#  $4$-Regular partitions and the pod function

Cristina Ballantine
Department of Mathematics and Computer Science
College of The Holy Cross
Worcester, MA 01610, USA
<EMAIL_ADDRESS>Mircea Merca
Department of Mathematics
University of Craiova
200585 Craiova, Romania
<EMAIL_ADDRESS>
###### Abstract

The partition function $pod(n)$ enumerates the partitions of $n$ wherein odd
parts are distinct and even parts are unrestricted. Recently, a number of
properties for $pod(n)$ have been established. In this paper, for
$k\in\\{0,2\\}$ we consider the partitions of $n$ into distinct parts not
congruent to $k$ modulo $4$ and the $4$-regular partitions of $n$ in order to
obtain new properties for $pod(n)$. In this context, we derive two new
infinite families of linear inequalities involving the function $pod(n)$ and
obtain new identities of Watson type.

Keywords: partitions, theta series, theta products

MSC 2010: 11P81, 11P82, 05A19, 05A20

##  1 Introduction

A partition $\lambda$ of $n$ is a nonincreasing sequence
$(\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell})$ of positive integers (called
parts) such that $\sum_{i=1}^{\ell}\lambda_{i}=n$. We refer to $\lambda_{i}$
as the $i$th part of $\lambda$, and as usual denote by $p(n)$ the number of
integer partitions of $n$. We note that $p(x)=0$ if
$x\not\in\mathbb{Z}_{\geqslant 0}$, and since the empty partition $\emptyset$
is the only partition of $0$, we have that $p(0)=1$. The generating function
for $p(n)$ satisfies the identity

$\displaystyle\sum_{n=0}^{\infty}p(n)\,q^{n}=\frac{1}{(q;q)_{\infty}},$ (1)

where we use the customary $q$-series notation

$\displaystyle(a;q)_{n}=\begin{cases}1,&\text{for $n=0$,}\\\
(1-a)(1-aq)\cdots(1-aq^{n-1}),&\text{for $n>0$;}\end{cases}$
$\displaystyle(a;q)_{\infty}=\lim_{n\to\infty}(a;q)_{n}.$

Moreover, we use the short notation

$(a_{1},a_{2},\ldots,a_{n};q)_{\infty}=(a_{1};q)_{\infty}(a_{2};q)_{\infty}\cdots(a_{n};q)_{\infty}.$

Because the infinite product $(a;q)_{\infty}$ diverges when $a\neq 0$ and
$|q|\geqslant 1$, whenever $(a;q)_{\infty}$ appears in a formula, we shall
assume $|q|<1$.

We denote by $pod(n)$ the function which enumerates the partitions of $n$ with
odd parts distinct and even parts unrestricted. Elementary techniques in the
theory of partitions give the following equivalent expressions for the
generating function for $pod(n)$:

$\sum_{n=0}^{\infty}pod(n)\,q^{n}=\frac{(-q;q^{2})}{(q^{2};q^{2})_{\infty}}=\frac{(q^{2};q^{4})}{(q;q)_{\infty}}.$
(2)

The $pod(n)$ function has been studied widely. It appears, for example, in the
works of K. Alladi [1, 2], G. E. Andrews [3, 4], G. E. Andrews and M. Merca
[8], C. Ballantine, M. Merca, D. Passary, A. J. Yee [11], A. Berkovich and F.
Garvan [12], S.-P. Cui, W. X. Gu, Z. S. Ma [18], H. Fang, F. Xue, O. X. M. Yao
[20], M. D. Hirschhorn and J. A. Sellers [29], S. Radu and J. A. Sellers [48].
In this article, we consider the interpretation of the $pod(n)$ function given
by the last expression in (2) as the number of partitions of $n$ into parts
not congruent to $2$ modulo $4$.

######  Definition 1.

Let $n$ be a nonnegative integer. We define $pod_{e}(n)$ (respectively
$pod_{o}(n)$) to be the number of partitions of $n$ into an even (respectively
odd) number of parts which are not congruent to $2$ modulo $4$.

For example, since the partitions of $8$ into parts not congruent to $2$
modulo $4$ are

$\displaystyle(8),\ (7,1),\ (5,3),\ (5,1,1,1),\ (4,4),\ (4,3,1),\
(4,1,1,1,1),$ $\displaystyle(3,3,1,1),\ (3,1,1,1,1,1),\ (1,1,1,1,1,1,1,1),$

we have that $pod_{e}(8)=7$ and $pod_{o}(8)=3$.

For an integer $\ell>1$, a partition is called $\ell$-regular if none of its
parts is divisible by $\ell$. In classical representation theory,
$\ell$-regular partitions of $n$ parameterize the irreducible $\ell$-modular
representations of the symmetric group $\mathcal{S}_{n}$ when $\ell$ is prime
[30]. The arithmetic properties of the number $b_{\ell}(n)$ of $\ell$-regular
partitions of $n$ have been investigated extensively (see, for example, [13,
16, 19, 22, 28, 35, 45, 46, 55, 56, 60]). The generating function for
$b_{\ell}(n)$ satisfies the identity

$\sum_{n=0}^{\infty}b_{\ell}(n)\,q^{n}=\frac{(q^{\ell};q^{\ell})_{\infty}}{(q;q)_{\infty}}.$

In this article, we consider the $4$-regular partitions of $n$ and provide
connections to the partitions of $n$ into distinct parts not congruent to $2$
modulo $4$.

######  Definition 2.

Let $n$ be a nonnegative integer. We define $b_{4,e}(n)$ (respectively
$b_{4,o}(n)$) to be the number of $4$-regular partitions of $n$ into an even
(respectively odd) number of parts.

For example, the partitions of $7$ into parts that are not multiples of $4$
are

$\displaystyle(7),\ (6,1),\ (5,2),\ (5,1,1),\ (3,3,1),\ (3,2,2),\ (3,2,1,1),\
(3,1,1,1,1),$ $\displaystyle(2,2,2,1),\ (2,2,1,1,1),\ (2,1,1,1,1,1),\
(1,1,1,1,1,1,1),$

and so $b_{4}(7)=12$, $b_{4,e}(7)=5$ and $b_{4,o}(7)=7$. The sequences
$b_{4,e}(n)$ and $b_{4,o}(n)$ can be found in the On-Line Encyclopedia of
Integer Sequences [52, A339406, A339407].

######  Definition 3.

Let $n$ be a nonnegative integer and $k\in\\{0,2\\}$. We define $Q_{k}(n)$ to
be the number of partitions of $n$ into distinct parts which are not congruent
to $k$ modulo $4$.

For example, $Q_{0}(14)=11$ since the $4$-regular partitions of $14$ into
distinct parts are

$\displaystyle(14),\ (13,1),\ (11,3),\ (11,2,1),\ (10,3,1),\ (9,5),$
$\displaystyle(9,3,2),\ (7,6,1),\ (7,5,2),\ (6,5,3),\ (6,5,2,1),$

while $Q_{2}(14)=6$, the relevant partitions being

$\displaystyle(13,1),\ (11,3),\ (9,5),\ (9,4,1),\ (8,5,1),\ (7,4,3).$

The generating functions for these sequences satisfy the identities

$\displaystyle\sum_{n=0}^{\infty}Q_{0}(n)\,q^{n}=(-q,-q^{2},-q^{3};q^{4})_{\infty}$
(3)

and

$\displaystyle\sum_{n=0}^{\infty}Q_{2}(n)\,q^{n}=(-q,-q^{3},-q^{4};q^{4})_{\infty}.$
(4)

We remark that the sequence $Q_{0}(n)$ can be found in the On-Line
Encyclopedia of Integer Sequences [52, A070048]. Also on the page for
$A070048$, we find another combinatorial interpretation for $Q_{0}(n)$: the
number of partitions of $n$ into odd parts in which no part appears more than
thrice.

