that $p$ and $T$ are in the same order and also for fixed $k$, the part $\#\{\text{isomorphism class in}~ Q_3\}$ is of order $O(1)$. Then $$\begin{aligned}
\label{q1}
\sum_{t=1}^kp^{t+s-k-m/2-1}\leq \sum_{t=1}^kp^{m-k-m/2}\leq \sum_{t=1}^kp^{\frac{2k-1}{2}-k}=O(p^{-1/2})~,\end{aligned}$$ which is due to the assumption that $t+s-1\leq m <2k$, so the contribution of is $o(p^{-1/2})$.

Next, we consider the contribution of (b). Since all $a_l=2$, we have the part of expectation equals 1. Therefore, is bounded by $$\begin{aligned}
\label{q2}
\frac{1}{pT^{2k}}\cdot\sum_{t=1}^kT^tp^s\#\{\text{isomorphism class in}~ Q_3\}=O\left(\sum_{t=1}^kp^{t+s-2k-1}\right)=O(1/p)~,\end{aligned}$$ where the equation is due to the fact that $t+s\leq 2k$.

Therefore, combine , and , we finally have $$\begin{aligned}
\label{tr}
{\mathbb{E}}(m_k(A_T))=\sum_{t=1}^k y_T^{2k-t}f_{t-1}(k)+o(1)~.\end{aligned}$$ To end the proof of , we need to determine the value of $f_{t-1}(k)$. This involves complex combinatorics and analytic arguments and the details of the derivation is given in Section \[de\]. Finally, using derived in Remark \[fff\] that $$\begin{aligned}
f_m(k)=\frac 1k\begin{pmatrix}
                 2k \\
                 m \\
               \end{pmatrix}\begin{pmatrix}
                              k \\
                              m+1 \\
                            \end{pmatrix}~,\end{aligned}$$ we have $${\mathbb{E}}(m_k(A_T))=\sum_{i=0}^{k-1}\frac 1k \begin{pmatrix}
                               2k \\
                               i \\
                             \end{pmatrix}\begin{pmatrix}
                                            k \\
                                            i+1 \\
                                          \end{pmatrix}y_T^{2k-1-i}+o(1)~,$$ which is .

[**The proof of .**]{} Recall $$\begin{aligned}
&\quad~{\mathop{\text{Var}}}(m_k(A_T))\nonumber\\
&=\frac{1}{p^2T^{4k}}\sum_{\bf{i_1},\bf{j_1},\bf{i_2},\bf{j_2}}\left[{\mathbb{E}}\left(E_{Q_1(\bf{i_1},\bf{j_1})}E_{Q_2(\bf{i_2},\bf{j_2})}\right)-{\mathbb{E}}\left(E_{Q_1(\bf{i_1},\bf{j_1})}\right){\mathbb{E}}\left(E_{Q_2(\bf{i_2},\bf{j_2})}\right)\right]~.\end{aligned}$$ If $Q_1$ has no edges coincident with edges of $Q_2$, then $${\mathbb{E}}\left(E_{Q_1(\bf{i_1},\bf{j_1})}E_{Q_2(\bf{i_2},\bf{j_2})}\right)-{\mathbb{E}}\left(E_{Q_1(\bf{i_1},\bf{j_1})}\right){\mathbb{E}}\left(E_{Q_2(\bf{i_2},\bf{j_2})}\right)=0$$ by independence between $E_{Q_1}$ and $E_{Q_2}$. If $Q=Q_1\bigcup Q_2$ has an overall single 