# The Number of Solutions to the Trinomial Thue Equation

Greg Knapp

###### Abstract.

In this paper, we study the number of integer pair solutions to the equation
$|F(x,y)|=1$ where $F(x,y)\in\mathbb{Z}[x,y]$ is an irreducible (over
$\mathbb{Z}$) binary form with degree $n\geqslant 3$ and exactly three nonzero
summands. In particular, we improve Thomas’ explicit upper bounds on the
number of solutions to this equation (see [13]). For instance, when
$n\geqslant 219$, we show that there are no more than 32 integer pair
solutions to this equation when $n$ is odd and no more than 40 integer pair
solutions to this equation when $n$ is even, an improvement on Thomas’ work in
[13], where he shows that there are no more than 38 such solutions when $n$ is
odd and no more than 48 such solutions when $n$ is even.

##  1\. Introduction

Axel Thue, in [14], showed that the equation

$\displaystyle F(x,y)=h$ (1)

has only finitely many integer pair solutions when $F(x,y)\in\mathbb{Z}[x,y]$
is an irreducible (over $\mathbb{Z}$) homogeneous polynomial of degree
$n\geqslant 3$ and $h\in\mathbb{Z}$ (for the purposes of this paper, only
integer pair solutions will be considered). A polynomial which is homogeneous
in two variables is called a _binary form_ and the equation (1) is called
_Thue’s equation_. As a consequence, the equation

$\displaystyle|F(x,y)|\leqslant h$ (2)

also has finitely many solutions. The equation (2) is called _Thue’s
Inequality_. Note that we are using all of the stated hypotheses here: if
$F(x,y)$ has a linear factor for instance (say $rx-sy$ divides $F(x,y)$ for
some relatively prime $r,s\in\mathbb{Z}$), then there are infinitely many
integer pair solutions to $F(x,y)=0$—namely, the integral multiples of the
pair $(s,r)$—and hence, there are infinitely many integer pair solutions to
(2). If $n=\deg(F)$ has $n\leqslant 2$, then the family of Pell equations show
that there can be infinitely many solutions to (2) even when $F(x,y)$ is
irreducible. Thue’s conclusion does hold if every irreducible factor of
$F(x,y)$ has degree at least three, but we assume that $F(x,y)$ is irreducible
for simplicity. Several natural questions arise such as

  1. 1.

How many solutions are there to (2)?

  2. 2.

How large are solutions to (2)?

  3. 3.

On which features of $F$ and $h$ do the solutions to (2) depend?

This paper largely handles the first question, though of course the second and
third questions are related. In particular, the number of nonzero summands of
$F(x,y)$ significantly impacts the number of solutions to (2).***The rough
reasoning for this is as follows: a solution $(p,q)$ to (2) corresponds to a
good rational approximation $p/q$ to a root of $f(X):=F(X,1)$. The only roots
of $f(X)$ which ought to allow good rational approximations are the real roots
of $f(X)$ and it is the number of nonzero summands of $f(X)$ that controls the
number of real roots of $f(X)$, as seen in Lemma 1 of [12]

######  Definition 1.

If a polynomial has exactly two nonzero summands, it is called a _binomial_ ;
if it has exactly three nonzero summands, it is called a _trinomial_ ; and if
it has exactly four nonzero summands, it is called a _tetranomial_.

Since the number of nonzero summands plays such an important role, authors
such as Bennett [2], Evertse [5], Grundman and Wisniewski [6], Hyyrö [7],
Mueller [8], Mueller and Schmidt [9], and Thomas [13] have examined binomial,
trinomial, and tetranomial Thue equations in hopes to get a better handle on
how the number of nonzero summands of $F(x,y)$ affect the number of solutions
to Thue equations.

In this paper, we focus on improving explicit bounds for the number of
solutions to the Thue equation

$\displaystyle|F(x,y)|=1$ (3)

in the particular case that $F(x,y)$ is a trinomial. In this setting, Thomas,
in [13], showed that there are no more than $2v(n)w(n)+8$ distinct integer
pair solutions to $|F(x,y)|=1$ when $F(x,y)\in\mathbb{Z}[x,y]$ is a trinomial
irreducible binary form of degree $n\geqslant 3$ and $w(n)$ and $v(n)$ are
piecewise defined as follows:

$\displaystyle v(n)=\begin{cases}3&\text{if }n\text{ is odd}\\\ 4&\text{if
}n\text{ is even}\end{cases}$

and

$n$ | 5†††There is an error in the proof of Lemma 4.1 in [13]: it is claimed that $\frac{b^{t}-1}{b-1}<b^{t}$ which is not the case for the choice of $b=1.5$ when $n=5$. Tracing this error through to its conclusion, the author believes that this is not correctable. | 6 | 7 | 8 | 9 | 10–11 | 12–16 | 17–37 | $\geqslant 38$
---|---|---|---|---|---|---|---|---|---
$w(n)$ | $27^{\dagger}$ | 16 | 13 | 11 | 9 | 8 | 7 | 6 | 5

We are able to improve the bounds that Thomas provides and we have the
following theorem.

######  Theorem 1.1.

Let $F(x,y)=h_{n}x^{n}+h_{k}x^{k}y^{n-k}+h_{0}y^{n}$ where
$h_{n},h_{k},h_{0},n,k\in\mathbb{Z}$ with $0<k<n$. Suppose that $F(x,y)$ is
irreducible over $\mathbb{Z}[x,y]$ and $n\geqslant 6$. Then there are at most
$2v(n)z(n)+8$ distinct integer pair solutions to the equation $|F(x,y)|=1$
where $v(n)=3$ if $n$ is odd, $v(n)=4$ if $n$ is even, and $z(n)$ is defined
by the following table.

$n$ | $6$ | $7$ | $8$ | $9$ |  $10$–$11$ |  $12$–$16$ |  $17$–$38$ |  $39$–$218$ | $\geqslant 219$
---|---|---|---|---|---|---|---|---|---
$z(n)$ | $15$ | $12$ | $11$ | $9$ | $8$ | $7$ | $6$ | $5$ | $4$

This result is primarily derived from an improvement in efficiency to a
counting technique associated to what is known as the gap principle (see Lemma
2.1).

##  2\. Counting With Gaps

The main technical accomplishment of this paper is the following version of a
counting technique often used in conjunction with the “gap principle.”

