# Mock theta functions and related combinatorics

Cristina Ballantine, Hannah Burson, Amanda Folsom,
Chi-Yun Hsu, Isabella Negrini, and Boya Wen  Department of Mathematics and
Computer Science, College of the Holy Cross, Worcester, MA 01610 USA
<EMAIL_ADDRESS>School of Mathematics, University of Minnesota, Twin
Cities, 127 Vincent Hall 206 Church St. SE, Minneapolis, MN 55455, USA
<EMAIL_ADDRESS>Department of Mathematics and Statistics, Amherst College,
Amherst, MA 01002, USA<EMAIL_ADDRESS>Department of Mathematics,
University of California, Los Angeles, Math Sciences Building, 520 Portola
Plaza, Box 951555, Los Angeles, CA 90095, USA<EMAIL_ADDRESS>Mathematics
and Statistics, McGill University, Burnside Hall 805 Sherbrooke Street West,
Montreal, Quebec H3A 0B9, Canada<EMAIL_ADDRESS>Department
of Mathematics, University of Wisconsin-Madison, 480 Lincoln Drive, Madison,
WI 53706, USA<EMAIL_ADDRESS>
###### Abstract.

In this paper we add to the literature on the combinatorial nature of the mock
theta functions, a collection of curious $q$-hypergeometric series introduced
by Ramanujan in his last letter to Hardy in 1920, which we now know to be
important examples of mock modular forms. Our work is inspired by Beck’s
conjecture, now a theorem of Andrews, related to Euler’s identity: the excess
in the number of parts in all partitions of $n$ into odd parts over the number
of partitions of $n$ into distinct parts is equal to the number of partitions
with only one (possibly repeated) even part and all other parts odd. We
establish Beck-type identities associated to partition identities due to
Andrews, Dixit, and Yee for the third order mock theta functions
$\omega(q),\nu(q)$, and $\phi(q)$. Our proofs are both analytic and
combinatorial in nature, and involve mock theta generating functions and
combinatorial bijections.

##  1\. Introduction

### Mock theta functions

In Ramanujan’s last letter to Hardy from 1920, he presented his _mock theta
functions,_ a collection of 17 curious $q$-hypergeometric series including

$\displaystyle\omega(q)$
$\displaystyle:=\sum_{k=0}^{\infty}\frac{q^{2k(k+1)}}{(q;q^{2})^{2}_{k+1}},$
$\displaystyle\nu(q)$
$\displaystyle:=\sum_{k=0}^{\infty}\frac{q^{k(k+1)}}{(-q;q^{2})_{k+1}},$
$\displaystyle\phi(q)$
$\displaystyle:=\sum_{k=0}^{\infty}\frac{q^{k^{2}}}{(-q^{2};q^{2})_{k}},$

of the _third order_. Here and throughout, the $q$-Pochhammer symbol is
defined for $n\in\mathbb{N}_{0}\cup\\{\infty\\}$ by

$\displaystyle(a;q)_{n}:=\prod_{j=0}^{n-1}(1-aq^{j})=(1-a)(1-aq)(1-aq^{2})\cdots(1-aq^{n-1}).$

Ramanujan didn’t define what he meant by the order of a mock theta function,
nor did he precisely define a mock theta function. However, we have since been
able to extract a definition from his own writing [14] (see also the recent
works [23, 27]):

> _“Suppose there is a function in the Eulerian form and suppose that all or
> an infinity of points $q=e^{2i\pi m/n}$ are exponential singularities and
> also suppose that at these points the asymptotic form of the function closes
> neatly…The question is: is the function taken the sum of two functions one
> of which is an ordinary theta function and the other a (trivial) function
> which is $O(1)$ at all the points $e^{2i\pi m/n}$? The answer is it is not
> necessarily so. When it is not so I call the function Mock
> $\theta$-function. I have not proved rigorously that it is not necessarily
> so. But I have constructed a number of examples…”_

Ramanujan’s reference to theta functions, a class of modular forms, and
Eulerian forms, which are $q$-series similar in shape to $\omega(q),\nu(q),$
and $\phi(q)$, expressible in terms of _$q$ -hypergeometric_ series ([19,
22]), indirectly points back to earlier examples of Eulerian modular forms.
For example, Dedekind’s $\eta$-function is an important modular theta function
of weight $1/2$ which can be expressed in terms of a $q$-hypergeometric series
as follows:

(1) $\displaystyle
q^{\frac{1}{24}}\eta^{-1}(\tau)=\sum_{n=0}^{\infty}\frac{q^{n^{2}}}{(q;q)_{n}^{2}},$

where $q=e^{2\pi i\tau}$ is the usual modular variable, with $\tau$ in the
upper half complex plane. Ramanujan’s letter on his mock theta functions
claimed that mock theta functions behave like (weakly holomorphic) modular
forms near roots of unity but are not themselves modular, hence the adjective
_mock_.

The precise roles played by the mock theta functions within the theory of
modular forms remained unclear in the decades following Ramanujan’s death
shortly after he wrote his last letter to Hardy. However, the importance of
these functions was clear – they have been shown to play meaningful roles in
the diverse subjects of combinatorics, $q$-hypergeometric series, mathematical
physics, elliptic curves and traces of singular moduli, Moonshine and
representation theory, and more. Within the last 20 years we have also finally
understood, with thanks to key work by Zwegers, Bruinier-Funke, and others
including Bringmann-Ono and Zagier [16], that the mock theta functions turn
out to be examples of _mock modular forms_ , which are holomorphic parts of
_harmonic Maass forms_ , modern relatives to ordinary Maass forms and modular
forms. This context has also allowed us to make more sense of the notion of
the order of a mock theta function. For more background and information on
these aspects of the mock theta functions, see, e.g., [16, 18, 20, 30].

Turning to the first application of mock theta functions mentioned above,
combinatorics, we recall that Dedekind’s modular $\eta$-function may also be
viewed as the reciprocal of the generating function for integer partitions.
That is, (1) may also be written as

(2)
$\prod_{n=1}^{\infty}\frac{1}{1-q^{n}}=\sum_{n=0}^{\infty}p(n)q^{n}=1+q+2q^{2}+3q^{3}+5q^{4}+7q^{5}+\cdots,$

where $p(n)$ is the number of partitions of $n$. That (2) is simultaneously a
modular form and a combinatorial generating function has led to some deep and
important results and theory. Namely, Hardy–Ramanujan introduced their famous
_Circle Method_ in analytic number theory, which combined with the modularity
of Dedekind’s $\eta$-function, led to the following exact formula for the
partition numbers [26]

$p(n)=2\pi(24n-1)^{-\frac{3}{4}}\sum_{k=1}^{\infty}\frac{A_{k}(n)}{k}I_{\frac{3}{2}}\left(\frac{\pi\sqrt{24n-1}}{6k}\right),$

an infinite sum in terms of Kloosterman sums $A_{k}$ and Bessel functions
$I_{s}$.

Like the modular $\eta$-function, the mock theta functions may also be viewed
as combinatorial generating functions. For example, we have that

$\displaystyle q\omega(q)=\sum_{n=1}^{\infty}a_{\omega}(n)q^{n},\ \ \ \ \ \ \
\nu(-q)=\sum_{n=0}^{\infty}a_{\nu}(n)q^{n},\ \ \ \ \ \ \
\phi(q)=\sum_{n=0}^{\infty}a_{\phi}(n)q^{n},$

where $a_{\omega}(n)$ counts the number of partitions of $n$ whose parts,
except for one instance of the largest part, form pairs of consecutive non-
negative integers [19, (26.84)]; $a_{\nu}(n)$ counts the number of partitions
of $n$ whose even parts are distinct, and if $m$ occurs as a part, then so
does every positive even number less than $m$; and
$a_{\phi}(n):=sc_{e}(n)-sc_{o}(n),$ where $sc_{o/e}(n)$ counts the number of
self-conjugate partitions $\lambda$ of $n$ with $L(\lambda)$ odd/even. Here,
$L(\lambda)$ is the number of parts of $\lambda$ minus the side length of its
Durfee square. (See, e.g., [2] and Section 2 for more background on integer
partitions.)

Using the newer theory of mock modular forms, we have results analogous to the
celebrated Hardy–Ramanujan–Rademacher exact formula for $p(n)$; for example,
due to Garthwaite [21] we have that

$a_{\omega}(n)=\frac{\pi}{2\sqrt{2}}(3n+2)^{-\frac{1}{4}}\mathop{\sum_{k=1}^{\infty}}_{(k,2)=1}\frac{(-1)^{\frac{k-1}{2}}A_{k}(\frac{n(k+1)}{2}-\frac{3(k^{2}-1)}{8})}{k}I_{\frac{1}{2}}\left(\frac{\pi\sqrt{3n+2}}{3k}\right).$

Numerous other papers, some of which we discuss in the sections that follow,
have established further meaningful combinatorial results pertaining to the
mock theta functions, including congruence properties, asymptotic properties,
and more, adding to broader and older theories which rest at the intersection
of combinatorics and modular forms.

### Beck-type partition identities

In this paper we seek to add to the growing literature on understanding the
combinatorial nature of the mock theta functions. Precisely, we study the
number of parts in all partitions interpolated by the third order mock theta
functions $\omega(q),\nu(q),$ and $\phi(q)$. In general, identities on the
number of parts in all partitions of a certain type have been of interest in
the literature, dating back to work of Beck and Andrews. Their work was
motivated by Euler’s famous partition identity, which states that for any
positive integer $n$,

$p(n\ |\text{ odd parts})=p(n\ |\ \text{distinct parts}),$

and which may be immediately deduced from the identity

$\prod_{n=1}^{\infty}\frac{1}{1-q^{2n-1}}=\prod_{n=1}^{\infty}(1+q^{n})$

upon realizing that the “modular” products appearing are generating functions
for the partition functions in Euler’s identity.

While the natural number-of-parts refinement of Euler’s identity is not true,
namely the number of partitions of $n$ into exactly $m$ odd parts is not in
general equinumerous with the number of partitions of $n$ into exactly $m$
distinct parts, Beck conjectured and Andrews proved [3] that the excess in the
number of parts in all partitions of $n$ into odd parts over the number of
parts in all partitions of $n$ into distinct parts is equal to the number of
partitions with only one (possibly repeated) even part and all other parts
odd. Andrews also showed that this excess is also equal to the number of
partitions with only one repeated part and all other parts distinct. Andrews
provided an analytic proof of this theorem using generating functions, and
Yang [28] and Ballantine–Bielak [9] later independently provided combinatorial
proofs.

Since Beck made the first conjecture of this type, combinatorial identities on
the excess between the number of parts in all partitions arising from a
partition identity like Euler’s are now fairly commonly referred to as “Beck-
type identities.” In recent past, a number of other interesting Beck-type
companions to other important identities have been established – see, e.g.,
[5], [10], [11], [24], [28].

Here, we establish Beck-type identities associated to the third order mock
theta functions $\omega(q),\nu(q),$ and $\phi(q)$ in Theorem 3.2, Theorem 4.2,
and Theorem 5.1, respectively. Our results may be viewed as Beck-type
companion identities to partition identities for the third order mock theta
functions $\omega(q),\nu(q),$ and $\phi(q)$ due to Andrews, Dixit and Yee in
[6]. We devote Section 2 to preliminaries on partitions, and state and prove
our main results on $\omega(q),\nu(q),$ and $\phi(q)$ in Section 3, Section 4,
and Section 5, respectively. As a Corollary to our main results, we also
establish mock theta pentagonal-number-theorem-type results in Theorem 4.3 and
Corollary 4.4. Generally speaking, our proofs are both analytic and
combinatorial in nature, and involve mock theta generating functions and
combinatorial bijections.

Throughout, we assume $|q|<1$, so that all series converge absolutely.

