\;\mbox{ (mod $n$)}\}, \nonumber
\\A_p'^{(k)}&=\{(j_1,\ldots,j_p)\in A_p\suchthat j_1+\cdots+j_k - j_{k+1}-\cdots -j_p=0 \;\mbox{ (mod $n$)}, j_1\neq \cdots \neq j_k \}, \nonumber
\\A_{p,s}^{(k)}&=\{(j_1,\ldots,j_p)\in A_p\suchthat j_1+\cdots+j_k - j_{k+1}-\cdots -j_p=sn\}. \nonumber\end{aligned}$$ In set $A_p$ and $\tilde{A}_p$, we collect $(j_1,\ldots,j_{p})$ according to their multiplicity.

Now we derive a convenient formula of trace for symmetric circulant matrices. First suppose $n$ is odd positive integers. We write $ n/2$ instead of $\lfloor n/2\rfloor $, as asymptotic is same as $n\to\infty$. Then $$\begin{aligned}
\Tr(SC_n^p)&=\sum_{\ell=0}^{n-1}\lambda_\ell^p
=\sum_{\ell=0}^{n-1}\left(X_0+2\sum_{j=1}^{n/2}X_j\cos(\omega_\ell j)\right)^p
\\&=\sum_{k=0}^{p}\binom{p}{k}X_0^{p-k}\sum_{\ell=0}^{n-1} \left(\sum_{j=1}^{n/2}X_j(e^{i\omega_\ell j}+e^{-i\omega_\ell j})\right)^{k},\end{aligned}$$ where $\omega_\ell=\frac{2\pi \ell}{n}$. Since $\sum_{\ell=0}^{n-1}e^{i\omega_\ell j}=0$ for $j\in \mathbb Z\backslash \{0\}$, we have $$\begin{aligned}
\label{trace formula_SC_odd}
\Tr(SC_n^p)=n\sum_{k=0}^{p}\binom{p}{\ell}X_0^{p-k}\sum_{A_{k}} X_{j_1}X_{j_2}\ldots X_{j_{k}},\end{aligned}$$ where $A_{k}$ for $k=1,\ldots,p$ is given by $$\begin{aligned}
A_{k}:=\l\{(j_1,\ldots,j_{k})\suchthat \sum_{i=1}^{k}\epsilon_i j_i=0\; \mbox{(mod n)}, \epsilon_i\in\{+1,-1\}, 1\le j_1,\ldots,j_{k}\le \frac{n}{2}\r\}\end{aligned}$$ and $A_0$ is an empty set with the understanding that the contribution from the sum corresponding to $A_0$ is 1. Note, in $A_{k}$, $(j_1,\ldots,j_{k})$ are collected according to their multiplicity.

Now suppose $n$ is even positive integers. We write $ n/2$ instead of $ n/2-1$, as asymptotic is same as $n\to\infty$. Then $$\begin{aligned}
\Tr(SC_n^p)=\sum_{\ell=0}^{n-1}\lambda_\ell^p
&=\sum_{\ell=0}^{n-1} \Big\{X_0 + (-1)^\ell X_{\frac{n}{2}}+2\sum_{j=1}^{n/2}X_j\cos(\omega_\ell j) \Big\}^p\\
&=\sum_{\ell=0}^{n-1} \sum_{k=0}^{p}\binom{p}{k} {(X_0 + (-1)^\ell X_{\frac{n}{2}})}^{p-k} \Big\{ \sum_{j=1}^{n/2}X_j(e^{i\omega_\ell j}+e^{-i\omega_\ell j}) \Big\}^{k} \\
&=\sum_{k=0}^{p}\binom{p}{k} \Big\{ {(X_0 +  X_{\frac{n}{2}})}^{p-k} \sum_{\ell=0, even}^{n-1} \Big[ \sum_{j=1}^{n/2}X_j (e^{i\omega_\ell 