A simple mathematical expression for the periodic sequence $(1, -2, -1, 2, 1, -2, -1, 2,\dots)$ The sequence is just $S_n=1, -2, -1, 2$ repeated indefinitely. The best I can do is:
$$S_n= \frac{(i - 2)\,i^n}{2i}  + \frac{(i+2)\,(-i)^n}{2i}  $$
where $i$ is the imaginary unit. In fact, this expression is where the sequence comes from. Can it be simplified, like using $(-1)^n$ for an alternating sequence of $1$ and $-1$? These values appear as part of a larger formula. Currently, I am simply using a more computational expression: V(n) = the (n modulo 4)'th element of [+1,-2,-1,+2], but this is rather hacky.
 A: Per the OP's request, given the sequence,
$$\alpha_n=1,-2,-1,2,1,-2,-1,1,\dots$$
and $\alpha_0=1$, we are to generate it using $(-1)^n$ and well-defined sequences of integers. The unsigned version is easy to do,
$$\beta_n = \tfrac{3-(-1)^n}2=1,2,1,2,1,2,1,2,\dots$$
So the problem really is to generate the signs,
$$U_n = 1,-1,-1,1,1,-1,-1,1,\dots$$
and fortunately answered by this post and this,
$$U_n =(-1)^{n(n+1)/2} = \sqrt{2}\cos\tfrac{(2n+1)\pi}4$$

$\color{blue}{Update:}$ We also have the rather exotic,
  $$U_n = (-1)^{T_n}$$
  with tribonacci numbers $T_n$,
  $$T_n=\sum_{k=0}^{n-1}\sum_{j=0}^{n-k-1}\tbinom{n-k-1}{j}\tbinom{j}{k-j}=\color{blue}0,1, 1, \color{blue}{2, 4}, 7, 13, \color{blue}{24, 44}, 81, 149, \color{blue}{274, 504},\dots$$
  and $T_0=0$.

Alternatively, we seek,
$$V_n = \tfrac{2n+1-(-1)^n}4= 0,1,1,2,2,3,3,4,\dots$$
which this post has considered. It then gives us three ways to form $\alpha_n$ as,

$$\alpha_n = \beta_n \,U_n=\tfrac{3-(-1)^n}2\,(-1)^{n(n+1)/2}\,=\,\tfrac{3-(-1)^n}2\,(-1)^{T_n}\tag1$$

and

$$\alpha_n = \beta_n \,(-1)^{V_n} = \tfrac{3-(-1)^n}2\,(-1)^{\frac14\big(2n+1-(-1)^n\big)}\tag2$$

P.S. For a signed sequence with period $5$, see here.
A: This can be done easily with Discrete Fourier Transform. Observe that
$$ \operatorname{DFT}\begin{bmatrix}1\\-2\\-1\\2\end{bmatrix} = \begin{bmatrix} 0 \\ 1-2i \\ 0 \\ 1 + 2i \end{bmatrix}
$$
Using the convention
$$
\operatorname{DFT}(\mathbf{v})(\xi) = \frac{1}{\sqrt{n}} \sum_{x=0}^{n-1} \mathit{v}_x e^{2\pi i \xi x/n}
$$
Hence the inverse DFT gives a periodic interpolation for your sequence:
$$ f(x) = \frac{1 - 2i}{2} e^{-2\pi i x/4} + \frac{1 + 2i}{2} e^{-6\pi i x/4}
$$

A: Your expression can be slightly simplified if $n=2k$ we have,
$$\frac{i^{2k}}{2i}[(i-2)+(2+i)]=i^{2k}=(-1)^k.$$
If $n=2k+1$ we have, $$\frac{1}{2i}[(i-2)i^{2k+1}-(2+i)i^{2k+1}]=2(-1)^{k+1}.$$  So 
$$a_n= \begin{cases} 
      (-1)^k & n=2k \\
    \\
      2(-1)^{k+1} & n=2k+1 
   \end{cases}
,$$ where k is an integer.
A: You cannot write it using a function depending only of $(-1)^n$, because such a function takes only two values, and you need four. On the other hand a function of $i^n$ takes four values, so what you did is OK.
A: You need a period of $4$, so powers of $-1$ won't help.  As you select expressions you need to think about what the objective is.  If you just want to compute the value, your $V(n)$ is a fine expression.  It returns the value you want.  What is wrong with that?  Any other formula, like the one early in your post, is only valuable if it does something more.  Maybe it simplifies with stuff later in your application, or maybe it impresses some reader.  You could do powers of $-1$ and $-1^{n/2}$ where that is integer division, but that will be logically equivalent to what you found already.
A: In Python you can do this with ($n$ starting at $1$):
(2-n%2)*(-1)**(n//2)

Which doesn't require the math module.
A: Expanding on my comment:
$$\cos(\pi n/2) -2\sin(\pi n/2)$$
A: You can simply interpolate$(1,-2,-1,2)$, and then use $x=n\bmod4$:
$$S_n=-\frac{1}{3}(n\bmod4)^3+4(n\bmod4)^2-\frac{38}{3}(n\bmod4)+10$$
