)^{m-2}\\
                           &=O(\epsilon n^m)+O\biggl({n^{m-1}\over\epsilon^3}\biggr).\qedhere
        \end{align*}
    \end{proof}
    \begin{claim}\label{claim:smallbound}
        $\abs{\tp{Small}}\leq O(\epsilon n^m)$.
    \end{claim}
    \begin{proof}
        Fix $(u_1,\dots,u_{2m})\in\tp{Small}$; without loss of generality, we may suppose that $u_{2m-2},u_1\notin\wtilde V$.
        Observe that $u_1,\dots,u_{2m-2}$ forms a copy of $P_{2m-2}$ and that the edge $u_{2m-1}u_{2m}$ has both end-points in $N(u_1)\cup N(u_{2m-2})$.
        Therefore, by applying \Cref{prop:oddpath} and using the fact that $G\in\gcl C$, we see that
        \begin{align*}
            \abs{\tp{Small}} &\leq 2\cdot\numb(G,P_{2m-2})\cdot\max_{u\neq v\in V\setminus\wtilde V}\abss{E\bigl(G[N(u)\cup N(v)]\bigr)}\\
                             &\leq (2\abs{E})^{m-1}\cdot \max_{u\neq v\in V\setminus\wtilde V}C\cdot\abss{N(u)\cup N(v)}\\
                             &\leq (2Cn)^{m-1}\cdot C\cdot 2\epsilon n =O(\epsilon n^m).\qedhere
        \end{align*}
    \end{proof}
    We now deal with $\tp{Good}$.
    Define $S\eqdef\sum_{uv\in{\wtilde V\choose 2}}\deg(u,v)$ and let $\mu$ be the probability mass on ${\wtilde V\choose 2}$ defined by $\mu(uv)=\deg(u,v)/S$.
    If $m=2$, then we can bound
    \begin{align*}
        \abs{\tp{Good}} &\leq\sum_{uv\in {\wtilde V\choose 2}}{\deg(u,v)\choose 2}\leq {1\over 2}\sum_{uv\in{\wtilde V\choose 2}}\deg(u,v)^2={S^2\over 2}\sum_{uv\in{\wtilde V\choose 2}}\mu(uv)^2\\
                        &={S^2\over 2}\sum_{H\in\cp(\wtilde V,K_2)}\mu(H)^2={\optb(\mu;K_2,2)\over 2}\cdot S^2\leq{\optb(K_2,2)\over 2}\cdot S^2.
    \end{align*}
    Similarly, if $m\geq 3$, then we can bound
    \begin{align*}
        \abs{\tp{Good}} &\leq\sum_{H\in\cp(\wtilde V,C_m)}\ \prod_{xy\in E(H)}\deg(x,y)=S^m\sum_{H\in\cp(\wtilde V,C_m)}\ \prod_{xy\in E(H)}\mu(xy)\\
                        &=S^m\sum_{H\in\cp(\wtilde V,C_m)}\mu(H)=\optb(\mu;C_m,1)\cdot S^m\leq\optb(C_m,1)\cdot S^m.
    \end{align*}
    Thus, by applying \Cref{prop:codegreebound} and setting $B_2=\optb(K_2,2)/2$ and $B_m=\optb(C_m,1)$ for all $m\geq 3$, we have shown that
    \[
        \abs{\tp{Good}}\leq B_m\cdot S^m\leq B_m\bigl(n+O(1/\epsilon^4)\bigr)^m= B_m\cdot n^m+O\biggl({n^{m-1}\over\epsilon^4}\biggr),
    \]
    provided $\epsilon^4 n\to\infty$.
    Therefore, by selecting $\epsilon=n^{-1/5}$ and applying \Cref{claim:bigbound,claim:smallbound}, we bound
    \begin{align*}
        \numb(G,C_{2m}) &=\abs{\tp{Good}}+\abs{\tp{Bad}}\leq \abs{\tp{Good}}+\abs{\tp{Big}}+\abs{\tp{Small}}\\
                        &\leq B_m\cdot n^m+O\biggl({n^{m-1}\over\epsilon^4}\biggr)+O(\epsilon n^m)+O\biggl({n^{m-1}\over\epsilon^3}\biggr) =B_m\cdot n^m+O(n^{m-1/5}).\qedhere
    \end{align*}
\end{proof}

\section{Odd paths}\label{sec:paths}
Thanks to \Cref{lem:tumorp}, in order to bound $\numb_\plan(G,P_{2m+1})$ from above, it suffices to find upper bounds on $\optp(m)$.
Recall that for a finite set $X$ and a probability mass $\mu$ on ${X\choose 2}$,
\[
    \optp(\mu;m)=\sum_{\mbf x\in (X)_m}\bar\mu(x_1)\biggl(\prod_{i=1}^{m-1}\mu(x_ix_{i+1})\biggr)\bar\mu(x_m),
\]
where $\bar\mu(x)=\sum_{y\in X\setminus\{x\}}\mu(xy)$.
\medskip

First, we handle the case of $m=2$.
\begin{prop}\label{prop:p4}
    $\optp(2)=2$.
\end{prop}
\begin{proof}
    The lower bound is realized if $\abs{\supp\mu}=1$.
    \medskip

    For the upper bound, fix a finite set $X$ and a probability mass $\mu$ on ${X\choose 2}$.
    Define the matrix $M\in\R^{X\times X}$ by $M_{xy}=\mu(xy)$ under the convention that $\mu(xx)=0$.
    Observe that $M$ is a symmetric, non-negative matrix all of whose row-sums are bounded above by $1$.
    In particular, the largest eigenvalue of $M$ is at most $1$ (c.f.~\cite[Lemma 8.1.21]{horn_matrix}).
    Thus, by applying standard facts about the Rayleigh quotient (c.f.~\cite[Theorem 4.2.2]{horn_matrix}) and using the fact that $\sum_{x\in X}\bar\mu(x)=2$, we bound
    \[
        \optp(\mu;2)=\sum_{x\neq y\in X}\bar\mu(x)\mu(xy)\bar\mu(y)=\langle\bar\mu, M\bar\mu\rangle\leq\langle\bar\mu,\bar\mu\rangle=\sum_{x\in X}\bar\mu(x)^2\leq\sum_{x\in X}\bar\mu(x)=2.\qedhere
    \]
\end{proof}
From here, we have a quick proof of the asymptotic result of Ghosh et al.~\cite{ghosh_planarp5}, albeit with a worse error term.
\begin{proof}[Proof of \cref{known:p5} of \Cref{thm:alreadyknown}]
    By applying \Cref{lem:tumorp} and \Cref{prop:p4}, we bound
    \[
        \numb_\plan(n,P_5)\leq\numb_{\gcl 3}(n,P_5)\leq{\optp(2)\over 2}\cdot n^3+O(n^{14/5})=n^3+O(n^{14/5}).\qedhere
    \]
\end{proof}

Next, we establish a general upper bound on $\optp(m)$.

\begin{theorem}\label{thm:probp}
    For any $m\geq 3$,
    \[
        \optp(m)\leq{1\over(m-1)!}.
    \]
\end{theorem}
\begin{proof}
    Fix a finite set $X$ and a probability mass $\mu$ on ${X\choose 2}$.
    The key to this bound is to interpret $\optp(\mu;m)$ as the probability of some event in a probability space defined by $\mu$.
    Intuitively $\optp(\mu;m