# Asymptotics for $t$-Core Partitions and Stanton’s Conjecture

Matt Tyler

###### Abstract

A partition is a $t$-core partition if $t$ is not one of its hook lengths. Let
$c_{t}(N)$ be the number of $t$-core partitions of $N$. In 1999, Stanton
conjectured $c_{t}(N)\leq c_{t+1}(N)$ if $4\leq t\neq N-1$. This was proved
for $t$ fixed and $N$ sufficiently large by Anderson, and for small values of
$t$ by Kim and Rouse. In this paper, we prove Stanton’s conjecture in general.

Our approach is to find a saddle point asymptotic formula for $c_{t}(N)$,
valid in all ranges of $t$ and $N$. This includes the known asymptotic
formulas for $c_{t}(N)$ as special cases, and shows that the behavior of
$c_{t}(N)$ depends on how $t^{2}$ compares in size to $N$. For example, our
formula implies that if $t^{2}=\kappa N+o(t)$, then
$c_{t}(N)=\frac{\exp\left(2\pi\sqrt{AN}\right)}{BN}\left(1+o(1)\right)$ for
suitable constants $A$ and $B$ defined in terms of $\kappa$.

##  1 Introduction and statement of results

Let $\lambda=(\lambda_{1},\dots,\lambda_{k})$ with
$\lambda_{1}\geq\dots\geq\lambda_{k}>0$ be a partition of
$|\lambda|=\lambda_{1}+\dots+\lambda_{k}$. The Ferrers–Young diagram of
$\lambda$ consists of $k$ rows of left-justified boxes, with $\lambda_{i}$
boxes in row $i$. Let $\lambda_{j}^{\prime}$ be the number of boxes in column
$j$ (so
$\lambda^{\prime}=(\lambda_{1}^{\prime},\dots,\lambda_{\ell}^{\prime})$ is the
conjugate partition to $\lambda$). The hook length for the box at row $i$ and
column $j$ is $(\lambda_{i}-j)+(\lambda_{j}^{\prime}-i)+1$. In other words,
the hook length is the number of boxes directly to the right, plus the number
of boxes directly below, plus $1$ for the box itself. We say $\lambda$ is a
$t$-core partition if $t$ is not one of its hook lengths. See Figure 1.1 for
an example.

$8$ $1$ $5$ $1$ $2$ $1$ Figure 1.1: The Ferrers–Young diagram for the
partition $(6,4,2)$ of $12$, labelled with hook lengths. This is a $t$-core
partition for $t\not\in\\{1,2,4,5,7,8\\}$.

We write $c_{t}(N)$ for the number of $t$-core partitions of $N$, and $p(N)$
for the number of partitions of $N$.

Much of the motivation for studying $t$-core partitions comes from the
representation theory of the symmetric group $S_{N}$. The irreducible complex
representations of $S_{N}$ are parameterized by the partitions $\lambda$ of
$N$. The representation associated to $\lambda$ has dimension
$\frac{N!}{\prod_{h\in\mathcal{H}(\lambda)}h}$ where $\mathcal{H}(\lambda)$ is
the multiset of hook lengths of $\lambda$.

When $p$ is prime, a $p$-core partition of $N$ is therefore a partition for
which the power of $p$ dividing $|S_{N}|$ is the same as the power of $p$
dividing the dimension of the associated representation. These are the
projective irreducible representations in characteristic $p$, i.e. the
$p$-blocks of defect $0$. More generally, by Nakayama’s conjecture (proved by
Brauer and Robinson), $p$-core partitions parameterize the $p$-blocks of
representations of $S_{n}$. See [JK84, sections 2.7 and 6.1] for details.

Another application of $t$-core partitions is due to Garvan, Kim, and Stanton
[GKS90], who used $t$-core partitions for $t=5,7,11$ to find combinatorial
proofs of Ramanujan’s congruences

$\displaystyle p(5N+4)$ $\displaystyle\equiv 0\ (\text{mod}\ 5),$ (1.1)
$\displaystyle p(7N+5)$ $\displaystyle\equiv 0\ (\text{mod}\
7),\quad\text{and}$ $\displaystyle p(11N+6)$ $\displaystyle\equiv 0\
(\text{mod}\ 11).$

The $t$-core partition conjecture states that $c_{t}(N)>0$ for all $t\geq 4$
and $N\geq 1$. This was proved for $t=5,7$ by Erdmann and Michler [EM96], when
$t$ has a prime factor $\leq 11$ by Ono [Ono94, Ono95], and in general by
Granville and Ono [GO00]. In 1999, Stanton [Sta99] made the “possibly rash”
conjecture that in fact much more is true.

######  Conjecture 1.1 (Stanton).

For all $4\leq t\neq N-1$, $c_{t}(N)\leq c_{t+1}(N)$.

We always have $c_{t}(N)\leq p(N)$, with equality if $t>N$, so the conjecture
is obvious if $t\geq N$. The restriction $t\neq N-1$ is necessary because
$c_{N-1}(N)=p(N)-N+1=c_{N}(N)+1$.

In the case $t>\frac{N}{2}$, Stanton’s conjecture was quickly proved by Craven
[Cra06] using combinatorial arguments. In 1999, Lulov and Pittel [LP99] used
the circle method to show that if $t\geq 6$, then

$c_{t}(N)=A_{t}(N)\frac{(2\pi)^{\frac{t-1}{2}}}{t^{\frac{t}{2}}\Gamma\left(\frac{t-1}{2}\right)}\left(N+\frac{t^{2}-1}{24}\right)^{\frac{t-3}{2}}+O_{t}\left(N^{\frac{t-1}{4}}\right)$
(1.2)

where $A_{t}(N)$ is an arithmetic quantity satisfying
$\left|A_{t}(N)-1\right|\leq\left|\zeta(\frac{t-3}{2})-1\right|$. This was
rediscovered in 2008 by Anderson [And08], who used it to prove the conjecture
when $t$ is fixed and $N$ is sufficiently large. This work was made explicit
by Kim and Rouse [KR14] in 2014, who proved the conjecture for certain values
of $t$.222There may be a technical error in their proof, but it works without
modification for $t\ll\log N$, and may be extendable up to $t\ll
N^{\frac{1}{4}}$. For completeness, we will independently treat the case
$t\geq 8$ here, and rely on their work for $4\leq t<8$, which is unaffected by
the error.

In this paper, we will prove Stanton’s conjecture.

######  Theorem 1.2.

Stanton’s conjecture 1.1 is true.

######  Remark 1.3.

Our proof shows that the inequality is strict for $4\leq t<N-1$, except for
$c_{5}(10)=c_{6}(10)=12$.

Previously, $c_{t}(N)$ has only been understood when $t$ is of constant size,
as in (1.2), or when $t>(1+\epsilon)\frac{\sqrt{6}}{2\pi}\sqrt{N}\log N$, in
which case $c_{t}(N)\sim p(N)$ [Mor19]. Our contribution is a saddle point
asymptotic formula for $c_{t}(N)$ that is valid in all ranges of $t$ and $N$
and includes these formulas as special cases. Before stating the theorem, we
will need a few definitions.

Let $e(z)=e^{2\pi iz}$, and let $\eta(z)$ be the Dedekind eta function

$\eta(z)=e\left(\frac{z}{24}\right)\prod_{n=1}^{\infty}(1-e(nz)).$ (1.3)

Let

$f_{t}(z)=\frac{\eta(tz)^{t}}{\eta(z)},$ (1.4)

which is a modular form of weight $\frac{t-1}{2}$ and level $t$, and which
will be of central importance in this paper. The generating function for
$t$-core partitions is [JK84, (2.7.17)]

$\sum_{N=0}^{\infty}c_{t}(N)e(Nz)=\prod_{n=1}^{\infty}\frac{(1-e(ntz))^{t}}{1-e(nz)}=e\left(\tfrac{1-t^{2}}{24}z\right)f_{t}(z).$
(1.5)

We also define

$\mu_{k}(z)=-\frac{z^{k+1}}{2\pi i}\left(\frac{d}{dz}\right)^{k}\log\eta(z).$
(1.6)

When $k\geq 1$, $\mu_{k}(z)=-\frac{z^{k+1}}{24}E_{2}^{(k-1)}(z)$ where
$E_{2}(z)=1-24\sum_{n=1}^{\infty}\sigma(n)e(nz)$ is the Eisenstein series of
weight $2$ and $\sigma(n)=\sum_{d\mid n}d$.

Our main theorem is the following asymptotic formula for $c_{t}(N)$. Unless
otherwise indicated with a subscript, all implied constants are universal.

######  Theorem 1.4.

Suppose $N\geq 0$ and $t\geq 6$.

  1. (i)

There is a unique value of $y>0$ satisfying

$\frac{\mu_{1}(ity)-\mu_{1}(iy)}{y^{2}}=N+\frac{t^{2}-1}{24}.$

Moreover, $y$ lies in the range

$\frac{t-1}{4\pi\left(N+\frac{t^{2}-1}{24}\right)}<y<\frac{1}{\frac{3}{\pi}+\sqrt{24N-1+\frac{9}{\pi^{2}}}}.$

  2. (ii)

For this value of $y$,

$c_{t}(N)=\frac{y^{\frac{3}{2}}e^{2\pi
y\left(N+\frac{t^{2}-1}{24}\right)}f_{t}(iy)}{\sqrt{\mu_{2}(iy)-\mu_{2}(ity)}}\left(1+O\left(\frac{1}{\min(t,\sqrt{N})}\right)\right).$

  3. (iii)

For this value of $y$, if $ty\geq\frac{1}{2}$, then

$c_{t+1}(N+t)-c_{t}(N+t)=(2\pi ty-1)c_{t}(N)\left(1+O\left(y+tye^{-2\pi
ty}\right)\right).$

Part ii of Theorem 1.4 shows that $c_{t}(N)$ exhibits a change in behavior
depending on how $t^{2}$ compares in size to $N$.

In the range $t^{2}\log t<\frac{8\pi^{2}}{1+\epsilon}N$ where $\epsilon>0$, we
have $y\approx\frac{t-1}{4\pi\left(N+\frac{t^{2}-1}{24}\right)}$, and part ii
of Theorem 1.4 implies

$c_{t}(N)=\frac{(2\pi)^{\frac{t-1}{2}}}{t^{\frac{t}{2}}\Gamma\left(\frac{t-1}{2}\right)}\left(N+\frac{t^{2}-1}{24}\right)^{\frac{t-3}{2}}\left(1+O\left(t^{-\min\left(\epsilon,1\right)}\right)\right).$
(1.7)

This expands the range of validity of Lulov and Pittel’s formula (1.2) from
fixed $t$ to $t^{2}\log t<\frac{8\pi^{2}}{1+\epsilon}N$. When $t^{2}\log
t<\frac{8\pi^{2}}{2+\epsilon}N$, a slightly different approach allows us to
refine the error term in (1.7) to be exponentially small.

######  Theorem 1.5.

For all $\epsilon>0$, if $t\geq 6$ and $t^{2}\log
t<\frac{8\pi^{2}}{2+\epsilon}N$, then

$c_{t}(N)=\frac{(2\pi)^{\frac{t-1}{2}}}{t^{\frac{t}{2}}\Gamma\left(\frac{t-1}{2}\right)}\left(N+\frac{t^{2}-1}{24}\right)^{\frac{t-3}{2}}\left(1+O\left(t^{-\frac{1}{2}}(1+\epsilon)^{-\frac{1}{4}t}+e^{-\frac{1}{6}t}+te^{-\frac{8\pi^{2}}{t(t-1)}N}\right)\right).$

In the range $t>(1+\epsilon)\frac{\sqrt{6}}{2\pi}\sqrt{N}\log N$, we have
$y\approx\frac{1}{\frac{3}{\pi}+\sqrt{24N-1+\frac{9}{\pi^{2}}}}$, and part ii
of Theorem 1.4 tells us only that

$c_{t}(N)=p(N)\left(1+O\left(N^{-\frac{1}{2}}+N^{-\frac{\epsilon}{2}}\log
N\right)\right).$ (1.8)

In this range, almost all partitions are $t$-cores, and we may once again
refine (1.8), this time to evaluate the remainder.

######  Theorem 1.6.

For all $\epsilon>0$, if $t>(1+\epsilon)\frac{\sqrt{6}}{2\pi}\sqrt{N}\log N$,
then

$p(N)-c_{t}(N)=tp(N-t)+O_{\epsilon}\left(t^{2}p(N-2t)\right).$ (1.9)

In the intermediary range $t\asymp\sqrt{N}$, $t$-core partitions have not
previously been understood. It turns out that $c_{t}(N)$ has an asymptotic
formula of the same shape as the Hardy–Ramanujan asymptotic formula for
partitions.

######  Corollary 1.7.

Suppose $t^{2}=\kappa N+o(t)$. Then,

$c_{t}(N)=\frac{e^{2\pi\sqrt{A(\kappa)N}}}{B(\kappa)N}\left(1+o(1)\right)$

where $A(\kappa)$ and $B(\kappa)$ are constants defined by choosing $v$ so
that

$\frac{1}{24v^{2}}-\frac{1}{24}+\frac{1}{v^{2}}\mu_{1}(iv)=\frac{1}{\kappa},$

and letting

$A(\kappa)=\frac{\kappa}{v^{2}}\left(\frac{1}{12}-\mu_{0}(iv)+\mu_{1}(iv)\right)^{2}\quad\text{and}\quad
B(\kappa)=\frac{\kappa}{v^{2}}\sqrt{\frac{1}{12}-\mu_{2}(iv)}.$

We will show in §4.3 that the function
$\frac{1}{24v^{2}}-\frac{1}{24}+\frac{1}{v^{2}}\mu_{1}(iv)$ is a monotonically
decreasing bijection $(0,\infty)\to(0,\infty)$, so $v$ always exists and is
unique.

0.040.080.120.1610203040 Figure 1.2: A graph of $A(\kappa)$ with an asymptote
at $\frac{1}{6}$.

48121610203040 Figure 1.3: A graph of $B(\kappa)$ with an asymptote at
$4\sqrt{3}$.

The constants $A(\kappa)$ and $B(\kappa)$ of Corollary 1.7 satisfy

$\lim_{\kappa\to\infty}A(\kappa)=\frac{1}{6}\quad\text{and}\quad\lim_{\kappa\to\infty}B(\kappa)=4\sqrt{3},$
(1.10)

which makes sense in light of (1.8) because
$p(N)=\frac{e^{2\pi\sqrt{N/6}}}{4\sqrt{3}N}\left(1+o(1)\right)$.

When $\frac{t^{2}}{N}$ is bounded from above, the ratio of main terms for
$c_{t+1}(N)$ and $c_{t}(N)$ in part ii of Theorem 1.4 is uniformly bounded
from below by a constant larger than $1$. This will prove Stanton’s conjecture
in this range for sufficiently large $t$. When $\frac{t^{2}}{N}$ grows larger,
(1.8) presents an obstruction — both $c_{t+1}(N)$ and $c_{t}(N)$ are
approximately the same size as $p(N)$. In this case, it is better to consider
their difference $c_{t+1}(N)-c_{t}(N)$ rather than their ratio, and we will
use part iii of Theorem 1.4 to complete the proof of Stanton’s conjecture.

The organization of this paper is as follows. In §2, we will bound the
contribution away from the saddle point. This is the heart of the paper, and
is the main input that allows us to evaluate $c_{t}(N)$ in new ranges. In §3,
we will prove some technical lemmas that will be useful in §4, mostly
involving bounding the functions $\mu_{k}$ for $k\leq 4$. In §4, we will use
the saddle point method to prove Theorems 1.4, 1.5, and 1.6 and Corollary 1.7.
Since our goal is to prove Stanton’s conjecture for all $t$, not just $t$
sufficiently large, we will actually keep track of the constants involved and
prove an explicit version of Theorem 1.4. In §5, we will prove Stanton’s
conjecture.

## Notation

Because our goal is to prove an exact inequality, we will require the use of
an explicit version of big $O$ notation. To that end, we use the symbol
$\vartheta$ to denote a complex number of magnitude $\leq 1$ which is allowed
to depend on all parameters, and which may be different at each occurrence.
For example, $a=b+\vartheta c$ means $|a-b|\leq|c|$. As usual with big $O$
notation, equations involving $\vartheta$ should be read from left to right,
so for example $1=2\vartheta$, but it would not be correct to write
$2\vartheta=1$.

