Question: I need a set-builder notation for $\{1 , 3 , 5 , -7 , -9 , -11 , 13 , 15 , 17 , \ldots\}$. It is an infinite set of odd positive integers, which are multiplied by $-1$ three by three. One of the terms of this question is not to use floor function.

Answer: 
Well, no doubt you are hoping for something cleverer than this, but:

$$ \{\,a\mid\exists k\in \mathbb {N}\_{0}:(\exists m \in \{1,3,5\}
:a = m+12k)\vee (\exists m \in \{-7,-9,-11\}
:a = m-12k)\,\}$$ 

which I suddenly realized is actually

$$ \{\,a\mid\exists k\in \mathbb {Z}:(\exists m \in \{1,3,5\}
:a = m+12k)\,\}$$
For $n\in\mathbb{N}$, $\cos(n\pi/3)$ is never $0$. It oscillates from being positive for three consecutive terms, then negative for three. Using this, one way to express your sequence is $$a\_n=\frac{\cos\left(\frac{(n-2)\pi}{3}\right)}{\left\lvert\cos\left(\frac{(n-2)\pi}{3}\right)\right\rvert}(2n-1)$$ As a set in set-builder notation, you have $$\left\{\frac{\cos\left(\frac{(n-2)\pi}{3}\right)}{\left\lvert\cos\left(\frac{(n-2)\pi}{3}\right)\right\rvert}(2n-1)\middle| n\in\mathbb{N}\right\}$$

Of course there is nothing more fundamental about the absolute value function than the signum function, so another way to express this is:$$\left\{\operatorname{sign}\left(\cos\left(\frac{(n-2)\pi}{3}\right)\right)(2n-1)\middle| n\in\mathbb{N}\right\}$$