactly one rational.\nConversely, it is conjectured that every rational with an odd denominator has an eventually cyclic parity sequence (Periodicity Conjecture ).\nIf a parity cycle has length n and includes odd numbers exactly m times at indices\nk0 < ⋯ < km−1\n, then the unique rational which generates immediately and periodically this parity cycle is\nFor 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.\nIt is repeatedly generated by the fraction\nas the latter leads to the rational cycle\nAny cyclic permutation of (1 0 1 1 0 0 1) is associated to one of the above fractions.\nFor instance, the cycle (0 1 1 0 0 1 1) is produced by the fraction\nFor a one-to-one correspondence, a parity cycle should be irreducible, that is, not partitionable into identical sub-cycles.\nAs 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.\nIn 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).\nIf 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 \"\n3n + d\n\" generalization of the Collatz function\n\n## 2-adic extension\nThe function\nis well-defined on the ring of [2-adic integers](https://bliptext.com/articles/2-adic-integers), where it is continuous and [measure-preserving](https://bliptext.com/articles/measure-preserving-transformation) with respect to the 2-adic measure.\nMoreover, its dynamics is known to be [ergodic](https://bliptext.com/articles/ergodic-theory).\nDefine the parity vector function Q acting on as\nThe function Q is a 2-adic [isometry](https://bliptext.com/articles/isometry).\nConsequently, every infinite parity sequence occurs for exactly one 2-adic integer, so that [almost all](https://bliptext.com/articles/almost-all) trajectories are acyclic in.\nAn equivalent formulation of the Collatz conjecture is that\n\n## Iterating on real or complex numbers\nThe Collatz map can be extended to the [real line](https://bliptext.com/articles/real-line) by choosing any function which evaluates to x/2 when x is an even integer, and to either 3x + 1 or (3x + 1)/2 (for the \"shortcut\" version) when x is an odd integer.\nThis is called an [interpolating](https://bliptext.com/articles/interpolating) function.\nA simple way to do this is to pick two functions g_1 and g_2, where:\nand use them as switches for our desired values:\nOne such choice is and.\nThe [iterations](https://bliptext.com/articles/iterations) of this map lead to a [dynamical system](https://bliptext.com/articles/dynamical-system), further investigated by Marc Chamberland.\nHe showed that the conjecture does not hold for positive real numbers since there are infinitely many [fixed points](https://bliptext.com/articles/fixed-point-mathematics), as well as [orbits](https://bliptext.com/articles/orbit-dynamics) escaping [monotonically](https://bliptext.com/articles/monotonic-function) to infinity.\nThe function f has two [attracting](https://bliptext.com/articles/attractor) cycles of period 2: (1;\\,2) and.\nMoreover, the set of unbounded orbits is conjectured to be of [measure](https://bliptext.com/articles/lebesgue-measure) 0.\nLetherman, Schleicher, and Wood extended the study to the [complex plane](https://bliptext.com/articles/complex-plane).\nThey used Chamberland's function for [complex sine and cosine](https://bliptext.com/articles/trigonometric-functions) and added the extra term\n, where h(z) is any [entire function](https://bliptext.com/articles/entire-function).\nSince this expression evaluates to zero for real integers, the extended function\nis an interpolation of the Collatz map to the complex plane.\nThe reason for adding the extra term is to make all integers [critical points](https://bliptext.com/articles/critical-point-mathematics) of f. With this, they show that no integer is in a [Baker domain](https://bliptext.com/articles/classification-of-fatou-components), which implies that any integer is either eventually periodic or belongs to a [wandering domain](https://bliptext.com/articles/wandering-set).\nThey conjectured that the latter is not the case, which would make all integer orbits finite.\nMost of the points have orbits that diverge to infinity.\nColoring these points based on how fast they diverge produces the image on the left, for.\nThe inner black regions and the outer region are the [Fatou components](https://bliptext.com/articles/classification-of-fatou-components), and the boundary between them is the [Julia set](https://bliptext.com/articles/julia-set) of f, which forms a [fractal](https://bliptext.com/articles/fractal) pattern, sometimes called a \"Collatz [fractal](https://bliptext.com/articles/fractal)\".\nThere are many other ways to define a complex interpolating function, such as using the [complex exponential](https://bliptext.com/articles/exponential-function) instead of sine and cosine:\nwhich exhibit different dynamics.\nIn this case, for instance, if, then.\nThe corresponding Julia set, shown on the right, consists of uncountably many curves, called hairs, or rays.\n# [Cobweb plot of the orbit 10 → 5 → 8 → 4 → 2 → 1 → ... in an extension of the Collatz map to the real line. | upload.wikimedia.org/wikipedia/commons/b/b4/Collatz///Cobweb.svg]\n# [A Collatz fractal centered at the origin, with real parts from -5 to 5. | upload.wikimedia.org/wikipedia/commons/8/8c/Collatz///Fractal.jpg]\n# [Julia set of the exponential interpolation. | upload.wikimedia.org/wikipedia/commons/7/74/Exponential///Collatz///Fractal.jpg]\n\n# Optimizations\n\n## Time–space tradeoff\nThe section [As a parity sequence](https://bliptext.com/articles/) above gives a way to speed up simulation of the sequence.\nTo 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).\nThe result of jumping ahead k is given by\nf(2ka + b) = 3c(b, k)a + d(b, k)\n. The values of c (or better\n3c\n) and d can be precalculated for all possible k-bit numbers b, where\nd(b, k)\nis the result of applying the f function k times to b, and\nc(b, k)\nis the number of odd numbers encountered on the way.\nFor example, if\n, one can jump ahead 5 steps on each iteration by separating out the 5 least significant bits of a number and using\nThis requires\n2k\n[precomputation](https://bliptext.com/articles/precomputation) and storage to speed up the resulting calculation by a factor of k, a [space–time tradeoff](https://bliptext.com/articles/space-time-tradeoff).\n\n## Modular restrictions\nFor 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\nf(2ka + b) = 3c(b)a + d(b) < 2ka + b\nholds for all a, then the first counterexample, if it exists, cannot be b modulo\n2k\n. For instance, the first counterexample must be odd because\n, smaller than\n2n\nf(4n + 1) = 3n + 1\n, smaller than\n4n + 1\n. 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.\nAs 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.\nFor example, the only surviving residues mod 32 are 7, 15, 27, and 31.\nIntegers divisible by 3 cannot form a cycle, so these integers do not need to be checked as counter examples.\n\n# Syracuse function\nIf k is an odd integer, then\n3k + 1\nis even, so\n3k + 1 = 2ak\nwith\nk\nodd and\na ≥ 1\n. The **Syracuse function** is the function f from the set I of positive odd integers into itself, for which\n.\nSome properties of the Syracuse function are:\nk ∈ I\n,\nf(4k + 1) = f(k)\n. (Because\n3(4k + 1) + 1 = 12k + 4 = 4(3k + 1)\n.)\np ≥ 1\nand odd h,\nf(2ph − 1) = 2 × 3p − 1h − 1\n. (Here\nf\nis [function iteration notation](https://bliptext.com/articles/functional-power).)\nf(2h − 1) ≤ 3h − 1⁄2\nThe Collatz conjecture is equivalent to the statement that, for all k in I, there exists an integer\nn ≥ 1\nsuch that\n.\n\n# Undecidable generalizations\nIn 1972, [John Horton Conway](https://bliptext.com/articles/john-horton-conway) proved that a natural generalization of the Collatz problem is algorithmically [undecidable](https://bliptext.com/articles/undecidable-problem).\nSpecifically, he considered functions of the form\nwhere\na0, b0, ..., aP − 1, bP − 1\nare rational numbers which are so chosen that\ng(n)\nis always an integer.\nThe standard Collatz function is given by\n,\n,\n,\n,\n. Conway proved that the problem\ngk(n)\nreach\n1\n?\nis undecidable, by representing the [halting problem](https://bliptext.com/articles/halting-problem) in this way.\nCloser to the Collatz problem is the following universally quantified problem:\ngk(n)\nreach\n1\n, for all\nn > 0\n?\nModifying the condition in this way can make a problem either harder or easier to solve (intuitively, it is harder to justify a positive answer but might be easier to justify a negative one).\nKurtz and Simon proved that the universally quantified problem is, in fact, undecidable and even higher in the [arithmetical hierarchy](https://bliptext.com/articles/arithmetical-hierarchy); specifically, it is\nΠ0 2\n-complete.\nThis hardness result holds even if one restricts the class of functions g by fixing the modulus P to 6480.