# Unimodality of ranks and a proof of Stanton’s conjecture

Kathrin Bringmann  ,  Siu Hang Man  University of Cologne, Department of
Mathematics and Computer Science, Weyertal 86-90, 50931 Cologne, Germany
<EMAIL_ADDRESS><EMAIL_ADDRESS>Charles University, Faculty
of Mathematics and Physics, Department of Algebra, Sokolovská 83, 186 00 Praha
8, Czech Republic<EMAIL_ADDRESS>and  Larry Rolen  Department of
Mathematics, 1420 Stevenson Center, Vanderbilt University, Nashville, TN 37240
<EMAIL_ADDRESS>
###### Abstract.

Recently, much attention has been given to various inequalities among
partition functions. For example, Nicolas, and later DeSavlvo–Pak, proved that
$p(n)$ is eventually log-concave, and Ji–Zang showed that the cranks are
eventually unimodal. This has led to a flurry of recent activity generalizing
such results in different directions. At the same time, Stanton recently made
deep conjectures on the positivity of certain polynomials associated to ranks
and cranks of partitions, with the ultimate goal of pointing the way to
“deeper” structure refining ranks and cranks. These have been shown to be
robust in recent works, which have identified further infinite families of
such conjectures in the case of colored partitions. In this paper, we employ
the Circle Method to prove unimodality for ranks. As a corollary, we prove
Stanton’s original conjecture. This points to future study of the positive,
integral coefficients Stanton conjectured to exist, hinting at new
combinatorial structure yet to be uncovered.

The first and the second author have received funding from the European
Research Council (ERC) under the European Union’s Horizon 2020 research and
innovation programme (grant agreement No. 101001179). The second author was
also supported by the Czech Science Foundation GAČR, grant 21-00420M. The
third author was supported by a grant from the Simons Foundation (853830) and
is also grateful for support from a 2021-2023 Dean’s Faculty Fellowship from
Vanderbilt University and to the Max Planck Institute for Mathematics in Bonn
for its hospitality and financial support.

##  1\. Introduction and statement of results

The study of the sizes and of various inequalities among partitions has a long
history, going back to Hardy and Ramanujan’s original development of the
Circle Method to prove their famous asymptotic for $p(n)$, the number of
integer partitions of $n$. Recently, there has been a reinvigoration of this
subject in several directions. Nicolas proved, and later DeSalvo-Pak
independently reproved [9, 16] that $p(n)$ is log-concave starting at $n=26$.
That is,

$p(n)^{2}\geq p(n-1)p(n+1),\quad\text{ for }n\geq 26.$

An infinite family of generalizations of log-concavity was conjectured by
Chen–Jia–Wang [7] and proven by Griffin, Ono, Zagier, and the third author
[12]. This was in direct analogy with a program of Jensen–Polá on the Riemann
Hypothesis, and uncovered a general phenomenon which gave new evidence for it.
Another striking multiplicative-additive inequality recently was proven by
Bessenrodt–Ono [5]

$p(a)p(b)\geq p(a+b)\qquad\text{ for }a,b\geq 2\ a+b>8.$

In this paper we prove further inequalities in the world of partitions, namely
for rank counting functions. Although analytic in nature, the motivation
behind these inequalities, based on ideas of Stanton, hints at combinatorial
structure “refining” partition ranks. In conjunction with recent works
extending Stanton’s conjectures, this suggests that the deeper combinatorial
structure Stanton posited is waiting to be discovered. We highlight this in
Question 1 below.

To set up this picture, we recall the classical story of ranks and cranks.
Ramanujan [19, 20] initiated the arithmetic study of $p(n)$ by stating and
proving his famous three congruences

$\displaystyle p(5n+4)$ $\displaystyle\equiv 0\,\,\left({\rm
mod}\,\,{5}\right),$ $\displaystyle p(7n+5)$ $\displaystyle\equiv
0\,\,\left({\rm mod}\,\,{7}\right),$ $\displaystyle p(11n+6)$
$\displaystyle\equiv 0\,\,\left({\rm mod}\,\,{11}\right).$

His proofs relied on $q$-series manipulation. Given the combinatorial nature
of the statement, it is natural to ask for a combinatorial proof. In the
search for such a combinatorial explanation, Dyson [10] defined the rank of a
partition to be its largest part minus its number of parts, and conjectured
that the partitions of $5n+4$ (resp. $7n+5$) are equidistributed with respect
to the rank modulo $5$ (resp. $7$). Dyson’s conjecture was proved by Atkin and
Swinnerton-Dyer [4]. Dyson also conjectured the existence of another statistic
which he called the crank, which would explain all of Ramanujan’s congruences.
More than four decades later, Andrews and Garvan [1] found such a statistic.
For a partition $\lambda$, let $\ell(\lambda)$ be the largest part of
$\lambda$, $o(\lambda)$ the number of ones in $\lambda$, and $\mu(\lambda)$
the number of parts in $\lambda$ larger than $o(\lambda)$. The crank of
$\lambda$ is then defined as

$\operatorname{crank}(\lambda):=\begin{cases}\ell(\lambda)&\text{ if
}o(\lambda)=0,\\\ \mu(\lambda)-o(\lambda)&\text{ if }o(\lambda)>0.\end{cases}$

Here, we are motivated by a recent search, spearheaded by Stanton, for
combinatorial structure refining ranks and cranks. While the rank and crank
split partitions into classes of equal size, there is no known direct map
between these classes. That is, while ranks and cranks provide a combinatorial
explanation for Ramanujan’s congruences, the proofs of the rank and crank
equidistribution properties are also done via $q$-series, and thus beg for a
combinatorial explanation themselves. In the search of such a map, Stanton
proposed several conjectures in his unpublished notes [23] regarding the rank
and crank of partitions. Atkin and Swinnerton-Dyer [4] showed that we have the
following generating function

$R(\zeta;q):=\sum_{\begin{subarray}{c}m\in\mathbb{Z}\\\ n\geq
0\end{subarray}}N(m,n)\zeta^{m}q^{n}=:\sum_{n=0}^{\infty}\operatorname{rank}_{n}(\zeta)q^{n}=\sum_{n=0}^{\infty}\frac{q^{n^{2}}}{(\zeta
q;q)_{n}\left(\zeta^{-1}q;q\right)_{n}}.$ (1.1)

Here, $(a;q)_{n}:=(a)_{n}:=\prod_{j=0}^{n-1}(1-aq^{j})$
($n\in\mathbb{N}_{0}\cup\\{\infty\\}$) is the usual $q$-Pochhammer symbol.
Meanwhile, Andrews and Garvan [1] proved that the crank generating function

$\displaystyle C(\zeta;q):=\sum_{\begin{subarray}{c}m\in\mathbb{Z}\\\ n\geq
0\end{subarray}}M(m,n)\zeta^{m}q^{n}=:\sum_{n=0}^{\infty}\operatorname{crank}_{n}(\zeta)q^{n}=\prod_{n=1}^{\infty}\frac{1-q^{n}}{\left(1-\zeta
q^{n}\right)\left(1-\zeta^{-1}q^{n}\right)}$ (1.2)

gives the right crank $M(m,n)$ except for $n=1$ (the correct values are
$M(0,1)=1$, and $M(m,1)=0$ for $m\neq 0$). We need a modification of the
Laurent polynomials in (1.1) and (1.2).

######  Definition.

For $n\in\mathbb{N}_{0}$, $\ell\in\left\\{{5,7,11}\right\\}$, and
$\beta:=\ell-\frac{\ell^{2}-1}{24}$, the  modified rank and crank are

$\displaystyle\operatorname{rank}_{\ell,n}^{*}(\zeta)$
$\displaystyle:=\operatorname{rank}_{\ell n+\beta}(\zeta)+\zeta^{\ell
n+\beta-2}-\zeta^{\ell n+\beta-1}+\zeta^{-\ell n-\beta+2}-\zeta^{-\ell
n-\beta+1},$ $\displaystyle\operatorname{crank}_{\ell,n}^{*}(\zeta)$
$\displaystyle:=\operatorname{crank}_{\ell n+\beta}(\zeta)+\zeta^{\ell
n+\beta-\ell}-\zeta^{\ell n+\beta}+\zeta^{-\ell n-\beta+\ell}-\zeta^{-\ell
n-\beta}.$

######  Example.

These modifications simply redefine the rank or crank of two partitions each.
In the case of the rank, there is always one partition of $n$ of rank $n-1$,
namely, $n$. The term $-\zeta^{\ell n+\beta-1}$ deletes this, and assigns it
the rank $n-2$. Similarly, the third and fourth monomials in the definition of
$\operatorname{rank}_{\ell,n}^{*}(\zeta)$ “reassigns” the partition
$1+1\ldots+1$ with rank $1-n$ to have rank $2-n$. While it may seem strange at
first glance, this choice is natural as it turns out it fixes the “trivial”
lack of unimodality of the number of partitions of $n$ with rank $m$ at the
tails. The main result of this paper show that for $n\geq 39$, there are no
other failures of unimodality, as shown in this table:

Table 1. Table of $N(m,n)$ for $n=1,2,3,4,5,6,\ldots,38,39$ |  |  |  |  | 1 |  |  |  |  |
---|---|---|---|---|---|---|---|---|---|---
|  |  |  | 1 | 0 | 1 |  |  |  |
|  |  | 1 | 0 | 1 | 0 | 1 |  |  |
|  | 1 | 0 | 1 | 1 | 1 | 0 | 1 |  |
| 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 2 | 1 | 2 | 1 | 1 | 0 | 1
⋮ | ⋮ | ⋮ | ⋮ | ⋮ | ⋮ | ⋮ | ⋮ | ⋮ | ⋮ | ⋮
… | 1169 | 1273 | 1331 | 1390 | 1389 | 1390 | 1331 | 1273 | 1169 | …
… | 1404 | 1494 | 1592 | 1627 | 1661 | 1627 | 1592 | 1494 | 1404 | …

To state Stanton’s conjectures on ranks and cranks we let
$\Phi_{\ell}(\zeta):=1+\zeta+\ldots+\zeta^{\ell-1}$ denote the $\ell$-th
cyclotomic polynomial, where $\zeta_{\ell}:=e^{\frac{2\pi i}{\ell}}$.

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

Let $n\in\mathbb{N}_{0}$.

  1. (1)

The following expressions are Laurent polynomials with non-negative
coefficients:

$\frac{\operatorname{rank}_{5,n}^{*}(\zeta_{\ell})}{\Phi_{5}(\zeta_{\ell})}\quad\text{
and
}\quad\frac{\operatorname{rank}_{7,n}^{*}(\zeta_{\ell})}{\Phi_{7}(\zeta_{\ell})}.$

  2. (2)

The following expressions are Laurent polynomials with non-negative
coefficients:

$\frac{\operatorname{crank}_{5,n}^{*}(\zeta_{\ell})}{\Phi_{5}(\zeta_{\ell})},\quad\quad\frac{\operatorname{crank}_{7,n}^{*}(\zeta_{\ell})}{\Phi_{7}(\zeta_{\ell})}\quad\text{
and
}\quad\frac{\operatorname{crank}_{11,n}^{*}(\zeta_{\ell})}{\Phi_{11}(\zeta_{\ell})}.$

These quotients being Laurent polynomials is equivalent to the
equidistribution of ranks and cranks. In [6], the first and the third author,
along with Gomez and Tripp reduced the conjecture to a problem on the
unimodality of the modified rank and crank. For the crank, Ji and Zang [13]
proved a related result, via a careful analysis of the generating function.

The main goal of this paper is to prove Conjecture 1.1. To state our methods,
we first recall the previous proof of part (1), and then outline our methods
for proving part (2). It turns out that the key phenomenon beyond Stanton’s
Conjecture is the eventual near-unimodality of ranks and cranks. In the case
of cranks, let $M(m,n)$ denote the number of partitions of $n$ with crank $m$.
Eventual unimodality in this case was recently proven by Ji and Zang [13] (as
$M(m,n)=M(-m,n)$, unimodality is the same as monotonicity starting at $m=0$).

######  Theorem 1.2 ([13], Theorem 1.7).

For $n\geq 44$ and $0\leq m\leq n-2$, we have

$M(m,n)\geq M(m+1,n).$

Using Theorem 1.2, the first and the third author, Gomez, and Tripp [6]
established Stanton’s conjecture for the crank.

######  Theorem 1.3 ([6], Theorem 1.3).

1.1 (2) is true.

To prove Stanton’s conjecture for the rank, we need a result analogous to
Theorem 1.2 for the rank, which was conjectured in [6]. The methods used to
prove Theorem 1.2 were the judicious use of clever $q$-series manipulations
and comparisons. These methods do not, however, generalize to other
situations. Here, we give a more general analytic proof based on the Circle
Method.

The monotonicity result we aim to show here is the following, where $N(m,n)$
denote the number of partitions of $n$ with rank $m$.

######  Conjecture 1.4.

For $n\geq 39$ and $0\leq m\leq n-3$, we have

$N(m,n)\geq N(m+1,n).$

As shown in [6], this is the key analytic result needed to prove Stanton’s
Conjecture.

######  Theorem 1.5 ([6], Theorem 1.5).

1.4 implies 1.1 (1).

1.4 is known to be true for $n$ sufficiently large; in fact, Zhou [25] showed
that $\left\\{{N(m,n)}\right\\}_{\left|{m}\right|\leq n-73}$ is log-concave.
However, such proofs are quite complicated and required different arguments
depending on the size of $m$, and no explicit bounds are known in the
literature. The main result of the paper is the following.

######  Theorem 1.6.

1.4, and hence 1.1 is true.

Returning to Stanton’s original goals for his conjecture, this suggests the
following.

######  Question 1.

What is the combinatorial meaning of the positive numbers occurring in
Conjecture 1.1? What does this suggest about bijections between ranks and
cranks mod $5,7,11$?

This question is further bolstered by recent work in [21] which gave a
procedure for finding infinite families of crank-type functions explaining
“most” Ramanujan-type congruences for all $k$-colored partition functions, and
in [6] which reworked these families to give new families which explained the
same Ramanujan-type congruences but also satisfy Stanton-type conjectures.
This suggests that the phenomenon is very general, and we plan to address this
in the future.

We outline the strategy for proving Theorem 1.6. Our proof for $m\ll
n^{\frac{3}{4}}$ is based on Zhou’s work on the eventual log-concavity of the
rank [25], which in turn is based on Wright’s Circle Method. Typical for such
approaches, this only shows that there exists $n_{0}\in\mathbb{N}$ (which may
be very large), such that the statement is true for $n\geq n_{0}$. In this
paper, we fine-tune the arguments in [25], establish strong error bounds, so
that the constant $n_{0}$ is sufficiently small to make explicit checking of
the remaining cases feasible. For $m\gg n^{\frac{1}{2}}$, we have another
argument, which is based on the convexity of the partition function $p(n)$.
Combining these arguments, we find an explicit $n_{0}$ such that the statement
is true for $n\geq n_{0}$, and check the remaining cases by computer.