##  2\. Preliminaries on partitions

Let $n\in\mathbb{N}_{0}$. A _partition_ of $n$, denoted
$\lambda=(\lambda_{1},\lambda_{2},\ldots,\lambda_{j})$, is a non-increasing
sequence of positive integers
$\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{j}$ called _parts_ that add
up to $n$. We refer to $n$ as the _size_ of $\lambda$. The length of $\lambda$
is the number of parts of $\lambda$, denoted by $\ell(\lambda)$. We denote by
$\ell_{o}(\lambda)$ and $\ell_{e}(\lambda)$ the number of odd, respectively
even parts of $\lambda$. For convenience, we abuse notation and use $\lambda$
to denote either the multiset of its parts or the non-increasing sequence of
parts. We write $a\in\lambda$ to mean the positive integer $a$ is a part of
$\lambda$. As mentioned in the introduction, we denote by $p(n)$ the number of
partitions of $n$. The empty partition is the only partition of size $0$.
Thus, $p(0)=1$. We write $|\lambda|$ for the size of $\lambda$ and
$\lambda\vdash n$ to mean that $\lambda$ is a partition of size $n$. For a
pair of partitions $(\lambda,\mu)$ we also write $(\lambda,\mu)\vdash n$ to
mean $|\lambda|+|\mu|=n$. We use the convention that $\lambda_{k}=0$ for all
$k>\ell(\lambda)$. When convenient we will also use the exponential notation
for parts in a partition: the exponent of a part is the multiplicity of the
part in the partition. This notation will be used mostly for rectangular
partitions. We write $(a^{b})$ for the partition consisting of $b$ parts equal
to $a$. Further, we denote by calligraphy style capital letters the set of
partitions enumerated by the function denoted by the same letter. For example,
we denote by $q_{o}(n)$ the number of partitions of $n$ into distinct odd
parts and by $\mathcal{Q}_{o}(n)$ the set of partitions of $n$ into distinct
odd parts.

The _Ferrers diagram_ of a partition
$\lambda=(\lambda_{1},\lambda_{2},\ldots,\lambda_{j})$ is an array of left
justified boxes such that the $i$th row from the top contains $\lambda_{i}$
boxes. We abuse notation and use $\lambda$ to mean a partition or its Ferrers
diagram. The $2$-_modular Ferrers diagram_ of $\lambda$ is a Ferrers diagram
in which row $i$ has $\lceil\frac{\lambda_{i}}{2}\rceil$ boxes, all but the
first filled with $2$. The first box of row $i$ is filled with $2$,
respectively $1$, if $\lambda_{i}$ is even, respectively odd.

######  Example 1.

The Ferrers diagram and the $2$-modular Ferrers diagram of
$\lambda=(5,4,3,3,2,2)$ are shown in Figure 1.

$\small\ydiagram{5,4,3,3,2,2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \small\ytableau
1&22\\\ 22\\\ 12\\\ 12\\\ 2\\\ 2$ Figure 1. A Ferrers diagram and 2-modular
Ferrers diagram

Given a partition $\lambda$, its _conjugate_ $\lambda^{\prime}$ is the
partition for which the rows in its Ferrers diagram are precisely the columns
in the Ferrers diagram of $\lambda$. For example, the conjugate of
$\lambda=(5,4,3,3,2,2)$ is $\lambda^{\prime}=(6,6,4,2,1)$. A partition is
called _self-conjugate_ if it is equal to its conjugate.

The Durfee square of a partition $\lambda$ is the largest square that fits
inside the Ferrers diagram of $\lambda$, i.e., the partition $(a^{a})$, where
$a$ is such that $\lambda_{a}\geq a$ and $\lambda_{a+1}\leq a$. For example,
the Durfee square of $\lambda=(5,4,3,3,2,2)$ is $(3^{3})=(3,3,3)$.

For more details on partitions, we refer the reader to [2].

An _odd Ferrers diagram_ $F$ is a Ferrers diagram of a partition filled with
$1$ and $2$ such that the first row is filled with $1$ and the remaining rows
form the $2$-modular Ferrers diagram of a partition $\lambda$ with all parts
odd. If the first row has length $k$, we identify the odd Ferrers diagram $F$
with the pair $(k,\lambda)$. The _size_ of an odd Ferrers diagram ${F}$ is the
sum of all entries in the boxes of diagram and is denoted by $|{F}|$. The
length of $F$ is the number of rows in the diagram.

######  Example 2.

Figure 2 shows the odd Ferrers diagram of size $44$ and length $7$
corresponding to the pair $(k,\lambda)$ with $k=8$ and
$\lambda=(11,7,7,5,5,1)$.

1&11111 11
122222
1222
1222
122
122
1

Figure 2. An odd Ferrers diagram

The rank of a partition $\lambda$, denoted $r(\lambda)$, is defined as
$r(\lambda)=\lambda_{1}-\ell(\lambda)$, i.e., the number of columns minus the
number of rows in its Ferrers diagram. In [12], the $M_{2}$-rank of a
partition is defined as the number of columns minus the number of rows in its
$2$-modular diagram. The rank of an odd Ferrers diagram $F=(k,\lambda)$,
denoted $\operatorname{rank}(F)$, is defined as the number of columns minus
the number of rows of $F$, or equivalently,
$\operatorname{rank}(F)=k-\ell(\lambda)-1$.

##  3\. The mock theta function $\omega$

Recall from Section 1 that Ramanujan’s third order mock theta function
$\omega$ is defined by

$\omega(q):=\sum_{k=0}^{\infty}\frac{q^{2k(k+1)}}{(q;q^{2})_{k+1}^{2}}.$

It is known [19, (26.84)] that

$q\omega(q)=A_{\omega}(q):=\sum_{k=1}^{\infty}\frac{q^{k}}{(q;q^{2})_{k}}=\sum_{n=1}^{\infty}a_{\omega}(n)q^{n},$

where $a_{\omega}(n)$ counts the number of partitions of $n$ whose parts,
except for one instance of the largest part, form pairs of consecutive non-
negative integers. We are allowing pairs of consecutive integers to be
$(0,1)$, but we are not considering $0$ as a part of the partition. There is
also the (highly) non-trivial identity by Andrews–Dixit–Yee [6]:

$q\omega(q)=B_{\omega}(q):=\sum_{k=1}^{\infty}\frac{q^{k}}{(q^{k};q)_{k+1}(q^{2k+2};q^{2})_{\infty}}=\sum_{n=1}^{\infty}b_{\omega}(n)q^{n},$

where $b_{\omega}(n)$ counts the number of partitions of $n$ such that all odd
parts are less than twice the smallest part. Hence
$a_{\omega}(n)=b_{\omega}(n)$.

We define two variable generalizations of $A_{\omega}(q)$ and $B_{\omega}(q)$
as follows. Let

(3) $\displaystyle
A_{\omega}(z;q):=\sum_{k=1}^{\infty}\dfrac{zq^{k}}{(1-zq)(z^{2}q^{3};q^{2})_{k-1}}=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}a_{\omega}(m,n)z^{m}q^{n},$

where $a_{\omega}(m,n)$ counts the number of partitions of $n$ with $m$ parts,
which except for one instance of the largest part, form pairs of consecutive
non-negative integers. Let

(4) $\displaystyle
B_{\omega}(z;q):=\sum_{k=1}^{\infty}\frac{zq^{k}}{(zq^{k};q)_{k+1}(zq^{2k+2};q^{2})_{\infty}}=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}b_{\omega}(m,n)z^{m}q^{n},$

where $b_{\omega}(m,n)$ counts the number of partitions of $n$ with $m$ parts,
whose odd parts are less than twice the smallest part. In particular, we have
that $A_{\omega}(1;q)=A_{\omega}(q)$, and $B_{\omega}(1;q)=B_{\omega}(q)$.

Following the notation convention introduced in Section 2,
$\mathcal{A}_{\omega}(n)$ is the set of partitions of $n$ whose parts, except
for one instance of the largest part, form pairs of consecutive non-negative
integers. We denote by $\mathcal{A}_{\omega,2}(n)$ the set of odd Ferrers
diagrams of size $n$, and then $a_{\omega,2}(n)=|\mathcal{A}_{\omega,2}(n)|$.

We next define two generating functions, $A_{\omega,2}$ and
$\widetilde{A}_{\omega,2}$, for odd Ferrers diagrams, which we later show are
related to $A_{\omega}$ and $B_{\omega}$. Namely, we let

(5) $\displaystyle
A_{\omega,2}(z;q):=\sum_{k=1}^{\infty}\frac{zq^{k}}{(zq;q^{2})_{k}}=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}a_{\omega,2}(m,n)z^{m}q^{n},$

where $a_{\omega,2}(m,n)$ counts the number of odd Ferrers diagrams of size
$n$ with $m$ rows. We note that this interpretation was introduced by Andrews
in [4]. We also let

(6)
$\displaystyle\widetilde{A}_{\omega,2}(z;q):=\sum_{k=1}^{\infty}\frac{z^{k}q^{k}}{(q;q^{2})_{k}}=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\widetilde{a}_{\omega,2}(m,n)z^{m}q^{n},$

where $\tilde{a}_{\omega,2}(m,n)$ counts the number of odd Ferrers diagrams of
size $n$ with $m$ columns. The combinatorial interpretation of
$\widetilde{A}_{\omega,2}(z;q)$ was first described by Li–Yang in [25,
(2.22)].

######  Lemma 3.1.

There is an explicit bijection
$\mathcal{A}_{\omega}(n)\xrightarrow{\sim}\mathcal{A}_{\omega,2}(n)$.
Moreover, if $\mu\mapsto{F}_{\mu}$ under this bijection, then the number of
parts of $\mu$ is equal to the number of rows of ${F}_{\mu}$ plus the number
of rows of ${F}_{\mu}$ containing at least a $2$, i.e.,

(7) $\displaystyle
A_{\omega}(z;q)=A_{\omega,2}(z^{2};q)\cdot\frac{1-z^{2}q}{z(1-zq)}.$

###### Proof.

Start with $\mu\in\mathcal{A}_{\omega}(n)$, remove one instance of the largest
part $\mu_{1}$, and merge the (consecutive) pairs of parts of the remaining
partition to obtain a partition $\lambda$ into odd parts. Then the
corresponding odd Ferrers diagram is ${F}_{\mu}=(\mu_{1},\lambda)$. This
transformation is invertible: given $(k,\lambda)\in\mathcal{A}_{\omega,2}(n)$,
each part of $\lambda$ is odd and hence the sum of a pair of consecutive non-
negative integers. The corresponding partition has parts $k$ and all pairs of
parts obtained by splitting the parts of $\lambda$ into consecutive integers.
The connection between the number of parts of $\mu$ and the number of rows of
${F}_{\mu}$ is clear from this explicit bijection. ∎

######  Theorem 3.2.

The excess of the number of parts in all partitions in
$\mathcal{A}_{\omega}(n)$ over the number of parts in all partitions in
$\mathcal{B}_{\omega}(n)$ equals the number of rows containing at least a $2$
in all odd Ferrers diagrams $F=(k,\lambda)$ of size $n$, which is the same as
the number of parts greater than $1$ in $\lambda$.

We will provide four proofs of this theorem.

We first introduce some useful identities. From equation (8) of [7], we have

(8) $B_{\omega}(z;q)=\widetilde{A}_{\omega,2}(z;q).$

Moreover, from equation (16) of [7], we have

(9) $\widetilde{A}_{\omega,2}(z;q)=A_{\omega,2}(z;q).$

(This can also be seen from the fact that conjugation provides a bijection
$\widetilde{\mathcal{A}}_{\omega,2}(m,n)\xrightarrow{\sim}\mathcal{A}_{\omega,2}(m,n)$
[25, p.539].) From (8) and (9), we have that

(10) $\displaystyle B_{\omega}(z;q)=A_{\omega,2}(z;q).$

By Lemma 3.1 and Theorem 3.2, or, after differentiating (10) at $z=1$, we have
the following result.