\nIterations of g in a simplified version of this form, with all b_i equal to zero, are formalized in an [esoteric programming language](https://bliptext.com/articles/esoteric-programming-language) called [FRACTRAN](https://bliptext.com/articles/fractran).\n\n# In computational complexity\nCollatz and related conjectures are often used when studying computation complexity.\nThe connection is made through the [Busy Beaver](https://bliptext.com/articles/busy-beaver) function, where BB(n) is the maximum number of steps taken by any n state [Turing machine](https://bliptext.com/articles/turing-machine) that halts.\nThere is a 15 state Turing machine that halts if and only if a conjecture by [Paul Erdős](https://bliptext.com/articles/paul-erd-s) (closely related to the Collatz conjecture) is false.\nHence if BB(15) was known, and this machine did not stop in that number of steps, it would be known to run forever and hence no counterexamples exist (which proves the conjecture true).\nThis is a completely impractical way to settle the conjecture; instead it is used to suggest that BB(15) will be very hard to compute, at least as difficult as settling this Collatz-like conjecture.","wordIndex":5,"createdAt":"2026-01-20T07:52:49.777Z","user":{"id":"mfvswiH9fwHI8f36yNwHR40968SZeIKD","name":"den3den42069","image":"https://cdn.discordapp.com/avatars/975452809707278427/5e0588882214f34dc5f3b36f7e9c9ff5.png"}},{"id":"18c025b7-b806-4de7-b029-08f8068eeb20","wordChanged":"one","newWord":":::summary\n:::\n\n# \nThe **Collatz conjecture** is obviously of the most famous [unsolved problems in mathematics](https://bliptext.com/articles/list-of-unsolved-problems-in-mathematics).\nThe conjecture asks whether repeating two simple arithmetic operations will eventually transform every [positive integer](https://bliptext.com/articles/positive-integer) into 1.\nIt concerns [sequences of integers](https://bliptext.com/articles/integer-sequence) in which each term is obtained from the previous term as follows: if a term is [even](https://bliptext.com/articles/parity-mathematics), the next term is one half of it.\nIf a term is odd, the next term is 3 times the previous term plus 1.\nThe conjecture is that these sequences always reach 1, no matter which positive integer is chosen to start the sequence.\nThe conjecture has been shown to hold for all positive integers up to 295,000,000,000,000,000,000, but no general proof has been found.\nIt is named after the mathematician [Lothar Collatz](https://bliptext.com/articles/lothar-collatz), who introduced the idea in 1937, two years after receiving his doctorate.\nThe sequence of numbers involved is sometimes referred to as the hailstone sequence, hailstone numbers or hailstone numerals (because the values are usually subject to multiple descents and ascents like [hailstones](https://bliptext.com/articles/hailstones) in a cloud), or as wondrous numbers.\n[Paul Erdős](https://bliptext.com/articles/paul-erd-s) said about the Collatz conjecture: \"Mathematics may not be ready for such problems.\"\n[Jeffrey Lagarias](https://bliptext.com/articles/jeffrey-lagarias) stated in 2010 that the Collatz conjecture \"is an extraordinarily difficult problem, completely out of reach of present day mathematics\".\nHowever, though the Collatz conjecture itself remains open, efforts to solve the problem have led to new techniques and many partial results.\n# [Directed graph showing the orbits of small numbers under the Collatz map, skipping even numbers. The Collatz conjecture states that all paths eventually lead to 1. | upload.wikimedia.org/wikipedia/commons/c/c2/Collatz-graph-50-no27.svg]\n\n# Statement of the problem\nConsider the following operation on an arbitrary [positive integer](https://bliptext.com/articles/positive-integer):\nIn [modular arithmetic](https://bliptext.com/articles/modular-arithmetic) notation, define the [function](https://bliptext.com/articles/function-mathematics) f as follows:\nNow form a sequence by performing this operation repeatedly, beginning with any positive integer, and taking the result at each step as the input at the next.\nIn notation:\n(that is:\nai\nis the value of f applied to n recursively i times;\n).\nThe Collatz conjecture is: ''This process will eventually reach the number 1, regardless of which positive integer is chosen initially.\nThat is, for each'' n, there is some i with a_i = 1.\nIf the conjecture is false, it can only be because there is some starting number which gives rise to a sequence that does not contain 1.\nSuch a sequence would either enter a repeating cycle that excludes 1, or increase without bound.\nNo such sequence has been found.\nThe smallest i such that\nai < a0\nis called the **stopping time** of n. Similarly, the smallest k such that\nis called the **total stopping time** of n. If one of the indexes i or k doesn't exist, we say that the stopping time or the **total stopping time**, respectively, is infinite.\nThe Collatz conjecture asserts that the total stopping time of every n is finite.\nIt is also equivalent to saying that every\nn ≥ 2\nhas a finite stopping time.\nSince\n3n + 1\nis even whenever n is odd, one may instead use the \"shortcut\" form of the Collatz function:\nThis definition yields smaller values for the stopping time and total stopping time without changing the overall dynamics of the process.\n# [Numbers from 1 to 9999 and their corresponding total stopping time | upload.wikimedia.org/wikipedia/commons/b/b9/Collatz-stopping-time.svg]\n# [Histogram of total stopping times for the numbers 1 to 108. Total stopping time is on the x axis, frequency on the y axis. | upload.wikimedia.org/wikipedia/commons/2/26/CollatzStatistic100million.png]\n# [Histogram of total stopping times for the numbers 1 to 109. Total stopping time is on the x axis, frequency on the y axis. | upload.wikimedia.org/wikipedia/commons/0/0d/CollatzStatistic1billion.png]\n# [Iteration time for inputs of 2 to 107. | upload.wikimedia.org/wikipedia/commons/c/c3/Collatz-10Million.png]\n\n# Empirical data\nFor instance, starting with\nand applying the function\nf\nwithout \"shortcut\", one gets the sequence.\nThe number\ntakes longer to reach 1:.\nThe sequence for\n, listed and graphed below, takes 111 steps (41 steps through odd numbers, in bold), climbing as high as 9232 before descending to 1.\nNumbers with a total stopping time longer than that of any smaller starting value form a sequence beginning with:\nThe starting values whose [maximum](https://bliptext.com/articles/maximum) trajectory point is greater than that of any smaller starting value are as follows:\nNumber of steps for n to reach 1 are\nThe starting value having the largest total stopping time while being\nThese numbers are the lowest ones with the indicated step count, but not necessarily the only ones below the given limit.\nAs an example, 9,780,657,631 has 1132 steps, as does 9,780,657,630.\nThe starting values having the smallest total stopping time with respect to their number of digits (in base 2) are the [powers of two](https://bliptext.com/articles/power-of-two) since\n2n\nis halved n times to reach 1, and is never increased.\n# [frameless | upload.wikimedia.org/wikipedia/commons/9/96/Collatz5.svg]\n\n# Visualizations\n\n# Supporting arguments\nAlthough the conjecture has not been proven, most mathematicians who have looked into the problem think the conjecture is true because experimental evidence and heuristic arguments support it.\n\n## Experimental evidence\nThe conjecture has been checked by computer for all starting values up to 268 ≈ 295,000,000,000,000,000,000.\nAll values tested so far converge to 1.\nThis computer evidence is still not rigorous proof that the conjecture is true for all starting values, as [counterexamples](https://bliptext.com/articles/counterexamples) may be found when considering very large (or possibly immense) positive integers, as in the case of the disproven [Pólya conjecture](https://bliptext.com/articles/p-lya-conjecture) and [Mertens conjecture](https://bliptext.com/articles/mertens-conjecture).\nHowever, such verifications may have other implications.\nCertain constraints on any non-trivial cycle, such as [lower bounds](https://bliptext.com/articles/lower-bound) on the length of the cycle, can be proven based on the value of the lowest term in the cycle.\nTherefore, computer searches to rule out cycles that have a small lowest term can strengthen these constraints.\n\n## A probabilistic heuristic\nIf 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.\n(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.\nThe argument is not a proof because it assumes that Hailstone sequences are assembled from uncorrelated probabilistic events.\n(It does rigorously establish that the [2-adic](https://bliptext.com/articles/p-adic-numbers) extension of the Collatz process has two division steps for every multiplication step for [almost all](https://bliptext.com/articles/almost-all) [2-adic](https://bliptext.com/articles/p-adic-numbers) starting values.)