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| {{unsolved|mathematics|Does the Collatz sequence from initial value ''n'' eventually reach 1, for all ''n'' > 0?}}
| | In the simplest analysis yeast infections are at least uncomfortable. The herb Echinacea is one of the most effective herbs for the candida fungus because of its strong immune stimulating actions that help in warding off infections. Avoid using scented feminine hygiene products in and around the genital area. - White, thick discharge that has a chunky texture. When faced with a life-threatening systemic yeast infection, Amphotericin B will commonly be used intravenously, but the side effects can be severe. <br><br>Make sure you get adequate sleep and continue with regular exercise as that will definitely help in the healing process. You should also be aware that it can take longer than usual for your symptoms to resolve, so you may need to take the treatment for a week, or perhaps even longer. Before concluding that you need a cure for yeast infection, you need to begin by determining first if you actually are suffering from genuine Candida (another name for this condition). We all possess a variety of kinds of bacteria in our bodies and nearly all of them are required for all round good health. It gets to the root cause of the infection and cures this rather than simply dealing with the symptoms making you believe it's gone, until it comes back. <br><br>If you would like to know 3 household items that are frequently recommended for natural ways to get rid of yeast infections BUT also hazardous that you may have already used then you should read yeast infection cure cautions to find out which items may be harming you. This article is based on the book, "Yeast Infection No More" by Linda Allen. The Candida microorganisms can subsequently start to grow unrestrained. Next; on the following page you'll discover a fantastic, proven method for treating candida:. A burning sensation, mainly during intercourse or while urinating. <br><br>Though the sudden outbreak often makes one believe they have "caught" a yeast infection, one doesn. The upsetting of our normal inner balance can be due to a multitude of factors including:. A woman's body goes through a number of changes during pregnancy. I am an article writter for a variety of well being issues, i like speaking, browsing and writing about wellness, so i could assist people to be more aware of their wellness. General Differences For a start, yeast infection is an overgrowth a 'friendly' bacteria called Candida Albicans which part of the ecology of the genital area. <br><br>First, it's worth looking at the symptoms of male yeast infection. Usually theres no pain but infection in mouth is visible and looks odd. The moist areas near the Vaginas are also big contributor of this infection. Ask your doctor if there are other natural ways to cure a yeast infection. Yogurt yeast infection which is heat treated will not yield any wonders as it will kill the bacteria.<br><br>If you have any issues relating to exactly where and how to use [http://www.naturalcureyeastinfection.info/sitemap/ natural treatment for yeast infection], you can contact us at our page. |
| The '''Collatz conjecture''' is a [[conjecture]] in [[mathematics]] named after [[Lothar Collatz]], who first proposed it in 1937. The conjecture is also known as the '''3''n'' + 1 conjecture''', the '''Ulam conjecture''' (after [[Stanislaw Ulam]]), '''Kakutani's problem''' (after [[Shizuo Kakutani]]), the '''Thwaites conjecture''' (after Sir Bryan Thwaites), '''Hasse's algorithm''' (after [[Helmut Hasse]]), or the '''Syracuse problem''';<ref>{{cite book |title=Logo: A Retrospective |last=Maddux |first=Cleborne D. |authorlink= |coauthors=Johnson, D. Lamont |year=1997 |publisher=Haworth Press |location=New York |isbn=0-7890-0374-0 |page=160 |pages= |quote=The problem is also known by several other names, including: Ulam's conjecture, the Hailstone problem, the Syracuse problem, Kakutani's problem, Hasse's algorithm, and the Collatz problem. }}</ref><ref>According to Lagarias (1985, p.4), the name "Syracuse problem" was proposed by Hasse in the 1950s, during a visit to [[Syracuse University]].</ref> the sequence of numbers involved is referred to as the '''hailstone sequence''' or '''hailstone numbers''' (because the values are usually subject to multiple descents and ascents like hailstones in a cloud),<ref>{{cite book |title=Wonders of Numbers |last=Pickover |first=Clifford A. |authorlink= |coauthors= |year=2001 |publisher=Oxford University Press |location=Oxford |isbn=0-19-513342-0 |page= |pages=116–118 |url= }}</ref><ref name=hn>{{cite web|title=Hailstone Number|url=http://mathworld.wolfram.com/HailstoneNumber.html|work=MathWorld|publisher=Wolfram Research, Inc.}}</ref> or as '''wondrous numbers'''.<ref>{{cite book |title=[[Gödel, Escher, Bach]] |last=Hofstadter |first=Douglas R.|authorlink=Douglas Hofstadter|coauthors= |year=1979 |publisher=Basic Books |location=New York |isbn=0-465-02685-0 |page= |pages=400–402 |url= }}</ref> | |
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| Take any [[natural number]] ''n''. If ''n'' is even, divide it by 2 to get ''n'' / 2. If ''n'' is odd, multiply it by 3 and add 1 to obtain 3''n'' + 1. Repeat the process (which has been called "Half Or Triple Plus One", or '''HOTPO'''<ref>{{cite book |title=Advanced Logo: A Language for Learning |last=Friendly |first=Michael |authorlink= |coauthors= |year=1988 |publisher=Lawrence Erlbaum Associates |location=Hillsdale, NJ |isbn=0-89859-933-4 |page= |pages= |url= }}</ref>) indefinitely. The conjecture is that no matter what number you start with, you shall always eventually reach 1. The property has also been called '''oneness'''.<ref>{{Cite web |date=December 1992 |last=Bourke |first=Paul |publisher=University of West Alabama |title=Decision Procedure for 'Oneness' |url=http://paulbourke.net/fractals/oneness/}}</ref>
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| [[Paul Erdős]] said, allegedly, about the Collatz conjecture: "Mathematics is not yet ripe for such problems." He also offered $500 for its solution.<ref>R. K. Guy: Don't try to solve these problems, Amer. Math. Monthly, '''90'''(1983), 35–41. By this Erdos means that there aren't powerful tools for manipulating such objects.</ref>
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| In 1972, [[John Horton Conway|J.H. Conway]] proved that a natural generalization of the Collatz problem is algorithmically [[undecidable problem|undecidable]].<ref>"J. H. Conway proved the remarkable result that a simple generalization of the problem is algorithmically undecidable." Quoting Lagarias 1985:
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| *{{cite book |first=J. H. | last=Conway | author-link=John Horton Conway | contribution=Unpredictable Iterations | title=Proceedings of the 1972 Number Theory Conference : University of Colorado, Boulder, Colorado, August 14–18, 1972 | publisher=[[University of Colorado]] | location=[[Boulder, CO]] | year=1972 | pages=49–52 | oclc=4181683 | zbl=0337.10041}}</ref>
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| ==Statement of the problem==
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| [[File:CollatzStatistic100million.png|thumb|300px|Histogram of stopping times for the numbers 1 to 100 million. Stopping time is on the x-axis, frequency on the y-axis.]]
