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36-649: The "3 x + 1" problem is also known as the Collatz conjecture , named after him and still unsolved. The Collatz–Wielandt formula for the Perron–Frobenius eigenvalue of a positive square matrix was also named after him. Collatz's 1957 paper with Ulrich Sinogowitz, who had been killed in the bombing of Darmstadt in World War II , founded the field of spectral graph theory . Collatz studied at universities in Germany including

72-448: A i is the value of f applied to n recursively i times; a i = f   ( n ) ). The Collatz conjecture is: This process will eventually reach the number 1, regardless of which positive integer is chosen initially. That is, for each n {\displaystyle n} , there is some i {\displaystyle i} with a i = 1 {\displaystyle a_{i}=1} . If

108-1185: A tree except for the 1–2–4 loop (the inverse of the 4–2–1 loop of the unaltered function f defined in the Statement of the problem section of this article). When the relation 3 n + 1 of the function f is replaced by the common substitute "shortcut" relation ⁠ 3 n + 1 / 2 ⁠ , the Collatz graph is defined by the inverse relation, R ( n ) = { { 2 n } if  n ≡ 0 , 1 { 2 n , 2 n − 1 3 } if  n ≡ 2 ( mod 3 ) . {\displaystyle R(n)={\begin{cases}\{2n\}&{\text{if }}n\equiv 0,1\\[4px]\left\{2n,{\frac {2n-1}{3}}\right\}&{\text{if }}n\equiv 2\end{cases}}{\pmod {3}}.} For any integer n , n ≡ 1 (mod 2) if and only if ⁠ 3 n + 1 / 2 ⁠ ≡ 2 (mod 3) . Equivalently, ⁠ 2 n − 1 / 3 ⁠ ≡ 1 (mod 2) if and only if n ≡ 2 (mod 3) . Conjecturally, this inverse relation forms

144-477: A finite number of bits. It is only in binary that this occurs. Conjecturally, every binary string s that ends with a '1' can be reached by a representation of this form (where we may add or delete leading '0's to  s ). Repeated applications of the Collatz function can be represented as an abstract machine that handles strings of bits . The machine will perform the following three steps on any odd number until only one 1 remains: The starting number 7

180-523: A point that is strictly below its initial value. The proof is based on the distribution of parity vectors and uses the central limit theorem . In 2019, Terence Tao improved this result by showing, using logarithmic density , that almost all (in the sense of logarithmic density) Collatz orbits are descending below any given function of the starting point, provided that this function diverges to infinity, no matter how slowly. Responding to this work, Quanta Magazine wrote that Tao "came away with one of

216-500: 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. In notation: a i = { n for  i = 0 , f ( a i − 1 ) for  i > 0 {\displaystyle a_{i}={\begin{cases}n&{\text{for }}i=0,\\f(a_{i-1})&{\text{for }}i>0\end{cases}}} (that is:

252-466: A tree except for a 1–2 loop (the inverse of the 1–2 loop of the function f(n) revised as indicated above). Alternatively, replace the 3 n + 1 with ⁠ n ′ / H ( n ′ ) ⁠ where n ′ = 3 n + 1 and H ( n ′ ) is the highest power of 2 that divides n ′ (with no remainder ). The resulting function f maps from odd numbers to odd numbers. Now suppose that for some odd number n , applying this operation k times yields

288-521: Is a sequence ( a 0 , a 1 , ..., a q ) of distinct positive integers where f ( a 0 ) = a 1 , f ( a 1 ) = a 2 , ..., and f ( a q ) = a 0 . The only known cycle is (1,2) of period 2, called the trivial cycle. The length of a non-trivial cycle is known to be at least 114 208 327 604 (or 186 265 759 595 without shortcut). If it can be shown that for all positive integers less than 3 × 2 69 {\displaystyle 3\times 2^{69}}

324-426: Is based on the simple continued fraction expansion of ⁠ ln 3 / ln 2 ⁠ . A k -cycle is a cycle that can be partitioned into k contiguous subsequences, each consisting of an increasing sequence of odd numbers, followed by a decreasing sequence of even numbers. For instance, if the cycle consists of a single increasing sequence of odd numbers followed by a decreasing sequence of even numbers, it

360-467: Is called a 1-cycle . Steiner (1977) proved that there is no 1-cycle other than the trivial (1; 2) . Simons (2005) used Steiner's method to prove that there is no 2-cycle. Simons and de Weger (2005) extended this proof up to 68-cycles; there is no k -cycle up to k = 68 . Hercher extended the method further and proved that there exists no k -cycle with k ≤ 91 . As exhaustive computer searches continue, larger k values may be ruled out. To state

396-412: Is 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. The Collatz conjecture asserts that the total stopping time of every n is finite. It is also equivalent to saying that every n ≥ 2 has a finite stopping time. Since 3 n + 1 is even whenever n is odd, one may instead use

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432-432: Is not a proof because it assumes that Hailstone sequences are assembled from uncorrelated probabilistic events. (It does rigorously establish that the 2-adic extension of the Collatz process has two division steps for every multiplication step for almost all 2-adic starting values.) As proven by Riho Terras , almost every positive integer has a finite stopping time. In other words, almost every Collatz sequence reaches

