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\documentclass[finalversion]{FPSAC2025}
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\title[]{On \texorpdfstring{$e$}{e}-positivity of trees and connected partitions}

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\author{\texorpdfstring{Foster Tom \thanks{\href{mailto:ftom@mit.edu}{ftom@mit.edu}.}\addressmark{1}}{Foster Tom }}

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\address{\addressmark{1}Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA}

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%% put your English abstract here, or comment this out if you don't have one yet
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\abstract{We prove that a tree with a vertex of degree at least five must be missing a connected partition of some type and therefore its chromatic symmetric function cannot be $e$-positive. We prove that this also holds for a tree with a vertex of degree four as long as it is not adjacent to any leaf. This brings us very close to the conjecture by Dahlberg, She, and van Willigenburg of non-$e$-positivity for all trees with a vertex of degree at least four. We also prove that spiders with four legs cannot have an $e$-positive chromatic symmetric function.}

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\keywords{chromatic symmetric function, connected partition, cut vertex, e-positive, proper colouring, spider graph, Stanley--Stembridge conjecture, tree isomorphism conjecture}

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\begin{document}

\maketitle
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\section{Introduction}\label{section:introduction}

Stanley \cite{chromsym} defined the \emph{chromatic symmetric function} of a graph $G=(V,E)$ to be 
\begin{equation}
X_G=\sum_{\substack{\kappa:V\to\{1,2,3,\ldots\}\\\text{ if }ij\in E,\text{ then }\kappa(i)\neq\kappa(j)}}\prod_{v\in V}x_{\kappa(v)}.
\end{equation}
There are two major problems in the study of chromatic symmetric functions. The first is the Stanley--Stembridge conjecture \cite{stanstem}, which asserts that unit interval graphs $G$ are \emph{$e$-positive}, meaning that $X_G$ is a positive sum of elementary symmetric functions. This problem is connected to positivity of immanants of Jacobi--Trudi matrices \cite{stanstem}, cohomology of Hessenberg varieties \cite{chromhesssplitting, dothessenberg, hessenberghopf, stanstemhess, chromquasihessenberg}, and characters of Hecke algebras \cite{chromhecke, chromcharacters}. Hikita \cite{stanstemproof} announced a proof of the Stanley--Stembridge conjecture by finding a positive $e$-expansion in terms of probabilities associated to standard Young tableaux. However, a $q$-analogue of this problem proposed by Shareshian and Wachs \cite{chromposquasi} is still open.

The second is the tree isomorphism conjecture, which asserts that nonisomorphic trees $T$ and $T'$ must have $X_T\neq X_{T'}$. This is intriguing because all trees with $n$ vertices have the same chromatic polynomial. Many authors proved that $X_T$ determines certain invariants of a tree and therefore it distinguishes particular subclasses of trees \cite{chromtreepolyinv, markedgraphs, propcatchrom, propqcatchrom, algochromtrees, catchromiso, distinguishingtrees, chromequal}. For example, \emph{spiders} $S(\lambda)$, which consist of paths of lengths $\lambda_1,\ldots,\lambda_\ell$ joined at a single vertex, are distinguished by their chromatic symmetric function \cite{distinguishingtrees}. 

Dahlberg, She, and van Willigenburg \cite{chromspostrees} considered a hybrid of these two questions by asking which trees $T$ are $e$-positive. Many authors proved $e$-positivity or non-$e$-positivity for particular subclasses of trees \cite{eposclawcon, chromspostrees, spider4m22m1, spiderbroom, spiderepos}. Dahlberg, She, and van Willigenburg checked the following conjecture for all trees with at most $12$ vertices.

\begin{conjecture}[{\cite[Conjecture 6.1]{chromspostrees}}] \label{conj:four} 
    If $T$ is a tree with a vertex of degree $d\geq 4$, then it is not $e$-positive.
\end{conjecture} 

Wolfgang found a powerful necessary condition for $e$-positivity. Let $G=(V,E)$ be an $n$-vertex graph and let $\lambda\vdash n$ be an integer partition of size $n$. A \emph{connected partition of $G$ of type $\lambda$} is a set partition $\mathcal S=\{S_1,\ldots,S_\ell\}$ of $V$ such that 
\begin{itemize}
\item the induced subgraph $G[S_i]$ is connected for every $1\leq i\leq\ell$, and
\item $|S_i|=\lambda_i$ for every $1\leq i\leq\ell$.
\end{itemize} 

\begin{lemma}[{\cite[Proposition 1.3.3]{conparts}}] A connected $n$-vertex graph $G$ must have a connected partition of type $\lambda$ for every $\lambda\vdash n$ in order to be $e$-positive.
\end{lemma}

\begin{example}
\Cref{fig:spider} shows some spiders and their chromatic symmetric functions. In particular, Wolfgang's necessary condition for $e$-positivity is not sufficient.
\end{example}

Dahlberg, She, and van Willigenburg proved the following result. 

\begin{theorem}[{\cite[Theorem 4.1]{chromspostrees}}] If $G$ is an $n$-vertex connected graph with a cut vertex whose deletion produces a graph with $d\geq\log_2 n+1$ connected components, then $G$ is missing a connected partition of some type and therefore is not $e$-positive.
\end{theorem}

In particular, if $T$ is an $n$-vertex tree with a vertex of degree $d\geq\log_2n+1$, then $T$ is not $e$-positive. Zheng strengthened this result considerably.

\begin{theorem}[{\cite[Theorem 3.13]{spiderepos}}] If $G$ is a connected graph with a cut vertex whose deletion produces a graph with at least $6$ connected components, then $G$ is missing a connected partition of some type.
\end{theorem}