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| [[Image:Collatz-stopping-time.svg|thumb|300px|right|Numbers from 1 to 9999 and their corresponding total stopping time.]]
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| Consider the following operation on an arbitrary positive [[integer]]:
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| * If the number is even, divide it by two.
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| * If the number is odd, triple it and add one.
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| In [[modular arithmetic]] notation, define the [[function (mathematics)|function]] ''f'' as follows:
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| : <math> f(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod{2}\\ 3n+1 & \text{if } n\equiv 1 \pmod{2} .\end{cases} </math>
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| Now, 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.
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| In notation:
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| : <math> a_i = \begin{cases}n & \text{for } i = 0 \\ f(a_{i-1}) & \text{for } i > 0 \end{cases}</math>
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| (that is: <math>a_i</math> is the value of <math>f</math> applied to <math>n</math> recursively <math>i</math> times; <math>a_i = f^i(n)</math>).
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| The Collatz conjecture is: ''This process will eventually reach the number 1, regardless of which positive integer is chosen initially.''
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| That smallest ''i'' such that ''a''<sub>''i''</sub> = 1 is called the '''total stopping time''' of ''n''.<ref name="lag85">* {{cite journal |doi=10.2307/2322189 |author=Jeffrey C. Lagarias |title=The 3''x'' + 1 problem and its generalizations |journal=American Mathematical Monthly |volume=92 |issue=1 |date=January 1985 |pages=3–23 |jstor=2322189}}
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| </ref> The conjecture asserts that every ''n'' has a well-defined total stopping time. If, for some ''n'', such an ''i'' doesn't exist, we say that ''n'' has infinite total stopping time and the conjecture is false.
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| If the conjecture is false, it can only be because there is some starting number which gives rise to a sequence which does not contain 1. Such a sequence might enter a repeating cycle that excludes 1, or increase without bound. No such sequence has been found.
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| ==Examples==
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| For instance, starting with ''n'' = 6, one gets the sequence 6, 3, 10, 5, 16, 8, 4, 2, 1.
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| ''n'' = 11, for example, takes longer to reach 1: 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.
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| <!-- NOTICE TO EDITORS... Please note that the number of steps is one less than the number of elements of the sequence! So 111 steps is CORRECT for n=27. Thanks for paying attention to this factoid! -->
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| The sequence for ''n'' = 27, listed and graphed below, takes 111 steps, climbing to 9232 before descending to 1.
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| :{ 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, '''7288''', 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, '''''9232''''', 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 }
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| [[Image:Collatz5.svg|500px|center]]
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| Numbers with a total stopping time longer than any smaller starting value form a sequence beginning with:
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| : 1, 2, 3, 6, 7, 9, 18, 25, 27, 54, 73, 97, … {{OEIS|A006877}}.
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| The longest progression for any initial starting number less than 100 million is 63,728,127, which has 949 steps. For starting numbers less than 1 billion it is 670,617,279, with 986 steps, and for numbers less than 10 billion it is 9,780,657,630, with 1132 steps.<ref>{{cite journal
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| |last1=Leavens |first1=Gary T. |last2=Vermeulen |first2=Mike
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| |date=December 1992 |title=3x+1 Search Programs |journal=Computers & Mathematics with Applications
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| |volume=24 |issue=11 |pages=79–99 |doi=10.1016/0898-1221(92)90034-F}}</ref><ref name=Roosendaal>{{cite web|last=Roosendaal | first=Eric |title=3x+1 Delay Records |url=http://www.ericr.nl/wondrous/delrecs.html |accessdate=27 November 2011}} (Note: "Delay records" are total stopping time records.)</ref>
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| The [[Power of two|powers of two]] converge to one quickly because <math>2^n</math> is halved <math>n</math> times to reach one, and is never increased.
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| == Visualizations ==
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| <gallery>
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| Collatz-graph-all-30-no27.svg|[[Directed graph]] showing the orbits of small numbers under the Collatz map. The Collatz conjecture is equivalent to the statement that all paths eventually lead to 1.
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| Collatz1000mathematica.png|Directed graph showing the orbits of the first 1000 numbers.
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| CollatzConjectureGraphMaxValues.jpg|''x''-axis represents starting number, ''y''-axis represents the highest number reached during the chain to 1
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| </gallery>
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| ==Cycles==
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| Any counterexample to the Collatz conjecture would have to consist either of an infinite divergent trajectory or a cycle different from the trivial (4,2,1) cycle. Thus, if one could prove that neither of these types of counterexample could exist, then all natural numbers would have a trajectory that reaches the trivial cycle.
