While I was experimenting with cyclotomic polynomials , denoted as $\Phi_n(x)$ , I came up with the following formula for $\Phi_{3p}(x)$ , where $p$ is a prime number greater than $3$ :
$$\Phi_{3p}(x)=\displaystyle\sum_{i=0}^{p-1}\left(i+1-3\left\lfloor \frac{i+2}{3} \right\rfloor\right) \cdot \left(x^i+x^{2(p-1)-i}\right)-(-1)^{\left\lfloor \frac{1}{3}\left(p-6\left\lfloor \frac{p}{6} \right\rfloor\right) \right\rfloor} \cdot x^{p-1}$$
I have verified this formula for $p \in [5,4999]$ .
Question. Is there an easy way to justify this formula?
For prime $p>3$, $$\Phi_{3p}(x)=\frac{1+x^p+x^{2p}}{1+x+x^2}=\frac{1-x+x^p-x^{p+1}+x^{2p}-x^{2p+1}}{1-x^3}.$$If $p\equiv1\pmod 3$ then \begin{align} \Phi_{3p}(x)&=\frac{1-x^{2p+1}}{1-x^3} -\frac{x-x^p}{1-x^3}-\frac{x^{p+1}-x^{2p}}{1-x^3}\\ &=(1+x^3+\cdots+x^{2p-2})-(x+x^4+\cdots+x^{p-3})-(x^{p+1}+x^{p+4}+\cdots+x^{2p-3}) \end{align} and if $p\equiv2\pmod 3$ then \begin{align} \Phi_{3p}(x)&=\frac{1-x^{p+1}}{1-x^3} -\frac{x-x^{2p}}{1-x^3}+\frac{x^p-x^{2p+1}}{1-x^3}\\ &=(1+x^3+\cdots+x^{p-2})-(x+x^4+\cdots+x^{2p-3})+(x^p+x^{p+3}+\cdots+x^{2p-2}). \end{align}
Do these accord with your formula?