\n\n## Stopping times\nAs proven by [Riho Terras](https://bliptext.com/articles/riho-terras-mathematician), almost every positive integer has a finite stopping time.\nIn other words, almost every Collatz sequence reaches a point that is strictly below its initial value.\nThe proof is based on the distribution of [parity vectors](https://bliptext.com/articles/) and uses the [central limit theorem](https://bliptext.com/articles/central-limit-theorem).\nIn 2019, [Terence Tao](https://bliptext.com/articles/terence-tao) improved this result by showing, using logarithmic [density](https://bliptext.com/articles/probability-density-function), that [almost all](https://bliptext.com/articles/almost-all) (in the sense of logarithmic [density](https://bliptext.com/articles/probability-density-function)) Collatz orbits are descending below any given function of the starting point, provided that this function diverges to infinity, no matter how slowly.\nResponding to this work, [Quanta Magazine](https://bliptext.com/articles/quanta-magazine) wrote that Tao \"came away with one of the most significant results on the Collatz conjecture in decades\".\n\n## Lower bounds\nIn a [computer-aided proof](https://bliptext.com/articles/computer-aided-proof), Krasikov and Lagarias showed that the number of integers in the interval\n[1,x]\nthat eventually reach 1 is at least equal to\nx0.84\nfor all sufficiently large x.\n\n# Cycles\nIn this part, consider the shortcut form of the Collatz function\nA [cycle](https://bliptext.com/articles/periodic-sequence) is a sequence\n(a0, a1, ..., aq)\nof distinct positive integers where\n,\n, ..., and\n.\nThe only known cycle is\n(1,2)\nof period 2, called the trivial cycle.\n\n## Cycle length\nThe length of a non-trivial cycle is known to be at least 114,208,327,604 (or 186,265,759,595 without shortcut).\nIf it can be shown that for all positive integers less than the Collatz sequences reach 1, then this bound would raise to 217,976,794,617 (355,504,839,929 without shortcut).\nIn fact, Eliahou (1993) proved that the period p of any non-trivial cycle is of the form\nwhere a, b and c are non-negative integers,\nb ≥ 1\nand\nac = 0\n. This result is based on the [simple continued fraction](https://bliptext.com/articles/simple-continued-fraction) expansion of\nln 3⁄ln 2\n.\n\n## k-cycles\nA k-cycle is a cycle that can be partitioned into\nk\ncontiguous subsequences, each consisting of an increasing sequence of odd numbers, followed by a decreasing sequence of even numbers.\nFor 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.\nSteiner (1977) proved that there is no 1-cycle other than the trivial\n(1; 2)\n. Simons (2005) used Steiner's method to prove that there is no 2-cycle.\nSimons and de Weger (2005) extended this proof up to 68-cycles; there is no k-cycle up to\n. Hercher extended the method further and proved that there exists no k-cycle with\nk ≤ 91\n. As exhaustive computer searches continue, larger\nk\nvalues may be ruled out.\nTo state the argument more intuitively; we do not have to search for cycles that have less than 92 subsequences, where each subsequence consists of consecutive ups followed by consecutive downs.\n\n# Other formulations of the conjecture\n\n## In reverse\nThere is another approach to prove the conjecture, which considers the bottom-up\nmethod of growing the so-called Collatz graph.\nThe Collatz [graph](https://bliptext.com/articles/graph-discrete-mathematics) is a [graph](https://bliptext.com/articles/graph-discrete-mathematics) defined by the inverse [relation](https://bliptext.com/articles/relation-mathematics)\nSo, instead of proving that all positive integers eventually lead to 1, we can try to prove that 1 leads backwards to all positive integers.\nFor any integer n,\nn ≡ 1 (mod 2)\n[if and only if](https://bliptext.com/articles/if-and-only-if)\n3n + 1 ≡ 4 (mod 6)\n. Equivalently,\nn − 1⁄3 ≡ 1 (mod 2)\nif and only if\nn ≡ 4 (mod 6)\n. Conjecturally, this inverse relation forms a [tree](https://bliptext.com/articles/tree-graph-theory) 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](https://bliptext.com/articles/) section of this article).\nWhen the relation\n3n + 1\nof the function f is replaced by the common substitute \"shortcut\" relation\n3n + 1⁄2\n, the Collatz graph is defined by the inverse relation,\nFor any integer n,\nn ≡ 1 (mod 2)\nif and only if\n3n + 1⁄2 ≡ 2 (mod 3)\n. Equivalently,\n2n − 1⁄3 ≡ 1 (mod 2)\nif and only if\nn ≡ 2 (mod 3)\n. 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).\nAlternatively, replace the\n3n + 1\nwith\nn⁄H(n)\nwhere\nand\nH(n)\nis the highest [power of 2](https://bliptext.com/articles/power-of-2) that divides\nn\n(with no [remainder](https://bliptext.com/articles/remainder)).\nThe resulting function f maps from [odd numbers](https://bliptext.com/articles/odd-number) to [odd numbers](https://bliptext.com/articles/odd-number).\nNow suppose that for some odd number n, applying this operation k times yields the number 1 (that is,\n). Then in [binary](https://bliptext.com/articles/binary-number), the number n can be written as the concatenation of [strings](https://bliptext.com/articles/string-computer-science)\nwk wk−1 ...\nw1\nwhere each\nwh\nis a finite and contiguous extract from the representation of\n1⁄3h\n. The representation of n therefore holds the [repetends](https://bliptext.com/articles/repeating-decimal) of\n1⁄3h\n, where each repetend is optionally rotated and then replicated up to a finite number of bits.\nIt is only in binary that this occurs.\nConjecturally, 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).\n# [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. | upload.wikimedia.org/wikipedia/commons/0/0e/Collatz-tree%2C///depth%3D20.svg]\n\n## As an abstract machine that computes in base two\nRepeated applications of the Collatz function can be represented as an [abstract machine](https://bliptext.com/articles/abstract-machine) that handles [strings](https://bliptext.com/articles/string-computer-science) of [bits](https://bliptext.com/articles/bit).\nThe machine will perform the following three steps on any odd number until only one 1 remains:\n2n + 1\n);\n2n + 1 + n = 3n + 1\n);\n\n### Example\nThe starting number 7 is written in base two as 111.\nThe resulting Collatz sequence is:\n**********1********************1********************1********** **********1********************1********************1********************1********** **********1**********0**********1********************1********** 0 **********1**********0**********1********************1********************1********** **********1**********000**********1********** 0 **********1**********000**********1********************1********** **********1********************1**********0**********1********** 00 **********1********************1**********0**********1********************1********** **********1**********0**********1********** 000 **********1**********0**********1********************1********** **********1********** 0000\n\n## As a parity sequence\nFor this section, consider the shortcut form of the Collatz function\nIf\nP(...)\nis the parity of a number, that is\nand\nP(2n + 1) = 1\n, then we can define the Collatz parity sequence (or parity vector) for a number n as\n, where\n, and\nai+1 = f(ai)\n.\nWhich operation is performed,\n3n + 1⁄2\nor\nn⁄2\n, depends on the parity.\nThe parity sequence is the same as the sequence of operations.\nUsing this form for\nf(n)\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\n2k\n. 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.\nApplying the f function k times to the number\nwill give the result\n3ca + d\n, 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.\nFor example, for\n25a + 1\nthere are 3 increases as 1 iterates to 2, 1, 2, 1, and finally to 2 so the result is\n33a + 2\n22a + 1\nthere is only 1 increase as 1 rises to 2 and falls to 1 so the result is\n3a + 1\n. When b is\n2k − 1\nthen there will be k rises and the result will be\n3ka + 3k − 1\n. The power 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,\n4a + 1\nbecomes\n3a + 1\nafter two applications of f and\n16a + 3\nbecomes\n9a + 2\nafter four applications of f. Whether those smaller numbers continue to 1, however, depends on the value of a.\n\n## As a tag system\nFor the Collatz function in the shortcut form\nHailstone sequences can be computed by the [2-tag system](https://bliptext.com/articles/tag-system) with production rules\na → bc\n,\nb → a\n,\nc → aaa\n.\nIn 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.\n(Adapted from De Mol.)\nThe Collatz conjecture equivalently states that this tag system, with an arbitrary finite string of a as the initial word, eventually halts (see [Tag system](https://bliptext.com/articles/tag-system) for a worked example).