######  Lemma 2.1.

Suppose that $L,M,T,p,y_{0},\dots,y_{\ell}\in\mathbb{R}_{>0}$ satisfy the
following conditions:

  1. (1)

$L\leqslant y_{0}\leqslant\dots\leqslant y_{\ell}\leqslant M$

  2. (2)

$p>2$

  3. (3)

$L^{p-2}>T$

  4. (4)

$y_{i+1}\geqslant T^{-1}y_{i}^{p-1}$ for each $0\leqslant i<\ell$

Then

$\ell\leqslant\frac{\log\left[\frac{\log(MT^{-1/(p-2)})}{\log\left(LT^{-1/(p-2)}\right)}\right]}{\log(p-1)}.$

This lemma is comparable to Lemma 1 in [11]. However, by fixing any $L>0$,
$p>2$, $\ell\in\mathbb{Z}_{>0}$, and $0<T<L^{p-2}$, then setting $y_{0}=L$,
$y_{i}=T^{-1}y_{i-1}^{p-1}$, and $M=y_{\ell}$, one can see that this upper
bound is sharp where the upper bound in Lemma 1 of [11] is not.

###### Proof.

By induction, we have

$\displaystyle M$ $\displaystyle\geqslant
y_{\ell}\geqslant\frac{y_{\ell-1}^{p-1}}{T}\geqslant\frac{\left(\frac{y_{\ell-2}^{p-1}}{T}\right)^{p-1}}{T}=\frac{y_{\ell-2}^{(p-1)^{2}}}{T\cdot
T^{p-1}}\geqslant\cdots$
$\displaystyle\cdots\geqslant\frac{y_{0}^{(p-1)^{\ell}}}{T^{\sum_{j=0}^{\ell-1}(p-1)^{j}}}=\frac{y_{0}^{(p-1)^{\ell}}}{T^{\frac{(p-1)^{\ell}-1}{p-2}}}\geqslant\frac{L^{(p-1)^{\ell}}}{T^{\frac{(p-1)^{\ell}-1}{p-2}}}$

and we multiply both sides of

$M\geqslant\frac{L^{(p-1)^{\ell}}}{T^{\frac{(p-1)^{\ell}-1}{p-2}}}$

by $T^{-1/(p-2)}$ to get

$MT^{-1/(p-2)}\geqslant\left(LT^{-1/(p-2)}\right)^{(p-1)^{\ell}}.$

Taking a log on both sides (and using the fact that $LT^{-1/(p-2)}>1$) yields

$\frac{\log\left(MT^{-1/(p-2)}\right)}{\log\left(LT^{-1/(p-2)}\right)}\geqslant(p-1)^{\ell}$

and taking logs again and using the fact that $p>2$ yields the desired
inequality. ∎

##  3\. The Trinomial Thue Equation

In this section, we follow Thomas in [13] for much of our reasoning. However,
we use different notation: the parameters which Thomas calls $u$ and $v$, we
call $a$ and $b$ (this aligns with similar notation used in [1] for instance,
and also makes clear the difference between these parameters—whose choice will
depend on $n$—and the values $u_{n}$ to be defined later and $v(n)$ defined in
Theorem 1.1). The parameters which Thomas calls $b$ and $b_{0}$, we will call
$d$ and $d_{0}$ due to their relation to the degree of $F(x,y)$ and also to
avoid conflict with the newly named $b$.

Throughout the remainder of this section, suppose that
$F(x,y)=h_{n}x^{n}+h_{k}x^{k}y^{n-k}+h_{0}y^{n}$ where
$h_{n},h_{k},h_{0},n,k\in\mathbb{Z}$, $0<k<n$, and $n\geqslant 6$. Suppose
further that $F(x,y)$ is irreducible over $\mathbb{Z}[x,y]$. Let
$H=\max(|h_{n}|,|h_{k}|,|h_{0}|)$ be the naïve height of $F(x,y)$. Any time we
refer to a “solution,” we specifically mean a solution to equation (3) in
$\mathbb{Z}^{2}$.

We will not give a sophisticated bound on the number of solutions $(p,q)$ with
$|pq|\leqslant 1$ and we will consider $(p,q)$ and $(-p,-q)$ to be equivalent
solutions, spurring the following definition.

######  Definition 2.

A pair $(p,q)\in\mathbb{Z}^{2}$ is called _regular_ if $p\neq 0$, $q>0$, and
$|p|\neq q$.

If there are $r$ regular solutions to (3), then there will be at most $2r+8$
distinct solutions since for every solution $(p,q)$ with $|pq|>1$, either
$(p,q)$ or $(-p,-q)$ is regular and there are at most 8 solutions with
$|pq|\leqslant 1$. From this fact and Theorem 3.1 below, Theorem 1.1 will
follow.

######  Theorem 3.1.

Equation (3) has at most $v(n)z(n)$ regular solutions where $v(n)$ and $z(n)$
are defined in Theorem 1.1.

More specifically, let $f(x):=F(x,1)$ and set $R_{F}$ to be the number of real
roots of $f$. We also wish to include certain critical points, so we make the
following definition:

######  Definition 3.

A critical point $\tau\in\mathbb{R}$ of $g(x)\in\mathbb{R}[x]$ is _proper_ if
there exists a neighborhood $U$ of $\tau$ for which
$g^{\prime\prime}(x)g(x)>0$ for all $x\in U\setminus\\{\tau\\}$.

Now let $C_{F}$ to be the number of proper critical points of $f(x)$. Setting
$N_{F}$ to be the number of regular solutions to (3), we will show the
following theorem.

######  Theorem 3.2.

Let $F(x,y)$ be a trinomial of degree $n\geqslant 6$. Then

$N_{F}\leqslant z(n)R_{F}+\ell(n)C_{F}$

where $\ell(n)$ is defined by the following table.

$n$ |  $6$–$7$ | $8$ | $\geqslant 9$
---|---|---|---
$\ell(n)$ | $4$ | $3$ | $2$

We first show that Theorem 3.2 implies Theorem 3.1. Since $\ell(n)$ is less
than $z(n)$, we have that $z(n)R_{F}+\ell(n)C_{F}\leqslant z(n)(R_{F}+C_{F})$
and one can check that that $R_{F}+C_{F}\leqslant v(n)$ with calculus, so we
get Theorem 3.1.

To prove Theorem 3.2, we need some additional setup.

######  Definition 4.

For a polynomial $g(x)\in\mathbb{R}[x]$, an _exceptional point_ of $f$ is
either a real root or a proper critical point of $g(x)$

Let $\mathcal{E}(f)$ be the set of exceptional points,
$\tau_{1}<\tau_{2}<\dots<\tau_{c}$, of $f$. Note that there exist improper
critical points $\eta_{1}<\eta_{2}<\dots<\eta_{c-1}$ so that
$\tau_{1}<\eta_{1}<\tau_{2}<\eta_{2}<\dots<\eta_{c-1}<\tau_{c}$. Setting
$\eta_{0}=-\infty$ and $\eta_{c}=+\infty$, we can define
$J_{1}=(-\infty,\eta_{1})$ and $J_{i}=[\eta_{i},\eta_{i+1})$ for $1\leqslant
i\leqslant c$.

