Based on this question, using the same $z_0$: $$z_0 = e^{2\pi i / 8}$$ if we modify the sequence from previous question to look like this ($*$ denotes discrete convolution): $$\left(z_0^{[-2k,3k]} * z_0^{[2k,-3k]}\right)$$ we will get for $k \in \{0,1,2,3\}$: $$k=0 \rightarrow \left[\begin{array}{ccc}1&2&1\end{array}\right], k=1 \rightarrow \left[\begin{array}{ccc}1&-1&1\end{array}\right]$$ $$k=2 \rightarrow \left[\begin{array}{ccc}1&0&1\end{array}\right], k=3 \rightarrow \left[\begin{array}{ccc}1&1&1\end{array}\right]$$ These four are all integer combinations for the middle pixel but the end pixels stay 1 and has no imaginary part at all. This one I came up with by trial and error.
So to my question:
How to systematically make filter factorizations which achieve this? I.e. that some points stay the same (the end points in this case are both 1 irregardless of $k$), but other points cycle through different values?