The following result introduces new combinatorial interpretations for the
partition functions $Q_{0}(n)$ and $Q_{2}(n)$.

######  Theorem 1.1.

For $n\geqslant 0$ the following hold.

  1. (i)

$(-1)^{n}\,Q_{0}(n)=pod_{e}(n)-pod_{o}(n)$

  2. (ii)

$(-1)^{n}\,Q_{2}(n)=b_{4,e}(n)-b_{4,o}(n)$

######  Corollary 1.2.

Let $n\geqslant 0$.

  1. (i)

$Q_{0}(n)$ and $pod(n)$ have the same parity.

  2. (ii)

$Q_{2}(n)$ and $b_{4}(n)$ have the same parity.

We also have the following result relating $Q_{0}(n)$ and $Q_{2}(n)$ to
$pod(n)$.

######  Theorem 1.3.

For $n\geqslant 0$ the following hold.

  1. (i)

$\displaystyle{Q_{0}(n)=pod(n)+2\sum_{k=1}^{\infty}(-1)^{k}\,pod\big{(}n-4k^{2}\big{)}}$

  2. (ii)

$\displaystyle{Q_{2}(n)=\sum_{k=0}^{\infty}(-1)^{k(k+1)/2}\,pod\big{(}n-k(k+1)\big{)}}$

In addition, $Q_{0}(n)$ and $Q_{2}(n)$ satisfy similar linear recurrence
relations involving the triangular numbers. To make these easier to state, we
first define

$\xi_{n}=\begin{cases}(-1)^{m}\cdot 2&\text{if $n=4m^{2}$ for some $m>0$,}\\\
1&\text{if $n=0$,}\\\ 0&\text{otherwise,}\end{cases}$

and

$\chi_{n}=\begin{cases}(-1)^{n/2}&\text{if $n=m(m+1)$ for some $m\geqslant
0$,}\\\ 0&\text{otherwise.}\end{cases}$

######  Theorem 1.4.

For $n\geqslant 0$ the following hold.

  1. (i)

$\displaystyle{\sum_{k=0}^{\infty}(-1)^{k(k+1)/2}\,Q_{0}\big{(}n-k(k+1)/2\big{)}=\xi_{n}}$

  2. (ii)

$\displaystyle{\sum_{k=0}^{\infty}(-1)^{k(k+1)/2}\,Q_{2}\big{(}n-k(k+1)/2\big{)}=\chi_{n}}$

The rest of this paper is organized as follows. In Section 2, we provide
analytic and combinatorial proofs of Theorem 1.1. In Section 3, we provide
proofs of Theorems 1.3 and 1.4 using generating functions and also give a
combinatorial proof of Theorem 1.3 (ii) (a combinatorial proof of Theorem
1.3(i) would be very welcome). In Section 4, we show that the identities of
Theorem 1.3 are limiting cases of much more general identities and use the
latter to derive two infinite families of linear inequalities involving
$pod(n)$. In Section 5, we present several Ramanujan type congruences for
$b_{4}(n)$ modulo $16$ and $64$. In Section 6, we obtain three identities of
Watson type and prove them analytically and combinatorially, and conclude with
conjectures for two infinite families of linear inequalities involving
$Q_{0}(n)$ and $Q_{2}(n)$.

##  2 Proof of Theorem 1.1

###  2.1 Analytic Proof

Define

$\displaystyle
F(z,q):=\prod_{k=0}^{\infty}\frac{1}{(1-zq^{4k+1})(1-zq^{4k+3})(1-zq^{4k+4})}.$

Then, using Euler’s identity $1/(q;q^{2})_{\infty}=(-q;q)_{\infty}$ and (3),
we have

$\displaystyle F(-1,q)$
$\displaystyle=\frac{1}{(-q,-q^{3},-q^{4};q^{4})_{\infty}}=\frac{(-q^{2};q^{4})_{\infty}}{(-q;q)_{\infty}}$
$\displaystyle=(q;q^{2})_{\infty}(-q^{2};q^{4})_{\infty}=\sum_{n=0}^{\infty}(-1)^{n}\,Q_{0}(n)\,q^{n}.$

On the other hand, the fact that

$F(z,q)=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}pod(n,m)\,z^{m}\,q^{n},$

where $pod(n,m)$ is equal to the number of partitions of $n$ with $m$ parts
none of which are congruent to $2$ modulo $4$, implies that

$F(-1,q)=\sum_{n=0}^{\infty}\big{(}pod_{e}(n)-pod_{o}(n)\big{)}\,q^{n},$

which establishes (i). To prove (ii), we define

$G(z,q):=\prod_{k=0}^{\infty}\frac{1}{(1-zq^{4k+1})(1-zq^{4k+2})(1-zq^{4k+3})}.$

Then, using Euler’s identity and (4), we have

$\displaystyle G(-1,q)$
$\displaystyle=\frac{1}{(-q,-q^{2},-q^{3};q^{4})_{\infty}}=\frac{(-q^{4};q^{4})_{\infty}}{(-q;q)_{\infty}}$
$\displaystyle=(q;q^{2})_{\infty}(-q^{4};q^{4})_{\infty}=\sum_{n=0}^{\infty}(-1)^{n}\,Q_{2}(n)\,q^{n}.$

Moreover, since

$G(z,q)=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}b_{4}(n,m)\,z^{m}\,q^{n},$

where $b_{4}(n,m)$ is the number of $4$-regular partitions of $n$ with $m$
parts, we also have

$G(-1,q)=\sum_{n=0}^{\infty}\big{(}b_{4,e}(n)-b_{4,o}(n)\big{)}\,q^{n},$

which establishes (ii).

###  2.2 Combinatorial Proof

For the rest of this article we will use calligraphy style capital letters to
denote the set of partitions enumerated by the function denoted by the same
letters. For example, $\mathcal{P}(n)$ denotes the set of partitions of $n$
and $\mathcal{POD}_{e}(n)$ is the set of partitions of $n$ into an even number
of parts which are not congruent to $2$ modulo $4$. Recalling that $pod(n)$
has more than one partition theoretic interpretation, in the sequel we assume
that $\mathcal{POD}(n)$ represents the set of partitions of $n$ having
distinct odd parts.

If $\lambda$ is a partition of $n$, we say that the size of $\lambda$ is $n$
and write $|\lambda|=n$. The length of $\lambda$, denoted by $\ell(\lambda)$,
is the number of parts in $\lambda$. We make the convention that
$\lambda_{\ell(\lambda)+1}=0$. Given two partitions $\lambda$ and $\mu$, we
denote by $\lambda\cup\mu$ the partition whose parts are precisely the parts
of $\lambda$ and $\mu$ (with multiplicity). If each part of $\mu$ is also a
part of $\lambda$ with equal or larger multiplicity, we denote by
$\lambda\setminus\mu$ the partition obtained from $\lambda$ by removing the
parts of $\mu$ (with multiplicity). Finally, we will often write a partition
$\lambda$ as $(\lambda^{e},\lambda^{o})$, where $\lambda^{e}$ (respectively
$\lambda^{o}$) consists of the even (respectively odd) parts of $\lambda$.

We start by proving the first identity of Theorem 1.1. Denote by
$\widetilde{\mathcal{Q}}_{0}(n)$ the set of partitions of $n$ into odd parts
repeated no more than three times. Glaisher’s transformation [24], which
iteratively splits each even part of a distinct partition into two equal parts
until all parts are odd, yields a bijection between $\mathcal{Q}_{0}(n)$ and
$\widetilde{\mathcal{Q}}_{0}(n)$.

For any partition $\lambda\in\widetilde{\mathcal{Q}}_{0}(n)$, we have
$\ell(\lambda)\equiv n\pmod{2}$. Thus, if $n$ is even (respectively odd),
$\widetilde{\mathcal{Q}}_{0}(n)$ is a subset of $\mathcal{POD}_{e}(n)$
(respectively $\mathcal{POD}_{o}(n)$).