Our main result lowers this bound further.

\begin{theorem}[{\cite[Theorem 1.5]{treefive}}]\label{thm:main}
Let $G$ be a connected graph with a cut vertex whose deletion produces a graph with either at least $5$ connected components, or $4$ connected components that each have at least two vertices. Then $G$ is missing a connected partition of some type.
\end{theorem}

\begin{corollary} If $T$ is a tree with a vertex of degree at least $5$, or with a vertex of degree $4$ that is not adjacent to any leaf, then $T$ is missing a connected partition of some type and therefore is not $e$-positive.
\end{corollary}

\begin{figure}\caption{\label{fig:spider} $S(1,1,1)$: missing a connected partition of type $(2,2)$, not $e$-positive. $S(3,2,1)$: has a connected partition of every type $\lambda\vdash 7$ and is $e$-positive. \textcolor{white}{aaaaaaaaaaaa} $S(4,1,1)$: has a connected partition of every type $\lambda\vdash 7$ but is not $e$-positive.}
$$
\begin{tikzpicture}
\draw (0.5,-0.5) node {$S(1,1,1)$};
\filldraw (0,0) circle (3pt) node[align=center,below] (1){};
\filldraw (0.5,0) circle (3pt) node[align=center,below] (2){};
\filldraw (0.5,0.5) circle (3pt) node[align=center,below] (3){};
\filldraw (1,0) circle (3pt) node[align=center,below] (4){};
\draw (0,0)--(1,0) (0.5,0)--(0.5,0.5);
\end{tikzpicture}
\hspace{30pt}
\begin{tikzpicture}
\draw (1.25,-0.5) node {$S(3,2,1)$};
\filldraw (0,0) circle (3pt) node[align=center,below] (1){};
\filldraw (0.5,0) circle (3pt) node[align=center,below] (2){};
\filldraw (1,0) circle (3pt) node[align=center,below] (3){};
\filldraw (1.5,0) circle (3pt) node[align=center,below] (4){};
\filldraw (1.5,0.5) circle (3pt) node[align=center,above] (5){};
\filldraw (2,0) circle (3pt) node[align=center,below] (6){};
\filldraw (2.5,0) circle (3pt) node[align=center,below] (7){};
\draw (0,0)--(2.5,0) (1.5,0)--(1.5,0.5);
\end{tikzpicture}
\hspace{30pt}
\begin{tikzpicture}
\draw (1.25,-0.5) node {$S(4,1,1)$};
\filldraw (0,0) circle (3pt) node[align=center,below] (1){};
\filldraw (0.5,0) circle (3pt) node[align=center,below] (2){};
\filldraw (1,0) circle (3pt) node[align=center,below] (3){};
\filldraw (1.5,0) circle (3pt) node[align=center,below] (4){};
\filldraw (2,0.5) circle (3pt) node[align=center,above] (5){};
\filldraw (2,0) circle (3pt) node[align=center,below] (6){};
\filldraw (2.5,0) circle (3pt) node[align=center,below] (7){};
\draw (0,0)--(2.5,0) (2,0)--(2,0.5);
\end{tikzpicture}$$
\begin{align*}
X_{S(1,1,1)}&=e_{211}\textcolor{red}{-2e_{22}}+5e_{31}+4e_4\\X_{S(3,2,1)}&=e_{2221}+2e_{3211}+5e_{322}+4e_{331}+12e_{421}+5e_{43}+4e_{511}+13e_{52}+11e_{61}+7e_7\\
X_{S(4,1,1)}&=e_{2221}+4e_{3211}\textcolor{red}{-3e_{322}}+10e_{331}+10e_{421}+17e_{43}+4e_{511}+3e_{52}+11e_{61}+7e_7
\end{align*}
\end{figure}

It turns out that there exists a tree $T$ with a vertex of degree $4$ that is adjacent to a leaf, and which has a connected partition of every type. Therefore, in this sense, \Cref{thm:main} cannot be improved. Specifically, for every $m\geq 1$ the spider graph $S(6m,6m-2,1,1)$ has a connected partition of every type \cite[Proposition 3.19]{treefive}, so there are infinitely many examples. This means that to prove non-$e$-positivity of trees with a vertex of degree $4$ in general, we need to know more about the tree to calculate particular coefficients. In the specific case of spiders, we will prove that \Cref{conj:four} holds.

\begin{theorem}[{\cite[Theorem 1.6]{treefive}}] \label{thm:spider4}
If $S$ is a spider with four legs, then it is not $e$-positive.
\end{theorem}

In the next section, we discuss how \Cref{thm:main} and \Cref{thm:spider4} can be proven by cleverly choosing the parts of our integer partitions.

\section{Proof of main result}\label{section:main}

We first fix notation that we will use for the remainder of this extended abstract. Let $G$ be an $n$-vertex connected graph with a cut vertex $v$ whose deletion produces a graph with at least $3$ connected components $A,B,C_1,\ldots,C_k$, which have sizes
\begin{equation*}
a\geq b\geq c_1\geq\cdots\geq c_k\geq 1,\text{ where }k\geq 1.
\end{equation*} This situation is illustrated in \Cref{fig:graphG}. Let $c=c_1+\cdots+c_k$. We will always assume that $c\geq 2$. We cannot say much when $c=1$ because there are infinite families of $e$-positive spiders of the form $S(a,b,1)$ \cite{eposclawcon, spider4m22m1, spiderbroom}. Note that
\begin{equation*}
n=a+b+c+1\geq 2b+c+1.
\end{equation*}
We now state our main result using this notation.

