1, 5, 9, 13, 17, 21,...
Emily Wong
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 $4n-3$.
$\endgroup$ 08 Answers
$\begingroup$$\{n \in \mathbb N\; :\; n \equiv 1 \mod 4\}$
$\endgroup$ 7 $\begingroup$$\{4k-3 \mid k \in \mathbb{N} \}$ or $\{4k+1 \mid k \in \mathbb{N} \}$, depending on whether you consider $0$ a natural number.
$\endgroup$ 1 $\begingroup$$$\{n \; \mid \; \exists k \in \mathbb N: n =4k-3\}$$
$\endgroup$ 14 $\begingroup$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...)
$\endgroup$ $\begingroup$$\{n\in\mathbb N:\frac{n+3}4\in\mathbb N\}$
$\endgroup$ $\begingroup$It's not exactly the form you're looking for, but close: $$ \{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\} $$ where $\mathbf1 _{\mathbb{N}}$ is the indicator function of the set $\mathbb N$.
$\endgroup$ $\begingroup$You can be as simple as this, given you don't insist on the $\{:\}$ set declaration:
$$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\}$$
if $0\in\mathbb{N}$ by your convention.
$\endgroup$ 0 $\begingroup$You can represent your set as the Range set of a Function from $\Bbb N$ to $\Bbb N$ itself defined as:
$$f(x)=4x+1$$
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