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| Such a strong result is not known, but certain types of cycles have been ruled out.
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| The type of a cycle may be defined with reference to the "shortcut" definition of the Collatz map, <math>f(n)=(3n+1)/2</math> for odd ''n'' and <math>f(n)=n/2</math> for even ''n''.
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| A ''cycle'' is a sequence <math>(a_0,a_1,\ldots,a_q)</math> where <math>f(a_0)=a_1</math>, <math>f(a_1)=a_2</math>, and so on, up to <math>f(a_q)=a_0</math> in a closed loop. The only known cycle is (1,2).
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| A ''k''-cycle is a cycle that can be partitioned into 2''k'' contiguous subsequences: ''k'' increasing sequences of odd numbers alternating with ''k'' decreasing sequences of even numbers. For instance, if the cycle consists of a single increasing sequence of odd numbers followed by a decreasing sequence of even numbers, it is called a ''1-cycle''.<ref name=Simons2005>Simons,J.;de Weger, B.; [http://deweger.xs4all.nl/papers/%5B35%5DSidW-3n+1-ActaArith%5B2005%5D.pdf "Theoretical and computational bounds for ''m''-cycles of the 3''n'' + 1 problem"], ''Acta Arithmetica'', (online version 1.0, November 18, 2003), 2005.</ref>
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| {{harvtxt|Steiner|1977}} proved that there is no 1-cycle other than the trivial (1,2). {{harvtxt|Simons|2000}} used Steiner's method to prove that there is no 2-cycle. {{harvtxt|Simons|de Weger|2003}} extended this proof up to 68-cycles: there is no ''k''-cycle up to ''k'' = 68.<ref name=Simons2005/> Beyond 68, this method gives upper bounds for the elements in such a cycle: for example, if there is a 75-cycle, then at least one element of the cycle is less than 2385×2<sup>50</sup>.<ref name=Simons2005/> Therefore as exhaustive computer searches continue, larger cycles may be ruled out.
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| ==Supporting arguments==
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| Although 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.
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| ===Experimental evidence===
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| The conjecture has been checked by computer for all starting values up to 5 × 2<sup>60</sup> ≈ 5.764{{e|18}}.<ref name=Silva>{{cite web |last=Silva | first=Tomás Oliveira e Silva |title=Computational verification of the 3x+1 conjecture |url=http://www.ieeta.pt/~tos/3x+1.html |accessdate=27 November 2011}}</ref> All initial values tested so far eventually end in the repeating cycle {4,2,1}, which has only three terms. From this lower bound on the starting value, a lower bound can also be obtained for the number of terms a repeating cycle other than {4,2,1} must have.<ref name=Garner/> When this relationship was established in 1981, the formula gave a lower bound of 35,400 terms.<ref name=Garner>{{cite journal |last=Garner |first=Lynn E. |authorlink= |coauthors= |year=1981 |month= |title=On the Collatz 3''n'' + 1 Algorithm |journal=Proceedings of the American Mathematical Society |volume=82 |issue=1 |pages=19–22 |doi=10.2307/2044308 |jstor=2044308|accessdate= |quote= }}</ref>
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| Such computer evidence is not a proof that the conjecture is true. As shown in the cases of the [[Pólya conjecture]], the [[Mertens conjecture]] and the [[Skewes' number]], sometimes a conjecture's only counterexamples are found when using very large numbers. Since sequentially examining all natural numbers is a process which can never be completed, such an approach can never demonstrate that the conjecture is true, merely that no counterexamples have yet been discovered.
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| ===A probabilistic heuristic===
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| If one considers only the ''odd'' numbers in the sequence generated by the Collatz process, then each odd number is on average 3/4 of the previous one.<ref>http://www.cecm.sfu.ca/organics/papers/lagarias/paper/html/node3.html</ref> (More precisely, the geometric mean of the ratios of outcomes is 3/4.) This yields a heuristic argument that every Hailstone sequence should decrease in the long run, although this is not evidence against other cycles, only against divergence. The argument is not a proof because it assumes that Hailstone sequences are assembled from uncorrelated probabilistic events. (It does rigorously establish that the [[p-adic numbers|2-adic]] extension of the Collatz process has two division steps for every multiplication step for almost all 2-adic starting values.)
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| ===Rigorous bounds===
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| Although it is not known rigorously whether all positive numbers eventually reach one according to the Collatz iteration, it is known that many numbers do so. In particular, Krasikov and Lagarias showed that the number of integers in the interval [1,''x''] that eventually reach one is at least proportional to ''x''<sup>0.84</sup>.<ref>{{Cite journal
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| | last1 = Krasikov | first1 = Ilia
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| | last2 = Lagarias | first2 = Jeffrey C. | authorlink2 = Jeffrey Lagarias
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| | doi = 10.4064/aa109-3-4
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| | issue = 3
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| | journal = Acta Arithmetica
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| | mr = 1980260
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| | pages = 237–258
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| | title = Bounds for the 3''x'' + 1 problem using difference inequalities
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| | volume = 109
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| | year = 2003
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| | postscript = <!-- Bot inserted parameter. Either remove it; or change its value to "." for the cite to end in a ".", as necessary. -->{{inconsistent citations}}}}.</ref>
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| ==Other formulations of the conjecture==
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| ===In reverse===
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| [[File:Collatz-tree, depth=20.svg|thumb|600px|The first 21 levels of the ''Collatz [[graph (mathematics)|graph]]'' generated in bottom-up fashion. The graph includes all numbers with an orbit length of 21 or less.]]