\begin{figure}\caption{\label{fig:graphG} A graph $G$ with a cut vertex $v$ joining at least $3$ connected components.}
$$
\begin{tikzpicture}
\filldraw (0,0) circle (3pt) node[align=center,below] (1){$v$};
\node[ellipse, minimum height=10mm,minimum width=20mm,draw] at (2,0) {$b$};
\draw (-1,1) circle (10pt) node {$c_1$};
\draw (0,1) circle (10pt) node {$\ldots$};
\draw (1,1) circle (10pt) node {$c_k$};
\node[ellipse, minimum height=10mm,minimum width=20mm,draw] at (-2,0) {$a$};
\node[ellipse, minimum height=10mm, minimum width=40mm,draw, dotted] at (0,1){};
\draw [dotted] (2,1)--(2.5,1);
\draw node at (2.5,1) {$c$};
\draw node at (-4,0) {$G=$};
\draw (0,0)--(1,0) (0,0)--(1.3,0.35) (0,0)--(1.3,-0.35) (0,0)--(-1,0) (0,0)--(-1.3,0.35) (0,0)--(-1.3,-0.35) (0,0)--(0,0.66) (0,0)--(0.1,0.68) (0,0)--(-0.1,0.68) (0,0)--(0.75,0.75) (0,0)--(0.66,0.84) (0,0)--(0.85,0.66) (0,0)--(-0.75,0.75) (0,0)--(-0.66,0.84) (0,0)--(-0.85,0.66);
\end{tikzpicture}
$$
\end{figure}

\begin{theorem}\label{thm:main2}
Suppose that $c\geq 2$. If any of the following hold, then $G$ is missing a connected partition of some type $\lambda\vdash n$.
\begin{enumerate}
\item We have $b\leq 2c-2$.
\item We have $b=2c-1$ and $c\geq c_1+1$.
\item We have $2c\leq b\leq\frac{c^2}2$.
\item We have $b\geq\frac{c^2}2$ and $c\geq c_1+2$.
\end{enumerate}
In particular, if $c\geq c_1+2$, then $G$ is missing a connected partition of some type $\lambda\vdash n$.
\end{theorem}

\begin{remark} The condition that $c\geq c_1+2$ is precisely that the deletion of $v$ produces either at least $5$ connected components, or $4$ connected components that each have at least two vertices, so \Cref{thm:main2} implies \Cref{thm:main}. We cannot remove this hypothesis because for every $m\geq 1$ the spider graph $S(6m,6m-2,1,1)$ has a connected partition of every type $\lambda\vdash 12m+3$. \end{remark}

\begin{remark} The condition that $c\geq c_1+1$ is precisely that the deletion of $v$ produces at least $4$ connected components. We cannot remove this hypothesis because the spider graph $S(5,3,2)$ has a connected partition of every type $\lambda\vdash 11$ and is in fact $e$-positive.\end{remark}

\begin{remark}
We cannot remove the hypothesis that $c\geq 2$ because for every $m\geq 1$ the spider graph $S(m+1,m,1)$ has a connected partition of every type $\lambda\vdash 2m+3$ and is in fact $e$-positive.
\end{remark}

The rest of this abstract is dedicated to proving \Cref{thm:main2}. It turns out that Part 4 is the easiest to prove because the total number of vertices $n$ is large compared to $c$.

\begin{proposition}\label{prop:largen}
If $c\geq c_1+2$ and $b\geq \frac{c^2}2$, then $G$ is missing a connected partition of some type $\lambda\vdash n$. 
\end{proposition}

\begin{proof}
The idea is to take a partition $\lambda\vdash n$ with parts equal to $(c-1)$ or $c$. We first check that this is possible. The well-known Frobenius coin problem \cite{frobeniuscoin} tells us that given integers $x$ and $y$ with $\gcd(x,y)=1$, every integer $n\geq (x-1)(y-1)$ can be written in the form $n=a_1x+a_2y$ for some integers $a_1,a_2\geq 0$. Because $b\geq\frac{c^2}2$, we have 
\begin{equation}
n\geq 2b+c+1\geq c^2+c+1\geq (c-1)(c-2),
\end{equation}
so we can write $n=a_1c+a_2(c-1)$ and take $\lambda$ consisting of $a_1$ $c$'s and $a_2$ $(c-1)$'s. We now show that $G$ is missing a connected partition of type $\lambda$. 

Suppose that $\mathcal S$ is a connected partition of $G$ of type $\lambda$ and let $S\in\mathcal S$ be the subset with $v\in S$. If there is any vertex $u\in C_i$ with $u\notin S$, then the subset $U\in\mathcal S$ with $u\in U$ would have $U\subseteq C_i$ because $v$ is a cut vertex of $G$, and therefore \begin{equation}
|U|\leq |C_i|\leq c_i\leq c_1\leq c-2,
\end{equation} which is impossible because every subset in $\mathcal S$ must have size $(c-1)$ or $c$. Therefore every $C_i\subseteq S$. However, this is also impossible because $v\in S$ and therefore 
\begin{equation}
|S|\geq |C_1|+\cdots+|C_k|+1=c+1,
\end{equation}
and again every subset in $\mathcal S$ must have size $(c-1)$ or $c$. So $G$ is missing a connected partition of type $\lambda$.
\end{proof}

We now describe another way to choose the parts of $\lambda$ so that $G$ is missing such a connected partition.