The paper is organized as follows. In Section 2 we give some preliminaries on
asymptotics, particularly for the reciprocal of $(q)_{\infty}$, the
Euler–Maclaurin summation formula, explicit estimates for Bessel functions,
and resulting bound for partitions and partition differences. Using these
analytic estimates, in Section 3, we set up the Circle Method and bound minor
and major arc contributions. To prove our main result, we require extensive
computer computations, which are summarized in a case-by-case basis in Section
4, culminating in the proof of Theorem 4.1 which proves Theorem 1.6 for small
$m$ (specifically, asymptotically up to about $n^{\frac{3}{4}}$) for $n$
sufficiently large. We conclude with the proof of Theorem 1.6, that is,
Stanton’s Conjecture, in Section 5.

##  2\. Preliminaries

###  2.1. Generating functions

We require the following generating function $N(m,n)$ (see [4, (2.12)]),

$\mathcal{N}_{m}(q):=\sum_{n=0}^{\infty}N(m,n)q^{n}=\frac{1}{(q;q)_{\infty}}\sum_{r=1}^{\infty}(-1)^{r+1}q^{\frac{r(3r-1)}{2}+mr}\left(1-q^{r}\right).$

Since $N(-m,n)=N(m,n)$ for $m\in\mathbb{Z}$, we may assume that
$m\in\mathbb{N}_{0}$. As we aim to prove monotonicity, we are interested in
$N(m,n)-N(m+1,n)$, which has the generating function

$\mathcal{N}_{m}(q)-\mathcal{N}_{m+1}(q)=\sum_{n=0}^{\infty}\left({N(m,n)-N(m+1,n)}\right)q^{n}=\frac{H_{m}(q)}{(q;q)_{\infty}},$
(2.1)

where

$H_{m}(q):=\sum_{r=1}^{\infty}(-1)^{r+1}q^{\frac{3r^{2}}{2}+\left(m+\frac{1}{2}\right)r}\left(q^{-\frac{r}{2}}-q^{\frac{r}{2}}\right)^{2}.$

###  2.2. Euler–Maclaurin summation

To produce further asymptotics in the Circle Method, we require the
Euler–Maclaurin summation. In general, it gives an exact formula for infinite
sums in terms of an associated integral. To this end, let
$f\colon\mathbb{R}\to\mathbb{R}$ be a differentiable function, such that $f$
and its derivative $f^{\prime}$ have rapid decay, and let $N\in\mathbb{Z}$. A
simple case of the Euler–Maclaurin summation formula, sufficient for our
purposes, states that

$\sum_{j=N}^{\infty}f(j)=\int_{N}^{\infty}f(u)du+\frac{f(N)}{2}+\int_{N}^{\infty}f^{\prime}(u)B_{1}(u-\lfloor
u\rfloor)du,$ (2.2)

where $B_{1}(u)=u-\frac{1}{2}$ is the first Bernoulli polynomial.111A more
detailed exposition on the Euler–Maclaurin summation formula can be found in
[24, p. 319].

###  2.3. Estimates and expansions for Bessel functions

We also require tight estimates for $I$-Bessel functions. Specifically, let
$u>0$ and $\kappa\geq 0$. Then we approximate $I_{-\kappa-1}(u)$, the
$I$-Bessel function of order $-\kappa-1$, by comparing it to the following
integral

$\mathcal{I}_{\kappa}(u):=\frac{1}{2\pi
i}\int_{1-i}^{1+i}w^{-\kappa-1}\exp\left({\frac{u}{2}\left({w+\frac{1}{w}}\right)}\right)dw$

This gives us the following, whose proof we omit as it is standard, following
from the integral representation of the $I$-Bessel function (see e.g. [3]).

######  Lemma 2.1.

For $u\in\mathbb{R}^{+}$ and $\kappa<0$, we have

$|I_{k}(u)-\mathcal{I}_{\kappa}(u)|\leq\frac{2^{\frac{1-\kappa}{2}}}{\pi}\left({\frac{2^{-\kappa-1}\Gamma(-\kappa)}{u^{-\kappa}}+\exp\left({\frac{3u}{4}}\right)}\right),$

We also require some facts on the asymptotic expansions of the $I$-Bessel
functions. First, we recall standard expansions. Let $N,M\in\mathbb{N}_{0}$,
$s\in\mathbb{C}$. Then we have for $w\in\mathbb{C}$

$I_{s}(w)=\frac{e^{w}}{\sqrt{2\pi
w}}\left({\sum_{j=0}^{N-1}(-1)^{j}\frac{a_{j}(s)}{w^{j}}+\delta_{N}(s,w)}\right)-ie^{-\pi
is}\frac{e^{-w}}{\sqrt{2\pi
w}}\left({\sum_{j=0}^{M-1}\frac{a_{j}(s)}{w^{j}}+\gamma_{M}(s,w)}\right)$

(see [17, (exercise 7.13.2)]), where

$\displaystyle a_{0}(s):=1,\qquad
a_{j}(s):=\frac{1}{8^{j}j!}{\prod_{k=1}^{j}\left(4s^{2}-(2k+1)^{2}\right)}.$

Moreover, the error terms $\delta_{N}$ and $\gamma_{M}$ satisfy the following
bounds:

$\displaystyle|\gamma_{M}(s,w)|$ $\displaystyle\leq
2\exp\left(\left|{\frac{s^{2}-\frac{1}{4}}{w}}\right|\right)\left|{\frac{a_{M}(s)}{w^{M}}}\right|,\quad|\delta_{N}(s,w)|\leq\frac{2\pi^{\frac{1}{2}}\Gamma\left(\frac{N}{2}+1\right)}{\Gamma\left(\frac{N}{2}+\frac{1}{2}\right)}\exp\left(\frac{\pi}{2}\left|{\frac{s^{2}-\frac{1}{4}}{w}}\right|\right)\left|{\frac{a_{N}(s)}{w^{N}}}\right|.$

For later use, we consider for $\nu\in\mathbb{N}_{0}$ the following linear
combination of $I$-Bessel functions:

$I_{s}^{[\nu]}(w):=\sum_{r=0}^{\nu}\binom{\nu}{r}(-1)^{\nu+r}I_{s-r}(w).$

Then $I_{s}^{[\nu]}(w)$ has an asymptotic expansion of the following form:

$I_{s}^{[\nu]}(w)=\frac{e^{w}}{\sqrt{2\pi
w}}\left({\sum_{j=0}^{N-1}(-1)^{j}\frac{a_{j}^{[\nu]}(s)}{w^{j}}+\delta_{N}^{[\nu]}(s,w)}\right)-ie^{-\pi
is}\frac{e^{-w}}{\sqrt{2\pi
w}}\left({\sum_{j=0}^{M-1}\frac{a_{j}^{[\nu]}(s)}{w^{j}}+\gamma_{M}^{[\nu]}(s,w)}\right),$
(2.3)

where

$\displaystyle a_{j}^{[\nu]}(s)$
$\displaystyle:=\sum_{r=0}^{\nu}\binom{\nu}{r}(-1)^{\nu+r}a_{j}(s-r),\quad\left|{\gamma_{M}^{[\nu]}(s,w)}\right|\leq\sum_{r=0}^{\nu}\binom{\nu}{r}\left|{\gamma_{M}(s-r,w)}\right|,$
$\displaystyle\left|{\delta_{N}^{[\nu]}(s,w)}\right|$
$\displaystyle\leq\sum_{r=0}^{\nu}\binom{\nu}{r}\left|{\delta_{N}(s-r,w)}\right|.$

Finally, a direct calculation gives the following.

######  Proposition 2.2.

Let $\kappa\geq 0$ and $\nu\in\mathbb{N}_{0}$. We have that
$a_{j}^{[\nu]}(\kappa)=0$ for $j<\frac{\nu}{2}$.

###  2.4. Estimates for partitions

We first give basic estimates for $(q;q)_{\infty}^{-1}$. A direct calculation,
using the modularity of the Dedekind $\eta$-function (see 5.8.1 of [8]) and
bounds for $p(n)$ (see [2, Theorem 14.5]) gives the following.222The explicit
constants given here (and for the rest of the paper) do not need to be
optimal.

######  Lemma 2.3.

Let $Z=x+iy$, where $x,y\in\mathbb{R}$, $0<x\leq\frac{1}{10}$,
$0\leq|y|\leq\pi$. Further let $q:=e^{-Z}$.

  1. (1)

For $|y|\leq x$ we have, with $|C|\leq 1$

$\frac{1}{(q;q)_{\infty}}=\sqrt{\frac{Z}{2\pi}}\exp\left({-\frac{Z}{24}+\frac{\pi^{2}}{6Z}}\right)+C.$

  2. (2)

For $x\leq|y|\leq\pi$, we have

$\left|{\frac{1}{(q;q)_{\infty}}}\right|\leq
2\sqrt{x}\exp\left(\frac{\pi^{2}}{12x}\right).$ (2.4)

We also require an approximation for $p(n)$ with bounded error. Using
Rademacher’s convergent series expansion for the partition function $p(n)$
[18] one can show the following asymptotic formula (see [25, Lemma 5.6] for a
non-explicit version).

######  Lemma 2.4.

For all $n\in\mathbb{N}$, there exists a constant
$|\mathcal{C}|\leq\frac{1}{8}$ for which

$p(n)=\frac{1}{4\sqrt{3}}\frac{e^{\pi\sqrt{\frac{2}{3}\left(n-\frac{1}{24}\right)}}}{n-\frac{1}{24}}-\frac{1}{4\sqrt{2}\pi}\frac{e^{\pi\sqrt{\frac{2}{3}\left(n-\frac{1}{24}\right)}}}{\left(n-\frac{1}{24}\right)^{\frac{3}{2}}}+\frac{\mathcal{C}e^{\frac{\pi}{2}\sqrt{\frac{2}{3}\left(n-\frac{1}{24}\right)}}}{n-\frac{1}{24}}.$

Note that Lemma 2.4 implies the following convenient upper bound for
$n\in\mathbb{N}$:

$p(n)\leq\frac{e^{\pi\sqrt{\frac{2}{3}\left(n-\frac{1}{24}\right)}}}{4\sqrt{3}\left(n-\frac{1}{24}\right)}.$
(2.5)

Finally, for $n\in\mathbb{N}_{0}$, $r\in\mathbb{N}$, we require strong
estimates on the following “second order” differences

$P(n,r):=p(n)-2p(n+r)+p(n+2r).$ (2.6)

Then one can use Lemma 2.4 directly to give the following key estimate.

######  Lemma 2.5.

Let $n\geq 100$, $r\in\mathbb{N}$. Then we have

$\frac{r^{2}e^{\pi\sqrt{\frac{2}{3}\left({n-\frac{1}{24}}\right)}}}{6\left({n+2r-\frac{1}{24}}\right)^{2}}\leq
P(n,r)\leq\frac{\pi^{2}r^{2}e^{\pi\sqrt{\frac{2}{3}\left({n+2r-\frac{1}{24}}\right)}}}{24\sqrt{3}\left({n-\frac{1}{24}}\right)^{2}}.$

##  3\. The Circle Method

We next recall a version of the Circle Method that we require for the proof of
our main result.

###  3.1. The setup

For $n\in\mathbb{N}$, we let

$\beta_{n}:=\frac{\pi}{\sqrt{6\left(n-\frac{1}{24}\right)}},\hskip
28.45274pt\mathcal{C}_{n}:=\left({n-\frac{1}{24}}\right)\beta_{n}=\frac{\pi}{\sqrt{6}}\sqrt{n-\frac{1}{24}}.$
(3.1)

Denote by $\mathcal{C}$ the circle with radius $e^{-\beta_{n}}$, centered at
the origin, traversed counter-clockwise exactly once. Cauchy’s Theorem gives

$N(m,n)-N(m+1,n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{H_{m}\left(e^{-Z}\right)}{\left(e^{-Z};e^{-Z}\right)_{\infty}}e^{nZ}dy.$
(3.2)

We split the integral into two parts: the major arc integral

$\mathfrak{M}:=\frac{1}{2\pi}\int_{\left|{y}\right|\leq\beta_{n}}\frac{H_{m}\left(e^{-Z}\right)}{(e^{-Z};e^{-Z})_{\infty}}e^{nZ}dy,$
(3.3)

and the minor arc integral

$\mathfrak{m}:=\frac{1}{2\pi}\int_{\beta_{n}\leq\left|{y}\right|\leq\pi}\frac{H_{m}\left(e^{-Z}\right)}{(e^{-Z};e^{-Z})_{\infty}}e^{nZ}dy.$
(3.4)

We bound the minor arc contributions and approximate the major arc
contributions, which give our main terms.

###  3.2. Minor arc bounds

We begin by expanding

$H_{m}\left(e^{-Z}\right)=\sum_{r=1}^{\infty}(-1)^{r+1}\left(e^{-\left(\frac{3r^{2}}{2}+\left(m-\frac{1}{2}\right)r\right)Z}-2e^{-\left(\frac{3r^{2}}{2}+\left(m+\frac{1}{2}\right)r\right)Z}+e^{-\left(\frac{3r^{2}}{2}+\left(m+\frac{3}{2}\right)r\right)Z}\right).$

Since $m\geq 0$, we have
$\frac{3r^{2}}{2}+(m+j)r\geq\frac{3r^{2}}{2}+jr\geq\frac{1}{2}(r-1)^{2}$ for
$j\in\\{-\frac{1}{2},\frac{1}{2},\frac{3}{2}\\}$ and $r\geq 1$. Hence

$\left|{H_{m}\left(e^{-Z}\right)}\right|\leq\sum_{r=1}^{\infty}\left(e^{-\left(\frac{3r^{2}}{2}+\left(m-\frac{1}{2}\right)r\right)\beta_{n}}+2e^{-\left(\frac{3r^{2}}{2}+\left(m+\frac{1}{2}\right)r\right)\beta_{n}}+e^{-\left(\frac{3r^{2}}{2}+\left(m+\frac{3}{2}\right)r\right)\beta_{n}}\right)\leq
4\sum_{r\geq 0}e^{-\frac{\beta_{n}r^{2}}{2}}.$

It is not hard to see that we may bound

$\left|{H_{m}\left(e^{-Z}\right)}\right|\leq
2\sqrt{\frac{2\pi}{\beta_{n}}}+4.$ (3.5)

Using this, we now turn to bounding the minor arc integral $\mathfrak{m}$.