We will always let $z=x+iy$ where $x$ and $y$ are real numbers and $y>0$.

We also make use of the shorthand

$M=N+\frac{t^{2}-1}{24}.$ (1.11)

## Acknowledgments

I am grateful to my advisor Kannan Soundararajan for suggesting this problem
and for offering invaluable advice. I would also like to thank Shintaro
Fushida-Hardy, Byungchan Kim, Milo Marsden, Jeremy Rouse, Dennis Stanton, and
Zachary Stier for many helpful comments. This paper is based upon work
supported by the National Science Foundation Graduate Research Fellowship
under Grant No. DGE-1656518.

##  2 Bounding the contribution away from the saddle point

In contrast to previous papers on this subject [GO00, And08, KR14], which
decompose $f_{t}(z)$ into a sum of Eisenstein series and cusp forms, our
approach is to use the saddle point method to evaluate the integral

$c_{t}(N)=\int_{-1/2}^{1/2}e\left(-Mz\right)f_{t}(z)dx.$ (2.1)

The Fourier coefficients $c_{t}(N)$ of $f_{t}(z)$ are all non-negative, so
$\left|f_{t}(z)\right|\leq f_{t}(iy)$, meaning the integrand in (2.1) is
maximized when $x=0$. Our goal in this section is to prove the following
proposition, which shows that this estimate can be substantially improved so
long as $x$ is not too close to $0$.

In subsequent sections, we will evaluate the integral (2.1) in the remaining
region where $x$ is close to $0$. We are free to make any choice of $y$, and
we will see that the $y$ specified in part i of Theorem 1.4 is a saddle point
that approximately reduces our integral to a Gaussian, which is easily
evaluated.

######  Proposition 2.1.

If $\min(t,\frac{1}{y})\geq 1000$, then

$\int_{\frac{y}{3}\leq\left|x\right|\leq\frac{1}{2}}\left|\frac{f_{t}(z)}{f_{t}(iy)}\right|dx<ye^{-\frac{1}{70}\min(t,\frac{1}{y})}.$

Recall that the group $\operatorname{SL}_{2}(\mathbb{Z})$ acts on the upper
half plane $\mathbb{H}=\\{z\mid y>0\\}$ by linear fractional transformations

$\gamma z=\frac{az+b}{cz+d}\text{ for }\gamma=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\in\operatorname{SL}_{2}(\mathbb{Z}).$ (2.2)

The Dedekind eta function (1.3) satisfies the transformation formula

$\eta\left(\frac{az+b}{cz+d}\right)=e\left(\frac{a+d}{24c}+\tfrac{1}{2}s(-d,c)\right)\sqrt{-i(cz+d)}\eta(z)\text{
for all }\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\in\operatorname{SL}_{2}(\mathbb{Z})$ (2.3)

where $s(h,k)$ is the Dedekind sum
$\sum_{r=1}^{k-1}\frac{r}{k}\left(\left\\{\frac{hr}{k}\right\\}-\tfrac{1}{2}\right)$.
This enables us to approximate $\eta(z)$ anywhere in the upper half plane, and
in particular implies

$\displaystyle\eta(-1/z)$ $\displaystyle=\sqrt{-iz}\eta(z)\quad\text{and}$
(2.4) $\displaystyle\left|\eta(\gamma z)\right|$
$\displaystyle=\left(\frac{y}{\operatorname{Im}\gamma
z}\right)^{\frac{1}{4}}\left|\eta(z)\right|\text{ for all
}\gamma\in\operatorname{SL}_{2}(\mathbb{Z}).$

We will need the following two lemmas for the proof of Proposition 2.1.

######  Lemma 2.2.

For all $x\in\mathbb{R}$ and $y\geq\frac{\sqrt{3}}{2}$,

$\left|\eta(z)\right|=e^{-\frac{\pi}{12}y+\vartheta 1.01e^{-2\pi y}}.$

###### Proof.

Suppose $y\geq\frac{\sqrt{3}}{2}$. Since $\log\eta(z)=\frac{\pi
i}{12}z-\sum_{n=1}^{\infty}\frac{\sigma(n)}{n}e(nz)$, we have

$\left|\log\eta(z)-\frac{\pi
i}{12}z\right|\leq\sum_{n=1}^{\infty}\frac{\sigma(n)}{n}e^{-2\pi
yn}<\sum_{n=1}^{\infty}ne^{-2\pi yn}\leq\frac{e^{-2\pi
y}}{(1-e^{-\sqrt{3}\pi})^{2}}<1.01e^{-2\pi y}.$ (2.5)

∎

######  Lemma 2.3.

For all $x\in\mathbb{R}$ and $y>0$,

$\left|\eta(z)\right|<\frac{7}{9}y^{-\frac{1}{4}}.$

In fact, $y^{\frac{1}{4}}\left|\eta(z)\right|$ is maximized when
$z=\frac{1+i\sqrt{3}}{2}$, where it equals
$\left(\frac{\sqrt{2}\pi}{3\Gamma(2/3)^{3}}\right)^{\frac{1}{2}}=0.7723\dots$
[Rou08].

###### Proof.

The function $(\operatorname{Im}z)^{\frac{1}{4}}\left|\eta(z)\right|$ is
$\operatorname{SL}_{2}(\mathbb{Z})$-invariant, so we may assume $z=x+iy$ is in
the standard fundamental domain $\left\\{z\in\mathbb{H}\mid\left|z\right|\geq
1,\left|x\right|\leq\tfrac{1}{2}\right\\}$ for
$\operatorname{SL}_{2}(\mathbb{Z})$, and in particular
$y\geq\frac{\sqrt{3}}{2}$. By Lemma 2.2,

$y^{\frac{1}{4}}\left|\eta(x+iy)\right|\leq
y^{\frac{1}{4}}e^{-\frac{\pi}{12}y+1.01e^{-\sqrt{3}\pi}}\leq\left(\frac{3}{\pi}\right)^{\frac{1}{4}}e^{-\frac{\pi}{12}\frac{3}{\pi}+1.01e^{-\sqrt{3}\pi}}<\frac{7}{9}.$
(2.6)

∎

###### Proof of Proposition 2.1.

Choose $\gamma=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\in\operatorname{SL}_{2}(\mathbb{Z})$ such that
$\operatorname{Im}\gamma z\geq\frac{\sqrt{3}}{2}$, and assume without loss of
generality that $c>0$ and $0\leq d<c$. By (2.4) and Lemma 2.2,

$\displaystyle\eta(iy)$ $\displaystyle=y^{-\frac{1}{2}}\eta(i/y)\leq
y^{-\frac{1}{2}}e^{-\frac{\pi}{12}\frac{1}{y}+1.01e^{-2000\pi}}\quad\text{and}$
(2.7) $\displaystyle\left|\eta(z)\right|$
$\displaystyle=\left(\frac{y}{\operatorname{Im}\gamma
z}\right)^{-\frac{1}{4}}\left|\eta(\gamma
z)\right|\geq\left(\frac{y}{\operatorname{Im}\gamma
z}\right)^{-\frac{1}{4}}e^{-\frac{\pi}{12}\operatorname{Im}\gamma
z-1.01e^{-\sqrt{3}\pi}},$

so we obtain

$\left|\frac{\eta(iy)}{\eta(z)}\right|<\left(y\operatorname{Im}\gamma
z\right)^{-\frac{1}{4}}e^{-\frac{\pi}{12}\left(\frac{1}{y}-\operatorname{Im}\gamma
z\right)+\frac{1}{228}}.$ (2.8)

It remains to bound $\left|\frac{\eta(tz)^{t}}{\eta(ity)^{t}}\right|$, for
which we split up into two cases depending on whether $ty\geq 1$ or $ty\leq
1$.

First, suppose $ty\geq 1$, so that by Lemma 2.2,

$\left|\frac{\eta(tz)^{t}}{\eta(ity)^{t}}\right|=e^{\vartheta 2.02te^{-2\pi
ty}}\leq e^{2.02e^{-2\pi}\frac{1}{y}}<e^{\frac{1}{265}\frac{1}{y}},$ (2.9)

and hence

$\left|\frac{f_{t}(z)}{f_{t}(iy)}\right|=\left|\frac{\eta(tz)^{t}}{\eta(ity)^{t}}\right|\left|\frac{\eta(iy)}{\eta(z)}\right|<\left(y\operatorname{Im}\gamma
z\right)^{-\frac{1}{4}}e^{-\frac{\pi}{12}\left(\frac{1}{y}-\operatorname{Im}\gamma
z-\frac{12}{\pi}\frac{1}{265}\frac{1}{y}\right)+\frac{1}{228}}$ (2.10)

Note that $y\operatorname{Im}\gamma z=\frac{y^{2}}{(cx+d)^{2}+c^{2}y^{2}}$. If
$c>1$, then $y\operatorname{Im}\gamma z\leq\frac{1}{c^{2}}\leq\frac{1}{4}$.
Otherwise, $c=1$ and $d=0$, and $y\operatorname{Im}\gamma
z=\frac{y^{2}}{x^{2}+y^{2}}\leq\frac{9}{10}$. The function $v\mapsto
v^{-\frac{1}{4}}e^{\frac{\pi}{12}\frac{1}{y}v}$ is bounded in the range
$\left[\frac{\sqrt{3}}{2}y,\frac{9}{10}\right]$ by its value at
$\frac{9}{10}$, so

$\left|\frac{f_{t}(z)}{f_{t}(iy)}\right|<\left(\frac{9}{10}\right)^{-\frac{1}{4}}e^{-\frac{\pi}{12}\frac{1}{y}\left(1-\frac{9}{10}-\frac{12}{\pi}\frac{1}{265}\right)+\frac{1}{228}}<e^{-\frac{1}{70}\frac{1}{y}}$
(2.11)

as claimed.

Now, suppose $ty\leq 1$, so that by Lemma 2.2,

$\eta(ity)^{t}\geq(ty)^{-\frac{t}{2}}e^{-\frac{\pi}{12}\frac{1}{y}-1.01e^{-2\pi}t},$
(2.12)

and hence

$\left|\frac{f_{t}(z)}{f_{t}(iy)}\right|\leq\eta(tz)^{t}(ty)^{\frac{t}{2}}(y\operatorname{Im}\gamma
z)^{-\frac{1}{4}}e^{\frac{\pi}{12}\operatorname{Im}\gamma z+\frac{1}{528}t}.$
(2.13)

We will split into two further cases, depending on whether
$\operatorname{Im}\gamma z\leq\frac{\sqrt{3}}{2}t$ or $\operatorname{Im}\gamma
z\geq\frac{\sqrt{3}}{2}t$.

In the first case, when $\operatorname{Im}\gamma z\leq\frac{\sqrt{3}}{2}t$, we
may use Lemma 2.3 to bound $\eta(tz)$ by $\frac{7}{9}(ty)^{-\frac{1}{4}}$ and
obtain

$\left|\frac{f_{t}(z)}{f_{t}(iy)}\right|\leq\left(\frac{7}{9}\right)^{t}(ty)^{\frac{t}{4}}(y\operatorname{Im}\gamma
z)^{-\frac{1}{4}}e^{\frac{\pi}{12}\operatorname{Im}\gamma z+\frac{1}{528}t},$
(2.14)

which is maximized in the range $\frac{\sqrt{3}}{2}\leq\operatorname{Im}\gamma
z\leq\frac{\sqrt{3}}{2}t$ when $\operatorname{Im}\gamma
z=\frac{\sqrt{3}}{2}t$, where it is equal to

$\left(\frac{7}{9}\right)^{t}(ty)^{\frac{t-1}{4}}\left(\frac{\sqrt{3}}{2}\right)^{-\frac{1}{4}}e^{\left(\frac{\sqrt{3}}{24}\pi+\frac{1}{528}\right)t}<1.04(ty)^{\frac{t-1}{4}}e^{-\frac{1}{45}t}.$
(2.15)

Let us consider now the second case, where $\operatorname{Im}\gamma
z\geq\frac{\sqrt{3}}{2}t$. Let $\gamma^{\prime}=\begin{pmatrix}*&*\\\
c/(t,c)&dt/(t,c)\end{pmatrix}\in\operatorname{SL}_{2}(\mathbb{Z})$, which
exists since $\frac{c}{(t,c)}$ and $\frac{dt}{(t,c)}$ are coprime. Then,

$\left|\eta(tz)\right|=\left(\frac{ty}{\operatorname{Im}\gamma^{\prime}tz}\right)^{-\frac{1}{4}}\left|\eta(\gamma^{\prime}tz)\right|,$
(2.16)

and

$\operatorname{Im}\gamma^{\prime}tz=\frac{ty}{\left|\frac{c}{(t,c)}tz+\frac{dt}{(t,c)}\right|^{2}}=\frac{(t,c)^{2}}{t}\operatorname{Im}\gamma
z\geq\frac{\sqrt{3}}{2}.$ (2.17)

Therefore,

$\left|\eta(tz)\right|\leq\sqrt{\frac{(t,c)}{t}}\left(\frac{y}{\operatorname{Im}\gamma
z}\right)^{-\frac{1}{4}}e^{-\frac{\pi}{12}\frac{(t,c)^{2}}{t}\operatorname{Im}\gamma
z+1.01e^{-\sqrt{3}\pi}},$ (2.18)

so

$\left|\frac{f_{t}(z)}{f_{t}(iy)}\right|\leq(t,c)^{\frac{t}{2}}(y\operatorname{Im}\gamma
z)^{\frac{t-1}{4}}e^{-\frac{\pi}{12}((t,c)^{2}-1)\operatorname{Im}\gamma
z+\frac{1}{159}t}.$ (2.19)

Suppose first that $(t,c)>1$. The function $v\mapsto
v^{\frac{t-1}{4}}e^{-\frac{\pi}{12}((t,c)^{2}-1)v}$ is maximized at
$v=\frac{3}{\pi}\frac{t-1}{(t,c)^{2}-1}<\frac{\sqrt{3}}{2}t$ and decreasing
for larger $v$, so the right-hand side of (2.19) is bounded in the range
$\operatorname{Im}\gamma z\geq\frac{\sqrt{3}}{2}t$ by its value when
$\operatorname{Im}\gamma z=\frac{\sqrt{3}}{2}t$. Hence,

$\displaystyle\left|\frac{f_{t}(z)}{f_{t}(iy)}\right|$
$\displaystyle\leq(t,c)^{\frac{t}{2}}\left(\frac{\sqrt{3}}{2}ty\right)^{\frac{t-1}{4}}e^{-\frac{\sqrt{3}}{24}\pi((t,c)^{2}-1)t+\frac{1}{159}t}$
(2.20)
$\displaystyle<2^{\frac{t}{2}}\left(\frac{\sqrt{3}}{2}ty\right)^{\frac{t-1}{4}}e^{-\frac{\sqrt{3}}{8}\pi
t+\frac{1}{159}t}$
$\displaystyle<\left(ty\right)^{\frac{t-1}{4}}e^{-\frac{1}{3}t}.$

What remains is the case $(t,c)=1$, where the right-hand side of (2.19)
collapses to become $(y\operatorname{Im}\gamma
z)^{\frac{t-1}{4}}e^{\frac{1}{159}t}$. Let

$\mathfrak{m}(c,d)=\left\\{x\mid\operatorname{Im}\gamma
z\geq\frac{\sqrt{3}}{2}t\right\\}=\left\\{x\mid\left|\frac{x+\frac{d}{c}}{y}\right|<\sqrt{\frac{2}{\sqrt{3}}\frac{1}{c^{2}ty}-1}\right\\}$
(2.21)

denote the relevant portion of the interval, so that

$\int_{\frac{y}{3}\leq\left|x\right|\leq\frac{1}{2}}\left|\frac{f_{t}(z)}{f_{t}(iy)}\right|dx\leq\int_{\begin{subarray}{c}\mathfrak{m}(1,0)\\\
\left|x\right|>\frac{y}{3}\end{subarray}}\left|\frac{f_{t}(z)}{f_{t}(iy)}\right|dx+\sum_{c=2}^{\infty}\sum_{\begin{subarray}{c}d=0\\\
(c,d)=1\end{subarray}}^{c-1}\int_{\mathfrak{m}(c,d)}\left|\frac{f_{t}(z)}{f_{t}(iy)}\right|dx+1.04(ty)^{\frac{t-1}{4}}e^{-\frac{1}{45}t}.$
(2.22)

