1, 5, 9, 13, 17, 21,... How would you describe the set $\{1, 5, 9, 13, 17, 21,\dots\}$ in the style of $x:P(x)=$? I know that the sequence is "the last number + 4" or $4n-3$. 
 A: $\{n \in \mathbb N\; :\; n \equiv 1 \mod 4\}$
A: $\{4k-3 \mid k \in \mathbb{N} \}$ or $\{4k+1 \mid k \in \mathbb{N} \}$, depending on whether you consider $0$ a natural number.
A: An arithmetic progression with first term 1 and common difference 4
(No I haven't described it in the form of $P(x)$, but this is how I would describe this number sequence...)
A: It's not exactly the form you're looking for, but close:
$$
\{n\in\mathbb N \text{ [or }\mathbb R \text{ or whatever relevant]}: (n-1)/4 \in \mathbb N\}
$$
If you really need an equality, then you might write
$$
\{n\in\mathbb N: \mathbf1 _{\mathbb{N}} (n-1)/4 = 1\}
$$
where $\mathbf1 _{\mathbb{N}}$ is the indicator function of the set $\mathbb N$.
A: You can be as simple as this, given you don't insist on the $\{:\}$ set declaration:
$$4\mathbb{N}-3 = \{1,5,9,13,17,\dotsc\}$$
if $0\notin\mathbb{N}$ by your convention; and 
$$4\mathbb{N}+1 = \{1,5,9,13,17,\dotsc\}$$
if $0\in\mathbb{N}$ by your convention.
A: You can represent your set as the Range set of a Function from $\Bbb N$ to $\Bbb N$ itself defined as:
$$f(x)=4x+1$$
A: $\{n\in\mathbb N:\frac{n+3}4\in\mathbb N\}$
A: $$\{n \;  \mid \; \exists k \in \mathbb N: n =4k-3\}$$
