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\begin{document}

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\theoremstyle{plain}
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\begin{center}
\vskip 1cm{\LARGE\bf
The Sixth Moment of 
Random Determinants}
\vskip 1cm
\large
Dominik Beck\\
Faculty of Mathematics and Physics\\
Charles University\\
Prague 11636\\
Czech Republic\\
\href{mailto:beckd@karlin.mff.cuni.cz}{\tt beckd@karlin.mff.cuni.cz}\\
\ \\
Zelin Lv and Aaron Potechin\\
The University of Chicago\\ 
Chicago, IL 60637 \\
USA \\
\href{mailto:zlv@uchicago.edu}{\tt zlv@uchicago.edu}\\
\href{mailto:potechin@uchicago.edu}{\tt potechin@uchicago.edu}
\end{center}
\begin{abstract}
We determine the sixth moment of the determinant of an asymmetric $n \times n$ random matrix where the entries are drawn independently from an arbitrary distribution $\Omega$ over $\mathbb{R}$ with mean $0$. Furthermore, we derive the asymptotic behavior of the sixth moment of the determinant as the size of the matrix tends to infinity. 
\end{abstract}

\section{Introduction}
The behavior of the determinant of a random matrix has been extensively studied. One line of work analyzed the $k$th moment of a random determinant, i.e., the expected value of the $k$th power of the determinant of a random matrix. Tur\'{a}n observed that the second moment of the determinant of an $n \times n$ matrix (where the entries have mean $0$ and variance $1$) is $n!$. The fourth moment of the determinant was determined by Nyquist, Rice, and Riordan \cite{NRR54}. For the special case when the entries of the random matrix are Gaussian, several works \cite{forsythe1952extent, NRR54} showed that the $k$th moment of the determinant is $\prod_{j=0}^{\frac{k}{2}-1}{\frac{(n + 2j)!}{(2j)!}}$ for even $k$.

Several variations of this question have also been analyzed. Dembo \cite{10.2307/43637554} generalized these results for $p \times n$ matrices (where we consider $E[\det (M\!M^\top )^{\!\frac{k}{2}}]$ rather than $E[\det (M)^k]$). Very recently, Beck \cite{https://doi.org/10.48550/arxiv.2207.09311} determined the fourth moment of the determinant of an $n \times n$ random matrix with independent entries from an arbitrary distribution (which may not have mean $0$). For symmetric random matrices, the second moment of the determinant was analyzed by Zhurbenko \cite{Zhu68}.

A second line of work analyzed the distribution of the determinant of a random matrix. Girko \cite{girko1980central, doi:10.1137/S0040585X97975939} showed that under some assumptions on the entries of the random matrix, the logarithm of the determinant obeys a central limit theorem, but his proofs are quite difficult. Nguyen and Vu \cite{10.1214/12-AOP791} gave a simpler proof of a stronger version of this central limit theorem.

A third line of work analyzed the probability that a random $n \times n$ matrix with $\pm{1}$ entries is singular. Koml\'{o}s \cite{komlos1967determinant, komlos1968determinant} was the first to prove that this probability is $o(1)$. Kahn, Koml{\'o}s, and Szemer{\'e}di \cite{kahn1995probability} proved that this probability is at most $.999^n$, which was the first exponential upper bound. A series of works \cite{https://doi.org/10.1002/rsa.20109, tao2007singularity,BOURGAIN2010559} improved this upper bound culminating in the work of Tikhomirov \cite{10.4007/annals.2020.191.2.6} who proved an upper bound of $(\frac{1}{2} + o(1))^n$, which is tight. For the symmetric case, Costello, Tao, and Vu \cite{CTV06} showed that the probability that a random symmetric matrix with $\pm{1}$ entries is singular is $o(1)$ and further work has improved this bound.

For a recent survey of results on random matrices, see \cite{Vu20}. In this paper, we extend the first line of work by analyzing the sixth moment of the determinant of a random matrix.

\subsection{The fourth moment of random determinants}
Before stating our main result, we describe the results of Nyquist, Rice, and Riordan \cite{NRR54} on the fourth moment of the determinant of a random matrix.
\begin{definition}
Given a distribution $\Omega$ over $\mathbb{R}$, we define $\mathcal{M}_{n \times n}(\Omega)$ to be the distribution of $n \times n$ matrices where each entry is drawn independently from $\Omega$.
\end{definition}
\begin{definition}
Given a distribution $\Omega$ over $\mathbb{R}$, we define $m_k$ to be the $k$th moment of $\Omega$, i.e., 
\[
m_k = E_{x \sim \Omega}[x^k].
\]
\end{definition}
\begin{definition}
We define $f_{k}(n) = E_{M \sim \mathcal{M}_{n \times n}(\Omega)}[\det (M)^{k}]$ to be the expected value of the $k$-th power of the determinant of a random $n\times n$ matrix. Similarly, we define $p_{k}(n)$ to be the expected value of the $k$-th power of the permanent of a random $n\times n$ matrix.
\end{definition}
\begin{remark}
Both $f_{k}(n)$ and $p_{k}(n)$ depend on the moments of $\Omega$, but we write $f_{k}(n)$ and $p_{k}(n)$ rather than $f_{k,\Omega}(n)$ and $p_{k,\Omega}(n)$ for brevity.
\end{remark}
Nyquist, Rice, and Riordan \cite{NRR54} showed that $f_4(n) = n!y_{n}$ where $y_n$ obeys the recurrence relation
\[
y_n = (n + m_4 - 1)y_{n-1} + (3-m_4)(n-1)y_{n-2}.
\]
where $y_0 = 1$ and $y_1 = m_4$. They further observed that if we take the generating function $Y(t) = \sum_{t = 0}^{\infty}{\frac{y_{n}t^{n}}{n!}}$ then $Y(t) = (1-t)^{-3}e^{(m_4-3)t}$. From this generating function, they found the equation
\[
f_4(n) = n!y_n =\frac{(n!)^2}{2}\sum^n_{k=0}\frac{(n-k+1)(n-k+2)}{k!}(m_4 - 3)^k.
\]
To prove their results, Nyquist, Rice, and Riordan \cite{NRR54} counted $4 \times n$ tables with certain properties. As we describe in Section \ref{sec:preliminaries}, we use the same general approach though our analysis is considerably more intricate.
\subsection{Our results}
Our main results are as follows.

\begin{theorem}\label{genf_thm}
For any distribution $\Omega$ over $\mathbb{R}$ such that $m_1 = m_3 = 0$ and $m_2 = 1$, the formal generating function $F_6(t) = \sum_{n=0}^\infty\frac{t^n}{(n!)^2} f_6(n)$ for $f_6(n)$ is
\begin{equation*}
F_6(t) = \frac{e^{t (m_6-15 m_4 + 30)}}{48 (1+3t - m_4 t)\!{}^{15}} \sum _{i=0}^{\infty }
   \frac{(1+i) (2+i) (4+i)! t^i}{(1+3t -m_4 t){}^{3 i}}.
\end{equation*}
\end{theorem}

Performing Taylor expansion on this generating function, we get a formula for computing the sixth moment of random determinants, namely:

\begin{corollary}\label{main_cor}
For any distribution $\Omega$ over $\mathbb{R}$ such that $m_1 = m_3 = 0$ and $m_2 = 1$, we have that $f_6(n)$ equals
\begin{align*}
n!^2 \sum_{j=0}^n \sum_{i=0}^j\frac{(1\!+\!i) (2\!+\!i) (4\!+\!i)! }{48 (n-j)!}\binom{14+j+2i}{j-i} (m_6\!-\!15m_4\!+\!30)^{n-j} (m_4\!-\!3)^{j-i}.
\end{align*}
\end{corollary}
\begin{remark}
If $m_2 \neq 1$ then we can scale the distribution $\Omega$ by $\frac{1}{\sqrt{m_2}}$ (which changes the determinant of matrices in $\mathcal{M}_{n \times n}(\Omega)$ by a factor of $(\frac{1}{\sqrt{m_2}})^{n}$) and then apply the result in Corollary \ref{main_cor}.
\end{remark}
\begin{remark}
If $\Omega = N(0,1)$ then $m_4 = 3$ and $m_6 = 15$ so $f_6(n) = P_n = \frac{n!(n+2)!(n+4)!}{48}$, which is a special case of the result that $f_k(n) = \prod_{j=0}^{\frac{k}{2}-1}{\frac{(n + 2j)!}{(2j)!}}$ when $\Omega = N(0,1)$ and $k$ is even.
\end{remark}
Another generalization is when $m_3 \neq 0$.
\begin{theorem}\label{m3nonzero_thm}
For any distribution $\Omega$ over $\mathbb{R}$ such that $m_1 = 0$ and $m_2 = 1$, the formal generating function $F_6(t) = \sum_{n=0}^\infty\frac{t^n}{(n!)^2} f_6(n)$ for $f_6(n)$ is
\begin{align*}
F_6(t) =(1+m_3^2t)\!{}^{10}\, \frac{e^{t (m_6-10 m_3^2-15m_4 + 30)}}{48 (1+3t-m_4t)^{15}} \sum _{i=0}^{\infty }
   \frac{(1+i) (2+i) (4+i)!t^i}{(1+3t-m_4t){}^{3 i}}.
\end{align*}
\end{theorem}
\begin{corollary}
For any distribution $\Omega$ over $\mathbb{R}$ such that $m_1 = 0$ and $m_2 = 1$,
\begin{align*}
f_6(n) = n!^2 \sum _{j=0}^n \sum _{i=0}^j \sum_{k=0}^{n-j} \frac{(1\!+\!i) (2\!+\!i) (4\!+\!i)! }{48 (n-j-k)!}\binom{10}{k}\binom{14+j+2i}{j-i} q_6^{n-j-k} q_4^{j-i}q_3^k,
\end{align*}
where $q_6 = m_6-10 m_3^2-15m_4 + 30$, $q_4 = m_4-3$, and $q_3 = m_3^2$.
\end{corollary}

Below, we show the values of $f_k(n)$ and $p_k(n)$ when $\Omega = \{-1,1\}$ for small values of $k$ and $n$. When $\Omega = \{-1,1\}$, we note that $f_4(n)$ is the integer sequence A052127 in the On-Line Encyclopedia of Integer Sequences \cite[\seqnum{A052127}]{oeis}. In the entry for this integer sequence, it is noted that $f_4(n) \sim (n!)^2\frac{(n^2+7n+10)}{(2e^2)}$ as $n \to \infty$.
\begin{table}[H]
\centering
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
n &amp; $f_2(n)$, \seqnum{A000142} &amp; $f_4(n)$, \seqnum{A052127} &amp; $f_6(n)$, \seqnum{A357571}	 &amp; $p_2(n)$ &amp; $p_4(n)$ &amp; $p_6(n)$  \\ \hline
1 &amp; 1        &amp; 1        &amp; 1        &amp; 1        &amp; 1        &amp; 1         \\ \hline
2 &amp; 2        &amp; 8        &amp; 32       &amp; 2        &amp; 8        &amp; 32        \\ \hline
3 &amp; 6        &amp; 96       &amp; 1536     &amp; 6        &amp; 96       &amp; 2976      \\ \hline
4 &amp; 24       &amp; 2112     &amp; 282624   &amp; 24       &amp; 2112     &amp; 513024    \\ \hline
5 &amp; 120      &amp; 68160    &amp; 66846720 &amp; 120      &amp; 68160    &amp; 157854720 \\ \hline
\end{tabular}
\caption{Values of $f_k(n)$ and $p_k(n)$ for $\Omega = \{-1,1\}$, $k \leq 6$, and $n \leq 5$.}
\label{values-det-per}
\end{table}
\begin{remark}
    Note that for all $n \in \mathbb{N}$, $p_2(n) = f_2(n)$ and $p_4(n) = f_4(n)$. However, for $n \geq 3$, $p_6(n) &gt; f_6(n)$.
\end{remark}
We can describe the asymptotic behavior of $f_6$ using the following asymptotic expansion.
\begin{theorem}\label{asym_thm}
For all $R \in \mathbb{N} \cup \{0\}$,
\[
f_6(n) = \frac{e^{3q_4}(n!)^2}{48} \bigg(\sum_{k=0}^R c_k (n+6-k)!\bigg) {\pm}O((n!)^2(n+6-R-1)!),
\]
where the coefficients $c_k$ are the Taylor expansion coefficients of the function $C(t) = \sum_{k\geq 0} c_k t^k$,
\begin{align*}
C(t) = \, &amp; e^{(q_6-3 q_4^2)t+ q_4^3 t^2 } (1+q_3 t)^{10} \bigg( 1-2 (3 q_4+4) t +3 (5 q_4^2+8 q_4+4) t^2 \\
&amp;-4 (q_4^2 (5 q_4+6)) t^3+q_4^3 (15 q_4+8) t^4-6
   q_4^5 t^5+q_4^6 t^6 \bigg).
\end{align*}
\end{theorem}

\begin{remark}
For the first terms in the expansion, we have
\begin{align*}
&amp; f_6(n) \sim \frac{e^{3 m_4-9}}{48} n!^3 \bigg{(}n^6 + (m_6-3 m_4^2-3 m_4+34) n^5 +\frac{1}{2} \left(m_6^2-10 m_3^4+9 m_4^4\right.\\
&amp; \left. +20 m_4^3-183 m_4^2-126
   m_4-6 m_4^2 m_6-6 m_4 m_6+56 m_6+905\right) n^4 + \cdots \bigg{)}. 
\end{align*}    
\end{remark}

\begin{remark}
Note that when $\Omega = \{-1,1\}$, as $n \to \infty$,
\begin{align*}
f_6(n) \sim \frac{(n!)^3}{48e^6} &amp; \left(  n^6+29 n^5+335 n^4+\frac{5861 n^3}{3}+\frac{17944 n^2}{3}+\frac{44036 n}{5} \right. \\
  &amp;\left. \quad + \frac{167536}{45}-\frac{210176}{63 n} \right).
\end{align*}
\end{remark}

\section{Preliminaries}\label{sec:preliminaries}
To prove Theorem \ref{genf_thm}, we need a few definitions and a key lemma. For this section and the next section, we assume that $m_1 = m_3 = 0$.
\begin{definition}
Given natural numbers $k$ and $n$ where $k$ is even, we define an even $k \times n$ table to be a $k \times n$ table where each row is a permutation of $[n]$ and each column contains each number  that appears in the column an even number of times. We define $T_{k,n}$ to be the set of all even $k \times n$ tables.
\end{definition}
\begin{definition}
Given an even table $t$ of size $k\times n$, we define its sign $\sgn(t)$ to be the product of the signs of its rows, which are permutations of $[n]$.
\end{definition}
\begin{definition}
Given a column $c$ where each element is in $[n]$, we define its weight $w(c)$ to be 
\[
w(c) = \prod_{j=1}^{n}{m_{\# \text{ of times } j \text{ appears in column c}}}
.\]
For even $6 \times n$ tables, we say that a column is a $6$-column if it contains some number $6$ times, a $4$-column if it contains one number four times and another number two times, and a $2$-column if it contains three different numbers two times. Observe that the weight of a $6$-column is $m_6$, the weight of a $4$-column is $m_4 m_2$, and the weight of a $2$-column is $m_2^3$.
\end{definition}
\begin{definition}
Given an even $k\times n$ table $t$, we define its weight $w(t)$ to be the product of the weights of its columns.

