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. | ||
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| + | [[Category:Number theory]] | ||
| + | [[Category:Conjectures]] | ||
Latest revision as of 18:45, 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: ![\[{6m+4\over 2}=3m+2\]](http://latex.artofproblemsolving.com/2/7/2/272ac48bcf29dca2e1d3233a567edbca76cc1c9f.png)
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.