######  Definition 5.

A real number $\rho$ _belongs to_ $\tau_{i}$ (and $\tau_{i}$ _belongs to_
$\rho$) if $\rho\in J_{i}$.

Observe that a regular pair $(p,q)\in\mathbb{Z}^{2}$ with $q\neq 0$ satisfying
(3) corresponds to a rational number $\frac{p}{q}$ satisfying

$\left|f\left(\frac{p}{q}\right)\right|=\frac{1}{q^{n}}.$

Moreover, because $|F(p,q)|=1$ and $F(x,y)\in\mathbb{Z}[x,y]$, $p$ and $q$
must be relatively prime, so this correspondence is one to one. Hence, rather
than counting solutions to (3), we instead count rational solutions to

$\left|f\left(\frac{x}{y}\right)\right|=\frac{1}{y^{n}}$

Thomas, in [13], shows that the number of regular solutions $(p,q)$ of (3) for
which there exists a critical point of $f(x)$, $\tau$, so that $\frac{p}{q}$
belongs to $\tau$ is no larger than $\ell(n)$ (see the completion of the proof
of Thomas’ Theorem 2.2, given after the statement of Theorem 7.1). So it only
remains to show

######  Lemma 3.1.

The number of regular solutions, $(p,q)$, of (3) for which $\frac{p}{q}$
belongs to a real root of $f$ is no larger than $z(n)$.

By Theorem 2.2 in [13], it suffices to show Lemma 3.2 for real roots of $f$
which are greater than 1. Then by Lemma 2.4 of [13], we conclude that any
regular $(p,q)$ for which $\frac{p}{q}$ belongs to an exceptional point
greater than 1 has $p>q\geqslant 1$ and so we may assume that $p>q\geqslant
1$. Defining

$p_{0}(n):=\begin{cases}3&\text{if }6\leqslant n\leqslant 8\\\ 2&\text{if
}n\geqslant 9\end{cases},$

we note that any regular solution $(p,q)$ with $p>q\geqslant 1$ must satisfy

$\displaystyle p\geqslant p_{0}(n)$ (4)

except for possibly $(2,1)$ when $n\leqslant 8$.

######  Definition 6.

A solution, $(p,q)$ to equation (3) with $p>q\geqslant 1$ and $p\geqslant
p_{0}(n)$ is called _special_.

Since at most one solution is not special in the case that $6\leqslant
n\leqslant 8$, it suffices to show the following lemma, which will be our
final reduction.

######  Lemma 3.2.

Let $\alpha>1$ be a real root of $f(x)$. Then the number of special solutions
$(p,q)$ of (3) for which $\frac{p}{q}$ belongs to $\alpha$ is no greater than
$z(n)-1$ if $6\leqslant n\leqslant 8$ and no greater than $z(n)$ if
$n\geqslant 9$.

To prove lemma 3.2, we split solutions into two cases: small and large. For
$F(x,y)$ of degree $n$ and naïve height $H$, we choose a constant
$Y_{F}=H^{\chi_{n}}\cdot e^{\pi_{n}}$ (for some values $\chi_{n}$ and
$\pi_{n}$ to be specified later, but which depend only on $n$) and make the
following definition.

######  Definition 7.

A special solution $(p,q)$ to (3) is _small_ if $q\leqslant Y_{F}$ and is
_large_ otherwise.

###  3.1. Small Special Solutions

One of Thomas’ main achievements in [13] is the following theorem (numbered
4.1 in [13]):

######  Theorem 3.3.

Suppose that $F(x,y)\in\mathbb{Z}[x,y]$ is an irreducible (over $\mathbb{Z}$)
trinomial binary form of degree $n\geqslant 5$ and naïve height $H$. Let
$(p,q)$ and $(p^{\prime},q^{\prime})$ be special solutions to (3) which belong
to a real root and suppose $q^{\prime}>q$. Then

$\displaystyle q^{\prime}>\frac{H^{d/n}p^{n^{*}-d}q^{d}}{K_{d}(n)}$ (5)

where $n^{*}:=\frac{n-2}{2}$, $d$ is chosen to be any real number satisfying
$0\leqslant d\leqslant n^{*}$, and

$K_{d}(n):=m_{n}(r_{n}(1+u_{n}))^{d}$

where

$\displaystyle m_{n}=2\sqrt{\frac{2n}{(n-1)(n-2)}},\qquad
r_{n}=(2.032)^{1/n},\qquad u_{n}=\sqrt{\frac{2}{(n-2)p_{0}^{n}}}.$

This approximation result will be helpful in proving the following
proposition:

######  Proposition 3.1.

Let $\alpha>1$ be a real root of $f(x)$. There are no more than

$\displaystyle
T:=\left\lfloor\max\left(\frac{\log\left(\frac{\chi_{n}}{\frac{d_{0}(d-1)+d}{n(d-1)}}+1\right)}{\log
d},\frac{\log\left(\frac{\pi_{n}}{\log
K_{d}(n)^{-\frac{1}{d-1}}Q_{1}}+1\right)}{\log d}\right)\right\rfloor+2$ (6)

small special solutions $(p,q)$ where $p/q$ belongs to $\alpha$.

###### Proof.

If there less than 2 special solutions $(p,q)$ where $p/q$ belongs to
$\alpha$, then we are done. Otherwise, suppose that there are exactly $t+2$
small special solutions $(p,q)$ where $p/q$ belongs to $\alpha$ and
$t\geqslant 0$. Label those $t+2$ solutions as
$(p_{0},q_{0}),\ldots,(p_{t+1},q_{t+1})$ ordered so that

$1\leqslant q_{0}<q_{1}<\ldots<q_{t+1}\leqslant Y_{F}$

(the strict inequality follows from the fact that the $\frac{p_{i}}{q_{i}}$
are principal convergents to $\alpha$ by corollary 3.2 in [13]).