\end{definition}
We can use the following proposition to reduce the problem of finding the sixth moment of a random determinant to a combinatorial problem.
\begin{proposition}\label{prop:tables}
For all even $k \in \mathbb{N}$, $f_k(n)=\sum_{t\in T_{k,n}}\sgn(t)w(t)$ and 
$p_k(n)=\sum_{t\in T_{k,n}}w(t)$.
\end{proposition}
\begin{proof} 
We observe that 
\[
f_k(n) = E_{A \sim \mathcal{M}_{n \times n}(\Omega)}\left[\sum_{\pi_1,\pi_2,...,\pi_k\in S_n}{\left(\prod_{i=1}^{k}\sgn (\pi_i)\right)\prod_{p=1}^{n}{ \left(\prod_{q=1}^{k}A_{p,\pi_q(p)}\right)}}\right]
\]
and 
\[
p_k(n) = E_{A \sim \mathcal{M}_{n \times n}(\Omega)}\left[\sum_{\pi_1,\pi_2,...,\pi_k\in S_n}{\prod_{p=1}^{n}{ \left(\prod_{q=1}^{k}A_{p,\pi_q(p)}\right)}}\right]
.\]
For each $p \in [n]$, we have that $E_{A \sim \mathcal{M}_{n \times n}(\Omega)}[\prod_{q=1}^{k}A_{p,\pi_q(p)}] = w(p)$ (i.e., the weight of column $p$), so $f_k(n)=\sum_{t\in T_{k,n}}\sgn(t)w(t)$ and $p_k(n)=\sum_{t\in T_{k,n}}w(t)$, as needed.

\end{proof}
Thus, computing the $k$th moment of a random determinant is equivalent to summing the signed weights of all even tables of size $k\times n$.
\begin{corollary}
If $\Omega$ is the uniform Bernoulli distribution (i.e., the uniform distribution on $\{-1,1\}$) then $f_k(n)=\sum_{t\in T_{k,n}}\sgn(t)$ and $p_k(n)=|T_{k,n}|$.
\end{corollary}
\begin{corollary}
If $k=2$, $k=4$, or $n \leq 2$ then $p_k(n)=f_k(n)$. If $n \geq 3$, $k \geq 6$, and $k$ is even then $p_k(n)&gt;f_k(n)$.
\end{corollary}
To analyze $f_6(n)$, it is useful to consider tables together with pairings of identical elements in each column.
\begin{definition}
Given an even $k \times n$ table $t$, we define a \emph{pairing} $P$ on $t$ to be a set of matchings $\{M_i: i \in [n]\}$, one for each column, where each matching $M_i$ pairs up identical elements of column $i$. We define $\mathcal{P}(t)$ to be the set of all pairings on $t$.
\end{definition}

\begin{example} The table on the left below is an even $6\times 4$ table with $27$ possible \textit{pairings}. The table on the right shows one of the $27$ possible pairings. 
\begin{table}[H]
\centering
\begin{tabular}{cc}
\begin{tabular}{|c|c|c|c|}
\hline
1 &amp; 2 &amp; 4 &amp; 3 \\ \hline
1 &amp; 2 &amp; 4 &amp; 3 \\ \hline
1 &amp; 3 &amp; 4 &amp; 2 \\ \hline
1 &amp; 3 &amp; 4 &amp; 2 \\ \hline
2 &amp; 4 &amp; 1 &amp; 3 \\ \hline
2 &amp; 4 &amp; 1 &amp; 3 \\ \hline
\end{tabular}&amp;

\begin{tabular}{|c|c|c|c|}
\hline
\cellcolor[HTML]{FD6864}1 &amp; \cellcolor[HTML]{FD6864}2 &amp; \cellcolor[HTML]{FD6864}4 &amp; \cellcolor[HTML]{FD6864}3 \\ \hline
\cellcolor[HTML]{FD6864}1 &amp; \cellcolor[HTML]{FD6864}2 &amp; \cellcolor[HTML]{FFFE65}4 &amp; \cellcolor[HTML]{FFFE65}3 \\ \hline
\cellcolor[HTML]{FFFE65}1 &amp; \cellcolor[HTML]{FFFE65}3 &amp; \cellcolor[HTML]{FFFE65}4 &amp; \cellcolor[HTML]{67FD9A}2 \\ \hline
\cellcolor[HTML]{FFFE65}1 &amp; \cellcolor[HTML]{FFFE65}3 &amp; \cellcolor[HTML]{FD6864}4 &amp; \cellcolor[HTML]{67FD9A}2 \\ \hline
\cellcolor[HTML]{67FD9A}2 &amp; \cellcolor[HTML]{67FD9A}4 &amp; \cellcolor[HTML]{67FD9A}1 &amp; \cellcolor[HTML]{FFFE65}3 \\ \hline
\cellcolor[HTML]{67FD9A}2 &amp; \cellcolor[HTML]{67FD9A}4 &amp; \cellcolor[HTML]{67FD9A}1 &amp; \cellcolor[HTML]{FD6864}3 \\ \hline
\end{tabular}
\end{tabular}
\end{table}
\end{example}
\begin{proposition}
For each even $6 \times n$ table $t$, 
$$
|\mathcal{P}(t)| = 15^{\# \mathrm{\ of\ 6-columns\ in\ t}}3^{\# \mathrm{\ of\ 4-columns\ in\ t}}
.$$
\end{proposition}

\begin{definition}
We define $P_n = \sum_{t \in T_{k,n}}{\sgn(t)|\mathcal{P}(t)|}$.
\end{definition}
It turns out that $P_n$ can be easily computed and this is crucial for our results.
\begin{lemma}\label{pairing}
For all $n \in \mathbb{N}$, $P_{n}=n(n+2)(n+4)P_{n-1}$ where $P_0=1$.
\end{lemma}
This lemma follows from the fact that when $\Omega = N(0,1)$ and $k$ is even, the $k$th moment of the determinant is $\prod_{j=0}^{\frac{k}{2}-1}{\frac{(n + 2j)!}{(2j)!}}$. We give a direct proof of this lemma in Appendix \ref{sec:directproof}.

Note that $P_n = \sum_{t \in T_{k,n}}{\sgn(t)15^{\# \text{ of 6-columns in t}}3^{\# \text{ of 4-columns in t}}}$ while \\
$f_6(n) = \sum_{t \in T_{k,n}}{\sgn(t){m_6}^{\# \text{ of 6-columns in t}}{m_4}^{\# \text{ of 4-columns in t}}}$. If $\Omega = N(0,1)$ (or we at least have that $m_1 = m_3 = 0$, $m_2 = 1$, $m_4 = 3$, and $m_6 = 15$) then $f_6(n) = P_n$.

In the next section, we show how to handle other distributions $\Omega$ using inclusion/exclusion.

\vspace{1em}
We finish our preliminaries section with the following generation function result on derangements:
\begin{lemma}\label{lem:derangement}
Let $S_n$ be the set of all permutations of order $n$ on $[n]=\{ 1,2,3,\ldots ,n\}$ and let $D_n$ be the set of all \emph{derangements} of the same order on $[n]$, i.e., the permutations in $S_n$ which have no fixed points. If we let $C(\pi)$ denote the number of cycles in a permutation $\pi$ and take $C_n(u) = \sum_{\pi \in D_n} u^{C(\pi)}$, then
\[
C_n(u) = (n-1)( C_{n-1}(u) + u\, C_{n-2}(u))
\]
and
\begin{equation*}
\sum_{n=0}^\infty \frac{x^n}{n!} C_n(u) = \frac{e^{-ux}}{(1-x)^u}.
\end{equation*}
\end{lemma}
\begin{proof}
See the chapter on Bivariate generating functions in the textbook ``Analytic Combinatorics'' by Flajolet and Sedgewick \cite{flajolet2009analytic}. For completeness, we present our own derivation. We proceed recursively based on the position of the node $n$. We can create a derangement $\pi \in D_n$ by either:
\begin{enumerate}
    \item Adding the node $n$ to one of the cycles of a derangement $\pi' \in D_{n-1}$. That is, if $i \rightarrow \pi(i)$, then we insert $n$ as $i \rightarrow n \rightarrow \pi(i)$. Since there are $n-1$ nodes in $\pi'$, there are $n-1$ different $\pi \in D_n$ we can create. In this case, the number of cycles is unchanged, i.e., $C(\pi) = C(\pi')$.
    \item Adding a cycle $(n,n-1)$ of length two to $\pi'' \in D_{n-2}$. We can then replace $n-1$ by any $i \in \pi''$. This gives $n-1$ new derangements $\pi \in D_n$ created from $\pi''$, all of them having $C(\pi) = C(\pi'')+1$.
\end{enumerate}
We can obtain all derangements $D_n$ in this way. These two possibilities are shown in the figures below.
\begin{figure}[H]
\centering
\begin{minipage}{0.45\textwidth}
\centering
\begin{tikzpicture}[thick, main/.style = {draw,circle}, large/.style = {draw, circle,minimum size=6em}, scale = 1.0]
\node[large] (D) at (0,0) {$D_{n-1}$};
\node[main] (n) at (3,0) {$n$};
\draw[-&gt;] (D) to [bend right=60] (n);
\draw[-&gt;] (n) to [bend right=60] (D);
\end{tikzpicture}
\caption{$D_{n-1} \rightarrow D_n$.}
\end{minipage}
\hfill
\begin{minipage}{0.45\textwidth}
\centering
\begin{tikzpicture}[thick, main/.style = {draw,circle}, large/.style = {draw, circle,minimum size=6em}, scale = 1.0]
\node[large] (D) at (0,0) {$D_{n-2}$};
\node[main] (n) at (3,1) {$n$};
\node[main] (i) at (3,-1) {$i$};
\draw[-&gt;] (n) to [bend right=40] (i);
\draw[-&gt;] (i) to [bend right=40] (n);
\end{tikzpicture}
\caption{$D_{n-2} \rightarrow D_n$.}
\end{minipage}
\end{figure}

\noindent
In terms of $C_n(u)$, we get the desired recurrence relation
\begin{align*}
C_n(u) &amp; = \sum_{\pi \in D_n} u^{C(\pi)} = (n-1)\sum_{\pi' \in D_n} u^{C(\pi')} + (n-1)\sum_{\pi'' \in D_{n-2}} u^{C(\pi'') + 1} \\
&amp; = (n-1)( C_{n-1}(u) + u\, C_{n-2}(u)),
\end{align*}
from which one can deduce its generating function easily.
\end{proof}

\section{Proof of Theorem \ref{genf_thm}}\label{sec:maintheoremproof}
Before preceding to the proof of Theorem \ref{genf_thm}, we first prove the following result on the sixth moment of random determinants. 
\begin{lemma} \label{main_lemma}
    For any distribution $\Omega$ over $\mathbb{R}$ such that $m_1 = m_3 = 0$ and $m_2 = 1$,
\[
f_6(n)=\sum_{j=0}^{n}{\sum_{a=0}^{j}\binom{n}{j}\binom{j}{a}(m_6 - 15)^{a} (m_4 - 3)^{(j-a)}D_{n,a,j-a}},
\]
where 
\begin{enumerate}
\item We have that $P_{n}=n(n+2)(n+4)P_{n-1}$ where $P_{0} = 1$. Equivalently, $P_{n} = \frac{n!(n+2)!(n+4)!}{2!4!}$.
\item We take $C_{n}=(n-1)(C_{n-1}+15C_{n-2})$ where  $C_0 = 1$ and $C_1 = 0$.
\item We take $H_{n,j,a,b}=\sum_{x=1}^{j}{\frac{j!}{x!}\binom{j-1}{x-1}\prod_{y=0}^{x-1}(3(n-a-b)-y)}$.
\item We take $D_{n,a,b}=\Big(\prod_{j=0}^{a+b-1}{(n-j)}\Big)\Big(\sum_{i=0}^{b}\binom{b}{i}C_{i}H_{n,b-i,a,b}\Big)P_{n-a-b}$.
\end{enumerate}
\end{lemma}
\begin{proof}
The idea behind the proof is as follows. We consider the tables where we know that some set $A \subseteq [n]$ of elements appear six times in a $6$-column and another set $B \subseteq [n] \setminus A$ of elements appear four times in a $4$-column and two times in a different column. We do not know whether the elements in $[n] \setminus (A \cup B)$ appear six times in a $6$-column, appear four times in a $4$-column and two times in a different column, or appear two times in three different columns. We consider these tables together with pairings for the columns which are unaccounted for by $A$ and $B$ (i.e., the columns which don't contain six of the same element in $A$ or four of the same element in $B$), which we call the center columns. 