Let
$\mathcal{A}(n)=(\mathcal{POD}_{e}(n)\cup\mathcal{POD}_{o}(n))\setminus\widetilde{\mathcal{Q}}_{0}(n)$.
Inspired by [27], we define an involution
$\varphi:\mathcal{A}(n)\to\mathcal{A}(n)$ that reverses the parity of the
length of partitions. First, we introduce more useful notation. For any
partition $\lambda$, if $d$ is a part of $\lambda$ with multiplicity $m_{d}$,
we denote by $s_{d}$ the nonnegative integer satisfying $2^{s_{d}}\leq
m_{d}<2^{s_{d}+1}$. If $\lambda\in\mathcal{A}(n)$ has an odd part $d$ with
$m_{d}\geq 4$, we let $r_{\lambda}$ be the largest value of $2^{s_{d}}d$ among
all such parts (else we let $r_{\lambda}=0$). If $\lambda^{e}=\emptyset$, it
follows from the definition of $\mathcal{A}(n)$ that $r_{\lambda}\neq 0$. We
define $\varphi(\lambda)$ as follows.

  1. (i)

If $r_{\lambda}\geq\lambda_{1}^{e}$, merge $2^{s_{d}}$ copies of $d$ into a
new part to obtain $\varphi(\lambda)$.

  2. (ii)

If $r_{\lambda}<\lambda_{1}^{e}$, split $\lambda_{1}^{e}=2^{k_{1}}c_{1}$,
where $k_{1}\geq 1$ and $c_{1}$ is odd, into $2^{k_{1}}$ parts equal to
$c_{1}$ obtain $\varphi(\lambda)$.

Partitions from case (i) are mapped by $\varphi$ to partitions from case (ii)
and vice-versa. Moreover, $\varphi$ is its own inverse and reverses the parity
of the length of partitions. This finishes the proof of the first identity of
Theorem 1.1.

For the second identity, denote by $\widetilde{\mathcal{Q}}_{2}(n)$ the set of
partitions $\lambda=(\lambda^{e},\lambda^{o})$ of $n$ such that $\lambda^{o}$
has distinct parts and each part of $\lambda^{e}$ is congruent to $2$ modulo
$4$ and has even multiplicity. We use the following variant of Glaisher’s
transformation to create a bijection between $\mathcal{Q}_{2}(n)$ and
$\widetilde{\mathcal{Q}}_{2}(n)$: given $\eta\in\mathcal{Q}_{2}(n)$, split
each part of $\eta$ having the form $2^{k}c$, with $k\geqslant 2$ and $c$ odd,
into $2^{k-1}$ parts equal to $2c$. The inverse of this transformation
iteratively merges equal parts of a partition in
$\widetilde{\mathcal{Q}}_{2}(n)$ until all parts are distinct. Since even
parts of partitions in $\widetilde{\mathcal{Q}}_{2}(n)$ have even
multiplicity, all obtained even parts are divisible by $4$.

For any partition $\lambda\in\widetilde{\mathcal{Q}}_{2}(n)$ we have
$\ell(\lambda)\equiv n\pmod{2}$. Thus, if $n$ is even (respectively odd),
$\widetilde{\mathcal{Q}}_{2}(n)$ is a subset of $\mathcal{B}_{4,e}(n)$
(respectively $\mathcal{B}_{4.o}(n)$).

Let
$\mathcal{B}(n)=(\mathcal{B}_{4,e}(n)\cup\mathcal{B}_{4,o}(n))\setminus\widetilde{\mathcal{Q}}_{2}(n)$.
Notice that for every partition $(\lambda^{e},\lambda^{o})\in\mathcal{B}(n)$,
if $\lambda^{o}$ has distinct parts, then there is at least one part in
$\lambda^{e}$ with odd multiplicity. We define a transformation
$\varepsilon:\mathcal{B}(n)\to\mathcal{B}(n)$ as follows.

  1. (i)

If $\lambda$ has a part $4k+2$ with odd multiplicity with the property that
all parts less than $2k+1$ in $\lambda^{o}$ have multiplicity one, let $4a+2$
be the smallest such part and define
$\varepsilon(\lambda)=(\lambda^{e}\setminus(4a+2),\lambda^{o}\cup(2a+1,2a+1))$.

  2. (ii)

Else, let $2c+1$ be the smallest repeated part in $\lambda^{o}$ and define
$\varepsilon(\lambda)=(\lambda^{e}\cup(4c+2),\lambda^{o}\setminus(2c+1,2c+1))$.

Then $\varepsilon$ is an involution on $\mathcal{B}(n)$ that maps a partition
from case (i) to a partition from case (ii) and vice-versa, and changes the
length of a partition by one. This finishes the proof of the second identity
of Theorem 1.1.

##  3 Proof of Theorems 1.3 and 1.4

###  3.1 Analytic proofs

The following theta identities are often attributed to Gauss [5, p.23, Eqs.
(2.2.12), (2.2.13)]:

$1+2\sum_{n=1}^{\infty}(-1)^{n}\,q^{n^{2}}=\frac{(q;q)_{\infty}}{(-q;q)_{\infty}}$
(5)

and

$\sum_{n=0}^{\infty}(-1)^{n(n+1)/2}\,q^{n(n+1)/2}=\frac{(q^{2};q^{2})_{\infty}}{(-q;q^{2})_{\infty}}.$
(6)

The Jacobi triple product identity (cf. [23, Eq. (1.6.1)]) states that

$(z;q)_{\infty}(q/z;q)_{\infty}(q;q)_{\infty}=\sum_{n=-\infty}^{\infty}(-z)^{n}\,q^{n(n-1)/2}.$
(7)

We can write

$\displaystyle(-q,-q^{2},-q^{3};q^{4})_{\infty}\cdot\frac{(q^{2};q^{2})_{\infty}}{(-q;q^{2})_{\infty}}$
$\displaystyle=(-q^{2};q^{4})_{\infty}(q^{2};q^{2})_{\infty}$
$\displaystyle=(q^{4};q^{8})_{\infty}(q^{4};q^{4})_{\infty}$
$\displaystyle=\frac{(q^{4};q^{4})_{\infty}}{(-q^{4};q^{4})_{\infty}}$
$\displaystyle=1+2\sum_{n=1}^{\infty}(-1)^{n}\,q^{4n^{2}},$ (8)

and

$\displaystyle(-q,-q^{3},-q^{4};q^{4})_{\infty}\cdot\frac{(q^{2};q^{2})_{\infty}}{(-q;q^{2})_{\infty}}$
$\displaystyle=(-q^{4};q^{4})_{\infty}(q^{2};q^{2})_{\infty}$
$\displaystyle=\frac{(q^{2};q^{2})_{\infty}}{(q^{4};q^{8})_{\infty}}$
$\displaystyle=(q^{2},q^{6},q^{8};q^{8})_{\infty}$
$\displaystyle=\sum_{n=-\infty}^{\infty}(-1)^{n}\,q^{2n(2n-1)}$
$\displaystyle=\sum_{n=0}^{\infty}(-1)^{n(n+1)/2}\,q^{n(n+1)},$ (9)

where we have invoked (7) with $q$ replaced by $q^{8}$ and $z$ replaced by
$q^{2}$. Thus, we deduce that

$\displaystyle\sum_{n=0}^{\infty}Q_{0}(n)\,q^{n}$
$\displaystyle=\frac{(-q;q^{2})_{\infty}}{(q^{2};q^{2})_{\infty}}\left(1+2\sum_{n=1}^{\infty}(-1)^{n}\,q^{4n^{2}}\right)$
$\displaystyle=\left(\sum_{n=0}^{\infty}pod(n)\,q^{n}\right)\left(1+2\sum_{n=1}^{\infty}(-1)^{n}\,q^{4n^{2}}\right)$
$\displaystyle=\sum_{n=0}^{\infty}\left(pod(n)+2\sum_{k=1}^{\infty}(-1)^{k}\,pod\big{(}n-4k^{2}\big{)}\right)q^{n},$

and

$\displaystyle\sum_{n=0}^{\infty}Q_{2}(n)\,q^{n}$
$\displaystyle=\frac{(-q;q^{2})_{\infty}}{(q^{2};q^{2})_{\infty}}\sum_{n=0}^{\infty}(-1)^{n(n+1)/2}\,q^{n(n+1)}$
$\displaystyle=\left(\sum_{n=0}^{\infty}pod(n)\,q^{n}\right)\left(\sum_{n=0}^{\infty}(-1)^{n(n+1)/2}\,q^{n(n+1)}\right)$
$\displaystyle=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{\infty}(-1)^{k(k+1)/2}\,pod\big{(}n-k(k+1)\big{)}\right)q^{n},$

from which Theorem 1.3 follows.

