$n$ odd case and $\tilde{A}_{k}$ for $k=1,\ldots,p$ is given by $$\begin{aligned}
\tilde{A}_{k}:=\l\{(j_1,\ldots,j_{k})\suchthat \sum_{i=1}^{k}\epsilon_i j_i=0\; \mbox{(mod } \frac{n}{2}) \mbox{ and } \sum_{i=1}^{k}\epsilon_i j_i \neq 0\; \mbox{(mod }n),  \epsilon_i\in\{+1,-1\}, 1\le j_1,\ldots,j_{k}\le \frac{n}{2}\r\}.\end{aligned}$$ Here note that $\tilde{A}_0$ is an empty set with the understanding that the contribution from the sum corresponding to $\tilde{A}_0$ is $1$ and in $\tilde{A}_k$, $(j_1,\ldots,j_{k})$ are collected according to their multiplicity. Also $$\begin{aligned}
 \label{eqn:Y_k}
Y_k & = {(X_0 +  X_{\frac{n}{2}})}^{p-k} + {(X_0 -  X_{\frac{n}{2}})}^{p-k}, \tilde{Y}_k = {(X_0 +  X_{\frac{n}{2}})}^{p-k} - {(X_0 -  X_{\frac{n}{2}})}^{p-k} \\
X_{J_{k}} & = X_{j_1}X_{j_2}\ldots X_{j_{k}}. \nonumber \end{aligned}$$ From the definition of $A_k$ and $\tilde{A}_k$, observe that $|A_k|= O(n^{k-1})$, because the entries of $A_k$ has one constraint, whereas $|\tilde{A}_k|= O(n^{k-2})$, because the entries of $\tilde{A}_k$ has two constraints. Therefore $$\label{eqn:A,tildeA}
|\tilde{A}_k| < |A_k|.$$

The following result will be used in the proof of Theorem \[thm:symcircovar\].

\[result:def\_h\] Suppose $|A_{p,s}|$ denotes the cardinality of $A_{p,s}$. Then $$h_p(k):=
\lim_{n\to \infty}\frac{|A_p^{(k)}|}{n^{p-1}}= \frac{1}{(p-1)!}\sum_{s=-\lceil\frac{p-k}{2}\rceil}^{\lfloor\frac{k}{2}\rfloor}\sum_{q=0}^{2s+p-k}(-1)^q\binom{p}{q}\l(\frac{2s+p-k-q}{2}\r)^{p-1},$$ where $\lceil x\rceil$ denotes the smallest integer not less than $x$.

For the proof of Result \[result:def\_h\], we refer to [@adhikari_saha2017 Lemma 14]. Now for a given vector $(j_1, j_2, \ldots, j_p)$, we define a term called [*opposite sign pair matched*]{} elements of the vector.

\[def:odd-even\] Suppose $(j_1, 