To obtain $f_6(n)$, we compute the contribution from each $A$ and $B$ and then take an appropriate linear combination of these contributions so that the contribution from each individual table $t$ is $\sgn(t)w(t)$.
\begin{definition}
Given $A \subseteq [n]$ and $B \subseteq [n] \setminus A$, we define $D_{n,A,B}$ to be the set of tables in $T_{6,n}$ such that the elements in $A$ appear six times in a 6-column and the elements in $B$ appear four times in a 4-column and two times in a different column.
\end{definition}
\begin{definition}
Given $t \in T_{6,n}$,  $A \subseteq [n]$, and $B \subseteq [n] \setminus A$ such that $t \in D_{n,A,B}$, we define the \emph{center columns} of $t$ to be the columns which do not contain six of the same element of $A$ and which do not contain four of the same element of $B$. We define $\mathcal{P}_{\mathrm{center}}(t)$ to be the set of pairings on the center columns of $t$.
\end{definition}
\begin{example}\label{ex:table} The following table is a $6 \times 9$ table $t \in D_{9,\{7\},\{1,5,6,8,9\}}$ together with a pairing $P \in \mathcal{P}_{\mathrm{center}}(t)$ on the center columns.
\begin{figure}[H]
\centering
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
7 &amp; 8 &amp; 5 &amp; \cellcolor[HTML]{FD6864}2 &amp; \cellcolor[HTML]{FD6864}4 &amp; \cellcolor[HTML]{FD6864}3 &amp; 6 &amp; 1 &amp; 9 \\ \hline
7 &amp; 9 &amp; 5 &amp; \cellcolor[HTML]{FD6864}2 &amp; \cellcolor[HTML]{FFFE65}4 &amp; \cellcolor[HTML]{FFFE65}3 &amp; 6 &amp; 1 &amp; 8 \\ \hline
7 &amp; 8 &amp; 3 &amp; \cellcolor[HTML]{FFFE65}5 &amp; \cellcolor[HTML]{FFFE65}4 &amp; \cellcolor[HTML]{67FD9A}2 &amp; 6 &amp; 1 &amp; 9 \\ \hline
7 &amp; 8 &amp; 3 &amp; \cellcolor[HTML]{FFFE65}5 &amp; \cellcolor[HTML]{FD6864}4 &amp; \cellcolor[HTML]{67FD9A}2 &amp; 6 &amp; 1 &amp; 9 \\ \hline
7 &amp; 9 &amp; 5 &amp; \cellcolor[HTML]{67FD9A}2 &amp; \cellcolor[HTML]{67FD9A}6 &amp; \cellcolor[HTML]{FFFE65}3 &amp; 1 &amp; 4 &amp; 8 \\ \hline
7 &amp; 8 &amp; 5 &amp; \cellcolor[HTML]{67FD9A}2 &amp; \cellcolor[HTML]{67FD9A}6 &amp; \cellcolor[HTML]{FD6864}3 &amp; 1 &amp; 4 &amp; 9 \\ \hline
\end{tabular}
\end{figure}
\end{example}


By symmetry, the contribution from each $D_{n,A,B}$ only depends on $|A|$ and $|B|$.
\begin{definition}
Given $n,a,b \in \mathbb{N} \cup \{0\}$ such that $a+b \leq n$, we define $D_{n,a,b}$ to be 
\[
D_{n,a,b} = \sum_{t \in D_{n,A,B}}{\sgn(t)|\mathcal{P}_{\mathrm{center}}(t)|}
\]
where $A \subseteq [n]$, $B \subseteq [n] \setminus A$, $|A| = a$, and $|B| = b$.
\end{definition}
\begin{lemma}\label{lem:inclusionexclusion}
For all $n \in \mathbb{N} \cup \{0\}$,
\[
f_6(n) = \sum_{j=0}^{n}{\sum_{a=0}^{j}\binom{n}{j}\binom{j}{a}(m_6 - 15)^{a}(m_4 - 3)^{(j-a)}D_{n,a,j-a}}.
\]
\end{lemma}
\begin{proof}
Observe that 
\begin{align*}
&amp;\sum_{j=0}^{n}{\sum_{a=0}^{j}\binom{n}{j}\binom{j}{a}(m_6 - 15)^{a}(m_4 - 3)^{(j-a)}D_{n,a,j-a}} \\
&amp;=\sum_{A \subseteq [n]}{\sum_{B \subseteq [n] \setminus A}{\sum_{t \in D_{n,A,B}}{(m_6 - 15)^{|A|}(m_4 - 3)^{|B|}\sgn(t)|\mathcal{P}_{\mathrm{center}}(t)|}}}.
\end{align*}
Given a table $t \in T_{6,n}$, let $A'$ be the set of elements in $[n]$ which appear six times in a $6$-column of $t$ and let $B'$ be the set of elements which appear four times in a $4$-column of $t$. Now consider the contribution from $t$ in 
\[
\sum_{A \subseteq [n]}{\sum_{B \subseteq [n] \setminus A}{\sum_{t \in D_{n,A,B}}{(m_6 - 15)^{|A|}(m_4 - 3)^{|B|}\sgn(t)|\mathcal{P}_{\mathrm{center}}(t)|}}}.
\]
We have that whenever $A \subseteq A'$ and $B \subseteq B'$, $t \in D_{n,A,B}$ and $|\mathcal{P}_{\mathrm{center}}(t)| = 15^{|A' \setminus A|}3^{|B' \setminus B|}$. Thus, the contribution from $t$ is
\begin{align*}
&amp;\sum_{A \subseteq A'}{\sum_{B \subseteq B'}(m_6 - 15)^{|A|}(m_4 - 3)^{|B|}15^{|A' \setminus A|}3^{|B' \setminus B|}\sgn(t)}\\
&amp;= {m_6}^{|A'|}{m_4}^{|B'|}\sgn(t) = \sgn(t)w(t).
\end{align*}
This implies that 
\[
\sum_{j=0}^{n}{\sum_{a=0}^{j}\binom{n}{j}\binom{j}{a}(m_6 - 15)^{a} (m_4 - 3)^{(j-a)}D_{n,a,j-a}} = \sum_{t \in T_{6,n}}{\sgn(t)w(t)} = f_6(n),
\]
as needed.
\end{proof}
We now compute $D_{n,a,b}$.
\begin{lemma}\label{lem:Dnab}
For all $n,a,b \in \mathbb{N} \cup \{0\}$ such that $a+b \leq n$,
\[
    D_{n,a,b}=\bigg(\prod_{j=0}^{a+b-1}{(n-j)}\bigg)\bigg(\sum_{i=0}^{b}\binom{b}{i}C_{i}H_{n,b-i,a,b}\bigg)P_{n-a-b}
\]
where $C_n$ is given by the recurrence relation $C_{n}=(n-1)(C_{n-1}+15C_{n-2})$, $C_0 = 1$, $C_1 = 0$ and 
\[
    H_{n,j,a,b}=\sum_{x=1}^{j}{\frac{j!}{x!}\binom{j-1}{x-1}\prod_{y=0}^{x-1}(3(n-a-b)-y)}.
\]
\end{lemma}
\begin{proof}
To prove this lemma, we group the tables in $D_{n,A,B}$ based on the structure of the $4$-columns containing four of the same element of $B$.
\begin{definition}
Given a table $t \in D_{n,A,B}$, we define the directed graph $G(t)$ to be the graph with vertices $V(G(t)) = B \cup \{v_{\mathrm{center}}\}$ and the following edges. For each $j \in B$, 
\begin{enumerate}
    \item If there is a $j' \in B \setminus \{b\}$ such that there are two $j$ in the $4$-column containing four $j'$ then we add an edge from $j$ to $j'$.
    \item If there is no such $j'$ then we add an edge from $j$ to $v_{\mathrm{center}}$.
\end{enumerate}
For each vertex $v \in V(G(t))$, we define the outdegree $\deg^{+}(v)$ of $v$ to be the number of edges going from $v$ to another vertex and we define the indegree $\deg^{-}(v)$  of $v$ to be the number of edges going from another vertex to $v$.
\end{definition}
\begin{proposition}
For all $t \in D_{n,A,B}$, $G(t)$ has the following properties.
\begin{enumerate}
    \item For all $j \in B$, $\deg^{+}(j)=1$ and $\deg^{-}(j) \leq 1$.
    \item $\deg^{+}(v_{\mathrm{center}}) = 0$.
\end{enumerate}
\end{proposition}
\begin{corollary}
For all $t \in D_{n,A,B}$, $G(t)$ consists of directed cycles and paths which end at $v_{\mathrm{center}}$, all of which are disjoint except for their common endpoint.
\end{corollary}

\begin{example}\label{ex:graph} The figure on the left below shows a table $t$ together with a pairing $P \in \mathcal{P}_{\mathrm{center}}(t)$ on the center columns of $t$. The figure on the right shows the resulting graph $G(t)$.
\begin{figure}[H]
\centering
\begin{minipage}[c]{0.55\textwidth}
\centering
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
7 &amp; 8 &amp; 5 &amp; \cellcolor[HTML]{FD6864}2 &amp; \cellcolor[HTML]{FD6864}4 &amp; \cellcolor[HTML]{FD6864}3 &amp; 6 &amp; 1 &amp; 9 \\ \hline
7 &amp; 9 &amp; 5 &amp; \cellcolor[HTML]{FD6864}2 &amp; \cellcolor[HTML]{FFFE65}4 &amp; \cellcolor[HTML]{FFFE65}3 &amp; 6 &amp; 1 &amp; 8 \\ \hline
7 &amp; 8 &amp; 3 &amp; \cellcolor[HTML]{FFFE65}5 &amp; \cellcolor[HTML]{FFFE65}4 &amp; \cellcolor[HTML]{67FD9A}2 &amp; 6 &amp; 1 &amp; 9 \\ \hline
7 &amp; 8 &amp; 3 &amp; \cellcolor[HTML]{FFFE65}5 &amp; \cellcolor[HTML]{FD6864}4 &amp; \cellcolor[HTML]{67FD9A}2 &amp; 6 &amp; 1 &amp; 9 \\ \hline
7 &amp; 9 &amp; 5 &amp; \cellcolor[HTML]{67FD9A}2 &amp; \cellcolor[HTML]{67FD9A}6 &amp; \cellcolor[HTML]{FFFE65}3 &amp; 1 &amp; 4 &amp; 8 \\ \hline
7 &amp; 8 &amp; 5 &amp; \cellcolor[HTML]{67FD9A}2 &amp; \cellcolor[HTML]{67FD9A}6 &amp; \cellcolor[HTML]{FD6864}3 &amp; 1 &amp; 4 &amp; 9 \\ \hline
\end{tabular}
\caption{A $6 \times 9$ table $t \in D_{9,\{7\},\{1,5,6,8,9\}}$.}
\end{minipage}
\hfill
\begin{minipage}{0.4\textwidth} \centering
\begin{tikzpicture}[thick, main/.style = {draw,circle}, scale = 0.9]
\node[main] (v) at (0,0) {$v_{\mathrm{center}}$};
\node[main] (5) at (1.5,-0.7) {$5$};
\node[main] (6) at (1.5,0.7) {$6$};
\node[main] (8) at (-2,1.4) {$8$};
\node[main] (9) at (-2,-0.7) {$9$};
\node[main] (1) at (2.8,1.3) {$1$};
\draw[-&gt;] (8) to [bend right=40] (9);
\draw[-&gt;] (9) to [bend right=40] (8);
\draw[-&gt;] (6) to [bend right=0] (v);
\draw[-&gt;] (1) to [bend right=0] (6);
\draw[-&gt;] (5) to [bend right=0] (v);
\end{tikzpicture}
\caption{The associated $G(t)$.}
\end{minipage}
\end{figure}
\end{example}

Note that as the cycles and paths are disjoint, we can split any table $t \in D_{n,A,B}$ into $6$-columns corresponding to $A$, $4$-columns corresponding to $B_{\mathrm{cycles}}$, $4$-columns corresponding to $B_{\mathrm{paths}}$, and $\mathrm{center}$ columns, respectively.
\begin{example} The figures below show the split of table $t\in D_{9,\{7\},\{1,5,6,8,9\}}$ from Example \ref{ex:table}.
\begin{figure}[H]
\centering
\begin{minipage}[c]{0.2\textwidth}
\centering
\begin{tabular}{|c|}
\hline
7 \\ \hline
7 \\ \hline
7 \\ \hline
7 \\ \hline
7 \\ \hline
7 \\ \hline
\end{tabular}
\caption{$A$.}
\end{minipage}
\hfill
\begin{minipage}[c]{0.2\textwidth}
\centering
\begin{tabular}{|c|c|}
\hline
8 &amp; 9 \\ \hline
9 &amp; 8 \\ \hline
8 &amp; 9 \\ \hline
8 &amp; 9 \\ \hline
9 &amp; 8 \\ \hline
8 &amp; 9 \\ \hline
\end{tabular}
\caption{$B_{\mathrm{cycles}}$.}
\end{minipage}
\hfill
\begin{minipage}[c]{0.2\textwidth}
\centering
\begin{tabular}{|c|c|c|}
\hline
5 &amp; 6 &amp; 1 \\ \hline
5 &amp; 6 &amp; 1 \\ \hline
3 &amp; 6 &amp; 1 \\ \hline
3 &amp; 6 &amp; 1 \\ \hline
5 &amp; 1 &amp; 4 \\ \hline
5 &amp; 1 &amp; 4 \\ \hline
\end{tabular}
\caption{$B_{\mathrm{paths}}$.}
\end{minipage}
\hfill
\begin{minipage}[c]{0.25\textwidth}
\centering
\begin{tabular}{|c|c|c|}
\hline
\cellcolor[HTML]{FD6864}2 &amp; \cellcolor[HTML]{FD6864}4 &amp; \cellcolor[HTML]{FD6864}3 \\ \hline
\cellcolor[HTML]{FD6864}2 &amp; \cellcolor[HTML]{FFFE65}4 &amp; \cellcolor[HTML]{FFFE65}3\\ \hline
\cellcolor[HTML]{FFFE65}5 &amp; \cellcolor[HTML]{FFFE65}4 &amp; \cellcolor[HTML]{67FD9A}2\\ \hline
\cellcolor[HTML]{FFFE65}5 &amp; \cellcolor[HTML]{FD6864}4 &amp; \cellcolor[HTML]{67FD9A}2\\ \hline
\cellcolor[HTML]{67FD9A}2 &amp; \cellcolor[HTML]{67FD9A}6 &amp; \cellcolor[HTML]{FFFE65}3\\ \hline
\cellcolor[HTML]{67FD9A}2 &amp; \cellcolor[HTML]{67FD9A}6 &amp; \cellcolor[HTML]{FD6864}3\\ \hline
\end{tabular}
\caption{$\mathrm{center}$.}
\end{minipage}

\end{figure}
\end{example}



We now consider how many ways there are to start with a table $t_{\mathrm{center}} \in T_{6,n-a-b}$ together with a pairing $P \in \mathcal{P}(t_{\mathrm{center}})$ and construct a table $t \in D_{n,A,B}$ (we will automatically have that $\sgn(t) = \sgn(t_{\mathrm{center}})$ and $\mathcal{P}_{\mathrm{center}}(t) = P$). Before giving the entire analysis, we describe the parts of the analysis corresponding to the cycles and paths of $G(t)$ as these are the trickiest parts of the analysis.