######  Corollary 3.3.

The total number of parts in all partitions in $\mathcal{B}_{\omega}(n)$
equals the total number of rows in all odd Ferrers diagrams of size $n$.

All four proofs of Theorem 3.2 make use of the fact that

(11)
$\displaystyle\left.\frac{\partial(A_{\omega}(z;q)-B_{\omega}(z;q))}{\partial
z}\right|_{z=1}$

is the generating function for the excess of the number of parts in all
partitions in $\mathcal{A}_{\omega}(n)$ over the number of parts in all
partitions in $\mathcal{B}_{\omega}(n)$.

###### First proof.

We compute the derivative difference (11), using (7), (8), and (9):

$\displaystyle\left.\frac{\partial(A_{\omega}(z;q)-B_{\omega}(z;q))}{\partial
z}\right|_{z=1}$ $\displaystyle=\left.\frac{\partial}{\partial
z}\right|_{z=1}\left(\frac{1-z^{2}q}{z(1-zq)}\cdot\widetilde{A}_{\omega,2}(z^{2};q)-\widetilde{A}_{\omega,2}(z;q)\right)$
$\displaystyle=\left.\frac{\partial\widetilde{A}_{\omega,2}(z;q)}{\partial
z}\right|_{z=1}-\frac{1}{1-q}A_{\omega}(q)$
$\displaystyle=\sum_{k=1}^{\infty}\frac{kq^{k}}{(q;q^{2})_{k}}-\frac{1}{1-q}\sum_{k=1}^{\infty}\frac{q^{k}}{(q;q^{2})_{k}}.$

The second term $\frac{1}{1-q}\sum_{k=1}^{\infty}\frac{q^{k}}{(q;q^{2})_{k}}$
is the generating function for the number of pairs $({F},(1^{b}))\vdash n$,
where ${F}$ is an odd Ferrers diagram and $b\geq 0$ is an integer.

By mapping a pair $({F},(1^{b}))$ to an odd Ferrers diagram with at least $b$
rows of size $1$ and coloring the final $b$ rows of size $1$, we can see that
$\frac{1}{1-q}\sum_{k=1}^{\infty}\frac{q^{k}}{(q;q^{2})_{k}}$ is also the
generating function for the number of odd Ferrers diagrams ${F}=(k,\lambda)$
weighted by $m_{\lambda}(1)+1$, where $m_{\lambda}(1)$ is the number of parts
equal to $1$ in $\lambda$.

Hence $\left.\frac{\partial(A_{\omega}(z;q)-B_{\omega}(z;q))}{\partial
z}\right|_{z=1}$ is the generating function for the number of odd Ferrers
diagrams ${F}=(k,\lambda)$ weighted by $k-(m_{\lambda}({1})+1)$.

Note that conjugation provides a bijection between odd Ferrers diagrams of
size $n$ with $m$ rows and odd Ferrers diagrams of size $n$ with $m$ columns.
Hence for a conjugate pair ${F}=(k,\lambda)$ and ${F^{\prime}}=(j,\mu)$, we
have

$\displaystyle
k-(m_{\lambda}({1})+1)+j-(m_{\mu}({1})+1)=j-(m_{\lambda}({1})+1)+k-(m_{\mu}({1})+1)$
$\displaystyle=$
$\displaystyle(\ell(\lambda)+1)-(m_{\lambda}({1})+1)+(\ell(\mu)+1)-(m_{\mu}({1})+1).$

Therefore, summing over all odd Ferrers diagrams of size $n$, the generating
function stays the same if we replace the weight by
$(\ell(\lambda)+1)-(m_{\lambda}({1})+1)$, which is the number of rows
containing at least a $2$ in ${F}$. ∎

###### Second proof.

We compute the derivative difference (11), using (7) and (10):

$\displaystyle\left.\frac{\partial(A_{\omega}(z;q)-B_{\omega}(z;q))}{\partial
z}\right|_{z=1}$ $\displaystyle=\left.\frac{\partial}{\partial
z}\right|_{z=1}\left(\frac{1-z^{2}q}{z(1-zq)}\cdot
A_{\omega,2}(z^{2};q)-A_{\omega,2}(z;q)\right)$
$\displaystyle=\left.\frac{\partial A_{\omega,2}(z;q)}{\partial
z}\right|_{z=1}-\frac{1}{1-q}A_{\omega}(q).$

We have seen in the first proof that the second term
$\frac{1}{1-q}A_{\omega}(q)$ is the generating function for the number of odd
Ferrers diagrams ${F}=(k,\lambda)$ weighted by $m_{\lambda}({1})+1$. Hence
$\left.\frac{\partial(A_{\omega}(z;q)-B_{\omega}(z;q))}{\partial
z}\right|_{z=1}$ is the generating function for the number of rows containing
at least a $2$ in all odd Ferrers diagrams of size $n$. ∎

###### Third proof.

We compute the derivative difference (11), using (10):

(12)
$\displaystyle\left.\frac{\partial(A_{\omega}(z;q)-B_{\omega}(z;q))}{\partial
z}\right|_{z=1}$ $\displaystyle=\left.\frac{\partial}{\partial
z}\right|_{z=1}\left(A_{\omega}(z;q)-A_{\omega,2}(z;q)\right)$ (13)
$\displaystyle=\sum_{k=1}^{\infty}\frac{q^{k}}{(q;q^{2})_{k}}\left(\sum_{j=1}^{k-1}\frac{q^{2j+1}}{1-q^{2j+1}}\right).$

This is the generating function for the number of pairs
$({F},((2j+1)^{b}))\vdash n$, where ${F}$ is an odd Ferrers diagram and
$j,b\geq 1$ are integers. For each pair $({F},((2j+1)^{b}))\vdash n$, we
insert $b$ copies of $(2j+1)$ as $2$-modular rows into ${F}$ and color the
final $b$ rows of size $(2j+1)$ to obtain a colored odd Ferrers diagram. The
number of such colored odd Ferrers diagrams of size $n$ is equal to the number
of rows containing at least a $2$ in all odd Ferrers diagrams of size $n$. ∎

###### Fourth proof..

From (12), we have that

(14)
$\sum_{m=1}^{\infty}(ma_{\omega}(m,n)-mb_{\omega}(m,n))=\sum_{m=1}^{\infty}(ma_{\omega}(m,n)-ma_{\omega,2}(m,n)),$

for each $n\geq 1$. The left hand side of (13) is the excess in the statement
of Theorem 3.2, whereas the right hand side is the excess of the number of
parts in all partitions in $\mathcal{A}_{\omega}(n)$ over the number of rows
in all odd Ferrers diagrams in $\mathcal{A}_{\omega,2}(n)$. By Lemma 3.1,
$\mathcal{A}_{\omega}(n)\xrightarrow{\sim}\mathcal{A}_{\omega,2}(n)$, and the
excess is precisely the number of rows containing at least a $2$ in all odd
Ferrers diagrams of size $n$. ∎

From the third proof of Theorem 3.2, we obtain new interpretations of the
derivative difference (12) in Corollaries 3.4 and 3.5 below. These are
analogous to the original Beck identity which can be reinterpreted as follows.
The excess in the total number of parts in all partitions of $n$ into distinct
parts over the total number of parts in all partitions of $n$ into odd parts
equals the number of pairs $(\xi,\eta)\vdash n$, where $\xi$ is a partition
into odd parts and $\eta$ is a rectangular partition into equal even parts.
This is also the number of pairs $(\xi,\eta)\vdash n$, where $\xi$ is a
partition into distinct parts and $\eta$ is a rectangular partition with at
least two parts.

######  Corollary 3.4.

The excess in the total number of parts in all partitions in
$\mathcal{A}_{\omega}(n)$ over the total number of parts in all odd Ferrers
diagrams of size $n$ equals the number of pairs $(\xi,\eta)\vdash n$ where,
$\xi$ is an odd Ferrers diagram and $\eta$ is a rectangular partition into odd
parts of size at least $3$.

######  Corollary 3.5.

The excess of the number of parts in all partitions in
$\mathcal{A}_{\omega}(n)$ over the number of parts in all partitions in
$\mathcal{B}_{\omega}(n)$ equals the number of pairs $(\lambda,\eta)\vdash n$,
where $\lambda\in\mathcal{A}_{\omega}$ and $\eta$ is a rectangular partition
into odd parts of size at least $3$.

##  4\. The mock theta function $\nu$

Recall from Section 1 the mock theta function

$\nu(q):=\sum_{k=0}^{\infty}\frac{q^{k(k+1)}}{(-q;q^{2})_{k+1}}.$

We write

$\nu(-q)=A_{\nu}(q)=\sum_{n=0}^{\infty}a_{\nu}(n)q^{n}.$

Since $k(k+1)=2+4+\cdots+2k$, $a_{\nu}(n)$ counts the number of partitions of
$n$ whose even parts are distinct, and if $m$ occurs as a part, then so does
every positive even number less than $m$.

Let

$A_{\nu,2}(q):=\sum_{k=0}^{\infty}(-q;q^{2})_{k}q^{k}=:\sum_{n=0}^{\infty}a_{\nu,2}(n)q^{n}.$

Then it follows from [1, Corollary 1], with $x=s=q$, or from [6, (44)], with
$x=y=q$, that

$\nu(-q)=A_{\nu,2}(q).$

Note that $A_{\nu,2}(q)$ is the generating function for the number of odd
Ferrers diagrams $(k,\lambda)$ where the partition $\lambda$ has distinct
parts.

From [6, Theorem 4.1] we also have

(15)
$\displaystyle\nu(-q)=B_{\nu}(q):=\sum_{k=0}^{\infty}q^{k}(-q^{k+1};q)_{k}(-q^{2k+2};q^{2})_{\infty}=\sum_{n=0}^{\infty}b_{\nu}(n)q^{n},$

where $b_{\nu}(n)$ counts the number of partitions of $n$ into distinct parts,
in which each odd part is less than twice the smallest part, and zero can be a
part (note that this is different from our usual convention). For example,
$(6,4,3,2)$ and $(6,4,3,2,0)$ are counted as different partitions, the former
from the term $q^{k}(-q^{k+1};q)_{k}(-q^{2k+2};q^{2})_{\infty}$ for $k=2$
while the latter from the term for $k=0$.

######  Lemma 4.1.

$\mathcal{A}_{\nu}(n)$ is in (explicit) one-to-one correspondence with
$\mathcal{A}_{\nu,2}(n)$. Moreover, if $\pi\in\mathcal{A}_{\nu}(n)$
corresponds to $(k,\lambda)\in\mathcal{A}_{\nu,2}(n)$, then $\ell(\pi)=k$ and
$\ell(\lambda)$ is the number of even parts in $\pi$.

###### Proof.

We adapt the bijection in [25, Theorem 1.3]. We start with an odd Ferrers
diagram $F=(k,\lambda)\in\mathcal{A}_{\nu,2}(n)$, where $\lambda$ has distinct
parts and length $\ell$. We will associate to $F$ a partition $\pi$ in
$\mathcal{A}_{\nu}(n)$. Consider the subdiagram
$T=(\ell,(2\ell-1,2\ell-1,\ldots,3,1))$ of $F$. We map $T$ to the partition
$\varepsilon=(2\ell,2\ell-2,\ldots,4,2)$. We remove $T$ from $F$ and shift all
remaining boxes to the left to obtain a diagram $R$. The conjugate of $R$ is
the $2$-modular diagram of a partition $\rho$ with odd parts. Define
$\pi:=\varepsilon\cup\rho$.

From the procedure above, we see that $\pi$ has $k$ parts and the number of
parts of $\lambda$ is equal to the number of even parts in $\pi$. ∎

######  Example 3.

Let $F=(k,\lambda)=(5,(9,5,1))$. As in Lemma 4.1, we decompose $F=T+R$ as
shown in Figure 3.