468-431: Is one of the most famous unsolved problems in mathematics . The conjecture asks whether repeating two simple arithmetic operations will eventually transform every positive integer into 1. It concerns sequences of integers in which each term is obtained from the previous term as follows: if a term is even , the next term is one half of it. If a term is odd, the next term is 3 times the previous term plus 1. The conjecture

504-419: Is that these sequences always reach 1, no matter which positive integer is chosen to start the sequence. The conjecture has been shown to hold for all positive integers up to 2.95 × 10 , but no general proof has been found. It is named after the mathematician Lothar Collatz , who introduced the idea in 1937, two years after receiving his doctorate. The sequence of numbers involved is sometimes referred to as

540-408: Is the parity of a number, that is P(2 n ) = 0 and P(2 n + 1) = 1 , then we can define the Collatz parity sequence (or parity vector) for a number n as p i = P( a i ) , where a 0 = n , and a i +1 = f ( a i ) . Which operation is performed, ⁠ 3 n + 1 / 2 ⁠ or ⁠ n / 2 ⁠ , depends on the parity. The parity sequence is the same as

576-564: Is written in base two as 111 . The resulting Collatz sequence is: For this section, consider the shortcut form of the Collatz function f ( n ) = { n 2 if  n ≡ 0 3 n + 1 2 if  n ≡ 1 ( mod 2 ) . {\displaystyle f(n)={\begin{cases}{\frac {n}{2}}&{\text{if }}n\equiv 0\\[4px]{\frac {3n+1}{2}}&{\text{if }}n\equiv 1\end{cases}}{\pmod {2}}.} If P(...)

612-449: The OEIS ) Numbers with a total stopping time longer than that of any smaller starting value form a sequence beginning with: The starting values whose maximum trajectory point is greater than that of any smaller starting value are as follows: Number of steps for n to reach 1 are The starting value having the largest total stopping time while being These numbers are the lowest ones with

648-865: The Technische Hochschule Karlsruhe (now Karlsruhe Institute of Technology ) in 1935 where he remained through 1937. From 1938 to 1943, he worked as a Privatdozent in Karlsruhe. In the war years he worked with Alwin Walther at the Institute for Practical Mathematics of the Technische Hochschule Darmstadt . From 1943 to 1952, Collatz held a chair at the Technische Hochschule Hannover (now Leibniz University Hanover ) . From 1952 until his retirement in 1978, Collatz worked at

684-528: The University of Greifswald and the University of Berlin , where he was supervised by Alfred Klose , receiving his doctorate in 1935 for a dissertation entitled Das Differenzenverfahren mit höherer Approximation für lineare Differentialgleichungen (The finite difference method with higher approximation for linear differential equations). He then worked as an assistant at the University of Berlin, before moving to

720-520: The University of Hamburg , where he founded the Institute of Applied Mathematics in 1953. After retirement as professor emeritus , he continued to be very active at mathematical conferences. For his many contributions to the field, Collatz had many honors bestowed upon him in his lifetime, including: He died unexpectedly from a heart attack in Varna , Bulgaria , while attending a mathematics conference. Collatz conjecture The Collatz conjecture

756-439: The f function k times to the number n = 2 a + b will give the result 3 a + d , where d is the result of applying the f function k times to b , and c is how many increases were encountered during that sequence. For example, for 2 a + 1 there are 3 increases as 1 iterates to 2, 1, 2, 1, and finally to 2 so the result is 3 a + 2 ; for 2 a + 1 there is only 1 increase as 1 rises to 2 and falls to 1 so

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792-435: The odd numbers in the sequence generated by the Collatz process, then each odd number is on average ⁠ 3 / 4 ⁠ of the previous one. (More precisely, the geometric mean of the ratios of outcomes is ⁠ 3 / 4 ⁠ .) This yields a heuristic argument that every Hailstone sequence should decrease in the long run, although this is not evidence against other cycles, only against divergence. The argument

828-543: The "shortcut" form of the Collatz function: f ( n ) = { n 2 if  n ≡ 0 ( mod 2 ) , 3 n + 1 2 if  n ≡ 1 ( mod 2 ) . {\displaystyle f(n)={\begin{cases}{\frac {n}{2}}&{\text{if }}n\equiv 0{\pmod {2}},\\[4px]{\frac {3n+1}{2}}&{\text{if }}n\equiv 1{\pmod {2}}.\end{cases}}} This definition yields smaller values for

864-700: The Collatz conjecture itself remains open, efforts to solve the problem have led to new techniques and many partial results. Consider the following operation on an arbitrary positive integer : In modular arithmetic notation, define the function f as follows: f ( n ) = { n / 2 if  n ≡ 0 ( mod 2 ) , 3 n + 1 if  n ≡ 1 ( mod 2 ) . {\displaystyle f(n)={\begin{cases}n/2&{\text{if }}n\equiv 0{\pmod {2}},\\[4px]3n+1&{\text{if }}n\equiv 1{\pmod {2}}.\end{cases}}} Now form