To handle the cycles of $G(t)$, we observe that the columns of $t$ corresponding to these cycles are independent from the rest of $t$. This means that once the locations and elements of these columns are chosen, it is sufficient to count the number of possible tables for these columns. This can be done as follows.
\begin{definition}
Define $C_n$ to be the number of tables $t_{\mathrm{cycles}} \in D_{n,\emptyset,[n]}$ such that $G(t_{\mathrm{cycles}})$ consists of directed cycles and for each $i \in [n]$, column $i$ contains four $i$.
\end{definition}

\begin{lemma}\label{lem:countingcycletables}
For all $n \geq 2$, $C_{n}=(n-1)(C_{n-1}+15C_{n-2})$ where $C_0 = 1$ and $C_1 = 0$ and as a consequence,
\[
\sum_{n = 0}^\infty \frac{t^n}{n!} C_n = \frac{e^{-15x}}{(1-x)^{15}}.
\]
\end{lemma}
\begin{proof}
Notice that $G(t_{\mathrm{cycles}})$ is a derangement of order $n$. On the other hand, given a derangement $\pi \in D_n$, we determine the number of tables $t_{\mathrm{cycles}} \in D_{n,\emptyset,[n]}$ with the property that $G(t_{\mathrm{cycles}}) = \pi$ and for each $i \in [n]$, column $i$ contains four $i$. Note that for each $i$ and $j$ in a given cycle of $\pi$, the four rows of column $i$ which contain $i$ are the same as the four rows of column $j$ which contain $j$. Thus, for each cycle $C$, to determine the entries of the part of $t_{\mathrm{cycles}}$ corresponding to $C$, it is sufficient to take an arbitrary $i$ in $C$ and choose which four rows of column $i$ contain $i$. There are $\binom{6}{4} = 15$ choices for this, so there are ${15}^{C(\pi)}$ distinct tables $t_{\mathrm{cycles}}$ which we can construct from $\pi$. Thus,
\[
C_n = \sum_{\pi \in D_n} {15}^{C(\pi)} = C_n(15).
\]
By Lemma \ref{lem:derangement}, we get the recurrence relation and the generating function.
\end{proof}
\begin{example} The figures below show a table $t_{\mathrm{cycles}}$ and the corresponding derangement $G(t_{\mathrm{cycles}})$.
\begin{figure}[H]
\centering
\begin{minipage}[c]{0.5\textwidth}
\centering
\begin{tabular}{|l|l|l|l|l|}
\hline
1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 \\ \hline
1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 \\ \hline
1 &amp; 2 &amp; 5 &amp; 3 &amp; 4 \\ \hline
1 &amp; 2 &amp; 5 &amp; 3 &amp; 4 \\ \hline
2 &amp; 1 &amp; 3 &amp; 4 &amp; 5 \\ \hline
2 &amp; 1 &amp; 3 &amp; 4 &amp; 5 \\ \hline
\end{tabular}
\caption{A $6\times 5$ table $t_{cylces} \in D_{5,\emptyset,[5]}$.}
\end{minipage}
\hfill
\begin{minipage}{0.45\textwidth}
\centering
\begin{tikzpicture}[thick, main/.style = {draw,circle}, scale = 0.8]
\node[main] (1) at (-5,0.8) {$1$};
\node[main] (2) at (-3,0.8) {$2$};
\node[main] (3) at (-1.5,0) {$3$};
\node[main] (4) at (0.5,0) {$4$}; 
\node[main] (5) at (-0.5,1.8) {$5$};
\draw[-&gt;] (1) to [bend right=40] (2);
\draw[-&gt;] (2) to [bend right=40] (1);
\draw[-&gt;] (3) to [bend right=20] (4);
\draw[-&gt;] (4) to [bend right=20] (5);
\draw[-&gt;] (5) to [bend right=20] (3);
\end{tikzpicture}
\caption{The associated $G(t_{\mathrm{cycles}})$.
}
\end{minipage}
\end{figure}
\end{example}

To handle the paths of $G(t)$, we observe that if we want to construct $t$ so that $G(t)$ has a path $b_1 \to b_2 \to \cdots \to b_{l} \to v_{\mathrm{center}}$, we can do this as follows:
\begin{enumerate}
\item We start with a table $t_{\mathrm{center}}$ for the center columns and a pairing $P \in \mathcal{P}(t_{\mathrm{center}})$.
\item For each $j \in [l]$, we add a column containing six $b_j$.
\item We choose a pair of entries of $t_{\mathrm{center}}$ which are paired together by $P$. Let $z \in [n]$ be the element in these entries and let $r_1$ and $r_2$ be the two rows of these entries.
\item For each $j \in \{0,1,\ldots,l-1\}$, we swap this pair of $z$ with the two $b_{l-j}$ in rows $r_1$ and $r_2$ and then move on to the next $j$. When we are done, 
\begin{enumerate}
\item[1.] The column which started with six $b_1$ now has two $z$ in rows $r_1$ and $r_2$.
\item[2.] For each $j \in [l-1]$, the column which started with six $b_{j+1}$ now has two $b_j$ in rows $r_1$ and $r_2$.
\item[3.] The original pair of $z$ are replaced by a pair of $b_l$.
\end{enumerate}
\end{enumerate}

\begin{example} The figures below show how to extend a table $t_{\mathrm{center}}$ and a pairing $P \in \mathcal{P}(t_{\mathrm{center}})$ based on a path to $v_{\mathrm{center}}$ and an endpoint for this path in $t_{\mathrm{center}}$ (i.e., a pair of entries in $t_{\mathrm{center}}$ which are paired up by $P$).

\begin{figure}[H]
\centering
\begin{minipage}[b]{0.27\textwidth}
\centering
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\cellcolor[HTML]{FD6864}3 &amp; \cellcolor[HTML]{FD6864}2 &amp; \cellcolor[HTML]{FD6864}6 &amp; \cellcolor[HTML]{FD6864}4 &amp; \cellcolor[HTML]{FD6864}1 &amp; \cellcolor[HTML]{FD6864}5\\ \hline
\cellcolor[HTML]{FFFE65}3 &amp; \cellcolor[HTML]{FD6864}2 &amp; \cellcolor[HTML]{FFFE65}5 &amp; \cellcolor[HTML]{FFFE65}6 &amp; \cellcolor[HTML]{FFFE65}1 &amp; \cellcolor[HTML]{FFFE65}4\\ \hline
\cellcolor[HTML]{FFFE65}3 &amp; \cellcolor[HTML]{FFFE65}4 &amp; \cellcolor[HTML]{67FD9A}2 &amp; \cellcolor[HTML]{FFFE65}6 &amp; \cellcolor[HTML]{FD6864}1 &amp; \cellcolor[HTML]{67FD9A}\circled{\textbf{5}}\\ \hline
\cellcolor[HTML]{FD6864}3 &amp; \cellcolor[HTML]{FFFE65}4 &amp; \cellcolor[HTML]{FD6864}6 &amp; \cellcolor[HTML]{67FD9A}2 &amp; \cellcolor[HTML]{FFFE65}1 &amp; \cellcolor[HTML]{FD6864}5\\ \hline
\cellcolor[HTML]{67FD9A}3 &amp; \cellcolor[HTML]{67FD9A}1 &amp; \cellcolor[HTML]{FFFE65}5 &amp; \cellcolor[HTML]{67FD9A}2 &amp; \cellcolor[HTML]{67FD9A}6 &amp; \cellcolor[HTML]{FFFE65}4\\ \hline
\cellcolor[HTML]{67FD9A}3 &amp; \cellcolor[HTML]{67FD9A}1 &amp; \cellcolor[HTML]{67FD9A}2 &amp; \cellcolor[HTML]{FD6864}4 &amp; \cellcolor[HTML]{67FD9A}6 &amp; \cellcolor[HTML]{67FD9A}\circled{\textbf{5}}\\ \hline
\end{tabular}
\caption{A table $t_{\mathrm{center}}$ with a pairing $P$.}
\end{minipage}
\hfill
\begin{minipage}[b]{0.27\textwidth} \centering
\begin{tikzpicture}[thick, main/.style = {draw,circle}, scale = 0.32]
\node[main] (v) at (0,0) {$v_{\mathrm{center}}$};
\node[main] (9) at (-10,5) {$9$};
\node[main] (8) at (-7,3.5) {$8$};
\node[main] (7) at (-4,2) {$7$};
\draw[-&gt;] (9) to [bend right=0] (8);
\draw[-&gt;] (8) to [bend right=0] (7);
\draw[-&gt;] (7) to [bend right=0] (v);
\end{tikzpicture}
\caption{A path to $v_{\mathrm{center}}$.
}
\end{minipage}
\hfill
\begin{minipage}[b]{0.4\textwidth}
\centering
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
9 &amp; 8 &amp; 7 &amp; \cellcolor[HTML]{FD6864}3 &amp; \cellcolor[HTML]{FD6864}2 &amp; \cellcolor[HTML]{FD6864}6 &amp; \cellcolor[HTML]{FD6864}4 &amp; \cellcolor[HTML]{FD6864}1 &amp; \cellcolor[HTML]{FD6864}5\\ \hline
9 &amp; 8 &amp; 7 &amp; \cellcolor[HTML]{FFFE65}3 &amp;  \cellcolor[HTML]{FD6864}2 &amp; \cellcolor[HTML]{FFFE65}5 &amp; \cellcolor[HTML]{FFFE65}6 &amp; \cellcolor[HTML]{FFFE65}1 &amp; \cellcolor[HTML]{FFFE65}4\\ \hline
\textbf{5} &amp; 9 &amp; 8 &amp; \cellcolor[HTML]{FFFE65}3 &amp; \cellcolor[HTML]{FFFE65}4 &amp; \cellcolor[HTML]{67FD9A}2 &amp; \cellcolor[HTML]{FFFE65}6 &amp; \cellcolor[HTML]{FD6864}1 &amp; \cellcolor[HTML]{67FD9A}\circled{7}\\ \hline
9 &amp; 8 &amp; 7 &amp; \cellcolor[HTML]{FD6864}3 &amp; \cellcolor[HTML]{FFFE65}4 &amp; \cellcolor[HTML]{FD6864}6 &amp; \cellcolor[HTML]{67FD9A}2 &amp; \cellcolor[HTML]{FFFE65}1 &amp; \cellcolor[HTML]{FD6864}5\\ \hline
9 &amp; 8 &amp; 7 &amp; \cellcolor[HTML]{67FD9A}3 &amp; \cellcolor[HTML]{67FD9A}1 &amp; \cellcolor[HTML]{FFFE65}5 &amp; \cellcolor[HTML]{67FD9A}2 &amp; \cellcolor[HTML]{67FD9A}6 &amp; \cellcolor[HTML]{FFFE65}4\\ \hline
\textbf{5} &amp; 9 &amp; 8 &amp; \cellcolor[HTML]{67FD9A}3 &amp; \cellcolor[HTML]{67FD9A}1 &amp; \cellcolor[HTML]{67FD9A}2 &amp; \cellcolor[HTML]{FD6864}4 &amp; \cellcolor[HTML]{67FD9A}6 &amp; \cellcolor[HTML]{67FD9A}\circled{7}\\ \hline
\end{tabular}
\caption{A $6 \times 9$ table $t$ where $G(t)$ has the given path to $v_{\mathrm{center}}$.}
\end{minipage}
\end{figure}
\end{example}