1&1111

12222

122

1 = 1&11

122

12

1 \+ 1& 1

2 2

2

Figure 3. A decomposition of an odd Ferrers diagram as in Lemma 4.1

Then, $\varepsilon=(6,4,2)$, $\rho=(5,3)$ and $\pi=(6,5,4,3,2)$.

As in Section 3, we introduce two variable generalizations of
$A_{\nu}(q),A_{\nu,2}(q),$ and $B_{\nu}(q)$ in which the exponent of $z$ keeps
track of the number of parts in partitions.

Let

(16)
$A_{\nu}(z;q):=\sum_{k=0}^{\infty}\frac{z^{k}q^{k^{2}+k}}{(zq;q^{2})_{k+1}}=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{\nu}(m,n)z^{m}q^{n},$

where $a_{\nu}(m,n)$ counts the number of partitions in $\mathcal{A}_{\nu}(n)$
with $m$ parts.

Let

(17)
$A_{\nu,2}(z;q):=\sum_{k=0}^{\infty}(-zq;q^{2})_{k}zq^{k}=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{\nu,2}(m,n)z^{m}q^{n},$

where $a_{\nu,2}(m,n)$ counts the number of odd Ferrers diagrams $(k,\lambda)$
in $\mathcal{A}_{\nu,2}(n)$ with $m$ rows.

Let

(18)
$B_{\nu}(z;q):=\sum_{k=0}^{\infty}zq^{k}(-zq^{k+1};q)_{k}(-zq^{2k+2};q^{2})_{\infty}=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}b_{\nu}(m,n)z^{m}q^{n},$

where $b_{\nu}(m,n)$ counts the number of partitions in $\mathcal{B}_{\nu}(n)$
with $m$ parts. Recall that partitions in $\mathcal{B}_{\nu}(n)$ can have $0$
as a part, which we do count in the number of parts – for example, the
partition $(6,4,2)$ has three parts, while $(6,4,2,0)$ has four parts.

######  Theorem 4.2.

The excess in the total number of parts in all partitions in
$\mathcal{A}_{\nu}(n)$ over the total number of parts in all partitions in
$\mathcal{B}_{\nu}(n)$ equals the sum of the number of odd parts minus $1$
over all partitions in $\mathcal{A}_{\nu}(n)$, or equivalently the sum of
ranks over all odd Ferrers diagrams in $\mathcal{A}_{\nu,2}(n)$. If $n\geq 1$,
the excess is non-negative.

We provide two proofs of this theorem.

###### Proof 1.

We have that

$\displaystyle\dfrac{\partial}{\partial z}$
$\displaystyle\left.\left(A_{\nu}(z;q)-B_{\nu}(z;q)\right)\right|_{z=1}$
$\displaystyle=\dfrac{\partial}{\partial
z}\left.\left(\sum_{k=0}^{\infty}\dfrac{z^{k}q^{k^{2}+k}}{(zq;q^{2})_{k+1}}-\sum_{k=0}^{\infty}zq^{k}(-zq^{k+1};q)_{k}(-zq^{2k+2};q^{2})_{\infty}\right)\right|_{z=1}$
$\displaystyle=\dfrac{\partial}{\partial
z}\left.\left(\sum_{k=0}^{\infty}\dfrac{z^{k}q^{k^{2}+k}}{(zq;q^{2})_{k+1}}-\sum_{k=0}^{\infty}\frac{z^{k+1}q^{k^{2}+k}}{(q;q^{2})_{k+1}}\right)\right|_{z=1}$
$\displaystyle=\sum_{k=0}^{\infty}\dfrac{q^{k^{2}+k}}{(q;q^{2})_{k+1}}\left(\left(\sum_{j=0}^{k}\frac{q^{2j+1}}{1-q^{2j+1}}\right)-1\right)$
$\displaystyle=:\sum_{n=0}^{\infty}c(n)q^{n},$

where we use [7, (7)] in the second line above. Note that $c(n)$ counts the
number of odd parts in all partitions in $\mathcal{A}_{\nu}(n)$ minus the
number of partitions in $\mathcal{A}_{\nu}(n)$. Rephrased, we have

$\sum_{n=0}^{\infty}c(n)q^{n}=\sum_{n=0}^{\infty}\sum_{\pi\in\mathcal{A}_{\nu}(n)}(\ell_{o}(\pi)-1)q^{n}.$

To show that $c(n)\geq 0$ for $n\geq 1$, notice that the only partition $\pi$
with $\ell_{o}(\pi)-1<0$ is $\pi=(2m,2m-2,\ldots 2)$. For this $\pi$, we can
split the largest part $2m$ into two odd parts $(2m-1)+1$ to obtain
$\widetilde{\pi}=(2m-1,2m-2,\ldots,2,1)$, another partition of the same size
in $\mathcal{A}_{\nu}$, such that
$(\ell_{o}(\pi)-1)+(\ell_{o}(\widetilde{\pi})-1)=-1+1=0$. All the other
partitions in $\mathcal{A}_{\nu}$ have at least one odd part. Therefore
$c(n)=\sum_{\pi\in\mathcal{A}_{\nu}(n)}(\ell_{o}(\pi)-1)\geq 0$. ∎

###### Proof 2.

From Lemma 4.1, if the partition $\pi\in\mathcal{A}_{\nu}(n)$ corresponds to
the odd Ferrers diagram $F=(k,\lambda)\in\mathcal{A}_{\nu,2}(n)$, then the
number of parts of $\pi$ is $k$. In terms of generating functions, we have the
identity

$A_{\nu}(z;q)=\sum_{k=0}^{\infty}\frac{z^{k}q^{k^{2}+k}}{(zq;q^{2})_{k+1}}=\sum_{k=0}^{\infty}(-q;q^{2})_{k}z^{k}q^{k}.$

Thus, $\left.\frac{\partial}{\partial z}\right|_{z=1}A_{\nu}(z;q)$ is the
generating function for the total number of columns in all odd Ferrers
diagrams in $\mathcal{A}_{\nu,2}(n)$.

On the other hand, from [7, Theorem 1] we have

$B_{\nu}(z;q)=\sum_{k=0}^{\infty}\frac{z^{k+1}q^{k^{2}+k}}{(q;q^{2})_{k+1}}.$

Again from Lemma 4.1, if the partition $\pi\in\mathcal{A}_{\nu}(n)$
corresponds to the odd Ferrers diagram
$F=(k,\lambda)\in\mathcal{A}_{\nu,2}(n)$, then the number of even parts in
$\pi$ is equal to $\ell(\lambda)$. In terms of generating functions, we have

$\sum_{k=0}^{\infty}\frac{z^{k+1}q^{k^{2}+k}}{(q;q^{2})_{k+1}}=\sum_{k=0}^{\infty}(-zq;q^{2})zq^{k}=A_{\nu,2}(z;q).$

(This is also [7, Theorem 2].) Hence $B_{\nu}(z;q)=A_{\nu,2}(z;q)$ and so
$\left.\frac{\partial}{\partial z}\right|_{z=1}B_{\nu}(z;q)$ is the generating
function for the total number of rows in all odd Ferrers diagrams in
$\mathcal{A}_{\nu,2}(n)$.

Combining these, we conclude that $\left.\frac{\partial}{\partial
z}\right|_{z=1}(A_{\nu}(z;q)-B_{\nu}(z;q))$ is the generating function for the
sum of ranks of all odd Ferrers diagrams in $\mathcal{A}_{\nu,2}(n)$.

Given an odd Ferrers diagram $F=(m,\lambda)\in\mathcal{A}_{\nu,2}(n)$, we have
$m\geq\ell(\lambda)$ since the parts of $\lambda$ are distinct. Then,
$\operatorname{rank}(F)=m-\ell(\lambda)-1\geq-1$ and
$\operatorname{rank}(F)=-1$ if and only if $m=\ell(\lambda)$, in which case
$F=(m,(2m-1,2m-3,\ldots,1))$. Hence, there is at most one odd Ferrers diagram
with rank $-1$ in $\mathcal{A}_{\nu,2}(n)$. If $\operatorname{rank}(F)=-1$,
the conjugate of $F$ is
$F^{\prime}=(m+1,(2m-1,2m-3,\ldots,3))\in\mathcal{A}_{\nu,2}(n)$,
$\operatorname{rank}(F^{\prime})=(m+1)-m=1$, and thus
$\operatorname{rank}(F)+\operatorname{rank}(F^{\prime})=0$. Since all other
odd Ferrers diagrams in $\mathcal{A}_{\nu,2}(n)$ have non-negative rank, it
follows that $c(n)\geq 0$.∎

We end this section by investigating the parity of $c(n)$. To this end, we
first prove a result similar to Euler’s Pentagonal Number Theorem.

######  Theorem 4.3.

For any non-negative integer $n$ we have

$|A_{\nu}(n\mid\mbox{even \\# of parts })|=|A_{\nu}(n\mid\mbox{odd \\# of
parts })|+e(n),$

where

$e(n)=\begin{cases}1&\mbox{ if }n=3j^{2}+2j\mbox{ for some }j\geq 0\\\
-1&\mbox{ if }n=3j^{2}+4j+1\mbox{ for some }j\geq 0\\\ 0&\mbox{
otherwise.}\end{cases}$

###### Proof.

We write a partition $\pi\in A_{\nu}(n)$ as $\pi=(\pi^{e},\pi^{o})$, where
$\pi^{e}$, respectively $\pi^{o}$, is the partition consisting of the even,
respectively odd parts of $\pi$. As usual, the largest part of $\pi^{e}$ is
$\pi_{1}^{e}$. We denote by $\pi^{o}_{s}$ the smallest part of $\pi^{o}$ and
by $m^{o}(\pi)$ the multiplicity of $\pi^{e}_{1}+1$ in $\pi^{o}$. We have
$m^{o}(\pi)\geq 0$.

Let $\tilde{A}_{\nu}(n)=\\{\pi\in
A_{\nu}(n)\mid\pi^{o}=(\pi^{e}_{1}+1)^{m^{o}(\pi)},\mbox{ with
}m^{o}(\pi)\in\\{\frac{\pi^{e}_{1}}{2},\frac{\pi^{e}_{1}}{2}+1\\}\\}$. Then,
$|\tilde{A}_{\nu}(n)|=0$ or $1$.

We define an involution on $A_{\nu}(n)\setminus\tilde{A}_{\nu}(n)$ as follows.

(i) If $\pi^{o}_{s}\geq 2m^{o}(\pi)+1$, remove $\pi^{e}_{1}$ from $\pi^{e}$
and the last two columns (of length $m^{o}(\pi)$) from $\pi^{o}$, and add
parts $\pi^{e}_{1}-1$ and $2m^{o}(\pi)+1$ to $\pi^{o}$.

(ii) If $\pi^{o}_{s}<2m^{o}(\pi)+1$, remove from $\pi^{o}$ one part equal to
$\pi^{e}_{1}+1$ (largest part) and one part equal to $\pi^{o}_{s}$, and add a
part equal to $\pi^{e}_{1}+2$ to $\pi^{e}$ and two columns equal to
$\frac{\pi^{o}_{s}-1}{2}$.

Note that the transformations in (i) and (ii) are inverses of each other.

We have $|\tilde{A}_{\nu}(n)|=1$ if and only if $n=3j^{2}+2j$ or
$3j^{2}+4j+1$. Moreover, for $\pi\in\tilde{A}_{\nu}(n)$,
$j=\ell(\pi^{e})=\ell_{e}(\pi)$ which completely determines the parity of
$\ell(\pi)$.

∎

######  Corollary 4.4.

Let $n\in\mathbb{N}$. Then $c(n)$ is odd if and only if $n$ is eight times a
generalized pentagonal number.

