j}+e^{-i\omega_\ell j}) \Big] ^{k} \\
& + {(X_0 -  X_{\frac{n}{2}} )}^{p-k} \sum_{\ell=0, odd}^{n-1}  
\Big[ \sum_{j=1}^{n/2}X_j (e^{i\omega_\ell j}+e^{-i\omega_\ell j}) \Big] ^{k} \Big\},\end{aligned}$$ where $\omega_\ell=\frac{2\pi \ell}{n}$. Since we know $$\begin{aligned}
    \sum_{\ell=0, even}^{n-1}   e^{i\omega_\ell j}
&=  \left\{\begin{array}{ccc}    
         \frac{n}{2} & \text{if}& j =0\; \mbox{(mod } \frac{n}{2}) \\\\
            0 & \text{if}&  j \neq0\; \mbox{(mod } \frac{n}{2}),     
              \end{array}\right.          \end{aligned}$$ and $$\begin{aligned}
    \sum_{\ell=0, odd}^{n-1}   e^{i\omega_\ell j}
&=  \left\{\begin{array}{ccc}    
         \frac{n}{2} & \text{if}& j =0\; \mbox{(mod } \frac{n}{2}) \mbox{ and }   j = 0\; \mbox{(mod n)} \\\\
         -\frac{n}{2} & \text{if}& j =0\; \mbox{(mod } \frac{n}{2}) \mbox{ and }   j \neq0\; \mbox{(mod n)} \\\\
            0 & \text{if}&  j \neq0\; \mbox{(mod } \frac{n}{2}) ,    
              \end{array}\right.          \end{aligned}$$ Therefore from the above last two observations, $\Tr(SC_n^p)$ will be $$\begin{aligned}
 \label{trace formula_SC_even}
 \Tr(SC_n^p) &=\frac{n}{2} \sum_{k=0}^{p}\binom{p}{k} \Big[ \Big\{ {(X_0 +  X_{\frac{n}{2}})}^{p-k} + {(X_0 -  X_{\frac{n}{2}})}^{p-k} \Big\} \sum_{A_k} X_{j_1}X_{j_2}\ldots X_{j_{k}} \nonumber\\
& \qquad + \Big\{ {(X_0 +  X_{\frac{n}{2}})}^{p-k} - {(X_0 -  X_{\frac{n}{2}})}^{p-k} \Big\} \sum_{ \tilde{A}_k} X_{j_1}X_{j_2}\ldots X_{j_{k}} \Big] \nonumber \\
%& = \frac{n}{2} \sum_{k=0}^{p}\binom{p}{k} \Big[ Y_k  \sum_{A_k} X_{j_1}X_{j_2}\ldots X_{j_{k}} + \tilde{Y}_k \sum_{ \tilde{A}_k} X_{j_1}X_{j_2}\ldots X_{j_{k}} \Big] 
& = \frac{n}{2} \sum_{k=0}^{p}\binom{p}{k} \Big[ Y_k  \sum_{A_k} X_{J_{k}} + \tilde{Y}_k \sum_{ \tilde{A}_k}  X_{J_{k}} \Big] , \mbox{ say}\end{aligned}$$ where for each $k=0, 1, 2,\ldots, p$, $A_{k}$ is same as $A_{k}$ of 