Following this logic, we can handle the paths in $G(t)$ as follows:
\begin{enumerate}
\item Choose the paths in $G(t)$.
\item For each path in $G(t)$, choose a pair of entries in $t_{\mathrm{center}}$ which are paired up by $P$ to act as an endpoint for the path. Note that no two paths can have the same pair of entries as an endpoint. 
\end{enumerate}
\begin{remark}
This argument for analyzing the paths of $G(t)$ is the reason why it is important to consider the table $t_{\mathrm{center}}$ for the center columns together with a pairing $P \in \mathcal{P}(t_{\mathrm{center}})$.
\end{remark}
To count the number of ways to choose the paths in $G(t')$ and their endpoints, we use the following lemma which counts the number of possible graphs $G_{\mathrm{paths}}$ with a given number of paths to $v_{\mathrm{center}}$ and no cycles.
\begin{lemma}\label{lem:countingpathgraphs}
Let $B' \subseteq [n]$ and take $j = |B'|$. For all $x \in [j]$, there are $\frac{j!}{x!}\binom{j-1}{x-1}$ possible graphs $G_{\mathrm{paths}}$ on the vertices $B' \cup \{v_{\mathrm{center}}\}$ which consist of $x$ disjoint paths to $v_{\mathrm{center}}$ and no cycles.
\end{lemma}
\begin{proof}
We can specify each such graph as follows:
\begin{enumerate}
    \item Choose an ordering for the elements of $B'$. There are $j!$ possibilities for this ordering.
    \item Choose the $x$ paths by putting $x-1$ dividing lines among the elements of $B'$. Between each neighbouring elements of $B'$, there must be at most one dividing line, since each path must have at least one vertex in $B'$. Hence, there are $\binom{j-1}{x-1}$ possibilities for this.
\end{enumerate}
However, if we do this, each graph $G_{\mathrm{paths}}$ is counted $x!$ times, one for each possible ordering of the $x$ paths. Thus, the number of possible graphs $G_{\mathrm{paths}}$ on the vertices $B' \cup \{v_{\mathrm{center}}\}$ which consist of $x$ disjoint paths to $v_{\mathrm{center}}$ and no cycles is $\frac{j!}{x!}\binom{j-1}{x-1}$, as needed.
\end{proof}

\begin{proposition}
Letting $x$ be the number of paths in $G_{\mathrm{paths}}$, there are $\prod_{y=0}^{x-1}{(3(n-a-b)-y)}$ ways to choose the endpoints for these paths.   
\end{proposition}

We now give the entire analysis for $D_{n,a,b}$. Given $A \subseteq [n]$ and $B \subseteq [n] \setminus A$ such that $|A| = a$ and $|B| = b$, we can compute our $D_{n,a,b} = \sum_{t \in D_{n,A,B}} \sgn(t)| \mathcal{P}_{\mathrm{center}}(t)|$ as follows.
\begin{enumerate}
    \item For each $i \in A$, we choose which column contains six copies of $i$. Similarly, for each $j \in B$, we choose which column contains four $j$. The number of choices for this is $\prod_{j=0}^{a+b-1}{(n-j)}$.
    \item After choosing these columns, we choose a table $t_{\mathrm{center}} \in T_{6,n-a-b}$ and a pairing $P \in \mathcal{P}(t_{\mathrm{center}})$ to fill in the remaining columns. The number of choices for this is $P_{n-a-b}$.
    \item We split into cases based on the number of vertices $i$ in $G(t)$ which are contained in cycles. For each $i$, we choose which $\binom{b}{i}$ of the elements in $B$ are contained in cycles. By Lemma \ref{lem:countingcycletables}, once these elements are chosen there are $C_i$ possibilities for the columns containing these elements.
    \item There are now $j = b - i$ elements of $B$ which are contained in paths. We further split into cases based on the number $x$ of paths in $G(t)$. By Lemma \ref{lem:countingpathgraphs}, there are $\frac{j!}{x!}\binom{j-1}{x-1}$ possibilities for what these paths are in $G(t)$.
    
    As discussed above, for each of the $x$ paths, we need to choose a different pair of entries which are paired up by $P$ to be an endpoint for the path and the number of choices for these pairs is $\prod_{y=0}^{x-1}{(3(n-a-b)-y)}$. Summing all of these possibilities up gives a factor of
    \[
        H_{n,j,a,b}=\sum_{x=1}^{j}{\frac{j!}{x!}\binom{j-1}{x-1}\prod_{y=0}^{x-1}(3(n-a-b)-y)}.
    \]
\end{enumerate}
Putting everything together, we have that 
\[
    D_{n,a,b}=\bigg(\prod_{j=0}^{a+b-1}{(n-j)}\bigg)\bigg(\sum_{i=0}^{b}\binom{b}{i}C_{i}H_{n,b-i,a,b}\bigg)P_{n-a-b},
\]
as needed.

Since this argument is intricate and is a central argument for our results, we make this argument formal with the following bijection. 
\begin{lemma}
Given $A = \{i_1,\ldots,i_a\} \subseteq [n]$ and $B = \{j_1,\ldots,j_b\} \subseteq [n] \setminus A$, there is a bijection between the following sets of data:
\begin{enumerate}
\item A table $t \in D_{n,A,B}$ together with a pairing $P \in \mathcal{P}_{\mathrm{center}}(t)$ on the center columns.
\item A tuple $(C_A,C_B,t_{\mathrm{center}},P,i,B_{\mathrm{cycles}},t_{\mathrm{cycles}},x,G_{\mathrm{paths}},E)$
where 
\begin{enumerate}
\item The indices $C_A$ specify the locations of the $6$-columns corresponding to the elements in $A$. More precisely, $C_A = (c_{i_1},c_{i_2},\ldots,c_{i_a})$ where for each $k \in [a]$, column $c_{i_k}$ contains six $i_k$.
\item The indices $C_B$ specify the locations of the $4$-columns corresponding to the elements in $B$. More precisely, $C_B = (c_{j_1},c_{j_2},\ldots,c_{j_b})$ where for each $k \in [b]$, column $c_{j_k}$ contains four $j_k$.
\item We let $t_{\mathrm{center}} \in T_{6,n-|A|-|B|}$ and $P \in \mathcal{P}(t_{\mathrm{center}})$ denote a table for the center columns and a pairing on this table.
\item We let $i \in [0,|B|]$ denote the number of elements of $B$ which are contained in cycles in $G(t)$.
\item We let $B_{\mathrm{cycles}} \subseteq B$ denote the $i$ elements of $B$ which are contained in cycles in $G(t)$.
\item We let $t_{\mathrm{cycles}} \in D_{i,\emptyset,[i]}$ denote a table such that $G(t_{\mathrm{cycles}})$ contains only cycles.
\item We let $x \in [0,|B|-i]$ denote the number of paths in $G(t)$.
\item Letting $B_{\mathrm{paths}} = B \setminus B_{\mathrm{cycles}}$, we let $G_{\mathrm{paths}}$ denote a graph on the vertices $B_{\mathrm{paths}} \cup v_{\mathrm{center}}$ which consists of $x$ disjoint paths to $v_{\mathrm{center}}$.
\item We let $E = (p_1,\ldots,p_x)$ denote the pairs in $P$ which are the endpoints for the $x$ paths in $G_{\mathrm{paths}}$.
\end{enumerate}
\end{enumerate}
\end{lemma}
\begin{proof}
Given a table $t \in D_{n,A,B}$ together with a pairing $P \in \mathcal{P}_{\mathrm{center}}(t)$ on the center columns, it is easy to find $C_A$, $C_B$, $i$, $B_{\mathrm{cycles}}$, $x$, and $G_{\mathrm{paths}}$. We obtain $t_{\mathrm{center}}$, $t_{\mathrm{cycles}}$, and $E$ as follows.
\begin{enumerate}
\item We obtain $t_{\mathrm{cycles}}$ from the entries of the columns containing the elements in $B_{\mathrm{cycles}}$. More precisely, let $c'_1,\ldots,c'_i$ be the columns containing the elements of $B_{\mathrm{cycles}}$ and let $j'_k$ be the element which appears four times in column $c'_k$ (note that $j'_1,\ldots,j'_i$ are not necessarily in increasing order). If the entry in row $r$ and column $c'_k$ is $j'_y$ then we take the entry in row $r$ and column $k$ of $t_{\mathrm{cycles}}$ to be $y$. 
\item In order to obtain $t_{\mathrm{center}}$ and $E$, we first modify $t$ as follows. We order the paths in $G_{\mathrm{paths}}$ based on the order of their starting vertices. Let $j_{k1} \to j_{k2} \to \cdots \to j_{k{l_k}} \to v_{\mathrm{center}}$ be the $k$th path in $G_{\mathrm{paths}}$. Let $r_{k1}$ and $r_{k2}$ be the rows of column $c_{j_{k1}}$ which contain an element $z_k \notin A \cup B$ rather than $j_{k1}$. Finally, let $c^{*}_k$ be the center column containing two $j_{k{l_k}}$. We now make the following modifications to $t$.
\begin{enumerate}
\item[1.] We replace the two $z_k$ in rows $r_{k1}$ and $r_{k2}$ of column $c_{j_{k1}}$ with $j_{k1}$.
\item[2.] For all $i \in [l_k-1]$, the two $j_{ki}$ in column $c_{j_{k(i+1)}}$ must be in rows $r_{k1}$ and $r_{k2}$. We replace these two $j_{ki}$ with $j_{k(i+1)}$.
\item[3.] The two $j_{k{l_k}}$ in column $c^{*}_k$ must be in rows $r_{k1}$ and $r_{k2}$. We replace these $j_{k{l_k}}$ with $z_k$. We take $p_k$ to be this pair of elements.
\end{enumerate}
Let $t'$ be the table we obtain after making these modifications for each of the $k$ paths. The center columns of $t'$ only contain elements in $[n] \setminus (A \cup B)$, so we can obtain $t_{\mathrm{center}}$ from $t'$. More precisely, letting $c''_1,\ldots,c''_{n-a-b}$ be the center columns of $t'$ and letting $i'_1,\ldots,i'_{n-a-b}$ be the elements of $[n] \setminus (A \cup B)$, if $t'$ has an element $i'_y$ in row $r$
and column $c'_j$ for some $r \in [6]$ and $j,y \in [n-a-b]$ then we put the element $y$ in row $r$ and column $j$ of $t_{\mathrm{center}}$. We translate $P \in \mathcal{P}_{\mathrm{center}}(t')$ and $E = (p_1,\ldots,p_k)$ from the center columns of $t'$ to $t_{\mathrm{center}}$ accordingly.
\end{enumerate}
Conversely, given a tuple $(C_A,C_B,t_{\mathrm{center}},P,i,B_{\mathrm{cycles}},t_{\mathrm{cycles}},x,G_{\mathrm{paths}},E)$, we construct $t \in D_{n,A,B}$ as follows.
\begin{enumerate}
\item For each $i_k \in A$, We start by putting six $i_k$ in column $c_{i_k}$. Similarly, for each $j_k \in B$, we start by putting six $j_k$ in column $c_{j_k}$.
\item We use $t_{\mathrm{center}}$ to fill in the center columns. More precisely, letting $c''_1 , \ldots, c''_{n-a-b}$ be the center columns and letting $i'_1,\ldots,i'_{n-a-b}$ be the elements of $[n] \setminus (A \cup B)$, if $t_{\mathrm{center}}$ has an element $y \in [n-a-b]$ in row $r$
and column $j$ then we put the element $i'_y$ in row $r$ and column $c''_j$ of $t$.

We translate the pairing $P \in \mathcal{P}(t_{\mathrm{center}})$ and the endpoints $E = (p_1, \ldots, p_x)$ to the center columns accordingly.
\item We use $t_{\mathrm{cycles}}$ to replace the columns of $t$ containing the $i$ elements of $B_{\mathrm{cycles}}$. More precisely, if $c''_1,\ldots,c''_i$ are the columns containing the elements of $B_{\mathrm{cycles}}$ and $j'_k$ is the element in column $c''_k$ (note that $j'_1,\ldots,j'_i$ are not necessarily in increasing order) then if $t_{cylces}$ has an element $y \in [i]$ in row $r$ and column $k$ then we put the element $j'_y$ in row $r$ and column $c''_k$.