######  Proposition 3.1.

For $\beta_{n}\leq\frac{1}{10}$, we have

$\left|{\mathfrak{m}}\right|\leq 13e^{\frac{3\mathcal{C}_{n}}{2}}.$

###### Proof.

By (2.4), for $\beta_{n}\leq\frac{1}{10}$ and
$\beta_{n}\leq\left|{y}\right|\leq\pi$ we have

$\left|{\frac{1}{\left(e^{-Z};e^{-Z}\right)_{\infty}}}\right|\leq
2\sqrt{\beta_{n}}e^{\frac{\pi^{2}}{12\beta_{n}}}.$

Hence

$|\mathfrak{m}|\leq
4\left(\sqrt{2\pi}+2\sqrt{\beta_{n}}\right)e^{\frac{\pi^{2}}{12\beta_{n}}+n\beta_{n}}.$

Now a direct calculation gives the claim. ∎

###  3.3. Major arc bounds

Next we consider the major arc integral $\mathfrak{M}$. Again we assume
$\beta_{n}\leq\frac{1}{10}$. On the major arc, we have $|y|\leq\beta_{n}$, so
we can use Lemma 2.3 and obtain

$\mathfrak{M}=\frac{1}{2\pi}\int_{\left|{y}\right|\leq\beta_{n}}\frac{H_{m}\left({e^{-Z}}\right)e^{nZ}}{(e^{-Z};e^{-Z})_{\infty}}dy=\frac{1}{2\pi}\int_{\left|{y}\right|\leq\beta_{n}}\left({\sqrt{\frac{Z}{2\pi}}e^{-\frac{Z}{24}+\frac{\pi^{2}}{6Z}}+C}\right)H_{m}\left({e^{-Z}}\right)e^{nZ}dy,$

for some $|C|\leq 1$. Let

$\mathcal{M}^{[1]}:=\frac{1}{2\pi}\int_{\left|{y}\right|\leq\beta_{n}}\sqrt{\frac{Z}{2\pi}}e^{\left(n-\frac{1}{24}\right)Z+\frac{\pi^{2}}{6Z}}H_{m}\left({e^{-Z}}\right)dy,\quad\mathcal{E}^{[1]}:=\mathfrak{M}-\mathcal{M}^{[1]}.$
(3.6)

Then it follows from (3.5) that for $\beta_{n}\leq\frac{1}{10}$, we have

$\left|{\mathcal{E}^{[1]}}\right|\leq\frac{1}{2\pi}\int_{|y|\leq\beta_{n}}\left|{H_{m}\left(e^{-Z}\right)e^{nZ}}\right|dy\leq\frac{2\beta_{n}}{\pi}\left(\sqrt{\frac{2\pi}{\beta_{n}}}+2\right)e^{n\beta_{n}}\leq
e^{\mathcal{C}_{n}}.$ (3.7)

To give an estimate to $\mathcal{M}^{[1]}$, we need a bound for
$H_{m}(e^{-Z})$. Note that $|Z|\leq\sqrt{2}\beta_{n}$. For any
$L\in\mathbb{N}$, we approximate $H_{m}(e^{-Z})$ by a sum of $L-1$ terms.
Specifically, we define

$\displaystyle\mathcal{M}_{\ell,m}(Z)$
$\displaystyle:=\frac{2Z^{2\ell}}{(2\ell)!}\sum_{r=1}^{\infty}(-1)^{r+1}r^{2\ell}e^{-\left(\frac{3r^{2}}{2}+\left(m+\frac{1}{2}\right)r\right)Z},$
$\displaystyle\mathcal{E}_{L,m}(Z)$
$\displaystyle:=H_{m}\left(e^{-Z}\right)-\sum_{\ell=1}^{L-1}\mathcal{M}_{\ell,m}(Z).$
(3.8)

Moreover, we set

$C_{1}(L,x_{0}):=\frac{2^{L+1}e^{\frac{x_{0}}{2}}}{(2L-1)!}\left({\frac{\Gamma\left({L+\frac{1}{2}}\right)}{2}+e^{-L}L^{L}\sqrt{x_{0}}}\right).$
(3.9)

In terms of these quantities, we obtain the following estimate for the error
term $\mathcal{E}_{L,m}(Z)$.

######  Lemma 3.2.

Let $x_{0}\in\mathbb{R}^{+}$. Then we have, for $\beta_{n}\leq x_{0}$,

$\left|{\mathcal{E}_{L,m}(Z)}\right|\leq
C_{1}(L,x_{0})\beta_{n}^{L-\frac{1}{2}}e^{-\left(m+\frac{1}{2}\right)\beta_{n}}.$

###### Proof.

Using Taylor’s Theorem, we obtain for $L\in\mathbb{N}$

$\left(e^{\frac{w}{2}}-e^{-\frac{w}{2}}\right)^{2}=w^{2\ell}\sum_{\ell=1}^{L-1}\frac{2}{(2\ell)!}+\frac{1}{(2L-1)!}\int_{0}^{w}(w-u)^{2L-1}\left(\frac{\partial{}}{\partial{u}}\right)^{2L}\left(e^{-\frac{u}{2}}-e^{\frac{u}{2}}\right)^{2}du.$

Letting $w=rZ$ yields the bound

$\left|{\left(e^{\frac{rZ}{2}}-e^{-\frac{rZ}{2}}\right)^{2}-2\sum_{\ell=1}^{L-1}\frac{(rZ)^{2\ell}}{(2\ell)!}}\right|\leq\frac{2}{(2L-1)!}(r|Z|)^{2L}e^{r\beta_{n}}.$

Note that

$|\mathcal{E}_{L,m}(Z)|\leq\frac{2^{L+1}\beta_{n}^{2L}e^{-\left(m+\frac{1}{2}\right)\beta_{n}}}{(2L-1)!}\sum_{r=1}^{\infty}r^{2L}e^{-\left(\frac{3r^{2}}{2}+\left(m+\frac{1}{2}\right)(r-1)\right)\beta_{n}+r\beta_{n}}.$
(3.10)

Next we estimate the sum in (3.10). Since $m\geq 0$, we have
$\frac{3r^{2}}{2}+(m+\frac{1}{2})(r-1)-r\geq\frac{3r^{2}}{2}+\frac{1}{2}(r-1)-r\geq
r^{2}-\frac{1}{2}$ for $r\geq 1$. We deduce that

$\sum_{r=1}^{\infty}r^{2L}e^{-\left(\frac{3r^{2}}{2}+\left(m+\frac{1}{2}\right)(r-1)\right)\beta_{n}+r\beta_{n}}\leq
e^{\frac{\beta_{n}}{2}}\sum_{r=1}^{\infty}r^{2L}e^{-r^{2}\beta_{n}}.$ (3.11)

The Euler–Maclaurin summation formula (2.2) gives (using that $L>0$)

$\sum_{r=1}^{\infty}r^{2L}e^{-r^{2}\beta_{n}}=\int_{0}^{\infty}u^{2L}e^{-\beta_{n}u^{2}}du+\int_{0}^{\infty}\frac{\partial{}}{\partial{u}}\left(u^{2L}e^{-\beta_{n}u^{2}}\right)B_{1}(u-\lfloor
u\rfloor)du.$ (3.12)

For the first integral in (3.12), we evaluate (see [11, 3.326.2])

$\int_{0}^{\infty}u^{2L}e^{-\beta_{n}u^{2}}du=\frac{\Gamma\left({L+\frac{1}{2}}\right)}{2}\beta_{n}^{-L-\frac{1}{2}}.$
(3.13)

Since $\left|{B_{1}(u-\lfloor u\rfloor)}\right|\leq\frac{1}{2}$, the absolute
value of the second integral in (3.12) is bounded by

$\frac{1}{2}\int_{0}^{\infty}\left|{\frac{\partial}{\partial
u}\left(u^{2L}e^{-\beta_{n}u^{2}}\right)}\right|du=\left(\frac{L}{e\beta_{n}}\right)^{L},$
(3.14)

using the Fundamental Theorem of Calculus. Plugging (3.13) and (3.14) into
(3.12), we obtain

$\sum_{r=1}^{\infty}r^{2L}e^{-r^{2}\beta_{n}}\leq\frac{\Gamma\left(L+\frac{1}{2}\right)}{2}\beta_{n}^{-L-\frac{1}{2}}+\left(\frac{L}{e\beta_{n}}\right)^{L}.$

Putting this into (3.11) then yields

$\sum_{r=1}^{\infty}r^{2L}e^{-\left(\frac{3r^{2}}{2}+\left(m+\frac{1}{2}\right)(r-1)\right)\beta_{n}+r\beta_{n}}\leq
e^{\frac{\beta_{n}}{2}}\left(\frac{\Gamma\left(L+\frac{1}{2}\right)}{2}\beta_{n}^{-L-\frac{1}{2}}+\left(\frac{L}{e\beta_{n}}\right)^{L}\right).$

Finally, inserting this into (3.10) yields

$\left|{\mathcal{E}_{L,m}(Z)}\right|\leq\frac{2^{L+1}\beta_{n}^{2L}e^{-\left(m+\frac{1}{2}\right)\beta_{n}+\frac{\beta_{n}}{2}}}{(2L-1)!}\left({\frac{\Gamma\left({L+\frac{1}{2}}\right)}{2}\beta_{n}^{-L-\frac{1}{2}}+\left({\frac{L}{e\beta_{n}}}\right)^{L}}\right).$

This gives the claim. ∎

Thus, we may estimate $H_{m}(e^{-Z})$ by
$\sum_{\ell=1}^{L-1}\mathcal{M}_{\ell,m}(Z)$ up to a known error bound. We
next estimate the terms $\mathcal{M}_{\ell,m}(Z)$. We first recall a result
from [15].

######  Proposition 3.3 ([15, Proposition 2.3]).

Let $w,Z\in\mathbb{C}$, $\operatorname{Re}(Z)>0$, $w\not\in 2\pi
i(\mathbb{Z}+\frac{1}{2})$, $\ell\in\mathbb{N}_{0}$, and $K\in\mathbb{N}$.
Then we have

$\sum_{r=1}^{\infty}(-1)^{r+1}r^{\ell}e^{-r^{2}Z-rw}=(-1)^{\ell}\sum_{k=0}^{K-1}\frac{(-Z)^{k}}{k!}\left(\frac{\partial}{\partial
w}\right)^{\ell+2k}{\frac{1}{1+e^{w}}}+Z^{K}R_{K}(\ell;w,Z),$

where

$R_{K}(\ell;w,Z):=\frac{(-1)^{\ell+K}}{(K-1)!}\int_{0}^{1}(1-t)^{K-1}\left(\frac{\partial{}}{\partial{w}}\right)^{\ell+2K}\sum_{r=1}^{\infty}(-1)^{r+1}e^{-r^{2}Zt-
rw}dt.$ (3.15)

We would like to apply Proposition 3.3 to estimate $\mathcal{M}_{\ell,m}(Z)$,
following the proof of [15, Proposition 2.6]. To this end, we show the
following.

######  Proposition 3.4.

Assuming the notation above, we have

$\left|{\mathcal{M}_{\ell,m}(Z)-\frac{2}{(2\ell)!}\sum_{k=0}^{K-1}\frac{\left(-\frac{3}{2}\right)^{k}Z^{2\ell+k}}{k!}\left[\left(\frac{\partial{}}{\partial{w}}\right)^{2\ell+2k}\frac{1}{1+e^{w}}\right]_{w=\left(m+\frac{1}{2}\right)Z}}\right|\\\
\leq
C_{2}(\ell,K,J,x_{0},\varrho)\beta_{n}^{2\ell+K}e^{-\left(m+\frac{1}{2}\right)\beta_{n}},$

where the constant $C_{2}$ is explicitly constructed in the proof.

Before beginning the proof, we also require basic facts about difference
operator. To this end for $j\in\mathbb{N}_{0}$, let $\Delta^{j}$ be the $j$-th
order forward difference operator:

$\displaystyle\Delta^{0}(h(r)):=h(r),\quad\Delta(h(r)):=h(r+1)-h(r),\quad\Delta^{j+1}(h(r)):=\Delta^{j-1}(h(r+1))-\Delta^{j-1}(h(r)).$

For any arithmetic function $h$, we have the generating function.

######  Lemma 3.5 ([15, Lemma 2.4]).

Let $z\in\mathbb{C}\setminus\\{1\\}$, and let
$h\colon\mathbb{N}_{0}\to\mathbb{C}$ satisfy
$\sum_{r=0}^{\infty}|h(r)z^{r}|<\infty$. Then we have, for all
$J\in\mathbb{N}$,

$\sum_{r=0}^{\infty}h(r)z^{r}=\sum_{j=0}^{J-1}\frac{\Delta^{j}(h(0))z^{j}}{(1-z)^{j+1}}+\frac{z^{J}}{(1-z)^{J}}\sum_{r=0}^{\infty}\Delta^{J}(h(r))z^{r}.$

###### Proof of Proposition 3.4.

By Proposition 3.3, we have

$\sum_{r\geq
1}(-1)^{r+1}r^{2\ell}e^{-\left({\frac{3r^{2}}{2}+\left(m+\frac{1}{2}\right)r}\right)Z}\\\
=\sum_{k=0}^{K-1}\frac{\left(-\frac{3Z}{2}\right)^{k}}{k!}\left[\left(\frac{\partial}{\partial
w}\right)^{2\ell+2k}{\frac{1}{1+e^{w}}}\right]_{w=\left(m+\frac{1}{2}\right)Z}+\left({\frac{3Z}{2}}\right)^{K}R_{K}\left({2\ell;\left({m+\frac{1}{2}}\right)Z,\frac{3Z}{2}}\right).$
(3.16)

Now we bound $R_{K}(2\ell;bZ,Z)$ for $b\geq 0$, and $Z=x+iy$ with $x\geq 0$,
and $\left|{y}\right|\leq x$. We start with some technical preparations. For
$0\leq x\leq x_{0}$, $r\in\mathbb{N}$, define $E_{Z}(r):=e^{-r^{2}Z}$, let
$J\in\mathbb{N}$, and $x_{0}\leq 1$. For $0\leq x\leq x_{0}$, we would like to
bound

$\Delta^{J}(E_{Z}(r))=(-1)^{J}\sum_{j=0}^{J}(-1)^{j}\binom{J}{j}e^{-(r+j)^{2}Z}.$

We do so in two different ways. First assume that $J$ is even and observe that

$\left|{\Delta^{J}(E_{Z}(r))}\right|=e^{-r^{2}x}\left|{\sum_{j=0}^{J}(-1)^{j}\binom{J}{j}e^{-\left(2rj+j^{2}\right)Z}}\right|\leq
C_{2a}(J,x_{0})e^{-r^{2}x},$ (3.17)

where $C_{2a}(J,x_{0})$ is a constant such that for all $r\geq 0$ and $Z=x+iy$
with $0\leq x\leq x_{0}$ and $|y|\leq x$

$\left|{\sum_{j=0}^{J}(-1)^{j}\binom{J}{j}e^{-\left(2rj+j^{2}\right)Z}}\right|\leq
C_{2a}(J,x_{0}).$ (3.18)

Next we give another bound, which works better if $r$ is small. Note that

$e^{r^{2}Z}\Delta^{J}(E_{Z}(r))=\sum_{j=0}^{J}(-1)^{j}\binom{J}{j}e^{-\left(2rj+j^{2}\right)Z}.$
(3.19)