Choose numbers $d_{0},d\in\mathbb{R}_{>0}$ and make the following definitions:

$\displaystyle c_{0}$ $\displaystyle:=n^{*}-d_{0},$ $\displaystyle K_{0}$
$\displaystyle:=K_{d_{0}}(n),$ $\displaystyle Q_{1}$
$\displaystyle:=\frac{p_{0}(n)^{c_{0}}}{K_{0}}.$

In particular, choose $d$ and $d_{0}$ so that

$\displaystyle 0\leqslant d_{0}\leqslant n^{*}-1.4,$ $\displaystyle
1<d\leqslant n^{*},$ $\displaystyle Q_{1}^{d-1}>\max(1,K_{d}(n)).$ (7)

In the proof of Proposition 3.3 and in the computations in section 3.4, we
show by example that choosing such $d$ and $d_{0}$ are possible.

First, observe that by Theorem 3.3 applied to $q_{1}>q_{0}\geqslant 1$ (and
using the observation that $p_{0}\geqslant p_{0}(n)$), we get
$q_{1}>H^{d_{0}/n}Q_{1}$.

Here is where we depart from Thomas’ method. We now aim to apply Lemma 2.1 to
$H^{d_{0}/n}Q_{1}<q_{1}<q_{2}<\dots<q_{t+1}\leqslant Y_{F}$. In the notation
of Lemma 2.1, we have $L=H^{d_{0}/n}Q_{1}$, $M=Y_{F}$, $p=d+1$, and
$T=\frac{K_{d}(n)}{H^{d/n}}$ (technically, $T$ should have a factor of
$1/p^{n^{*}-d}$, but the result of Lemma 2.1 still holds if $T$ is replaced by
something larger and moreover, we will choose $d=n^{*}$ later, rendering the
difference moot). To apply the conclusion of Lemma 2.1, we need to check that
$p>2$ (trivial based on the fact that $d$ is chosen to be greater than 1) and
we need to check that $L^{p-2}>T$. But this occurs if and only if

$\left(H^{d_{0}/n}Q_{1}\right)^{d-1}>\frac{K_{d}(n)}{H^{d/n}},$

i.e. $H^{(d_{0}(d-1)+d)/n}Q_{1}^{d-1}>K_{d}(n)$, which is guaranteed by (7).

Now applying Lemma 2.1 and using the fact that $t$ is an integer yields

$\displaystyle t$
$\displaystyle\leqslant\Bigg{\lfloor}\frac{\log\left[\frac{\log\left(Y_{F}\left(\frac{K_{d}(n)}{H^{d/n}}\right)^{-\frac{1}{d-1}}\right)}{\log\left(H^{d_{0}/n}Q_{1}\left(\frac{K_{d}(n)}{H^{d/n}}\right)^{-\frac{1}{d-1}}\right)}\right]}{\log
d}\Bigg{\rfloor}$
$\displaystyle=\bigg{\lfloor}\frac{\log\left[\frac{\log\left(Y_{F}K_{d}(n)^{-\frac{1}{d-1}}H^{\frac{d}{n(d-1)}}\right)}{\log\left(K_{d}(n)^{-\frac{1}{d-1}}H^{\frac{d_{0}}{n}+\frac{d}{n(d-1)}}Q_{1}\right)}\right]}{\log
d}\bigg{\rfloor}$
$\displaystyle\leqslant\bigg{\lfloor}\frac{\log\left[\frac{\log(Y_{F})}{\log\left(K_{d}(n)^{-\frac{1}{d-1}}H^{\frac{d_{0}}{n}+\frac{d}{n(d-1)}}Q_{1}\right)}+1\right]}{\log
d}\bigg{\rfloor}.$

where in the last step, we use the fact that $Q_{1}>1$. Now using the
definition $Y_{F}=H^{\chi_{n}}\cdot e^{\pi_{n}}$, we have

$\displaystyle t$
$\displaystyle\leqslant\Bigg{\lfloor}\frac{\log\left[\frac{\chi_{n}\log
H+\pi_{n}}{\frac{d_{0}(d-1)+d}{n(d-1)}\log
H+\log\left(K_{d}(n)^{-\frac{1}{d-1}}Q_{1}\right)}+1\right]}{\log
d}\Bigg{\rfloor}$
$\displaystyle\leqslant\left\lfloor\max\left(\frac{\log\left(\frac{\chi_{n}}{\frac{d_{0}(d-1)+d}{n(d-1)}}+1\right)}{\log
d},\frac{\log\left(\frac{\pi_{n}}{\log
K_{d}(n)^{-\frac{1}{d-1}}Q_{1}}+1\right)}{\log d}\right)\right\rfloor$
$\displaystyle=T-2$

Therefore, the number of small special solutions $(p,q)$ for which $p/q$
belongs to $\alpha$ is $t+2\leqslant T$. ∎

###  3.2. Large Special Solutions

Here we follow Thomas in [13] as he follows Bombieri-Schmidt in [4]. If we
choose numbers $a$ and $b$ satisfying

$\displaystyle 0<a<b<1-\sqrt{2\cdot\frac{n+a^{2}}{n^{2}}}$ (8)

then we can define

$\displaystyle L$ $\displaystyle=\frac{\sqrt{2(n+a^{2})}}{1-b}$ $\displaystyle
D$ $\displaystyle=\frac{L}{n-L}$ $\displaystyle A$
$\displaystyle=\frac{1}{a^{2}}$ $\displaystyle E$
$\displaystyle=\frac{1}{2(b^{2}-a^{2})}.$

Now we choose

$\displaystyle\chi_{n}$ $\displaystyle=D(A+1)+1$ (9) $\displaystyle\pi_{n}$
$\displaystyle=(D(4+A)+2)\log(2)+\frac{(D+1)\log(n)}{2}+\frac{nAD}{2}.$ (10)

With these choices of $\pi_{n}$ and $\chi_{n}$, we aim to apply Lemma 2 of [4]
and conclude the following:

######  Proposition 3.2.

Suppose $\alpha>1$ is a real root of $f(x)$. If $\chi_{n}\geqslant 2$ and
$\pi_{n}\geqslant 5\log(2)+2\log(n)$, then there are at most

$\displaystyle Z:=\left\lfloor\frac{\log
E+2\log(n)-\log(L-2)}{\log(n-1)}\right\rfloor+2$ (11)

large special solutions belonging to $\alpha$.

The proof of this proposition relies on the two following lemmas:

######  Lemma 3.3.

$Y_{F}$ as defined here is greater than or equal to $Y_{0}$ as defined in [4].

Lemma 3.3 ensures that any large solution in the sense of this paper is a
large solution in the sense of Bombieri and Schmidt.

######  Lemma 3.4.

Suppose that $\chi_{n}\geqslant 2$ and $\pi_{n}\geqslant 5\log(2)+2\log(n)$.
If $\alpha>1$ is a real root of $f(x)$ and $(p,q)$ is a large special solution
of (3) so that $p/q$ belongs to $\alpha$, then $\alpha$ is the closest
(complex) root of $f(x)$ to $p/q$.