We let $t'$ be the table which we obtain after this replacement.
\item We order the paths in $G_{\mathrm{paths}}$ based on their starting vertex. For each $k \in [x]$, let $j_{k1} \to j_{k2} \to \cdots \to j_{kl_{k}} \to v_{\mathrm{center}}$ be the $k$th path in $G_{\mathrm{paths}}$. After translating $E$ to the center columns, $p_k$ is a pair of elements $z_k$ in rows $r_{k1}$ and $r_{k2}$ of column $c^{*}_k$ for some $z_k \in [n] \setminus (A \cup B)$, $c^{*}_k \in \{c''_1,\ldots,c''_{n-a-b}\}$, and $r_{k1},r_{k2} \in [6]$. We obtain $t$ by performing the following modifications to $t'$ for each of the $k$ paths.
\begin{enumerate}
\item[1.] We replace the two $z_k$ in rows $r_{k1}$ and $r_{k2}$ of column $c^{*}_k$ with $j_{k{l_k}}$.
\item[2.] For all $i \in [l_k-1]$, we replace the two $j_{k(i+1)}$ in rows $r_{k1}$ and $r_{k2}$ of column $c_{j_{k(i+1)}}$ with $j_{ki}$.
\item[3.] We replace the two $j_{k1}$ in rows $r_{k1}$ and $r_{k2}$ of column $c_{j_{k1}}$ with $z_k$.
\end{enumerate}
\end{enumerate}
\begin{example}\label{ex:bijection} The figures below illustrate the bijection when $A = \{7\}$, $B = \{1,5,6,8,9\}$, $C_A = (1)$, $C_B = (8,4,7,2,9)$ (as these are the 4-columns for $1,5,6,8,9$ respectively), $i = 2$, $B_{\mathrm{cycles}} = \{8,9\}$, $x = 2$, and $t_{\mathrm{center}}$ $t_{\mathrm{cycles}}$, $G_{\mathrm{paths}}$, and $E$ are as shown below.
\begin{figure}[H]
\centering
\begin{minipage}[b]{0.25\textwidth}
\centering
\begin{tabular}{|c|c|c|}
\hline
\cellcolor[HTML]{FD6864}1 &amp; \cellcolor[HTML]{FD6864}3 &amp; \cellcolor[HTML]{FD6864}2 \\ \hline
\cellcolor[HTML]{FD6864}1 &amp; \cellcolor[HTML]{FFFE65}\circled{\textbf{3}} &amp; \cellcolor[HTML]{FFFE65}2\\ \hline
\cellcolor[HTML]{FFFE65}\circled{\textbf{2}} &amp; \cellcolor[HTML]{FFFE65}\circled{\textbf{3}} &amp; \cellcolor[HTML]{67FD9A}1\\ \hline
\cellcolor[HTML]{FFFE65}\circled{\textbf{2}} &amp; \cellcolor[HTML]{FD6864}3 &amp; \cellcolor[HTML]{67FD9A}1\\ \hline
\cellcolor[HTML]{67FD9A}1 &amp; \cellcolor[HTML]{67FD9A}3 &amp; \cellcolor[HTML]{FFFE65}2\\ \hline
\cellcolor[HTML]{67FD9A}1 &amp; \cellcolor[HTML]{67FD9A}3 &amp; \cellcolor[HTML]{FD6864}2\\ \hline
\end{tabular}
\begin{tabular}{|c|c|}
\hline 
1 &amp; 2 \\
\hline 
2 &amp; 1 \\
\hline 
1 &amp; 2 \\
\hline 
1 &amp; 2 \\
\hline 
2 &amp; 1 \\
\hline 
1 &amp; 2 \\
\hline
\end{tabular}
\caption{$t_{\mathrm{center}}$ and $t_{\mathrm{cycles}}$.}
\end{minipage}
\hfill
\begin{minipage}[b]{0.3\textwidth} \centering
\begin{tikzpicture}[thick, main/.style = {draw,circle}, scale = 0.84]
\node[main] (v) at (0,0) {$v_{\mathrm{center}}$};
\node[main] (5) at (1.5,-0.7) {$5$};
\node[main] (6) at (1.5,0.7) {$6$};
\node[main] (8) at (-1.8,1.4) {$8$};
\node[main] (9) at (-1.8,-0.7) {$9$};
\node[main] (1) at (2.8,1.3) {$1$};
\draw[-&gt;] (8) to [bend right=40] (9);
\draw[-&gt;] (9) to [bend right=40] (8);
\draw[-&gt;] (6) to [bend right=0] (v);
\draw[-&gt;] (1) to [bend right=0] (6);
\draw[-&gt;] (5) to [bend right=0] (v);
\end{tikzpicture}
\caption{$G(t)$.}
\end{minipage}
\hfill
\begin{minipage}[b]{0.4\textwidth}
\centering
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
7 &amp; 8 &amp; \cellcolor[HTML]{FD6864}2  &amp; 5 &amp; \cellcolor[HTML]{FD6864}4 &amp; \cellcolor[HTML]{FD6864}3 &amp; 6 &amp; 1 &amp; 9 \\ \hline
7 &amp; 9 &amp; \cellcolor[HTML]{FD6864}2 &amp; 5 &amp; \cellcolor[HTML]{FFFE65}\circled{\textbf{6}} &amp; \cellcolor[HTML]{FFFE65}3 &amp; 1 &amp; 4 &amp; 8 \\ \hline
7 &amp; 8 &amp; \cellcolor[HTML]{FFFE65}\circled{\textbf{5}} &amp; 3 &amp; \cellcolor[HTML]{FFFE65}\circled{\textbf{6}} &amp; \cellcolor[HTML]{67FD9A}2 &amp; 1 &amp; 4 &amp; 9 \\ \hline
7 &amp; 8 &amp; \cellcolor[HTML]{FFFE65}\circled{\textbf{5}} &amp; 3 &amp; \cellcolor[HTML]{FD6864}4 &amp; \cellcolor[HTML]{67FD9A}2 &amp; 6 &amp; 1 &amp; 9 \\ \hline
7 &amp; 9 &amp; \cellcolor[HTML]{67FD9A}2 &amp; 5 &amp; \cellcolor[HTML]{67FD9A}4 &amp; \cellcolor[HTML]{FFFE65}3 &amp; 6 &amp; 1 &amp; 8 \\ \hline
7 &amp; 8 &amp; \cellcolor[HTML]{67FD9A}2 &amp; 5 &amp; \cellcolor[HTML]{67FD9A}4 &amp; \cellcolor[HTML]{FD6864}3 &amp; 6 &amp; 1 &amp; 9 \\ \hline
\end{tabular}
\caption{The resulting table $t$.}
\end{minipage}
\end{figure}
\end{example}

It is not hard to check that these maps are inverses of each other.
\end{proof}
\end{proof}
This completes the proof of Lemma \ref{lem:Dnab}. Combining Lemmas \ref{lem:inclusionexclusion} and \ref{lem:Dnab} completes the proof of Lemma \ref{main_lemma}.
\end{proof}

We now simplify the terms in Lemma \ref{main_lemma}.
\begin{proposition}\label{prop:H}
\begin{equation*}
    H_{n,j,a,b} = \frac{(3(n-a-b)+j-1)!}{(3(n-a-b)-1)!}.
\end{equation*}
\end{proposition}
\begin{proof}
Originally,
\[
H_{n,j,a,b} = \sum_{x=1}^{j}{\frac{j!}{x!}\binom{j-1}{x-1}\prod_{y=0}^{x-1}(3(n-a-b)-y)}.
\]
Let $z = 3(n-a-b)$. For the inner product, we can write
\[
\prod_{y=0}^{x-1}(3(n-a-b)-y) = \frac{z!}{(z-x)!},
\]
so
\[
H_{n,j,a,b} = \sum_{x=1}^{j}\frac{j!}{x!}\binom{j-1}{x-1}\frac{z!}{(z-x)!} = j! \sum_{x=1}^{j}\binom{j-1}{x-1} \binom{z}{x} = j! \binom{z+j-1}{j}.
\]
The last equality is a special case of the \emph{Chu-Vandermonde Identity}.
\end{proof}
With these simplifications, we can derive an expression for the generating function
\[
F_6(t) = \sum_{n=0}^\infty \frac{t^n}{(n!)^2} f_6(n).
\]
By Lemma \ref{main_lemma},
\[
F_6(t) = \sum_{0 \leq a \leq j \leq n} \frac{t^n}{(n!)^2} \binom{n}{j}\binom{j}{a}(m_6 - 15)^{a} (m_4 - 3)^{(j-a)}D_{n,a,j-a}.
\]
Summing with respect to $b=j-a$ instead of $a$ and observing that 
\begin{align*}
D_{n,a,b}&amp;=\bigg(\prod_{k=0}^{a+b-1}{(n-k)}\bigg)\bigg(\sum_{i=0}^{b}\binom{b}{i}C_{i}H_{n,b-i,a,b}\bigg)P_{n-a-b}\\ 
&amp;= \frac{n!}{(n-j)!}\bigg(\sum_{i=0}^{b}\binom{b}{i}C_{i}H_{n,b-i,j-b,b}\bigg)P_{n-j}\\
&amp;=n!\bigg(\sum_{i=0}^{b}\binom{b}{i}C_{i}H_{n,b-i,j-b,b}\bigg)\frac{(n\!-\!j\!+\!2)!(n\!-\!j\!+\!4)!}{48},
\end{align*}
we have that
\begin{align*}
F_6(t) = \sum_{0 \leq i \leq b \leq j \leq n} &amp;\frac{t^n}{n!} \binom{n}{j}\binom{j}{b}\binom{b}{i}(m_6 - 15)^{(j-b)} (m_4 - 3)^{b}\\
&amp;\frac{(n-j+2)!(n-j+4)!}{48}H_{n,b-i,j-b,b} C_i.
\end{align*}
By Proposition \ref{prop:H}, $H_{n,b-i,j-b,b} = (3 n-3 j+ b-i-1)!/(3 n-3j-1)!$. Using the reparametrization $b = i + s, j = b + r, n = j + q$, where $s,r,q$ goes from $0$ to $\infty$, we get
\begin{align*}
F_6(t) = \sum_{q=0}^\infty \sum_{r=0}^\infty \sum_{s=0}^\infty \sum_{i=0}^\infty \, &amp; \frac{t^{i+s+r+q}}{q!r!s!i!} (m_6 - 15)^r (m_4 - 3)^{(i+s)} \\
&amp;\frac{(q+2)!(q+4)!}{48} \frac{(3q+s-1)!}{(3q-1)!} C_i.
\end{align*}

Grouping the terms to separate the dependence on $r$, $s$, and $i$, we have that $F_6(t)$ equals
\begin{align*}
 &amp;\left(\sum_{q=0}^\infty \frac{t^q}{q!}\frac{(q+2)!(q+4)!}{48}\right) \left(\sum_{r=0}^\infty \frac{t^r (m_6-15)^r}{r!}  \right) \\
 &amp;\left( \sum_{s=0}^\infty \frac{t^s}{s!} \frac{(3q+s-1)!}{(3q-1)!} (m_4 - 3)^s \right) \left(\sum_{i=0}^\infty \frac{t^i}{i!} (m_4 - 3)^i  C_i \right).
\end{align*}
Summing all the inner sums (the rightmost using Lemma \ref{lem:countingcycletables}),
\[
F_6(t) = \sum_{q=0}^\infty \frac{t^q}{q!} \frac{(q+2)!(q+4)!}{48} e^{t(m_6-15)} \frac{1}{(1-t(m_4-3))^{3q}} \frac{e^{-15t(m_4-3)}}{(1-t(m_4-3))^{15}}.
\]

\section{Generalization for an arbitrary third moment}
Restating Proposition \ref{prop:tables}, we can write
\begin{equation*}
    f_6(n) = \sum_{t \in T_{6,n}} \operatorname{sgn}(t)w(t),
\end{equation*}
where $T_{6,n}$ is the set of all permutation tables of length $n$ with six rows whose columns fall in one of the following categories:
\begin{itemize}
    \item $6$-columns: six copies of a single number (weight $m_6$)
    \item $4$-columns: four copies of one number and two copies of a distinct number (weight $m_4$)
    \item $2$-columns: three pairs of distinct numbers (weight $1$)
\end{itemize}
The weight $w(t)$ of a table $t$ is simply the product of the weights of its columns.

When $m_3 \neq 0$, we can have the following type of column in addition to $2$-columns, $4$-columns, and $6$-columns:
\begin{itemize}
    \item $3$-columns: three copies of one number and three copies of a distinct number (weight $m_3^2$)
\end{itemize}
To handle this, we make the following definitions.
\begin{definition}
We define $f^*_{6}(n) = E_{M \sim \mathcal{M}_{n \times n}(\Omega)}[\det (M)^{6}]$ to be the expected value of the sixth power of the determinant of a matrix $M \sim \mathcal{M}_{n \times n}(\Omega)$ (i.e., an $n \times n$ random matrix with entries drawn from $\Omega$) where $\Omega$ is a distribution such that $m_1 = 0$.

We define the formal generating function $F^*_6(t)$ for $f^*_6(n)$ to be $F^*_6(t) = \sum_{n=0}^\infty\frac{t^n}{(n!)^2} f^*_6(n)$.
\end{definition}
\begin{remark}
We write $f^*_{6}(n)$ instead of $f_{6}(n)$ to distinguish this case from the case we previously analyzed where $m_1 = m_3 = 0$.
\end{remark}
\begin{definition}
We define $T^*_{6,n}$ to be the set of all $6 \times n$ tables whose rows are permutations of $[n]$ and whose columns are $2$-columns, $3$-columns, $4$-columns, or $6$-columns.
\end{definition}
Similar to before, we can write $f^*_6(n)$ as the sum over tables in $T^*_{6,n}$ of the contribution from each table.
\begin{proposition}
For all $n \in \mathbb{N}$, $f^*_6(n) = \sum_{t \in T^*_{6,n}} \operatorname{sgn}(t)w(t)$.
\end{proposition}
We can evaluate $f^*_6(n)$ using our expression for $f_6(n)$.
\begin{definition}
We define $f_6(n) = \sum_{t \in T_{6,n}} \operatorname{sgn}(t)w(t) = E_{M \sim \mathcal{M}_{n \times n}(\Omega')}[\det (M)^{6}]$ where $\Omega'$ is the distribution where we first sample $x$ from $\Omega$ and then randomly choose $x$ or $-x$ (so $\Omega$ and $\Omega'$ have the same even moments but the odd moments of $\Omega'$ are $0$).
\end{definition}
\begin{lemma}
For all $n \in \mathbb{N}$,
\begin{equation*}
f^*_6(n) = \sum_{j = 0}^n \binom{n}{j}^2 f_6(n-j) \, j! m_3^{2j} (-1)^j \sum_{\pi \in D_j} (-10)^{C(\pi)}.
\end{equation*}
\end{lemma}
\begin{proof}
We first break each table $t \in T^*_{6,n}$ into tables $s$ and $t'$, where $s$ contains all of the 3-columns of $t$. The signs of these tables are related as
\begin{equation*}
 \operatorname{sgn}(t) = \operatorname{sgn}(s) \operatorname{sgn}(t').
\end{equation*}
\begin{definition}
Given $J \subset [n]$, letting $j = |J|$, we define $T_{6,J}$ to be the set of all $6 \times |J|$ tables whose rows are permutations of $J$ and whose columns are all $2$-columns, $4$-columns, or $6$-columns. We define $Q_{6,J}$ to be the set of all $6 \times |J|$ tables whose rows are permutations of $J$ and whose columns are all $3$-columns.
\end{definition}  
Since the selection of the subset $J \subseteq [n]$ does not depend on the positions of the columns containing the elements of $J$ in table $t$, we can write our sum as
\begin{equation*}
    f^*_6(n) = \sum_{J \subset [n]} \binom{n}{j} \sum_{t' \in T_{6,[n]/J}} \operatorname{sgn}(t')w(t') \sum_{s \in Q_{6,J}} \operatorname{sgn}(s)w(s).
\end{equation*}
No matter which numbers $J$ are selected, as long as we select the same amount of them, the contribution is the same. Hence,
\begin{equation*}
    f^*_6(n) = \sum_{j = 0}^n \binom{n}{j}^2 \sum_{t' \in T_{6,n-j}} \operatorname{sgn}(t')w(t') \sum_{s \in Q_{6,j}} \operatorname{sgn}(s)w(s),
\end{equation*}
where $Q_{6,j} = Q_{6,[j]}$. The first inner sum is simply $f_6(n-j)$. For the second inner sum, by symmetry, we can fix the first permutation in $s$ to be the identity. Since $w(s) = m_3^{2j}$, we get
\begin{equation*}
\sum_{s \in Q_{6,j}} w(s)\operatorname{sgn}(s) = j! m_3^{2j} \sum_{\substack{s \in Q_{6,j} \\ s_1 = \mathrm{id}}} \operatorname{sgn}(s).
\end{equation*}
Similarly as before, we construct a graph $G(s)$ from the numbers in table $s$ and then show $G(s)$ is a derangement. Let $c$ be a number in the first row of a given column of table $s$. Since it is a $3$-column, we denote the other number in the column as $c'$. We construct a graph $G(s)$ for a given table $s$ whose edges are all of these pairs $c \rightarrow c'$.