Letting $u=\frac{x+\frac{d}{c}}{y}$, we find

$\displaystyle\int_{\mathfrak{m}(c,d)}\left|\frac{f_{t}(z)}{f_{t}(iy)}\right|dx$
$\displaystyle\leq
e^{\frac{1}{159}t}\int_{\mathfrak{m}(c,d)}\left(\frac{y^{2}}{\left|cz+d\right|^{2}}\right)^{\frac{t-1}{4}}dx$
(2.23)
$\displaystyle=yc^{\frac{1-t}{2}}e^{\frac{1}{159}t}\int_{\left|u\right|<\sqrt{\frac{2}{\sqrt{3}}\frac{1}{c^{2}ty}-1}}(u^{2}+1)^{\frac{1-t}{4}}du$
$\displaystyle<0.2yc^{\frac{1-t}{2}}e^{\frac{1}{159}t},$

so

$\displaystyle\sum_{c=2}^{\infty}\sum_{\begin{subarray}{c}d=0\\\
(c,d)=1\end{subarray}}^{c-1}\int_{\mathfrak{m}(c,d)}\left|\frac{f_{t}(z)}{f_{t}(iy)}\right|dx<0.2y\left(\zeta\left(\tfrac{t-3}{2}\right)-1\right)e^{\frac{1}{159}t}<ye^{-\frac{1}{3}t}.$
(2.24)

Letting $v=1+(\frac{x}{y})^{2}$, we also have

$\displaystyle\int_{\begin{subarray}{c}\mathfrak{m}(1,0)\\\
\left|x\right|>\frac{y}{3}\end{subarray}}\left|\frac{f_{t}(z)}{f_{t}(iy)}\right|dx$
$\displaystyle\leq
e^{\frac{1}{159}t}\int_{\begin{subarray}{c}\mathfrak{m}(1,0)\\\
\left|x\right|>\frac{y}{3}\end{subarray}}\left(\frac{y^{2}}{\left|z\right|^{2}}\right)^{\frac{t-1}{4}}dx$
(2.25)
$\displaystyle=ye^{\frac{1}{159}t}\int_{\frac{10}{9}}^{\frac{2}{\sqrt{3}}\frac{1}{ty}}v^{\frac{1-t}{4}}\frac{1}{\sqrt{v-1}}dv$
$\displaystyle<3ye^{\frac{1}{159}t}\int_{\frac{10}{9}}^{\infty}v^{\frac{1-t}{4}}dv$
$\displaystyle<0.02ye^{-\frac{1}{50}t}.$

Altogether, we have

$\int_{\frac{y}{3}\leq\left|x\right|\leq\frac{1}{2}}\left|\frac{f_{t}(z)}{f_{t}(iy)}\right|dx<0.02ye^{-\frac{1}{50}t}+ye^{-\frac{1}{3}t}+1.04(ty)^{\frac{t-1}{4}}e^{-\frac{1}{45}t}<ye^{-\frac{1}{70}t},$
(2.26)

as claimed. ∎

##  3 Technical lemmas

###  3.1 Explicit Gaussian integrals

In §4, we will use the following explicit bound for the error term in a
Gaussian integral. The proof is elementary, but somewhat long.

######  Lemma 3.1.

Suppose $\alpha>38$, $\beta\in\left(-\frac{2}{25},\frac{2}{25}\right)$, and
$\varepsilon\in(-1,1)$. Then,

$\displaystyle\int_{-1/3}^{1/3}e(\beta x)e^{-\pi\alpha
x^{2}\left(1+2i\varepsilon x+\vartheta 3x^{2}\right)}dx$
$\displaystyle=\frac{1}{\sqrt{\alpha}}\left(1+\vartheta\frac{3.45}{\alpha}\right)$
$\displaystyle\left|\int_{-1/3}^{1/3}xe^{-\pi\alpha x^{2}\left(1+2i\varepsilon
x+\vartheta 3x^{2}\right)}dx\right|$ $\displaystyle\leq
2\alpha^{-\frac{3}{2}}.$

###### Proof.

Let

$I=\int_{-1/3}^{1/3}e(\beta x)e^{-\pi\alpha x^{2}\left(1+2i\varepsilon
x+\vartheta 3x^{2}\right)}dx,$ (3.1)

which we decompose as $I=I_{1}-\varepsilon I_{2}+\vartheta I_{3}$ where

$\displaystyle I_{1}$ $\displaystyle=\int_{-1/3}^{1/3}e(\beta x)e^{-\pi\alpha
x^{2}}dx,$ (3.2) $\displaystyle I_{2}$ $\displaystyle=\int_{-1/3}^{1/3}2\pi
i\alpha x^{3}e(\beta x)e^{-\pi\alpha x^{2}}dx,\quad\text{and}$ $\displaystyle
I_{3}$ $\displaystyle=\int_{-1/3}^{1/3}\left|(1-2\pi i\varepsilon\alpha
x^{3})e(\beta x)e^{-\pi\alpha x^{2}}-e(\beta x)e^{-\pi\alpha
x^{2}\left(1+2i\varepsilon x+\vartheta 3x^{2}\right)}\right|dx.$

The main term $I_{1}$ is simply

$\displaystyle I_{1}$ $\displaystyle=\int_{-\infty}^{\infty}e(\beta
x)e^{-\pi\alpha x^{2}}dx+\vartheta 2\int_{1/3}^{\infty}e^{-\pi\alpha x^{2}}dx$
(3.3)
$\displaystyle=e^{-\pi\frac{\beta^{2}}{\alpha}}\frac{1}{\sqrt{\alpha}}+\vartheta
2\int_{1/3}^{\infty}\frac{2\pi\alpha x}{2\pi\alpha\frac{1}{3}}e^{-\pi\alpha
x^{2}}dx$
$\displaystyle=e^{-\pi\frac{\beta^{2}}{\alpha}}\frac{1}{\sqrt{\alpha}}+\vartheta\frac{3}{\pi\alpha}e^{-\frac{\pi}{9}\alpha}.$

Similarly, the term $I_{2}$ is

$\displaystyle I_{2}$
$\displaystyle=2\pi\alpha\int_{-\infty}^{\infty}ix^{3}e(\beta x)e^{-\pi\alpha
x^{2}}dx+\vartheta 4\pi\alpha\int_{1/3}^{\infty}x^{3}e^{-\pi\alpha x^{2}}dx$
(3.4)
$\displaystyle=e^{-\pi\frac{\beta^{2}}{\alpha}}\frac{1}{\sqrt{\alpha}}\left(-3\frac{\beta}{\alpha}+2\pi\frac{\beta^{3}}{\alpha^{2}}\right)+\vartheta\left(\frac{2}{9}+\frac{2}{\pi\alpha}\right)e^{-\frac{\pi}{9}\alpha}$

As for $I_{3}$, using Taylor’s theorem in the form
$e^{z}=1+ze^{sz}=1+z+\frac{1}{2}z^{2}e^{s^{\prime}z}$ for some
$s,s^{\prime}\in(0,1)$, we find that the integrand in $I_{3}$ is equal to

$\displaystyle e^{-\pi\alpha x^{2}}\left|(1-2\pi i\varepsilon\alpha
x^{3})-e^{-\pi\alpha x^{2}\left(2i\varepsilon x+\vartheta
3x^{2}\right)}\right|$ (3.5) $\displaystyle\leq e^{-\pi\alpha
x^{2}}\left(\left|(1-2\pi i\varepsilon\alpha x^{3})-e^{-2\pi
i\varepsilon\alpha x^{3}}\right|+\left|e^{-2\pi i\varepsilon\alpha
x^{3}}-e^{-\pi\alpha x^{2}\left(2i\varepsilon x+\vartheta
3x^{2}\right)}\right|\right)$ $\displaystyle\leq e^{-\pi\alpha
x^{2}}\left(2\pi^{2}\alpha^{2}x^{6}+3\pi\alpha x^{4}e^{3\pi\alpha
x^{4}}\right).$

Therefore,

$\displaystyle I_{3}$
$\displaystyle\leq\int_{-1/3}^{1/3}\left(2\pi^{2}\alpha^{2}x^{6}e^{-\pi\alpha
x^{2}}+3\pi\alpha x^{4}e^{-\pi\alpha x^{2}(1-3x^{2})}\right)dx$ (3.6)
$\displaystyle\leq 2\pi^{2}\alpha^{2}\int_{-\infty}^{\infty}x^{6}e^{-\pi\alpha
x^{2}}dx+3\pi\alpha\int_{-\infty}^{\infty}x^{4}e^{-\frac{2}{3}\pi\alpha
x^{2}}dx$
$\displaystyle=\left(\frac{15}{4\pi}+\frac{(3/2)^{\frac{9}{2}}}{\pi}\right)\alpha^{-\frac{3}{2}}.$

Altogether, we have

$\displaystyle I$
$\displaystyle=\frac{e^{-\pi\frac{\beta^{2}}{\alpha}}}{\sqrt{\alpha}}\left(1+3\varepsilon\frac{\beta}{\alpha}-2\pi\varepsilon\frac{\beta^{3}}{\alpha^{2}}\right)+\vartheta
e^{-\frac{\pi}{9}\alpha}\left(\frac{5}{\pi\alpha}+\frac{2}{9}\right)+\vartheta\left(\frac{15}{4\pi}+\frac{(3/2)^{\frac{9}{2}}}{\pi}\right)\alpha^{-\frac{3}{2}}$
(3.7)
$\displaystyle=\frac{1}{\sqrt{\alpha}}\left(1+\vartheta\frac{3.45}{\alpha}\right),$

as claimed.

We will now treat the second integral in the lemma. Let

$J=\int_{-1/3}^{1/3}xe^{-\pi\alpha x^{2}\left(1+2i\varepsilon x+\vartheta
3x^{2}\right)}dx.$ (3.8)

Since $xe^{-\pi\alpha x^{2}}$ is odd, we can write

$J=\int_{-1/3}^{1/3}x\left(e^{-\pi\alpha x^{2}\left(1+2i\varepsilon
x+\vartheta 3x^{2}\right)}-e^{-\pi\alpha x^{2}}\right)dx.$ (3.9)

The integrand has magnitude bounded by

$\displaystyle\left|x\right|e^{-\pi\alpha x^{2}}\left|e^{-2\pi
i\varepsilon\alpha x^{3}+\vartheta 3\pi\alpha x^{4}}-1\right|$ (3.10)
$\displaystyle\leq\left|x\right|e^{-\pi\alpha x^{2}}\left|-2\pi
i\varepsilon\alpha x^{3}+\vartheta 3\pi\alpha x^{4}\right|e^{3\pi\alpha
x^{4}}$ $\displaystyle\leq\left(2\pi\alpha
x^{4}+3\pi\alpha\left|x\right|^{5}\right)e^{-\pi\alpha x^{2}(1-3x^{2})},$

so we find

$\displaystyle\left|J\right|$
$\displaystyle\leq\int_{-1/3}^{1/3}\left(2\pi\alpha
x^{4}+3\pi\alpha\left|x\right|^{5}\right)e^{-\pi\alpha x^{2}(1-3x^{2})}dx$
(3.11) $\displaystyle\leq\int_{-\infty}^{\infty}\left(2\pi\alpha
x^{4}+3\pi\alpha\left|x\right|^{5}\right)e^{-\frac{2}{3}\pi\alpha x^{2}}dx$
$\displaystyle=\frac{(3/2)^{\frac{7}{2}}}{\pi}\alpha^{-\frac{3}{2}}+\frac{81}{4\pi^{2}}\alpha^{-2}$
$\displaystyle<2\alpha^{-\frac{3}{2}}.$

∎

###  3.2 The functions $\mu_{k}$

Recall the functions

$\mu_{k}(z)=-\frac{z^{k+1}}{2\pi i}\left(\frac{d}{dz}\right)^{k}\log\eta(z)$
(3.12)

from the introduction (1.6). These functions appear in the Taylor expansion

$\frac{1}{2\pi i}\log
f_{t}(z)=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}\frac{\mu_{k}(iy)-\mu_{k}(ity)}{(iy)^{k+1}},$
(3.13)

and our main goal in this section is to bound the third and fourth terms in
this Taylor expansion in terms of the second.

Using the functional equation $\eta(z)=(-iz)^{-\frac{1}{2}}\eta(-1/z)$, we
find two different expansions for $\mu_{k}(z)$, useful depending on whether
$\operatorname{Im}z$ or $\operatorname{Im}\frac{-1}{z}$ is large. We have

$\displaystyle\mu_{k}(z)$ $\displaystyle=\sum_{n=1}^{\infty}z^{k+1}(2\pi
in)^{k-1}\sigma(n)e(nz)-\begin{cases}z^{2}/24&\text{if $k=0,1$}\\\ 0&\text{if
$k\geq 2$}\end{cases}$ (3.14)
$\displaystyle=\sum_{n=1}^{\infty}P_{k}\left(\frac{2\pi
in}{z}\right)\sigma(n)e(\tfrac{-n}{z})+\frac{(-1)^{k}k!}{24}+\frac{z}{4\pi
i}\begin{cases}\log(-iz)&\text{if $k=0$}\\\ (-1)^{k-1}(k-1)!&\text{if $k\geq
1$}\end{cases}$

where $P_{k}(r)$ is the degree $k-1$ polynomial (Laurent in the case $k=0$)
defined by the recurrence $P_{0}(r)=r^{-1}$ and

$P_{k}(r)=(r-k)P_{k-1}(r)-rP_{k-1}^{\prime}(r).$ (3.15)

We will also need expansions for $\mu_{k}^{\prime}(z)$. We have

$\displaystyle\mu_{k}^{\prime}(z)$
$\displaystyle=\sum_{n=1}^{\infty}\left((2\pi inz)^{k+1}+(k+1)(2\pi
inz)^{k}\right)\frac{\sigma(n)}{2\pi in}e(nz)-\begin{cases}z/12&\text{if
$k=0,1$}\\\ 0&\text{if $k\geq 2$}\end{cases}$ (3.16)
$\displaystyle=\sum_{n=1}^{\infty}Q_{k}\left(\frac{2\pi
in}{z}\right)\frac{\sigma(n)}{2\pi in}e(\tfrac{-n}{z})+\frac{1}{4\pi
i}\begin{cases}1+\log(-iz)&\text{if $k=0$}\\\ (-1)^{k-1}(k-1)!&\text{if $k\geq
1$}\end{cases}$

where $Q_{k}(r)$ is the degree $k+1$ polynomial defined by the recurrence
$Q_{0}(r)=r+1$, and

$\displaystyle Q_{k}(r)$
$\displaystyle=(r-k+1)Q_{k-1}(r)-rQ_{k-1}^{\prime}(r)$ (3.17)
$\displaystyle=r^{2}(P_{k}(r)-P_{k}^{\prime}(r)).$

The first few values of $P_{k}(r)$ and $Q_{k}(r)$ are given in the table
(3.18) below.

$\begin{array}[]{c|c|c}k&P_{k}(r)&Q_{k}(r)\\\ \hline\cr 0&r^{-1}&r+1\\\
1&1&r^{2}\\\ 2&r-2&r^{3}-3r^{2}\\\ 3&r^{2}-6r+6&r^{4}-8r^{3}+12r^{2}\\\
4&r^{3}-12r^{2}+36r-24&r^{5}-15r^{4}+60r^{3}-60r^{2}\end{array}$ (3.18)

As a first observation, we can see directly from (3.16) and (3.18) that

$\mu_{2}^{\prime}(z)+\mu_{2}^{\prime}(\tfrac{-1}{z})=-\frac{1}{4\pi i}.$
(3.19)

######  Lemma 3.2.

  1. (i)

The function $y\mapsto i\mu_{2}^{\prime}(iy)$ is a monotonically increasing
bijection $(0,\infty)\to(-\frac{1}{4\pi},0)$.