Given an algebraic $\beta$, Lemma 2 of [4] only counts the number of rational
numbers which are nearest to $\beta$ out of all of the conjugates of $\beta$
and which form good approximations of $\beta$. If there were a real root
$\alpha>1$ of $f(x)$ and a large special solution $(p,q)$ of (3) for which
$p/q$ belonged to $\alpha$ yet there was a root $\beta$ of $f(x)$ with $\beta$
closer to $p/q$ than $\alpha$, Lemma 2 of [4] would not count $p/q$. However,
Lemma 3.4 confirms that this is not the case.

We first prove these two lemmas:

###### Proof of Lemma 3.3.

$Y_{0}$ depends on the Mahler measure $M(F)$ rather than the height $H(F)$.
These are related (for trinomials $F(x,y)$) by $M(F)\leqslant 3^{1/2}H(F)$,
which follows from the fact that $M(F)\leqslant\ell_{2}(F)$ (see Lemma 1.6.7
in [3]). Now, using Thomas’ notation, we have that

$Y_{0}:=\left(2C\right)^{\frac{1}{n-\lambda}}\left(4e^{A_{1}}\right)^{\frac{\lambda}{n-\lambda}}$

where

$\displaystyle C$ $\displaystyle=(2n^{1/2}M(F))^{n}$ $\displaystyle t$
$\displaystyle=\sqrt{\frac{2}{n+a^{2}}}$ $\displaystyle A_{1}$
$\displaystyle=\frac{t^{2}}{2-nt^{2}}\left(\log M(F)+\frac{n}{2}\right)$
$\displaystyle\lambda$ $\displaystyle=\frac{2}{t(1-b)}.$

Some of our other constants regularly appear in the estimation which shows
$Y_{0}<Y_{F}$ and we list them here for simplicity:

$\displaystyle
A=\frac{1}{a^{2}}=\frac{2}{2(n+a^{2})-2n}=\frac{2}{n+a^{2}}\cdot\frac{1}{2-\frac{2n}{n+a^{2}}}=\frac{t^{2}}{2-nt^{2}}$
$\displaystyle
L=\frac{\sqrt{2(n+a^{2})}}{1-b}=\frac{2}{1-b}\sqrt{\frac{n+a^{2}}{2}}=\frac{2}{t(1-b)}=\lambda$
$\displaystyle D=\frac{L}{n-L}=\frac{\lambda}{n-\lambda}$

Note also that this implies that $D+1=\frac{n}{n-\lambda}$.

Before making the final estimate, we take a moment to observe that

$\left(\sqrt{3}\right)^{\frac{n}{n-\lambda}+AD}<2^{\frac{n-1}{n-\lambda}}.$

This estimate is tedious, but not difficult. One can show that

$\left(\sqrt{3}\right)^{\frac{n}{n-\lambda}+AD}<2^{\frac{n-1}{n-\lambda}}$

occurs if and only if

$2<\left(\frac{2}{\sqrt{3}}\right)^{n+A\lambda}.$

Estimating $A$ from below by

$A>\frac{1}{(1-\sqrt{2(n+1)/n^{2}})^{2}}$

and estimating $\lambda$ from below by $\lambda>\sqrt{2n}$ gives that

$2<\left(\frac{2}{\sqrt{3}}\right)^{n+A\lambda}$

is implied by

$2<\left(\frac{2}{\sqrt{3}}\right)^{n+\frac{n^{2}\sqrt{2n}}{n^{2}-2n\sqrt{2n+2}+2n+2}}.$

Upon observing that

$n+\frac{n^{2}\sqrt{2n}}{n^{2}-2n\sqrt{2n+2}+2n+2}\geqslant 20$

when $n\geqslant 6$ for instance, one can now see that

$2<\left(\frac{2}{\sqrt{3}}\right)^{n+\frac{n^{2}\sqrt{2n}}{n^{2}-2n\sqrt{2n+2}+2n+2}}$

and as a result, we must have

$\left(\sqrt{3}\right)^{\frac{n}{n-\lambda}+AD}<2^{\frac{n-1}{n-\lambda}}.$

We can now conclude

$\displaystyle Y_{0}$
$\displaystyle=(2C)^{\frac{1}{n-\lambda}}(4e^{A_{1}})^{\frac{\lambda}{n-\lambda}}$
$\displaystyle=(2(2n^{1/2}M(F))^{n})^{\frac{1}{n-\lambda}}(4e^{A(\log
M(F)+\frac{n}{2})})^{\frac{\lambda}{n-\lambda}}$
$\displaystyle=2^{\frac{1+n+2\lambda}{n-\lambda}}\cdot
n^{\frac{n}{2(n-\lambda)}}\cdot M(F)^{\frac{n}{n-\lambda}}e^{AD(\log
M(F)+\frac{n}{2})}$ $\displaystyle=2^{\frac{1+n+2\lambda}{n-\lambda}}\cdot
n^{\frac{n}{2(n-\lambda)}}\cdot M(F)^{\frac{n}{n-\lambda}+AD}\cdot
e^{\frac{ADn}{2}}$ $\displaystyle\leqslant
2^{\frac{1+n+2\lambda}{n-\lambda}}\cdot
n^{\frac{n}{2(n-\lambda)}}\cdot(\sqrt{3}H(F))^{\frac{n}{n-\lambda}+AD}\cdot
e^{\frac{ADn}{2}}.$

Next we use the fact that
$\left(\sqrt{3}\right)^{\frac{n}{n-\lambda}+AD}<2^{\frac{n-1}{n-\lambda}}$ to
find that

$\displaystyle Y_{0}$
$\displaystyle<2^{\frac{1+n+2\lambda}{n-\lambda}+\frac{n-1}{n-\lambda}+AD}\cdot
n^{\frac{n}{2(n-\lambda)}}\cdot H(F)^{\frac{n}{n-\lambda}+AD}\cdot
e^{\frac{ADn}{2}}$ $\displaystyle=2^{\frac{2n+2\lambda}{n-\lambda}+AD}\cdot
n^{\frac{D+1}{2}}\cdot H(F)^{1+D+AD}\cdot e^{\frac{ADn}{2}}$
$\displaystyle=H(F)^{\chi_{n}}\cdot\exp\left(\left(\frac{2n+2\lambda}{n-\lambda}+AD\right)\log(2)+\frac{D+1}{2}\log
n+\frac{ADn}{2}\right)$
$\displaystyle=H(F)^{\chi_{n}}\cdot\exp\left(\left(\frac{4\lambda+2(n-\lambda)}{n-\lambda}+AD\right)\log(2)+\frac{D+1}{2}\log
n+\frac{ADn}{2}\right)$
$\displaystyle=H(F)^{\chi_{n}}\cdot\exp\left(\left(4D+2+AD\right)\log(2)+\frac{D+1}{2}\log
n+\frac{ADn}{2}\right)$ $\displaystyle=H(F)^{\chi_{n}}\cdot e^{\pi_{n}}$
$\displaystyle=Y_{F}.$