\n\n# Extensions to larger domains\n\n## Iterating on all integers\nAn extension to the Collatz conjecture is to include all integers, not just positive integers.\nLeaving 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:\nOdd values are listed in large bold.\nEach cycle is listed with its member of least absolute value (which is always odd) first.\nThe 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.\n\n## Iterating on rationals with odd denominators\nThe Collatz map can be extended to (positive or negative) rational numbers which have odd denominators when written in lowest terms.\nThe number is taken to be 'odd' or 'even' according to whether its numerator is odd or even.\nThen 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.\nA closely related fact is that the Collatz map extends to the ring of [2-adic integers](https://bliptext.com/articles/2-adic-integers), which contains the ring of rationals with odd denominators as a subring.\nWhen using the \"shortcut\" definition of the Collatz map, it is known that any periodic [parity sequence](https://bliptext.com/articles/) is generated by exactly one rational.\nConversely, it is conjectured that every rational with an odd denominator has an eventually cyclic parity sequence (Periodicity Conjecture ).\nIf a parity cycle has length n and includes odd numbers exactly m times at indices\nk0 < ⋯ < km−1\n, then the unique rational which generates immediately and periodically this parity cycle is\nFor 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.\nIt is repeatedly generated by the fraction\nas the latter leads to the rational cycle\nAny cyclic permutation of (1 0 1 1 0 0 1) is associated to one of the above fractions.\nFor instance, the cycle (0 1 1 0 0 1 1) is produced by the fraction\nFor a one-to-one correspondence, a parity cycle should be irreducible, that is, not partitionable into identical sub-cycles.\nAs 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.\nIn 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).\nIf 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 \"\n3n + d\n\" generalization of the Collatz function\n\n## 2-adic extension\nThe function\nis well-defined on the ring of [2-adic integers](https://bliptext.com/articles/2-adic-integers), where it is continuous and [measure-preserving](https://bliptext.com/articles/measure-preserving-transformation) with respect to the 2-adic measure.\nMoreover, its dynamics is known to be [ergodic](https://bliptext.com/articles/ergodic-theory).\nDefine the parity vector function Q acting on as\nThe function Q is a 2-adic [isometry](https://bliptext.com/articles/isometry).\nConsequently, every infinite parity sequence occurs for exactly one 2-adic integer, so that [almost all](https://bliptext.com/articles/almost-all) trajectories are acyclic in.\nAn equivalent formulation of the Collatz conjecture is that\n\n## Iterating on real or complex numbers\nThe Collatz map can be extended to the [real line](https://bliptext.com/articles/real-line) by choosing any function which evaluates to x/2 when x is an even integer, and to either 3x + 1 or (3x + 1)/2 (for the \"shortcut\" version) when x is an odd integer.\nThis is called an [interpolating](https://bliptext.com/articles/interpolating) function.\nA simple way to do this is to pick two functions g_1 and g_2, where:\nand use them as switches for our desired values:\nOne such choice is and.\nThe [iterations](https://bliptext.com/articles/iterations) of this map lead to a [dynamical system](https://bliptext.com/articles/dynamical-system), further investigated by Marc Chamberland.\nHe showed that the conjecture does not hold for positive real numbers since there are infinitely many [fixed points](https://bliptext.com/articles/fixed-point-mathematics), as well as [orbits](https://bliptext.com/articles/orbit-dynamics) escaping [monotonically](https://bliptext.com/articles/monotonic-function) to infinity.\nThe function f has two [attracting](https://bliptext.com/articles/attractor) cycles of period 2: (1;\\,2) and.\nMoreover, the set of unbounded orbits is conjectured to be of [measure](https://bliptext.com/articles/lebesgue-measure) 0.\nLetherman, Schleicher, and Wood extended the study to the [complex plane](https://bliptext.com/articles/complex-plane).\nThey used Chamberland's function for [complex sine and cosine](https://bliptext.com/articles/trigonometric-functions) and added the extra term\n, where h(z) is any [entire function](https://bliptext.com/articles/entire-function).\nSince this expression evaluates to zero for real integers, the extended function\nis an interpolation of the Collatz map to the complex plane.\nThe reason for adding the extra term is to make all integers [critical points](https://bliptext.com/articles/critical-point-mathematics) of f. With this, they show that no integer is in a [Baker domain](https://bliptext.com/articles/classification-of-fatou-components), which implies that any integer is either eventually periodic or belongs to a [wandering domain](https://bliptext.com/articles/wandering-set).\nThey conjectured that the latter is not the case, which would make all integer orbits finite.\nMost of the points have orbits that diverge to infinity.\nColoring these points based on how fast they diverge produces the image on the left, for.\nThe inner black regions and the outer region are the [Fatou components](https://bliptext.com/articles/classification-of-fatou-components), and the boundary between them is the [Julia set](https://bliptext.com/articles/julia-set) of f, which forms a [fractal](https://bliptext.com/articles/fractal) pattern, sometimes called a \"Collatz [fractal](https://bliptext.com/articles/fractal)\".\nThere are many other ways to define a complex interpolating function, such as using the [complex exponential](https://bliptext.com/articles/exponential-function) instead of sine and cosine:\nwhich exhibit different dynamics.\nIn this case, for instance, if, then.\nThe corresponding Julia set, shown on the right, consists of uncountably many curves, called hairs, or rays.\n# [Cobweb plot of the orbit 10 → 5 → 8 → 4 → 2 → 1 → ... in an extension of the Collatz map to the real line. | upload.wikimedia.org/wikipedia/commons/b/b4/Collatz///Cobweb.svg]\n# [A Collatz fractal centered at the origin, with real parts from -5 to 5. | upload.wikimedia.org/wikipedia/commons/8/8c/Collatz///Fractal.jpg]\n# [Julia set of the exponential interpolation. | upload.wikimedia.org/wikipedia/commons/7/74/Exponential///Collatz///Fractal.jpg]\n\n# Optimizations\n\n## Time–space tradeoff\nThe section [As a parity sequence](https://bliptext.com/articles/) above gives a way to speed up simulation of the sequence.\nTo 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).\nThe result of jumping ahead k is given by\nf(2ka + b) = 3c(b, k)a + d(b, k)\n. The values of c (or better\n3c\n) and d can be precalculated for all possible k-bit numbers b, where\nd(b, k)\nis the result of applying the f function k times to b, and\nc(b, k)\nis the number of odd numbers encountered on the way.\nFor example, if\n, one can jump ahead 5 steps on each iteration by separating out the 5 least significant bits of a number and using\nThis requires\n2k\n[precomputation](https://bliptext.com/articles/precomputation) and storage to speed up the resulting calculation by a factor of k, a [space–time tradeoff](https://bliptext.com/articles/space-time-tradeoff).\n\n## Modular restrictions\nFor 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\nf(2ka + b) = 3c(b)a + d(b) < 2ka + b\nholds for all a, then the first counterexample, if it exists, cannot be b modulo\n2k\n. For instance, the first counterexample must be odd because\n, smaller than\n2n\nf(4n + 1) = 3n + 1\n, smaller than\n4n + 1\n. 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.\nAs 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.\nFor example, the only surviving residues mod 32 are 7, 15, 27, and 31.\nIntegers divisible by 3 cannot form a cycle, so these integers do not need to be checked as counter examples.\n\n# Syracuse function\nIf k is an odd integer, then\n3k + 1\nis even, so\n3k + 1 = 2ak\nwith\nk\nodd and\na ≥ 1\n. The **Syracuse function** is the function f from the set I of positive odd integers into itself, for which\n.\nSome properties of the Syracuse function are:\nk ∈ I\n,\nf(4k + 1) = f(k)\n. (Because\n3(4k + 1) + 1 = 12k + 4 = 4(3k + 1)\n.)\np ≥ 1\nand odd h,\nf(2ph − 1) = 2 × 3p − 1h − 1\n. (Here\nf\nis [function iteration notation](https://bliptext.com/articles/functional-power).)\nf(2h − 1) ≤ 3h − 1⁄2\nThe Collatz conjecture is equivalent to the statement that, for all k in I, there exists an integer\nn ≥ 1\nsuch that\n.