  2. (ii)

The function $y\mapsto\mu_{2}(iy)$ is a monotonically decreasing bijection
$(0,\infty)\to(0,\frac{1}{12})$.

Suppose $0<y\leq\frac{1}{10}$ and $t>1$.

  1. (iii)

If $ty\leq 1$, then

$\frac{1}{8\pi}<\frac{\mu_{2}(iy)-\mu_{2}(ity)}{ty-y}<\frac{1}{4\pi}.$

  2. (iv)

If $ty\geq 1$, then

$\frac{1}{16}<\mu_{2}(iy)-\mu_{2}(ity)<\frac{1}{12}.$

###### Proof.

  1. (i)

From (3.16), we see $\lim_{y\to 0}i\mu_{2}^{\prime}(iy)=-\frac{1}{4\pi}$ and
$\lim_{y\to\infty}i\mu_{2}^{\prime}(iy)=0$, so it remains to show that
$i\mu_{2}^{\prime}(iy)$ is monotonically increasing. By (3.19), we may assume
$y\geq 1$, in which case each term in the sum

$i\mu_{2}^{\prime}(iy)=\sum_{n=1}^{\infty}\left(-(2\pi ny)^{3}+3(2\pi
ny)^{2}\right)\frac{\sigma(n)}{2\pi n}e^{-2\pi ny}$ (3.20)

is monotonically increasing.

  2. (ii)

From (3.14), we see $\lim_{y\to 0}\mu_{2}(iy)=\frac{1}{12}$ and
$\lim_{y\to\infty}\mu_{2}(iy)=0$, and $\mu_{2}(iy)$ is monotonically
decreasing by part (i).

  3. (iii)

By the mean value theorem,
$\frac{\mu_{2}(iy)-\mu_{2}(iv)}{y-ty}=i\mu_{2}^{\prime}(iv)$ for some
$v\in(y,ty)\subset(0,1]$. By part (i),

$-\frac{1}{4\pi}<i\mu_{2}^{\prime}(iv)<i\mu_{2}^{\prime}(i)=-\frac{1}{8\pi}.$
(3.21)

  4. (iv)

By part (ii),

$0.0635\approx\mu_{2}\left(\frac{i}{10}\right)-\mu_{2}(i)\leq\mu_{2}(iy)-\mu_{2}(ity)<\frac{1}{12}-0.$
(3.22)

∎

######  Lemma 3.3.

  1. (i)

For all $y>0$, $0<i\mu_{3}^{\prime}(iy)<\frac{3}{4\pi}$.

  2. (ii)

The function $y\mapsto\mu_{3}(iy)$ is a monotonically increasing bijection
$(0,\infty)\to(-\frac{1}{4},0)$.

  3. (iii)

If $0<y\leq\frac{1}{10}$ and $t>1$, then

$\left|\frac{\mu_{3}(iy)-\mu_{3}(ity)}{\mu_{2}(iy)-\mu_{2}(ity)}\right|<6.$

###### Proof.

  1. (i)

First, suppose $y\leq 1$. The function $Q_{3}(r)e^{-r}=r^{2}(r-2)(r-6)e^{-r}$
is positive and bounded by $0.3$ for $r>6$, and decreasing for $r>8$, so

$\displaystyle 0<i\mu_{3}^{\prime}(iy)$
$\displaystyle=\frac{1}{2\pi}+\sum_{n=1}^{\infty}Q_{3}\left(\frac{2\pi
n}{v}\right)\frac{\sigma(n)}{2\pi n}e^{-\frac{2\pi n}{y}}$ (3.23)
$\displaystyle<\frac{1}{2\pi}+\frac{0.3}{2\pi}+\sum_{n=2}^{\infty}Q_{3}(2\pi
n)\frac{n}{2\pi}e^{-2\pi n}$ $\displaystyle<\frac{3}{4\pi}.$

Now, suppose $y\geq 1$. The function $(r^{4}-4r^{3})e^{-r}$ is positive and
decreasing for $r>6$, so

$\displaystyle 0<i\mu_{3}^{\prime}(iy)$
$\displaystyle=\sum_{n=1}^{\infty}\left((2\pi ny)^{4}-4(2\pi
ny)^{3}\right)\frac{\sigma(n)}{2\pi n}e^{-2\pi ny}$ (3.24)
$\displaystyle<\sum_{n=1}^{\infty}\left((2\pi n)^{4}-4(2\pi
n)^{3}\right)\frac{n}{2\pi}e^{-2\pi n}$ $\displaystyle<\frac{3}{4\pi}.$

  2. (ii)

From (3.14), we see $\lim_{y\to 0}\mu_{3}(iy)=-\frac{1}{4}$ and
$\lim_{y\to\infty}\mu_{3}(iy)=0$, and $\mu_{3}(iy)$ is monotonically
increasing by part (i).

  3. (iii)

If $ty\leq 1$, then by part iii of Lemma 3.2,

$\left|\frac{\mu_{3}(iy)-\mu_{3}(ity)}{\mu_{2}(iy)-\mu_{2}(ity)}\right|<8\pi\left|\frac{\mu_{3}(iy)-\mu_{3}(ity)}{y-ty}\right|=8\pi\left|\mu_{3}^{\prime}(isy)\right|<6.$
(3.25)

for some $s\in(1,t)$.

If $ty\geq 1$, then by part iv of Lemma 3.2,

$\left|\frac{\mu_{3}(iy)-\mu_{3}(ity)}{\mu_{2}(iy)-\mu_{2}(ity)}\right|<\frac{1/4-0}{1/16}=4.$
(3.26)

∎

######  Lemma 3.4.

Suppose $z=x+iy$ with $0<y\leq\frac{1}{10}$ and $\left|x\right|<\frac{y}{3}$.
For any $t>1$,

$\left|\frac{\mu_{4}(z)-\mu_{4}(tz)}{\mu_{2}(iy)-\mu_{2}(ity)}\right|<36.$

###### Proof.

We will treat separately the cases $ty\leq\frac{9}{10}$ and
$ty\geq\frac{9}{10}$.

First, suppose $ty\leq\frac{9}{10}$. By the mean value theorem,

$\left|\mu_{4}(z)-\mu_{4}(tz)\right|=(t-1)\left|z\right|\left|\mu_{4}^{\prime}(sz)\right|$
(3.27)

for some $s\in(1,t)$. Note that

$\mu_{4}^{\prime}(sz)=-\frac{3}{2\pi
i}+\sum_{n=1}^{\infty}Q_{4}\left(\frac{2\pi
in}{sz}\right)\frac{\sigma(n)}{2\pi in}e^{-\frac{2\pi in}{sz}}.$ (3.28)

Let $r=\frac{2\pi i}{sz}$, and note that
$\left|\operatorname{Im}r\right|<\frac{\operatorname{Re}r}{3}$ and
$\operatorname{Re}r>2\pi$. For fixed $u$, the function
$\left|Q_{4}(u+iv)\right|$ is increasing in $\left|v\right|$ since $Q_{4}(r)$
has real roots. Moreover, the function
$\left|Q_{4}(u(1+\frac{i}{3}))\right|e^{-u}$ is decreasing in $u$ for $u>3$.
Therefore, in the region
$\left|\operatorname{Im}r\right|<\frac{\operatorname{Re}r}{3}$ and
$\operatorname{Re}r>2\pi$, the function $\left|Q_{4}(nr)e^{-nr}\right|$ is
bounded by $\left|Q_{4}(2\pi n(1+\frac{i}{3}))\right|e^{-2\pi n}$, which means

$\left|\mu_{4}^{\prime}(sz)\right|\leq\frac{3}{2\pi}+\sum_{n=1}^{\infty}\left|Q_{4}(2\pi
n(1+i/3)\right|\frac{\sigma(n)}{2\pi n}e^{-2\pi n}<\frac{4}{\pi}.$ (3.29)

Therefore, by part iii of Lemma 3.2,

$\left|\frac{\mu_{4}(z)-\mu_{4}(tz)}{\mu_{2}(iy)-\mu_{2}(ity)}\right|<\frac{4/\pi}{1/(8\pi)}\frac{|z|}{y}<32\sqrt{\frac{10}{9}}<36.$
(3.30)

Now, suppose $ty\geq\frac{9}{10}$. Then,

$\displaystyle\left|\mu_{4}(tz)\right|$
$\displaystyle=\left|\sum_{n=1}^{\infty}z^{5}(2\pi
in)^{3}\sigma(n)e(nz)\right|$
$\displaystyle\leq\left(\frac{10}{9}\right)^{\frac{5}{2}}\sum_{n=1}^{\infty}\left(\frac{9}{10}\right)^{5}(2\pi
n)^{3}n^{2}e^{-2\pi\frac{9}{10}n}$ $\displaystyle<1,$ (3.31)

and since $\left|P_{4}\left(\frac{2\pi
in}{z}\right)\right|<10^{5}(\frac{n}{y})^{3}$,

$\displaystyle\left|\mu_{4}(z)\right|$
$\displaystyle\leq\left|1-\frac{3z}{2\pi
i}\right|+\sum_{n=1}^{\infty}\left|P_{4}\left(\frac{2\pi
in}{z}\right)\sigma(n)e(\tfrac{-n}{z})\right|\leq
1+10^{5}\sum_{n=1}^{\infty}n^{5}e^{-18\pi n}$ $\displaystyle\leq 1.01.$ (3.32)

Therefore, by part ii of Lemma 3.2,

$\left|\frac{\mu_{4}(z)-\mu_{4}(tz)}{\mu_{2}(iy)-\mu_{2}(ity)}\right|<\frac{2.01}{\mu_{2}(\frac{1}{10}i)-\mu_{2}(\frac{9}{10}i)}<36.$
(3.33)

∎

##  4 Asymptotic formulas for $c_{t}(N)$

Recall that $M=N+\frac{t^{2}-1}{24}$.

###  4.1 Proof of part i of Theorem 1.4

By (3.14), we have

$\mu_{1}(iy)=\frac{y^{2}}{24}-\sum_{n=1}^{\infty}y^{2}\sigma(n)e^{-2\pi
ny}=-\frac{1}{24}+\frac{y}{4\pi}+\sum_{n=1}^{\infty}\sigma(n)e^{-2\pi n/y},$
(4.1)

which means

$\lim_{y\to\infty}\frac{\mu_{1}(ity)-\mu_{1}(iy)}{y^{2}}=\frac{t^{2}-1}{24}\quad\text{and}\quad\lim_{y\to
0^{+}}\frac{\mu_{1}(ity)-\mu_{1}(iy)}{y^{2}}=\infty,$ (4.2)

and hence there exists $y>0$ such that
$\frac{\mu_{1}(ity)-\mu_{1}(iy)}{y^{2}}=M$.

Applying part ii of Lemma 3.2, we have

$\frac{d}{dy}\left(\frac{\mu_{1}(ity)-\mu_{1}(iy)}{y^{2}}\right)=\frac{\mu_{2}(ity)-\mu_{2}(iy)}{y^{3}}<0,$
(4.3)

so this value of $y$ is unique.

Moreover,

$-\frac{1}{24}+\frac{y}{4\pi}<\mu_{1}(iy)<\frac{y^{2}}{24},$ (4.4)

so

$\frac{\mu_{1}(ity)-\mu_{1}(iy)}{y^{2}}<\frac{t^{2}}{24}+\frac{1}{24y^{2}}-\frac{1}{4\pi
y}.$ (4.5)

By (3.16), we also have

$i\mu_{1}^{\prime}(iy)=\frac{1}{4\pi}+\sum_{n=1}^{\infty}\left(\frac{2\pi
n}{y}\right)^{2}\frac{\sigma(n)}{2\pi n}e^{-2\pi n/y}>\frac{1}{4\pi},$ (4.6)

so by the mean value theorem,

$\frac{\mu_{1}(ity)-\mu_{1}(iy)}{y^{2}}>\frac{t-1}{4\pi y}.$ (4.7)

Together, (4.5) and (4.7) imply that if
$\frac{\mu_{1}(ity)-\mu_{1}(iy)}{y^{2}}=M$, then

$\frac{t-1}{4\pi M}<y<\frac{1}{\frac{3}{\pi}+\sqrt{24N-1+\frac{9}{\pi^{2}}}}.$
(4.8)

###  4.2 Proof of part ii of Theorem 1.4

Since $c_{t}(N)=p(N)$ if $t>N$, we may assume $N$ is sufficiently large. By
Lulov and Pittel’s formula (1.2), we may also assume $t$ is sufficiently
large. Moreover, parts iii and iv of Lemma 3.2 and (4.5) imply that if
$\left|\frac{\mu_{1}(ity)-\mu_{1}(iy)}{y^{2}}-M\right|\ll\frac{1}{y}$, then

$\frac{y}{\mu_{2}(iy)-\mu_{2}(ity)}\ll\frac{1}{\min(t,\frac{1}{y})}\ll\frac{1}{\min(t,\sqrt{N})}.$
(4.9)

Therefore, the following theorem implies (and makes explicit) part ii of
Theorem 1.4.

######  Theorem 4.1.

Choose $y$ so that

$\left|\frac{\mu_{1}(ity)-\mu_{1}(iy)}{y^{2}}-M\right|<\frac{2}{25y}$

(e.g. choosing $y$ as in §4.1 sets this quantity to $0$), and suppose
$t,\frac{1}{y}\geq 1000$. Then,

$c_{t}(N)=\frac{y^{\frac{3}{2}}e^{2\pi
My}f_{t}(iy)}{\sqrt{\mu_{2}(iy)-\mu_{2}(ity)}}\left(1+\vartheta\frac{3.5y}{\mu_{2}(iy)-\mu_{2}(ity)}\right).$

###### Proof.

Consider the integral

$\displaystyle c_{t}(N)$ $\displaystyle=\int_{-1/2}^{1/2}e(-Mz)f_{t}(z)dx$
(4.10) $\displaystyle=e^{2\pi
My}f_{t}(iy)\left(\int_{-y/3}^{y/3}e\left(-Mx+\frac{1}{2\pi
i}\log\frac{f_{t}(z)}{f_{t}(iy)}\right)dx+\vartheta
ye^{-\frac{1}{70}\min(t,\frac{1}{y})}\right),$

where we have applied Proposition 2.1 to obtain the second line from the
first.

We have the Taylor expansion

$\displaystyle\frac{1}{2\pi i}\log\frac{f_{t}(z)}{f_{t}(iy)}$
$\displaystyle=x\frac{\mu_{1}(iy)-\mu_{1}(ity)}{(iy)^{2}}+\frac{x^{2}}{2}\frac{\mu_{2}(iy)-\mu_{2}(ity)}{(iy)^{3}}$
(4.11)
$\displaystyle+\frac{x^{3}}{6}\frac{\mu_{3}(iy)-\mu_{3}(ity)}{(iy)^{4}}+\frac{x^{4}}{24}\frac{\mu_{4}(z^{\prime})-\mu_{4}(tz^{\prime})}{z^{\prime
5}}$

for some $z^{\prime}=x^{\prime}+iy$ with $x^{\prime}$ between $0$ and $x$.
Lemmas 3.3 and 3.4 allow us to bound the third and fourth terms above in terms
of the quadratic term, and we can rewrite (4.11) as

$\frac{1}{2\pi
i}\log\frac{f_{t}(z)}{f_{t}(iy)}=x\frac{\mu_{1}(iy)-\mu_{1}(ity)}{(iy)^{2}}+\frac{x^{2}}{2}\frac{\mu_{2}(iy)-\mu_{2}(ity)}{(iy)^{3}}\left(1+2i\varepsilon\frac{x}{y}+\vartheta
3\frac{x^{2}}{y^{2}}\right)$ (4.12)

where $\varepsilon\in(-1,1)$.