###### Proof.

We have that

$c(n)=\sum_{\pi\in A_{\nu}(n)}(\ell_{o}(\pi)-1).$

With the notation in the proof of Theorem 4.3, we have
$\ell_{o}(\pi)=\ell(\pi^{o})\equiv n\pmod{2}$ because the number of parts in a
partition with odd parts has the same parity as its size. Therefore, if $n$ is
odd, $\ell_{o}(\pi)-1$ is even for every $\pi\in A_{\nu}(n)$ and $c(n)$ is
even.

If $n$ is even, $c(n)\equiv|A_{\nu}(n)|\pmod{2}$. From Theorem 4.3, it follows
that $|A_{\nu}(n)|\equiv 1\pmod{2}$ if and only if $n=3j^{2}+2j$ or
$3j^{2}+4j+1$ for some $j\geq 0$. Since $n$ is even, if $n=3j^{2}+2j$, then
$j$ must be even, and if $n=3j^{2}+4j+1$, then $j$ must be odd. Therefore,
$|A_{\nu}(n)|\equiv 1\pmod{2}$ if and only if $n$ is eight times a generalized
pentagonal number. ∎

######  Remark 1.

Theorem 4.3 also follows from setting $a=1$ in Entry 3.7 of [15]. Using Lemma
4.1, Theorem 4.3 can be adapted to a pentagonal number theorem for
$A_{\nu,2}(n)$. Then, [25, Section 3.2] leads to a combinatorial proof of [6,
Theorem 5.4].

##  5\. The mock theta function $\phi$

Recall from Section 1 that the third order mock theta function $\phi$ is
defined by

(19)
$\displaystyle\phi(q):=\sum_{n=0}^{\infty}\frac{q^{n^{2}}}{(-q^{2},q^{2})_{n}}=\sum_{n=0}^{\infty}(sc_{e}(n)-sc_{o}(n))q^{n},$

where $sc_{o/e}(n)$ counts the number of self-conjugate partitions $\lambda$
of $n$ with $L(\lambda)$ odd/even. Here, $L(\lambda)$ is the number of parts
of $\lambda$ minus the side length of its Durfee square. From [6, Proof of
Theorem 4.2], we have

$\phi(q)=1+\sum_{n=0}^{\infty}(-1)^{n}q^{2n+1}(q;q^{2})_{n}.$

We first define the following generalization of $\phi(q)$:

$\displaystyle B_{\phi}(z;q)$
$\displaystyle:=1+\sum_{n=0}^{\infty}z^{n}q^{2n+1}(q;q^{2})_{n}=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}b_{\phi}(m,n)z^{m}q^{n},$

where $b_{\phi}(0,0):=1$, and for $(m,n)\neq(0,0)$, $b_{\phi}(m,n)$ equals the
difference between the number of partitions of $n$ into distinct, odd parts
with largest part $2m+1$ and an odd number of parts, and the number of such
partitions with an even number of parts. Note that $B_{\phi}(-1,q)=\phi(q),$
and that this gives rise to a different combinatorial interpretation for the
coefficients of $\phi$ than the one given in (19). Namely, the coefficient of
$q^{n}$ in the $q$-series expansion for $\phi(q)$ also equals
$do_{e}(n)-do_{o}(n)$, where $do_{e}(n)$, resp. $do_{o}(n)$, counts the number
of partitions of $n$ into distinct odd parts with $M_{2}$-rank even,
respectively odd.

Next we define another bivariate function, which we later explain is related
to $\phi$ when $z=1$ (see (21)):

$\displaystyle A_{\phi}(z;q)$
$\displaystyle:=q\sum_{n=0}^{\infty}zq^{n}(-zq^{n+1};q)_{n}(-zq^{2n+1};q^{2})_{\infty}=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}a_{\phi}(m,n)z^{m}q^{n+1},$

where $a_{\phi}(m,n)$ is the number of partitions of $n$ into distinct parts,
with $m$ parts, such that each even part is at most twice the smallest part.
The function $A_{\phi}(z;q)$ is related to $\phi(q)$ by the following
identity:

(21) $\displaystyle
A_{\phi}(1;q)=1-\phi(q)+2(-q;q^{2})_{\infty}\sum_{n=1}^{\infty}q^{n^{2}}.$

Using the Jacobi triple product [2, (2.2.10) with $z=1$], identity (21) is
essentially [6, Theorem 4.2] with some minor typographical errors corrected.
Unlike the mock theta functions $\omega(q)$ and $\nu(-q)$ studied in Sections
3 and 4, the $q$-series coefficients of $\phi(q)$ are not uniformly non-
negative, e.g.,

$\phi(q)=1+q-q^{3}+q^{4}+q^{5}-q^{6}-q^{7}+2q^{9}-2q^{11}+q^{12}+q^{13}-q^{14}-2q^{15}+q^{16}+O(q^{17}).$

However, the authors of [6] present (21) for $\phi(q)$ as a companion identity
to their similar result [6, Theorem 4.1] (see (15)) which shows that the mock
theta function $\nu(-q)$ is equal to the generating function for partitions
into distinct parts, in which each odd part is less than twice the smallest
part. Identity (21) similarly relates the mock theta function $\phi(q)$ to the
generating function for partitions into distinct parts in which each even part
is at most twice the smallest part, but up to a theta function. Indeed it is
identity (21) that leads to our “Beck-type” Theorem 5.1 for the mock theta
function $\phi(q)$ below. To state it, we introduce the functions

(22) $\displaystyle F_{1}(q)$
$\displaystyle:=F_{3}(q)\Big{(}1+2\sum_{n=1}^{\infty}q^{n^{2}}\Big{)}+2(-q;q^{2})_{\infty}\sum_{n=1}^{\infty}q^{n^{2}},$
(23) $\displaystyle F_{2}(q)$
$\displaystyle:=(-q;q^{2})_{\infty}\sum_{m=1}^{\infty}\frac{q^{2m-1}}{1+q^{2m-1}},\
$ (24) $\displaystyle F_{3}(q)$
$\displaystyle:=(-q;q^{2})_{\infty}\sum_{m=1}^{\infty}\frac{q^{2m}}{1+q^{2m}}.$

The functions $F_{2}(q)$ and $F_{3}(q)$, including their combinatorial
interpretations, are studied in [10].

######  Theorem 5.1.

We have that

(25) $\displaystyle\frac{\partial}{\partial
z}\Big{|}_{z=1}(A_{\phi}(z;q)+B_{\phi}(-z^{-1};q))=F_{1}(q)-F_{2}(q).$

Moreover, we have that

$\frac{\partial}{\partial
z}\Big{|}_{z=1}(A_{\phi}(z;q)+B_{\phi}(-z^{-1};q))\succeq 0.$

######  Remark 2.

A combinatorial interpretation of (25) in Theorem 5.1 can be deduced from the
combinatorial definitions of $a_{\phi}(m,n)$ and $b_{\phi}(m,n)$ provided
above, together with combinatorial interpretations of the $q$-series
coefficients of in the functions $F_{2}(q)$ and $F_{3}(q)$ provided in [10],
and the definition of $F_{1}(q)$. While this combinatorial intepretation
involves partition differences, Theorem 5.1 establishes the non-negativity of
the $q$-series coefficients of (25). On the other hand, it is of interest to
find another proof of this fact by finding a different and _manifestly_
positive combinatorial interpretation of the $q$-series coefficients of
$F_{1}(q)-F_{2}(q)$ (i.e., one which does not involve a combinatorial
difference). We leave this as an open problem.

###  5.1. Proof of Theorem 5.1

In this section, we prove Theorem 5.1, assuming the truth of Proposition 5.2
and Proposition 5.4 stated below. We provide a combinatorial proof of
Proposition 5.2 in Section 5.2, and provide both combinatorial and analytic
proofs of Proposition 5.4 in Section 5.3. In what follows, we use the notation
$G(q)\succeq_{S}0$, where $S\subseteq\mathbb{N}$, to mean that when expanded
as a $q$-series, the coefficients of $G(q)$ are non-negative, with the
exception of the coefficients of $q^{n}$ for $n\in S$.

######  Proposition 5.2.

We have that

$2F_{3}(q)-F_{2}(q)\succeq_{S}0,$

where $S:=\\{1,4,8,16\\}.$ Moreover, the coefficients of $2F_{3}(q)-F_{2}(q)$
are at least 4, with the exception of the coefficients of $q^{n}$ with $n$ in
the set $U:=\\{1,2,3,4,5,8,9,12,13,16,17\\}$.

######  Corollary 5.3.

We have that

(26)
$\displaystyle(2F_{3}(q)-F_{2}(q))\sum_{n=1}^{\infty}q^{n^{2}}\succeq_{T}0,$

where $T:=\\{2,5,9,13,17\\}$.

###### Proof of Corollary 5.3.

Writing $2F_{3}(q)-F_{2}(q)=:\sum_{n=1}^{\infty}a_{n}q^{n},$ we have that the
coefficient of $q^{n}$ in $(2F_{3}(q)-F_{2}(q))\sum_{j=1}^{\infty}q^{j^{2}}$
is equal to

(27) $\displaystyle\mathop{\sum_{k=1}^{n-1}a_{k}}_{k+m^{2}=n,\ 1\leq m^{2}\leq
n-1}.$

By Proposition 5.2, we have that $a_{k}\geq 0$ for any $k\not\in
S=\\{1,4,8,16\\}$, and $a_{k}\geq 4$ for any $k\not\in U$.

Let $n\notin T$ be a positive integer. We can directly compute the $q$-series
in (26) to $O(q^{65})$ to see the coefficients are non-negative. Now assume
that $n>65$. To prove that (27) is non-negative, it suffices to show that for
any $k\in S$ with $k+m^{2}=n$, there is another $k^{\prime}\neq k$ with
$k^{\prime}+m^{\prime 2}=n$ such that $a_{k^{\prime}}+a_{k}\geq 0$. Because
$n>65$, we must have $m\geq 2$. Let $k^{\prime}:=k+2m-1$. Then $1\leq
k^{\prime}\leq n-1$, $k^{\prime}>k$, and $k^{\prime}+(m-1)^{2}=n$, where
$1\leq(m-1)^{2}\leq n-1$. Note that $k^{\prime}\not\in U$. This is because for
$k\in S$ and $k^{\prime}\in U$, we have
$n=k+m^{2}=k+(\frac{k^{\prime}-k+1}{2})^{2}\leq 65$. By direct calculation, we
find that $\min\\{a_{k}\\}_{k\in S}=-4,$ and hence $a_{k}+a_{k^{\prime}}\geq
0$. ∎

######  Proposition 5.4.

Let $F_{2}(q)=:\sum_{n=1}^{\infty}b_{n}q^{n}$. For $n\geq 9$, we have that

$b_{n}\leq b_{n-1}+b_{n-4}.$

######  Corollary 5.5.

We have that

(28) $\displaystyle
F_{2}(q)\sum_{n=1}^{\infty}q^{n^{2}}-F_{2}(q)\succeq_{V}0,$

where $V:=\\{1,3,4,6,8\\}$.

