# 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 $$4n-3$$.

$\{n \in \mathbb N\; :\; n \equiv 1 \mod 4\}$

• How do you distinguish between (n≡1) mod 4 and n≡(1 mod 4) ? – Joshua Aug 31 '15 at 20:38
• If you're going to put in parentheses at all, it should be $n \equiv 1 \ (\text{mod}\; 4)$. See en.wikipedia.org/wiki/Modular_arithmetic#Congruence_relation – Robert Israel Aug 31 '15 at 21:01
• (1 mod 4) is valid. (7 mod 4) + 3 = 6. – Joshua Aug 31 '15 at 21:23
• @Joshua You're confusing modular congruence with the binary modulus operation. The usage above is designating modular congruence. – apnorton Sep 1 '15 at 0:19
• @MarioCarneiro "Correct" here is a very vague and difficult notion. Once you state clearly at the beginnin, that you work in $\mathbb{Z}/n\mathbb{Z}$ a lot, you can safely use $k\equiv 4\operatorname{mod} n$ and everybody knows what you mean. – yo' Sep 1 '15 at 8:17

$\{4k-3 \mid k \in \mathbb{N} \}$ or $\{4k+1 \mid k \in \mathbb{N} \}$, depending on whether you consider $0$ a natural number.

• I think of the options presented so far, this is the nicest in that: (i) it doesn't require an understanding of modular arithmetic; (ii) it avoids confusion with quantifiers; and (iii) it most clearly captures the sequence via the mapping $\{1\to 1, 2\to 5, 3\to 9,\dotsc\}$. – Joshua Taylor Sep 1 '15 at 2:29
• Thanks! This is how I write it in MAGMA, I feel it is very intuitive. – Morgan Rodgers Sep 1 '15 at 5:23

$$\{n \; \mid \; \exists k \in \mathbb N: n =4k-3\}$$

• No. Try to find such $k$ for $n=2$, please. Also, if I write $\forall k \in \mathbb N : n=4k-3$, and n is in this set, then we have n=4*1-3=4*2-3=4*3-3=4*4-3=.... which is clearly not true for any $n$. – wythagoras Aug 31 '15 at 18:07
• Secondly, if I write $\forall k \in \mathbb N : n=4k-3$, then I am searching for $n$ that statistify $n=4k-3$ for ALL natural $k$. So we have $n=4\cdot1-3=1$ because it must hold for $k=1$. We also have $n=4\cdot2-3=5$ because it must hold for $k=2$. So there can't be such n. – wythagoras Aug 31 '15 at 18:42
• @El'endiaStarman In that route, it'd be even easier to say $\{4n-3 \mid n \in \mathbb{N}\}$ – Joshua Taylor Sep 1 '15 at 2:27
• @JoshuaTaylor: That's the best. – El'endia Starman Sep 1 '15 at 2:29
• @El'endiaStarman I didn't see it before my comment, but Morgan Rodgers posted it as an answer, too. – Joshua Taylor Sep 1 '15 at 2:31

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...)

$\{n\in\mathbb N:\frac{n+3}4\in\mathbb N\}$

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 $\mathbb N$.

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.

You can represent your set as the Range set of a Function from $\Bbb N$ to $\Bbb N$ itself defined as:

$$f(x)=4x+1$$

• The codomain of a function $\Bbb N\to\Bbb N$ is always $\Bbb N$. The notion you mean is called image or (sometimes) range. Please don't use the term "codomain" which is specifically meant to be something different, namely the set $Y$ for a function that was specified as being $X\to Y$, regardless of the actual values the function takes. – Marc van Leeuwen Sep 2 '15 at 11:49
• Yeah sry for the ambiguity, now it should be clearer @MarcvanLeeuwen – Renato Faraone Sep 2 '15 at 11:51