Therefore, if we let

$\alpha=\frac{\mu_{2}(iy)-\mu_{2}(ity)}{y}\quad\text{and}\quad\beta=y\left(\frac{\mu_{1}(iy)-\mu_{1}(ity)}{(iy)^{2}}-M\right),$
(4.13)

which satisfy

$\frac{1}{26}\leq\frac{\alpha}{\min(t,1/y)}\leq\frac{1}{12}\quad\text{and}\quad\left|\beta\right|<\frac{2}{25}$
(4.14)

by parts iii and iv of Lemma 3.2 and the assumption in Theorem 4.1, then

$e\left(-Mx+\frac{1}{2\pi i}\log\frac{f_{t}(z)}{f_{t}(iy)}\right)=e\left(\beta
x/y\right)e^{-\alpha(x/y)^{2}\left(1+2i\epsilon(x/y)+\vartheta
3(x/y)^{2}\right)},$ (4.15)

and by Lemma 3.1, making the change of variables $u=x/y$, the integral in the
second line of (4.10) is simply

$y\int_{-1/3}^{1/3}e(\beta u)e^{-\pi\alpha u^{2}\left(1+2i\varepsilon
u+\vartheta
3u^{2}\right)}du=\frac{y}{\sqrt{\alpha}}\left(1+\frac{3.45}{\alpha}\right).$
(4.16)

Together with (4.10), this shows

$\displaystyle c_{t}(N)$ $\displaystyle=e^{2\pi
My}f_{t}(iy)\left(\frac{y}{\sqrt{\alpha}}\left(1+\vartheta\frac{3.45}{\alpha}\right)+\vartheta
ye^{-\frac{1}{70}\min(t,\frac{1}{y})}\right)$ (4.17)
$\displaystyle=\frac{y^{\frac{3}{2}}e^{2\pi
My}f_{t}(iy)}{\sqrt{\mu_{2}(iy)-\mu_{2}(ity)}}\left(1+\vartheta\frac{3.45}{\alpha}+\vartheta\sqrt{\alpha}e^{-\frac{1}{70}\min(t,\frac{1}{y})}\right)$
$\displaystyle=\frac{y^{\frac{3}{2}}e^{2\pi
My}f_{t}(iy)}{\sqrt{\mu_{2}(iy)-\mu_{2}(ity)}}\left(1+\vartheta\frac{3.5}{\alpha}\right),$

which completes the proof of Theorem 4.1. ∎

Note also that

$\int_{-y/3}^{y/3}\left|f_{t}(z)\right|dx\leq
f_{t}(iy)\int_{-y/3}^{y/3}e^{-\pi\alpha(x/y)^{2}\left(1+\vartheta
3(x/y)^{2}\right)}dx\leq\sqrt{\frac{3}{2\alpha}}yf_{t}(iy),$ (4.18)

a fact which will be useful in §4.4.

###  4.3 Proof of Corollary 1.7

First, note that

$\displaystyle\lim_{v\to\infty}\left(\frac{1}{24v^{2}}-\frac{1}{24}+\frac{1}{v^{2}}\mu_{1}(iv)\right)$
$\displaystyle=0\quad\text{and}$ (4.19) $\displaystyle\lim_{v\to
0^{+}}\left(\frac{1}{24v^{2}}-\frac{1}{24}+\frac{1}{v^{2}}\mu_{1}(iv)\right)$
$\displaystyle=\infty,$

and by part ii of Lemma 3.2,

$\frac{d}{dv}\left(\frac{1}{24v^{2}}-\frac{1}{24}+\frac{1}{v^{2}}\mu_{1}(iv)\right)=\frac{1}{v^{3}}\left(\mu_{2}(iv)-\frac{1}{12}\right)<0.$
(4.20)

Hence, the function
$\frac{1}{24v^{2}}-\frac{1}{24}+\frac{1}{v^{2}}\mu_{1}(iv)$ is a monotonically
decreasing bijection $(0,\infty)\to(0,\infty)$, and $v$ exists and is unique.

Let $y=\frac{v}{t}$. We begin by showing that the hypotheses of Theorem 4.1
are satisfied for $t$ sufficiently large. Indeed,

$\displaystyle\left|\frac{\mu_{1}(ity)-\mu_{1}(iy)}{y^{2}}-M\right|$
$\displaystyle=\left|\frac{1}{y^{2}}\mu_{1}(iv)+\frac{1}{24y^{2}}-\frac{1}{4\pi
y}+O\left(y^{-2}e^{-2\pi/y}\right)-\frac{t^{2}}{\kappa}-\frac{t^{2}-1}{24}+o(t)\right|$
(4.21)
$\displaystyle=\left|\frac{v^{2}}{y^{2}}\left(\frac{1}{24v^{2}}-\frac{1}{24}+\frac{1}{v^{2}}\mu_{1}(iv)-\frac{1}{\kappa}\right)-\frac{1}{4\pi
y}\right|+o(t)$ $\displaystyle=\frac{1}{4\pi y}\left(1+o(1)\right).$

Therefore, by Theorem 4.1,

$c_{t}(N)=\frac{y}{\sqrt{\alpha}}e^{2\pi
My}f_{t}(iy)\left(1+O\left(\frac{1}{\alpha}\right)\right)$ (4.22)

where

$\displaystyle\alpha=\frac{\mu_{2}(iy)-\mu_{2}(ity)}{y}=\frac{1}{y}\left(\frac{1}{12}-\mu_{2}(iv)\right)+O(1).$
(4.23)

We also have

$\displaystyle\log
f_{t}(iy)=\frac{2\pi}{y}\left(\mu_{0}(iy)-\mu_{0}(ity)\right)=\frac{2\pi}{y}\left(\frac{1}{24}-\mu_{0}(iv)\right)+\frac{1}{2}\log
y+O\left(y^{-2}e^{-2\pi/y}\right),$ (4.24)

so (4.22) becomes

$\displaystyle c_{t}(N)$
$\displaystyle=\frac{y^{2}}{\sqrt{\frac{1}{12}-\mu_{2}(iv)}}\exp\left(\frac{2\pi}{y}\left(\left(N+\frac{t^{2}-1}{24}\right)y^{2}+\frac{1}{24}-\mu_{0}(iv)\right)\right)\left(1+O\left(y\right)\right).$
(4.25)

We conclude by noting

$\frac{y^{2}}{\sqrt{\frac{1}{12}-\mu_{2}(iv)}}=\frac{\kappa}{t^{2}B(\kappa)}=\frac{1}{B(\kappa)N}\left(1+o\left(y\right)\right)$
(4.26)

and

$\displaystyle\frac{2\pi}{y}\left(\left(N+\frac{t^{2}-1}{24}\right)y^{2}+\frac{1}{24}-\mu_{0}(iv)\right)$
$\displaystyle=2\pi\sqrt{\frac{\kappa
N}{v^{2}}}\left(\frac{v^{2}}{\kappa}+\frac{v^{2}}{24}+\frac{1}{24}-\mu_{0}(iv)\right)+o(1)$
(4.27) $\displaystyle=2\pi\sqrt{A(\kappa)N}+o(1).$

###  4.4 Proof of part iii of Theorem 1.4

We will prove the following explicit version of part iii of Theorem 1.4, from
which part iii of Theorem 1.4 immediately follows.

######  Theorem 4.2.

Choose $y$ as in §4.1, and suppose $t,\frac{1}{y}\geq 1000$, and
$ty\geq\frac{1}{2}$. Then,

$c_{t+1}(N+t)-c_{t}(N+t)=c_{t}(N)\left(2\pi ty-1+\vartheta
ty\left(705y+120tye^{-2\pi ty}\right)\right).$

We begin with the following lemma, which will allow us to interpolate between
$c_{t+1}(N)$ and $c_{t}(N)$.

######  Lemma 4.3.

If $ty\geq\frac{1}{2}$, then

$\displaystyle\left|\log\left(e\left(-\tfrac{2t+1}{24}z\right)\frac{f_{t+1}(z)}{f_{t}(z)}\right)\right|$
$\displaystyle\leq 7.5\left|tz\right|e^{-2\pi ty}\quad\text{and}$
$\displaystyle\left|\log\left(e\left(-\tfrac{2t+1}{24}z\right)\frac{f_{t+1}(z)}{f_{t}(z)}\right)+e(tz)(2\pi
itz+1)\right|$ $\displaystyle\leq\left(40\left|z\right|+22e^{-2\pi
ty}\right)\left|tz\right|e^{-2\pi ty}.$

###### Proof.

By the mean value theorem,

$\displaystyle\frac{1}{2\pi
i}\log\left(e\left(-\tfrac{2t+1}{24}z\right)\frac{f_{t+1}(z)}{f_{t}(z)}\right)$
$\displaystyle=\frac{1}{z}\left(\mu_{0}(tz)+\frac{(tz)^{2}}{24}-\left(\mu_{0}((t+1)z)+\frac{((t+1)z)^{2}}{24}\right)\right)$
(4.28) $\displaystyle=-\mu_{0}^{\prime}((t+\delta)z)-\frac{(t+\delta)z}{12}$

for some $\delta\in(0,1)$. Let $w=(t+\delta)z$, $u=(t+\delta)x$, and
$v=(t+\delta)y$ (so $w=u+iv$).

Consider the function

$\mu_{0}^{\prime}(w)+\frac{w}{12}=\sum_{n=1}^{\infty}(2\pi
inw+1)\frac{\sigma(n)}{2\pi in}e(nw)$ (4.29)

and note that if $v\geq\frac{1}{2}$, then

$\left|\mu_{0}^{\prime}(w)+\frac{w}{12}\right|\leq\sum_{n=1}^{\infty}\left|2\pi
inw+1\right|\frac{\sigma(n)}{2\pi n}e^{-2\pi
nv}\leq\left|w\right|\sum_{n=1}^{\infty}\sigma(n)e^{-2\pi
nv}<\frac{15}{4\pi}\left|w\right|e^{-2\pi v}.$ (4.30)

Therefore,

$\left|\log\left(e\left(-\tfrac{2t+1}{24}z\right)\frac{f_{t+1}(z)}{f_{t}(z)}\right)\right|\leq
2\pi\left|\mu_{0}^{\prime}(w)+\frac{w}{12}\right|\leq
7.5\left|w\right|e^{-2\pi v}\leq 7.5\left|tz\right|e^{-2\pi ty},$ (4.31)

which is the first inequality claimed. For the second, we have

$\displaystyle{\left|\log\left(e\left(-\tfrac{2t+1}{24}z\right)\frac{f_{t+1}(z)}{f_{t}(z)}\right)+(2\pi
iw+1)e(w)\right|}$ (4.32)
$\displaystyle=2\pi\left|-\mu_{0}^{\prime}(w)-\frac{w}{12}+(2\pi
iw+1)\frac{1}{2\pi i}e(w)\right|$
$\displaystyle=2\pi\left|\sum_{n=2}^{\infty}(2\pi inw+1)\frac{\sigma(n)}{2\pi
in}e(nw)\right|$ $\displaystyle\leq
2\pi\left|w\right|\sum_{n=2}^{\infty}\sigma(n)e^{-2\pi nv}$
$\displaystyle<22\left|w\right|e^{-4\pi v}$ $\displaystyle\leq
22\left|tz\right|e^{-4\pi ty}$

and, applying the mean value theorem,

$\left|(2\pi iw+1)e(w)-(2\pi itz+1)e(tz)\right|\leq
40\left|tz^{2}\right|e^{-2\pi ty}.$ (4.33)

Adding (LABEL:eqn:func_0-mvt-1) and (4.33) completes the proof. ∎

###### Proof of Theorem 4.2.

Recall that we assume $t,\frac{1}{y}\geq\frac{1}{1000}$ and
$ty\geq\frac{1}{2}$.

Consider the integral

$c_{t+1}(N+t)-c_{t}(N+t)=\int_{-1/2}^{1/2}e(-(M+t)z)f_{t}(z)\left(e\left(-\tfrac{2t+1}{24}z\right)\frac{f_{t+1}(z)}{f_{t}(z)}-1\right)dx.$
(4.34)

We will begin by bounding the contribution to this integral where
$\left|x\right|\geq\frac{y}{3}$. By splitting up into two cases corresponding
to $ty\leq\frac{1}{2\pi}\log(4t)$ and $ty\geq\frac{1}{2\pi}\log(4t)$, we will
show

$\int_{\frac{y}{3}\leq\left|x\right|\leq\frac{1}{2}}\left|e(-(M+t)z)f_{t}(z)\left(e\left(-\tfrac{2t+1}{24}z\right)\frac{f_{t+1}(z)}{f_{t}(z)}-1\right)dx\right|\leq
10tye^{-\frac{1}{70}\min(t,\frac{1}{y})}e^{2\pi My}f_{t}(iy).$ (4.35)

First, suppose $ty\leq\frac{1}{2\pi}\log(4t)$. By Proposition 2.1, under the
assumptions above on $t$ and $y$, we have

$\displaystyle\int_{\frac{y}{3}\leq\left|x\right|\leq\frac{1}{2}}\left|f_{t}(z)\right|dx$
$\displaystyle\leq
ye^{-\frac{1}{70}\min(t,\frac{1}{y})}f_{t}(iy)\quad\text{and}$ (4.36)
$\displaystyle\int_{\frac{y}{3}\leq\left|x\right|\leq\frac{1}{2}}\left|f_{t+1}(z)\right|dx$
$\displaystyle\leq ye^{-\frac{1}{70}\min(t,\frac{1}{y})}f_{t+1}(iy).$

By Lemma 4.3,

$e^{2\pi\frac{2t+1}{24}y}\frac{f_{t+1}(iy)}{f_{t}(iy)}\leq e^{7.5tye^{-2\pi
ty}}<1.2,$ (4.37)

which means,

$\displaystyle\int_{\frac{y}{3}\leq\left|x\right|\leq\frac{1}{2}}\left|e(-(M+t)z)f_{t}(z)\left(e\left(-\tfrac{2t+1}{24}z\right)\frac{f_{t+1}(z)}{f_{t}(z)}-1\right)dx\right|$
(4.38) $\displaystyle\quad\leq
2.2ye^{-\frac{1}{70}\min(t,\frac{1}{y})}e^{2\pi(M+t)y}f_{t}(iy).$

Since $e^{2\pi ty}\leq 4t$, this proves (4.35) in this case.

Now, suppose $ty\geq\frac{1}{2\pi}\log(4t)$. By Lemma 4.3, we find

$\left|e\left(-\tfrac{2t+1}{24}z\right)\frac{f_{t+1}(z)}{f_{t}(z)}-1\right|\leq
7.5\left|tz\right|e^{-2\pi ty}e^{7.5\left|tz\right|e^{-2\pi ty}}\leq
10te^{-2\pi ty},$ (4.39)

In this case, Proposition 2.1 implies

$\int_{\frac{y}{3}\leq\left|x\right|\leq\frac{1}{2}}\left|e(-(M+t)z)f_{t}(z)\left(e\left(-\tfrac{2t+1}{24}z\right)\frac{f_{t+1}(z)}{f_{t}(z)}-1\right)dx\right|\leq
10tye^{-\frac{1}{70}\min(t,\frac{1}{y})}e^{2\pi My}f_{t}(iy),$ (4.40)

as in (4.35).