Using (6), the relations (8) and (9) can be written as

$\left(\sum_{n=0}^{\infty}Q_{0}(n)\,q^{n}\right)\left(\sum_{n=0}^{\infty}(-1)^{n(n+1)/2}\,q^{n(n+1)/2}\right)=1+2\sum_{n=1}^{\infty}(-1)^{n}\,q^{4n^{2}}$

and

$\left(\sum_{n=0}^{\infty}Q_{2}(n)\,q^{n}\right)\left(\sum_{n=0}^{\infty}(-1)^{n(n+1)/2}\,q^{n(n+1)/2}\right)=\sum_{n=0}^{\infty}(-1)^{n(n+1)/2}\,q^{n(n+1)}.$

We rewrite these identities as

$\sum_{n=0}^{\infty}\left(\sum_{k=0}^{\infty}(-1)^{k(k+1)/2}\,Q_{0}\big{(}n-k(k+1)/2\big{)}\right)q^{n}=1+2\sum_{n=1}^{\infty}(-1)^{n}\,q^{4n^{2}}$

and

$\sum_{n=0}^{\infty}\left(\sum_{k=0}^{\infty}(-1)^{k(k+1)/2}\,Q_{2}\big{(}n-k(k+1)/2\big{)}\right)q^{n}=\sum_{n=0}^{\infty}(-1)^{n(n+1)/2}\,q^{n(n+1)},$

from which Theorem 1.4 follows.

###  3.2 Combinatorial proof of Theorem 1.3 (ii)

First, we introduce a graphical representation of partitions. The Ferrers
diagram of a partition $\lambda=(\lambda_{1},\lambda_{2},\ldots,\lambda_{i})$
is an array of left justified boxes such that the $i$th row from the top
contains $\lambda_{i}$ boxes.

If $n$ and $k$ are nonnegative integers, we define a transformation on
$\mathcal{POD}(n-k(k+1))$ as follows. Start with
$(\lambda^{e},\lambda^{o})\in\mathcal{POD}(n-k(k+1))$. The $2$-modular Ferrers
diagram of $\lambda^{e}$ is obtained from the ordinary Ferrers diagram of
$\lambda^{e}$ by replacing two boxes at a time in each row and placing a $2$
in the resulting box. To this diagram we append at the top the rotated Ferrers
diagram of the staircase of length $k$ with each box filled with $2$. Next,
starting in the upper left corner of the obtained diagram, we draw a zig-zag
line beginning with a right step and continuing with pairs of alternating down
and right steps for as long as both step segments border boxes of the diagram.

For example, if $\lambda^{e}=(14,14,12,12,8,4)$ and $k=3$, we obtain

2 2&2 222 2222222 2222222 222222 222222 2222 2 2

The diagram defines two partitions into distinct even parts: the partition
$\alpha$ whose $2$-modular Ferrers diagram is made up of the columns below the
zig-zag line, and the partition $\beta$ whose $2$-modular Ferrers diagram is
formed by the rows to the right of the zig-zag line. In the example above,
$\alpha=(18,16,12,10,6,4)$ and $\beta=(6,4)$.

The triple $(\lambda^{o},\alpha,\beta)$ is completely determined by
$(\lambda^{e},\lambda^{o})$ and $k$. Moreover
$|\lambda^{o}|+|\alpha|+|\beta|=n$ and
$k\leqslant\ell(\alpha)-\ell(\beta)\leqslant k+1$. Denote by
$\mathcal{A}_{k}(n)$ the set of triples of partitions
$(\lambda^{o},\alpha,\beta)$, where $\lambda^{o}$ is a partition into distinct
odd parts, $\alpha$ and $\beta$ are partitions into distinct even parts,
$|\lambda^{o}|+|\alpha|+|\beta|=n$, and
$k\leqslant\ell(\alpha)-\ell(\beta)\leqslant k+1$. Then the transformation
described above is a bijection between $\mathcal{POD}(n-k(k+1))$ and
$\mathcal{A}_{k}(n)$ (for the inverse transformation we refer the reader to
[9, Section 2.1], where a similar transformation for ordinary Ferrers diagrams
is defined). Thus, it remains to show that

$\displaystyle{Q_{2}(n)=\sum_{k=0}^{\infty}(-1)^{k(k+1)/2}\,|\mathcal{A}_{k}(n)|}.$
(10)

Let $\mathcal{A}(n)$ be the set of triples $(\lambda^{o},\alpha,\beta)$, where
$\lambda^{o}$ is a partition into distinct odd parts and $\alpha$ and $\beta$
are partitions into distinct even parts such that
$|\lambda^{o}|+|\alpha|+|\beta|=n$ and $\ell(\alpha)-\ell(\beta)\geqslant 0$.

Each triple $(\lambda^{o},\alpha,\beta)\in\mathcal{A}(n)$ with
$\ell(\alpha)>\ell(\beta)$ appears in exactly two of the sets
$\mathcal{A}_{k}(n)$, namely when $k=\ell(\alpha)-\ell(\beta)$ and when
$k=\ell(\alpha)-\ell(\beta)-1$. Triples
$(\lambda^{o},\alpha,\beta)\in\mathcal{A}(n)$ with $\ell(\alpha)=\ell(\beta)$
appear only in $\mathcal{A}_{0}(n)$. Thus, from the parity of triangular
numbers, the contribution of $(\lambda^{o},\alpha,\beta)$ with
$|\lambda^{o}|+|\alpha|+|\beta|=n$ to the righthand side of (10) is

$\begin{cases}1&\mbox{ if }\ell(\alpha)=\ell(\beta),\\\ 0&\mbox{ if
}\ell(\alpha)-\ell(\beta)\equiv 1\pmod{2},\\\ 2&\mbox{ if
}\ell(\alpha)-\ell(\beta)\equiv 0\pmod{4},\\\ -2&\mbox{ if
}\ell(\alpha)-\ell(\beta)\equiv 2\pmod{4}.\end{cases}$

We denote by $\mathcal{M}\mathcal{A}_{0}(n)$ be the multiset of triples
$(\lambda^{o},\alpha,\beta)\in\mathcal{A}(n)$ satisfying $\alpha\neq\beta$,
$\ell(\alpha)-\ell(\beta)\equiv 0\pmod{4}$, and if
$\ell(\alpha)-\ell(\beta)>0$, the triple has multiplicity $2$ in
$\mathcal{M}\mathcal{A}_{0}(n)$, and if $\ell(\alpha)-\ell(\beta)=0$, the
triple has multiplicity $1$. Similarly, we denote by
$\mathcal{M}\mathcal{A}_{2}(n)$ be the multiset of triples
$(\lambda^{o},\alpha,\beta)\in\mathcal{A}(n)$ with
$\ell(\alpha)-\ell(\beta)\equiv 2\pmod{4}$ and each triple has multiplicity
$2$.