\begin{lemma}\label{lem:q}
Let $q$ be a positive integer and let $J=\{x,x+1,\ldots,y-1,y\}$, where
\begin{equation}\label{eq:xy}
x=\left\lceil\frac{b+1}q\right\rceil\text{ and }y=\left\lfloor\frac{b+c}q\right\rfloor.
\end{equation}
Also suppose that $x\geq c+1$. If $\lambda\vdash n$ is a partition with all parts in the interval $J$, then $G$ is missing a connected partition of type $\lambda$.
\end{lemma}

\begin{proof}
Suppose that $\mathcal S$ is a connected partition of $G$ of type $\lambda$ and let $S\in\mathcal S$ be the subset with $v\in S$. As before, because $|S|\geq x\geq c+1\geq c_1+1$, we must have every $C_i\subseteq S$. Thus $\mathcal S$ is of the form 
\begin{equation*}
\mathcal S=\{A_1,\ldots,A_i,S,B_1,\ldots,B_j\},
\end{equation*}
where every $A_{i'}\subseteq A$ and every $B_{j'}\subseteq B$. But now, if $j\geq q$, we would have
\begin{equation*}
|B|\geq |B_1|+\cdots+|B_j|\geq jx\geq q\left\lceil\frac{b+1}q\right\rceil\geq b+1,
\end{equation*} 
which is a contradiction. Similarly, if $j\leq q-1$, we would have 
\begin{equation*}|B|+|C_1|+\cdots+|C_k|+1\leq |S|+|B_1|+\cdots+|B_j|\leq(j+1)y\leq q\left\lfloor\frac{b+c}q\right\rfloor\leq b+c,
\end{equation*} again a contradiction. So $G$ is missing a connected partition of type $\lambda$.
\end{proof}

To apply \Cref{lem:q}, it will be useful to generalize the result of the Frobenius coin problem by seeing which numbers can be written as a sum of numbers in an interval $J$.

\begin{lemma}\label{lem:frobint}
Let $x$ and $y$ be positive integers with $x<y$. Then there exists a partition $\lambda\vdash n$ with all parts in the interval $J=\{x,\ldots,y\}$ as long as 
\begin{equation}\label{eq:intervalcoin}
n\geq\left\lceil\frac{x-1}{y-x}\right\rceil x.
\end{equation}
\end{lemma}

\begin{proof}
For a positive integer $t$, let $tJ=\{tx,tx+1,\ldots,ty-1,ty\}$. First note that every integer $n\in tJ$ has a partition $\lambda\vdash n$ consisting of $t$ integers in $J$. Indeed, for $n=ty$ we can take $\lambda$ to be $t$ $y$'s, and for $tx\leq n\leq ty-1$ we can write $n=tq+r$ for some integers $x\leq q\leq y-1$ and $0\leq r\leq t-1$, and take $\lambda$ consisting of $r$ $(q+1)$'s and $(t-r)$ $q$'s.

We now see that as $t$ increases, the intervals $tJ$ become closer to each other. Specifically, for $t\geq\left\lceil\frac{x-1}{y-x}\right\rceil\geq\frac{x-1}{y-x}$, we have $ty+1\geq (t+1)x$, which means that as soon as we exit the interval $tJ$, we are already in the interval $(t+1)J$. This means that every integer $n$ satisfying \eqref{eq:intervalcoin} is in some interval $tJ$ and has a partition $\lambda\vdash n$ with all parts in $J$.
\end{proof}

In fact, this argument shows that we can find an \emph{equitable} partition $\lambda\vdash n$, which means that all of the parts of $\lambda$ are $i$ or $(i+1)$ for some $i$. 

\begin{example}
Let $x=6$ and $y=9$, so that $J=\{6,7,8,9\}$ and we can write the numbers
\begin{equation*}
\{\underbrace{6,7,8,9}_J\}\cup\{\underbrace{12,13,14,15,16,17,18\}}_{2J}\cup\{\underbrace{18,19,20,21,22,23,24,25,26,27}_{3J}\}\cup\cdots
\end{equation*}
as a sum of numbers in $J$. We see that once $t\geq\left\lceil\frac{x-1}{y-x}\right\rceil=2$, we no longer have a gap between the intervals $tJ$ and $(t+1)J$, so we are able to express any integer $n\geq 2x=12$.
\end{example}

By considering the special case of $q=1$ in \Cref{lem:q}, we can now prove Part 1 of \Cref{thm:main2}. The proof of Part 2 is similar, instead using $q=2$, and is omitted.

\begin{proposition}
If $c\geq 2$ and $b\leq 2c-2$, then $G$ is missing a connected partition of some type $\lambda\vdash n$.
\end{proposition}

\begin{proof}
Because $b\leq 2c-2$, we have
\begin{equation*}
\left\lceil\frac b{c-1}\right\rceil(b+1)\leq 2(b+1)\leq 2b+c+1\leq n,
\end{equation*}
so \Cref{lem:frobint} tells us that there exists a partition $\lambda\vdash n$ with all parts in the interval $J=\{b+1,\ldots,b+c\}$. By \Cref{lem:q} with $q=1$ and noting that $x=b+1\geq c+1$, $G$ is missing a connected partition of type $\lambda$.
\end{proof}