Our next goal is to replace
$e\left(-\tfrac{2t+1}{24}z\right)\frac{f_{t+1}(z)}{f_{t}(z)}-1$ in the
integrand with $-(2\pi itz+1)e(tz)$ in the range $\left|x\right|<\frac{y}{3}$.
To that end, let
$\Upsilon(z)=\frac{f_{t+1}(z)}{f_{t}(z)}e\left(-\tfrac{2t+1}{24}z\right)$, and
apply Lemma 4.3 to find

$\displaystyle{\left|\Upsilon(z)-1+e(tz)(2\pi itz+1)\right|}$ (4.41)
$\displaystyle\leq{\left|\log\Upsilon(z)+e(tz)(2\pi
itz+1)\right|}+{\left|\Upsilon(z)-1-\log\Upsilon(z)\right|}$
$\displaystyle\leq(40\left|z\right|+22e^{-2\pi ty})\left|tz\right|e^{-2\pi
ty}+\frac{1}{2}\left|7.5tze^{-2\pi ty}\right|^{2}e^{7.5\left|tz\right|e^{-2\pi
ty}}$ $\displaystyle\leq\left(45y+85tye^{-2\pi ty}\right)tye^{-2\pi ty}.$

Therefore, if we let

$\alpha=\frac{\mu_{2}(iy)-\mu_{2}(ity)}{y}\in\left(\frac{\min(t,\frac{1}{y})}{26},\frac{\min(t,\frac{1}{y})}{12}\right)$
(4.42)

as in §4.2, then applying (4.18), we have

$\displaystyle\int_{-y/3}^{y/3}e(-(M+t)z)f_{t}(z)\left(e\left(-\tfrac{2t+1}{24}z\right)\frac{f_{t+1}(z)}{f_{t}(z)}-1\right)dx$
(4.43) $\displaystyle=\int_{-y/3}^{y/3}e(-(M+t)z)f_{t}(z)e(tz)(-2\pi itz-1)dx$
$\displaystyle\qquad+\vartheta\sqrt{\frac{3}{2\alpha}}ye^{2\pi(M+t)y}f_{t}(iy)\left(45y+85tye^{-2\pi
ty}\right)tye^{-2\pi ty}$ $\displaystyle=\int_{-y/3}^{y/3}e(-Mz)f_{t}(z)(-2\pi
itz-1)dx$ $\displaystyle\qquad+\left(56y+105tye^{-2\pi
ty}\right)\frac{ty^{2}}{\sqrt{\alpha}}e^{2\pi My}f_{t}(iy).$

By Lemma 3.1 and (4.15), we have

$\displaystyle\left|\int_{-y/3}^{y/3}e(-Mz)f_{t}(z)(-2\pi itx)dx\right|$
$\displaystyle=2\pi ty^{2}e^{2\pi
My}f_{t}(iy)\left|\int_{-1/3}^{1/3}xe^{-\pi\alpha x^{2}\left(1+2i\varepsilon
x+\vartheta 3x^{2}\right)}dx\right|$ (4.44) $\displaystyle\leq 2\pi
ty^{2}e^{2\pi My}f_{t}(iy)\left(2\alpha^{-\frac{3}{2}}\right)$
$\displaystyle\leq 327\frac{ty^{3}}{\sqrt{\alpha}}e^{2\pi My}f_{t}(iy).$

What remains is

$\displaystyle\int_{-y/3}^{y/3}e(-Mz)f_{t}(z)(2\pi ty-1)dx$ (4.45)
$\displaystyle=(2\pi ty-1)\left(\int_{-1/2}^{1/2}e(-Mz)f_{t}(z)+\vartheta
3ye^{-\frac{1}{140}\frac{1}{y}}e^{2\pi My}f_{t}(iy)\right)$
$\displaystyle=(2\pi ty-1)\left(c_{t}(N)+\vartheta
ye^{-\frac{1}{70}\min(t,\frac{1}{y})}e^{2\pi My}f_{t}(iy)\right)$

Altogether, combining (4.35, LABEL:eqn:diff-simpler-integrand, 4.44,
LABEL:eqn:diff-y-integral), we have

$\displaystyle c_{t+1}(N+t)-c_{t}(N+t)$ $\displaystyle=(2\pi
ty-1)c_{t}(N)+\vartheta tye^{2\pi My}f_{t}(iy)$ (4.46)
$\displaystyle\qquad\cdot\left(11e^{-\frac{1}{70}\min(t,\frac{1}{y})}+383\frac{y^{2}}{\sqrt{\alpha}}+105\frac{ty^{2}}{\sqrt{\alpha}}e^{-2\pi
ty}\right).$

Finally, note that by Theorem 4.1,

$c_{t}(N)=\frac{y}{\sqrt{\alpha}}e^{2\pi
My}f_{t}(iy)\left(1+\vartheta\frac{3.51}{\alpha}\right)>0.9\frac{y}{\sqrt{\alpha}}e^{2\pi
My}f_{t}(iy),$ (4.47)

so we may write (4.46) as

$\displaystyle c_{t+1}(N+t)-c_{t}(N+t)$ (4.48)
$\displaystyle\quad=c_{t}(N)\left(2\pi
ty-1+\vartheta\frac{ty}{0.9}\left(11\frac{\sqrt{\alpha}}{y}e^{-\frac{1}{70}\min(t,\frac{1}{y})}+383y+105tye^{-2\pi
ty}\right)\right)$ $\displaystyle\quad=c_{t}(N)\left(2\pi ty-1+\vartheta
ty\left(705y+120tye^{-2\pi ty}\right)\right),$

as claimed. ∎

###  4.5 Proof of Theorem 1.5

As in §4.2, we may assume $N$ and $t$ are sufficiently large. Let
$\hat{\epsilon}=\min(\epsilon,3)$ and choose $y=\frac{t-1}{4\pi M}$, which we
may assume is sufficiently small. The hypotheses in Theorem 1.5 imply

$ty<\frac{2\pi}{(2+\hat{\epsilon})\log t}.$ (4.49)

We may repeat the proof of Proposition 2.1 to find

$\int_{y\sqrt{1+\hat{\epsilon}}\leq\left|x\right|\leq\frac{1}{2}}\left|\frac{f_{t}(z)}{f_{t}(iy)}\right|dx\ll\frac{y}{\sqrt{t-1}}e^{-\frac{1}{6}t}$
(4.50)

The hypothesis (4.49) also implies that if $\left|x\right|\leq
y\sqrt{1+\hat{\epsilon}}$, then

$\operatorname{Im}\frac{-1}{tz}\geq\frac{1}{(2+\hat{\epsilon})ty}>\frac{\log
t}{2\pi},$ (4.51)

which means

$\displaystyle f_{t}(z)$
$\displaystyle=t^{-\frac{t}{2}}(-iz)^{\frac{1-t}{2}}\frac{\eta\left(\frac{-1}{tz}\right)^{t}}{\eta\left(\frac{-1}{z}\right)}$
(4.52) $\displaystyle=t^{-\frac{t}{2}}(-iz)^{\frac{1-t}{2}}e^{\vartheta
2.02te^{-2\pi\operatorname{Im}\frac{-1}{tz}}}$
$\displaystyle=t^{-\frac{t}{2}}(-iz)^{\frac{1-t}{2}}\left(1+O\left(te^{-2\pi\operatorname{Im}\frac{-1}{tz}}\right)\right),$

and in particular $f_{t}(iy)\leq 2t^{-\frac{t}{2}}y^{\frac{1-t}{2}}$.

Our goal is to evaluate

$\displaystyle c_{t}(N)$
$\displaystyle=t^{-\frac{t}{2}}\int_{-y\sqrt{1+\hat{\epsilon}}}^{y\sqrt{1+\hat{\epsilon}}}e(-Mz)(-iz)^{\frac{1-t}{2}}dx$
(4.53)
$\displaystyle\quad+O\left(t^{\frac{2-t}{2}}\int_{-y\sqrt{1+\hat{\epsilon}}}^{y\sqrt{1+\hat{\epsilon}}}\left|e(-Mz)(-iz)^{\frac{1-t}{2}}e^{-2\pi\operatorname{Im}\frac{-1}{tz}}\right|dx\right)$
$\displaystyle\quad+O\left(e^{2\pi
My}t^{-\frac{t}{2}}y^{\frac{1-t}{2}}\frac{y}{\sqrt{t-1}}e^{-\frac{1}{6}t}\right).$

Let $\mathcal{L}$ be the contour that traverses in straight lines from
$-y\sqrt{1+\hat{\epsilon}}-i\infty$ to $-y\sqrt{1+\hat{\epsilon}}+iy$ to
$y\sqrt{1+\hat{\epsilon}}+iy$ to $y\sqrt{1+\hat{\epsilon}}-i\infty$. Hankel’s
formula states

$\int_{\mathcal{L}}e(-Mz)(-iz)^{\frac{1-t}{2}}dz=\frac{(2\pi)^{\frac{t-1}{2}}}{\Gamma\left(\frac{t-1}{2}\right)}M^{\frac{t-3}{2}},$
(4.54)

which means that the first line in the right-hand side of (4.53) is

$\frac{(2\pi)^{\frac{t-1}{2}}}{t^{\frac{t}{2}}\Gamma\left(\frac{t-1}{2}\right)}M^{\frac{t-3}{2}}+O\left(e^{2\pi
My}t^{-\frac{t}{2}}y^{\frac{1-t}{2}}\frac{(1+\hat{\epsilon})^{\frac{1-t}{4}}}{M}\right).$
(4.55)

By widening the domain of integration from $\left|x\right|\leq
y\sqrt{1+\hat{\epsilon}}$ to $\left|x\right|\leq 2y$ and making the change of
variables $u=\frac{x}{y}$, we find that the second line in the right-hand side
of (4.53) is

$\ll e^{2\pi
My}t^{\frac{2-t}{2}}y^{\frac{3-t}{2}}\int_{-2}^{2}(1+u^{2})^{\frac{1-t}{4}}e^{-\frac{2\pi}{ty}\frac{1}{1+u^{2}}}du.$
(4.56)

We will now show

$\int_{-2}^{2}(1+u^{2})^{\frac{1-t}{4}}e^{-\frac{2\pi}{ty}\frac{1}{1+u^{2}}}du\ll\max\left(e^{-\frac{1}{5}t},\frac{1}{\sqrt{t-1}}e^{-\frac{2\pi}{ty}}\right).$
(4.57)

Indeed, the function $v\mapsto v^{\frac{t-1}{4}}e^{-\frac{2\pi}{ty}v}$ is
maximized at $v=\frac{t(t-1)y}{8\pi}$. Therefore, if $\frac{t(t-1)y}{8\pi}\leq
1$, then the integrand in (4.57) is bounded by $e^{-\frac{1}{4}t}$, while if
$\frac{t(t-1)y}{8\pi}\geq 1$, then the integrand in (4.57) is bounded by
$e^{-\frac{2\pi}{ty}}$ and decreasing in $u$. This is sufficient to prove
(4.57) when $\frac{t(t-1)y}{8\pi}\leq\frac{5}{4}$. When
$\frac{t(t-1)y}{8\pi}\geq\frac{5}{4}$, note that $\frac{1}{1+u^{2}}\geq
1-u^{2}$ and $1+u^{2}\geq e^{\frac{5}{6}u^{2}}$ if
$\left|u\right|\leq\frac{1}{2}$. Therefore,

$\displaystyle\int_{-2}^{2}(1+u^{2})^{\frac{1-t}{4}}e^{-\frac{2\pi}{ty}\frac{1}{1+u^{2}}}du$
$\displaystyle\ll\int_{-1/2}^{1/2}e^{-\frac{5}{24}(t-1)u^{2}}e^{-\frac{2\pi}{ty}(1-u^{2})}du$
(4.58) $\displaystyle\leq
e^{-\frac{2\pi}{ty}}\int_{-\infty}^{\infty}e^{-\frac{1}{120}(t-1)u^{2}}du$
$\displaystyle\ll\frac{1}{\sqrt{t-1}}e^{-\frac{2\pi}{ty}},$

proving (4.57).

Applying (4.55, 4.56, 4.57) to (4.53), we find

$\displaystyle c_{t}(N)$
$\displaystyle=\frac{(2\pi)^{\frac{t-1}{2}}}{t^{\frac{t}{2}}\Gamma\left(\frac{t-1}{2}\right)}M^{\frac{t-3}{2}}$
(4.59) $\displaystyle\quad+O\left(e^{2\pi
My}t^{-\frac{t}{2}}y^{\frac{1-t}{2}}\left(\frac{(1+\hat{\epsilon})^{\frac{1-t}{4}}}{M}+ty\max\left(e^{-\frac{1}{5}t},\frac{1}{\sqrt{t-1}}e^{-\frac{2\pi}{ty}}\right)+\frac{y}{\sqrt{t-1}}e^{-\frac{1}{6}t}\right)\right).$

Recalling the choice $y=\frac{t-1}{4\pi M}$ (which minimizes $e^{2\pi
My}y^{\frac{1-t}{2}}$) and applying Stirling’s formula, we find

$e^{2\pi
My}t^{-\frac{t}{2}}y^{\frac{1-t}{2}}=t^{-\frac{t}{2}}\left(\frac{4e\pi
M}{t-1}\right)^{\frac{t-1}{2}}\ll\frac{\sqrt{t-1}}{y}\frac{(2\pi)^{\frac{t-1}{2}}}{t^{\frac{t}{2}}\Gamma\left(\frac{t-1}{2}\right)}M^{\frac{t-3}{2}},$
(4.60)

and hence

$c_{t}(N)=\frac{(2\pi)^{\frac{t-1}{2}}}{t^{\frac{t}{2}}\Gamma\left(\frac{t-1}{2}\right)}M^{\frac{t-3}{2}}\left(1+O\left(t^{-\frac{1}{2}}(1+\epsilon)^{-\frac{1}{4}t}+e^{-\frac{1}{6}t}+te^{-\frac{2\pi}{ty}}\right)\right),$
(4.61)

as claimed.

######  Remark 4.4.

In §5, we will require an explicit version of Theorem 1.5 to prove Stanton’s
conjecture for $8\leq t\leq 1000$. By keeping careful track of constants and
repeating the proof above with $\epsilon=3$, we can show that if $t\geq 8$,
$M\geq 100\,000$, and $\frac{t(t-1)}{4\pi
M}<\frac{1}{5}\min\left(\frac{2}{\sqrt{3}},\frac{2\pi}{\log t}\right)$, then

$c_{t}(N)=\frac{(2\pi)^{\frac{t-1}{2}}}{t^{\frac{t}{2}}\Gamma\left(\frac{t-1}{2}\right)}M^{\frac{t-3}{2}}\left(1+\vartheta\left(2.5e^{-\frac{1}{8}t}+20t^{-4}\right)\right).$
(4.62)

###  4.6 Proof of Theorem 1.6

We may assume $N\geq 3t$, since otherwise
$c_{t}(N)=p(N)-tp(N-t)+\frac{1}{2}(t^{2}-3t)p(N-2t)$.

Let

$y=\frac{1}{\frac{3}{\pi}+\sqrt{24(N-2t)-1+\frac{9}{\pi^{2}}}},$ (4.63)

so that our assumption $t>(1+\epsilon)\frac{\sqrt{6}}{2\pi}\sqrt{N}\log N$
implies $\left|te(tz)\right|=te^{-2\pi
ty}\ll_{\epsilon}N^{-\frac{\epsilon}{2}}\log N$.

Consider the integrals

$\displaystyle p(N)$
$\displaystyle=\int_{-1/2}^{1/2}e\left(-\left(N-\tfrac{1}{24}\right)z\right)\frac{1}{\eta(z)}dx\quad\text{and}$
(4.64) $\displaystyle c_{t}(N)$
$\displaystyle=\int_{-1/2}^{1/2}e\left(-\left(N-\tfrac{1}{24}\right)z\right)\frac{1}{\eta(z)}e\left(-\tfrac{t^{2}}{24}z\right)\eta(tz)^{t}dx.$

Note that

$\log\left(e\left(-\tfrac{t^{2}}{24}z\right)\eta(tz)^{t}\right)=-\frac{2\pi
i}{z}\left(\mu_{0}(tz)+\frac{(tz)^{2}}{24}\right)=-te(tz)+O_{\epsilon}\left(te(2tz)\right),$
(4.65)

so

$\displaystyle c_{t}(N)$
$\displaystyle=\int_{-1/2}^{1/2}e\left(-\left(N-\tfrac{1}{24}\right)z\right)\frac{1}{\eta(z)}e^{-te(tz)+O_{\epsilon}\left(te(2tz)\right)}dx$
(4.66)
$\displaystyle=\int_{-1/2}^{1/2}e\left(-\left(N-\tfrac{1}{24}\right)z\right)\frac{1}{\eta(z)}\left(1-te(tz)+O_{\epsilon}\left(t^{2}e(2tz)\right)\right)dx$
$\displaystyle=p(N)-tp(N-t)+O_{\epsilon}\left(t^{2}\int_{-1/2}^{1/2}\left|e\left(-\left(N-2t-\tfrac{1}{24}\right)z\right)\frac{1}{\eta(z)}\right|dx\right)$
$\displaystyle=p(N)-tp(N-t)+O_{\epsilon}\left(t^{2}p(N-2t)\right),$

with this last equality implied by e.g. (4.18) applied to
$c_{N}(N-2t)=p(N-2t)$.

##  5 Proof of Stanton’s conjecture

In this section, we will prove Stanton’s conjecture 1.1

Recall from the introduction that we may assume $4\leq t<N-1$. To begin, we
check using SageMath [Sage] that Stanton’s conjecture holds for all $t<N-1$
and $N<100\,000$, and also that it holds for all $t<1000$ and $N<500\,000$.