For any triple $(\lambda^{o},\alpha,\beta)\in\mathcal{A}(n)$ let $i$ be the
smallest positive integer such that $\alpha_{i}\neq\beta_{i}$ (we make the
convention that $\beta_{\ell(\beta)+1}=0$).

We now define a map from $\mathcal{M}\mathcal{A}_{0}(n)$ to
$\mathcal{M}\mathcal{A}_{2}(n)$ as follows. Start with
$(\lambda^{o},\alpha,\beta)\in\mathcal{M}\mathcal{A}_{o}(n)$ and suppose
$\ell(\alpha)-\ell(\beta)=k$.

Case 1: $\alpha_{i}<\beta_{i}$. Let $\tilde{\alpha}=\alpha\cup(\beta_{i})$ and
$\tilde{\beta}=\beta\setminus(\beta_{i})$. Then,
$\ell(\tilde{\alpha})-\ell(\tilde{\beta})=k+2$,
$\tilde{\alpha}_{i}>\tilde{\beta}_{i}$, and the first $i-1$ parts of
$\tilde{\alpha}$ and $\tilde{\beta}$ are equal. Moreover, the triple
$(\lambda^{o},\tilde{\alpha},\tilde{\beta})$ lies in
$\mathcal{M}\mathcal{A}_{2}(n)$.

Case 2: $\beta_{i}<\alpha_{i}$. Let
$\tilde{\alpha}=\alpha\setminus(\alpha_{i})$ and
$\tilde{\beta}=\beta\cup(\alpha_{i})$. Then,
$\ell(\tilde{\alpha})-\ell(\tilde{\beta})=k-2$,
$\tilde{\beta}_{i}>\tilde{\alpha}_{i}$, and the first $i-1$ parts of
$\tilde{\alpha}$ and $\tilde{\beta}$ are equal. If $k>0$, the triple
$(\lambda^{o},\tilde{\alpha},\tilde{\beta})$ lies in
$\mathcal{M}\mathcal{A}_{2}(n)$. If $k=0$, since
$\ell(\tilde{\beta})-\ell(\tilde{\alpha})=2$, the triple
$(\lambda^{o},\tilde{\beta},\tilde{\alpha})$ lies in
$\mathcal{M}\mathcal{A}_{2}(n)$. Notice that this partition is also obtained
from $(\lambda^{o},\beta,\alpha)$ which is in Case 1.

For the inverse of the transformation, start with
$(\lambda^{o},\tilde{\alpha},\tilde{\beta})\in\mathcal{M}\mathcal{A}_{2}(n)$
and suppose $\ell(\tilde{\alpha})-\ell(\tilde{\beta})=k$.

Case I: $\tilde{\beta}_{i}<\tilde{\alpha}_{i}$. Let
$\alpha=\tilde{\alpha}\cup(\tilde{\beta}_{i})$ and
$\beta=\tilde{\beta}\setminus(\tilde{\beta}_{i})$. Then
$\ell(\alpha)-\ell(\beta)=k+2$ and
$(\lambda^{o},\alpha,\beta)\in\mathcal{M}\mathcal{A}_{o}(n)$.

Case II: $\tilde{\alpha}_{i}<\tilde{\beta}_{i}$. Let
$\alpha=\tilde{\alpha}\setminus(\tilde{\alpha}_{i})$ and
$\beta=\tilde{\beta}\cup(\tilde{\alpha}_{i})$ Then
$\ell(\alpha)-\ell(\beta)=k-2$ and
$(\lambda^{o},\alpha,\beta)\in\mathcal{M}\mathcal{A}_{o}(n)$. If $k=2$, one
copy of $(\lambda^{o},\tilde{\alpha},\tilde{\beta})$ is mapped to
$(\lambda^{o},\alpha,\beta)$ and the second copy is mapped to
$(\lambda^{o},\beta,\alpha)$.

Therefore, the transformation defined above is a bijection between
$\mathcal{M}\mathcal{A}_{0}(n)$ and $\mathcal{M}\mathcal{A}_{2}(n)$. Then, the
righthand side of (10) is equal to the number of triples
$(\lambda^{o},\alpha,\beta)\in\mathcal{A}(n)$ with $\alpha=\beta$. These
triples are in bijection with the partitions in $\mathcal{Q}_{2}(n)$ via the
mapping that takes $(\lambda^{o},\alpha,\alpha)$ to $\lambda^{o}\cup 2\alpha$,
where $2\alpha$ is the partition obtained from $\alpha$ by doubling each of
its parts.

##  4 Linear inequalities involving $pod(n)$

Linear inequalities involving partition functions have been studied
extensively [7, 8, 26, 31, 37, 38, 39, 40, 41]. For example, V. J. W. Guo and
J. Zeng [26] proved that

$\displaystyle(-1)^{k-1}\sum_{j=0}^{2k-1}(-1)^{j(j+1)/2}\,pod\big{(}n-j(j+1)/2\big{)}\geqslant
0,$

for $n,k>0$. Recently, G. E. Andrews and M. Merca [8, Corollary 10]
established a partition theoretic interpretation of this inequality by showing
that

$\displaystyle(-1)^{k-1}\sum_{j=0}^{2k-1}(-1)^{j(j+1)/2}\,pod\big{(}n-j(j+1)/2\big{)}=MP_{k}(n),$

where $MP_{k}(n)$ is the number of partitions of $n$ in which the first part
larger than $2k-1$ is odd and appears exactly $k$ times and all other odd
parts appear at most once. For example, $MP_{2}(19)=10$, and the partitions in
question are

$\displaystyle(9,9,1),\ (9,5,5),\ (8,5,5,1),\ (7,7,3,2),\ (7,7,2,2,1),\
(7,5,5,2),$ $\displaystyle(6,5,5,3),\ (6,5,5,2,1),\ (5,5,3,2,2,2),\
(5,5,2,2,2,2,1).$

Shortly after that, C. Ballantine, M. Merca, D. Passary and A. J. Yee [11]
gave combinatorial proofs of this interpretation.

In this section, inspired by Theorem 1.3, we obtain new infinite families of
linear inequalities for $pod(n)$. To this end, we recall that an overpartition
of $n$ is a nonincreasing sequence of natural numbers whose sum is $n$ in
which the first occurrence of a number may be overlined [15]. For example, the
eight overpartitions of $3$ are

$(3),\ (\overline{3}),\ (2,1),\ (2,\overline{1}),\ (\overline{2},1),\
(\overline{2},\overline{1}),\ (1,1,1),\ (\overline{1},1,1).$

G. E. Andrews and M. Merca [8] introduced the function $\overline{M}_{k}(n)$
which counts the number of overpartitions of $n$ in which the first part
larger than $k$ appears at least $k+1$ times. For example,
$\overline{M}_{2}(12)=16$, with the relevant overpartions being

$\displaystyle(4,4,4),\,(\overline{4},4,4),\,(3,3,3,3),\,(\overline{3},3,3,3),\,(3,3,3,2,1),\,(3,3,3,\overline{2},1),$
$\displaystyle(3,3,3,2,\overline{1}),\,(3,3,3,\overline{2},\overline{1}),\,(\overline{3},3,3,2,1),\,(\overline{3},3,3,\overline{2},1),\,(\overline{3},3,3,2,\overline{1}),$
$\displaystyle(\overline{3},3,3,\overline{2},\overline{1}),\,(3,3,3,1,1,1),\,(3,3,3,\overline{1},1,1),\,(\overline{3},3,3,1,1,1),\,(\overline{3},3,3,\overline{1},1,1).$

We now prove an identity that has Theorem 1.3.(i) as its limiting case when
$k\to\infty$.