∎

###### Proof of Lemma 3.4.

Since $\frac{p}{q}$ is a large special solution, we have

$\displaystyle p>q\geqslant Y_{F}$ $\displaystyle=H^{\chi_{n}}e^{\pi_{n}}$

Since $\frac{p}{q}$ belongs to $\alpha$, Thomas’ Corollary 3.1 in [13]
indicates that

$\displaystyle\left|\frac{p}{q}-\alpha\right|$
$\displaystyle<\frac{1}{p^{n^{*}}q}$ $\displaystyle<\frac{1}{Y_{F}^{n/2}}$

Suppose, by contradiction, that there exists $\beta\in\mathbb{C}$ with
$f(\beta)=0$ and
$\left|\frac{p}{q}-\beta\right|<\left|\frac{p}{q}-\alpha\right|$. Then by the
triangle equality, we find that

$\displaystyle\left|\alpha-\beta\right|$
$\displaystyle\leqslant\left|\frac{p}{q}-\beta\right|+\left|\frac{p}{q}-\alpha\right|$
$\displaystyle<\frac{2}{Y_{F}^{n/2}}$ (12)

Since $\alpha$ and $\beta$ are distinct roots of $f$, Theorem 4 in [10]
indicates that

$\displaystyle|\alpha-\beta|>\frac{1}{2n^{n/2+2}(4H)^{n}}=\frac{1}{2^{2n+1}n^{n/2+2}H^{n}}$
(13)

Combining (12) and (13), we find that

$\frac{1}{2^{2n+1}n^{n/2+2}H^{n}}<\frac{2}{Y_{F}^{n/2}}$

and rearranging yields

$\frac{Y_{F}^{n/2}}{2^{2n+2}n^{n/2+2}H^{n}}<1.$

From here, we can use the fact that $Y_{F}=H^{\chi_{n}}e^{\pi_{n}}$ to find

$\displaystyle 1$
$\displaystyle>\frac{H^{n(\chi_{n}/2-1)}e^{n\pi_{n}/2}}{2^{2n+2}n^{n/2+2}}$
$\displaystyle\geqslant\frac{e^{n\pi_{n}/2}}{2^{2n+2}n^{n/2+2}}$

where the last inequality follows because $\chi_{n}\geqslant 2$. After
rearranging, this implies that

$\displaystyle\pi_{n}$ $\displaystyle<\frac{(4n+4)\log(2)+(n+4)\log(n)}{n}$
$\displaystyle\leqslant 5\log(2)+2\log(n)$

where the last inequality follows from the fact that $n\geqslant 6$. However,
the last inequality contradicts our hypothesis that $\pi_{n}\geqslant
5\log(2)+2\log(n)$, so no such $\beta$ can exist and the closest root of
$f(x)$ to $p/q$ is $\alpha$. ∎

Finally, we prove proposition 3.2.

###### Proof of Proposition 3.2..

Let $\alpha>1$ be a real root of $f(x)$. By lemma 3.3, every large special
solution $(p,q)$ so that $p/q$ belongs to $\alpha$ is large in the sense of
[4]. Moreover, by lemma 3.4, any large special solution $(p,q)$ so that $p/q$
belongs to $\alpha$ has

$\left|\frac{p}{q}-\alpha\right|=\min_{f(\beta)=0}\left|\frac{p}{q}-\beta\right|.$

Hence, every large special solution $(p,q)$ so that $p/q$ belongs to $\alpha$
is large (in the sense of [4]) and is nearest to $\alpha$ among all the roots
of $f(X)$. Lemma 2 of [4] indicates that there are no more than

$Z=\left\lfloor\frac{\log E+2\log(n)-\log(L-2)}{\log(n-1)}\right\rfloor+2$

large solutions $(p,q)$ so that $p/q$ is nearest to $\alpha$ among all the
roots of $f(X)$ and so we conclude that there are no more than $Z$ large
special solutions $(p,q)$ with $p/q$ belonging to $\alpha$. ∎

###  3.3. Choosing Parameters for Large Degrees

Begin by assuming $n\geqslant 507$. We handle all smaller instances of $n$
computationally.

######  Proposition 3.3.

For $n\geqslant 507,$ we can take $d_{0}=\frac{n^{*}}{2}$, $d=n^{*}$,
$a=\frac{1}{4}$, $C=7/6$, $c=\frac{8}{9C^{2}-1}$,
$b=1-\frac{\sqrt{2n+\frac{1}{8}}}{\frac{cn^{2}}{n-1}+2}$ and obtain $T=2$ and
$Z=2$.

Observe first that these are the smallest possible values of $T$ and $Z$.

###### Proof.

To show this, we first must show that these choices of $d_{0},d,a,$ and $b$
are valid.

Certainly $0\leqslant d_{0}\leqslant n^{*}-1.4$ and $1<d\leqslant n^{*}$. All
that remains to show for $d_{0}$ and $d$ is (7). We have

$\displaystyle Q_{1}^{d-1}$
$\displaystyle=\left(\frac{p_{0}^{c_{0}}}{K_{0}}\right)^{d-1}$
$\displaystyle\geqslant\left(\frac{2^{n^{*}/2}}{K_{n^{*}/2}(n)}\right)^{\frac{n^{*}}{2}-1}.$

But observe that

$\displaystyle K_{n^{*}/2}(n)$
$\displaystyle=2\sqrt{\frac{2n}{(n-1)(n-2)}}\left(2.032^{1/n}\left(1+\sqrt{\frac{2}{(n-2)p_{0}^{n}}}\right)\right)^{n^{*}/2}$
$\displaystyle\leqslant 2\cdot
2.032\left(1+\frac{1}{\sqrt{(n-2)2^{n-1}}}\right)^{\frac{n-2}{4}}$
$\displaystyle\leqslant 5\left(1+\frac{1}{n-2}\right)^{\frac{n-2}{4}}$
$\displaystyle\leqslant 5e^{1/4}$

so $Q_{1}^{d-1}$ is certainly greater than 1. Similar reasoning shows that
$K_{d}(n)\leqslant 5e^{1/2}$, so it is certainly also the case that

$K_{d}(n)\leqslant
5e^{1/2}<\left(\frac{2^{n^{*}/2}}{5e^{1/4}}\right)^{\frac{n^{*}}{2}-1}\leqslant\left(\frac{2^{n^{*}/2}}{K_{n^{*}/2}(n)}\right)^{\frac{n^{*}}{2}-1}=Q_{1}^{b-1}.$

Hence, our choices of $d$ and $d_{0}$ are valid.