\begin{example} Example showing the correspondence between a table $s$  where $s_1 = id$ and the corresponding derangement $G(s)$.
\begin{figure}[H]
\centering
\begin{minipage}[c]{0.45\textwidth}
\centering
    \begin{tabular}{|c|c|c|c|c|c|c|c|c|}
    \hline
        1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6 &amp; 7 &amp; 8 &amp; 9 \\ \hline
        3 &amp; 6 &amp; 1 &amp; 4 &amp; 5 &amp; 7 &amp; 2 &amp; 8 &amp; 9 \\
        1 &amp; 2 &amp; 3 &amp; 9 &amp; 4 &amp; 6 &amp; 7 &amp; 5 &amp; 8 \\
        1 &amp; 6 &amp; 3 &amp; 9 &amp; 4 &amp; 7 &amp; 2 &amp; 5 &amp; 8 \\ 
        3 &amp; 2 &amp; 1 &amp; 9 &amp; 4 &amp; 6 &amp; 7 &amp; 5 &amp; 8 \\ 
        3 &amp; 6 &amp; 1 &amp; 4 &amp; 5 &amp; 7 &amp; 2 &amp; 8 &amp; 9 \\
    \hline
    \end{tabular}
\caption{A $6\times 9$ table $s \in Q_{6,9}$.}
\end{minipage}
\hfill
\begin{minipage}{0.5\textwidth}
\centering
\begin{tikzpicture}[thick, main/.style = {draw,circle}, scale = 0.85]
\node[main] (1) at (-2.5,1) {$1$};
\node[main] (3) at (-2.5,-1) {$3$};
\node[main] (2) at (-1,0.8) {$2$};
\node[main] (6) at (1,0.8) {$6$}; 
\node[main] (7) at (0,-1) {$7$};
\node[main] (4) at (2,0) {$4$};
\node[main] (9) at (3,1) {$9$};
\node[main] (8) at (4,0) {$8$};
\node[main] (5) at (3,-1) {$5$};
\draw[-&gt;] (1) to [bend left=40] (3);
\draw[-&gt;] (3) to [bend left=40] (1);
\draw[-&gt;] (2) to [bend left=40] (6);
\draw[-&gt;] (6) to [bend left=40] (7);
\draw[-&gt;] (7) to [bend left=40] (2);
\draw[-&gt;] (9) to [bend left=20] (8);
\draw[-&gt;] (8) to [bend left=20] (5);
\draw[-&gt;] (5) to [bend left=20] (4);
\draw[-&gt;] (4) to [bend left=20] (9);
\end{tikzpicture}

\caption{The associated $G(s)$.
}
\end{minipage}
\end{figure}
\end{example}

\noindent
For the signs, letting $\pi = G(s)$, we have
\begin{equation*}
\operatorname{sgn}(s) = \operatorname{sgn}(\pi) = (-1)^{j-C(\pi)}.
\end{equation*}
Note that since $c$ and $c'$ are always different, the set of all $G(s)$ corresponds to the set $D_j$ of all derangements. Since there are $\binom{5}{2} =10$ possibilities how to arrange the leftover 5 numbers $c$ and $c'$ in each of the 3-columns corresponding to a given cycle of $\pi$, we get
\begin{equation*}
\sum_{s \in Q_{6,j}} w(s)\operatorname{sgn}(s) = j! m_3^{2j} (-1)^j \sum_{\pi \in D_j} (-1)^{C(\pi)} {10}^{C(\pi)}.
\end{equation*}
Putting everything together, we obtain that
\begin{equation*}
    f^*_6(n) = \sum_{j = 0}^n \binom{n}{j}^2 f_6(n-j) \, j! m_3^{2j} (-1)^j \sum_{\pi \in D_j} (-10)^{C(\pi)}.
\end{equation*}
\end{proof}
\begin{corollary}
\begin{equation*}
F_6^*(t) = (1+m_3^2t)^{10} e^{-10m_3^2t} \, F_6(t).
\end{equation*}
\end{corollary}
\begin{proof}
In terms of generating functions,
\begin{equation*}
\begin{split}
    F_6^*(t) &amp; = \sum_{n=0}^\infty \frac{t^n}{n!^2} f^*_6(n) = \sum_{n=0}^\infty\sum_{j = 0}^n \frac{t^{n-j}}{(n-j)!^2} f_6(n-j) \frac{(-m_3^2t)^j}{j!} \sum_{\pi \in D_j} (-10)^{C(\pi)} \\
    &amp; = F_6(t) \sum_{j=0}^\infty\frac{(-m_3^2t)^j}{j!} \sum_{\pi \in D_j} (-10)^{C(\pi)} = F_6(t) \frac{e^{-10m_3^2t}}{(1+m_3^2t)^{-10}}.
\end{split}
\end{equation*}
The final equality is a special case of Lemma \ref{lem:derangement}. Theorem \ref{m3nonzero_thm} follows.
\end{proof}

\section{Asymptotics}
The proof relies directly on the calculus developed by Borinsky \cite{borinsky2018}, enabling us to extract the asymptotic behaviour of coefficients from their factorially divergent generating function. We use the following result from Borinsky \cite{borinsky2018}:
\begin{definition}
We say a formal power series $f(t) = \sum_{n \geq 0} f_n t^n$ is factorially divergent of type $(\alpha,\beta)$, if $f_n \sim \sum_{k=0}^R c_k \alpha^{n+\beta - k} \Gamma(n+\beta-k) $ as $n \to \infty$ for any fixed $R$ integer. We also define an operator $\mathcal{A}^\alpha_\beta$ acting of $f(t)$ such that $(\mathcal{A}^\alpha_\beta f)(t) = \sum_{k\geq 0} c_k t^k$. If moreover $f(t)$ is analytic at $0$, then $(\mathcal{A}^\alpha_\beta f)(t) = 0$.
\end{definition}
\begin{lemma}\label{lem:borinsky}
Let $f(t)$ and $g(t)$ be two factorially divergent power series of type $(\alpha,\beta)$, then
\begin{align*}
(\mathcal{A}^\alpha_\beta (fg))(t) &amp; =  (\mathcal{A}^\alpha_\beta f)(t) g(t) + f(t) (\mathcal{A}^\alpha_\beta g)(t),\text{ and}\\
(\mathcal{A}^\alpha_\beta (f \circ g) )(t) &amp; = f'(g(t)) (\mathcal{A}^\alpha_\beta g)(t) + \Big(\frac{t}{g(t)}\Big)^\beta e^{\frac{1}{\alpha}(\frac{1}{t}-\frac{1}{g(t)})} (\mathcal{A}^\alpha_\beta f )(g(t)),
\end{align*}
where the second equality holds when $g(t) = 1 + t + O(t^2)$.
\end{lemma}

Recall Theorem \ref{m3nonzero_thm}, which states that
\begin{align*}
F_6(t) =(1+m_3^2t)\!{}^{10}\, \frac{e^{t (m_6-10 m_3^2-15m_4 + 30)}}{48 (1+3t-m_4t)^{15}} \sum _{i=0}^{\infty}
   \frac{(1+i) (2+i) (4+i)!t^i}{(1+3t-m_4t){}^{3 i}}.
\end{align*}
Hence, we can write $F_6(t) = h(t)f(g(t))$, where
\begin{align*}
f(t) &amp;= \sum_{i=0}^\infty (1+i)(2+i)(4+i)! t^i,\\ 
g(t) &amp;= \frac{t}{(1+3t-m_4t)^3},\\
h(t) &amp;= (1+m_3^2t)\!{}^{10}\, \frac{e^{t (m_6-10 m_3^2-15m_4 + 30)}}{48 (1+3t-m_4t)^{15}} \\
\end{align*}
are factorially divergent of type $(1,7)$ since
\[
(1+i)(2+i)(4+i)! = \Gamma(i+7)-8\Gamma(i+6)+12\Gamma(i+5)
\]
and $g(t)$ and $h(t)$ are analytic. Thus, by Lemma \ref{lem:borinsky},
\begin{align*}
(\mathcal{A}^1_7 F_6)(t) &amp; = h(t)\Big(\frac{t}{g(t)}\Big)^7 e^{\frac{1}{t}-\frac{1}{g(t)}} (\mathcal{A}^1_7 f )(g(t)) \\
&amp;= h(t) \Big(\frac{t}{g(t)}\Big)^7 e^{\frac{1}{t}-\frac{1}{g(t)}} (1-8 g(t) + 12 g^2(t)).
\end{align*}
Apart from a factor $(n!)^2 e^{3(m_4-3)}/48$, this is our function $C(t)$ from the original statement of Theorem \ref{asym_thm}.

For $\Omega = \{-1,1\}$, the asymptotic expression 
\begin{align*}
f_6(n) \sim \frac{(n!)^3}{48e^6} &amp; \bigg(  n^6+29 n^5+335 n^4+\frac{5861 n^3}{3}+\frac{17944 n^2}{3}+\frac{44036 n}{5}  \\
  &amp; \quad + \frac{167536}{45}-\frac{210176}{63 n} \bigg)
\end{align*}
gives an excellent approximation to $f_6(n)$ for $n \geq 10$. The following figure shows the ratio of this asymptotic expression to the actual value of $f_6(n)$ for $n$ up to $20$.

\begin{figure}[h]
\centering
\includegraphics[scale=1.0]{20ver2.pdf}
\caption{The ratio between the asymptotic expression and $f_6(n)$ for $\Omega = \{-1,1\}$.}
\end{figure}

\section{Acknowledgments}
Zelin Lv and Aaron Potechin were supported by NSF grant CCF-2008920.

\begin{appendix}
\section{Direct proof of Lemma \ref{pairing}}\label{sec:directproof}
\begin{no-lemma} (\textit{Restatement of Lemma \ref{pairing}}).
For all $n \in \mathbb{N}$, $P_{n}=n(n+2)(n+4)P_{n-1}$ where $P_0=1$.
\end{no-lemma}
\begin{proof}
We recursively compute $P_{n} = \sum_{t \in T_{k,n}}{\sgn(t)|\mathcal{P}(t)|}$ based on where the six $n$ are located in $t$. 

We can count the cases where all of the $n$ are in a $6$-column as follows. Given a table $t \in T_{k,n-1}$ and a pairing $P \in \mathcal{P}(t)$, we can obtain a table $t' \in T_{k,n}$ and a pairing $P' \in \mathcal{P}(t)$ by choosing a location for the $6$-column, choosing a pairing for this column, and using $t$ and $P$ to fill in the remainder of $t'$ and $P'$. There are $n$ possible places for the $6$-column, it has $15$ possible pairings, and $\sgn(t') = \sgn(t)$, so this gives a contribution of $15nP_{n-1}$. 

We can count the cases where four of the $n$ are in a $4$-column and two of the $n$ appear in a different column as follows. Given a table $t \in T_{k,n-1}$ and a pairing $P \in \mathcal{P}(t)$, we can obtain a table $t' \in T_{k,n}$ and a pairing $P' \in \mathcal{P}(t)$ with the following steps:
\begin{enumerate}
    \item Choose which column will be the $4$-column containing four of the $n$. We initially put all six $n$ in this column.
    \item Fill in the remaining columns using $t$ and $P$.
    \item Choose one of the $3(n-1)$ pairs in $P$ and swap two of the $n$ with this pair.
    \item Choose a pairing for the remaining four $n$.
\end{enumerate}
There are $n$ possible places for the $4$-column containing four of the $n$, there are $3(n-1)$ pairs in $P$ which can be swapped with two of the $n$, there are $3$ different pairings for the remaining four $n$, and $\sgn(t') = \sgn(t)$, so this gives a contribution of $3(3)n(n-1)P_{n-1} = 9n(n-1)P_{n-1}$.

The trickiest case to analyze is the case when the six $n$ are split into three different columns. The idea for this case is that there is a correspondence between sets of two columns containing pairs of the elements $a,b,c,d,e,f$ and sets of three columns containing pairs of the elements $a,b,c,d,e,f$ where each column also contains a pair of $n$. This correspondence is highly non-trivial and relies on the signs of the permutations.
\begin{definition}
Let $S_1,S_2,S_3,S_4,S_5,S_6$ be six sets such that each set $S_{i}$ contains two of the elements $\{a,b,c,d,e,f\}$ and each element in $\{a,b,c,d,e,f\}$ is contained in two of the sets $S_1,S_2,S_3,S_4,S_5,S_6$.

We define $T_2(S_1,S_2,S_3,S_4,S_5,S_6)$ to be the set of $6 \times 2$ tables $t$ such that the $i$th row contains the elements in $S_i$ and each element appears an even number of times in each column. Similarly, we define $T_3(S_1,S_2,S_3,S_4,S_5,S_6)$ to be the set of $6 \times 3$ tables $t$ such that the $i$th row contains the elements in $S_i \cup \{n\}$ and each element appears an even number of times in each column.