######  Theorem 4.1.

For $n,k>0$, we have

$(-1)^{k}\left(pod(n)+2\sum_{j=1}^{k}(-1)^{j}\,pod\big{(}n-4j^{2}\big{)}-Q_{0}(n)\right)=\sum_{j=0}^{\left\lfloor
n/4\right\rfloor}Q_{0}(n-4j)\,\overline{M}_{k}(j).$

###### Proof.

According to G. E. Andrews and M. Merca [8, Theorem 7], we have the following
truncated version of (5):

$\displaystyle\frac{(-q;q)_{\infty}}{(q;q)_{\infty}}\left(1+2\sum_{j=1}^{k}(-1)^{j}\,q^{j^{2}}\right)$
(11)
$\displaystyle\qquad=1+(-1)^{k}\frac{(-q;q)_{k}}{(q;q)_{k}}\sum_{j=k+1}^{\infty}\frac{2\,q^{j(k+1)}}{1-q^{j}}\cdot\frac{(-q^{j+1};q)_{\infty}}{(q^{j+1};q)_{\infty}}.$

As explained in [8, Proof of Corollary 8], the series on the right hand side
of this identity is the generating function for $\overline{M}_{k}(n)$, i.e.,

$\sum_{n=0}^{\infty}\overline{M}_{k}(n)\,q^{n}=\frac{(-q;q)_{k}}{(q;q)_{k}}\sum_{j=k+1}^{\infty}\frac{2\,q^{j(k+1)}}{1-q^{j}}\cdot\frac{(-q^{j+1};q)_{\infty}}{(q^{j+1};q)_{\infty}}.$

Therefore, replacing $q$ by $q^{4}$ in (11) yields

$\displaystyle\frac{(-q^{4};q^{4})_{\infty}}{(q^{4};q^{4})_{\infty}}\left(1+2\sum_{j=1}^{k}(-1)^{j}\,q^{4j^{2}}\right)=1+(-1)^{k}\sum_{n=0}^{\infty}\overline{M}_{k}(n)\,q^{4n}.$

Multiplying both sides of this identity by $(-q,-q^{2},-q^{3};q^{4})_{\infty}$
we obtain

$\displaystyle\frac{(-q;q^{2})_{\infty}}{(q^{2};q^{2})_{\infty}}\left(1+2\sum_{j=1}^{k}(-1)^{j}\,q^{4j^{2}}\right)-(-q,-q^{2},-q^{3};q^{4})_{\infty}$
$\displaystyle\qquad\qquad=(-1)^{k}\,(-q,-q^{2},-q^{3};q^{4})_{\infty}\sum_{n=0}^{\infty}\overline{M}_{k}(n)\,q^{4n}.$

Then, using (2) and (3), the identity becomes

$\displaystyle\left(\sum_{n=0}^{\infty}pod(n)\,q^{n}\right)\left(1+2\sum_{j=1}^{k}(-1)^{j}\,q^{4j^{2}}\right)-\sum_{n=0}^{\infty}Q_{0}(n)\,q^{n}$
$\displaystyle\qquad\qquad=(-1)^{k}\left(\sum_{n=0}^{\infty}Q_{0}(n)\,q^{n}\right)\left(\sum_{n=0}^{\infty}\overline{M}_{k}(n)\,q^{4n}\right)$

The theorem follows by comparing coefficients of $q^{n}$ on the two sides of
this equation. ∎

As a consequence of Theorem 4.1 we obtain the following infinite family of
linear inequalities involving $pod(n)$.

######  Corollary 4.2.

For $n,k>0$, we have

$(-1)^{k}\left(pod(n)+2\sum_{j=1}^{k}(-1)^{j}\,pod\big{(}n-4j^{2}\big{)}-Q_{0}(n)\right)\geqslant
0,$

with strict inequality if and only if $n\geqslant 4(k+1)^{2}$.

######  Remark.

For example,

$\displaystyle pod(n)-2pod(n-4)\leqslant Q_{0}(n),$ $\displaystyle
pod(n)-2pod(n-4)+2pod(n-16)\geqslant Q_{0}(n),$ $\displaystyle
pod(n)-2pod(n-4)+2pod(n-16)-2pod(n-36)\leqslant Q_{0}(n),$ $\displaystyle
pod(n)-2pod(n-4)+2pod(n-16)-2pod(n-36)+2pod(n-64)\geqslant Q_{0}(n).$

Our next result has Theorem 1.3.(ii) as its limiting case when $k\to\infty$.

######  Theorem 4.3.

For $n,k>0$, we have

$(-1)^{k}\left(Q_{2}(n)-\sum_{j=0}^{2k-1}(-1)^{j(j+1)/2}\,pod\big{(}n-j(j+1)\big{)}\right)=\sum_{j=0}^{\left\lfloor
n/2\right\rfloor}Q_{2}(n-2j)\,MP_{k}(j).$

###### Proof.

According to G. E. Andrews and M. Merca [8, Theorem 9], we have the following
truncated version of (6):

$\displaystyle\frac{(-q;q^{2})_{\infty}}{(q^{2};q^{2})_{\infty}}\sum_{j=0}^{2k-1}(-q)^{j(j+1)/2}$
$\displaystyle\qquad=1+(-1)^{k-1}\frac{(-q;q^{2})_{k}}{(q^{2};q^{2})_{k-1}}\sum_{j=k}^{\infty}\frac{q^{k(2j+1)}\,(-q^{2j+3};q^{2})_{\infty}}{(q^{2j+2};q^{2})_{\infty}}.$
(12)

As explained in [8, Proof of Corollary 10], the series on the right hand side
of this identity is the generating function for $MP_{k}(n)$, i.e.,

$\sum_{n=0}^{\infty}MP_{k}(n)\,q^{n}=\frac{(-q;q^{2})_{k}}{(q^{2};q^{2})_{k-1}}\sum_{j=k}^{\infty}\frac{q^{k(2j+1)}\,(-q^{2j+3};q^{2})_{\infty}}{(q^{2j+2};q^{2})_{\infty}}.$

Therefore, replacing $q$ by $q^{2}$ in (12) yields

$\displaystyle\frac{(-q^{2};q^{4})_{\infty}}{(q^{4};q^{4})_{\infty}}\sum_{j=0}^{2k-1}(-1)^{j(j+1)/2}\,q^{j(j+1)}=1+(-1)^{k-1}\sum_{n=0}^{\infty}MP_{k}(n)\,q^{2n}.$

Multiplying both sides of this identity by $(-q,-q^{3},-q^{4};q^{4})_{\infty}$
we obtain

$\displaystyle\frac{(-q;q^{2})_{\infty}}{(q^{2};q^{2})_{\infty}}\sum_{j=0}^{2k-1}(-1)^{j(j+1)/2}\,q^{j(j+1)}-(-q,-q^{3},-q^{4};q^{4})_{\infty}$
$\displaystyle\qquad\qquad=(-1)^{k-1}\,(-q,-q^{3},-q^{4};q^{4})_{\infty}\sum_{n=0}^{\infty}MP_{k}(n)\,q^{2n}.$

Then, using (2) and (4), the identity becomes

$\displaystyle\left(\sum_{n=0}^{\infty}pod(n)\,q^{n}\right)\left(\sum_{j=0}^{2k-1}(-1)^{j(j+1)/2}\,q^{j(j+1)}\right)-\sum_{n=0}^{\infty}Q_{2}(n)\,q^{n}$
$\displaystyle\qquad\qquad=(-1)^{k-1}\left(\sum_{n=0}^{\infty}Q_{2}(n)\,q^{n}\right)\left(\sum_{n=0}^{\infty}MP_{k}(n)\,q^{2n}\right).$

The theorem follows by comparing coefficients of $q^{n}$ on the two sides of
this equation. ∎

As a consequence of Theorem 4.3 we obtain the following infinite family of
linear inequalities involving $pod(n)$.

######  Corollary 4.4.

For $n,k>0$, we have

$(-1)^{k}\left(Q_{2}(n)-\sum_{j=0}^{2k-1}(-1)^{j(j+1)/2}\,pod\big{(}n-j(j+1)\big{)}\right)\geqslant
0,$

with strict inequality if and only if $n\geqslant 2k(2k+1)$.

