# Difference between revisions of "Collatz Problem"

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Define the following [[function]] on <math>\mathbb{N}</math>: <math>f(n)=\begin{cases} 3n+1 & 2\nmid n, \\ \frac{n}{2} & 2\mid n.\end{cases}</math> The Collatz conjecture says that, for any [[positive integer]] <math>n</math>, the sequence <math>\{n,f(n),f(f(n)),f(f(f(n))),\ldots\}</math> contains 1. This conjecture is still open. Some people have described it as the easiest unsolved problem in mathematics. | Define the following [[function]] on <math>\mathbb{N}</math>: <math>f(n)=\begin{cases} 3n+1 & 2\nmid n, \\ \frac{n}{2} & 2\mid n.\end{cases}</math> The Collatz conjecture says that, for any [[positive integer]] <math>n</math>, the sequence <math>\{n,f(n),f(f(n)),f(f(f(n))),\ldots\}</math> contains 1. This conjecture is still open. Some people have described it as the easiest unsolved problem in mathematics. | ||

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+ | ==Properties of <math>f(n)</math> == | ||

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+ | Self similarity of <math>f(n)</math> follows from generalizing <math>n</math> to an integral, integer coefficient polynomial. If <math>n=2^zx+b</math> for example, it can be shown by parity argument, that <math>n</math> has the same parity as <math>b</math>. It then follows, that same conditional path will be followed by <math>n</math> as it was for <math>b</math>; any time the lead coefficient still has a factor of 2. | ||

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+ | Observing that if <math>n=2m+1</math> then <math>3n+1=6m+4</math>, as well as: <cmath>{6m+4\over 2}=3m+2</cmath> we can then observe that; only if <math>m</math> is even will another division by 2 be possible. | ||

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+ | The above observation leads to 2 important points about <math>n=4c+3</math>; namely they are the only possible lowest elements of a non-trivial cycle, and also the only possible lowest elements of an infinitely increasing sequence. | ||

[[Category:Number theory]] | [[Category:Number theory]] | ||

[[Category:Conjectures]] | [[Category:Conjectures]] |

## Latest revision as of 19:45, 24 February 2020

Define the following function on : The Collatz conjecture says that, for any positive integer , the sequence contains 1. This conjecture is still open. Some people have described it as the easiest unsolved problem in mathematics.

## Properties of

Self similarity of follows from generalizing to an integral, integer coefficient polynomial. If for example, it can be shown by parity argument, that has the same parity as . It then follows, that same conditional path will be followed by as it was for ; any time the lead coefficient still has a factor of 2.

Observing that if then , as well as: we can then observe that; only if is even will another division by 2 be possible.

The above observation leads to 2 important points about ; namely they are the only possible lowest elements of a non-trivial cycle, and also the only possible lowest elements of an infinitely increasing sequence.