\n\n# Undecidable generalizations\nIn 1972, [John Horton Conway](https://bliptext.com/articles/john-horton-conway) proved that a natural generalization of the Collatz problem is algorithmically [undecidable](https://bliptext.com/articles/undecidable-problem).\nSpecifically, he considered functions of the form\nwhere\na0, b0, ..., aP − 1, bP − 1\nare rational numbers which are so chosen that\ng(n)\nis always an integer.\nThe standard Collatz function is given by\n,\n,\n,\n,\n. Conway proved that the problem\ngk(n)\nreach\n1\n?\nis undecidable, by representing the [halting problem](https://bliptext.com/articles/halting-problem) in this way.\nCloser to the Collatz problem is the following universally quantified problem:\ngk(n)\nreach\n1\n, for all\nn > 0\n?\nModifying the condition in this way can make a problem either harder or easier to solve (intuitively, it is harder to justify a positive answer but might be easier to justify a negative one).\nKurtz and Simon proved that the universally quantified problem is, in fact, undecidable and even higher in the [arithmetical hierarchy](https://bliptext.com/articles/arithmetical-hierarchy); specifically, it is\nΠ0 2\n-complete.\nThis hardness result holds even if one restricts the class of functions g by fixing the modulus P to 6480.\nIterations of g in a simplified version of this form, with all b_i equal to zero, are formalized in an [esoteric programming language](https://bliptext.com/articles/esoteric-programming-language) called [FRACTRAN](https://bliptext.com/articles/fractran).\n\n# In computational complexity\nCollatz and related conjectures are often used when studying computation complexity.\nThe connection is made through the [Busy Beaver](https://bliptext.com/articles/busy-beaver) function, where BB(n) is the maximum number of steps taken by any n state [Turing machine](https://bliptext.com/articles/turing-machine) that halts.\nThere is a 15 state Turing machine that halts if and only if a conjecture by [Paul Erdős](https://bliptext.com/articles/paul-erd-s) (closely related to the Collatz conjecture) is false.\nHence if BB(15) was known, and this machine did not stop in that number of steps, it would be known to run forever and hence no counterexamples exist (which proves the conjecture true).\nThis is a completely impractical way to settle the conjecture; instead it is used to suggest that BB(15) will be very hard to compute, at least as difficult as settling this Collatz-like conjecture.","wordIndex":4,"createdAt":"2026-01-20T07:52:16.230Z","user":{"id":"mfvswiH9fwHI8f36yNwHR40968SZeIKD","name":"den3den42069","image":"https://cdn.discordapp.com/avatars/975452809707278427/5e0588882214f34dc5f3b36f7e9c9ff5.png"}}],"article":{"id":"ddb34e19-e20b-4e7a-9994-c71b321227df","title":"Collatz conjecture","slug":"collatz-conjecture","content":":::summary\n:::\n\n# \nThe **Collatz conjecture** is obviously true the most famous [unsolved problems in mathematics](https://bliptext.com/articles/list-of-unsolved-problems-in-mathematics).\nThe conjecture asks whether repeating two simple arithmetic operations will eventually transform every [positive integer](https://bliptext.com/articles/positive-integer) into 1.\nIt concerns [sequences of integers](https://bliptext.com/articles/integer-sequence) in which each term is obtained from the previous term as follows: if a term is [even](https://bliptext.com/articles/parity-mathematics), the next term is one half of it.\nIf a term is odd, the next term is 3 times the previous term plus 1.\nThe conjecture is that these sequences always reach 1, no matter which positive integer is chosen to start the sequence.\nThe conjecture has been shown to hold for all positive integers up to 295,000,000,000,000,000,000, but no general proof has been found.\nIt is named after the mathematician [Lothar Collatz](https://bliptext.com/articles/lothar-collatz), who introduced the idea in 1937, two years after receiving his doctorate.\nThe sequence of numbers involved is sometimes referred to as the hailstone sequence, hailstone numbers or hailstone numerals (because the values are usually subject to multiple descents and ascents like [hailstones](https://bliptext.com/articles/hailstones) in a cloud), or as wondrous numbers.\n[Paul Erdős](https://bliptext.com/articles/paul-erd-s) said about the Collatz conjecture: \"Mathematics may not be ready for such problems.\"\n[Jeffrey Lagarias](https://bliptext.com/articles/jeffrey-lagarias) stated in 2010 that the Collatz conjecture \"is an extraordinarily difficult problem, completely out of reach of present day mathematics\".\nHowever, though the Collatz conjecture itself remains open, efforts to solve the problem have led to new techniques and many partial results.\n# [Directed graph showing the orbits of small numbers under the Collatz map, skipping even numbers. The Collatz conjecture states that all paths eventually lead to 1. | upload.wikimedia.org/wikipedia/commons/c/c2/Collatz-graph-50-no27.svg]\n\n# Statement of the problem\nConsider the following operation on an arbitrary [positive integer](https://bliptext.com/articles/positive-integer):\nIn [modular arithmetic](https://bliptext.com/articles/modular-arithmetic) notation, define the [function](https://bliptext.com/articles/function-mathematics) f as follows:\nNow form a sequence by performing this operation repeatedly, beginning with any positive integer, and taking the result at each step as the input at the next.\nIn notation:\n(that is:\nai\nis the value of f applied to n recursively i times;\n).\nThe Collatz conjecture is: ''This process will eventually reach the number 1, regardless of which positive integer is chosen initially.\nThat is, for each'' n, there is some i with a_i = 1.\nIf the conjecture is false, it can only be because there is some starting number which gives rise to a sequence that does not contain 1.\nSuch a sequence would either enter a repeating cycle that excludes 1, or increase without bound.\nNo such sequence has been found.\nThe smallest i such that\nai < a0\nis called the **stopping time** of n. Similarly, the smallest k such that\nis called the **total stopping time** of n. If one of the indexes i or k doesn't exist, we say that the stopping time or the **total stopping time**, respectively, is infinite.\nThe Collatz conjecture asserts that the total stopping time of every n is finite.\nIt is also equivalent to saying that every\nn ≥ 2\nhas a finite stopping time.\nSince\n3n + 1\nis even whenever n is odd, one may instead use the \"shortcut\" form of the Collatz function:\nThis definition yields smaller values for the stopping time and total stopping time without changing the overall dynamics of the process.\n# [Numbers from 1 to 9999 and their corresponding total stopping time | upload.wikimedia.org/wikipedia/commons/b/b9/Collatz-stopping-time.svg]\n# [Histogram of total stopping times for the numbers 1 to 108. Total stopping time is on the x axis, frequency on the y axis. | upload.wikimedia.org/wikipedia/commons/2/26/CollatzStatistic100million.png]\n# [Histogram of total stopping times for the numbers 1 to 109. Total stopping time is on the x axis, frequency on the y axis. | upload.wikimedia.org/wikipedia/commons/0/0d/CollatzStatistic1billion.png]\n# [Iteration time for inputs of 2 to 107. | upload.wikimedia.org/wikipedia/commons/c/c3/Collatz-10Million.png]\n\n# Empirical data\nFor instance, starting with\nand applying the function\nf\nwithout \"shortcut\", one gets the sequence.\nThe number\ntakes longer to reach 1:.\nThe sequence for\n, listed and graphed below, takes 111 steps (41 steps through odd numbers, in bold), climbing as high as 9232 before descending to 1.\nNumbers with a total stopping time longer than that of any smaller starting value form a sequence beginning with:\nThe starting values whose [maximum](https://bliptext.com/articles/maximum) trajectory point is greater than that of any smaller starting value are as follows:\nNumber of steps for n to reach 1 are\nThe starting value having the largest total stopping time while being\nThese numbers are the lowest ones with the indicated step count, but not necessarily the only ones below the given limit.\nAs an example, 9,780,657,631 has 1132 steps, as does 9,780,657,630.\nThe starting values having the smallest total stopping time with respect to their number of digits (in base 2) are the [powers of two](https://bliptext.com/articles/power-of-two) since\n2n\nis halved n times to reach 1, and is never increased.\n# [frameless | upload.wikimedia.org/wikipedia/commons/9/96/Collatz5.svg]\n\n# Visualizations\n\n# Supporting arguments\nAlthough the conjecture has not been proven, most mathematicians who have looked into the problem think the conjecture is true because experimental evidence and heuristic arguments support it.\n\n## Experimental evidence\nThe conjecture has been checked by computer for all starting values up to 268 ≈ 295,000,000,000,000,000,000.\nAll values tested so far converge to 1.