######  Remark.

For example,

$\displaystyle pod(n)-pod(n-2)\geqslant Q_{2}(n),$ $\displaystyle
pod(n)-pod(n-2)-pod(n-6)+pod(n-12)\leqslant Q_{2}(n),$ $\displaystyle
pod(n)-pod(n-2)-pod(n-6)+pod(n-12)$
$\displaystyle\qquad\qquad\qquad\qquad+pod(n-20)-pod(n-30)\geqslant Q_{2}(n),$
$\displaystyle pod(n)-pod(n-2)-pod(n-6)+pod(n-12)$
$\displaystyle\qquad\qquad+pod(n-20)-pod(n-30)-pod(42)+pod(56)\leqslant
Q_{2}(n).$

##  5 Ramanujan type congruences

In recent years, many congruences for the number of $\ell$-regular partitions
have been discovered by G. E. Andrews, M. D. Hirschhorn and J. A. Sellers [6],
S. C. Chen [14], S.-P. Cui and N. S. S. Gu [16, 17], B. Dandurand and D.
Penniston [19], D. Furcy and D. Penniston [22], B. Gordon and K. Ono [25], W.
J. Keith [32], B. L. S. Lin and A. Y. Z. Wang [34], J. Lovejoy and D.
Penniston [35], D. Penniston [45, 46], E. X. W. Xia [54, 55], E. X. W. Xia and
O. X. M. Yao [56, 57], O. X. M. Yao [58], and J. J. Webb [60].

For example, for $\alpha\geqslant 1$ and $n\geqslant 0$, from [6] we have

$\displaystyle b_{4}\left(3^{2\alpha+1}n+\frac{17\cdot
3^{2\alpha}-1}{8}\right)\equiv 0\pmod{2},$ $\displaystyle
b_{4}\left(3^{2\alpha+2}n+\frac{11\cdot 3^{2\alpha+1}-1}{8}\right)\equiv
0\pmod{2},$ $\displaystyle b_{4}\left(3^{2\alpha+2}n+\frac{19\cdot
3^{2\alpha+1}-1}{8}\right)\equiv 0\pmod{2},$

and for $r\in\\{13,21,29,37\\}$, from [14] we have

$b_{4}\left(5^{2\alpha+2}n+\frac{r\cdot 5^{2\alpha+1}-1}{8}\right)\equiv
0\pmod{4}.$

From [54] , for $\alpha\geqslant 0$, $n\geqslant 0$ and $r\in\\{11,19\\}$, we
have

$b_{4}\left(3^{4\alpha+4}n+\frac{r\cdot 3^{4\alpha+3}-1}{8}\right)\equiv
0\pmod{8}.$

To facilitate the study of partition functions, S. Radu [49, 50] considered a
class of functions $a(n)$ defined by

$\displaystyle\sum_{n=0}^{\infty}a(n)\,q^{n}=\prod_{\delta|M}(q^{\delta};q^{\delta})_{\infty}^{r_{\delta}},$
(13)

where the product is over the positive divisors of $M>0$ and
$r_{\delta}\in\mathbb{Z}$. Using the ideas of H. Rademacher [47], M. Newman
[42, 43] and O. Kolberg [33], Radu [49] discovered a method for establishing
congruences of the form

$a(mn+t)\equiv 0\pmod{u},$

for fixed $m$, $t$ and $u$, and any $n\geqslant 0$, and developed the so-
called Ramanujan–Kolberg algorithm [50] for deriving identities involving the
generating functions of $a(mn+t)$ and modular functions for $\Gamma_{0}(N)$ (a
description of this algorithm can be found in P. Paule and S. Radu [44]). Very
recently, N. A. Smoot [53] provided a successful Mathematica implementation of
Radu’s algorithm. The package is called RaduRK and requires 4ti2, a software
package for algebraic, geometric and combinatorial problems on linear spaces.
Instructions for the proper installation of these packages can be found in
[53].

The procedure

RK[N,M,r,m,j]

take as input an integer $N\geqslant 2$ which defines the congruence subgroup
$\Gamma_{0}(N)$, a generating function defined by $M$ and
$r=(r_{\delta})_{\delta|M}$ as in (13), and an arithmetic progression $mn+j$
with $0\leqslant j<m$. The algorithm decides if there exists an identity of
the form

$f(q)\cdot\prod_{j^{\prime}\in
P}\left(\sum_{n=0}^{\infty}a(mn+j^{\prime})\,q^{n}\right)=\sum_{g\in A}g\cdot
p_{g}(t),$

where

$A=\\{1,g_{1},\ldots,g_{u}\\}\qquad\text{and}\qquad\\{p_{g}(t)\\}_{g\in
A}=\\{p_{1},p_{g_{1}},\ldots,p_{g_{u}}\\}.$

We remark that $t,g_{1},\ldots,g_{u}$ are modular function for
$\Gamma_{0}(N)$. For the definition of these notions and a general
introduction to RaduRK algorithm, see [44]. For the correctness proof and
details of the algorithm, see [53].

In this section, we use the implemented RaduRK algorithm to prove several
congruences modulo $16$ for $b_{4}(n)$. Letting $M=4$ and
$r_{1}=-1,r_{2}=0,r_{4}=1$ in (13) yields the generating function for
$b_{4}(n)$.

######  Theorem 5.1.

Let $\alpha\in\\{8,13,18,23\\}$. For all $n\geqslant 0$, we have

$b_{4}(25n+\alpha)\equiv 0\pmod{16}.$

Since, $b_{4}(8)=16$, $b_{4}(25+13)=8528$, $b_{4}(18)=208$ and
$b_{4}(23)=592$, we have that

$\sum_{n=0}^{\infty}b_{4}(25n+\alpha)\,q^{n}\not\equiv 0\pmod{32},$

for all $\alpha\in\\{8,13,18,23\\}$. Thus, Theorem 5.1 follows directly from
the following two lemmas.

######  Lemma 5.2.

$\displaystyle\left(\sum_{n=0}^{\infty}b_{4}(25n+8)\,q^{n}\right)\left(\sum_{n=0}^{\infty}b_{4}(25n+23)\,q^{n}\right)\equiv
0\pmod{256}.$

###### Proof.

To establish this congruence identity, we consider the RaduRK program with

$\texttt{RK[20,4,\\{-1,0,1\\},25,8]}.$

The algorithm returns:

$\displaystyle P$ $\displaystyle=\\{8,23\\}$ $\displaystyle f(q)$
$\displaystyle=\frac{(q;q)_{\infty}^{52}\,(q^{4};q^{4})_{\infty}^{12}\,(q^{10};q^{10})_{\infty}^{32}}{q^{31}\,(q^{2};q^{2})_{\infty}^{32}\,(q^{5};q^{5})_{\infty}^{10}\,(q^{20};q^{20})_{\infty}^{54}}$
$\displaystyle t$
$\displaystyle=\frac{(q^{4};q^{4})^{4}_{\infty}\,(q^{10};q^{10})^{2}_{\infty}}{q^{2}\,(q^{2};q^{2})^{2}_{\infty}\,(q^{20};q^{20})^{4}_{\infty}}$
$\displaystyle A$
$\displaystyle=\left\\{1,\frac{(q^{4};q^{4})_{\infty}\,(q^{5};q^{5})_{\infty}^{5}}{q^{3}\,(q;q)_{\infty}\,(q^{20};q^{20})_{\infty}^{5}}-\frac{(q^{4};q^{4})_{\infty}^{4}\,(q^{10};q^{10})_{\infty}^{2}}{q^{2}\,(q^{2};q^{2})_{\infty}^{2}\,(q^{20};q^{20})_{\infty}^{4}}\right\\}$
$\displaystyle\\{p_{g}(t)\\}_{g\in A}$
$\displaystyle=\left\\{-387500000000\,t^{2}-4722300000000\,t^{3}+19755240000000\,t^{4}\right.$
$\displaystyle\qquad-13492968000000\,t^{5}-28902996000000\,t^{6}+51723282400000\,t^{7}$
$\displaystyle\qquad-27746680960000\,t^{8}-697717120000\,t^{9}+7036326368000\,t^{10}$
$\displaystyle\qquad-2875996422400\,t^{11}+195120171520\,t^{12}+113806525952\,t^{13}$
$\displaystyle\qquad+2380696832\,t^{14}+2340096\,t^{15},$ $\displaystyle\qquad
12500000000\,t+262500000000\,t^{2}+1449800000000\,t^{3}$
$\displaystyle\qquad-9119240000000\,t^{4}+13179468000000\,t^{5}-2546388000000\,t^{6}$
$\displaystyle\qquad-9465334400000\,t^{7}+8751301760000\,t^{8}-2237298720000\,t^{9}$
$\displaystyle\qquad-628733856000\,t^{10}+318550950400\,t^{11}+22754516480\,t^{12}$
$\displaystyle\qquad\left.+119739648\,t^{13}+9472\,t^{14}\right\\}.$

