So i have this sequence
$\dfrac{\pi}{3}-k,\dfrac{\pi}{3},\dfrac{\pi}{3}+k,\dfrac{2\pi}{3}-k,\dfrac{2\pi}{3},\dfrac{2\pi}{3}+k,\dfrac{3\pi}{3}-k,\dfrac{3\pi}{3},\dfrac{3\pi}{3}+k,...$
I want to make a formula for this sequence, here is what i made myself for $n>=1$
$ceiling(\dfrac{n}{3})*\dfrac{\pi}{3}+ [(floor(n) \mod 3) - 1]*k$
But i think its too complicated (too long) for this easy sequence.
If it makes it easier, $n$ can start from $0$ or any other index as well. things after $2\pi$ repeat, since its used in polar coordinates and $k$ is constant.
$k = \arctan(\dfrac{\sin(\dfrac{\pi}{3})}{\cos(\dfrac{\pi}{3})+2})$
The first three sequence values are given by: $$\frac{\pi}{3}-k,\frac{\pi}{3},\frac{\pi}{3}+k$$ The second set is given by $$2\frac{\pi}{3}-k,2\frac{\pi}{3},2\frac{\pi}{3}+k$$ The third set is given by: $$3\frac{\pi}{3}-k,3\frac{\pi}{3},3\frac{\pi}{3}+k$$ Within each set of three sequence, the common difference is $k$. Also, for each triple, the value of $\pi$ is multiplied by an incremental integer. Therefore, as one solution, it is possible to write a general term of the sequence as $$a(m,n)=m\frac{\pi}{3}+(n-m-1)k$$ where $m=1,2,3,...$ and $n=m, m+1, m+2$.