###### Proof of Corollary 5.5.

To prove (28), we show

(29) $\displaystyle\sum_{1\leq m^{2}\leq n-1}b_{n-m^{2}}\geq b_{n}$

for $n\in\mathbb{N}\setminus V$. By Proposition 5.4, and the non-negativity of
$b_{n}$, for $n\geq 9$, we have

$\sum_{1\leq m^{2}\leq n-1}b_{n-m^{2}}\geq b_{n-1}+b_{n-4}\geq b_{n}.$

For $n\in\\{2,5,7\\}$, the inequality (29) can be verified directly. ∎

####  5.1.1. Proof of Theorem 5.1

First, by straightforward manipulations, we find that [25, (2.4)] leads to the
identity

(30) $\displaystyle
B_{\phi}(-z^{-1};q)=1+\sum_{n=0}^{\infty}\frac{z^{-n}q^{(n+1)^{2}}}{(-z^{-1}q^{2};q^{2})_{n+1}}.$

Next we multiply [13, Theorem 6.11, $z\mapsto zq$] by $zq$ and use (30) to
find that

${A}_{\phi}(z;q)+{B}_{\phi}(-z^{-1};q)=D_{\phi}(z;q),$

where

$\displaystyle D_{\phi}(z;q)$
$\displaystyle:=1+z(-zq;q^{2})_{\infty}\left(-1+\frac{(-q;q)_{\infty}(q^{2};q^{2})_{\infty}(-q/z;q^{2})_{\infty}}{(-q^{2}/z;q^{2})_{\infty}}\right)$
$\displaystyle=1-z(-zq;q^{2})_{\infty}+z\frac{(-q;q)_{\infty}}{(-q^{2}/z;q^{2})_{\infty}}\left(1+\sum_{n=1}^{\infty}(z^{n}+z^{-n})q^{n^{2}}\right).$

Above, we have also used the Jacobi triple product [2, (2.2.10)]. Thus, the
derivative difference on the left hand side of (25) equals
$\frac{\partial}{\partial z}\big{|}_{z=1}{D}_{\phi}(z;q).$ After a direct
calculation using the definition of $D_{\phi}(z;q)$, and some simplification,
we obtain that this equals $F_{1}(q)-F_{2}(q)$.

To prove the second assertion of the theorem, it now suffices to show that
$F_{1}(q)-F_{2}(q)\succeq 0.$ From [10], we have that

$F_{3}(q)\succeq 0.$

From this and (22), it is not difficult to see that

(31) $\displaystyle F_{1}(q)-2F_{3}(q)\sum_{n=1}^{\infty}q^{n^{2}}\succeq 0.$

Thus, we have from (26), (28), and (31) that $F_{1}(q)-F_{2}(q)\succeq_{W}0$
for some explicit, finite, set $W$. The proof is complete after a direct
calculation of the $q$-series for $F_{1}(q)-F_{2}(q)$ up to $O(q^{n_{W}})$,
where $n_{W}:=\max W$, which reveals that in fact $F_{1}(q)-F_{2}(q)\succeq 0$
as claimed.

###  5.2. Proof of Proposition 5.2

###### Proof of Proposition 5.2.

We prove that for $n\geq 45$, the coefficient of $q^{n}$ in
$2F_{3}(q)-F_{2}(q)$ is at least $4$. For $n<45$ and $n\not\in U$, it can be
verified directly that the coefficient of $q^{n}$ is at least $4$. From [10],
$F_{2}(q)$ is the generating function for $|A(n)|$, where

$A(n):=\\{(\lambda,(a))\vdash n\big{|}\lambda\in\mathcal{Q}_{o},a\mbox{
odd},a\not\in\lambda\\}.$

From [10, Theorem 1.10] with $r=1$, $\ell=2$, $F_{3}(q)$ is the generating
function for $|B(n)|-\varepsilon(n)$, where

$\varepsilon(n):=\begin{cases}1&\text{if }n\equiv 0\pmod{4},\\\
0&\text{else,}\end{cases}$

and

$B(n):=\\{(\lambda,(c^{d}))\vdash n\big{|}\lambda\in\mathcal{Q}_{o},c\text{
even, }d\text{ odd, }\lambda_{1}-\lambda_{2}\leq c,\text{ and
}\lambda\neq\mu(c)\\}.$

Here, $\mu(c)$ is defined for even $c\geq 4$ to be the partition

$\mu(c)=\begin{cases}(\frac{c}{2}+1,\frac{c}{2}-1)&\mbox{ if }c\equiv
0\pmod{4}\\\ (\frac{c}{2}+2,\frac{c}{2}-2)&\mbox{ if }c\equiv
2\pmod{4}.\end{cases}$

Thus, $\mu(c)$ is a partition of $c$ into two distinct odd parts with minimum
difference between the parts.

For the remainder of the proof, let $n\geq 45$. We will show combinatorially
that $2(|B(n)|-1)\geq|A(n)|+4$, i.e., $2|B(n)|\geq|A(n)|+6$. Creating a
mapping from $A(n)$ to $B(n)$ that would achieve this turns out to be very
intricate. Instead, we define a mapping $\psi$ on $A(n)$ so that at most two
elements of $A(n)$ have equal image. The image of $\psi$ is a subset of
$U(n):=\\{(\lambda,(c^{d}))\vdash n\mid\lambda\text{ has odd parts, }c\text{
even, }d\text{ odd},\lambda_{1}-\lambda_{2}\leq c\\}$. After defining $\psi$,
we determine its image and the multiplicity of elements in the image. Then we
determine the elements in the image of $\psi$ that are not in $B(n)$ as well
as a subset of elements of $B(n)$ that are not in the image of $\psi$. We use
these sets to complete the proof.

We write $\psi(A(n))$ for the image of $A(n)$ under $\psi$ as a _multiset_ ,
each element apearing with multiplicity equal to the size of its pre-image
under $\psi$. Thus each element in $\psi(A(n))$ has multiplicity $1$ or $2$
and $|A(n)|=|\psi(A(n))|$. We write $B^{\prime}(n)$ for the multiset whose
elements are precisely those of $B(n)$ each appearing with multiplicity $2$.
Our goal is to show that $|B^{\prime}(n)|\geq|\psi(A(n))|+6$.

To define $\psi$, we start with
$(\lambda,(a))=((\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell(\lambda)}),(a))\in
A(n)$, and perform two steps.
Step 1: Define

$(\eta,(c))=\begin{cases}((\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell(\lambda)-1}),(a+1))&\mbox{
if }1\in\lambda,\\\
((\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell(\lambda)},1),(a-1))&\mbox{ if
}1\not\in\lambda,a\neq 1,\\\
((\lambda_{2},\ldots,\lambda_{\ell(\lambda)}),(\lambda_{1}+1))&\mbox{ if
}1\not\in\lambda,a=1.\end{cases}$

We obtain pairs $(\eta,(c))$, with $c$ even and $\eta\in\mathcal{Q}_{o}$
satisfying one of the following conditions:

$\begin{array}[]{l}1\not\in\eta,{c\geq 4,}\text{ and }c-1\not\in\eta;\\\
1\in\eta\text{ and }c+1\not\in\eta.\end{array}$

Moreover, pairs such that $1\not\in\eta$ and $\eta_{1}\leq c-3$ occur twice.

If $\eta=\emptyset$ (which occurs only if $n$ is even), define
$\psi(\lambda,(a))=(\eta,(c))=(\emptyset,(n))\in B(n)$. Thus, if $n$ is even,
the pair $(\emptyset,(n))$ appears in $\psi(A(n))$ with multiplicity $2$ since
it is the image of $((1),(n-1))$ and $((n-1),(1))$.

If $\eta\neq\emptyset$, continue with Step 2.
Step 2: Write $\eta_{1}-\eta_{2}=qc+r$ with $q,r\in\mathbb{Z}$, $q\geq 0$,
$0<r\leq c$.
We define $\eta-(qc):=(\eta_{1}-qc,\eta_{2},\ldots,\eta_{\ell(\eta)})$. Note
that $\eta-(qc)\neq\emptyset$.

If $q$ is even, define

$\psi(\lambda,(a))=(\xi,(c^{d}))=(\eta-(qc),(c^{q+1})).$

If $q$ is odd, define

$\psi(\lambda,(a))=(\xi,(c^{d}))=\begin{cases}((\eta-(qc))\cup(c-1)\cup(1),(c^{q}))&\mbox{
if }1\not\in\eta,\\\ ((\eta-(qc))\setminus\\{1\\}\cup(c+1),(c^{q}))&\mbox{ if
}1\in\eta.\end{cases}$

All pairs obtained satisfy $\xi_{1}-\xi_{2}\leq c$.

Next, we determine the image $\psi(A(n))$ of $\psi$.

From Step 1, if $n$ is even, $(\emptyset,(n))\in\psi(A(n))$.

From Step 2 above, following pairs $(\xi,(c^{d}))\in U(n)$ are in
$\psi(A(n))$.

From the case $q$ even: $(\xi,(c^{d}))\in U(n)$, $\xi\in\mathcal{Q}_{o}$, and
satisfying one of

$\begin{array}[]{l}1\not\in\xi,{c\geq 4,}\text{ and }c-1\not\in\xi;\\\
1\not\in\xi,{c\geq 4,}\ \xi_{1}=c-1\text{ and }d>1;\\\ 1\in\xi\text{ and
}c+1\not\in\xi,\text{ and }(\xi,(c^{d}))\neq((1),(2^{d})),d>1;\\\ 1\in\xi,\
\xi_{1}=c+1\text{ and }d>1.\\\ \end{array}$

(Note that if $c+1$ and $c-1$ are both in $\xi$, then they are the two largest
parts and we must have $1\in\xi$ and $d>1$.)

From the case $q$ odd: $(\xi,(c^{d}))\in U(n)$, and satisfying one of

$\begin{array}[]{l}1\in\xi,{c\geq 4,}\text{ and }c-1\in\xi\text{ and
}\ell(\xi)\geq 3;\\\ 1\not\in\xi\text{ and }c+1\in\xi\text{ and }\ell(\xi)\geq
2.\end{array}$

Moreover, $\xi$ has at most one repeated part and this can happen as follows:
$\xi_{1}=\xi_{2}=c-1{\geq 3}$, $\xi_{3}<\xi_{2}$ and $1\in\xi$;
$\xi_{1}=\xi_{2}=c+1$, $\xi_{3}<\xi_{2}$ and $1\not\in\xi$; or $\xi=(c-1,1,1)$
and $c\geq 4$.

Each of the following pairs $(\xi,(c^{d}))\in\psi(A(n))$ occurs with
multiplicity $2$.

$\begin{array}[]{l}\xi=\emptyset\text{ and }d=1;\\\ 1\not\in\xi,\ \xi_{1}\leq
c-3,\ c\geq 4,\text{ and }d=1;\\\ 1\in\xi,{c\geq 4,}\ \xi_{1}=c+1,\
\xi_{2}=c-1,\text{ and }d>1;\\\ 1\in\xi,c\geq 4,\ c-1\in\xi,\text{ and
}c+1\not\in\xi\text{ and }\ell(\xi)\geq 3;\\\ 1\not\in\xi,c\geq 4,\
c+1\in\xi,\text{ and }c-1\not\in\xi\text{ and }\ell(\xi)\geq 2.\end{array}$

The pairs $(\xi,(c^{d}))\in B(n)$ satisfying the conditions below are not in
$\psi(A(n))$.

$\begin{array}[]{l}1\in\xi,\ c+1\in\xi,\ c-1\not\in\xi,\ \xi_{1}>c+1;\\\ c=2,\
1\in\xi,\ 3\in\xi,\ \xi_{1}>3;\\\ 1\not\in\xi,{c\geq 4,}\ c-1\in\xi,\
c+1\not\in\xi,\ \xi_{1}>c+1.\end{array}$

The list above is not exhaustive but it is sufficient for our purposes.

Pairs $(\xi,(c^{d}))$ that are in $\psi(A(n))\setminus B^{\prime}(n)$ satisfy
one of the following conditions:

  * (i)

$\xi=\mu(c)$;

  * (ii)

$\xi_{1}=\xi_{2}=c-1$ and $1\in\xi$ and $c\geq 4$; or $\xi_{1}=\xi_{2}=c+1$
and $1\not\in\xi$;

  * (iii)

$\xi=(c-1,1,1)$, $c\geq 4$;

Next, we map most pairs in $\psi(A(n))\setminus B^{\prime}(n)$ to pairs in
$B^{\prime}(n)\setminus\psi(A(n))$, i.e. to pairs in $B(n)$ whose pre-image
under $\psi$ has size at most one. Specifically, to simplify the argument, a
small number of pairs in $\psi(A(n))\setminus B^{\prime}(n)$ will not be
mapped below and will be collected in a multiset $\mathcal{E}(n)$.

