Getting the sequence $\{1, 0, -1, 0, 1, 0, -1, 0, \ldots\}$ without trig?

What's the simplest way to write a function that outputs the sequence:

{1, 0, -1, 0, 1, 0, -1, 0, ...}

... without using any trig functions?

I was able to come up with a very complex sequence involving -1 to some complicated formula, but I was hoping there is a more simple solution.

$n$ should start at 0 and go to infinity.

Update:

All the solutions you guys provided are great! I wasn't aware there were so many of them. I should have mentioned that I prefer a solution which doesn't use recursion; imaginary numbers; matrices; functions with if statements; or functions such as ceil, floor, or mod. I'm looking for something using basic algebra: addition/subtraction, multiplication/division, exponents, etc. However, I will accept anything since I didn't include this clause originally.

This is what I came up with:

$$a_n=\frac{\left(-1\right)^n+1}{2}\cdot \left(-1\right)^{\left(\frac{n}{2}-\frac{\left(-1\right)^{n+1}+1}{4}\right)}$$

Is there a less complicated way (i.e. fewer terms) to get this same sequence?

Solution 1:

Let's try too : $|n \bmod 4-2|-1$

Solution 2:

How about $\dfrac{i^n + (-i)^n}{2}$? (Of course, that is arguably just trigonometry in disguise).

Or as a recurrence: $a_n = -a_{n-2}$ with $(a_0,a_1)=(1,0)$.

Or $\begin{bmatrix}1 & 0\end{bmatrix}\begin{bmatrix}0&-1\\1&0\end{bmatrix}^n\begin{bmatrix}1\\0\end{bmatrix}$? (Which can be viewed as a better-disguised version of either of the two previous suggestions).

Solution 3:

Whether this is simplest will depend on exactly what you mean, but the following is a pretty simple description. It's certainly simpler than anything involving trig functions.

$$a_n=\begin{cases} 0 & \text{if n is odd} \\ 1 & \text{if n is divisible by 4} \\ -1 & \text{otherwise} \end{cases}$$