Next, we wish to check that our choices for $a$ and $b$ are valid. To check
$0<a<b$, note that

$\displaystyle b$
$\displaystyle=1-\frac{\sqrt{2n+\frac{1}{8}}}{\frac{cn^{2}}{n-1}+2}\geqslant
1-\frac{2\sqrt{n}}{\frac{cn^{2}}{n-1}}=1-\frac{2(n-1)\sqrt{n}}{cn^{2}}\geqslant
1-\frac{2}{c\sqrt{n}}$ $\displaystyle\geqslant
1-\frac{2}{c\sqrt{507}}>\frac{1}{4}=a.$ (14)

To check that $b<1-\frac{\sqrt{2n+2a^{2}}}{n}$, it suffices to show that
$n>\frac{cn^{2}}{n-1}+2$. But this occurs if and only if $(1-c)n^{2}-3n+2>0$,
i.e. $n>\frac{3+\sqrt{9-8(1-c)}}{2(1-c)}\approx 9.66$, which we certainly
have.

To show that $T=2$, we claim that we have the following two inequalities (and
from equation (6), it will follow that $T=2$):

$\displaystyle\frac{\chi_{n}}{\frac{d_{0}(d-1)+d}{n(d-1)}}+1<d$ (15)
$\displaystyle\frac{\pi_{n}}{\log K_{d}(n)^{-\frac{1}{d-1}}Q_{1}}+1<d$ (16)

We first show (15). Substituting $d_{0}=\frac{n^{*}}{2}$ and $d=n^{*}$,
observe that (15) is equivalent to

$\chi_{n}<\frac{\left(\frac{n-2}{4}\right)\left(\frac{n-4}{2}-1\right)+\frac{n-2}{2}}{n}=\frac{n-2}{8}=\frac{n^{*}}{4}.$

Keeping an eye on the definition of $\chi_{n}$ given in equation (9), we have
that

$\displaystyle A$ $\displaystyle=16$ $\displaystyle C$
$\displaystyle=\frac{7}{6}$ $\displaystyle c$
$\displaystyle=\frac{8}{9C^{2}-1}=\frac{32}{45}$ $\displaystyle b$
$\displaystyle=1-\frac{\sqrt{2n+\frac{1}{8}}}{\left(\frac{cn^{2}}{n-1}+2\right)}$

All of these together yield

$\displaystyle
L=\frac{cn^{2}}{n-1}+2=\frac{32}{45}\left(n+1+\frac{1}{n-1}\right)+2$ (17)

and it is now easy to check that

$\frac{32}{45}n\leqslant L\leqslant\frac{32}{45}n+3.$

From here we have

$\displaystyle D$
$\displaystyle=\frac{L}{n-L}\leqslant\frac{\frac{32}{45}n+3}{n-\left(\frac{32}{45}n+3\right)}=\frac{\frac{32}{45}n+3}{\frac{13}{45}n-3}=\frac{32}{13}+\frac{6075}{13(13n-135)}\leqslant
2.54$ $\displaystyle D$
$\displaystyle\geqslant\frac{\frac{32}{45n}}{n-\frac{32}{45}n}=\frac{32}{13}\approx
2.46$

when we use the fact that $n\geqslant 507$. To convert these into estimates on
$\chi_{n}$, we have

$\displaystyle\chi_{n}$
$\displaystyle=17D+1\leqslant\frac{94853}{2152}\leqslant 44.08$ (18)
$\displaystyle\chi_{n}$ $\displaystyle=17D+1\geqslant\frac{557}{13}\geqslant
42.8.$ (19)

Since $n\geqslant 507$, we now have $\chi_{n}\leqslant 44.08<\frac{n-2}{8}$
which confirms equation (15).

Equation (16) is more complicated to handle. Observe that by equation (10), we
have

$\displaystyle\pi_{n}$ $\displaystyle=(D(4+A)+2)\log 2+\frac{(D+1)\log
n}{2}+\frac{ADn}{2}$ $\displaystyle\leqslant 36.6+1.77\log n+20.28n$
$\displaystyle\leqslant 37+21n.$ (20)

For reference later, we will also note

$\displaystyle\pi_{n}$ $\displaystyle\geqslant 35.5+1.7\log n+19.6n$
$\displaystyle\geqslant 46+19n$ (21)

It will additionally be helpful for us to have an estimate on $K_{d}(n)$. We
have

$\displaystyle K_{d}(n)$
$\displaystyle=2\sqrt{\frac{2n}{(n-1)(n-2)}}\left(2.032^{1/n}\left(1+\sqrt{\frac{2}{(n-2)p_{0}^{n}}}\right)\right)^{d}$
$\displaystyle\leqslant
4\sqrt{\frac{3n-3}{(n-1)(n-2)}}\left(1+\sqrt{\frac{2n-4}{(n-2)p_{0}^{n}}}\right)^{d}$
$\displaystyle\leqslant
4\sqrt{\frac{3}{n-2}}\left(1+\sqrt{\frac{2}{p_{0}^{n}}}\right)^{n^{*}}.$

Now, since $p_{0}\geqslant 2$, we have

$\displaystyle K_{d}(n)$ $\displaystyle\leqslant
4\sqrt{\frac{3}{n-2}}\left(1+\sqrt{\frac{2}{2^{n}}}\right)^{\frac{n-2}{2}}\leqslant
4\sqrt{\frac{3}{n-2}}\left(1+\frac{1}{2^{\frac{n-2}{2}}}\right)^{\frac{n-2}{2}}$
$\displaystyle\leqslant
4\sqrt{\frac{3}{n-2}}\left(1+\frac{1}{\left(\frac{n-2}{2}\right)}\right)^{\frac{n-2}{2}}\leqslant
4e\sqrt{\frac{3}{n-2}}$ (22)

and similar reasoning yields

$\displaystyle K_{d_{0}}(n)\leqslant
4\sqrt{\frac{3}{n-2}}\left(1+\frac{1}{2^{n^{*}}}\right)^{n^{*}/2}.$ (23)

Combining the upper bounds in equations (3.3), (3.3), and (23) with the fact
that for $n\geqslant 270$,