By Taylor’s Theorem, we obtain

$\left|{e^{-j^{2}Z}-\sum_{\nu=0}^{\frac{J}{2}-1}\frac{\left(-j^{2}Z\right)^{\nu}}{\nu!}}\right|\leq\frac{\left(j^{2}|Z|\right)^{\frac{J}{2}}}{\left(\frac{J}{2}\right)!}\leq\frac{2^{\frac{J}{4}}j^{J}x^{\frac{J}{2}}}{\left(\frac{J}{2}\right)!}.$
(3.20)

We apply (3.20) to (3.19) and get

$\left|{e^{r^{2}Z}\Delta^{J}(E_{Z}(r))-\sum_{j=0}^{J}(-1)^{j}\binom{J}{j}e^{-2rjZ}\sum_{\nu=0}^{\frac{J}{2}-1}\frac{(-j^{2}Z)^{\nu}}{\nu!}}\right|\leq\frac{2^{\frac{J}{4}}x^{\frac{J}{2}}}{\left(\frac{J}{2}\right)!}\sum_{j=0}^{J}\binom{J}{j}j^{J}.$

We deduce that

$\displaystyle\left|e^{r^{2}Z}\Delta^{J}(E_{Z}(r))\right|$
$\displaystyle\leq\left|\sum_{\nu=0}^{\frac{J}{2}-1}\frac{(-Z)^{\nu}}{\nu!}\sum_{j=0}^{J}(-1)^{j}\binom{J}{j}j^{2\nu}e^{-2rjZ}\right|+\frac{2^{\frac{J}{4}}x^{\frac{J}{2}}}{\left(\frac{J}{2}\right)!}\sum_{j=0}^{J}\binom{J}{j}j^{J}$
$\displaystyle\leq\sum_{\nu=0}^{\frac{J}{2}-1}\frac{2^{\frac{\nu}{2}}x^{\nu}}{\nu!}\left|\left[\left(\frac{\partial{}}{\partial{w}}\right)^{2\nu}\left(1-e^{-w}\right)^{J}\right]_{w=2rZ}\right|+\frac{2^{\frac{J}{4}}x^{\frac{J}{2}}}{\left(\frac{J}{2}\right)!}\sum_{j=0}^{J}\binom{J}{j}j^{J}.$
(3.21)

For $0\leq\nu<\frac{J}{2}$, choose $C_{2b}(J,\nu)$ such that for all
$w\in\mathbb{C}$ with $\operatorname{Re}(w)\geq 0$ and
$|\operatorname{Im}(w)|\leq\operatorname{Re}(w)$

$\left|{w^{2\nu-J}\left(\frac{\partial}{\partial
w}\right)^{2\nu}\left(1-e^{-w}\right)^{J}}\right|\leq C_{2b}(J,\nu).$ (3.22)

Plugging (3.22) into (3.21) gives, using that $|Z|\leq\sqrt{2}x$,

$\displaystyle\left|{e^{r^{2}Z}\Delta^{J}(E_{Z}(r))}\right|$
$\displaystyle\leq\sum_{\nu=0}^{\frac{J}{2}-1}C_{2b}(J,\nu)\frac{2^{\frac{\nu}{2}}x^{\nu}}{\nu!}|2rZ|^{J-2\nu}+\frac{2^{\frac{J}{4}}x^{\frac{J}{2}}}{\left(\frac{J}{2}\right)!}\sum_{j=0}^{J}\binom{J}{j}j^{J}$
$\displaystyle\leq\sum_{\nu=0}^{\frac{J}{2}-1}C_{2b}(J,\nu)\frac{2^{\frac{3J}{2}-\frac{5\nu}{2}}}{\nu!}r^{J-2\nu}x^{J-\nu}+\frac{2^{\frac{J}{4}}x^{\frac{J}{2}}}{\left(\frac{J}{2}\right)!}\sum_{j=0}^{J}\binom{J}{j}j^{J}$
$\displaystyle=:\sum_{\nu=0}^{\frac{J}{2}-1}C_{2c}(J,\nu)r^{J-2\nu}x^{J-\nu}+C_{2d}(J)x^{\frac{J}{2}}.$
(3.23)

Now we estimate $R_{K}(2\ell;bZ,Z)$. From (3.15), we have the trivial bound

$|R_{K}(2\ell;bZ,Z)|\leq\frac{1}{(K-1)!}\int_{0}^{1}\left|{\left[\left(\frac{\partial{}}{\partial{w}}\right)^{2\ell+2K}\sum_{r=1}^{\infty}(-1)^{r}e^{-r^{2}Zt-
rw}\right]_{w=bZ}}\right|dt.$ (3.24)

Applying Lemma 3.5 with $z=-e^{w}$ and $h(r)=E_{Z}(r)$, we get

$\sum_{r=1}^{\infty}(-1)^{r}e^{-r^{2}Z-rw}=-1+\sum_{j=0}^{J-1}\frac{(-1)^{j}e^{-jw}\Delta^{j}(E_{Z}(0))}{\left(1+e^{-w}\right)^{j+1}}+\frac{e^{-Jw}}{\left(1+e^{-w}\right)^{J}}\sum_{r=1}^{\infty}(-1)^{r}e^{-rw}\Delta^{J}(E_{Z}(r)).$

Plugging this into (3.24) with $Z\mapsto Zt$ and multiplying both sides by
$e^{bZ}$, we get a bound

$\left|{e^{bZ}R_{K}(2\ell;bZ,Z)}\right|\leq\frac{1}{(K-1)!}\left(\sum_{j=0}^{J-1}\int_{0}^{1}\left|{e^{bZ}\left[\left(\frac{\partial{}}{\partial{w}}\right)^{2\ell+2K}\frac{(-1)^{j}e^{-jw}\Delta^{j}E_{Zt}(0)}{\left(1+e^{-w}\right)^{j+1}}\right]_{w=bZ}}\right|dt\right.\\\
\left.{\vphantom{\sum_{j=0}^{J-1}}}+\int_{0}^{1}\left|{e^{bZ}\left[\left(\frac{\partial{}}{\partial{w}}\right)^{2\ell+2K}\left(\frac{e^{-Jw}}{\left(1+e^{-w}\right)^{J}}\sum_{r=1}^{\infty}(-1)^{r}e^{-rw}\Delta^{J}(E_{Zt}(r))\right)\right]_{w=bZ}}\right|dt\right).$
(3.25)

We now consider the first integral in (3.25). For $k\in\mathbb{N}$ and
$j\in\mathbb{N}_{0}$, let $C_{2e}(k,j)$ and $C_{2f}(j,x_{0})$ be constants
such that for all $w\in\mathbb{C}$ with $\operatorname{Re}(w)\geq 0$,
$|\operatorname{Im}(w)|\leq\operatorname{Re}(w)$, and $Z=x+iy$ with $0\leq
x\leq x_{0}$, $|y|\leq x$, we have

$\left|{e^{w}\left(\frac{\partial{}}{\partial{w}}\right)^{k}\frac{(-1)^{j}e^{-jw}}{\left(1+e^{-w}\right)^{j+1}}}\right|\leq
C_{2e}(k,j),\qquad\left|{\Delta^{j}(E_{Z}(0))}\right|\leq C_{2f}(j,x_{0}).$
(3.26)

Then we have, for $Z=x+iy$ with $0\leq x\leq x_{0}$ and $\left|{y}\right|\leq
x$,

$\sum_{j=0}^{J-1}\int_{0}^{1}\left|{e^{bZ}\left[\left(\tfrac{\partial}{\partial
w}\right)^{2\ell+2K}\tfrac{(-1)^{j}e^{-jw}}{\left(1+e^{-w}\right)^{j+1}}\right]_{w=bZ}}\right|\Delta^{j}(E_{Zt}(0))dt\leq\sum_{j=0}^{J-1}C_{2e}(2\ell+2K,j)C_{2f}(j,x_{0}).$
(3.27)

For the second term in (3.25), we use the product rule, and bound it by

$\sum_{j=0}^{2\ell+2K}\binom{2\ell+2K}{j}\left|{e^{bZ}\left[\left(\tfrac{\partial}{\partial
w}\right)^{2\ell+2K-j}\tfrac{e^{-Jw}}{\left(1+e^{-w}\right)^{J}}\right]_{w=bZ}}\right|\int_{0}^{1}\left|\sum_{r=1}^{\infty}(-1)^{r}r^{j}e^{-rbZ}\Delta^{J}(E_{Zt}(r))\right|dt.$
(3.28)

Now we estimate the inner sum in (3.28). Clearly, we have (as
$\operatorname{Re}(Z)=x\geq 0$)

$\Bigg{|}\sum_{r=1}^{\infty}(-1)^{r}r^{j}e^{-rbZ}\Delta^{J}(E_{Zt}(r))\Bigg{|}\leq\sum_{r=1}^{\infty}r^{j}\left|{\Delta^{J}(E_{Zt}(r))}\right|.$
(3.29)

Let $\varrho>\frac{1}{2}$ (chosen below). We split the sum in (3.29) into two
parts and write

$\sum_{r=1}^{\infty}r^{j}\left|{\Delta^{J}(E_{Zt}(r))}\right|=\sum_{r=0}^{\left\lfloor(xt)^{-\varrho}\right\rfloor}r^{j}\left|{\Delta^{J}(E_{Zt}(r))}\right|+\sum_{r=\left\lfloor(xt)^{-\varrho}\right\rfloor+1}^{\infty}r^{j}\left|{\Delta^{J}(E_{Zt}(r))}\right|.$
(3.30)

For the first sum in (3.30), we use (3.23) to bound it against

$\sum_{\nu=0}^{\frac{J}{2}-1}C_{2c}(J,\nu)(xt)^{J-\nu-\varrho(J+j-2\nu+1)}+C_{2d}(J)\left((xt)^{\frac{J}{2}-\varrho(j+1)}+\delta_{j=0}(xt)^{\frac{J}{2}}\right).$
(3.31)

We choose $\varrho$ such that all the exponents of $xt$ in (3.31) are non-
negative. If $J$ is sufficiently large (say $J\geq 2\ell+2K+4$), then we can
make this choice. Then (3.31) is bounded by a constant independent of $x$ and
$t$. For the second sum in (3.30), we use the bound in (3.17) and obtain

$\sum_{r=\left\lfloor(xt)^{-\varrho}\right\rfloor+1}^{\infty}r^{j}\left|{\Delta^{J}(E_{Zt}(r))}\right|\leq
C_{2a}(J,x_{0})\sum_{r=\left\lfloor(xt)^{-\varrho}\right\rfloor+1}^{\infty}r^{j}e^{-r^{2}xt}.$
(3.32)

Now assume that
$(xt)^{-1}\geq(\frac{j}{2})^{\frac{1}{2}(\varrho-\frac{1}{2})^{-1}}$. Note
that $u^{j}e^{-u^{2}xt}$ is decreasing for $u>(xt)^{-\varrho}$. Thus use the
integral comparison criterion, to obtain

$\sum_{r=\left\lfloor(xt)^{-\varrho}\right\rfloor+1}^{\infty}r^{j}e^{-r^{2}xt}\leq\int_{(xt)^{-\varrho}}^{\infty}u^{j}e^{-xtu^{2}}du\leq
C_{2g}(j,\varrho),$ (3.33)

for a constant $C_{2g}(j,\varrho)$ independent of $xt$. Plugging (3.33) into
(3.32) yields

$\sum_{r=\left\lfloor(xt)^{-\varrho}\right\rfloor+1}^{\infty}r^{j}\left|{\Delta^{J}(E_{Zt}(r))}\right|\leq
C_{2a}(J,x_{0})C_{2g}(j,\varrho)$ (3.34)

for $x_{0}\leq(\frac{j}{2})^{-\frac{1}{2}(\varrho-\frac{1}{2})^{-1}}$.
Inserting (3.31) and (3.34) into (3.30), we get

$\sum_{r=0}^{\infty}r^{j}\left|{\Delta^{J}(E_{Zt}(r))}\right|\leq\sum_{\nu=0}^{\frac{J}{2}-1}C_{2c}(J,\nu)(x_{0}t)^{J-\nu-\varrho(J+j-2\nu+1)}\\\
+C_{2d}(J)\left({(x_{0}t)^{\frac{J}{2}-\varrho(j+1)}+\delta_{j=0}(x_{0}t)^{\frac{J}{2}}}\right)+C_{2a}(J,x_{0})C_{2g}(j,\varrho).$

Integrating over $t$, we obtain

$\int_{0}^{1}\sum_{r=0}^{\infty}r^{j}\left|{\Delta^{J}(E_{Zt}(r))}\right|dt\leq
C_{2h}(J,j,x_{0},\varrho),$ (3.35)

for $x_{0}\leq(\frac{j}{2})^{-\frac{1}{2}(\varrho-\frac{1}{2})^{-1}}$, where

$C_{2h}(J,j,x_{0},\varrho):=\sum_{\nu=0}^{\frac{J}{2}-1}\frac{C_{2c}(J,\nu)}{J-\nu+1-\varrho(J+j-2\nu+1)}x_{0}^{J-\nu-\varrho(J+j-2\nu+1)}\\\
+\frac{C_{2d}(J)}{\frac{J}{2}+1-\varrho(j+1)}x_{0}^{\frac{J}{2}-\varrho(j+1)}+\frac{\delta_{j=0}C_{2d}(J)}{\frac{J}{2}+1}x_{0}^{\frac{J}{2}}+C_{2a}(J,x_{0})C_{2g}(j,\varrho).$

For $k\in\mathbb{N}_{0}$, let $C_{2i}(J,k)$ be constants such that for
$w\in\mathbb{C}$ with $\operatorname{Re}(w)\geq 0$ and
$|\operatorname{Im}(w)|\leq\operatorname{Re}(w)$

$\left|{e^{w}\left(\frac{\partial}{\partial
w}\right)^{k}{\frac{e^{-Jw}}{\left(1+e^{-w}\right)^{J}}}}\right|\leq
C_{2i}(J,k).$ (3.36)

Using (3.28), (3.35), and (3.36), the second integral in (3.25) is bounded by

$\sum_{j=0}^{2\ell+2K}\binom{2\ell+2K}{j}C_{2i}(J,2\ell+2K-j)C_{2h}(J,j,x_{0},\varrho)$
(3.37)

for $x_{0}\leq(\ell+K)^{-\frac{1}{2}(\varrho-\frac{1}{2})^{-1}}$ (this
assumption is needed for (3.35) to hold). Combining the error estimates (3.27)
and (3.37), and plugging into (3.25), we obtain that

