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I'm thinking about this question in the sense that we often have a term $(-1)^n$ for an integer $n$, so that we get a sequence $1,-1,1,-1...$ but I'm trying to find an expression that only gives every 3rd term as positive, thus it would read; $-1,-1,1,-1,-1,1,-1,-1...$

Alternatively a sequence yielding $1,1,2,1,1,2,1,1,2...$ could also work, as $n$ could just be substituted by it in $(-1)^n$

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    $\begingroup$ Check this out. $\endgroup$
    – Eff
    Aug 21, 2016 at 22:10
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    $\begingroup$ For $-1,-1,1,-1,-1,1,\ldots$ you can use $(-1)^{n^2\bmod 3}$. $\endgroup$
    – user236182
    Aug 21, 2016 at 22:11
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    $\begingroup$ Well, f(n) = 1 if $3|n$ and f(n) = -1 if $3 \not \mid n$ is a perfectly legitimate function. For something less "if then" $(-1)^{\lceil (n/3 - \lfloor n/3\rfloor)\rceil}$ but that just point the "if then" burden where no-one sees it. Still there are better ones. $\endgroup$
    – fleablood
    Aug 21, 2016 at 23:13
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    $\begingroup$ Why do you need such a function? If for fun, then the answers are good. If for short notation (for whatever reason), then some are not very convenient. If for easy readability, then most answers are horrible. $\endgroup$
    – JiK
    Aug 22, 2016 at 12:00
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    $\begingroup$ @GEdgar Yes, I find Creating an alternating sequence of positive and negative numbers. $\endgroup$ Aug 22, 2016 at 18:22

18 Answers 18

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Let $F_n$ be the Fibonacci sequence $0,1,1,2,3,5,8,13,21,34,...$ then a possible sequence is $$(-1)^{F_{n+1}}$$

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    $\begingroup$ nice observation! $\endgroup$
    – BigbearZzz
    Aug 21, 2016 at 22:35
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    $\begingroup$ I like the specific use of Fibonacci. I think a recursive function might even give the OP a direct route, say starting with $1,-1,...$ and then using product of previous two entries. $\endgroup$
    – Nij
    Aug 22, 2016 at 0:26
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    $\begingroup$ $(-1)^\frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}}$ $\endgroup$
    – EMBLEM
    Aug 22, 2016 at 19:12
  • $\begingroup$ I think it should be $F_n$ not $F_{n+1}$. $\endgroup$ Aug 23, 2016 at 8:09
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    $\begingroup$ @Nij This is essentially just writing the sequence $(-1)^{F_{n}}$ using the recursive definition of Fibonacci numbers. $\endgroup$ Aug 23, 2016 at 18:31
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How about $$a_n=\frac 23\cos\left(\frac{2\pi n}3\right)+\frac 43\ $$

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    $\begingroup$ Was wondering for a few minutes now why I did not see how this gets to -1 and why it returns 2 until I re-read the question, in particular the second paragraph. I thought my math was failing me completely... $\endgroup$
    – luk2302
    Aug 22, 2016 at 20:57
  • $\begingroup$ @luk2302 That's actually interesting because the first time I read the question I completely missed the first paragraph :P $\endgroup$
    – BigbearZzz
    Aug 22, 2016 at 21:00
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If you want something purely in terms of elementary operations, you could use the closed form:

$$\frac{2}{3}\left(-\frac{1}{2} + \omega^n + \omega^{2n}\right)$$

Where $\omega=\frac{-1+\sqrt{3}i}{2}$ is a complex cube root of unity. When $3|n$ we get $\omega^n =\omega^{2n} = 1$. On the other hand, when $n$ is not divisible by $3$, $\omega^n$ and $\omega^{2n}$ will be the two roots of the polynomial:

$$z^2 + z + 1$$

And from Vieta we get $\omega^n + \omega^{2n} = -1$.

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    $\begingroup$ This is not obvious from the form of the expression, but this is the same function (extended to complexes) as @BigbearZzz's. $\endgroup$ Aug 23, 2016 at 17:43
  • $\begingroup$ @MarioCarneiro Indeed it is, nice observation. $\endgroup$
    – sbares
    Aug 23, 2016 at 20:11
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$$-1+2\left \lfloor {\frac n 3} \right \rfloor -2\left\lfloor \frac {n-1}3 \right\rfloor $$ , where $n$ starts from $1$.