Kim-Rouse [KR14] verified the conjecture for $4\leq t<8$. In §5.1, we will use
Theorem 1.5 (or rather, the explicit version mentioned in Remark 4.4) to
verify the conjecture for $8\leq t<1000$ and $N\geq 500\,000$.

If $t\geq 1000$ and $N\geq 100\,000$, then as in §4.1, choose $y$ in the range

$\frac{t-1}{4\pi
M}<y<\frac{1}{\frac{3}{\pi}+\sqrt{24N-1+\frac{9}{\pi^{2}}}}<\frac{1}{1000}$
(5.1)

such that

$\frac{\mu_{1}(ity)-\mu_{1}(iy)}{y^{2}}=M.$ (5.2)

We will complete the proof of Stanton’s conjecture by treating the case
$ty\leq\frac{1}{2}$ in §5.2 using Theorem 4.1, and the case
$ty\geq\frac{1}{2}$ in §5.3 using Theorem 4.2.

######  Remark 5.1.

For any $\epsilon>0$ and $t$ and $N$ sufficiently large depending on
$\epsilon$, we could prove Stanton’s conjecture using Theorem 1.5 if
$ty<(1-\epsilon)\frac{\pi}{\log t}$, using Theorem 4.1 if
$ty<(1-\epsilon)\frac{\log t}{2\pi}$, using Theorem 4.2 if
$ty\geq\frac{1}{2}$, and using Theorem 1.6 if $ty>(1+\epsilon)\frac{\log
t}{\pi}$, giving us some leeway in deciding which theorem to use for which
range. Our decision was based on an attempt to minimize the number of cases
that must be checked by hand.

###  5.1 Stanton’s conjecture for $8\leq t<1000$

Suppose $8\leq t\leq 1000$ and $N\geq 500\,000$. This implies
$\frac{t(t-1)}{4\pi
M}<\frac{1}{5}\min\left(\frac{2}{\sqrt{3}},\frac{2\pi}{\log t}\right)$, so by
Remark 4.4,

$c_{t}(N)=\frac{(2\pi)^{\frac{t-1}{2}}}{t^{\frac{t}{2}}\Gamma\left(\frac{t-1}{2}\right)}M^{\frac{t-3}{2}}\left(1+\vartheta
E(t)\right)$ (5.3)

where $E(t)=2.5e^{-\frac{1}{8}t}+20t^{-4}$.

Applying the inequalities

$\frac{t^{\frac{t}{2}}}{(t+1)^{\frac{t+1}{2}}}>\frac{1}{\sqrt{e(t+1)}}\quad\text{and}\quad\frac{\Gamma\left(\frac{t-1}{2}\right)}{\Gamma\left(\frac{t}{2}\right)}>\sqrt{\frac{2}{t}}$
(5.4)

yields

$\frac{c_{t+1}(N)}{c_{t}(N)}>\sqrt{\frac{4\pi
M}{et(t+1)}}\frac{1-E(t+1)}{1+E(t)}.$ (5.5)

For all $8\leq t<1000$, we have

$\sqrt{\frac{4\pi\cdot 500\,000}{et(t+1)}}\frac{1-E(t+1)}{1+E(t)}>1.5,$ (5.6)

which proves Stanton’s conjecture in this range.

###  5.2 Stanton’s conjecture for $t\geq 1000$ and $ty\leq\frac{1}{2}$

Here, we assume $t\geq 1000$, $N\geq 100\,000$, and $ty\leq\frac{1}{2}$. By
Theorem 4.1,

$c_{t}(N)=\frac{y^{\frac{3}{2}}e^{2\pi
My}f_{t}(iy)}{\sqrt{\mu_{2}(iy)-\mu_{2}(ity)}}\left(1+\vartheta\frac{3.5y}{\mu_{2}(iy)-\mu_{2}(ity)}\right).$
(5.7)

This same value of $y$ also satisfies the hypotheses in Theorem 4.1 for
$c_{t+1}(N)$. Indeed, by the mean value theorem,

$\displaystyle\left|y\left(\frac{\mu_{1}(i(t+1)y)-\mu_{1}(iy)}{y^{2}}-\left(N+\frac{(t+1)^{2}-1}{24}\right)\right)\right|$
$\displaystyle=\left|\frac{\mu_{1}(i(t+1)y)-\mu_{1}(ity)}{y}-\frac{2t+1}{24}y\right|$
(5.8) $\displaystyle=\left|i\mu_{1}^{\prime}(isy)-\frac{sy}{12}\right|$

for some $s\in(t,t+1)$, and by (3.16),

$\displaystyle\left|i\mu_{1}^{\prime}(isy)-\frac{sy}{12}\right|=\left|\frac{1}{4\pi}-\frac{sy}{12}+\sum_{n=1}^{\infty}\left(\frac{2\pi
n}{sy}\right)^{2}\frac{\sigma(n)}{2\pi n}e^{-2\pi
n/(sy)}\right|<\frac{1}{4\pi}.$ (5.9)

Therefore, Theorem 4.1 also tells us

$c_{t+1}(N)=\frac{y^{\frac{3}{2}}e^{2\pi\left(M+\frac{2t+1}{24}\right)y}f_{t+1}(iy)}{\sqrt{\mu_{2}(iy)-\mu_{2}(i(t+1)y)}}\left(1+\vartheta\frac{3.5y}{\mu_{2}(iy)-\mu_{2}(i(t+1)y)}\right).$
(5.10)

By part i of Lemma 3.2, we have

$\frac{3.5y}{\mu_{2}(iy)-\mu_{2}(ity)}\leq\frac{3.5}{-i(t-1)\mu_{2}^{\prime}(\frac{i}{2})}\leq\frac{45}{t}$
(5.11)

and

$\frac{3.5y}{\mu_{2}(iy)-\mu_{2}(i(t+1)y)}\leq\frac{3.5}{-it\mu_{2}^{\prime}(\frac{501i}{1000})}\leq\frac{45}{t},$
(5.12)

so

$\frac{c_{t+1}(N)}{c_{t}(N)}=e^{2\pi
y\frac{2t+1}{24}}\frac{f_{t+1}(iy)}{f_{t}(iy)}\sqrt{\frac{\mu_{2}(iy)-\mu_{2}(ity)}{\mu_{2}(iy)-\mu_{2}(i(t+1)y)}}\frac{1+\vartheta\frac{45}{t}}{1+\vartheta\frac{45}{t}}.$
(5.13)

As in (4.28), there is some $s\in(t,t+1)$ such that

$\displaystyle\log\left(e^{2\pi\left(\frac{2t+1}{24}\right)y}\frac{f_{t+1}(iy)}{f_{t}(iy)}\right)$
$\displaystyle=-2\pi i\left(\mu_{0}^{\prime}(isy)+\frac{isy}{12}\right)$
(5.14)
$\displaystyle=-\frac{1}{2}\log(esy)+\frac{\pi}{6}sy-\sum_{n=1}^{\infty}\left(\frac{2\pi
n}{sy}+1\right)\frac{\sigma(n)}{n}e^{-2\pi n/(sy)}$ $\displaystyle>0.1.$

We also have

$\frac{\mu_{2}(iy)-\mu_{2}(ity)}{\mu_{2}(iy)-\mu_{2}(i(t+1)y)}=1-\frac{\mu_{2}(ity)-\mu_{2}(i(t+1)y)}{\mu_{2}(iy)-\mu_{2}(i(t+1)y)}=1+\vartheta\frac{1.1}{t},$
(5.15)

so altogether (5.13, 5.14, 5.15) imply

$\frac{c_{t+1}(N)}{c_{t}(N)}>e^{0.1}\sqrt{1-\frac{1.1}{1000}}\frac{1-\frac{45}{1000}}{1+\frac{45}{1000}}>1.$
(5.16)

###  5.3 Stanton’s conjecture for $t\geq 1000$ and $ty\geq\frac{1}{2}$

Here, we assume $t\geq 1000$, $N\geq 100\,000$, and $ty\geq\frac{1}{2}$.
Choose $y^{\prime}$ such that

$\frac{\mu_{1}(ity^{\prime})-\mu_{1}(iy^{\prime})}{y^{\prime
2}}=N-t+\frac{t^{2}-1}{24}.$ (5.17)

Since $\frac{\mu_{1}(ity)-\mu_{1}(iy)}{y^{2}}$ is monotonically decreasing in
$y$ by (4.3), we have $y^{\prime}>y\geq\frac{1}{2t}$.

If $t>\frac{N}{2}$ (so $c_{t}(N)=p(N)-tp(N-t)$), then Stanton’s conjecture was
proved by Craven [Cra06]. Otherwise, $N-t\geq\frac{N}{2}\geq 50\,000$, so

$\frac{1}{y^{\prime}}>\frac{3}{\pi}+\sqrt{24\cdot
50\,000-1+\frac{9}{\pi^{2}}}>1000,$ (5.18)

and Theorem 4.2 applies, which means

$\displaystyle c_{t+1}(N)-c_{t}(N)$ $\displaystyle=c_{t}(N-t)\left(2\pi
ty^{\prime}-1+\vartheta ty^{\prime}\left(705y^{\prime}+120ty^{\prime}e^{-2\pi
ty^{\prime}}\right)\right),$ (5.19) $\displaystyle\geq c_{t}(N-t)\left(2\pi
ty^{\prime}-1-ty^{\prime}\left(\frac{705}{1000}+60e^{-\pi}\right)\right)$
$\displaystyle>0.$

## References

  * [And08] Jaclyn Anderson  “An asymptotic formula for the $t$-core partition function and a conjecture of Stanton”  In _Journal of Number Theory_ 128.9, 2008, pp. 2591–2615  DOI: https://doi.org/10.1016/j.jnt.2007.10.006
  * [Cra06] David Craven  “The Number of $t$-Cores of Size $n$” [Online; accessed 25-February-2024], https://web.mat.bham.ac.uk/D.A.Craven/docs/papers/tcores0608.pdf, 2006
  * [EM96] Karin Erdmann and Gerhard O. Michler  “Blocks for symmetric groups and their covering groups and quadratic forms.”  In _Beiträge zur Algebra und Geometrie_ 37.1 Springer, 1996, pp. 103–118  URL: http://eudml.org/doc/227689
  * [GKS90] Frank Garvan, Dongsu Kim and Dennis Stanton  “Cranks and $t$-cores.”  In _Inventiones mathematicae_ 101.1, 1990, pp. 1–18  URL: http://eudml.org/doc/143794
  * [GO00] Andrew Granville and Ken Ono  “Defect zero blocks for finite simple groups”  In _Transactions of the American Mathematical Society_ 348, 2000  DOI: 10.1090/S0002-9947-96-01481-X
  * [JK84] Gordon James and Adalbert Kerber  “The Representation Theory of the Symmetric Group”, Encyclopedia of Mathematics and its Applications  Cambridge University Press, 1984
  * [KR14] Bhungchan Kim and Jeremy Rouse  “EXPLICIT BOUNDS FOR THE NUMBER OF $p$-CORE PARTITIONS”  In _Transactions of the American Mathematical Society_ 366.2 American Mathematical Society, 2014, pp. 875–902  URL: http://www.jstor.org/stable/23812973
  * [LP99] Nathan Lulov and Boris Pittel  “On the Random Young Diagrams and Their Cores”  In _Journal of Combinatorial Theory, Series A_ 86.2, 1999, pp. 245–280  DOI: https://doi.org/10.1006/jcta.1998.2939
  * [Mor19] Lucia Morotti  “On divisibility by primes in columns of character tables of symmetric groups”  In _Archiv der Mathematik_ 114, 2019, pp. 361–365  URL: https://api.semanticscholar.org/CorpusID:119610647
  * [Ono94] Ken Ono  “On the positivity of the number of $t$-core partitions”  In _Acta Arithmetica_ 66.3, 1994, pp. 221–228  URL: http://eudml.org/doc/206601
  * [Ono95] Ken Ono  “A Note on the Number of $t$-Core Partitions”  In _Rocky Mountain Journal of Mathematics_ 25.3 Rocky Mountain Mathematics Consortium, 1995, pp. 1165–1169  DOI: 10.1216/rmjm/1181072212
  * [Rou08] Jeremy A. Rouse  “Bounds for the coefficients of powers of the $\Delta$‐function”  In _Bulletin of the London Mathematical Society_ 40, 2008  URL: https://api.semanticscholar.org/CorpusID:9276621
  * [Sage] The Sage Developers  “SageMath, the Sage Mathematics Software System (Version 9.1)” https://www.sagemath.org, 2020
  * [Sta99] Dennis Stanton  “OPEN POSITIVITY CONJECTURES FOR INTEGER PARTITIONS”  In _Trends in Mathematics_ 2, 1999, pp. 19–25


# Slogatron: Advanced Wealthiness Generator

Bryant Chandler and Aadesh Neupane

###### Abstract

Creating catchy slogans is a demanding and clearly creative job for ad
agencies. The process of slogan creation by humans involves finding key
concepts of the company and its products, and developing a memorable short
phrase to describe the key concept. We attempt to follow the same sequence,
but with an evolutionary algorithm. A user inputs a paragraph describing
describing the company or product to be promoted. The system randomly samples
initial slogans from a corpus of existing slogans. The initial slogans are
then iteratively mutated and improved using an evolutionary algorithm.
Mutation randomly replaces words in an individual with words from the input
paragraphs. Internal evaluation measures a combination of grammatical
correctness, and semantic similarity to the input paragraphs. Subjective
analysis of output slogans leads to the conclusion that the algorithm
certainly outputs valuable slogans. External evaluation found that the slogans
were somewhat successful in conveying a message, because humans were generally
able to select the correct promoted item given a slogan.

## Introduction

Slogans (also called taglines) are considered to be an important part of a
company’s identity and recognition. It provides the brand with its own unique
image, which in turn helps in the brand recognition and recall in consumer’s
minds. Although they are vital for a company’s brand value, there is very
little consensus on what makes a good slogan. Researchers have identified
various factors which contribute to make a good slogan. (?) reported that for
a slogan to be successful it should be part of a strategic view of brand
identity as it is capable of telling where the brand is going, and it must
emphasize points of differentiation that are not only meaningful, but
congruent with existing brand perception. Even given these attributes, some
slogans fail to win the hearts of consumers. (?) found that the popularity for
a slogan may be unrelated to media expenditure, and driven largely by the
clarity of the message, the exposition of the benefits, rhymes, and
creativity. Moreover, they found that jingles or brevity have no direct
correlation on the like-ability of slogans.

Companies spend enormous sums of money to develop clever and impressive
slogans, because slogans contribute to impression of a brand among people. The
creation of a good slogan requires significant time, money and creativity for
people experienced in that particular area. The creative team starts by
researching the company’s mission, vision, services, and projects. Then they
start brainstorming about major topics related to the company. Lastly, they
filter out slogans that don’t fit the company’s profile, and using trial and
error, they select among the remaining lists of slogans.

”Connecting people”- Nokia, ”Diamonds are forever”- De Beers, ”Melts in your
mouth, not it your hands”- M&M’s are some examples of popular slogans. They
are simple, consistent, highlight the company key characteristics, and provide
credibility. Attempts have been made to embed these characteristics into
creative computation models (?; ?), but the results leave room for
improvement.

Our system reads the company’s mission, vision, and product description as
input, and generates slogans using an evolutionary algorithm with a fitness
function that attempts to favor the best slogans. The system outputs some of
its highest internally rated slogans to the user. The goal is for the human to
find slogans that they can use with little to no modification.

## Related Work

Generation of creative artifacts has remained an interesting problem for
computer scientists. They have experimented with machines which can create
music, lyrics, visual art, paintings, recipes, jokes, poems, stories, news
articles, games, and much more. We are most interested in studying machines
that generate creative and catchy sentences.

(?) produced work about incremental sentence generation and described various
constraints on the representation of the grammar for sentence generation. This
generation method doesn’t fall in the realm of creativity because of the
strict use of a grammar, but it is an initial step toward sentence generation.
The Poevolve (?) system was able to generate poems using evolutionary
techniques with neural networks for a fitness function, but the output was not
syntactically cohesive. Poem generation can be complex, because it needs to
follow many strict constraints such as: phonetics, linguistics, and semantics.
(?) developed a robust system for riddle/joke generation specifically designed
for children with complex communication needs using the ideas from the JAPE
(?) system. It uses a large-scale multimedia lexicon and a set of symbolic
rules to generate jokes.