$\left|{e^{bZ}R_{K}(2\ell;bZ,Z)}\right|\leq\frac{1}{(K-1)!}\left(\sum_{j=0}^{J-1}C_{2e}(2\ell+2K,j)C_{2f}(j,x_{0})\right.\\\
+\left.\sum_{j=0}^{2\ell+2K}\binom{2\ell+2K}{j}C_{2i}(J,2\ell+2K-j)C_{2h}(J,j,x_{0},\varrho)\right),$

which holds for $Z=x+iy$, $0\leq x\leq
x_{0}\leq(\ell+K)^{-\frac{1}{2}(\varrho-\frac{1}{2})^{-1}}$, $|y|\leq x$.
Plugging back into (3.16), we obtain, for $\beta_{n}\leq
x_{0}\leq\frac{2}{3}(\ell+K)^{-\frac{1}{2}(\varrho-\frac{1}{2})^{-1}}$,

$\left|{\sum_{r=1}^{\infty}(-1)^{r+1}r^{2\ell}e^{-\left(\frac{3r^{2}}{2}+\left(m+\frac{1}{2}\right)r\right)Z}-\sum_{k=0}^{K-1}\frac{\left(-\frac{3Z}{2}\right)^{k}}{k!}\left[\left(\frac{\partial}{\partial
w}\right)^{2\ell+2k}{\frac{1}{1+e^{w}}}\right]_{w=\left(m+\frac{1}{2}\right)Z}}\right|\\\
\leq\frac{3^{K}\beta_{n}^{K}e^{-\left(m+\frac{1}{2}\right)\beta_{n}}}{2^{\frac{K}{2}}(K-1)!}\left(\sum_{j=0}^{J-1}C_{2e}(2\ell+2K,j)C_{2f}\left(j,\frac{3x_{0}}{2}\right)\right.\\\
+\left.\sum_{j=0}^{2\ell+2K}\binom{2\ell+2K}{j}C_{2i}(J,2\ell+2K-j)C_{2h}\left({J,j,\frac{3x_{0}}{2},\varrho}\right)\right).$
(3.38)

Multiplying both sides of (3.38) by $\frac{2Z^{2\ell}}{(2\ell)!}$ we obtain,
again using that $|Z|\leq\sqrt{2}x$

$\displaystyle\left|{\mathcal{M}_{\ell,m}(Z)-\frac{2}{(2\ell)!}\sum_{k=0}^{K-1}\frac{\left(-\frac{3}{2}\right)^{k}Z^{2\ell+k}}{k!}\left[\left(\frac{\partial{}}{\partial{w}}\right)^{2\ell+2k}\frac{1}{1+e^{w}}\right]_{w=\left(m+\frac{1}{2}\right)Z}}\right|$
$\displaystyle\hskip
28.45274pt\leq\frac{2^{\ell-\frac{K}{2}+1}3^{K}}{(2\ell)!(K-1)!}\beta_{n}^{2\ell+K}e^{-\left(m+\frac{1}{2}\right)\beta_{n}}\left(\sum_{j=0}^{J-1}C_{2e}(2\ell+2K,j)C_{2f}\left(j,\frac{3x_{0}}{2}\right)\right.$
$\displaystyle\hskip
170.71652pt\left.+\sum_{j=0}^{2\ell+2K}\binom{2\ell+2K}{j}C_{2i}(J,2\ell+2K-j)C_{2h}\left(J,j,\frac{3x_{0}}{2},\varrho\right)\right)$
$\displaystyle\hskip
28.45274pt=:C_{2}(\ell,K,J,x_{0},\varrho)\beta_{n}^{2\ell+K}e^{-\left(m+\frac{1}{2}\right)\beta_{n}}.\qed$

###  3.4. Back to the Circle Method

We now combine the above results and use the Circle Method. By (3.2), we have
a splitting

$N(m,n)-N(m+1,n)=\mathfrak{M}+\mathfrak{m},$

where $\mathfrak{M}$ and $\mathfrak{m}$ are given in (3.3) and (3.4),
respectively. From Proposition 3.1 we know that

$\left|{\mathfrak{m}}\right|\leq 13e^{\frac{3\mathcal{C}_{n}}{2}}.$

Meanwhile, by (3.6), we have a splitting

$\mathfrak{M}=\mathcal{M}^{[1]}+\mathcal{E}^{[1]},$

with (see (3.7))

$\left|{\mathcal{E}^{[1]}}\right|\leq e^{\mathcal{C}_{n}}.$

A straightforward calculation then shows the following.

######  Lemma 3.6.

Let $n\in\mathbb{N}$. Define $\beta_{n}$ and $\mathcal{C}_{n}$ as in (3.1) and
let $Z=\beta_{n}+iy$. For $\nu\in\mathbb{N}_{0}$, we have

$\int_{\left|{y}\right|\leq\beta_{n}}\left|{y}\right|^{\nu}\left|{e^{\left(n-\frac{1}{24}\right)Z+\frac{\pi^{2}}{6Z}}}\right|dy\leq\frac{12^{\frac{\nu+1}{2}}}{\pi^{\nu+1}}\Gamma\left({\frac{\nu+1}{2}}\right)\beta_{n}^{\frac{3}{2}(\nu+1)}e^{2\mathcal{C}_{n}}.$

Next we give a decomposition of $\mathcal{M}^{[1]}$ which follows directly
from (3.6) and Lemmas 3.2 and 3.6.

######  Lemma 3.7.

For $L\geq 2$, we have a decomposition

$\mathcal{M}^{[1]}=\sum_{\ell=1}^{L-1}\mathcal{M}_{\ell}^{[2]}+\mathcal{E}_{L}^{[2]},$

where

$\displaystyle\mathcal{M}_{\ell}^{[2]}$
$\displaystyle:=\;\frac{1}{(2\pi)^{\frac{3}{2}}}\int_{\left|{y}\right|\leq\beta_{n}}\sqrt{Z}e^{\left(n-\frac{1}{24}\right)Z+\frac{\pi^{2}}{6Z}}\mathcal{M}_{\ell,m}(Z)dy,$
(3.39) $\displaystyle\mathcal{E}_{L}^{[2]}$
$\displaystyle:=\;\frac{1}{(2\pi)^{\frac{3}{2}}}\int_{\left|{y}\right|\leq\beta_{n}}\sqrt{Z}e^{\left(n-\frac{1}{24}\right)Z+\frac{\pi^{2}}{6Z}}\mathcal{E}_{L,m}(Z)dy.$
(3.40)

Moreover, the error term $\mathcal{E}_{L}^{[2]}$ satisfies, for $\beta_{n}\leq
x_{0}$

$\left|{\mathcal{E}_{L}^{[2]}}\right|\leq\frac{\sqrt{3}}{2^{\frac{1}{4}}\pi^{2}}C_{1}(L,x_{0})\beta_{n}^{L+\frac{3}{2}}e^{2\mathcal{C}_{n}-\left(m+\frac{1}{2}\right)\beta_{n}}=:\mathcal{D}^{[2]}_{L}(x_{0})\beta_{n}^{L+\frac{3}{2}}e^{2\mathcal{C}_{n}-\left(m+\frac{1}{2}\right)\beta_{n}}.$
(3.41)

From Proposition 3.4 and Lemma 3.6 we obtain the following lemma.

######  Lemma 3.8.

For $\ell,K\in\mathbb{N}$, we have a decomposition

$\displaystyle\mathcal{M}_{\ell}^{[2]}=\sum_{k=0}^{K-1}\mathcal{M}_{\ell,k}^{[3]}+\mathcal{E}_{\ell,K}^{[3]},$

where

$\mathcal{M}_{\ell,k}^{[3]}:=\frac{2\left(-\frac{3}{2}\right)^{k}}{(2\pi)^{\frac{3}{2}}(2\ell)!k!}\int_{\left|{y}\right|\leq\beta_{n}}Z^{2\ell+k+\frac{1}{2}}e^{\left({n-\frac{1}{24}}\right)Z+\frac{\pi^{2}}{6Z}}\left[{\left({\frac{\partial}{\partial
w}}\right)^{2\ell+2k}{\frac{1}{1+e^{w}}}}\right]_{w=\left(m+\frac{1}{2}\right)Z}dy.$

The error term $\mathcal{E}_{\ell,K}^{[3]}$ satisfies, for $\beta_{n}\leq
x_{0}\leq\frac{2}{3}(\ell+K)^{-\frac{1}{2}(\varrho-\frac{1}{2})^{-1}}$

$\displaystyle\left|{\mathcal{E}_{\ell,K}^{[3]}}\right|\leq\frac{\sqrt{3}}{2^{\frac{1}{4}}\pi^{2}}C_{2}(\ell,K,J,x_{0},\varrho)\beta_{n}^{2\ell+K+2}e^{2\mathcal{C}_{n}-\left(m+\frac{1}{2}\right)\beta_{n}}=:\mathcal{D}^{[3]}_{\ell,K}(J,x_{0},\varrho)\beta_{n}^{2\ell+K+2}e^{2\mathcal{C}_{n}-\left(m+\frac{1}{2}\right)\beta_{n}}.$

Next we give a decomposition of $\mathcal{M}_{\ell,k}^{[3]}$.

######  Lemma 3.9.

For $\ell,R\in\mathbb{N}$, $k\in\mathbb{N}_{0}$, we have a decomposition

$\mathcal{M}_{\ell,k}^{[3]}=\sum_{r=0}^{R-1}\mathcal{M}_{\ell,k,r}^{[4]}+\mathcal{E}_{\ell,k,R}^{[4]},$

where

$\mathcal{M}_{\ell,k,r}^{[4]}:=\tfrac{2\left(-\frac{3}{2}\right)^{k}\left(m+\frac{1}{2}\right)^{r}}{\sqrt{2\pi}(2\ell)!k!r!}\left[\left(\tfrac{\partial}{\partial
w}\right)^{2\ell+2k+r}\tfrac{1}{1+e^{w}}\right]_{w=\left(m+\frac{1}{2}\right)\beta_{n}}\tfrac{1}{2\pi}\int_{|y|\leq\beta_{n}}Z^{2\ell+k+\frac{1}{2}}e^{\left(n-\frac{1}{24}\right)Z+\frac{\pi^{2}}{6Z}}(iy)^{r}dy.$
(3.42)

The error term $\mathcal{E}_{\ell,k,R}^{[4]}$ satisfies

$\displaystyle\left|{\mathcal{E}_{\ell,k,R}^{[4]}}\right|$
$\displaystyle\leq\frac{2^{\ell-\frac{k}{2}+R+\frac{3}{4}}3^{k+\frac{R+1}{2}}\Gamma\left(\frac{R+1}{2}\right)\left(m+\frac{1}{2}\right)^{R}}{\pi^{R+\frac{5}{2}}(2\ell)!k!(R-1)!}C_{2e}(2\ell+2k+R,0)\beta_{n}^{2\ell+k+\frac{3R}{2}+2}e^{2\mathcal{C}_{n}-\left(m+\frac{1}{2}\right)\beta_{n}}.$
$\displaystyle=:\mathcal{D}_{\ell,k,R}^{[4]}{\textstyle\left(m+\frac{1}{2}\right)^{R}}\beta_{n}^{2\ell+k+\frac{3R}{2}+2}e^{2\mathcal{C}_{n}-\left(m+\frac{1}{2}\right)\beta_{n}}.$
(3.43)

###### Proof.

For $j\in\mathbb{N}_{0}$, we have Taylor series expansion in $Z$ at
$Z=\beta_{n}$

$\left[\left(\frac{\partial{}}{\partial{w}}\right)^{j}\frac{1}{1+e^{w}}\right]_{w=\left(m+\frac{1}{2}\right)Z}=\sum_{r=0}^{R-1}\left[\left(\frac{\partial{}}{\partial{w}}\right)^{j+r}\frac{1}{1+e^{w}}\right]_{w=\left(m+\frac{1}{2}\right)\beta_{n}}\frac{\left(m+\frac{1}{2}\right)^{r}(iy)^{r}}{r!}\\\
+\frac{i^{R}\left(m+\frac{1}{2}\right)^{R}}{(R-1)!}\int_{0}^{y}(y-u)^{R-1}\left[\left(\frac{\partial{}}{\partial{w}}\right)^{j+R}\frac{1}{1+e^{w}}\right]_{w=\left(m+\frac{1}{2}\right)(\beta_{n}+iu)}du.$
(3.44)

It follows by (3.26) that the error term in (3.44) is bounded by

$\frac{\left(m+\frac{1}{2}\right)^{R}|y|^{R}}{(R-1)!}C_{2e}(j+R,0)e^{-\left(m+\frac{1}{2}\right)\beta_{n}}.$
(3.45)

The lemma then follows from (3.45) and Lemma 3.6. ∎

We need yet another decomposition.

######  Lemma 3.10.

For $\ell,\Lambda\in\mathbb{N}$ and $k,r\in\mathbb{N}_{0}$, we have a
decomposition

$\mathcal{M}_{\ell,k,r}^{[4]}=\sum_{l=\left\lceil\frac{r}{2}\right\rceil}^{\Lambda-1}\mathcal{M}_{\ell,k,r,l}^{[5]}+\mathcal{E}_{\ell,k,r,\Lambda}^{[5]},$

where

$\mathcal{M}_{\ell,k,r,l}^{[5]}:=\frac{(-1)^{l}\left(-\frac{3}{2}\right)^{k}3^{l+\frac{1}{2}}}{\pi^{2l+2}(2\ell)!k!r!}a_{l}^{[r]}\left(-2\ell-k-\tfrac{3}{2}\right)\left(m+\tfrac{1}{2}\right)^{r}\\\
\times\left[\left(\frac{\partial}{\partial
w}\right)^{2\ell+2k+r}{\frac{1}{1+e^{w}}}\right]_{w=\left(m+\frac{1}{2}\right)\beta_{n}}\beta_{n}^{2\ell+k+r+l+2}e^{2\mathcal{C}_{n}},$
(3.46)

and the error term $\mathcal{E}_{\ell,k,r,\Lambda}^{[5]}$ satisfies for
$\beta_{n}\leq x_{0}$ and some constant
$\mathcal{D}_{\ell,k,r,\Lambda}^{[5]}(x_{0})$

$\left|{\mathcal{E}_{\ell,k,r,\Lambda}^{[5]}}\right|\leq\mathcal{D}_{\ell,k,r,\Lambda}^{[5]}(x_{0}){\left(m+\tfrac{1}{2}\right)}^{r}\beta_{n}^{2\ell+k+r+\Lambda+2}e^{2\mathcal{C}_{n}-\left(m+\frac{1}{2}\right)\beta_{n}}.$
(3.47)

###### Proof.

We make the change of variable $Z\mapsto\beta_{n}Z$ to obtain that the
integral in (3.42) (including the previous $\frac{1}{2\pi}$) equals

$\beta_{n}^{2\ell+k+r+\frac{3}{2}}\sum_{j=0}^{r}\binom{r}{j}(-1)^{r+j}\mathcal{I}_{-2\ell-
k-j-\frac{3}{2}}(2\mathcal{C}_{n}).$ (3.48)

