Formula for a periodic sequence of 1s and -1s with period 5 I've been playing with periodic sequences of 1s and -1s lately. This is what I came up with:
\begin{eqnarray*}
-(-1)^n& = &1, -1, 1, -1,\ldots\quad\textrm{(Period 2)}\\
\left(-(-1)^n\right)^{\frac{n+2}{2}} & = &1, 1, 1, -1, 1, 1, 1, -1,\ldots\quad\textrm{(Period 4)}\\
 \left(\left(-(-1)^n\right)^{\frac{n+2}{2}}\right)^{\frac{n+4}{4}}& = &1, 1, 1, 1, 1, 1, 1, -1, 1, 1, 1, 1, 1, 1, 1, -1, \ldots\quad\textrm{(Period 8)}
\end{eqnarray*}
One can easily find a similar formula for a periodic sequence with period $ 2^n, n\in\mathbb{N} $.
I also found a formula:
$$
(-1)^{2 \sin\left(\frac{(2n-1)\pi}{6}\right)},
$$
which gives a sequence $ -1, 1, -1, -1, 1, -1,\ldots $ with period 3.
My question is: Is there a formula for a periodic sequence of 1s and -1s with period 5? If there is, what is it?
I know about this formula for a periodic zero and one sequence with period $ N $:
$$
\sum\limits_{k = 1}^N\cos\left(-2\pi\frac{n(k-1)}{N}\right)/N = 0, 0, 0, \ldots, 1.
$$
However, it requires to sum up $ N $ expressions to count the $ n\textrm{th} $ term, which is why I don't like it. I would also like the formula not to contain functions like floor or modulus.
 A: We have
$$
\frac{4\cos\left(\frac{2\pi k}5\right)+4\cos\left(\frac{4\pi k}5\right)-3}5=
\left\{\begin{array}{}
-1&\text{if }k\not\equiv0\pmod5\\
+1&\text{if }k\equiv0\pmod5\\
\end{array}\right.\tag{1}
$$

Explanation
The roots of $z^5-1$ are $z=e^{2\pi ik/5}$ for $k\in\{0,1,2,3,4\}$. Vieta says that the coefficient of $z^4$ in $z^5-1$ the sum of the roots of $z^5-1$. That is, the sum of the roots is $0$. Taking the real part of the roots yields
$$
1+\cos\left(\frac{2\pi}5\right)+\cos\left(\frac{4\pi}5\right)+\cos\left(\frac{6\pi}5\right)+\cos\left(\frac{8\pi}5\right)=0\tag{2}
$$
When $k\not\equiv0\pmod5$, $k$ is invertible $\bmod5$. Therefore,
$$
\left(\cos\left(\frac{2\pi k}5\right),\cos\left(\frac{4\pi k}5\right),\cos\left(\frac{6\pi k}5\right),\cos\left(\frac{8\pi k}5\right)\right)\tag{3}
$$
is a permutation of
$$
\left(\cos\left(\frac{2\pi}5\right),\cos\left(\frac{4\pi}5\right),\cos\left(\frac{6\pi}5\right),\cos\left(\frac{8\pi}5\right)\right)\tag{4}
$$
Therefore,
$$
1+\cos\left(\frac{2\pi k}5\right)+\cos\left(\frac{4\pi k}5\right)+\cos\left(\frac{6\pi k}5\right)+\cos\left(\frac{8\pi k}5\right)=0\tag{5}
$$
when $k\not\equiv0\pmod5$. When $k\equiv0\pmod5$,
$$
1+\cos\left(\frac{2\pi k}5\right)+\cos\left(\frac{4\pi k}5\right)+\cos\left(\frac{6\pi k}5\right)+\cos\left(\frac{8\pi k}5\right)=5\tag{6}
$$
Since $\cos(x)$ is an even function with period $2\pi$, $(5)$ and $(6)$ become
$$
1+2\cos\left(\frac{2\pi k}5\right)+2\cos\left(\frac{4\pi k}5\right)
=\left\{\begin{array}{}
0&\text{if }k\not\equiv0\pmod5\\
5&\text{if }k\equiv0\pmod5\\
\end{array}\right.\tag{7}
$$
Equation $(1)$ is simply a scaled and translated version of $(7)$.
A: Surprisingly, you can use Fibonacci numbers and its generalizations for periods $3,4,5,\dots$
Period 3. Fibonacci numbers $F_n= 1, 1,  \color{blue}2, 3, 5,  \color{blue}8, 13, 21,  \color{blue}{34},\dots$
$$A_n = -(-1)^{F_n}=1, 1, -1, 1, 1, -1,\dots$$
Period 4. Tribonacci numbers $T_n = 1, 1,  \color{blue}{2, 4}, 7, 13,  \color{blue}{24, 44}, 81, 149,  \color{blue}{274, 504},\dots$
$$B_n = -(-1)^{T_n}=1, 1, -1, -1, 1, 1, -1, -1\dots$$
Period 5. Tetranacci numbers $W_n = 1, 1, 1, 1, \color{blue}4, 7, 13, 25, 49,  \color{blue}{94}, 181, 349, 673, 1297, \color{blue}{2500},\dots$
$$C_n = -(-1)^{W_n}=1, 1, 1, 1, -1, 1, 1, 1, 1, -1,\dots$$
and so on.
A: This is close:
${|\cos{\frac{\pi*x}{5}-\frac{\pi}{2}}|}^{0.0001}$
