Is there a simple function that generates the series; 1,1,2,1,1,2,1,1,2… or -1,-1,1,-1,-1,1…

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

How about $$a_n=\frac 23\cos\left(\frac{2\pi n}3\right)+\frac 43\ $$

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 $\omega^n + \omega^{2n} = -1$.

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

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.

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 $1, -1, -1, 1, -1, -1, \cdots$.

$$-1+2\left \lfloor {\frac n 3} \right \rfloor -2\left\lfloor \frac {n-1}3 \right\rfloor $$ , where $n$ starts from $1$.

$a(0) = a(1) = -1$
$a(n) = a(n-1) * a(n-2)$

How about $\operatorname{Re}\left(e^{\tfrac{2}{3}\pi i n}\right)$?

How about:

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

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

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

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

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: $\{1,1,2,1,1,2,...\}$