By checking the list of conditions above, one can see that of the pairs listed
in (i)-(iii), only pairs of the form $(\mu(c),c)$ with $c\geq 8$ appear with
multiplicity $2$. The pre-image of this pair under $\psi$ is
$\\{((\mu(c),1),(c-1)),((c-1,\mu(c)),(1))\\}$. For each $n\equiv 0\pmod{4}$,
$n\geq 16$ there is only one such pair $(\mu(c),c)=(\mu(n/2),(n/2))$ and we
place two copies of $(\mu(n/2),(n/2))$ in $\mathcal{E}(n)$.

We describe this correspondence according to cases (i)-(iii) above.

(i) This case occurs only if $n\equiv 0\pmod{4}$. We consider pairs
$(\mu(c),(c^{d}))\neq(\mu(n/2),(n/2))$ and map them to $(\mu(3c),(c^{d-2}))\in
B^{\prime}(n)\setminus\psi(A(n))$. Note that $d>1$.

(ii) If $\xi_{1}=\xi_{2}=c\pm 1$, we consider two cases.

In the first case, if $\xi_{3}<\xi_{2}-2$, and
$(\xi,(c^{d}))\neq((3,3),(2^{d}))$, we map $(\xi,(c^{d}))$ to
$(\xi\setminus\\{\xi_{1},\xi_{2}\\}\cup\mu(2\xi_{1}),(c^{d}))$. Note that
$\mu(2\xi_{1})=\mu(2c\pm 2)$. If $n\equiv 0\pmod{4}$, we place
$((3,3),(2^{(n-6)/2}))$ in $\mathcal{E}(n)$.

In the second case, if $\xi_{1}=\xi_{2}=c\pm 1$, $\xi_{3}=\xi_{2}-2$, and
$(\xi,(c^{d}))\neq((3,3,1),(4^{d}))$, we map

$(\xi,(c^{d}))=((c-1,c-1,c-3,\ldots,\xi_{\ell(\xi)-1},1),(c^{d}))$

to

$(\xi\setminus\\{\xi_{1},\xi_{2},\xi_{3},1\\}\cup\mu(3c-4),(c^{d}));$

and map

$(\xi,(c^{d}))=((c+1,c+1,c-1,\ldots\xi_{\ell(\xi)}),(c^{d}))$

to

$(\xi\setminus\\{\xi_{1},\xi_{2},\xi_{3}\\}\cup\mu(3c)\cup\\{1\\},(c^{d})).$

Note that in this case $\xi_{\ell(\xi)}>1$ and $c\geq 4$. If $(n-7)/2$ is an
odd positive integer, we place $((3,3,1),(4^{(n-7)/4}))$ in $\mathcal{E}(n)$.

(iii) If $d>3$, we map $((c-1,1,1),(c^{d}))$ to
$(\mu(5c)\cup\\{1\\},(c^{d-4}))$. If $n\equiv 1\pmod{4}$ is odd, we place
$(((n-3)/2,1,1),((n-1)/2))$ in $\mathcal{E}(n)$. If $n\equiv 1\pmod{8}$, we
also place $(((n-5)/4,1,1),(((n-1)/4)^{3}))$ in $\mathcal{E}(n)$.

Denote by $\mathcal{S}(n)$ the image of the correspondence above.

Thus, if $n$ is even,

$\displaystyle\mathcal{E}(n)\subseteq\\{(\mu(n/2),(n/2)),$
$\displaystyle(\mu(n/2),(n/2)),((3,3),(2^{(n-6)/2}))\\},$

and if $n$ is odd,

$\displaystyle\mathcal{E}(n)\subseteq\\{((3,3,1),(4^{(n-7)/4})),$
$\displaystyle(((n-3)/2,1,1),((n-1)/2)),$
$\displaystyle(((n-5)/4,1,1),(((n-1)/4)^{3}))\\}.$

If $n\geq 45$ the following pairs in $B(n)$ do not occur in
$\psi(A(n))\cup\mathcal{S}(n)$ and appear with multiplicity $2$ in
$B^{\prime}(n)$.

If $n$ is even, for $k=2,3,4$,

$\begin{array}[]{l}((\mu(n-4k-2),2k+1,1),(2k)),\\\
((\mu(n-6k-2),2k+3,2k-1),(2k)).\end{array}$

If $n\equiv 3\pmod{4}$, for $1\leq k\leq 5$,

$\begin{array}[]{l}((\mu(n-4k-7),5,3,1),(2^{2k-1})).\end{array}$

If $n\equiv 1\pmod{4}$, for $1\leq k\leq 5$,

$\begin{array}[]{l}((\mu(n-4k-9),7,3,1),(2^{2k-1})).\end{array}$

Thus, we found a submultiset of $B^{\prime}(n)$ disjoint from
$\psi(A(n))\cup\mathcal{S}(n)$ and of size larger than $|\mathcal{E}(n)|+6$
and hence we have shown that $|B^{\prime}(n)|\geq|\psi(A(n)|+6$. ∎

###  5.3. Proof of Proposition 5.4

We provide two different proofs of Proposition 5.4 below, one of which is
combinatorial in nature (see Section 5.3.1), the other of which is analytic
(see Section 5.3.2).

####  5.3.1. Combinatorial Proof of Proposition 5.4

As shown in [10], if $F_{2}(q)=\sum_{n=1}^{\infty}b_{n}q^{n}$, then $b_{n}$
equals the number of parts in all partitions of $n$ with distinct odd parts,
i.e.

$b_{n}=\sum_{\lambda\in Q_{o}(n)}\ell(\lambda).$

We first create a length preserving bijection $\varphi$ from
$\\{\lambda\in\mathcal{Q}_{o}(n)\mid 1\not\in\lambda\\}$ to
$\\{\xi\in\mathcal{Q}_{o}(n-4)\mid\mbox{if }\ell(\xi)\geq 3,\mbox{ then
}\xi_{\ell(\xi)-2}-\xi_{\ell(\xi)-1}>2\\}$.

Let
$\lambda=(\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell(\lambda)})\in\mathcal{Q}_{o}(n)$,
$1\not\in\lambda$ and define

$\varphi(\lambda)=\begin{cases}(n-4)&\mbox{ if }\ell(\lambda)=1,\mbox{ i.e.,
}\lambda=(n)\\\
(\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell(\lambda)-1}-2,\lambda_{\ell(\lambda)}-2)&\mbox{
if }\ell(\lambda)\geq 2.\end{cases}$

Clearly, $\ell(\lambda)=\ell(\varphi(\lambda))$.

Next, we consider the bijection $\psi$ from
$\\{\lambda\in\mathcal{Q}_{o}(n)\mid 1\in\lambda\\}$ to
$\\{\xi\in\mathcal{Q}_{o}(n-1)\mid 1\not\in\lambda\\}$ given by
$\psi(\lambda)=\lambda\setminus\\{1\\}$. Clearly,
$\ell(\lambda)=\ell(\psi(\lambda))+1$.

It remains to show that the number of parts equal to $1$ in all partitions in
$\mathcal{Q}_{o}(n)$ is less than or equal to the total number of parts in all
partitions in $\mathcal{Y}:=\\{\xi\in\mathcal{Q}_{o}(n-1)\mid
1\in\lambda\\}\cup\\{\xi\in\mathcal{Q}_{o}(n-4)\mid\ell(\xi)\geq 3\mbox{ and
}\xi_{\ell(\xi)-2}-\xi_{\ell(\xi)-1}=2\\}$. We create an injection $\xi$ from
$\\{\lambda\in\mathcal{Q}_{o}(n)\mid 1\in\lambda\\}$ to the set of partitions
in $\mathcal{Y}$ with exactly one marked part.

Let $\lambda\in\mathcal{Q}_{o}(n)$ be such that $1\in\lambda$. Since $n\geq
9$, we have $\ell(\lambda)\geq 2$. Let $a:=\lambda_{\ell(\lambda)-1}$.

If $a\geq 9$, $1\not\in\mu(a-1)$, and we define
$\xi(\lambda)=\lambda\setminus\\{a\\}\cup\mu(a-1)\in\mathcal{Q}_{o}(n-1)$ with
part $1$ marked. Note $\xi(\lambda)$ has at least three parts and if it has
exactly three parts, then the difference between the first and second part is
$2$ or $4$. The marked part of $\xi(\lambda)$ is the last part.

Next we consider the case when $a=3,5,$ or $7$. Since $n\geq 9$, we have
$\ell(\lambda)\geq 3$.

If $\ell(\lambda)\geq 4$, define
$\xi(\lambda)=\lambda\setminus\\{\lambda_{1},a\\}\cup\\{\lambda_{1}+a-1\\}\in\mathcal{Q}_{o}(n-1)$
with marked first, second, or third part according to $a=3,5$, or $7$,
respectively. Note that if $a=7$, the difference between the first and second
part in $\xi(\lambda)$ is at least $8$. Thus, when $\ell(\lambda)=4$ and
$a=7$, then $\xi(\lambda)$ has exactly three parts and the marked part is $1$
but the obtained marked partition is different from the marked partitions
obtained in the case $a\geq 9$.

If $\ell(\lambda)=3$, then $n$ is odd. We define
$\xi(n-8,7,1)=(\overline{n-2},1)$, $\xi(n-6,5,1)=(n-2,\overline{1})$ and

$\xi(n-6,5,1)=\begin{cases}(\mu(n-5),\overline{1})&\mbox{ if }n\geq 17,n\equiv
1\pmod{4}\\\ (\mu(n-7),\overline{3})&\mbox{ if }n\geq 17,n\equiv 3\pmod{4}\\\
(\overline{n-2},1)&\mbox{ if }9\leq n\leq 15.\end{cases}$

Note that $(n-8,7,1)$ occurs only when $n\geq 17$, and that
$(\mu(n-5),\overline{1}),(\mu(n-7),\overline{3})\in\mathcal{Q}_{o}(n-4)$.

####  5.3.2. Analytic Proof of Proposition 5.4

To prove Proposition 5.4, it suffices to show that

(32) $\displaystyle(q^{4}+q-1)F_{2}(q)\succeq_{S^{\prime}}0,$

where $S^{\prime}:=\\{0,1,2,3,4,5,6,7,8\\}$. Indeed, we will prove the
stronger result with $S^{\prime}=\\{1,3,4,6,8\\}$. Towards (32), we establish
Lemma 5.6 below, which is stated in terms of the polynomials

$\displaystyle f_{m}(q):=\begin{cases}-q+q^{2}-q^{4}+2q^{5}-q^{6},&m=1,\\\
-q^{3},&m=2,\\\ -q^{5}+q^{7}-q^{8}+q^{9}+2q^{10}+q^{13}-q^{15},&m=3,\\\
-q^{2m-1}+q^{2m+1}-q^{2m+2}+q^{2m+3}+q^{2m+4},&m\geq 4.\end{cases}$

######  Lemma 5.6.

For each integer $m\geq 1$, we have that

(33)
$\displaystyle(q^{4}+q-1)q^{2m-1}\mathop{\prod_{\ell=1}^{\infty}}_{\ell\neq
m}(1+q^{2\ell-1})=f_{m}(q)+g_{m}(q),$

where $g_{m}(q)\succeq 0,$ and

$g_{m}(q)=\begin{cases}O(q^{7}),&m=1,\\\ O(q^{5}),&m=2,\\\ O(q^{16}),&m=3,\\\
O(q^{2m+6}),&m\geq 4.\end{cases}$

Using Lemma 5.6, we give an analytic proof of Proposition 5.4 below. Following
its proof, the remainder of this section is devoted to proving Lemma 5.6.