Now it remains to prove Part 3, so we assume that $2c\leq b\leq\frac{c^2}2$. Our goal is to find a positive integer $q$ such that $x\geq c+1$ and there exists a partition $\lambda\vdash n$ with all parts in the interval $J$, so that we can apply \Cref{lem:q}. By \Cref{lem:frobint}, this will happen if
\begin{equation}\label{eq:analysisbound}
\left\lceil\frac{x-1}{y-x}\right\rceil x\leq 2b+c+1.\end{equation}

\begin{example}\label{ex:oneq}
Let $b=75$ and $c=14$, so that $n\geq 2b+c+1=165$. We will try to find a value of $q$ for which we can apply \Cref{lem:q}.
\begin{itemize}
\item For $q\geq 6$, we have $x\leq\left\lceil\frac{76}6\right\rceil=13\leq c$, so this will not work.
\item For $q=5$, we have $x=\left\lceil\frac{76}5\right\rceil=16$ and $y=\left\lfloor\frac{89}5\right\rfloor=17$. We have $x\geq c+1$, but 
\begin{equation*}
\left\lceil\frac{x-1}{y-x}\right\rceil x=240>165,
\end{equation*}
so this will not work. For example, we cannot write $n=239$ as a sum of $16$'s and $17$'s.
\item For $q=4$, we have $x=\left\lceil\frac{76}4\right\rceil=19$ and $y=\left\lfloor\frac{89}4\right\rfloor=22$. We have $x\geq c+1$ and 
\begin{equation*}
\left\lceil\frac{x-1}{y-x}\right\rceil x=114\leq 165,
\end{equation*} so this will work. \Cref{lem:frobint} tells us that there exists a partition $\lambda\vdash n$ with all parts in the interval $J=\{19,20,21,22\}$ and \Cref{lem:q} tells us that $G$ will be missing a connected partition of type $\lambda$. 
\end{itemize}
\end{example}

The next example will show that there may be no single value of $q$ for which the inequality \eqref{eq:analysisbound} holds. However, for every value of $n\geq 2b+c+1$, there will exist some $q$ such that there exists a partition $\lambda\vdash n$ with all parts in the corresponding interval $J$.

\begin{example}\label{ex:moreq}
Let $b=24$ and $c=7$, so that $n\geq 2b+c+1=56$.
\begin{itemize}
\item For $q\geq 4$, we have $x\leq 7\leq c$, so this will not work.
\item For $q=3$, we have $x=9$, $y=10$, and $\left\lceil\frac{x-1}{y-x}\right\rceil x=72>56$, so this will not work.
\item For $q=2$, we have $x=13$, $y=15$, and $\left\lceil\frac{x-1}{y-x}\right\rceil x=78>56$, so this will not work.
\item For $q=1$, we have $x=25$, $y=31$, and $\left\lceil\frac{x-1}{y-x}\right\rceil x=100>56$, so this will not work.
\end{itemize}
We find that no single value of $q$ satisfies \eqref{eq:analysisbound}. For $n\geq 72$, we can take $q=3$, but this will not work for every $n\geq 56$. Nevertheless, for each $56\leq n\leq 71$, there is some value of $q$ such that we can find a partition $\lambda\vdash n$ with all parts in the corresponding interval $J$.
\begin{align*}
q=3&:\{\underbrace{54,55,\textcolor{red}{56},\textcolor{red}{57},\textcolor{red}{58},\textcolor{red}{59},\textcolor{red}{60}}_{6J}\}\cup\{\underbrace{\textcolor{red}{63},\textcolor{red}{64},\textcolor{red}{65},\textcolor{red}{66},\textcolor{red}{67},\textcolor{red}{68},\textcolor{red}{69},\textcolor{red}{70}}_{7J}\}\cup\{\underbrace{\textcolor{red}{72},\textcolor{red}{73},\textcolor{red}{74},\textcolor{red}{\ldots}}_{8J}\}\cup\cdots\\
q=2&:\{\underbrace{52,53,54,55,56,57,58,59,60}_{4J}\}\cup\{\underbrace{65,66,67,68,69,70,\textcolor{red}{71},72,73,74,75}_{5J}\}\cup\cdots\\
q=1&:\{\underbrace{50,51,52,53,54,55,56,57,58,59,60,\textcolor{red}{61},\textcolor{red}{62}}_{2J}\}\cup\{\underbrace{75,76,77,\ldots}_{3J}\}\cup\cdots
\end{align*}
\end{example}

It turns out that there exist exactly $65$ $(b,c)$ pairs with $2\leq c\leq 40$ and $2c\leq b\leq\frac{c^2}2$ for which we cannot find a single value of $q$ such that $x\geq c+1$ and the inequality \eqref{eq:analysisbound} holds. Nevertheless, we checked by computer that in each case, for every $n\geq 2b+c+1$ there is some value of $q$ such that we can find a partition $\lambda\vdash n$ with all parts in the corresponding interval $J$.

\begin{proposition}[{ \cite[Proposition 3.13]{treefive}}]
    \label{prop:c40}
If $2\leq c\leq 40$ and $2c\leq b\leq\frac{c^2}2$, then $G$ is missing a connected partition of some type $\lambda\vdash n$.
\end{proposition}

When $c\geq 41$, we can always find a single value of $q$ that works. We make precise estimates, the details of which are omitted. The argument is technical but elementary.

\begin{lemma}[{\cite[Lemma 3.17]{treefive}}]\label{lem:analysis} 
Fix positive integers $b$ and $c$ with $c\geq 41$ and $2c\leq b\leq\frac{c^2}2$. There exists a positive integer $q$ such that, letting $x$ and $y$ be as in \eqref{eq:xy}, we have $x\geq c+1$ and the inequality \eqref{eq:analysisbound} holds.
\end{lemma}

\begin{remark}
We have a rough approximation
\begin{equation*}
\left\lceil\frac{x-1}{y-x}\right\rceil x\approx \frac{b/q}{c/q}\frac bq=\frac{b^2}{cq}.
\end{equation*}
To minimize this, we want $q$ to be large, but to have $x\geq c+1$, we must have $q\leq \frac {b+1}c\approx\frac bc$. For $q\approx\frac bc$, we have $\left\lceil\frac{x-1}{y-x}\right\rceil x\approx b$, so this is promising. However, the floors and ceilings mean that the denominator $(y-x)$ may in fact be closer to $(\frac cq-2)$, and this $(-2)$ term can really hurt us if $q$ is close to $\frac c2$. Therefore, when $b$ is close to $\frac{c^2}2$, we will need to make a clever choice of $q$ and perform careful estimates.
\end{remark}