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| There is another approach to prove the conjecture, which considers the bottom-up
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| method of growing the so-called ''Collatz graph''. The ''Collatz graph'' is a [[graph (mathematics)|graph]] defined by the inverse [[relation (mathematics)|relation]]
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| <math> R(n) = \begin{cases} \{2n\} & \text{if } n\equiv 0,1,2,3,5 \\ \{2n,(n-1)/3\} & \text{if } n\equiv 4 \end{cases} \pmod{6}. </math>
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| So, instead of proving that all natural numbers eventually lead to 1, we can prove that 1 leads to all natural numbers. For any integer ''n'', ''n'' ≡ 1 (mod 2) [[iff]] 3''n'' + 1 ≡ 4 (mod 6). Equivalently, (''n'' − 1)/3 ≡ 1 (mod 2) iff ''n'' ≡ 4 (mod 6). Conjecturally, this inverse relation forms a [[tree (graph theory)|tree]] except for the 1–2–4 loop (the inverse of the 1–4–2 loop of the unaltered function ''f'' defined in the statement of the problem above). 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,
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| <math> R(n) = \begin{cases} \{2n\} & \text{if } n\equiv 0,1 \\ \{2n,(2n-1)/3\} & \text{if } n\equiv 2 \end{cases} \pmod{3}. </math>
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| ===As an abstract machine that computes in base two===
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| Repeated applications of the Collatz function can be represented as an [[abstract machine]] that handles [[string (computer science)|strings]] of [[bit]]s. The machine will perform the following three steps on any odd number until only one "1" remains:
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| # Append 1 to the (right) end of the number in binary (giving 2''n'' + 1);
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| # Add this to the original number by binary addition (giving 2''n'' + 1 + ''n'' = 3''n'' + 1);
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| # Remove all trailing "0"s (i.e. repeatedly divide by two until the result is odd).
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| This prescription is plainly equivalent to computing a Hailstone sequence in base two.
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| ====Example====
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| The starting number 7 is written in base two as 111. The resulting Hailstone sequence is:
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| <div style="font-family:'Courier New', 'Lucida Console', 'Courier', Monospace">
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| 111
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| <u>1111</u>
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| 1011<s>0</s>
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| <u>10111</u>
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| 10001<s>0</s>
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| <u>100011</u>
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| 1101<s>00</s>
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| <u>11011</u>
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| 101<s>000</s>
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| <u>1011</u>
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| 1<s>0000</s>
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| </div>
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| ===As a parity sequence===
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| For this section, consider the Collatz function in the slightly modified form
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| : <math> f(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \\ (3n +1)/2 & \text{if } n \equiv 1. \end{cases} \pmod{2}</math>
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| This can be done because when ''n'' is odd, 3''n'' + 1 is always even.
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| If P(…) is the parity of a number, that is P(2''n'') = 0 and P(2''n'' + 1) = 1, then we can define the Hailstone parity sequence (or parity vector) for a number ''n'' as ''p<sub>i</sub>'' = P(''a<sub>i</sub>''), where ''a''<sub>0</sub> = ''n'', and ''a''<sub>''i''+1</sub> = ''f''(''a''<sub>''i''</sub>).
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| What operation is performed (3n + 1)/2 or n/2 depends on the parity. The parity sequence is the same as the sequence of operations.
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|
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| 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<sup>''k''</sup>. 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.<ref name="lag85"/><ref>{{citation
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| | last = Terras | first = Riho
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| | issue = 3
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| | journal = Polska Akademia Nauk
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| | mr = 0568274
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| | pages = 241–252
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| | title = A stopping time problem on the positive integers
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| | url = http://matwbn.icm.edu.pl/ksiazki/aa/aa30/aa3034.pdf
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| | volume = 30
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| | year = 1976}}.</ref>
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| Applying the ''f'' function ''k'' times to the number ''a''·2<sup>''k''</sup> + ''b'' will give the result ''a''·3<sup>''c''</sup> + ''d'', where ''d'' is the result of applying the ''f'' function ''k'' times to ''b'', and ''c'' is how many odd numbers were encountered during that sequence.
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| ===As a tag system===
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| For the Collatz function in the form
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| <math> f(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \\ (3n +1)/2 & \text{if } n \equiv 1. \end{cases} \pmod{2}</math>
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| Hailstone sequences can be computed by the extremely simple
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| [[Tag system#Example: Computation of Hailstone sequences|2-tag system]] with production rules
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| ''a'' → ''bc'', ''b'' → ''a'', ''c'' → ''aaa''. In this system, the positive integer ''n'' is represented by a string of ''n'' ''a'''s, and iteration of the tag operation halts on any word of length less than 2. (Adapted from De Mol.)
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| The Collatz conjecture equivalently states that this tag system, with an arbitrary finite string of ''a'''s as the initial word, eventually halts (see ''[[Tag system#Example: Computation of Collatz sequences|Example: Computation of Collatz sequences]]'' for a worked example).
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| ==Extensions to larger domains==
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| ===Iterating on all integers===
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| An obvious extension is to include all integers, not just positive integers. In this case there are a total of 5 known cycles, which all integers seem to eventually fall into under iteration of ''f''. These cycles are listed here, starting with the well-known cycle for positive ''n''.
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| Odd values are listed in bold. Each cycle is listed with its member of least absolute value (which is always odd or zero) first.