\nThis computer evidence is still not rigorous proof that the conjecture is true for all starting values, as [counterexamples](https://bliptext.com/articles/counterexamples) may be found when considering very large (or possibly immense) positive integers, as in the case of the disproven [Pólya conjecture](https://bliptext.com/articles/p-lya-conjecture) and [Mertens conjecture](https://bliptext.com/articles/mertens-conjecture).\nHowever, such verifications may have other implications.\nCertain constraints on any non-trivial cycle, such as [lower bounds](https://bliptext.com/articles/lower-bound) on the length of the cycle, can be proven based on the value of the lowest term in the cycle.\nTherefore, computer searches to rule out cycles that have a small lowest term can strengthen these constraints.\n\n## A probabilistic heuristic\nIf 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.\n(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.\nThe argument is not a proof because it assumes that Hailstone sequences are assembled from uncorrelated probabilistic events.\n(It does rigorously establish that the [2-adic](https://bliptext.com/articles/p-adic-numbers) extension of the Collatz process has two division steps for every multiplication step for [almost all](https://bliptext.com/articles/almost-all) [2-adic](https://bliptext.com/articles/p-adic-numbers) starting values.)\n\n## Stopping times\nAs proven by [Riho Terras](https://bliptext.com/articles/riho-terras-mathematician), almost every positive integer has a finite stopping time.\nIn other words, almost every Collatz sequence reaches a point that is strictly below its initial value.\nThe proof is based on the distribution of [parity vectors](https://bliptext.com/articles/) and uses the [central limit theorem](https://bliptext.com/articles/central-limit-theorem).\nIn 2019, [Terence Tao](https://bliptext.com/articles/terence-tao) improved this result by showing, using logarithmic [density](https://bliptext.com/articles/probability-density-function), that [almost all](https://bliptext.com/articles/almost-all) (in the sense of logarithmic [density](https://bliptext.com/articles/probability-density-function)) Collatz orbits are descending below any given function of the starting point, provided that this function diverges to infinity, no matter how slowly.\nResponding to this work, [Quanta Magazine](https://bliptext.com/articles/quanta-magazine) wrote that Tao \"came away with one of the most significant results on the Collatz conjecture in decades\".\n\n## Lower bounds\nIn a [computer-aided proof](https://bliptext.com/articles/computer-aided-proof), Krasikov and Lagarias showed that the number of integers in the interval\n[1,x]\nthat eventually reach 1 is at least equal to\nx0.84\nfor all sufficiently large x.\n\n# Cycles\nIn this part, consider the shortcut form of the Collatz function\nA [cycle](https://bliptext.com/articles/periodic-sequence) is a sequence\n(a0, a1, ..., aq)\nof distinct positive integers where\n,\n, ..., and\n.\nThe only known cycle is\n(1,2)\nof period 2, called the trivial cycle.\n\n## Cycle length\nThe length of a non-trivial cycle is known to be at least 114,208,327,604 (or 186,265,759,595 without shortcut).\nIf it can be shown that for all positive integers less than the Collatz sequences reach 1, then this bound would raise to 217,976,794,617 (355,504,839,929 without shortcut).\nIn fact, Eliahou (1993) proved that the period p of any non-trivial cycle is of the form\nwhere a, b and c are non-negative integers,\nb ≥ 1\nand\nac = 0\n. This result is based on the [simple continued fraction](https://bliptext.com/articles/simple-continued-fraction) expansion of\nln 3⁄ln 2\n.\n\n## k-cycles\nA k-cycle is a cycle that can be partitioned into\nk\ncontiguous subsequences, each consisting of an increasing sequence of odd numbers, followed by a decreasing sequence of even numbers.\nFor 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.\nSteiner (1977) proved that there is no 1-cycle other than the trivial\n(1; 2)\n. Simons (2005) used Steiner's method to prove that there is no 2-cycle.\nSimons and de Weger (2005) extended this proof up to 68-cycles; there is no k-cycle up to\n. Hercher extended the method further and proved that there exists no k-cycle with\nk ≤ 91\n. As exhaustive computer searches continue, larger\nk\nvalues may be ruled out.\nTo state the argument more intuitively; we do not have to search for cycles that have less than 92 subsequences, where each subsequence consists of consecutive ups followed by consecutive downs.\n\n# Other formulations of the conjecture\n\n## In reverse\nThere is another approach to prove the conjecture, which considers the bottom-up\nmethod of growing the so-called Collatz graph.\nThe Collatz [graph](https://bliptext.com/articles/graph-discrete-mathematics) is a [graph](https://bliptext.com/articles/graph-discrete-mathematics) defined by the inverse [relation](https://bliptext.com/articles/relation-mathematics)\nSo, instead of proving that all positive integers eventually lead to 1, we can try to prove that 1 leads backwards to all positive integers.\nFor any integer n,\nn ≡ 1 (mod 2)\n[if and only if](https://bliptext.com/articles/if-and-only-if)\n3n + 1 ≡ 4 (mod 6)\n. Equivalently,\nn − 1⁄3 ≡ 1 (mod 2)\nif and only if\nn ≡ 4 (mod 6)\n. Conjecturally, this inverse relation forms a [tree](https://bliptext.com/articles/tree-graph-theory) 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](https://bliptext.com/articles/) section of this article).\nWhen the relation\n3n + 1\nof the function f is replaced by the common substitute \"shortcut\" relation\n3n + 1⁄2\n, the Collatz graph is defined by the inverse relation,\nFor any integer n,\nn ≡ 1 (mod 2)\nif and only if\n3n + 1⁄2 ≡ 2 (mod 3)\n. Equivalently,\n2n − 1⁄3 ≡ 1 (mod 2)\nif and only if\nn ≡ 2 (mod 3)\n. 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).\nAlternatively, replace the\n3n + 1\nwith\nn⁄H(n)\nwhere\nand\nH(n)\nis the highest [power of 2](https://bliptext.com/articles/power-of-2) that divides\nn\n(with no [remainder](https://bliptext.com/articles/remainder)).\nThe resulting function f maps from [odd numbers](https://bliptext.com/articles/odd-number) to [odd numbers](https://bliptext.com/articles/odd-number).\nNow suppose that for some odd number n, applying this operation k times yields the number 1 (that is,\n). Then in [binary](https://bliptext.com/articles/binary-number), the number n can be written as the concatenation of [strings](https://bliptext.com/articles/string-computer-science)\nwk wk−1 ...\nw1\nwhere each\nwh\nis a finite and contiguous extract from the representation of\n1⁄3h\n. The representation of n therefore holds the [repetends](https://bliptext.com/articles/repeating-decimal) of\n1⁄3h\n, where each repetend is optionally rotated and then replicated up to a finite number of bits.\nIt is only in binary that this occurs.\nConjecturally, 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).\n# [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. | upload.wikimedia.org/wikipedia/commons/0/0e/Collatz-tree%2C///depth%3D20.svg]\n\n## As an abstract machine that computes in base two\nRepeated applications of the Collatz function can be represented as an [abstract machine](https://bliptext.com/articles/abstract-machine) that handles [strings](https://bliptext.com/articles/string-computer-science) of [bits](https://bliptext.com/articles/bit).\nThe machine will perform the following three steps on any odd number until only one 1 remains:\n2n + 1\n);\n2n + 1 + n = 3n + 1\n);\n\n### Example\nThe starting number 7 is written in base two as 111.\nThe resulting Collatz sequence is:\n**********1********************1********************1********** **********1********************1********************1********************1********** **********1**********0**********1********************1********** 0 **********1**********0**********1********************1********************1********** **********1**********000**********1********** 0 **********1**********000**********1********************1********** **********1********************1**********0**********1********** 00 **********1********************1**********0**********1********************1********** **********1**********0**********1********** 000 **********1**********0**********1********************1********** **********1********** 0000\n\n## As a parity sequence\nFor this section, consider the shortcut form of the Collatz function\nIf\nP(...)\nis the parity of a number, that is\nand\nP(2n + 1) = 1\n, then we can define the Collatz parity sequence (or parity vector) for a number n as\n, where\n, and\nai+1 = f(ai)\n.\nWhich operation is performed,\n3n + 1⁄2\nor\nn⁄2\n, depends on the parity.\nThe parity sequence is the same as the sequence of operations.\nUsing this form for\nf(n)\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\n2k\n. 