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Collatz conjecture
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==Statement of the problem== [[File:Collatz-stopping-time.svg|thumb|Numbers from 1 to 9999 and their corresponding total stopping time]] [[File:CollatzStatistic100million.png|thumb|Histogram of total stopping times for the numbers 1 to 10<sup>8</sup>. Total stopping time is on the {{mvar|x}} axis, frequency on the {{mvar|y}} axis.]] [[File:CollatzStatistic1billion.png|thumb|Histogram of total stopping times for the numbers 1 to 10<sup>9</sup>. Total stopping time is on the {{mvar|x}} axis, frequency on the {{mvar|y}} axis.]] [[File:Collatz-10Million.png|thumb|Iteration time for inputs of 2 to 10<sup>7</sup>.]] [[File:Collatz Gif.gif|alt=Total Stopping Time: numbers up to 250, 1000, 4000, 20000, 100000, 500000|thumb|Total stopping time of numbers up to 250, 1000, 4000, 20000, 100000, 500000]] Consider the following operation on an arbitrary [[positive integer]]: * If the number is even, divide it by two. * If the number is odd, triple it and add one. In [[modular arithmetic]] notation, define the [[function (mathematics)|function]] {{mvar|f}} as follows: <math display="block"> 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> 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. In notation: <math display="block"> a_i = \begin{cases}n & \text{for } i = 0, \\ f(a_{i-1}) & \text{for } i > 0 \end{cases}</math> (that is: {{math|''a<sub>i</sub>''}} is the value of {{mvar|f}} applied to {{mvar|n}} recursively {{mvar|i}} times; {{math|''a<sub>i</sub>'' {{=}} ''f''{{hsp}}{{isup|''i''}}(''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'' <math>n</math>, there is some <math>i</math> with <math>a_i = 1</math>. If 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 {{mvar|i}} such that {{math|''a<sub>i</sub>'' < ''a''<sub>0</sub> }} is called the '''stopping time''' of {{mvar|n}}. Similarly, the smallest {{mvar|k}} such that {{math|''a<sub>k</sub>'' {{=}} 1}} is called the '''total stopping time''' of {{mvar|n}}.<ref name="Lagarias (1985)"/> If one of the indexes {{mvar|i}} or {{mvar|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 {{mvar|n}} is finite. It is also equivalent to saying that every {{math|''n'' β₯ 2}} has a finite stopping time. Since {{math|3''n'' + 1}} is even whenever {{mvar|n}} is odd, one may instead use the "shortcut" form of the Collatz function: <math display = "block"> f(n) = \begin{cases} \frac{n}{2} &\text{if } n \equiv 0 \pmod{2},\\ \frac{3n+1}{2} & \text{if } n\equiv 1 \pmod{2}. \end{cases}</math> This definition yields smaller values for the stopping time and total stopping time without changing the overall dynamics of the process.
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