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| {| class="wikitable" style="text-align: center;"
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| ! Cycle !! Odd-value cycle length !! Full cycle length
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| |-
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| |style="text-align: left;"| '''1''' → 4 → 2 → '''1''' '''…''' || 1 || 3
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| |-
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| |style="text-align: left;"| 0 → 0 '''…''' || 0 || 1
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| |-
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| |style="text-align: left;"| '''−1''' → −2 → '''−1''' '''…''' || 1 || 2
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| |-
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| |style="text-align: left;"| '''−5''' → −14 → '''−7''' → −20 → −10 → '''−5''' '''…''' || 2 || 5
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| |-
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| |style="text-align: left;"| '''−17''' → −50 → '''−25''' → −74 → '''−37''' → −110 → '''−55''' → −164 → −82 → '''−41''' → −122 → '''−61''' → −182 → '''−91''' → −272 → −136 → −68 → −34 → '''−17''' '''…''' || 7 || 18
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| |}
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| The Generalized Collatz Conjecture is the assertion that every integer, under iteration by ''f'', eventually falls into one of these five cycles.
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| ===Iterating with odd denominators or 2-adic integers===
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| The standard Collatz map can be extended to (positive or negative) rational numbers which have odd denominators when written in lowest terms. The number is taken to be odd or even according to whether its numerator is odd or even. A closely related fact is that the Collatz map extends to the ring of [[2-adic integers]], which contains the ring of rationals with odd denominators as a subring.
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| The parity sequences as defined above are no longer unique for fractions. However, it can be shown that any possible parity cycle is the parity sequence for exactly one fraction: if a cycle has length ''n'' and includes odd numbers exactly ''m'' times at indices ''k''<sub>0</sub>, …, ''k''<sub>''m''−1</sub>, then the unique fraction which generates that parity cycle is
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| :<math>\frac{3^{m-1} 2^{k_0} + \cdots + 3^0 2^{k_{m-1}}}{2^n - 3^m}.</math>
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| For example, the parity cycle (1 0 1 1 0 0 1) has length 7 and has 4 odd numbers at indices 0, 2, 3, and 6. The unique fraction which generates that parity cycle is
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| :<math>\frac{3^3 2^0 + 3^2 2^2 + 3^1 2^3 + 3^0 2^6}{2^7 - 3^4} = \frac{151}{47},</math>
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| the complete cycle being: 151/47 → 250/47 → 125/47 → 211/47 → 340/47 → 170/47 → 85/47 → 151/47
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| Although the cyclic permutations of the original parity sequence are unique fractions, the cycle is not unique, each permutation's fraction being the next number in the loop cycle:
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| :(0 1 1 0 0 1 1) → <math>\frac{3^3 2^1 + 3^2 2^2 + 3^1 2^5 + 3^0 2^6}{2^7 - 3^4} = \frac{250}{47}</math>
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| :(1 1 0 0 1 1 0) → <math>\frac{3^3 2^0 + 3^2 2^1 + 3^1 2^4 + 3^0 2^5}{2^7 - 3^4} = \frac{{125}}{47}</math>
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| :(1 0 0 1 1 0 1) → <math>\frac{3^3 2^0 + 3^2 2^3 + 3^1 2^4 + 3^0 2^6}{2^7 - 3^4} = \frac{211}{47}</math>
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| :(0 0 1 1 0 1 1) → <math>\frac{3^3 2^2 + 3^2 2^3 + 3^1 2^5 + 3^0 2^6}{2^7 - 3^4} = \frac{340}{47}</math>
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| :(0 1 1 0 1 1 0) → <math>\frac{3^3 2^1 + 3^2 2^2 + 3^1 2^4 + 3^0 2^5}{2^7 - 3^4} = \frac{170}{47}</math>
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| :(1 1 0 1 1 0 0) → <math>\frac{3^3 2^0 + 3^2 2^1 + 3^1 2^3 + 3^0 2^4}{2^7 - 3^4} = \frac{85}{47}</math>
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| Also, for uniqueness, the parity sequence should be "prime", i.e., not partitionable into identical sub-sequences. For example, parity sequence (1 1 0 0 1 1 0 0) can be partitioned into two identical sub-sequences (1 1 0 0)(1 1 0 0). Calculating the 8-element sequence fraction gives
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| :(1 1 0 0 1 1 0 0) → <math>\frac{3^3 2^0 + 3^2 2^1 + 3^1 2^4 + 3^0 2^5}{2^8 - 3^4} = \frac{125}{175}</math>
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| But when reduced to lowest terms {5/7}, it is the same as that of the 4-element sub-sequence
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| :(1 1 0 0) → <math>\frac{3^1 2^0 + 3^0 2^1}{2^4 - 3^2} = \frac{5}{7}.</math>
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| And this is because the 8-element parity sequence actually represents two circuits of the loop cycle defined by the 4-element parity sequence.
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| In this context, the Collatz conjecture is equivalent to saying that (0 1) is the only cycle which is generated by positive whole numbers (i.e. 1 and 2).
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| ===Iterating on real or complex numbers===
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| [[Image:CobwebCollatz2.PNG|right|thumb|300px|[[Cobweb plot]] of the orbit 10-5-8-4-2-1-2-1-2-1-etc. in the real extension of the Collatz map (optimized by replacing "3''n'' + 1" with "(3''n'' + 1)/2" )]]
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| The Collatz map can be viewed as the restriction to the integers of the smooth real and complex map
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| :<math>f(z)=\frac 1 2 z \cos^2\left(\frac \pi 2 z\right)+(3z+1)\sin^2\left(\frac \pi 2 z\right),</math>
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| which simplifies to <math>\frac{1}{4}(2 + 7z - (2 + 5z)\cos(\pi z)).</math>
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| If the standard Collatz map defined above is optimized by replacing the relation 3''n'' + 1 with the common substitute "shortcut" relation (3''n'' + 1)/2, it can be viewed as the restriction to the integers of the smooth real and complex map
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| :<math>f(z)=\frac 1 2 z \cos^2\left(\frac \pi 2 z\right)+(3z+1)\sin^2\left(\frac \pi 2 z\right),</math>
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| which simplifies to <math>\frac{1}{4}(1 + 4z - (1 + 2z)\cos(\pi z))</math>.