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  • $\begingroup$ Computationally the most efficient function here I think. Nice one. $\endgroup$
    – tomsmeding
    Aug 22, 2016 at 5:52
  • $\begingroup$ @tomsmeding Thanks! $\endgroup$
    – Kay K.
    Aug 22, 2016 at 6:00
  • $\begingroup$ @tomsmeding I don't think it's computionally efficient, because you need 2 divisions, whereas many of the below solutions need only one modulo operation $\endgroup$
    – phuclv
    Aug 23, 2016 at 10:34
  • $\begingroup$ @LưuVĩnhPhúc Good point, I mentally skipped the ones using modulo because I thought them "cheating", but of course they're easier for a computer. $\endgroup$
    – tomsmeding
    Aug 23, 2016 at 16:23
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    $\begingroup$ computationally efficient is {1,1,-1}[n % 3] $\endgroup$ Aug 23, 2016 at 21:19
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$a(0) = a(1) = -1$
$a(n) = a(n-1) \times a(n-2)$

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You can check out OEIS which gives you several ways to generate your sequence.

For example:

$$a_n=2-((n+1)^2 \pmod 3)$$

$$a_n=\frac{4}{3} - \frac{\cos(2\pi n/3)}{3} - \frac{\sin(2\pi n/3)}{\sqrt3}$$

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You could use a simple trig function whose period is $3$. We can therefore try a function of the form $$f(x)=a\cos\left(\frac{2\pi x}{3}\right)+b$$

Substituting the coordinates $(0,2),(1,1)$ gives the values of $a,b$ and we end up with $$f(x)=\frac 23\cos\left(\frac{2\pi x}{3}\right)+\frac 43$$ and this generates the required sequence.

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No operators!

$f(x) = \begin{cases} -1 & \text{if}~x=0\\ -1 & \text{if}~x=1\\ 1 & \text{if}~x=2\\ f(x-3) & \text{otherwise}\\ \end{cases}$

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    $\begingroup$ Could also eliminate the last case and make the other two $x = 3k$, $x = 3k + 1$, etc. $\endgroup$ Aug 23, 2016 at 6:41
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    $\begingroup$ Fine except for f(-1) calling an infinite loop. :) $\endgroup$
    – Wildcard
    Aug 24, 2016 at 1:34
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    $\begingroup$ @Wildcard there is no such thing as an infinite loop. OP asked for a sequence and sequences have a beginning (typically 0 or 1). Several of the other answers have the same 'issue'. $\endgroup$
    – user63495
    Aug 24, 2016 at 5:08
  • $\begingroup$ @Wildcard: from $f(2-3)=f(2)$ you deduce that $f(-1)=1$. If you really want to programme it, you have to stop briefly and think. $\endgroup$
    – PJTraill
    Aug 24, 2016 at 22:08
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Let $\omega \neq 1$ be a cubic root of unity. We have that $1 + \omega + \omega^2 = 0$, i.e. $\omega + \omega^2 = -1$. Also, $w^{-1} = \omega^2$ and $\omega^3 = 1$.

Put $a_n = \omega^n + \omega^{-n}$. We get the sequences

\begin{eqnarray*} a_0 &=& \omega^{0} + \omega^{0} = 2 \\ a_1 &=& \omega^{1} + \omega^{-1} = \omega^{1} + \omega^{2} = -1 \\ a_2 &=& \omega^{2} + \omega^{-2} = \omega^{2} + \omega^{1} = -1 \\ \end{eqnarray*}

and then the sequence repeats.

If you put $b_n = \frac{2 a_n -1}{3}$, then you get $1, -1, -1, 1, -1, -1, \cdots$.

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How about:

$$(n\bmod3)^2-3(n\bmod3)+1$$

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Iverson bracket notation makes it easy to say exactly what you mean here:

$a_n = 2\cdot[\text{$n$ is a multiple of $3$}] - 1$

or, shorter but more (probably excessively) confusing:

$a_n = 2[3\mid n] - 1$

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$$(-1)^{(n\,\text{mod}\, 3)!}$$

Edit. a simpler solution, without a factorial and starting with $n=2$:

$$(-1)^{(n\,\text{mod}\, 3)+1}$$

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  • $\begingroup$ Why exponentiate when $(n\pmod 3)!$ gives an equivalent solution? $\endgroup$
    – Nij
    Aug 22, 2016 at 4:23
  • $\begingroup$ @Nij that's for the sequence 1,1,2,1,1. But I realized that you don't even need the factorial. $\endgroup$
    – Pulsar
    Aug 22, 2016 at 4:31
  • $\begingroup$ The factorial does matter if used as exponent, but not when used directly, to be clear. 0, 1, 2 gives 1, -1, 1. But 0!, 1!, 2! gives 1, 1, 4. Used as exponents they then return 1, -1, 1 versus -1, -1, 1. $\endgroup$
    – Nij
    Aug 22, 2016 at 5:57
  • $\begingroup$ @Nij $2!=2$, not $4$. $\endgroup$
    – Pulsar
    Aug 22, 2016 at 11:44
  • $\begingroup$ I knew I was making some mistake there. Fortunately it doesn't change the result! $\endgroup$
    – Nij
    Aug 22, 2016 at 18:05
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Yet another one:

$$1 + \left\lfloor \frac{n \bmod 3}{2} \right\rfloor$$

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  • $\begingroup$ You might consider specifying the meaning by "mod" in this context. In mathematics, "mod" is almost never a function of two arguments like it is in programming and in this answer; instead, it's an equivalence relation. A common alternative is to make a remainder function, $\text{rem}(n,3)$ which returns remainder of $n$ on division by $3$. $\endgroup$
    – 6005
    Aug 23, 2016 at 2:59
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    $\begingroup$ @6005 This is a genuine usage of the \bmod modulo function, which does appear in pure mathematics occasionally. This is not to be confused with the \pmod modulo qualifier on equivalence, which is the standard way to talk about modular equivalence. There's no need to make a separate notation for this usage. FYI, the binary modulo function is natural once you have floor: $a\bmod b=a-b\lfloor a/b\rfloor$. $\endgroup$ Aug 23, 2016 at 17:48
  • $\begingroup$ @MarioCarneiro Well, it's fine if you disagree. I don't see how the unmotivated formula "$a - b \lfloor a / b \rfloor$" is "natural". I have taught discrete math classes and you really have to stress, pedagogically at least, that mod is not a binary function. People write things like $a \text{ mod } b = c$ instead of $a \equiv c \pmod{b}$, which is wrong unless $c$ is actually between $0$ and $b-1$. So it does throw a lot of people off, and in my opinion it's bad notation. I'm sure it does appear in some math books, but not in any undergraduate or graduate course I took. $\endgroup$
    – 6005
    Aug 23, 2016 at 18:32
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    $\begingroup$ Binary modulo is the most common use. Half the answers on this page make use of it without hesitation or definition. Outside of a discrete mathematics classroom, $a \bmod b = c$ is meaningful and correct. $\endgroup$ Aug 24, 2016 at 2:00
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    $\begingroup$ @6005 I only mean that binary mod is within the set of functions generated by the field operations plus $\lfloor\cdot\rfloor$, so in a sense having one implies having the other as well. I don't see the expression $a\bmod b=c$ very often, but when I do I would understand it just as you have described - $c$ is the unique number equivalent to $a$ in the range $[0,b)$. Obviously this is quite different from $a\equiv c\pmod b$, but they both have their uses. $\endgroup$ Aug 24, 2016 at 3:58
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This works for positive and negative integers of $x$ and produces a sequence of $1,1,2,1,1,2,...$

$y = 2 - (x^2 \text{ mod } 3)$

Or for the sequence $-1,-1,1,-1,-1,1,...$

$y = 1 - 2(x^2 \text{ mod } 3)$

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Here is my proposal, for $n \ge 1$:

$$1+\left\lfloor \frac{1}{n^{(n \pmod {3})}} \right\rfloor - \left\lfloor \frac{1}{n}\right\rfloor$$

It will provide the sequence: $\{1,1,2,1,1,2,...\}$

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  • $\begingroup$ @CiaPan thanks for the change, I did not know that formatting option! $\endgroup$
    – iadvd
    Aug 22, 2016 at 6:50
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Lemma:fibonacci numbers generates:

$odd-odd-even-odd-odd-even-odd-odd-\dots$

proof:

If

$F_i=odd$

$F_{i+1}=odd$

Then we have:

$F_{i+2}=even$

$F_{i+3}=odd$

$F_{i+4}=odd$

then the system reapeates.

It gives the formula:

$(-1)^{F_{n}}$

This gives the closed form:

$$(-1)^{(\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^{n+1}-(\frac{1-\sqrt{5}}{2})^{n+1})}$$

Which is really ugly.I will introduce Lucas numbers to make this better.

Lucas numbers look like fibonacci numbers but $F_1=1$ and $F_2=3$.

Which gives the same odd and even resault.Then we can write:

$(-1)^{L_n}$

That gives the closed form:

$$(-1)^{(\frac{1+\sqrt{5}}{2})^n+(\frac{1-\sqrt{5}}{2})^n}$$

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Another one:

$$2\times((n+1\bmod3) \bmod 2)-1$$

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my 2 cents, how about

$a_{n+1}=(-1)^{r(a_n)(1-a_n)}$ where $a_0 = 1$ and $r \in \{1,2,3\}$ which is similar to the logistic map in general I think that using a one dimensional chaotic polynomial could help.

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