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Consider the function $\sigma : \mathbb{Z} \to \mathbb{Z}$ where $\sigma = n - 3$. The orbits are

$\{3n : n \in \mathbb{Z} \}, \{3n + 1 : n \in \mathbb{Z} \}, \{ 3n + 2: n \mathbb{Z} \}$

What exactly are they really doing? I tried listing out a few numbers $\sigma (1), \sigma(2), \sigma(3)$ etc... but they just keep going on. How are they finding these? What happens if the function is more complicated like $\rho = n^2 + 1$? What strategy are they using? What is their thinking towards this problem?

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By the orbit of a number $n$, they mean the numbers $n,\sigma(n),\sigma(\sigma(n)),\sigma(\sigma(\sigma(n)))$ and so on; but also $\sigma^{-1}(n),\sigma^{-1}(\sigma^{-1}(n))$ and so on. When $\sigma(n)=n-3$, that's the numbers $n,n-3,n-6,n-9$ and so on, and also $n+3,n+6,n+9$ and so on. So, depending on the value of $n$, you get all the multiples of 3, or all the numbers one more than a multiple of 3, or all the numbers two more than a multiple of 3.

It's not always this easy --- there are some fairly simple functions for which it's notoriously difficult to find all the orbits --- have a look for the Collatz problem to see what I mean.

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    But you obviously know that Collatz is not a permutation - which is what concerns the OP.2012-12-22
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    @alancalvitti, you are right. However, there are small variations on Collatz which *are* permutations, and which appear to be as hard to analyze as Collatz is.2012-12-23
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    Can you give an example please? I understand that group theory gets ever more complicated as time goes on (entire journals devoted to the subject) and all the arrows are iso's. but specific permutation on $\mathbb N$ or $\mathbb Z$?2012-12-23
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    @alancalvitti, consider the map given by $f(3n+1)=4n+1$, $f(3n-1)=4n-1$, $f(3n)=2n$. This map is a permutation on the positive integers. I believe it is unknown whether the orbit of 8 goes to infinity or is eventually periodic. It's discussed briefly in Guy's Unsolved Problems In Number Theory.2012-12-23
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    Interesting function, I plotted it w/ Mathematica. However, permutations are bijections, hence surjective, whereas your $f$ is not. For example, these numbers have no preimage: 1, 2, 3, 4, 8, 9, 10, 11... ($f$ is odd, so similarly for the negative counterparts)2012-12-23
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    @alancalvitti, $f(1)=1$, $f(3)=2$, $f(2)=3$, $f(6)=4$, $f(12)=8$, $f(7)=9$, $f(15)=10$, $f(8)=11$. It's a permutation. Every integer is of exactly one of the forms $4n+1$, $4n-1$, $2n$. It's one-one and onto --- it's a permutation.2012-12-24
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    how can for example, $f(3) = 2$ when above you defined $f(3 n) = 2 n$? Isn't 3 expressible as $3 n, n=1$?2012-12-24
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    @alancalvitti, yes, $3=3n$ with $n=1$, so $f(3)=2n=2$.2012-12-25
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    Ok got it, will recode accordingly. Happy xdays.2012-12-25