Recently, many text generation methods have emerged to solve problems which
are similar to our problem of slogan generation. One project (?) built a
system capable of generating Internet Memes with coherent text, image, and
humor values. Another (?) developed a twitter bot that generates riddles about
celebrities by combining information from both well-structured and poorly-
structured information sources. This (?) system used WordNet (?) and
ConceptNet (?) to modify existing textual expressions to obtain more
positively or negatively biased attitudes while keeping the original meaning.
The modification is performed by adding or deleting words from existing
sentences based on the their part-of-speech. Finally, (?) blended well-known
phrases with items from the news, in order to create interesting headlines. It
was a worthy effort to create headlines in a linguistically motivated
framework that accounts for syntactic aspects of language.

A precursor to the BrainSUP (?) framework for slogan generation was a project
working on creative naming (?). It created a system which combines several
linguistic resources and natural language processing techniques to generate
neologisms based on homophonic puns and metaphors. It allowed users to
determine the category of the service to be promoted together with the
features to be emphasized. The researchers made use of WordNet and ConceptNet
to get cognitively similar words.

Despite slogan generation being extremely socially relevant, the work already
done in this sub-domain is limited. BrainSUP (?) is a good attempt to
accomplish creative sentence generation. It makes heavy use of syntactic
information to enforce well-formed sentences, and constrain the search for a
solution. Moreover, it allows users to force several words to appear in final
generated sentences, and control the generation process across emotion, color,
domain relatedness, and phonetic properties. The BrainSUP framework is general
enough to be applied to other tasks. An extension of the framework is used in
a successive study (?), which seeks to automate and evaluate the keyword
method to teach secondary language vocabulary. Extrinsic evaluation confirmed
the effectiveness of the proposed solution in the study.

This effort (?) is closely related to our application, as we were inspired by
some of their approach to evolutionary slogan generation. Their method follows
the BrainSUP framework for the initial population generation, and many of the
evaluation schemes are also inspired from BrainSUP. The researchers have used
evolutionary algorithms to generate slogans using eight evaluation functions
to guide the generator towards best slogans. A new generation of slogans is
created by pairing parents and performing crossover, followed by mutation with
fixed probabilities. They have defined two varieties of crossover and
mutation. Both varieties of crossover involve swapping different parts of a
pair of parents in random locations based on POS tags. Mutation involves
deletion of a random word, addition of an adverb, or replacement of a random
word with its synonym, antonym, meronym, holonym, hypernym or hyponym.

Evolutionary Algorithms. Many solutions developed in the realm of
computational creativity employ evolutionary algorithms for the generation of
creative artifacts. One recipe generator (?) used an evolutionary algorithm
acting on a population of recipes, each composed of a list of ingredients to
produce novel recipes. Another (?) used evolutionary algorithms to generate
meaningful poetry with sophisticated linguistic formalism for semantic
representations and patterns of stress. The popularity of evolutionary
algorithms in the field of computational creativity is due to its general
architecture for generation and simple but effective mechanisms to control and
steer the generation process.

There are too many ideas within the domain of evolutionary algorithms to
enumerate, but we will review a few that have been used in this work. The
first is mutation, which loosely encapsulates the idea of randomly modifying
an individual within the population in some way. These mutations may or may
not improve the performance of an individual. The second is niching or
speciation, which allows an evolutionary algorithm to optimize for a problem
with multiple solutions (?). The approach allows multiple populations to
develop independently, which provides populations that have optimized for
different factors. As an addition to niching, occasional crossover can be done
between niches in an attempt to produce even stronger individuals. The third
is fitness function, which guides the algorithm towards producing refined
individuals. The fitness function is a objective or constraint function which
measures how well a given solution achieves the objective.

## Approach

Our solution to the slogan generation problem is based around a naive
evolutionary algorithm with a layout as seen in Figure 1. The population for
each generation after the initial generation is made up of $1/3$ newly sampled
individuals, $1/3$ mutated individuals ($1/6$ from the top scoring and $1/6$
from the bottom scoring), and $1/3$ unmodified top scoring individuals. In
this section we will discuss initial individual generation, mutation, and
evaluation. Finally, we will discuss our use of speciation/niching to generate
a variety of slogans.

Figure 1: System diagram

### Initial Generation

For our system to generate final slogans, it needs an initial population. In
an effort to give flexibility, we initially attempted to generate text using
RNNs and Markov text generation. We found that it was difficult to generate
satisfactory text using these methods. They can frequently repeat patterns,
and the generated text is often not grammatically correct. Instead, we decided
to initiate individuals by sampling from a corpus of existing text. We started
by using Wikipedia and the Brown University corpus, but we found that the
sampled sentences were generally overly complex and lacked a slogan-like
structure. Given these issues, we moved to sampling from a corpus of almost
6000 slogans (?). Using this approach creates better output, but it has the
downside of being highly correlated with existing slogans. Future work should
look at ways to break out of the restriction to slogans that look like
existing slogans.

### Mutation

The goal of mutation is to inject words related to the seed summary into the
slogan. During initialization of the algorithm we extract all nouns and verbs
from the summary, and use the Datamuse API to get related nouns and verbs.
From the related words we construct lists of seed nouns and verbs. Mutations
are applied to the top $1/6$ and bottom $1/6$ of individuals. A mutation is
performed by selecting a random word from the individual and replacing it with
one of the seed words that belongs to the same part of speech. A weakness of
this approach is that the conjugation of verbs might not match the original,
which could make the individual have incorrect grammar. Future work should
adjust conjugation before replacement.

### Evaluation

Internal evaluation is important for the algorithm to filter out individuals
that don’t possess the desired characteristics and promote exemplary
individuals. It is frequently represented as a fitness function in
evolutionary algorithm literature. We split the evaluation function into
englishness and semantic similarity. The score for each part is a number
between zero and one, with the total score being a weighted sum of the two
parts. Semantic similarity was weighted much less than englishness, because
weighting it highly cause the algorithm to output slogans with too many of the
seed words included. A low weight on semantic similarity allows the algorithm
to favor applicable slogans, while still having good English.

Englishness. As described earlier, slogans have different properties that make
them memorable and distinctive. Slogans with proper grammatical structure will
be easier to remember and more attractive. For the algorithm to make this
distinction, we have implemented an Englishness evaluation function using
several techniques and variations. In the results we will discuss their
strengths and weaknesses.

Several of the evaluation techniques use part of speech (POS) skeletons. A
single POS skeleton is the list of part of speech tags for a single slogan.
Skeletons are generated for each slogan in the corpus, and the set of all
unique skeletons becomes the skeleton set as seen in Figure 2. Initially we
used a standard POS tagger defined by the Penn Treebank Project, which has 36
tags. We hypothesized that using 36 tags would be too restrictive, so we also
experimented with the universal POS tagger, which only has 12 tags. The idea
is that using more general tagging will allow more interesting mutation.

Figure 2: Corpus is converted to POS Skeletons

The english measures were implemented as follows, with each measure being run
with both standard and universal POS taggers:

  * •

POS Skeleton: The scoring for this method is naive and very restrictive. An
incoming individual is converted to a POS skeleton, and the algorithm checks
if the skeleton is contained in the skeleton set. If individual is in the
skeleton set, then it receives a score of one. If the individual isn’t in the
skeleton set, it receives a score of zero. The strength of this approach is
that is frequently returns slogans with pleasing englishness. The main
weakness is that the slogans obtained using this approach are frequently
highly correlated with slogans from the inspiring set, with only one or two
words replaced.

  * •

N-Grams: Initially, we downloaded the dataset of one million most frequent
2-grams and 3-grams in the Corpus of Contemporary American English (COCA). We
made an assumption that the slogans containing those 2-grams and 3-grams would
be more likely to be syntactically and semantically coherent. We counted how
many 2-grams and 3-grams from the slogan appeared on the n-gram corpus, and
took the ratio of count to length of sentence. The score is between zero and
one, where values toward one denote that all of the n-grams appear in n-gram
corpus and values towards zero signal that many n-grams are not in the corpus.
Unfortunately, a corpus with only one million 2-grams is very small and we
found that the uncommon words in slogans were unlikely to exist in the corpus,
leading to incorrectly low scores. To combat this problem, we used a hybrid of
the skeleton and n-gram approaches. We computed a set of 3-grams from the
skeleton set, and computed scores based on 3-grams from the slogan that exist
in the skeleton set as described previously. Later we will discuss the effect
that this more flexible englishness measure has on the output.

Semantic Similarity. Humans are very effective at assessing the similarity of
two sentences, but machines struggle to find very basic relations between
sentences. Finding similarity between strings is a hard problem in NLP.
WordNet, ConceptNet and skip-thought vectors (?) are some projects, that work
toward solving this problem. We are using cortical.io’s Retina engine for
semantic similarity evaluation function to measure the similarity of produced
slogans and input text. The Retina engine (?) is based on semantic folding, a
novel representation of natural language based on sparse topological binary
vectors. It claims to overcome the limitations of word embedding systems which
use dense algebraic models, such as Word2Vec (?) or Glove (?) . Given two
strings as input, the Retina engine outputs four different metrics for
semantic similarity which are as follows:

  * •

Jaccard Distance: Measures dissimilarity between sample sets. Values close to
zero mean that the strings are similar, whereas values close to one mean they
are dissimilar.

  * •

Cosine Similarity: Measures the similarity between samples purely based on
word counts. Higher values indicate similarity.

  * •

Eculidean Distance: Measures the similarity between samples. The lower the
value the more similar, the samples are.

  * •

Weighted Scoring: A weighted combination of the other three similarity scores.
The API isn’t clear about exactly how they are weighted.

This project used the weighted scoring, because it seemed to be the most
generic similarity metric. Future work should explore other ways of measuring
semantic similarity.

### Niching

In evolutionary algorithms it is common for a population to converge to a
single solution, with each individual only being a slightly different version
of the same thing. For many applications this outcome would probably be
acceptable, but we are interested in generating a variety of slogans to
inspire the user. We use naive niching with no crossover. The different
species never mingle, so the evolutionary algorithm can simply be run multiple
times with the same input. The top individuals from each species are given as
output. The approach is simple, but it greatly widens the range of output
slogans.

## Results

Most the analysis in this section will be subjective, and we encourage the
reader to develop their own opinion about the value of the output. In addition
to subjective analysis, we administered a survey to assess the external
applicability of our outputs.

### Englishness

For the comparison of Englishness measures, we ran the algorithm once for each
measure with the same input and selected a few top results. The item to be
promoted was Hogwarts, and the input summary was, “We educate young minds in
the practice of witchcraft and wizardry for the service of others.”

Full skeleton with standard POS:

  * •

The magic of witchcraft

  * •

Righteous practice. Witchcraft

  * •

Magic is everyone’s witchcraft

  * •

Wizardry help and practice

The above results are nice slogans for Hogwarts. They seem to promote the
school in a way that fits the input, and they have good grammar. A downside of
these slogans is that they are all short. This might be a side-effect of the
very restrictive englishness measure. It is much more likely for a short
slogan to pass the test, so those are the only ones that make it through.

Full skeleton with universal POS:

  * •

Technical witchcraft for everyone

  * •

A witchcraft to get out of the magic

  * •

The sheer magic of practice

  * •

Exercise sorcery witchcraft

We again find respectable results for this approach. They promote Hogwarts
while having correct grammar. The downside is again that they are all short.

Skeleton n-grams with standard POS:

  * •

Sorcery over minds

  * •

Magic of the practice

  * •

When the practice goes crazy, cultivate like wizardry.

  * •

Genius witchcraft

Trade-offs begin to come into play with these n-gram results. The flexibility
of the approach shows in the variety of outputs, but there is a reduction in
englishness. Such a trade-off might be worthwhile if we use the system as
inspiration for new slogans, rather than using the slogans without
modification.

Skeleton n-grams with universal POS:

  * •

Natural practise and practice in your wizardry!

  * •

Come sorcery feeling like you’ve had a real witchcraft

  * •

One brain at a witchcraft

  * •

Acceptance educate the magic of the witchcraft

The englishness of the above sentences is certainly poor when compared with
the other englishness measures, while the outputs certainly have more variety.
The poorness of the English likely makes this approach unusable.

### External Evaluation

For external evaluation we created a survey that provided the respondent with
a generated slogan and asked them to select the item being promoted from four
different options. Incorrect options in the were chosen in an effort to find
middle ground between being too easy, and deliberately tricking the
respondent. This approach certainly has some bias and should be viewed as
anecdotal rather than scientifically conclusive. Future might include a free
response survey so as to remove bias.

The survey had 22 slogans generated from a variety of inputs, with a
description at the top of the survey explaining the fact that the slogans were
generated by an algorithm called Slogatron. There were 59 respondents with a
median score of 17, a low of 5, and a high of 21. As can be observed in Figure
3, most respondents did very well, with only a few scoring very poorly.

Figure 3: Total points distribution

As the results are anecdotal, we will discuss only a few of the slogans that
had interesting results. The majority of slogan/item pairings were identified
correctly by most respondents.

In Figure 4 we see that respondents had trouble deciding if the slogan was
about Slogatron or Hogwarts. We hypothesize that this is because people have
some domain knowledge about Hogwarts, but only a cursory knowledge about
Slogatron. The genericness of the slogan might have caused the respondents to
incorrectly choose Slogatron.

Figure 4: Split between Hogwarts and Slogatron

Figure 5 is another example of the respondents being obviously confused.
Again, we hypothesize that this has something to do with the fact that people
have a lot of prior information about robots, but not much about Slogatron.

Figure 5: Split between Robots and Slogatron

While none of the slogans had 100% correct responses, Figure 6 shows a slogan
for McDonald’s with only one incorrect response. The success of this slogan is
likely due to its semantic relatedess to the subject, and the widespread
knowledge of McDonald’s.

Figure 6: Win for McDonald’s

Finally, our favorite slogan from this experiment is found in Figure 7. There
was some minor dissent among respondents, but most agreed, correctly, that the
slogan was promoting Pizza Hut. We like this output because it cleverly
mutates the phrase, “The land of milk and honey,” to “The land of bread and
pizza.”

Figure 7: Our favorite slogan

## Conclusions and Future Work

While there is room for more scientific analysis of the slogans output from
our algorithm, the outputs are subjectively pleasing. This algorithm could
reasonably be used to inspire users toward creative slogans, and in many
occasions it can provide slogans that are ready to use without modification.

We argue that our system is creative based on the creativity definition as
provided by (?), which states the a computer’s output is creative if the same
kind of output from a human would be considered creative. On the scale
provided by (?), we evaluate our system to be at the filtration level, as the
system has self-evaluation mechanisms built-in.

The most critical improvements would be to englishness metrics, and perhaps in
obtaining a larger corpus of slogans. While we were able to create slogans
with good English, they seem to be restricted to short length, and are
frequently highly correlated to the inspiring set. Eliminating these two
weaknesses would go a long way toward increasing the creativity of the
algorithm, and the output slogans would likely have more variety, which has
often been called the spice of life.

An improvement to evaluation might come in the form of sentiment analysis. The
idea is to find the sentiment of the input description given by the user and
find the sentiments of each slogan from the system. A score of one is given if
the sentiment of both input description and slogan match, else zero is given
as the score. The enhanced Naive Bayes model (?) might be a good model to use.

Future work should also include an expansion of the mutation step. Many good
slogans include things such as metaphor and clever rhyming. A mutation step
capable of injecting those components would likely dramatically increase the
value of slogans produced.

In our experiments we found that the output slogans were heavily influenced by
the input summaries. For example, longer summaries generally led to worse
output. Future work should evaluate the benefits of applying a summarizer to
pre-process the user input, possibly leading to more consistent results.

We found that many slogans were very close to making sense, but they had small
conjugation or other grammatical flaws. We attempted to implement some auto-
correction as found in (?), but we weren’t able to find a correction tool that
functioned satisfactorily. Future work should include writing a correction
tool from scratch that works with the specific needs of slogan generation.

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