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| ====Collatz fractal====
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| [[Iterated function|Iterating]] the above optimized map in the complex plane produces the Collatz [[fractal]].
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| The point of view of iteration on the real line was investigated by Chamberland (1996), and on the complex plane by Letherman, Schleicher, and Wood (1999).
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| <div style="clear:both;"></div>
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| [[Image:CollatzFractal.png|thumb|center|500px|Collatz map [[fractal]] in a neighbourhood of the real line]]
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| ==Optimizations==
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| ===Time-space tradeoff===
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| The ''[[#As a parity sequence|As a parity sequence]]'' section above gives a way to speed up simulation of the sequence. To jump ahead ''k'' steps on each iteration (using the ''f'' function from that section), break up the current number into two parts, ''b'' (the ''k'' least significant bits, interpreted as an integer), and ''a'' (the rest of the bits as an integer). The result of jumping ahead ''k'' steps can be found as:
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| :''f'' <sup>''k''</sup>(''a'' 2<sup>''k''</sup> + ''b'') = ''a'' 3<sup>''c''(b)</sup> + ''d''(b).
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| The ''c'' and ''d'' arrays are precalculated for all possible ''k''-bit numbers ''b'', where ''d''(b) is the result of applying the ''f'' function ''k'' times to ''b'', and ''c''(b) is the number of odd numbers encountered on the way.<ref>{{citation|contribution=Looking for Class Records in the 3x+1 Problem by means of the COMETA Grid Infrastructure|url=http://www.ippari.unict.it/~scollo/papers/CR3x+1PAr2008.pdf|first=Giuseppe|last=Scollo|year=2007|title=Grid Open Days at the University of Palermo}}.</ref> For example, if k=5, you can jump ahead 5 steps on each iteration by separating out the 5 least significant bits of a number and using:
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| : ''c''(0..31) = {0,3,2,2,2,2,2,4,1,4,1,3,2,2,3,4,1,2,3,3,1,1,3,3,2,3,2,4,3,3,4,5}
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| : ''d''(0..31) = {0,2,1,1,2,2,2,20,1,26,1,10,4,4,13,40,2,5,17,17,2,2,20,20,8,22,8,71,26,26,80,242}.
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| This requires 2<sup>''k''</sup> [[precomputation]] and storage to speed up the resulting calculation by a factor of ''k'', a [[space-time tradeoff]].
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| ===Modular restrictions===
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| For the special purpose of searching for a counterexample to the Collatz conjecture, this precomputation leads to an even more important acceleration, used by [[Tomás Oliveira e Silva]] in his computational confirmations of the Collatz conjecture up to large values of ''n''. If, for some given ''b'' and ''k'', the inequality
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| :''f'' <sup>''k''</sup>(''a'' 2<sup>''k''</sup> + ''b'') = ''a'' 3<sup>''c''(b)</sup> + ''d''(b) < ''a'' 2<sup>''k''</sup> + ''b''
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| holds for all ''a'', then the first counterexample, if it exists, cannot be ''b'' modulo 2<sup>''k''</sup>.<ref>{{citation
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| | last = Garner | first = Lynn E.
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| | doi = 10.2307/2044308
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| | issue = 1
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| | journal = Proceedings of the American Mathematical Society
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| | mr = 603593
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| | pages = 19–22
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| | title = On the Collatz 3''n'' + 1 algorithm
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| | volume = 82
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| | year = 1981}}.</ref> For instance, the first counterexample must be odd because ''f''(2''n'') = ''n'', smaller than 2''n''; and it must be 3 mod 4 because ''f''<sup>2</sup>(4''n'' + 1) = 3''n'' + 1, smaller than 4''n'' + 1. For each starting value ''a'' which is not a counterexample to the Collatz conjecture, there is a ''k'' for which such an inequality holds, so checking the Collatz conjecture for one starting value is as good as checking an entire congruence class. As ''k'' increases, the search only needs to check those residues ''b'' that are not eliminated by lower values of ''k''. Only an exponentially small fraction of the residues survive.<ref>{{harvtxt|Lagarias|1985}}, Theorem D.</ref> For example, the only surviving residues mod 32 are 7, 15, 27, and 31.
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| ==Syracuse function==
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| If ''k'' is an odd integer, then 3''k'' + 1 is even, so we can write 3''k'' + 1 = 2<sup>''a''</sup>''k''′, with ''k''' odd and ''a'' ≥ 1. We define a function ''f'' from the set <math>I</math> of odd integers into itself, called the ''Syracuse function,'' by taking ''f'' (''k'') = ''k''′ {{OEIS|id=A075677}}.
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| Some properties of the Syracuse function are:
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| * ''f'' (4''k'' + 1) = ''f'' (''k'') for all ''k'' in <math>I</math>.
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| * For all ''p'' ≥ 2 and ''h'' odd, ''f'' <sup>''p''−1</sup>(2<sup>''p''</sup> ''h'' − 1) = 2·3<sup>''p'' − 1</sup>''h'' − 1 (here, ''f'' <sup>''p''−1</sup> is [[function composition#Functional powers|function iteration notation]]).