For each $t \in T_2(S_1,S_2,S_3,S_4,S_5,S_6)$, we define $\sgn(t)$ to be the product of the signs of the rows of $t$ where row $i$ of $t$ has sign $1$ if the elements of $S_i$ appear in order and sign $-1$ if the elements of $S_i$ appear out of order. Similarly, for each $t \in T_3(S_1,S_2,S_3,S_4,S_5,S_6)$, we define $\sgn(t)$ to be the product of the signs of the rows of $t$ where row $i$ of $t$ has sign $1$ if it takes an even number of swaps to transform it into $S_i \cup \{n\}$ and $-1$ if it takes an odd number of swaps to transform it into $S_i \cup \{n\}$.
\end{definition}
\begin{lemma}\label{lem:correspondence}
For all possible $S_1,S_2,S_3,S_4,S_5,S_6$, 
\[
\sum_{t \in T_3(S_1,S_2,S_3,S_4,S_5,S_6)}{\sgn(t)} = 6\sum_{t \in T_2(S_1,S_2,S_3,S_4,S_5,S_6)}{\sgn(t)}.
\]
\end{lemma}
\begin{corollary} For all $n \in \mathbb{N}$,
\[
\sum_{t \in T_{6,n}: n \text{ appears in } 3 \text{ different columns}}{\sgn(t)|\mathcal{P}(t)|} = n(n-1)(n-2)P_{n-1}.
\]
\end{corollary}
\begin{proof}
Recall that
\[
\sum_{t \in T_{6,n-1}}{\sgn(t)|\mathcal{P}(t)|} = P_{n-1}.
\]
We now apply Lemma \ref{lem:correspondence} to the first two columns of the pairs $(t,P)$ where $t \in T_{6,n-1}$ and $P \in \mathcal{P}(t)$. To do this, we use $P$ to relabel the elements in the first two columns as $a,b,c,d,e,f$. One way to do this is as follows. We go through the rows one by one and assign the next unused label(s) to the element(s) which whose pair has not yet appeared. If there are two such elements, we assign the first unused label to the lower element and the next unused label to the higher element. If both elements are the same, we assign the first unused label to the column where the pair of this element appears first. If there is still a tie, we assign the same label to both elements and skip the next label. Lemma \ref{lem:correspondence} still holds in this case as having $S_{i} = S_{j} = \{a,a\}$ instead of $S_{i} = S_{j} = \{a,b\}$ divides both sides by $2$.

After doing this relabeling, for each $i \in [6]$, we take $S_i$ to be the first two elements in row $i$. Applying Lemma \ref{lem:correspondence}, we obtain tables $t'$ and pairings $P'$ by taking $P'$ to be the unique pairing for each column and inverting the labeling of the elements in the first two columns of $t$ by $\{a,b,c,d,e,f\}$. This implies that whenever $n \geq 3$,
\[
\sum_{t \in T_{6,n}: n \text{ appears in the first three columns}}\!\!\!\!{\sgn(t)|\mathcal{P}(t)|} = 6\!\!\sum_{t \in T_{6,n-1}}{\sgn(t)|\mathcal{P}(t)|} = 6P_{n-1}.
\]
There are $\binom{n}{3} = \frac{n(n-1)n-2)}{6}$ possibilities for which three columns contain $n$ so we have that 
\[
\sum_{t \in T_{6,n}: n \text{ appears in } 3 \text{ different columns}}{\sgn(t)|\mathcal{P}(t)|} = n(n-1)(n-2)P_{n-1},
\]
as needed.
\end{proof}

Summing these three cases up, we have
\begin{align*}
  P_n&amp;=15nP_{n-1}+9n(n-1)P_{n-1}+n(n-1)(n-2)P_{n-1}\\  
  &amp;=(n^3 + 6 n^2 + 8 n)P_{n-1} = n(n+2)(n+4)P_{n-1}.
\end{align*}
\end{proof}
We now prove Lemma \ref{lem:correspondence}.
\begin{proof}[Proof of Lemma  \ref{lem:correspondence}]
Up to permutations of the rows and $\{a,b,c,d,e,f\}$, we have the following four cases for $S_1,S_2,S_3,S_4,S_5,S_6$:
\begin{enumerate}
    \item $S_1 = S_2 = \{a,b\}$, $S_3 = S_4 = \{c,d\}$, and $S_5 = S_6 = \{e,f\}$.
    \item $S_1 = S_2 = \{a,b\}$, $S_3 = \{c,d\}$, $S_4 = \{c,e\}$, $S_5 = \{d,f\}$, and $S_6 = \{e,f\}$.
    \item $S_1 = \{a,b\}$, $S_2 = \{a,c\}$, $S_3 = \{b,d\}$, $S_4 = \{d,e\}$, $S_5 = \{c,f\}$, and $S_6 = \{e,f\}$.
    \item $S_1 = \{a,b\}$, $S_2 = \{a,c\}$, $S_3 = \{b,c\}$, $S_4 = \{d,e\}$, $S_5 = \{d,f\}$, and $S_6 = \{e,f\}$.
\end{enumerate}
We can see that these are the only possibilities as follows. If we construct a multi-graph where the vertices are $\{a,b,c,d,e,f\}$ and the edges are $\{S_1, S_2, S_3, S_4, S_5, S_6\}$ then in this multi-graph, every vertex will have degree $2$. 
\begin{enumerate}
    \item If there is a cycle of length $2$ then for the remaining $4$ vertices, we will either have two more cycles of length $2$ or a cycle of length $4$. This gives cases 1 and 2.
    \item If there is a cycle of length $3$ then there must be another cycle of length $3$ on the remaining vertices. This gives case 4.
    \item If there are no cycles of length $2$ or $3$ then we must have a cycle of length $6$. This gives case 3.
\end{enumerate}

For the first three cases, we notice that $T_2(S_1,S_2,S_3,S_4,S_5,S_6)$ is nonempty as shown by the examples below. For the fourth case, we notice that $T_2(S_1,S_2,S_3,S_4,S_5,S_6)$ is empty.
\[
\begin{Bmatrix}
    a&amp;b\\a&amp;b\\c &amp; d\\c &amp; d\\e &amp; f\\e &amp; f \end{Bmatrix}, 
\begin{Bmatrix}
    a&amp;b\\a&amp;b\\c &amp; d\\c &amp; e\\f &amp; d\\f &amp; e \end{Bmatrix}, 
\begin{Bmatrix}
    a&amp;b \\a&amp;c \\d &amp; b\\d &amp; e\\f &amp; c\\f&amp; e
\end{Bmatrix}
\]
For all four cases, $T_3(S_1,S_2,S_3,S_4,S_5,S_6)$ is nonempty as shown by the examples below.
\[
\begin{Bmatrix}
    a&amp;b &amp; n\\a&amp;b &amp; n\\c&amp;n &amp; d\\c&amp;n &amp; d\\n&amp;e &amp; f\\n&amp;e &amp; f\end{Bmatrix}, 
\begin{Bmatrix}
    a&amp;b &amp; n\\a&amp;b &amp; n\\c&amp;n &amp; d\\c&amp;n &amp; e\\n&amp;f &amp; d\\n&amp;f&amp; e
\end{Bmatrix}, 
\begin{Bmatrix}
    a&amp;b &amp; n\\a&amp;c &amp; n\\n&amp;b &amp; d\\e&amp;n &amp; d\\n&amp;c &amp; f\\e&amp;n&amp; f
\end{Bmatrix},
\begin{Bmatrix}
    a&amp;b &amp; n\\a&amp;n &amp; c\\n&amp;b &amp; c\\d&amp;e &amp; n\\d&amp;n &amp; f\\n&amp;e&amp; f
\end{Bmatrix}
\]
We now show that for each of the four cases, 
\[
\sum_{t \in T_3(S_1,S_2,S_3,S_4,S_5,S_6)}{\sgn(t)} = 6\sum_{t \in T_2(S_1,S_2,S_3,S_4,S_5,S_6)}{\sgn(t)}.
\]
\begin{enumerate}
    \item For the first case, $|T_2(S_1,S_2,S_3,S_4,S_5,S_6)| = 8$ as we can choose the order of $\{a,b\}$ in row $1$, the order of $\{c,d\}$ in row $3$, and the order of $\{e,f\}$ in row $5$. All $t \in T_2(S_1,S_2,S_3,S_4,S_5,S_6)$ have positive sign as rows 2, 4, and 6 must be the same as rows 1, 3, and 5. Thus, $\sum_{t \in T_2(S_1,S_2,S_3,S_4,S_5,S_6)}{\sgn(t)} = 8$.
    
    To analyze $T_3(S_1,S_2,S_3,S_4,S_5,S_6)$, observe that there are $6$ choices for the positions of the $n$ in rows $1$, $3$, and $5$ and we can again choose the order of $\{a,b\}$ in row $1$, the order of $\{c,d\}$ in row $3$, and the order of $\{e,f\}$ in row $5$. Thus, $|T_3(S_1,S_2,S_3,S_4,S_5,S_6)| = 48$. All $t \in T_3(S_1,S_2,S_3,S_4,S_5,S_6)$ have positive sign as rows 2, 4, and 6 must be the same as rows 1, 3, and 5. Thus, $\sum_{t \in T_3(S_1,S_2,S_3,S_4,S_5,S_6)}{\sgn(t)} = 48$.
    \item For the second case, $|T_2(S_1,S_2,S_3,S_4,S_5,S_6)| = 4$ as we can choose the order of $\{a,b\}$ in row $1$ and the order of $\{c,d\}$ in row $3$ and this uniquely determines the rest of the table. It can be checked that all $t \in T_2(S_1,S_2,S_3,S_4,S_5,S_6)$ have positive sign. Hence, we have that $\sum_{t \in T_2(S_1,S_2,S_3,S_4,S_5,S_6)}{\sgn(t)} = 4$.
    
    To analyze $T_3(S_1,S_2,S_3,S_4,S_5,S_6)$, observe that there are $6$ choices for the order of $\{a,b,n\}$ in row $1$. Once this order is chosen, there are two choices for the position of the $n$ in row $3$ and two choices for the order of $\{c,d\}$ in row $3$. It can be checked that this uniquely determines the rest of the table and all $t \in T_3(S_1,S_2,S_3,S_4,S_5,S_6)$ have positive sign so we have that $\sum_{t \in T_3(S_1,S_2,S_3,S_4,S_5,S_6)}{\sgn(t)} = 24$.
    \item For the third case, $|T_2(S_1,S_2,S_3,S_4,S_5,S_6)| = 2$ as we can choose the order of $\{a,b\}$ in row $1$ and this uniquely determines the rest of the table. Both $t \in T_2(S_1,S_2,S_3,S_4,S_5,S_6)$ have negative sign so we have that $\sum_{t \in T_2(S_1,S_2,S_3,S_4,S_5,S_6)}{\sgn(t)} = -2$.
    
    For $T_3(S_1,S_2,S_3,S_4,S_5,S_6)$, there are $6$ choices for the order of $\{a,b,n\}$ in row $1$. When row $1$ is $a,b,n$, we have the following four tables:
    \[
    \begin{Bmatrix}
    a&amp;b &amp; n\\a&amp;c &amp; n\\n&amp;b &amp; d\\e&amp;n &amp; d\\n&amp;c &amp; f\\e&amp;n&amp; f
    \end{Bmatrix},
    \begin{Bmatrix}
    a&amp;b &amp; n\\a&amp;n &amp; c\\d&amp;b &amp; n\\d&amp;n &amp; e\\n&amp;f &amp; c\\n&amp;f&amp; e
    \end{Bmatrix},
    \begin{Bmatrix}
    a&amp;b &amp; n\\a&amp;n &amp; c\\n&amp;b &amp; d\\e&amp;n &amp; d\\n&amp;f &amp; c\\e&amp;f&amp; n
    \end{Bmatrix},
    \begin{Bmatrix}
    a&amp;b &amp; n\\a&amp;n &amp; c\\n&amp;b &amp; d\\n&amp;e &amp; d\\f&amp;n &amp; c\\f&amp;e&amp; n
    \end{Bmatrix}
    \]
    Of these tables, the first, second, and fourth table have negative sign while the third table has positive sign so the net contribution is $-2$. Multiplying this by $6$, we have that 
    \[
    \sum_{t \in T_3(S_1,S_2,S_3,S_4,S_5,S_6)}{\sgn(t)} = -12.
    \]
    \item For the fourth case, $T_2(S_1,S_2,S_3,S_4,S_5,S_6)$ is empty because each column can only contain one of $\{a,b,c\}$ and one of $\{b,c,d\}$.
    
    To analyze $T_3(S_1,S_2,S_3,S_4,S_5,S_6)$, observe that we can choose the order of $\{a,b,n\}$ in row 1 and the order of $\{d,e,n\}$ in row 4 and this uniquely determines the rest of the table. The sign of each table will be the product of the sign for row $1$ and the sign for row $4$, so we have the same number of tables with positive and negative sign and thus 
    $\sum_{t \in T_2(S_1,S_2,S_3,S_4,S_5,S_6)}{\sgn(t)} = \sum_{t \in T_3(S_1,S_2,S_3,S_4,S_5,S_6)}{\sgn(t)} = 0$.
\end{enumerate}
\end{proof}
\end{appendix}

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\bigskip
\hrule
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\noindent 2020 {\it Mathematics Subject Classification}:
Primary 05A15; Secondary 05A16, 05A19, 05C30. 

\noindent \emph{Keywords: } moment, random determinant,
random matrix theory, generating function, asymptotic analysis.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences 
\seqnum{A000142},
\seqnum{A052127}, and
\seqnum{A357571}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received October 3 2022; 
revised versions received February 24 2023; June 13 2023; June 15 2023.
Published in {\it Journal of Integer Sequences}, June 17 2023.

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\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
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