\begin{proof}[Proof sketch] There are finitely many cases where $41\leq c\leq 499$, which we checked by computer, so we may now assume that $c\geq 500$. If $2c\leq b\leq\frac{c^2}{4.2}$, we can take $q=\left\lfloor\frac bc\right\rfloor$ and we will have $x\geq c+1$ and \begin{equation*}\left\lfloor\frac{x-1}{y-x}\right\rfloor x\leq 1.955b+c\leq 2b+c+1.\end{equation*}
If $\frac{c^2}{4.2}\leq b\leq\frac{c^2}{3.4}$, we can take $q=\lfloor0.46\sqrt b\rfloor$ and we will have $x\geq c+1$ and 
\begin{equation*}\left\lfloor\frac{x-1}{y-x}\right\rfloor x\leq 1.997b\leq 2b+c+1.\end{equation*} If $\frac{c^2}{3.4}\leq b\leq\frac{c^2}2$, then one of $q=\left\lfloor\frac{\sqrt b}{\sqrt{3.5}}\right\rfloor$ or $q=\left\lfloor\frac{\sqrt b}{\sqrt{3.5}}\right\rfloor-1$ will have $x\geq c+1$ and 
\begin{equation*}
\left\lfloor\frac{x-1}{y-x}\right\rfloor x\leq 1.92b\leq 2b+c+1.\end{equation*}
\end{proof}

Now the remaining Part 3 of \Cref{thm:main2} follows from \Cref{prop:c40}, \Cref{lem:analysis}, \Cref{lem:q}, and \Cref{lem:frobint}.

\begin{proposition} If $c\geq 2$ and $2c\leq b\leq\frac{c^2}2$, then $G$ is missing a connected partition of some type $\lambda\vdash n$.
\end{proposition}

We can also prove that spiders with four legs are not $e$-positive.

\begin{proof}[Proof of \Cref{thm:spider4}] Because $S$ has four legs, we have $c\geq c_1+1\geq 2$. If $b\leq\frac{c^2}2$, the result follows from Parts 1, 2, and 3 of \Cref{thm:main2}. If $b\geq \frac{c^2}2$, then 
\begin{equation*}
n\geq 2b+c+1\geq c^2+c+1,
\end{equation*}
and Zheng showed \cite[Corollary 4.6]{spiderepos} that such a spider $S$ is not $e$-positive.
\end{proof}

Because a spider with two legs is a path and is well-known to be $e$-positive \cite[Proposition 5.3]{chromsym}, the problem of classifying $e$-positive spiders is now reduced to classifying $e$-positivity of spiders $S(a,b,c)$ with exactly three legs.

For $c=1$, there are infinite families of $e$-positive and non-$e$-positive spiders of the form $S(a,b,1)$ \cite{eposclawcon, spider4m22m1, spiderbroom}. Although some necessary and sufficient conditions are known in terms of divisibility properties of $a$ and $b$, \Cref{fig:sab1} suggests that a full classification would be difficult.

For $c\geq 2$, we can use \Cref{thm:main2} to rule out many cases and by calculating specific coefficients, we can check many cases by computer. We conjecture that there are in fact only finitely many $e$-positive spiders of the form $S(a,b,c)$ where $c\geq 2$.

\begin{conjecture}\label{conj:sabc}
Let $S=S(a,b,c)$ be a spider with three legs and suppose that $c\geq 2$. Then $S$ is $e$-positive if and only if
\begin{equation*}
(a,b,c)\in\{(5,3,2),(6,4,2),(8,6,2),(9,7,2),(12,4,2),(8,5,3),(14,9,5)\}.
\end{equation*}
\end{conjecture}

We checked \Cref{conj:sabc} up to $n=95$ vertices.

\begin{figure}\caption{\label{fig:sab1} The $e$-positivity of spiders $S(a,b,1)$. Blue: $e$-positive. Red: not $e$-positive.}

\begin{equation*}
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\fill[cyan] (9,3.6) rectangle (9.6,4.2);
\fill[magenta] (9.6,3.6) rectangle (10.2,4.2);
\fill[cyan] (10.2,3.6) rectangle (10.8,4.2);
\fill[magenta] (10.8,3.6) rectangle (11.4,4.2);

\fill[magenta] (0,3) rectangle (0.6,3.6);
\fill[magenta] (0.6,3) rectangle (1.2,3.6);
\fill[magenta] (1.2,3) rectangle (1.8,3.6);
\fill[cyan] (1.8,3) rectangle (2.4,3.6);
\fill[magenta] (2.4,3) rectangle (3,3.6);
\fill[magenta] (3,3) rectangle (3.6,3.6);
\fill[magenta] (3.6,3) rectangle (4.2,3.6);
\fill[magenta] (4.2,3) rectangle (4.8,3.6);
\fill[magenta] (4.8,3) rectangle (5.4,3.6);
\fill[cyan] (5.4,3) rectangle (6,3.6);
\fill[magenta] (6,3) rectangle (6.6,3.6);
\fill[cyan] (6.6,3) rectangle (7.2,3.6);
\fill[magenta] (7.2,3) rectangle (7.8,3.6);
\fill[magenta] (7.8,3) rectangle (8.4,3.6);
\fill[magenta] (8.4,3) rectangle (9,3.6);
\fill[magenta] (9,3) rectangle (9.6,3.6);
\fill[magenta] (9.6,3) rectangle (10.2,3.6);
\fill[magenta] (10.2,3) rectangle (10.8,3.6);
\fill[cyan] (10.8,3) rectangle (11.4,3.6);
\fill[magenta] (11.4,3) rectangle (12,3.6);