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| * For all odd ''h'', ''f''(2''h'' − 1) ≤ (3''h'' − 1)/2
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| The Syracuse conjecture is that for all ''k'' in <math>I</math>, there exists an integer ''n'' ≥ 1 such that ''f''<sup>''n''</sup>(''k'') = 1. Equivalently, let ''E'' be the set of odd integers ''k'' for which there exists an integer ''n'' ≥ 1 such that ''f''<sup>''n''</sup>(''k'') = 1. The problem is to show that ''E'' = <math>I</math>. The following is the beginning of an attempt at a proof by induction:
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| 1, 3, 5, 7, and 9 are known to be elements of ''E''. Let ''k'' be an odd integer greater than 9. Suppose that the odd numbers up to and including ''k'' − 2 are in ''E'' and let us try to prove that ''k'' is in ''E''. As ''k'' is odd, ''k'' + 1 is even, so we can write ''k'' + 1 = 2<sup>''p''</sup>''h'' for ''p'' ≥ 1, ''h'' odd, and ''k'' = 2<sup>''p''</sup>''h'' − 1. Now we have:
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| * If ''p'' = 1, then ''k'' = 2''h'' − 1. It is easy to check that ''f'' (''k'') < ''k '', so ''f'' (''k'') ∈ ''E''; hence ''k'' ∈ ''E''.
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| * If ''p'' ≥ 2 and ''h'' is a multiple of 3, we can write ''h'' = 3''h′''. Let ''k′'' = 2<sup>''p'' + 1</sup>''h′'' − 1; then ''f'' (''k′'') = ''k'' , and as ''k′'' < ''k'' , ''k′'' is in ''E''; therefore ''k'' = ''f'' (''k′'') ∈ ''E''.
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| * If ''p'' ≥ 2 and ''h'' is not a multiple of 3 but ''h'' ≡ (−1)<sup>''p''</sup> mod 4, we can still show that ''k'' ∈ ''E''.
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| The problematic case is that where ''p'' ≥ 2 , ''h'' not multiple of 3 and ''h'' ≡ (−1)<sup>''p'' + 1</sup> mod 4. Here, if we manage to show that for every odd integer ''k''′, 1 ≤ ''k''′ ≤ ''k'' − 2 ; 3''k''′ ∈ ''E'' we are done.
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| ==See also==
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| * [[Residue class-wise affine groups]]
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| * [[Modular arithmetic]]
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| * [[BOINC]]
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| * [[Distributed computing]]
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| ==Notes==
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| {{reflist|2}}
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| ==References and external links==
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| ===Papers===
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| * Jeffrey C. Lagarias. (1985). The 3x + 1 problem and its generalizations. ''[[The American Mathematical Monthly]]'' 92(1):3-23.
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| * {{SpringerEOM | urlname=S/s110330 | title=Syracuse problem | author=Jeffrey C. Lagarias}}
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| * {{cite arXiv |author=Jeffrey C. Lagarias |eprint=math.NT/0608208 |title=The 3''x'' + 1 problem: An annotated bibliography, II (2000–) |class=math.NT |year=2006}}
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| * Günther J. Wirsching. The Dynamical System Generated by the <math>3n+1</math> Function. Number 1681 in Lecture Notes in Mathematics. Springer-Verlag, 1998.
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| * [http://www.numbertheory.org/3x+1/ Keith Matthews' 3''x'' + 1 page: Review of progress, plus various programs]
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| * Marc Chamberland. A continuous extension of the 3''x'' + 1 problem to the real line. Dynam. Contin. Discrete Impuls Systems 2:4 (1996), 495–509.
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| * Simon Letherman, Dierk Schleicher, and Reg Wood: The (3''n'' + 1)-Problem and Holomorphic Dynamics. Experimental Mathematics 8:3 (1999), 241–252.
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| * Ohira, Reiko & Yamashita, Michinori [http://risweb2.ris.ac.jp/faculty/earth_env/yamasita/open/p-col.pdf A generalization of the Collatz problem] {{Link language|ja}}
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| * Urata, Toshio [http://web.archive.org/web/20080406061036/http://auemath.aichi-edu.ac.jp/~turata/Fall.files/CTZVI.pdf Some Holomorphic Functions connected with the Collatz Problem]
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| * Eliahou, Shalom, ''The 3x+1 problem: new lower bounds on nontrivial cycle lengths'', Discrete Mathematics 118 (1993) p. 45-56; [http://images.math.cnrs.fr/Le-probleme-3n-1-y-a-t-il-des.html ''Le problème 3n+1 : y a-t-il des cycles non triviaux ?''], ''[[Images des mathématiques]]'' (2011) {{Link language|fr}}
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| * Matti K. Sinisalo: [http://web.archive.org/web/20091024183537/http://geocities.com/mattiksinisalo/collatz.doc On the minimal cycle lengths of the Collatz sequences], Preprint, June 2003, University of Oulu
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| * Paul Stadfeld: [http://home.versatel.nl/galien8/blueprint/blueprint.html Blueprint for Failure: How to Construct a Counterexample to the Collatz Conjecture]
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| * De Mol, Liesbeth, "Tag systems and Collatz-like functions", ''Theoretical Computer Science'', '''390:1''', 92–101, January 2008 ([http://logica.ugent.be/liesbeth/TagColOK.pdf Preprint Nr. 314]).
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| * {{cite arXiv |author=Bruschi, Mario |eprint=0810.5169 |title=A generalization of the Collatz problem and conjecture |class=math.NT |year=2008}}
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| * {{cite journal |author=Andrei, Stefan; Masalagiu, Cristian |doi=10.1007/s002360050117 |title=About the Collatz conjecture |year=1998 |journal=Acta Informatica |volume=35 |issue=2 |pages=167}}
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| *Van Bendegem, Jean Paul, [http://compmath.files.wordpress.com/2008/08/jpvb_collatz.pdf "The Collatz Conjecture: A Case Study in Mathematical Problem Solving"], ''Logic and Logical Philosophy'', Volume 14 (2005), 7–23
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| *[[Edward Belaga|Belaga, Edward G.]], [http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.54.483 "Reflecting on the 3x+1 Mystery"], [[University of Strasbourg]] preprint, 1998.