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.\nApplying the f function k times to the number\nwill give the result\n3ca + d\n, 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.\nFor example, for\n25a + 1\nthere are 3 increases as 1 iterates to 2, 1, 2, 1, and finally to 2 so the result is\n33a + 2\n22a + 1\nthere is only 1 increase as 1 rises to 2 and falls to 1 so the result is\n3a + 1\n. When b is\n2k − 1\nthen there will be k rises and the result will be\n3ka + 3k − 1\n. The power 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,\n4a + 1\nbecomes\n3a + 1\nafter two applications of f and\n16a + 3\nbecomes\n9a + 2\nafter four applications of f. Whether those smaller numbers continue to 1, however, depends on the value of a.\n\n## As a tag system\nFor the Collatz function in the shortcut form\nHailstone sequences can be computed by the [2-tag system](https://bliptext.com/articles/tag-system) with production rules\na → bc\n,\nb → a\n,\nc → aaa\n.\nIn 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.\n(Adapted from De Mol.)\nThe Collatz conjecture equivalently states that this tag system, with an arbitrary finite string of a as the initial word, eventually halts (see [Tag system](https://bliptext.com/articles/tag-system) for a worked example).\n\n# Extensions to larger domains\n\n## Iterating on all integers\nAn extension to the Collatz conjecture is to include all integers, not just positive integers.\nLeaving 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:\nOdd values are listed in large bold.\nEach cycle is listed with its member of least absolute value (which is always odd) first.\nThe 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.\n\n## Iterating on rationals with odd denominators\nThe Collatz map can be extended to (positive or negative) rational numbers which have odd denominators when written in lowest terms.\nThe number is taken to be 'odd' or 'even' according to whether its numerator is odd or even.\nThen 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.\nA closely related fact is that the Collatz map extends to the ring of [2-adic integers](https://bliptext.com/articles/2-adic-integers), which contains the ring of rationals with odd denominators as a subring.\nWhen using the \"shortcut\" definition of the Collatz map, it is known that any periodic [parity sequence](https://bliptext.com/articles/) is generated by exactly one rational.\nConversely, it is conjectured that every rational with an odd denominator has an eventually cyclic parity sequence (Periodicity Conjecture ).\nIf a parity cycle has length n and includes odd numbers exactly m times at indices\nk0 < ⋯ < km−1\n, then the unique rational which generates immediately and periodically this parity cycle is\nFor 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.\nIt is repeatedly generated by the fraction\nas the latter leads to the rational cycle\nAny cyclic permutation of (1 0 1 1 0 0 1) is associated to one of the above fractions.\nFor instance, the cycle (0 1 1 0 0 1 1) is produced by the fraction\nFor a one-to-one correspondence, a parity cycle should be irreducible, that is, not partitionable into identical sub-cycles.\nAs 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.\nIn 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).\nIf 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 \"\n3n + d\n\" generalization of the Collatz function\n\n## 2-adic extension\nThe function\nis well-defined on the ring of [2-adic integers](https://bliptext.com/articles/2-adic-integers), where it is continuous and [measure-preserving](https://bliptext.com/articles/measure-preserving-transformation) with respect to the 2-adic measure.\nMoreover, its dynamics is known to be [ergodic](https://bliptext.com/articles/ergodic-theory).\nDefine the parity vector function Q acting on as\nThe function Q is a 2-adic [isometry](https://bliptext.com/articles/isometry).\nConsequently, every infinite parity sequence occurs for exactly one 2-adic integer, so that [almost all](https://bliptext.com/articles/almost-all) trajectories are acyclic in.\nAn equivalent formulation of the Collatz conjecture is that\n\n## Iterating on real or complex numbers\nThe Collatz map can be extended to the [real line](https://bliptext.com/articles/real-line) by choosing any function which evaluates to x/2 when x is an even integer, and to either 3x + 1 or (3x + 1)/2 (for the \"shortcut\" version) when x is an odd integer.\nThis is called an [interpolating](https://bliptext.com/articles/interpolating) function.\nA simple way to do this is to pick two functions g_1 and g_2, where:\nand use them as switches for our desired values:\nOne such choice is and.\nThe [iterations](https://bliptext.com/articles/iterations) of this map lead to a [dynamical system](https://bliptext.com/articles/dynamical-system), further investigated by Marc Chamberland.\nHe showed that the conjecture does not hold for positive real numbers since there are infinitely many [fixed points](https://bliptext.com/articles/fixed-point-mathematics), as well as [orbits](https://bliptext.com/articles/orbit-dynamics) escaping [monotonically](https://bliptext.com/articles/monotonic-function) to infinity.\nThe function f has two [attracting](https://bliptext.com/articles/attractor) cycles of period 2: (1;\\,2) and.\nMoreover, the set of unbounded orbits is conjectured to be of [measure](https://bliptext.com/articles/lebesgue-measure) 0.\nLetherman, Schleicher, and Wood extended the study to the [complex plane](https://bliptext.com/articles/complex-plane).\nThey used Chamberland's function for [complex sine and cosine](https://bliptext.com/articles/trigonometric-functions) and added the extra term\n, where h(z) is any [entire function](https://bliptext.com/articles/entire-function).\nSince this expression evaluates to zero for real integers, the extended function\nis an interpolation of the Collatz map to the complex plane.\nThe reason for adding the extra term is to make all integers [critical points](https://bliptext.com/articles/critical-point-mathematics) of f. With this, they show that no integer is in a [Baker domain](https://bliptext.com/articles/classification-of-fatou-components), which implies that any integer is either eventually periodic or belongs to a [wandering domain](https://bliptext.com/articles/wandering-set).\nThey conjectured that the latter is not the case, which would make all integer orbits finite.\nMost of the points have orbits that diverge to infinity.\nColoring these points based on how fast they diverge produces the image on the left, for.\nThe inner black regions and the outer region are the [Fatou components](https://bliptext.com/articles/classification-of-fatou-components), and the boundary between them is the [Julia set](https://bliptext.com/articles/julia-set) of f, which forms a [fractal](https://bliptext.com/articles/fractal) pattern, sometimes called a \"Collatz [fractal](https://bliptext.com/articles/fractal)\".\nThere are many other ways to define a complex interpolating function, such as using the [complex exponential](https://bliptext.com/articles/exponential-function) instead of sine and cosine:\nwhich exhibit different dynamics.\nIn this case, for instance, if, then.\nThe corresponding Julia set, shown on the right, consists of uncountably many curves, called hairs, or rays.\n# [Cobweb plot of the orbit 10 → 5 → 8 → 4 → 2 → 1 → ... in an extension of the Collatz map to the real line. | upload.wikimedia.org/wikipedia/commons/b/b4/Collatz///Cobweb.svg]\n# [A Collatz fractal centered at the origin, with real parts from -5 to 5. | upload.wikimedia.org/wikipedia/commons/8/8c/Collatz///Fractal.jpg]\n# [Julia set of the exponential interpolation. | upload.wikimedia.org/wikipedia/commons/7/74/Exponential///Collatz///Fractal.jpg]\n\n# Optimizations\n\n## Time–space tradeoff\nThe section [As a parity sequence](https://bliptext.com/articles/) above gives a way to speed up simulation of the sequence.\nTo 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).\nThe result of jumping ahead k is given by\nf(2ka + b) = 3c(b, k)a + d(b, k)\n. The values of c (or better\n3c\n) and d can be precalculated for all possible k-bit numbers b, where\nd(b, k)\nis the result of applying the f function k times to b, and\nc(b, k)\nis the number of odd numbers encountered on the way.\nFor example, if\n, one can jump ahead 5 steps on each iteration by separating out the 5 least significant bits of a number and using\nThis requires\n2k\n[precomputation](https://bliptext.com/articles/precomputation) and storage to speed up the resulting calculation by a factor of k, a [space–time tradeoff](https://bliptext.com/articles/space-time-tradeoff).