Applying Lemma 2.1, we can approximate the sum in (3.48) with $I$-Bessel
functions. To be more precise, we have, for $\beta_{n}\leq x_{0}$

$\beta_{n}^{2\ell+k+r+\frac{3}{2}}\left|{\sum_{j=0}^{r}\binom{r}{j}(-1)^{r+j}\left(\mathcal{I}_{-2\ell-
k-j-\frac{3}{2}}(2\mathcal{C}_{n})-I_{-2\ell-
k-j-\frac{3}{2}}(2\mathcal{C}_{n})\right)}\right|\\\
\leq\frac{\beta_{n}^{2\ell+k+r+\frac{3}{2}}}{\pi}\sum_{j=0}^{r}\binom{r}{j}2^{\ell+\frac{k}{2}+\frac{j}{2}+\frac{1}{4}}\left(\frac{\Gamma\left(2\ell+k+j+\frac{3}{2}\right)}{\mathcal{C}_{n}^{2\ell+k+j+\frac{3}{2}}}+2e^{\frac{3\mathcal{C}_{n}}{2}}\right)\leq
C_{3a}(\ell,k,r,x_{0})e^{\frac{3\mathcal{C}_{n}}{2}}$ (3.49)

for some constant $C_{3a}(\ell,k,r,x_{0})$. Now we apply (2.3) (with $M=0$)
and obtain for $\Lambda\in\mathbb{N}$

$\displaystyle\beta_{n}^{2\ell+k+r+\frac{3}{2}}\left|I_{-2\ell-k-\frac{3}{2}}^{[r]}(2\mathcal{C}_{n})-\frac{e^{2\mathcal{C}_{n}}}{2\sqrt{\pi\mathcal{C}_{n}}}\sum_{l=\left\lceil\frac{r}{2}\right\rceil}^{\Lambda-1}(-1)^{l}a_{l}^{[r]}\left(-2\ell-k-\tfrac{3}{2}\right)(2\mathcal{C}_{n})^{-l}\right|$
$\displaystyle\hskip
42.67912pt\leq\beta_{n}^{2\ell+k+r+\frac{3}{2}}\left(\frac{e^{2\mathcal{C}_{n}}}{2\sqrt{\pi\mathcal{C}_{n}}}\left|\delta_{\Lambda}^{[r]}\left(-2\ell-k-\tfrac{3}{2},2\mathcal{C}_{n}\right)\right|+\frac{e^{-2\mathcal{C}_{n}}}{2\sqrt{\pi\mathcal{C}_{n}}}\left|\gamma_{0}^{[r]}\left(-2\ell-k-\tfrac{3}{2},2\mathcal{C}_{n}\right)\right|\right)$
$\displaystyle\hskip 42.67912pt\leq
C_{3b}(\ell,k,r,\Lambda,x_{0})\beta_{n}^{2\ell+k+r+\Lambda+2}e^{2\mathcal{C}_{n}}$
(3.50)

for $\beta_{n}\leq x_{0}$ and some constant $C_{3b}(\ell,k,r,\Lambda,x_{0})$.
Plugging this back into (3.48), we conclude that

$\left|{\frac{1}{2\pi}\int_{|y|\leq\beta_{n}}Z^{2\ell+k+\frac{1}{2}}e^{\left(n-\frac{1}{24}\right)Z+\frac{\pi^{2}}{6Z}}(iy)^{r}dy-\sum_{l=\left\lceil\frac{r}{2}\right\rceil}^{\Lambda-1}\frac{(-1)^{l}3^{l+\frac{1}{2}}}{\sqrt{2}\pi^{2l+\frac{3}{2}}}a_{l}^{[r]}\left(-2\ell-k-\frac{3}{2}\right)\beta_{n}^{2\ell+k+r+l+2}e^{2\mathcal{C}_{n}}}\right|\\\
\leq
C_{3b}(\ell,k,r,\Lambda,x_{0})\beta_{n}^{2\ell+k+r+\Lambda+2}e^{2\mathcal{C}_{n}}+C_{3a}(\ell,k,r,x_{0})e^{\frac{3\mathcal{C}_{n}}{2}}$
(3.51)

for $\beta_{n}\leq x_{0}$, by (3.49) and (3.50). It then follows from (3.51)
that

$\displaystyle\left|{\mathcal{E}_{\ell,k,r,\Lambda}^{[5]}}\right|$
$\displaystyle\leq\frac{2\left(\frac{3}{2}\right)^{k}\left(m+\frac{1}{2}\right)^{r}}{\sqrt{2\pi}(2\ell)!k!r!}C_{3c}(2\ell+2k+r)$
$\displaystyle\times\left(C_{3b}(\ell,k,r,\Lambda,x_{0})\beta_{n}^{2\ell+k+r+\Lambda+2}e^{2\mathcal{C}_{n}-\left(m+\frac{1}{2}\right)\beta_{n}}+C_{3a}(\ell,k,r,x_{0})e^{\frac{3}{2}\mathcal{C}_{n}-\left(m+\frac{1}{2}\right)\beta_{n}}\right)$

for $\beta_{n}\leq x_{0}$, where $C_{3c}(j)$ is a constant such that, for
$w\in\mathbb{R}_{0}^{+}$,

$\left|{e^{w}\left(\frac{\partial}{\partial
w}\right)^{j}{\frac{1}{1+e^{w}}}}\right|\leq C_{3c}(j).$ (3.52)

Now let $\mathcal{D}_{\ell,k,r,\Lambda}^{[5]}(x_{0})$ be constants such that

$\displaystyle\frac{2\left(\frac{3}{2}\right)^{k}\left(m+\frac{1}{2}\right)^{r}}{\sqrt{2\pi}(2\ell)!k!r!}C_{3c}(2\ell+2k+r)$
$\displaystyle\hskip
42.67912pt\times\left(C_{3b}(\ell,k,r,\Lambda,x_{0})+C_{3a}(\ell,k,r,x_{0})e^{-\frac{\mathcal{C}_{n}}{2}}\beta_{n}^{-(2\ell+k+r+\Lambda+2)}\right)\beta_{n}^{2\ell+k+r+\Lambda+2}e^{2\mathcal{C}_{n}-\left(m+\frac{1}{2}\right)\beta_{n}}$
$\displaystyle\hskip
213.39566pt\leq\mathcal{D}_{\ell,k,r,\Lambda}^{[5]}(x_{0}){\textstyle\left(m+\frac{1}{2}\right)}^{r}\beta_{n}^{2\ell+k+r+\Lambda+2}e^{2\mathcal{C}_{n}-\left(m+\frac{1}{2}\right)\beta_{n}}$

for $\beta_{n}\leq x_{0}$. It is straightforward to see that
$\mathcal{D}_{\ell,k,r,\Lambda}^{[5]}(x_{0})$ exists. The lemma now follows. ∎

###  3.5. An extra decomposition

We conclude this section with one final decomposition we require in the proof
of our main result. In this subsection, assume that $m\ll\beta_{n}^{-1}$. For
$j\in\mathbb{N}_{0}$, consider

$f_{j}(u):=e^{u}\left(\frac{\partial}{\partial
u}\right)^{j}{\frac{1}{1+e^{u}}}.$ (3.53)

By Taylor’s Theorem, for $\gamma\geq 0$, there is a constant
$C_{4}(j,N,\gamma)$ such that, for $0\leq u\leq\gamma$,

$\left|f_{j}(u)-\sum_{\nu=0}^{N-1}f_{j}^{(\nu)}(0)\frac{u^{\nu}}{\nu!}\right|\leq
C_{4}(j,N,\gamma)u^{N}.$ (3.54)

Using the Taylor expansion above, we immediately obtain the following lemma.

######  Lemma 3.11.

Let $\ell,N\in\mathbb{N}$, $k,r,l\in\mathbb{N}_{0}$ Then we have a
decomposition

$\mathcal{M}_{\ell,k,r,l}^{[5]}=\sum_{\nu=0}^{N-1}\mathcal{M}_{\ell,k,r,l,\nu}^{[6]}+\mathcal{E}_{\ell,k,r,l,N}^{[6]},$

where

$\displaystyle\mathcal{M}_{\ell,k,r,l,\nu}^{[6]}$
$\displaystyle:=\frac{(-1)^{l}\left(-\frac{3}{2}\right)^{k}3^{l+\frac{1}{2}}}{\pi^{2l+2}(2\ell)!k!r!\nu!}a_{l}^{[r]}\left(-2\ell-k-\tfrac{3}{2}\right)\left(m+\tfrac{1}{2}\right)^{r+\nu}f_{2\ell+2k+r}^{(\nu)}(0)\beta_{n}^{2\ell+k+r+l+\nu+2}$
$\displaystyle\hskip 298.75394pt\times
e^{2\mathcal{C}_{n}-\left(m+\frac{1}{2}\right)\beta_{n}},$ (3.55)
$\displaystyle\mathcal{E}_{\ell,k,r,l,N}^{[6]}$
$\displaystyle:=\frac{\left(\tfrac{3}{2}\right)^{k}3^{l+\tfrac{1}{2}}}{\pi^{2l+2}(2\ell)!k!r!}\left|{a_{l}^{[r]}\left(-2\ell-k-\tfrac{3}{2}\right)}\right|\left(m+\tfrac{1}{2}\right)^{r+N}$
$\displaystyle\times\left(f_{2\ell+2k+r}\left(\left(m+\tfrac{1}{2}\right)\beta_{n}\right)-\sum_{\nu=0}^{N-1}f_{2\ell+2k+r}^{(\nu)}(0)\frac{\left(m+\frac{1}{2}\right)^{\nu}\beta_{n}^{\nu}}{\nu!}\right)\beta_{n}^{2\ell+k+r+l+N+2}e^{2\mathcal{C}_{n}-\left(m+\frac{1}{2}\right)\beta_{n}}.$

For $\gamma>0$ and $(m+\frac{1}{2})\beta_{n}\leq\gamma$, the error term
$\mathcal{E}_{\ell,k,r,l,N}^{[6]}$ satisfies

$\displaystyle\left|{\mathcal{E}_{\ell,k,r,l,N}^{[6]}}\right|$
$\displaystyle\leq\frac{\left(\tfrac{3}{2}\right)^{k}3^{l+\frac{1}{2}}}{\pi^{2l+2}(2\ell)!k!r!}\left|{a_{l}^{[r]}\left(-2\ell-k-\tfrac{3}{2}\right)}\right|\left(m+\tfrac{1}{2}\right)^{r+N}C_{4}(2\ell+2k+r,N,\gamma)$
$\displaystyle\hskip
85.35826pt\times\beta_{n}^{2\ell+k+r+l+N+2}e^{2\mathcal{C}_{n}-\left(m+\frac{1}{2}\right)\beta_{n}}$
$\displaystyle=:\mathcal{D}_{\ell,k,r,l,N}^{[6]}(\gamma)\beta_{n}^{2\ell+k+r+l+N+2}e^{2\mathcal{C}_{n}-\left(m+\frac{1}{2}\right)\beta_{n}}.$
(3.56)

##  4\. Numerical computations

In this section, we prove the following theorem which establishes our main
result for small $m$. Recall the definitions of $\beta_{n}$ and $\Lambda_{n}$
from (3.1).

######  Theorem 4.1.

Let $n\geq 925276$. Then we have, for $m\leq
3^{-1}\beta_{n}^{-\frac{3}{2}}-\frac{1}{2}$,

$N(m,n)-N(m+1,n)\geq 0.$ (4.1)

The strategy is to apply the decompositions above, and choose parameters such
that the sizes of the error terms are quite small. The parameters we choose
also depend on the size of $m$. Precisely, for the proof of the theorem, we
pick $(x_{0},L,K,R,\Lambda,N,J)=(\frac{1}{750},5,4,1,2,2,50)$ for
$m+\frac{1}{2}\leq\beta_{n}^{-1}$, and
$(x_{0},L,K,R,\Lambda,J)=(\frac{1}{750},5,4,4,2,50)$ for
$\frac{1}{\beta_{n}}\leq m+\frac{1}{2}\leq 3^{-1}\beta_{n}^{-\frac{3}{2}}$.
Once these parameters are chosen, we need to compute the following constants
to obtain the error bounds:

  * •

$C_{2a}$: We need $C_{2a}(50,\frac{1}{500})$.

  * •

$C_{2b}$: We need $C_{2b}(50,\nu)$ for $0\leq\nu\leq 24$.

  * •

$C_{2e}$: We need $C_{2e}(k,j)$ for $k\in\left\\{{10,12,14,16}\right\\}$,
$0\leq j\leq 49$, as well as $C_{2e}(k,0)$ for $2\leq k\leq 18$.

  * •

$C_{2f}$: We need $C_{2f}(j,\frac{1}{500})$ for $0\leq j\leq 49$.

  * •

$C_{2g}$: We need $C_{2g}(j,\varrho)$ for $0\leq j\leq 16$. Note that at this
stage the parameter $\varrho$ is not specified. In principle, we want to
choose $\varrho$ such that the overall error is as small as possible. This is
however quite tedious. Therefore we instead pick the parameter $\varrho$ such
that the overall error is quite small. Hiding the trial-and-error steps, we
are interested in $C_{2g}(j,\varrho)$ for $0\leq j\leq 16$, and
$\varrho\in\\{{0.66021,0.66950,0.67773,0.68510}\\}$.

  * •

$C_{2i}$: We need $C_{2i}(50,k)$ for $0\leq k\leq 16$.

  * •

$C_{3a}$: We need $C_{3a}(\ell,k,r,\frac{1}{750})$ for $1\leq\ell\leq 4$,
$0\leq k\leq 3$, $0\leq r\leq 3$.

  * •

$C_{3b}$: We need $C_{3b}(\ell,k,r,2,\frac{1}{750})$ for $1\leq\ell\leq 4$,
$0\leq k\leq 3$, $0\leq r\leq 3$.

  * •

$C_{3c}$: We need $C_{3c}(j)$ for $2\leq j\leq 17$.

  * •

$C_{4}$: We need to compute $C_{4}(j,2,1)$ for $2\leq j\leq 14$.

The constants $C_{1},C_{2},C_{2c},C_{2d}$, and $C_{2h}$ are defined
explicitly, possibly in terms of other constants. The constants
$C_{2b},C_{2e},C_{2f},C_{2g},C_{2i},C_{3a},C_{3b},C_{3c},$ and $C_{4}$ are
found using one-dimensional optimization. We now give an outline on how to
compute these.

  1. (1)

For $C_{2b}$, we have to compute $C_{2b}(50,\nu)$ for $0\leq\nu\leq 24$. As
the defining function (3.22) decays as $\operatorname{Re}(w)\to\infty$, it
suffices to consider the function over a (sufficiently large) compact region
in $Z$. By the maximum modulus principle, the maximal modulus of the function
occurs at the boundary of the region. So we may find an optimal value for
$C_{2b}(50,\nu)$ using a one-dimensional plot along the lines
$|\operatorname{Im}(w)|=\operatorname{Re}(w)$. The same argument applies for
$C_{2e}$ and $C_{2i}$ (see (3.26) and (3.36)) if the parameters are fixed. For
$C_{2f}$ (see (3.26)), the region under consideration is by definition
bounded, so we do not need to consider limits.

