Definition of an Upper Bound
Let me write the definition in a little bit more detail.

Suppose that $f(n),g(n)$ are two functions from $\mathbb{N}$ to $\mathbb{R}_+$, that is, they accept as input a natural number, and return as output a positive real number.
We say that $f(n) = O(g(n))$ if there exist $n_0 \in \mathbb{N}$ and $C \in \mathbb{R}_+$ such that for all $n \geq n_0$,
$$ f(n) \leq Cg(n). $$

In order to show that $f(n) = O(g(n))$, you have to find $n_0$ and $C$ such that for all $n \geq n_0$ it holds that $f(n) \leq Cg(n)$. You can choose whatever $n_0,C$ you want as long as the condition holds. It turns out that for this definition, we can always take $n_0 = 1$ (for other variants of the definition, this no longer holds). So you only have to find a positive constant $C &gt; 0$ such that for all $n \geq 1$ it holds that
$$ f(n) \leq Cg(n). $$
(This assumes that the first natural number is 1 rather than 0.)
Now we can answer your question:

$n$ is the argument of the functions $f,g$.
$n_0$ is the point beyond which the inequality $f(n) \leq C g(n)$ should hold. When proving that $f(n) = O(g(n))$, you get to choose $n_0$. In most cases you can just take $n_0 = 1$.


For the curious: why do we need $n_0$? Sometimes we would like to accommodate functions which are not strictly positive, but just eventually positive. For example, if 0 is a natural number (in contrast to the convention used above) then $n^2$ is not strictly positive, since it vanishes at $n = 0$. Therefore $n = O(n^2)$ is only true if we allow to &quot;skip&quot; $n = 0$, which we do by choosing $n_0 = 1$.