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| * Belaga, Edward G.; Mignotte, Maurice, [http://www-irma.u-strasbg.fr/~belaga/a8*BelagaMathInfo06Presentation060920.ppt "Walking Cautiously into the Collatz Wilderness: Algorithmically, Number Theoretically, Randomly"], Fourth Colloquium on Mathematics and Computer Science : Algorithms, Trees, Combinatorics and Probabilities, September 18–22, 2006, Institut Élie Cartan, Nancy, France
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| * Belaga, Edward G.; Mignotte, Maurice, [http://www.emis.de/journals/EM/expmath/volumes/7/7.html "Embedding the 3x+1 Conjecture in a 3x+d Context"], ''Experimental Mathematics'', Volume 7, issue 2, 1998.
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| * Steiner, R.P.; "A theorem on the syracuse problem", ''Proceedings of the 7th Manitoba Conference on Numerical Mathematics'',pages 553–559, 1977.
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| * Simons,J.;de Weger, B.; [http://deweger.xs4all.nl/papers/%5B35%5DSidW-3n+1-ActaArith%5B2005%5D.pdf "Theoretical and computational bounds for ''m''-cycles of the 3''n'' + 1 problem"], ''Acta Arithmetica'', (online version 1.0, November 18, 2003), 2005.
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| * Sinyor, J.; [http://downloads.hindawi.com/journals/ijmms/2010/458563.pdf "The 3x+1 Problem as a String Rewriting System"], ''International Journal of Mathematics and Mathematical Sciences'', Volume 2010 (2010), Article ID 458563, 6 pages.
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| ===Books===
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| * {{cite book | last1=Everest | first1=Graham | last2=van der Poorten | first2=Alf | author2-link=Alfred van der Poorten | last3=Shparlinski | first3=Igor | last4=Ward | first4=Thomas | title=Recurrence sequences | series=Mathematical Surveys and Monographs | volume=104 | location=[[Providence, RI]] | publisher=[[American Mathematical Society]] | year=2003 | isbn=0-8218-3387-1 | zbl=1033.11006 | at=Chapter 3.4 }}
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| * {{cite book |last=Guy | first=Richard K. | authorlink=Richard K. Guy | title=Unsolved problems in number theory | publisher=[[Springer-Verlag]] |edition=3rd | year=2004 |isbn=0-387-20860-7 | zbl=1058.11001 | at="E17: Permutation Sequences" }} Cf. [http://books.google.com/books?id=1AP2CEGxTkgC&pg=PA337&lpg=PA337&dq=%22Unpredictable+Iterations%22+conway&source=bl&ots=Tilvg0yIqF&sig=5Nz7invP8gpLKpkgd0Q7aX2njUU&hl=en&ei=WVITSp-fHdmLtgfJmKCTBA&sa=X&oi=book_result&ct=result&resnum=9#PPA336,M1 pp.336–337]
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| * {{cite book | title=The Ultimate Challenge: the 3x+1 problem | editor1-first=Jeffrey C. | editor1-last=Lagarias | editor1-link=Jeffrey Lagarias | publisher=[[American Mathematical Society]] | year=2010 | isbn=978-0-8218-4940-8 | zbl=1253.11003 }}
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| ==External links==
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| * [[Distributed Computing]] [http://boinc.thesonntags.com/collatz/ project] that verifies the Collatz conjecture for larger values.
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| * An ongoing [[distributed computing]] [http://www.ericr.nl/wondrous/index.html project] by Eric Roosendaal verifies the Collatz conjecture for larger and larger values.
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| * Another ongoing [[distributed computing]] [http://www.ieeta.pt/~tos/3x+1.html project] by Tomás Oliveira e Silva continues to verify the Collatz conjecture (with fewer statistics than Eric Roosendaal's page but with further progress made).
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| * [http://gfredericks.com/sandbox/arith/collatz An animated implementation] that uses [[arbitrary-precision arithmetic]].
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| * {{MathWorld | urlname=CollatzProblem | title=Collatz Problem}}
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| * [http://codeabbey.com/index/task_view/collatz-sequence Collatz Sequence] explanation and exercise
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| * {{PlanetMath | urlname=CollatzProblem | title=Collatz Problem}}
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| * [http://Mensanator.com/mensanator666/collatz/hailstone.htm Hailstone Patterns] discusses different resonators along with using important numbers in the problem (like 6 and 3^5) to discover patterns.
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| * {{youtube|id=BhR-iIjhTNM|title=Collatz Iterations on the Ulam Spiral grid}}
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| * [http://demonstrations.wolfram.com/CollatzPaths/ Collatz Paths] by Jesse Nochella, [[Wolfram Demonstrations Project]].
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| * [http://syracuse.elaunira.com Page allowing to study and to show the guess for a given number]
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| * [http://go.helms-net.de/math/collatz/aboutloop/collloopintro_main.htm Collatz cycles?] about cycles, very basic, contains also some unusual graphs(html)
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| * [http://go.helms-net.de/math/collatz/Collatz061102.pdf Collatz cycles? (pdf)] about loops, compacted text, rather basic (pdf)
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| * [http://www.nitrxgen.net/collatz.php Collatz sequence] for any number up to 500 digits in length.
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| {{DEFAULTSORT:Collatz Conjecture}}
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| [[Category:Articles with inconsistent citation formats]]
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| [[Category:Conjectures]]
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| [[Category:Number theory]]
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