###### Analytic proof of Proposition 5.4.

By Lemma 5.6, we have that

(34)
$\displaystyle(q^{4}+q-1)F_{2}(q)=\sum_{m=1}^{\infty}(f_{m}(q)+g_{m}(q)).$

We rewrite $\sum_{m=1}^{\infty}f_{m}(q)$ as

$\displaystyle\sum_{j=1}^{3}f_{j}(q)+\sum_{m\geq 4}q^{2m+3}-\sum_{m\geq
4}(q^{2m-1}+q^{2m+2})+\sum_{m\geq 4}(q^{2m+1}+q^{2m+4})$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{3}f_{j}(q)+\sum_{m\geq 4}q^{2m+3}-\sum_{m\geq
4}(q^{2m-1}+q^{2m+2})+\sum_{m\geq 5}(q^{2m-1}+q^{2m+2})$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{3}f_{j}(q)+\sum_{m\geq 4}q^{2m+3}-q^{7}-q^{10}$
$\displaystyle=$
$\displaystyle-q+q^{2}-q^{3}-q^{4}+q^{5}-q^{6}-q^{8}+q^{9}+q^{10}+q^{11}+2q^{13}+\sum_{m\geq
7}q^{2m+3}.$

Since $\sum_{m=1}^{\infty}g_{m}(q)\succeq 0$, we obtain the non-negativity of
coefficients stated (32). ∎

###### Proof of Lemma 5.6.

We divide the proof into cases, depending on $m$.

Throughout the proof, we make use of the following calculations. Let $a,i,j$
be positive integers. We express a product $\displaystyle\prod_{k\geq
i}(1+q^{2k-1})$ in terms of the smallest, respectively largest exponent
appearing in monomials as

(35) $1+\sum_{k\geq
i}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})=1+q^{2i-1}+\sum_{k\geq
i+1}q^{2k-1}\prod_{i\leq\ell<k}(1+q^{2\ell-1}).$

Then,

(36) $\displaystyle q^{a+2j}\sum_{k\geq
i}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})-q^{a}\sum_{k\geq
i+j}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})$ $\displaystyle=q^{a+2j}\sum_{k\geq
i}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})-q^{a+2j}\sum_{k\geq
i+j}q^{2(k-j)-1}\prod_{\ell>k}(1+q^{2\ell-1})$
$\displaystyle=q^{a+2j}\sum_{k\geq
i}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})-q^{a+2j}\sum_{k\geq
i}q^{2k-1}\prod_{\ell>k+j}(1+q^{2\ell-1})\succeq 0.$

Similarly

(37) $\displaystyle q^{a}\sum_{k\geq
i}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})-q^{a+2}\sum_{k\geq
i}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})$
$\displaystyle=q^{a+2i-1}+q^{a+2i+1}(1+q^{2i-1})+q^{a}\sum_{k\geq
i+2}q^{2k-1}\prod_{i\leq\ell<k}(1+q^{2\ell-1})$ $\displaystyle\ \ \
-q^{a+2+2i-1}-q^{a+2}\sum_{k\geq
i+1}q^{2k-1}\prod_{i\leq\ell<k}(1+q^{2\ell-1})\succeq 0.$

The non-negativity of coefficients follows from the fact that

$q^{a+2}\sum_{k\geq
i+1}q^{2k-1}\prod_{i\leq\ell<k}(1+q^{2\ell-1})=q^{a}\sum_{k\geq
i+2}q^{2k-1}\prod_{i\leq\ell<k-1}(1+q^{2\ell-1}).$

We continue with the proof of Lemma 5.6.

Case $m\geq 4.$ We rewrite the left hand side of (33) as

$P_{m}(q)Q_{m}(q),$

where

$\displaystyle P_{m}(q)$
$\displaystyle:=(q^{4}+q-1)q^{2m-1}(1+q)(1+q^{3})(1+q^{5})$ $\displaystyle\
=-q^{2m-1}+q^{2m+1}-q^{2m+2}+q^{2m+3}+q^{2m+4}+2q^{2m+6}+q^{2m+8}$
$\displaystyle\hskip 14.45377pt+2q^{2m+9}+q^{2m+11}+q^{2m+12},$ $\displaystyle
Q_{m}(q)$ $\displaystyle:=\mathop{\prod_{\ell=4}^{\infty}}_{\ell\neq
m}(1+q^{2\ell-1})=1+\mathop{\sum_{k\geq 4}}_{k\neq
m}q^{2k-1}\mathop{\prod_{\ell>k}}_{\ell\neq m}(1+q^{2\ell-1}).$

Then, $g_{m}(q)=$

(38) $\displaystyle=-(q^{2m-1}+q^{2m+2})\mathop{\sum_{k\geq 4}}_{k\neq
m}q^{2k-1}\mathop{\prod_{\ell>k}}_{\ell\neq m}(1+q^{2\ell-1})$ (39)
$\displaystyle+(q^{2m+1}+q^{2m+3}+q^{2m+4})\mathop{\sum_{k\geq 4}}_{k\neq
m}q^{2k-1}\mathop{\prod_{\ell>k}}_{\ell\neq m}(1+q^{2\ell-1})$ (40)
$\displaystyle+(2q^{2m+6}+q^{2m+8}+2q^{2m+9}+q^{2m+11}+q^{2m+12})Q_{m}(q).$

Thus, $g_{m}(q)=O(q^{2m+6})$. To show that $g_{m}(q)\succeq 0$, we show that
all terms in (38) appear with positive sign in (39) or (40).

We first consider the case $m=4$. Then (38) equals

$\displaystyle-(q^{16}+q^{19})\prod_{\ell>5}(1+q^{2\ell-1})-(q^{7}+q^{10})\sum_{k\geq
6}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})=:-C_{1}-C_{2}.$

All terms in $C_{1}$ appear in (40). Using (36) with $j=1$, $i=5$, and
$a=7,10$ respectively, terms in $C_{2}$ cancel with terms in (39). Thus,
$g_{4}(q)\succeq 0$.

For $m>4$, we rewrite (38) separating the terms according to $k=4$, $k\neq
4,m+1$, and $k=m+1$. When $k\neq 4,m+1$, we factor out $q^{2}$ and shift the
index of summation. Thus, (38) equals

$\displaystyle-(q^{2m+6}+q^{2m+9})\mathop{\prod_{\ell>4}}_{\ell\neq
m}(1+q^{2\ell-1})$ $\displaystyle-(q^{2m+1}+q^{2m+4})\mathop{\sum_{k\geq
4}}_{k\neq m-1,m}q^{2k-1}\mathop{\prod_{\ell>k+1}}_{\ell\neq
m}(1+q^{2\ell-1})$
$\displaystyle-(q^{4m}\mathop{\prod_{\ell>m+1}}(1+q^{2\ell-1})+q^{4m+3}\mathop{\prod_{\ell>m+1}}(1+q^{2\ell-1})).$

Writing $q^{4m}=q^{2m+3}\cdot q^{2m-3}$ and $q^{4m+3}=q^{2m+6}\cdot q^{2m-3}$,
we see that each term in (38) cancels with a corresponding positive term in
(39) or (40) (and terms in (39) and (40) are used at most once in this
cancellation). Hence, $g_{m}(q)\succeq 0$.
Case $m=1.$ Using (35), we rewrite the left hand side of (33) as

$\displaystyle(q^{4}+q-1)$ $\displaystyle q\prod_{\ell\geq 2}(1+q^{2\ell-1})$
$\displaystyle\hskip 7.22743pt=q^{5}+q^{2}-q+(q^{5}+q^{2}-q)\sum_{k\geq
2}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})$ $\displaystyle\hskip
7.22743pt=-q+q^{2}-q^{4}+2q^{5}-q^{6}+g_{1}(q),$

where

$\displaystyle g_{1}(q)=$ $\displaystyle q^{5}\sum_{k\geq
2}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})+q^{5}\sum_{k\geq
3}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})$ $\displaystyle+$ $\displaystyle
q^{2}\sum_{k\geq 3}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})-q^{4}\sum_{k\geq
3}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})$ $\displaystyle-$ $\displaystyle
q^{6}\sum_{k\geq 4}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})-q\sum_{k\geq
4}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})$ $\displaystyle=:$
$\displaystyle\,A_{1}+A_{2}+A_{3}-A_{4}-A_{5}-A_{6}.$

From this expression, it is clear that $g_{1}(q)=O(q^{7})$. To show that
$g_{1}(q)\succeq 0$, we first compute $A_{1}-A_{6}$. This equals

$\displaystyle q^{5}\sum_{k\geq
2}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})-q^{5}\sum_{k\geq
2}q^{2k-1}\prod_{\ell>k+2}(1+q^{2\ell-1})$ $\displaystyle=$
$\displaystyle\sum_{k\geq
2}(q^{4k+5}+q^{4k+7}+q^{6k+8})\prod_{\ell>k+2}(1+q^{2\ell-1})=:B_{1}+B_{2}+B_{3}.$

Next, separating terms by $k$ even and odd respectively, we rewrite $A_{5}$ as

$\sum_{j\geq 2}q^{4j+5}\prod_{\ell>2j}(1+q^{2\ell-1})+\sum_{j\geq
3}q^{4j+7}\prod_{\ell>2j+1}(1+q^{2\ell-1}).$

Since $k+2\leq 2k$ if $k\geq 2$, we have $B_{1}+B_{2}-A_{5}\succeq 0$. From
(37) with $a=2,i=3$, it follows that $A_{3}-A_{4}\succeq 0$. Hence,
$g_{1}(q)\succeq 0$.
Case $m=2.$ Using (35), we rewrite the left hand side of (33) as

$(q^{4}+q-1)q^{3}(1+q)\prod_{\ell\geq 3}(1+q^{2\ell-1})=-q^{3}+g_{2}(q),$

where

$g_{2}(q)=(q^{5}+q^{7}+q^{8})\prod_{\ell\geq
3}(1+q^{2\ell-1})-q^{3}\sum_{k\geq
3}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})=O(q^{5}).$

To show $g_{2}(q)\succeq 0$, we write

$\displaystyle q^{3}\sum_{k\geq
3}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})=q^{8}\prod_{\ell>3}(1+q^{2\ell-1})+q^{3}\sum_{k\geq
4}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})$

Using (35) and (36) with $a=3,j=1,i=3$, it follows that $g_{2}(q)\succeq 0$.
Case $m=3.$ Using (35), we rewrite the left hand side of (33) as

$\displaystyle(q^{4}+q-1)$ $\displaystyle q^{5}(1+q)(1+q^{3})\prod_{\ell\geq
4}(1+q^{2\ell-1})$
$\displaystyle=-q^{5}+q^{7}-q^{8}+q^{9}+2q^{10}+q^{13}-q^{15}+g_{3}(q),$

where

$\displaystyle g_{3}(q)=$ $\displaystyle q^{12}+q^{15}$
$\displaystyle+(-q^{5}+q^{7}-q^{8}+q^{9}+2q^{10}+q^{12}+q^{13})\sum_{k\geq
4}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})$ $\displaystyle=$
$\displaystyle(q^{9}+2q^{10}+q^{12}+q^{13})\sum_{k\geq
4}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})$
$\displaystyle+(q^{7}+q^{14})\sum_{k\geq
5}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})$
$\displaystyle-(q^{8}+q^{15}+q^{12})\sum_{k\geq
5}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})$
$\displaystyle-(q^{5}+q^{14})\sum_{k\geq
6}q^{2k-1}\prod_{\ell>k}(1+q^{2\ell-1})=O(q^{16}).$

Using (36) with $a=8,j=1,i=4$ and also with $a=5,j=1,i=5$, as well as (37)
with $a=13,i=5$, we obtain $g_{3}(q)\succeq 0$. ∎

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## Acknowledgements

The authors thank the Banff International Research Station (BIRS) and the
Women in Numbers 5 (WIN5) Program. The third author is partially supported by
National Science Foundation Grant DMS-1901791. The fifth author is partially
supported by a FRQNT scholarship by Fonds de Recherche du Québec, and an ISM
scholarship by Institut des Sciences Mathématiques.