  2. (2)

For $C_{2g}$, $C_{3a}$, $C_{3b}$, $C_{3c}$, and $C_{4}$, we can find their
optimal values (if the parameters are fixed) using real-valued functions (see
(3.33), (3.49), (3.50), (3.52), and (3.54), respectively).

It remains to compute $C_{2a}(50,\frac{1}{500})$, which is computationally
more difficult. We recall the definition in (3.18). We show that we may take

$C_{2a}\left(50,\tfrac{1}{500}\right)=\left|{\sum_{j=0}^{50}\binom{50}{j}(-1)^{j}e^{-\left(2\cdot
563j+j^{2}\right)\frac{1+i}{500}}}\right|\approx 30.02166509.$ (4.2)

We start with some observations.

  1. (1)

If $r$ is fixed, then

$\sum_{j=0}^{50}\binom{50}{j}(-1)^{j}e^{-\left(2rj+j^{2}\right)Z}$

is holomorphic in $Z$. By the maximum modulus principle, the left-hand side of
(3.18) obtains its maximum on the boundary of $0\leq\left|{y}\right|\leq
x\leq\frac{1}{500}$. That is, (3.18) is maximal if $y=\pm x$, or if
$0\leq\left|{y}\right|\leq x=\frac{1}{500}$. As (3.18) is invariant under
complex conjugation, we may assume that $y\geq 0$.

  2. (2)

First assume that $rx\geq\frac{3}{2}$. Then $|e^{-2rZ}|\leq e^{-3}$, and we
have the trivial bound

$\displaystyle\left|{\sum_{j=0}^{50}\binom{50}{j}(-1)^{j}e^{-\left(2rj+j^{2}\right)Z}}\right|\leq\sum_{j=0}^{50}\binom{50}{j}e^{-3j}=\left({1+e^{-3}}\right)^{50}\approx
11.35170075.$

So the function satisfies the bound (4.2) if $rx\geq\frac{3}{2}$. It suffices
to consider the case where $0\leq rx\leq\frac{3}{2}$.

First consider the easier case $0\leq y\leq x=\frac{1}{500}$ from observation
(1). By observation (2) we only have to consider the case
$r\leq\frac{3}{2}\cdot 500=750$. As $r\in\mathbb{N}_{0}$, we can verify that
our constant works in this case, with $x=\frac{1}{500}$ and $0\leq r\leq 750$
fixed, and let $0\leq y\leq\frac{1}{500}$ vary.

In the trickier case $0\leq y=x\leq\frac{1}{500}$ from observation (1), (3.18)
becomes

$\left|{\sum_{j=0}^{50}\binom{50}{j}(-1)^{j}e^{-\left(2rjx+j^{2}x\right)(1+i)}}\right|\leq
C_{2a}(J,x_{0}).$ (4.3)

Let

$\displaystyle
f(\rho,x):=\sum_{j=0}^{50}\binom{50}{j}(-1)^{j}e^{-\left(2j\rho+j^{2}x\right)(1+i)}.$

The function on the left-hand side of (4.3) is $|f(rx,x)|$. By observation (2)
we only have to consider the compact region $0\leq\rho\leq\frac{3}{2}$, $0\leq
x\leq\frac{1}{500}$. We split $f(\rho,x)$ into real and imaginary parts

$\displaystyle
f(\rho,x)=\sum_{j=0}^{50}\binom{50}{j}(-1)^{j}e^{-\left(2j\rho+j^{2}x\right)}\left({\cos\left({2j\rho+j^{2}x}\right)-i\sin\left({2j\rho+j^{2}x}\right)}\right),$

and use this to obtain an expression of $\left|{f(\rho,x)}\right|^{2}$

$\displaystyle\scalebox{0.9}{$\displaystyle\left|{f(\rho,x)}\right|^{2}=\left({\sum_{j=0}^{50}\binom{50}{j}(-1)^{j}e^{-\left(2j\rho+j^{2}x\right)}\cos\left({2j\rho+j^{2}x}\right)}\right)^{2}+\left({\sum_{j=0}^{50}\binom{50}{j}(-1)^{j}e^{-\left(2j\rho+j^{2}x\right)}\sin\left({2j\rho+j^{2}x}\right)}\right)^{2}.$}$

We consider the Taylor series expansion for $\left|{f(\rho,x)}\right|$ at
$x=0$ as a function of $x$

$\displaystyle\left|{f(\rho,x)}\right|=\sum_{\ell=0}^{L-1}\frac{c_{\ell}(\rho)}{\ell!}x^{\ell}+R_{L}.$

Here $c_{\ell}$ is a function of $\rho$, and for $x\geq 0$, $R_{L}$ is bounded
by

$\displaystyle\left|{R_{L}}\right|\leq\frac{x^{L}}{L!}\max_{0\leq u\leq
x}\left|{\frac{\partial{{}^{L}}}{\partial{u^{L}}}\left|{f(\rho,u)}\right|}\right|\leq\frac{x^{L}}{L!}\sum_{j=0}^{50}\binom{50}{j}j^{2L}.$

Let
$d_{\ell}:=\max\left\\{{c_{\ell}(\rho)\;:\;0\leq\rho\leq\frac{3}{2}}\right\\}$.
Then we have for $0\leq\rho\leq\frac{3}{2}$ and $x\geq 0$

$\left|{f(\rho,x)}\right|\leq\sum_{\ell=0}^{L-1}\frac{d_{\ell}}{\ell!}x^{\ell}+\frac{x^{L}}{L!}\sum_{j=0}^{50}\binom{50}{j}j^{2L}.$
(4.4)

We check that

$\displaystyle d_{0}$ $\displaystyle\leq 28.68558614,$ $\displaystyle d_{1}$
$\displaystyle\leq 1220.781415,$ $\displaystyle d_{2}$ $\displaystyle\leq
99722.52362,$ $\displaystyle d_{3}$ $\displaystyle\leq 12254165.90,$
$\displaystyle d_{4}$ $\displaystyle\leq 2040711473,$ $\displaystyle d_{5}$
$\displaystyle\leq 4.319895682\cdot 10^{11},$ $\displaystyle d_{6}$
$\displaystyle\leq 1.111629852\cdot 10^{14},$ $\displaystyle d_{7}$
$\displaystyle\leq 3.373057489\cdot 10^{16},$ $\displaystyle d_{8}$
$\displaystyle\leq 1.194739620\cdot 10^{19},$ $\displaystyle d_{9}$
$\displaystyle\leq 5.092346787\cdot 10^{21}.$

Plugging into to (4.4), we compute that

$\displaystyle\left|{f(\rho,x)}\right|\leq 29.70793207\text{ for
}0\leq\rho\leq\tfrac{3}{2},\;0\leq x\leq\tfrac{1}{6000}.$

So it remains to check the region $\frac{1}{6000}\leq x\leq\frac{1}{500}$.
Since we may assume from observation (2) that $0\leq rx\leq\frac{3}{2}$, and
$r\in\mathbb{N}_{0}$, we can verify that our constant works in this case by
considering $0\leq r\leq 9000$ fixed, and let $0\leq x\leq\frac{1}{500}$ vary.

######  Remark.

For our calculations, we always pick the optimal values for these named
constants, computed using numerical methods. As usual, values used for error
bounds are always rounded up, and values used for main terms are always
rounded down. For the convenience to the reader, we also sometimes display the
approximated values for the constants. These approximations are marked with
the approximation sign $\approx$.

With these observations in mind we now prove the main result of this section.

###### Proof of Theorem 4.1.

First suppose that $m+\frac{1}{2}\leq\frac{1}{\beta_{n}}$, i.e., we set
$\gamma=1$. We pick $x_{0}=\frac{1}{750}$, and $L=5$, $K=4$, $R=1$,
$\Lambda=2$, $N=2$, and $J=50$. Since the constants $C_{j}$ are now
determined, we are able to compute the error bounds $\mathcal{D}^{[j]}$. From
(3.55), the main term is given by

$\mathcal{M}_{1,0,0,0,1}^{[6]}\geq
0.1082531754\left(m+\tfrac{1}{2}\right)\beta_{n}^{5}e^{2\Lambda_{n}-\left(m+\frac{1}{2}\right)\beta_{n}}.$

Next we compute the error terms. We “optimize” the error
$\mathcal{D}_{\ell,K}^{[3]}$ with respect to the parameter $\varrho$

$\displaystyle\mathcal{D}_{1,4}^{[3]}(x_{0},J,{0.66021})$
$\displaystyle\approx{6.448427833}\cdot 10^{6},$
$\displaystyle\mathcal{D}_{2,4}^{[3]}(x_{0},J,{0.66950})$
$\displaystyle\approx{1.424375158}\cdot 10^{10},$
$\displaystyle\mathcal{D}_{3,4}^{[3]}(x_{0},J,{0.67773})$
$\displaystyle\approx{1.394394143}\cdot 10^{13},$
$\displaystyle\mathcal{D}_{4,4}^{[3]}(x_{0},J,{0.68510})$
$\displaystyle\approx{8.019576913}\cdot 10^{15}.$

Using these bounds, we compute for $\beta_{n}\leq x_{0}$ that the contribution
from $\mathcal{M}_{\ell,k,r,l,\nu}^{[6]}$ for
$(\ell,k,r,l,\nu)\neq(1,0,0,0,1)$ and all the error terms is bounded by

$\mathcal{D}_{a}:=\left({0.02839882501}\beta_{n}^{5}+0.01275733148\left(m+\frac{1}{2}\right)\beta_{n}^{5}+0.01145145086\left(m+\frac{1}{2}\right)^{2}\beta_{n}^{6}\right)e^{2\mathcal{C}_{n}-\left(m+\frac{1}{2}\right)\beta_{n}}.$

As $m\geq 0$ and $(m+\frac{1}{2})\beta_{n}\leq 1$, we conclude that for
$\beta_{n}\leq x_{0}$, and $m+\frac{1}{2}\leq\frac{1}{\beta_{n}}$,

$\frac{\mathcal{D}_{a}}{\mathcal{M}_{1,0,0,0,1}^{[6]}}\leq\frac{2\cdot{0.02839882501}+0.01275733148+0.01145145086}{0.1082531754}\leq{0.7483053690}.$

Now suppose $\frac{1}{\beta_{n}}\leq m+\frac{1}{2}\leq
c{\beta_{n}^{-\frac{3}{2}}}$. Again pick $x_{0}=\frac{1}{750}$. This time we
use $L=5$, $K=4$, $R=4$, and $\Lambda=2$. Recalling the definition of
$f_{j}(u)$ in (3.53), we compute that $f_{2}(u)\geq 0.2469769644$ for $u\geq
1$. It follows that for $m+\frac{1}{2}\geq\beta_{n}^{-1}$,

$\mathcal{M}_{1,0,0,0}^{[5]}\geq
0.02167141828\cdot\beta_{n}^{4}e^{2\mathcal{C}_{n}-\left(m+\frac{1}{2}\right)\beta_{n}}.$

Using the constant $C_{3c}$, we could bound the size of other terms
$\mathcal{M}_{\ell,k,r,l}^{[5]}$ and treat them as error. We compute for
$\beta_{n}\leq x_{0}$ that the contribution from
$\mathcal{M}_{\ell,k,r,l}^{[5]}$ and all the error terms is bounded by

$\mathcal{D}_{b}(c):=\left({0.0004276419495}+0.004001738087c+0.01799670557c^{2}+0.1370655930c^{3}\right.\\\
\left.+0.8767499596c^{4}\right)\beta_{n}^{4}e^{2\mathcal{C}_{n}-\left(m+\frac{1}{2}\right)\beta_{n}}.$

Picking $c=\frac{1}{3}$, we compute that for $\beta_{n}\leq x_{0}$, and
$\frac{1}{\beta_{n}}\leq m+\frac{1}{2}\leq{3^{-1}\beta_{n}^{-\frac{3}{2}}}$,

$\frac{\mathcal{D}_{b}\left(\frac{1}{3}\right)}{\mathcal{M}_{1,0,0,0}^{[5]}}\leq{0.9072671370}.$

We conclude that for $\beta_{n}\leq\frac{1}{750}$ and
$m+\frac{1}{2}\leq{3^{-1}\beta_{n}^{-\frac{3}{2}}}$, (4.1) holds. Using (3.1),
the condition $\beta_{n}\leq\frac{1}{750}$ translates into $n\geq 925276$.
Hence the theorem is established. ∎

##  5\. Partition inequalities and proof of Theorem 1.6

In this section, we prove our main theorem, and thus establish Stanton’s
Conjecture. In the previous section, we prove positivity of rank differences
for small values of $m$. The following theorem treats unimodality of the rank
function if $m$ is large.

######  Theorem 5.1.

Let $n\geq 11523$, and $2\sqrt{n}\leq m\leq n-3$. Then we have

$N(m,n)-N(m+1,n)\geq 0.$ (5.1)

###### Proof.

Suppose that $n\geq 11523$ and $m\geq 2\sqrt{n}$. Define

$\omega_{r}=\omega_{r}(m,n):=n-\tfrac{3r^{2}}{2}-\left(m+\tfrac{3}{2}\right)r.$

Using (2.1) and the generating function for partitions it follows immediately
that

$N(m,n)-N(m+1,n)=\sum_{r=1}^{\infty}(-1)^{r+1}(p(\omega_{r})-2p(\omega_{r}+r)+p(\omega_{r}+2r))=\sum_{r=1}^{\infty}(-1)^{r+1}P(\omega_{r},r),$
(5.2)

where $P(\omega_{r},r)$ is given in (2.6). Note that the sum above is actually
finite.

Now we split the theorem into a number of cases, which we consider
individually.

  1. (1)

Suppose that $n-102\leq m\leq n-3$. Then we have

$N(m,n)-N(m+1,n)=P(\omega_{1},1).$

We have $\omega_{1}\leq 99$ and we check directly that $P(k,1)\geq 0$ for
$0\leq k\leq 99$.

