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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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    Ok got it, will recode accordingly. Happy xdays.2012-12-25