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In my work I came across that sequence
$-1,-1,-1,-1,1,1,1,1,\dots$ and repeating this 8-block so on forever
Now I cant find an ( e.g. trigonometric/complex ) expression $f(n)$ ( e.g. $f(n) =(-1)^g(n)$ ) which gives me the sequence starting with $n=2,3,4,5,6,7,8,9,…$ and so on forever.
Consider the function $f(n)$ such that \begin{equation} f(n) = \left \{ \begin{array}[ll] (-1 & \textrm{ if } n \textrm{ mod } 8 \equiv 0,1,2,3 \\ 1 & \textrm{ if } n \textrm{ mod } 8 \equiv 4,5,6,7 \end{array} \right. \end{equation}
I will assume that you really are starting at $2$. If the start is different, modification is easy.
Let $g(n)=\left\lfloor\frac{n+2}{4}\right\rfloor$. Here $\lfloor x\rfloor$ is the "floor" function, the greatest integer $\le x$.
Then your sequence is given by $(-1)^{g(n)}$.
By symmetry, we can assume $$ f(n) = a_1 \sin\frac{\pi n}{4} +a_3 \sin\frac{3\pi n}{4} +b_1 \cos\frac{\pi n}{4} +b_3 \cos\frac{3\pi n}{4}. $$ Then $$ \begin{aligned} f(1) &= \frac{a_1}{\sqrt{2}} +\frac{a_3}{\sqrt{2}} + \frac{b_1}{\sqrt{2}} -\frac{b_3}{\sqrt{2}} = -1\\ f(2) &= a_1 - a_3 = -1\\ f(3) &= \frac{a_1}{\sqrt{2}} +\frac{a_3}{\sqrt{2}} - \frac{b_1}{\sqrt{2}} +\frac{b_3}{\sqrt{2}} = -1 \\ f(4) &= -b_1 - b_3 = -1. \end{aligned} $$ So $$ \begin{aligned} a_1 &= -\frac{1}{2} -\frac{1}{\sqrt 2}, \\ a_3 &= +\frac{1}{2} -\frac{1}{\sqrt 2}, \\ b_1 &= b_3 = \frac{1}{2}. \end{aligned} $$ So $$\begin{aligned} f(n) &= \left( -\frac{1}{2} -\frac{1}{\sqrt 2} \right) \sin\frac{\pi n}{4} + \left( \frac{1}{2} -\frac{1}{\sqrt 2} \right) \sin\frac{3 \pi n}{4} + \frac{1}{2} \left( \cos\frac{\pi n}{4} + \cos\frac{3 \pi n}{4} \right) \\ &= \sin\frac{\pi n}{4} \cos\frac{\pi n}{2} -\sqrt 2 \cos\frac{\pi n}{4} \sin\frac{\pi n}{2} + \cos\frac{\pi n}{4} \cos\frac{\pi n}{2} \\ &= \sqrt 2 \left[ \sin\frac{\pi (n+1)}{4} \cos\frac{\pi n}{2} - \cos\frac{\pi n}{4} \sin\frac{\pi n}{2} \right]. \end{aligned} $$
Let $\zeta = \mathrm{e}^{\mathrm{i} \pi/4}$ and $$f(z) = \frac{1}{4} z \left(\left(\sqrt{2}+(1-\mathrm{i})\right) z^6 +\left(-\sqrt{2}+(1+\mathrm{i})\right) z^4 \\ -\left(\sqrt{2}+(-1+\mathrm{i})\right) z^2 +\sqrt{2}+(1+\mathrm{i})\right).$$ Then $f(\zeta^n)$ is your sequence. (It even starts at $n=2$.)
Brought to you by the joys of Lagrange interpolation (on the eighth roots of $1$).
(I haven't worked through the details, but if we turn this into a trigonometric expression, it looks as if we should get something close to @user293511's answer.)
Modified to start at $n=2$:$$f(n)=(-1)^{\frac{(n+2)!}{(n-2)!4!}}$$ This will give you your desired sequence.
Here is a plot of this function.
Simply observe the 3rd bit
\begin{equation} f(n) = \left \{ \begin{array}[ll] (-1 & \textrm{ if } n \land 2^3 \equiv 0 \\ 1 & \textrm{ if } n \land 2^3 \equiv 1 \end{array} \right. \end{equation}
Or (if my mathematical symbols are correct)
\begin{equation} f(n) = 2 (n \land 2^3) - 1 \end{equation}
I'm trying to express the bit-wise logical and of n with the binary number 01000.
By adding a third harmonic to a sine wave I came up with $\sin{x}+(\sqrt2-1)\sin3x$ which has equal value at $x = \frac\pi8, 3\frac\pi8, 5\frac\pi8, 7\frac\pi8, 17\frac\pi8, \dots$ and the negative value at the corresponding missing odd coordinates. The formula does get more complicated when you normalise the values to be $\pm1$ for $n = 2\dots8$ though:
$$f(n) =\frac{\sin(\frac{2x+5}8\pi)+(\sqrt2-1)\sin(\frac{6x+15}8\pi)}{\sin\frac\pi{8}+(\sqrt2-1)\sin\frac{3\pi}8}$$
