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 say 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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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.

Revision as of 12:53, 18 July 2006

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