$\pi_{n}<37+21n<\frac{\log(1.9)}{8}(n-2)(n-4)$

yields

$\displaystyle(d-1)\log\left[K_{d}(n)^{-\frac{1}{d-1}}Q_{1}\right]$
$\displaystyle=\log\left[K_{d}(n)^{-1}Q_{1}^{d-1}\right]$
$\displaystyle\geqslant\log\left[\frac{1}{4}\cdot
e^{-1}\cdot\left(\frac{n-2}{3}\right)^{1/2}\cdot\left(\frac{p_{0}^{c_{0}}}{K_{0}}\right)^{n^{*}-1}\right]$
$\displaystyle\geqslant\log\left[\frac{1}{4}\cdot
e^{-1}\cdot\left(\frac{n-2}{3}\right)^{1/2}\left(\frac{2^{n^{*}/2}}{K_{d_{0}}(n)}\right)^{n^{*}-1}\right]$
$\displaystyle\geqslant\log\left[\frac{1}{4}\cdot
e^{-1}\cdot\left(\frac{n-2}{3}\right)^{1/2}\left(\frac{2^{n^{*}/2}}{\left(1+2^{-n^{*}}\right)^{n^{*}/2}}\right)^{\frac{n-4}{2}}\left(\frac{1}{4}\sqrt{\frac{n-2}{3}}\right)^{\frac{n-4}{2}}\right]$
$\displaystyle\geqslant\log\left[\left(\frac{1}{4}\right)^{\frac{n-2}{2}}\left(\frac{n-2}{3}\right)^{\frac{n-2}{4}}e^{-1}\left(\frac{2}{1+2^{-n^{*}}}\right)^{\left(\frac{n-2}{4}\right)\left(\frac{n-4}{2}\right)}\right]$
$\displaystyle\geqslant\log\left[\left(\frac{n-2}{48}\right)^{\frac{n-2}{4}}e^{-1}\cdot
1.9^{\frac{(n-2)(n-4)}{8}}\right]$ $\displaystyle\geqslant\frac{\log
1.9}{8}(n-2)(n-4)+\frac{n-2}{4}\log\left(\frac{n-2}{48}\right)-1$
$\displaystyle\geqslant\frac{\log(1.9)}{8}(n-2)(n-4)$ $\displaystyle>\pi_{n}$

which now implies that equation (16) is satisfied. Hence, we conclude that
$T=2$.

Finally, we check that $Z=2$. In order to use Proposition 3.2, we must verify
that $\chi_{n}\geqslant 2$ and $\pi_{n}\geqslant 5\log(2)+2\log(n)$. However,
these quickly follow from (19) and (3.3).

As before in equation (14), we have $b\geqslant
1-\frac{2}{c\sqrt{507}}>0.87509$, so

$\displaystyle E=\frac{1}{2(b^{2}-a^{2})}<0.711<c$

and so (also using (17))

$\displaystyle\frac{\log
E+2\log(n)-\log(L-2)}{\log(n-1)}<\frac{\log\left(\frac{cn^{2}}{L-2}\right)}{\log(n-1)}=1.$

Therefore, by (11), we note that $Z=2$. ∎

Note that Proposition 3.3 proves Lemma 3.2 for $n\geqslant 507$.

###  3.4. Choosing Parameters for Small Degrees

For $n\leqslant 506$, we make parameter choices listed in the Jupyter notebook
listed here:

https://pages.uoregon.edu/gknapp4/files/trinomial_computations.ipynb.

One can check (and the code checks this automatically) that the parameter
choices satisfy (7), (8) along with the necessary bounds on $\pi_{n}$ and
$\chi_{n}$ in order to use proposition 3.2, and yield the $T$ and $Z$ values
giving $z(n)=T+Z+1$ when $6\leqslant n\leqslant 8$ and $z(n)=T+Z$ for
$n\geqslant 9$. Therefore, these computations verify Lemma 3.2 for $n\leqslant
506$, which concludes our investigation.

In brief, the code picks a value of $n$, sets $d=n^{*}$, brute force loops
over a number of valid values for the parameters $d_{0},a,b$, computes the
corresponding $T$ and $Z$ values defined in equations (6) and (11), and
records the values of $d_{0},a,$ and $b$ which minimize $T+Z$. The following
table contains a summary of some of the more interesting data points listed in
the Jupyter notebook.

$n$ | $d_{0}$ | $d$ | $a$ | $b$ | $T$ | $Z$
---|---|---|---|---|---|---
6 | 0 | 2 | 0.18 | 0.29 | 10 | 4
7 | 0.55 | 2.5 | 0.2 | 0.28 | 7 | 4
8 | 0.992 | 3 | 0.16 | 0.41 | 7 | 3
9 | 0.882 | 3.5 | 0.17 | 0.4 | 6 | 3
10 | 1.196 | 4 | 0.23 | 0.41 | 5 | 3
11 | 1.705 | 4.5 | 0.14 | 0.37 | 5 | 3
12 | 2.088 | 5 | 0.27 | 0.41 | 4 | 3
13 | 2.255 | 5.5 | 0.2 | 0.37 | 4 | 3
14 | 2.53 | 6 | 0.16 | 0.35 | 4 | 3
15 | 3.009 | 6.5 | 0.13 | 0.34 | 4 | 3
16 | 3.136 | 7 | 0.11 | 0.32 | 4 | 3
17 | 3.965 | 7.5 | 0.33 | 0.43 | 3 | 3
18 | 4.29 | 8 | 0.27 | 0.39 | 3 | 3
$\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$
37 | 7.728 | 17.5 | 0.08 | 0.25 | 3 | 3
38 | 9.794 | 18 | 0.07 | 0.25 | 3 | 3
39 | 11.97 | 18.5 | 0.43 | 0.47 | 2 | 3
40 | 12.144 | 19 | 0.41 | 0.46 | 2 | 3
$\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$
218 | 57.564 | 108 | 0.03 | 0.87 | 3 | 2
219 | 68.2227 | 108.5 | 0.399258 | 0.883258 | 2 | 2
220 | 68.6488 | 109 | 0.408477 | 0.884477 | 2 | 2
$\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$
506 | 218.022 | 252 | 0.517076 | 0.927076 | 2 | 2
507 | 218.457 | 252.5 | 0.517138 | 0.927138 | 2 | 2

##  4\. Acknowledgments

I would like to thank Shabnam Akhtari for her guidance on this project and
paper, Ben Young for his suggestions on the computations and feedback on my
code, and Cathy Hsu for her mentorship and general research advice. This
research was supported by NSF grant number 2001281.

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