  2. (2)

Suppose that $\frac{n}{2}-2\leq m\leq n-103$. Again we have

$N(m,n)-N(m+1,n)=P(\omega_{1},1).$

In this case we have $\omega_{1}\geq 100$, so Lemma 2.5 applies and we have

$P(\omega_{1},1)\geq\frac{e^{\pi\sqrt{\frac{2}{3}\left(\omega_{1}-\frac{1}{24}\right)}}}{6\left(\omega_{1}+2-\frac{1}{24}\right)^{2}}>0.$

  3. (3)

Suppose that $\frac{n}{2}-54\leq m\leq\frac{n}{2}-\frac{5}{2}$. Then we have

$N(m,n)-N(m+1,n)=P(\omega_{1},1)-P(\omega_{2},2).$

We have $\omega_{1}\geq 100$ and $\omega_{2}\leq 99$ for $n\geq 11523$.
Applying Lemma 2.5 to $P(\omega_{1},1)$, we obtain

$P(\omega_{1},1)\geq\frac{e^{\pi\sqrt{\frac{2}{3}\left(\omega_{1}-\frac{1}{24}\right)}}}{6\left(\omega_{1}+2-\frac{1}{24}\right)^{2}}\geq\frac{e^{\pi\sqrt{\frac{2}{3}\left(\frac{n}{2}-\frac{1}{2}-\frac{1}{24}\right)}}}{6\left(\frac{n}{2}+53-\frac{1}{24}\right)^{2}}.$

Meanwhile, we directly compute that $|P(\omega_{2},2)|\leq 11516573$ for
$\omega_{2}\leq 99$. Using these bounds, we check the claimed inequality (5.1)
for $n\geq 11523$ by a computer verification.

  4. (4)

Suppose that $\frac{n}{3}-\frac{11}{3}\leq m\leq\frac{n}{2}-\frac{109}{2}$.
Again we have

$N(m,n)-N(m+1,n)=P(\omega_{1},1)-P(\omega_{2},2).$

In this case, we have $\omega_{1},\omega_{2}\geq 100$ so Lemma 2.5 applies for
both $P(\omega_{1},1)$ and $P(\omega_{2},2)$. We obtain

$\displaystyle P(\omega_{1},1)$
$\displaystyle\geq\frac{e^{\pi\sqrt{\frac{2}{3}\left(\omega_{1}-\frac{1}{24}\right)}}}{6\left(\omega_{1}+2-\frac{1}{24}\right)^{2}}\geq\frac{e^{\pi\sqrt{\frac{2}{3}\left(\frac{n}{2}+\frac{103}{2}-\frac{1}{24}\right)}}}{6\left(\frac{2n}{3}+\frac{8}{3}-\frac{1}{24}\right)^{2}},$
$\displaystyle P(\omega_{2},2)$
$\displaystyle\leq\frac{\pi^{2}e^{\pi\sqrt{\frac{2}{3}\left(\omega_{2}+4-\frac{1}{24}\right)}}}{6\sqrt{3}\left(\omega_{2}-\frac{1}{24}\right)^{2}}\leq\frac{\pi^{2}e^{\pi\sqrt{\frac{2}{3}\left(\frac{n}{3}+\frac{7}{3}-\frac{1}{24}\right)}}}{6\sqrt{3}\left(100-\frac{1}{24}\right)^{2}}.$

Using these bounds, we numerically check (5.1) for $n\geq 11523$.

We repeat these computations with $m$ in the ranges given below. The strategy
for $\omega_{r}\geq 100$ is as follows. We recall that we aim to give a lower
bound on (5.2), an alternating sum of the numbers $P(\omega_{r},r)$. Due to
the presence of these alternating signs, we either require an upper or lower
bound on $P(\omega_{r},r)$, which depends on the parity of $r$. Lemma 2.5
gives both an upper and lower bound on $P(\omega_{r},r)$, and for each value
of $r$ we use the appropriate one. That is, we use the lower bound for odd
values of $r$ and we use the upper bound there for the even values of $r$. For
the smaller values $\omega_{r}\leq 99$, we compute $\max_{-2r\leq k\leq
99}|P(\omega_{r},r)|$, and treat them as error terms (regardless of the parity
of $r$). Under the assumption that $n\geq 11523$, we check by computer that
the following ranges are all nonempty, and that there is at most one term for
which Lemma 2.5 does not apply. Then we check (5.1) by computer for $n\geq
11523$ for these cases.

  1. (1)

If $\frac{n}{3}-39\leq m\leq\frac{n}{3}-4$, then (5.2) has $3$ terms, and
Lemma 2.5 is not applicable to $P(\omega_{3},3)$.

  2. (2)

If $\frac{n}{4}-\frac{21}{4}\leq m\leq\frac{n}{3}-\frac{118}{3}$, then (5.2)
has $3$ terms, and Lemma 2.5 is always applicable.

  3. (3)

If $\frac{n}{4}-\frac{129}{4}\leq m\leq\frac{n}{4}-\frac{11}{2}$, then (5.2)
has $4$ terms, and Lemma 2.5 is not applicable to $P(\omega_{4},4)$.

  4. (4)

If $\frac{n}{5}-\frac{34}{5}\leq m\leq\frac{n}{4}-\frac{65}{2}$, then (5.2)
has $4$ terms, and Lemma 2.5 is always applicable.

  5. (5)

If $\frac{n}{5}-\frac{144}{5}\leq m\leq\frac{n}{5}-7$, then (5.2) has $5$
terms, and Lemma 2.5 is not applicable to $P(\omega_{5},5)$.

  6. (6)

If $\frac{n}{6}-\frac{25}{3}\leq m\leq\frac{n}{5}-29$, then (5.2) has $5$
terms, and Lemma 2.5 is always applicable.

  7. (7)

If $\frac{n}{6}-27\leq m\leq\frac{n}{6}-\frac{17}{2}$, then (5.2) has $6$
terms, and Lemma 2.5 is not applicable to $P(\omega_{6},6)$.

  8. (8)

If $\frac{n}{7}-\frac{69}{7}\leq m\leq\frac{n}{6}-\frac{163}{6}$, then (5.2)
has $6$ terms, and Lemma 2.5 is always applicable.

  9. (9)

If $\frac{n}{7}-\frac{183}{7}\leq m\leq\frac{n}{7}-10$, then (5.2) has $7$
terms, and Lemma 2.5 is not applicable to $P(\omega_{7},7)$.

  10. (10)

If $\frac{n}{8}-\frac{91}{8}\leq m\leq\frac{n}{7}-\frac{184}{7}$, then (5.2)
has $7$ terms, and Lemma 2.5 is always applicable.

For all the cases above, we verify (5.1) by computer if the said conditions
are satisfied.

It remains to consider the case where $2\sqrt{n}\leq
m\leq\frac{n}{8}-\frac{23}{2}$. We split the sum (5.2) into two parts:

$\sum_{r=1}^{\infty}(-1)^{r+1}P(\omega_{r},r)=\sum_{\begin{subarray}{c}r\geq
1\\\
\omega_{r}\geq\frac{3n}{4}+\frac{1}{24}\end{subarray}}(-1)^{r+1}P(\omega_{r},r)+\sum_{\begin{subarray}{c}r\geq
1\\\
\omega_{r}<\frac{3n}{4}+\frac{1}{24}\end{subarray}}(-1)^{r+1}P(\omega_{r},r).$
(5.3)

Consider the terms (and discarding at most one positive term), we have that

$\displaystyle\sum_{\begin{subarray}{c}r\geq 1\\\
\omega_{r}\geq\frac{3n}{4}+\frac{1}{24}\end{subarray}}(-1)^{r+1}P(\omega_{r},r)$
$\displaystyle\geq\sum_{\begin{subarray}{c}r\geq 1\\\
\omega_{2r}\geq\frac{3n}{4}+\frac{1}{24}\end{subarray}}(P(\omega_{2r-1},2r-1)-P(\omega_{2r},2r))$
(5.4) $\displaystyle\geq\sum_{\begin{subarray}{c}r\geq 1\\\
\omega_{2r}\geq\frac{3n}{4}+\frac{1}{24}\end{subarray}}\left(\frac{(2r-1)^{2}e^{\pi\sqrt{\frac{2}{3}\left(\omega_{2r-1}-\frac{1}{24}\right)}}}{6\left(\omega_{2r-1}+4r-2-\frac{1}{24}\right)^{2}}-\frac{\pi^{2}(2r)^{2}e^{\pi\sqrt{\frac{2}{3}\left(\omega_{2r}+4r-\frac{1}{24}\right)}}}{24\sqrt{3}\left(\omega_{2r}-\frac{1}{24}\right)^{2}}\right),$

applying Lemma 2.5 for the last inequality. Note that
$\omega_{2}=n-2m-9\geq\frac{3n}{4}+4\geq\frac{3n}{4}+\frac{1}{24}$, so this
sum is always nonempty. In order to have each summand being positive, we need

$e^{\pi\sqrt{\frac{2}{3}}\left({\sqrt{\omega_{2r-1}-\frac{1}{24}}-\sqrt{\omega_{2r}+4r-\frac{1}{24}}}\right)}\geq\frac{\pi^{2}}{4\sqrt{3}}\frac{(2r)^{2}}{(2r-1)^{2}}\frac{\left(\omega_{2r-1}+2(2r-1)-\frac{1}{24}\right)^{2}}{\left(\omega_{2r}-\frac{1}{24}\right)^{2}}.$
(5.5)

To see this, note that

$\frac{\left(\omega_{2r-1}+2(2r-1)-\frac{1}{24}\right)^{2}}{\left(\omega_{2r}-\frac{1}{24}\right)^{2}}<\frac{16}{9}.$
(5.6)

Since $\frac{(2r)^{2}}{(2r-1)^{2}}\leq 4$ for $r\geq 1$, we deduce that the
right-hand side of (5.5) is bounded from above by
$\frac{16\pi^{2}}{9\sqrt{3}}$. On the other hand, using the definition of
$\omega_{r}$, we compute

$\sqrt{\omega_{2r-1}-\frac{1}{24}}-\sqrt{\omega_{2r}+4r-\frac{1}{24}}\geq\frac{2r+m}{2\sqrt{n}}.$
(5.7)

Hence, if $m\geq 2\sqrt{n}$, then the left-hand side of (5.5) is bounded from
below by $e^{\pi\sqrt{\frac{2}{3}}}$. Meanwhile, by direct computation we have
that $e^{\pi\sqrt{\frac{2}{3}}}>\frac{16\pi^{2}}{9\sqrt{3}}$. So the
inequality (5.5) holds.

Now we look at the second sum in (5.3). In the following we show that this sum
is bounded by the term $r=1$ on the right-hand side in (5.4). This then
implies the non-negativity of (5.3), and proves the theorem in this case. We
first make some observations.

  1. (1)

If $n\geq 11523$, $m\geq 2\sqrt{n}$ and $\omega_{r}+2r>0$, then we have
$r\leq\sqrt{\frac{2n}{3}}$.

  2. (2)

The partition function $p(n)$ is monotonically increasing.

Using the observations above, we obtain a trivial bound

$\displaystyle\sum_{\begin{subarray}{c}r\geq 1\\\
\omega_{r}<\frac{3n}{4}+\frac{1}{24}\end{subarray}}\left|{P(\omega_{r},r)}\right|$
$\displaystyle\leq\sum_{\begin{subarray}{c}r\geq 1\\\
\omega_{r}<\frac{3n}{4}+\frac{1}{24}\end{subarray}}\left({p(\omega_{r})+{2}p(\omega_{r}+r)+p(\omega_{r}+2r)}\right)$
$\displaystyle\leq\sqrt{\frac{2n}{3}}\cdot 4\cdot
p\left({\left\lfloor\tfrac{3n}{4}+2\sqrt{\tfrac{2n}{3}}+\tfrac{1}{24}\right\rfloor}\right)\leq\frac{\sqrt{2n}\cdot
e^{\pi\sqrt{\frac{2}{3}\left(\frac{3n}{4}+2\sqrt{\frac{2n}{3}}\right)}}}{3\left(\frac{3n}{4}+2\sqrt{\frac{2n}{3}}\right)},$
(5.8)

the final inequality following from (2.5).

Now we show that this error is bounded by the term $r=1$ on the right-hand
side in (5.4). Using (5.6) and (5.7), the first term on the right-hand side of
(5.4) is bounded below by

$\frac{e^{\pi\sqrt{\frac{2}{3}\left(\omega_{1}-\frac{1}{24}\right)}}}{6\left(\omega_{1}+2-\frac{1}{24}\right)^{2}}\left(1-\frac{16\pi^{2}}{9\sqrt{3}}e^{-\pi\sqrt{\frac{2}{3}}}\right).$

Now recall the assumption $2\sqrt{n}\leq m\leq\frac{n}{8}-\frac{23}{2}$. By
minimizing the exponent and maximizing the denominator, we obtain a lower
bound

$\frac{e^{\pi\sqrt{\frac{2}{3}\left(\omega_{1}-\frac{1}{24}\right)}}}{6\left(\omega_{1}+2-\frac{1}{24}\right)^{2}}\left(1-\frac{16\pi^{2}}{9\sqrt{3}}e^{-\pi\sqrt{\frac{2}{3}}}\right)\geq\frac{e^{\pi\sqrt{\frac{2}{3}\left(\frac{7n}{8}+\frac{17}{2}-\frac{1}{24}\right)}}}{6\left(n-2\sqrt{n}-1-\frac{1}{24}\right)^{2}}\left(1-\frac{16\pi^{2}}{9\sqrt{3}}e^{-\pi\sqrt{\frac{2}{3}}}\right).$

By direct computation, we check that for $n\geq 11523$,

$\frac{\sqrt{2n}\cdot
e^{\pi\sqrt{\frac{{\frac{3n}{4}+2\sqrt{\frac{2n}{3}}}}{3}}}}{3\left(\frac{3n}{4}+2\sqrt{\frac{2n}{3}}\right)}\leq\frac{e^{\pi\sqrt{\frac{2}{3}\left(\frac{7n}{8}+\frac{17}{2}-\frac{1}{24}\right)}}}{6\left(n-2\sqrt{n}-1-\frac{1}{24}\right)^{2}}\left(1-\frac{16\pi^{2}}{9\sqrt{3}}e^{-\pi\sqrt{\frac{2}{3}}}\right).$

Hence the second sum in (5.3) is bounded by the first sum in (5.3), which
necessarily exists (see the line after (5.4)) and is positive. This yields the
claim. ∎

We now combine the small and large $m$ bounds to prove Theorem 1.6.

###### Proof of Theorem 1.6.

By Theorem 4.1 the claim holds for $n\geq 925276$ and $0\leq
m\leq{3^{-1}\beta_{n}^{-\frac{3}{2}}}-\frac{1}{2}$. On the other hand, by
Theorem 5.1 for $n\geq 11523$ and $2\sqrt{n}\leq m\leq n-3$. Observing that
$2\sqrt{n}\leq{3^{-1}\beta_{n}^{-\frac{3}{2}}}-\frac{1}{2}$ for $n\geq
925276$, we see that it remains to check the following cases:

  1. (1)

$n\leq 11522$, $0\leq m\leq n-3$,

  2. (2)

$11523\leq n\leq 925275$, $0\leq m<2\sqrt{n}$.

This is a large, but finite number of cases to check. This can be performed as
follows. An efficient way to do this is via the identity (5.2), which computes
the difference $N(m,n)-N(m+1,n)$ in $O(n^{\frac{1}{2}})$ operations from the
partition function $p(n)$, which is widely accessible as a list and easily
computable. Performing this computer check completes the proof of the theorem.
∎

## References

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