\fill[magenta] (0,2.4) rectangle (0.6,3);
\fill[magenta] (0.6,2.4) rectangle (1.2,3);
\fill[magenta] (1.2,2.4) rectangle (1.8,3);
\fill[magenta] (1.8,2.4) rectangle (2.4,3);
\fill[magenta] (2.4,2.4) rectangle (3,3);
\fill[cyan] (3,2.4) rectangle (3.6,3);
\fill[magenta] (3.6,2.4) rectangle (4.2,3);
\fill[magenta] (4.2,2.4) rectangle (4.8,3);
\fill[magenta] (4.8,2.4) rectangle (5.4,3);
\fill[magenta] (5.4,2.4) rectangle (6,3);
\fill[magenta] (6,2.4) rectangle (6.6,3);
\fill[cyan] (6.6,2.4) rectangle (7.2,3);
\fill[magenta] (7.2,2.4) rectangle (7.8,3);
\fill[cyan] (7.8,2.4) rectangle (8.4,3);
\fill[magenta] (8.4,2.4) rectangle (9,3);
\fill[magenta] (9,2.4) rectangle (9.6,3);
\fill[magenta] (9.6,2.4) rectangle (10.2,3);
\fill[magenta] (10.2,2.4) rectangle (10.8,3);
\fill[magenta] (10.8,2.4) rectangle (11.4,3);
\fill[cyan] (11.4,2.4) rectangle (12,3);
\fill[magenta] (12,2.4) rectangle (12.6,3);

\fill[magenta] (0,1.8) rectangle (0.6,2.4);
\fill[magenta] (0.6,1.8) rectangle (1.2,2.4);
\fill[magenta] (1.2,1.8) rectangle (1.8,2.4);
\fill[magenta] (1.8,1.8) rectangle (2.4,2.4);
\fill[magenta] (2.4,1.8) rectangle (3,2.4);
\fill[cyan] (3,1.8) rectangle (3.6,2.4);
\fill[magenta] (3.6,1.8) rectangle (4.2,2.4);
\fill[magenta] (4.2,1.8) rectangle (4.8,2.4);
\fill[magenta] (4.8,1.8) rectangle (5.4,2.4);
\fill[cyan] (5.4,1.8) rectangle (6,2.4);
\fill[magenta] (6,1.8) rectangle (6.6,2.4);
\fill[cyan] (6.6,1.8) rectangle (7.2,2.4);
\fill[magenta] (7.2,1.8) rectangle (7.8,2.4);
\fill[cyan] (7.8,1.8) rectangle (8.4,2.4);
\fill[magenta] (8.4,1.8) rectangle (9,2.4);
\fill[magenta] (9,1.8) rectangle (9.6,2.4);
\fill[magenta] (9.6,1.8) rectangle (10.2,2.4);
\fill[magenta] (10.2,1.8) rectangle (10.8,2.4);
\fill[magenta] (10.8,1.8) rectangle (11.4,2.4);
\fill[magenta] (11.4,1.8) rectangle (12,2.4);
\fill[cyan] (12,1.8) rectangle (12.6,2.4);
\fill[magenta] (12.6,1.8) rectangle (13.2,2.4);

\fill[magenta] (0,1.2) rectangle (0.6,1.8);
\fill[magenta] (0.6,1.2) rectangle (1.2,1.8);
\fill[magenta] (1.2,1.2) rectangle (1.8,1.8);
\fill[magenta] (1.8,1.2) rectangle (2.4,1.8);
\fill[magenta] (2.4,1.2) rectangle (3,1.8);
\fill[cyan] (3,1.2) rectangle (3.6,1.8);
\fill[magenta] (3.6,1.2) rectangle (4.2,1.8);
\fill[magenta] (4.2,1.2) rectangle (4.8,1.8);
\fill[magenta] (4.8,1.2) rectangle (5.4,1.8);
\fill[cyan] (5.4,1.2) rectangle (6,1.8);
\fill[magenta] (6,1.2) rectangle (6.6,1.8);
\fill[cyan] (6.6,1.2) rectangle (7.2,1.8);
\fill[magenta] (7.2,1.2) rectangle (7.8,1.8);
\fill[magenta] (7.8,1.2) rectangle (8.4,1.8);
\fill[magenta] (8.4,1.2) rectangle (9,1.8);
\fill[cyan] (9,1.2) rectangle (9.6,1.8);
\fill[magenta] (9.6,1.2) rectangle (10.2,1.8);
\fill[cyan] (10.2,1.2) rectangle (10.8,1.8);
\fill[magenta] (10.8,1.2) rectangle (11.4,1.8);
\fill[magenta] (11.4,1.2) rectangle (12,1.8);
\fill[magenta] (12,1.2) rectangle (12.6,1.8);
\fill[cyan] (12.6,1.2) rectangle (13.2,1.8);
\fill[magenta] (13.2,1.2) rectangle (13.8,1.8);