900-441: The Collatz sequences reach 1, then this bound would raise to 217 976 794 617 ( 355 504 839 929 without shortcut). In fact, Eliahou (1993) proved that the period p of any non-trivial cycle is of the form p = 301994 a + 17087915 b + 85137581 c {\displaystyle p=301994a+17087915b+85137581c} where a , b and c are non-negative integers, b ≥ 1 and ac = 0 . This result

936-1252: 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. There is another approach to prove the conjecture, which considers the bottom-up method of growing the so-called Collatz graph . The Collatz graph is a graph defined by the inverse relation R ( n ) = { { 2 n } if  n ≡ 0 , 1 , 2 , 3 , 5 { 2 n , n − 1 3 } if  n ≡ 4 ( mod 6 ) . {\displaystyle R(n)={\begin{cases}\{2n\}&{\text{if }}n\equiv 0,1,2,3,5\\[4px]\left\{2n,{\frac {n-1}{3}}\right\}&{\text{if }}n\equiv 4\end{cases}}{\pmod {6}}.} So, instead of proving that all positive integers eventually lead to 1, we can try to prove that 1 leads backwards to all positive integers. For any integer n , n ≡ 1 (mod 2) if and only if 3 n + 1 ≡ 4 (mod 6) . Equivalently, ⁠ n − 1 / 3 ⁠ ≡ 1 (mod 2) if and only if n ≡ 4 (mod 6) . Conjecturally, this inverse relation forms

972-431: The case of the disproven Pólya conjecture and Mertens conjecture . However, such verifications may have other implications. Certain constraints on any non-trivial cycle, such as lower bounds on the length of the cycle, can be proven based on the value of the lowest term in the cycle. Therefore, computer searches to rule out cycles that have a small lowest term can strengthen these constraints. If one considers only

1008-408: 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. Such a sequence would either enter a repeating cycle that excludes 1, or increase without bound. No such sequence has been found. The smallest i such that a i < a 0 is called the stopping time of n . Similarly, the smallest k such that a k = 1

1044-469: The hailstone sequence, hailstone numbers or hailstone numerals (because the values are usually subject to multiple descents and ascents like hailstones in a cloud), or as wondrous numbers. Paul Erdős said about the Collatz conjecture: "Mathematics may not be ready for such problems." Jeffrey Lagarias stated in 2010 that the Collatz conjecture "is an extraordinarily difficult problem, completely out of reach of present day mathematics". However, though

1080-452: The indicated step count, but not necessarily the only ones below the given limit. As an example, 9 780 657 631 has 1132 steps, as does 9 780 657 630 . The starting values having the smallest total stopping time with respect to their number of digits (in base 2) are the powers of two since 2 is halved n times to reach 1, and is never increased. Although the conjecture has not been proven, most mathematicians who have looked into

1116-801: The most significant results on the Collatz conjecture in decades". In a computer-aided proof , Krasikov and Lagarias showed that the number of integers in the interval [1, x ] that eventually reach 1 is at least equal to x for all sufficiently large x . In this part, consider the shortcut form of the Collatz function f ( n ) = { n 2 if  n ≡ 0 ( mod 2 ) , 3 n + 1 2 if  n ≡ 1 ( mod 2 ) . {\displaystyle f(n)={\begin{cases}{\frac {n}{2}}&{\text{if }}n\equiv 0{\pmod {2}},\\[4px]{\frac {3n+1}{2}}&{\text{if }}n\equiv 1{\pmod {2}}.\end{cases}}} A cycle

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1152-442: The number 1 (that is, f ( n ) = 1 ). Then in binary , the number n can be written as the concatenation of strings w k w k −1 ... w 1 where each w h is a finite and contiguous extract from the representation of ⁠ 1 / 3 ⁠ . The representation of n therefore holds the repetends of ⁠ 1 / 3 ⁠ , where each repetend is optionally rotated and then replicated up to

1188-457: The problem think the conjecture is true because experimental evidence and heuristic arguments support it. The conjecture has been checked by computer for all starting values up to 2 ≈ 2.95 × 10 . All values tested so far converge to 1. This computer evidence is still not rigorous proof that the conjecture is true for all starting values, as counterexamples may be found when considering very large (or possibly immense) positive integers, as in

1224-436: The result is 3 a + 1 . When b is 2 − 1 then there will be k rises and the result will be 3 a + 3 − 1 . 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, 4 a + 1 becomes 3 a + 1 after two applications of f and 16

1260-428: The sequence of operations. Using this form for f ( n ) , it can be shown that the parity sequences for two numbers m and n will agree in the first k terms if and only if m and n are equivalent modulo 2 . 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. Applying

1296-563: The stopping time and total stopping time without changing the overall dynamics of the process. For instance, starting with n = 12 and applying the function f without "shortcut", one gets the sequence 12, 6, 3, 10, 5, 16, 8, 4, 2, 1 . The number n = 19 takes longer to reach 1: 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 . The sequence for n = 27 , listed and graphed below, takes 111 steps (41 steps through odd numbers, in bold), climbing as high as 9232 before descending to 1. (sequence A008884 in

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