\n\n## Modular restrictions\nFor 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\nf(2ka + b) = 3c(b)a + d(b) < 2ka + b\nholds for all a, then the first counterexample, if it exists, cannot be b modulo\n2k\n. For instance, the first counterexample must be odd because\n, smaller than\n2n\nf(4n + 1) = 3n + 1\n, smaller than\n4n + 1\n. 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.\nAs 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.\nFor example, the only surviving residues mod 32 are 7, 15, 27, and 31.\nIntegers divisible by 3 cannot form a cycle, so these integers do not need to be checked as counter examples.\n\n# Syracuse function\nIf k is an odd integer, then\n3k + 1\nis even, so\n3k + 1 = 2ak\nwith\nk\nodd and\na ≥ 1\n. The **Syracuse function** is the function f from the set I of positive odd integers into itself, for which\n.\nSome properties of the Syracuse function are:\nk ∈ I\n,\nf(4k + 1) = f(k)\n. (Because\n3(4k + 1) + 1 = 12k + 4 = 4(3k + 1)\n.)\np ≥ 1\nand odd h,\nf(2ph − 1) = 2 × 3p − 1h − 1\n. (Here\nf\nis [function iteration notation](https://bliptext.com/articles/functional-power).)\nf(2h − 1) ≤ 3h − 1⁄2\nThe Collatz conjecture is equivalent to the statement that, for all k in I, there exists an integer\nn ≥ 1\nsuch that\n.\n\n# Undecidable generalizations\nIn 1972, [John Horton Conway](https://bliptext.com/articles/john-horton-conway) proved that a natural generalization of the Collatz problem is algorithmically [undecidable](https://bliptext.com/articles/undecidable-problem).\nSpecifically, he considered functions of the form\nwhere\na0, b0, ..., aP − 1, bP − 1\nare rational numbers which are so chosen that\ng(n)\nis always an integer.\nThe standard Collatz function is given by\n,\n,\n,\n,\n. Conway proved that the problem\ngk(n)\nreach\n1\n?\nis undecidable, by representing the [halting problem](https://bliptext.com/articles/halting-problem) in this way.\nCloser to the Collatz problem is the following universally quantified problem:\ngk(n)\nreach\n1\n, for all\nn > 0\n?\nModifying the condition in this way can make a problem either harder or easier to solve (intuitively, it is harder to justify a positive answer but might be easier to justify a negative one).\nKurtz and Simon proved that the universally quantified problem is, in fact, undecidable and even higher in the [arithmetical hierarchy](https://bliptext.com/articles/arithmetical-hierarchy); specifically, it is\nΠ0 2\n-complete.\nThis hardness result holds even if one restricts the class of functions g by fixing the modulus P to 6480.\nIterations of g in a simplified version of this form, with all b_i equal to zero, are formalized in an [esoteric programming language](https://bliptext.com/articles/esoteric-programming-language) called [FRACTRAN](https://bliptext.com/articles/fractran).\n\n# In computational complexity\nCollatz and related conjectures are often used when studying computation complexity.\nThe connection is made through the [Busy Beaver](https://bliptext.com/articles/busy-beaver) function, where BB(n) is the maximum number of steps taken by any n state [Turing machine](https://bliptext.com/articles/turing-machine) that halts.\nThere is a 15 state Turing machine that halts if and only if a conjecture by [Paul Erdős](https://bliptext.com/articles/paul-erd-s) (closely related to the Collatz conjecture) is false.\nHence if BB(15) was known, and this machine did not stop in that number of steps, it would be known to run forever and hence no counterexamples exist (which proves the conjecture true).\nThis is a completely impractical way to settle the conjecture; instead it is used to suggest that BB(15) will be very hard to compute, at least as difficult as settling this Collatz-like conjecture.","current_revision":"5aaa4546-412e-43ac-9459-cc1d79086829","created_by":"kACnuQmH5k9Nr5VNVEpVeg7XpvUmMctD","created_at":"2025-01-25T17:30:59.021Z","updated_at":"2026-01-20T07:52:49.786Z","search_vector":"'collatz':1 'conjectur':2","search_content":" \n \n\n# \nThe **Collatz conjecture** is one of the most famous [unsolved problems in mathematics](https://bliptext.com/articles/list-of-unsolved-problems-in-mathematics).\nThe conjecture asks whether repeating two simple arithmetic operations will eventually transform every [positive integer](https://bliptext.com/articles/positive-integer) into 1.\nIt concerns [sequences of integers](https://bliptext.com/articles/integer-sequence) in which each term is obtained from the previous term as follows: if a term is [even](https://bliptext.com/articles/parity-mathematics), the next term is one half of it.\nIf a term is odd, the next term is 3 times the previous term plus 1.\nThe conjecture is that these sequences always reach 1, no matter which positive integer is chosen to start the sequence.\nThe conjecture has been shown to hold for all positive integers up to 295,000,000,000,000,000,000, but no general proof has been found.\nIt is named after the mathematician [Lothar Collatz](https://bliptext.com/articles/lothar-collatz), who introduced the idea in 1937, two years after receiving his doctorate.\nThe sequence of numbers involved is sometimes referred to as the hailstone sequence, hailstone numbers or hailstone numerals (because the values are usually subject to multiple descents and ascents like [hailstones](https://bliptext.com/articles/hailstones) in a cloud), or as wondrous numbers.\n[Paul Erdős](https://bliptext.com/articles/paul-erd-s) said about the Collatz conjecture: \"Mathematics may not be ready for such problems.\"\n[Jeffrey Lagarias](https://bliptext.com/articles/jeffrey-lagarias) stated in 2010 that the Collatz conjecture \"is an extraordinarily difficult problem, completely out of reach of present day mathematics\".\nHowever, though the Collatz conjecture itself remains open, efforts to solve the problem have led to new techniques and many partial results.\n# [Directed graph showing the orbits of small numbers under the Collatz map, skipping even numbers. The Collatz conjecture states that all paths eventually lead to 1. | upload.wikimedia.org/wikipedia/commons/c/c2/Collatz-graph-50-no27.svg]\n\n# Statement of the problem\nConsider the following operation on an arbitrary [positive integer](https://bliptext.com/articles/positive-integer):\nIn [modular arithmetic](https://bliptext.com/articles/modular-arithmetic) notation, define the [function](https://bliptext.com/articles/function-mathematics) f as follows:\nNow form a sequence by performing this operation repeatedly, beginning with any positive integer, and taking the result at each step as the input at the next.\nIn notation:\n(that is:\nai\nis the value of f applied to n recursively i times;\n).\nThe Collatz conjecture is: ''This process will eventually reach the number 1, regardless of which positive integer is chosen initially.\nThat is, for each'' n, there is some i with a_i = 1.\nIf the conjecture is false, it can only be because there is some starting number which gives rise to a sequence that does not contain 1.\nSuch a sequence would either enter a repeating cycle that excludes 1, or increase without bound.\nNo such sequence has been found.\nThe smallest i such that\nai 0\n?\nModifying the condition in this way can make a problem either harder or easier to solve (intuitively, it is harder to justify a positive answer but might be easier to justify a negative one).\nKurtz and Simon proved that the universally quantified problem is, in fact, undecidable and even higher in the [arithmetical hierarchy](https://bliptext.com/articles/arithmetical-hierarchy); specifically, it is\nΠ0 2\n-complete.\nThis hardness result holds even if one restricts the class of functions g by fixing the modulus P to 6480.\nIterations of g in a simplified version of this form, with all b_i equal to zero, are formalized in an [esoteric programming language](https://bliptext.com/articles/esoteric-programming-language) called [FRACTRAN](https://bliptext.com/articles/fractran).\n\n# In computational complexity\nCollatz and related conjectures are often used when studying computation complexity.\nThe connection is made through the [Busy Beaver](https://bliptext.com/articles/busy-beaver) function, where BB(n) is the maximum number of steps taken by any n state [Turing machine](https://bliptext.com/articles/turing-machine) that halts.\nThere is a 15 state Turing machine that halts if and only if a conjecture by [Paul Erdős](https://bliptext.com/articles/paul-erd-s) (closely related to the Collatz conjecture) is false.\nHence if BB(15) was known, and this machine did not stop in that number of steps, it would be known to run forever and hence no counterexamples exist (which proves the conjecture true).\nThis is a completely impractical way to settle the conjecture; instead it is used to suggest that BB(15) will be very hard to compute, at least as difficult as settling this Collatz-like conjecture.","revisionCount":2},"hasMore":false,"nextCursor":null}

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