\fill[magenta] (0,0.6) rectangle (0.6,1.2);
\fill[magenta] (0.6,0.6) rectangle (1.2,1.2);
\fill[magenta] (1.2,0.6) rectangle (1.8,1.2);
\fill[magenta] (1.8,0.6) rectangle (2.4,1.2);
\fill[magenta] (2.4,0.6) rectangle (3,1.2);
\fill[cyan] (3,0.6) rectangle (3.6,1.2);
\fill[magenta] (3.6,0.6) rectangle (4.2,1.2);
\fill[magenta] (4.2,0.6) rectangle (4.8,1.2);
\fill[magenta] (4.8,0.6) rectangle (5.4,1.2);
\fill[cyan] (5.4,0.6) rectangle (6,1.2);
\fill[magenta] (6,0.6) rectangle (6.6,1.2);
\fill[cyan] (6.6,0.6) rectangle (7.2,1.2);
\fill[magenta] (7.2,0.6) rectangle (7.8,1.2);
\fill[magenta] (7.8,0.6) rectangle (8.4,1.2);
\fill[magenta] (8.4,0.6) rectangle (9,1.2);
\fill[magenta] (9,0.6) rectangle (9.6,1.2);
\fill[magenta] (9.6,0.6) rectangle (10.2,1.2);
\fill[cyan] (10.2,0.6) rectangle (10.8,1.2);
\fill[magenta] (10.8,0.6) rectangle (11.4,1.2);
\fill[magenta] (11.4,0.6) rectangle (12,1.2);
\fill[magenta] (12,0.6) rectangle (12.6,1.2);
\fill[magenta] (12.6,0.6) rectangle (13.2,1.2);
\fill[cyan] (13.2,0.6) rectangle (13.8,1.2);
\fill[magenta] (13.8,0.6) rectangle (14.4,1.2);

\fill[magenta] (0,0) rectangle (0.6,0.6);
\fill[magenta] (0.6,0) rectangle (1.2,0.6);
\fill[magenta] (1.2,0) rectangle (1.8,0.6);
\fill[magenta] (1.8,0) rectangle (2.4,0.6);
\fill[magenta] (2.4,0) rectangle (3,0.6);
\fill[cyan] (3,0) rectangle (3.6,0.6);
\fill[magenta] (3.6,0) rectangle (4.2,0.6);
\fill[magenta] (4.2,0) rectangle (4.8,0.6);
\fill[magenta] (4.8,0) rectangle (5.4,0.6);
\fill[cyan] (5.4,0) rectangle (6,0.6);
\fill[magenta] (6,0) rectangle (6.6,0.6);
\fill[magenta] (6.6,0) rectangle (7.2,0.6);
\fill[magenta] (7.2,0) rectangle (7.8,0.6);
\fill[cyan] (7.8,0) rectangle (8.4,0.6);
\fill[magenta] (8.4,0) rectangle (9,0.6);
\fill[cyan] (9,0) rectangle (9.6,0.6);
\fill[magenta] (9.6,0) rectangle (10.2,0.6);
\fill[cyan] (10.2,0) rectangle (10.8,0.6);
\fill[magenta] (10.8,0) rectangle (11.4,0.6);
\fill[cyan] (11.4,0) rectangle (12,0.6);
\fill[magenta] (12,0) rectangle (12.6,0.6);
\fill[cyan] (12.6,0) rectangle (13.2,0.6);
\fill[magenta] (13.2,0) rectangle (13.8,0.6);
\fill[cyan] (13.8,0) rectangle (14.4,0.6);
\fill[magenta] (14.4,0) rectangle (15,0.6);

\draw (0,0) -- (0,15) (0.6,0)--(0.6,15) (1.2,0)--(1.2,14.4) (1.8,0)--(1.8,13.8) (2.4,0)--(2.4,13.2) (3,0)--(3,12.6) (3.6,0)--(3.6,12) (4.2,0)--(4.2,11.4) (4.8,0)--(4.8,10.8) (5.4,0)--(5.4,10.2) (6,0)--(6,9.6) (6.6,0)--(6.6,9) (7.2,0)--(7.2,8.4) (7.8,0)--(7.8,7.8) (8.4,0)--(8.4,7.2) (9,0)--(9,6.6) (9.6,0)--(9.6,6) (10.2,0)--(10.2,5.4) (10.8,0)--(10.8,4.8) (11.4,0)--(11.4,4.2) (12,0)--(12,3.6) (12.6,0)--(12.6,3) (13.2,0)--(13.2,2.4) (13.8,0)--(13.8,1.8) (14.4,0)--(14.4,1.2);
\draw (0,0) -- (15,0) (0,0.6)--(15,0.6) (0,1.2)--(14.4,1.2) (0,1.8)--(13.8,1.8) (0,2.4)--(13.2,2.4) (0,3)--(12.6,3) (0,3.6)--(12,3.6) (0,4.2)--(11.4,4.2) (0,4.8)--(10.8,4.8) (0,5.4)--(10.2,5.4) (0,6)--(9.6,6) (0,6.6)--(9,6.6) (0,7.2)--(8.4,7.2) (0,7.8)--(7.8,7.8) (0,8.4)--(7.2,8.4) (0,9)--(6.6,9) (0,9.6)--(6,9.6) (0,10.2)--(5.4,10.2) (0,10.8)--(4.8,10.8) (0,11.4)--(4.2,11.4) (0,12)--(3.6,12) (0,12.6)--(3,12.6) (0,13.2)--(2.4,13.2) (0,13.8)--(1.8,13.8) (0,14.4)--(1.2,14.4);
\draw (15,15)--(0,15) (15,15)--(15,0);
\end{tikzpicture}
\end{equation*}
\end{figure}



\acknowledgements{The author would like to think Jos\'e Aliste-Prieto, Richard Stanley, Stephanie van Willigenburg, and Kai Zheng for helpful discussions.}

\printbibliography



\end{document}