Taking into account that $256$ is the common factor of all the coefficients of
the polynomials $p_{g}(t)$, we deduce the identity

$\displaystyle\frac{(q;q)_{\infty}^{52}\,(q^{4};q^{4})_{\infty}^{12}\,(q^{10};q^{10})_{\infty}^{32}}{q^{31}\,(q^{2};q^{2})_{\infty}^{32}\,(q^{5};q^{5})_{\infty}^{10}\,(q^{20};q^{20})_{\infty}^{54}}\left(\sum_{n=0}^{\infty}b_{4}(25n+8)\,q^{n}\right)\left(\sum_{n=0}^{\infty}b_{4}(25n+23)\,q^{n}\right)$
$\displaystyle=256\,Y_{1}+256\,Y_{2}\left(\frac{(q^{4};q^{4})_{\infty}\,(q^{5};q^{5})_{\infty}^{5}}{q^{3}\,(q;q)_{\infty}\,(q^{20};q^{20})_{\infty}^{5}}-\frac{(q^{4};q^{4})_{\infty}^{4}\,(q^{10};q^{10})_{\infty}^{2}}{q^{2}\,(q^{2};q^{2})_{\infty}^{2}\,(q^{20};q^{20})_{\infty}^{4}}\right),$

where

$\displaystyle Y_{1}$
$\displaystyle=-1513671875\,t^{2}-18446484375\,t^{3}+77168906250\,t^{4}$
$\displaystyle\quad-52706906250\,t^{5}-112902328125\,t^{6}+202044071875\,t^{7}$
$\displaystyle\quad-108385472500\,t^{8}-2725457500\,t^{9}+27485649875\,t^{10}$
$\displaystyle\quad-11234361025\,t^{11}+762188170\,t^{12}+444556742\,t^{13}$
$\displaystyle\quad+9299597\,t^{14}+9141\,t^{15},$ $\displaystyle Y_{2}$
$\displaystyle=48828125\,t+1025390625\,t^{2}+5663281250\,t^{3}-35622031250\,t^{4}$
$\displaystyle\quad+51482296875\,t^{5}-9946828125\,t^{6}-36973962500\,t^{7}$
$\displaystyle\quad+34184772500\,t^{8}-8739448125\,t^{9}-2455991625\,t^{10}$
$\displaystyle\quad+1244339650\,t^{11}+88884830\,t^{12}+467733\,t^{13}+37\,t^{14}.$

Lemma 5.2 follows immediately. ∎

######  Lemma 5.3.

$\displaystyle\left(\sum_{n=0}^{\infty}b_{4}(25n+13)\,q^{n}\right)\left(\sum_{n=0}^{\infty}b_{4}(25n+18)\,q^{n}\right)\equiv
0\pmod{256}.$

###### Proof.

To establish this congruence identity, we consider the RaduRK program with

$\texttt{RK[20,4,\\{-1,0,1\\},25,13]}.$

The algorithm returns:

$\displaystyle P$ $\displaystyle=\\{13,18\\}$ $\displaystyle f(q)$
$\displaystyle=\frac{(q;q)_{\infty}^{52}\,(q^{4};q^{4})_{\infty}^{12}\,(q^{10};q^{10})_{\infty}^{32}}{q^{31}\,(q^{2};q^{2})_{\infty}^{32}\,(q^{5};q^{5})_{\infty}^{10}\,(q^{20};q^{20})_{\infty}^{54}}$
$\displaystyle t$
$\displaystyle=\frac{(q^{4};q^{4})^{4}_{\infty}\,(q^{10};q^{10})^{2}_{\infty}}{q^{2}\,(q^{2};q^{2})^{2}_{\infty}\,(q^{20};q^{20})^{4}_{\infty}}$
$\displaystyle A$
$\displaystyle=\left\\{1,\frac{(q^{4};q^{4})_{\infty}\,(q^{5};q^{5})_{\infty}^{5}}{q^{3}\,(q;q)_{\infty}\,(q^{20};q^{20})_{\infty}^{5}}-\frac{(q^{4};q^{4})_{\infty}^{4}\,(q^{10};q^{10})_{\infty}^{2}}{q^{2}\,(q^{2};q^{2})_{\infty}^{2}\,(q^{20};q^{20})_{\infty}^{4}}\right\\}$
$\displaystyle\\{p_{g}(t)\\}_{g\in A}$
$\displaystyle=\left\\{62500000000\,t-650000000000\,t^{2}-4309800000000\,t^{3}\right.$
$\displaystyle\quad+19463740000000\,t^{4}-13372968000000\,t^{5}-29049996000000\,t^{6}$
$\displaystyle\quad+51949994400000\,t^{7}-27941584960000\,t^{8}-595662560000\,t^{9}$
$\displaystyle\quad+7002545088000\,t^{10}-2869711052800\,t^{11}+194775499520\,t^{12}$
$\displaystyle\quad+113785213952\,t^{13}+2380097792\,t^{14}+2273536\,t^{15},$
$\displaystyle\quad-50000000000\,t+587500000000\,t^{2}+712300000000\,t^{3}$
$\displaystyle\quad-8147740000000\,t^{4}+12342468000000\,t^{5}-2047288000000\,t^{6}$
$\displaystyle\quad-9669166400000\,t^{7}+8801213760000\,t^{8}-2240988480000\,t^{9}$
$\displaystyle\quad-629908896000\,t^{10}+318735916800\,t^{11}+22753946880\,t^{12}$
$\displaystyle\quad\left.+120139008\,t^{13}+13312\,t^{14}\right\\}.$

Taking into account that $256$ is the common factor of all the coefficients of
the polynomials $p_{g}(t)$, we deduce the identity

$\displaystyle\frac{(q;q)_{\infty}^{52}\,(q^{4};q^{4})_{\infty}^{12}\,(q^{10};q^{10})_{\infty}^{32}}{q^{31}\,(q^{2};q^{2})_{\infty}^{32}\,(q^{5};q^{5})_{\infty}^{10}\,(q^{20};q^{20})_{\infty}^{54}}\left(\sum_{n=0}^{\infty}b_{4}(25n+13)\,q^{n}\right)\left(\sum_{n=0}^{\infty}b_{4}(25n+18)\,q^{n}\right)$
$\displaystyle=256\,Y_{1}+256\,Y_{2}\left(\frac{(q^{4};q^{4})_{\infty}\,(q^{5};q^{5})_{\infty}^{5}}{q^{3}\,(q;q)_{\infty}\,(q^{20};q^{20})_{\infty}^{5}}-\frac{(q^{4};q^{4})_{\infty}^{4}\,(q^{10};q^{10})_{\infty}^{2}}{q^{2}\,(q^{2};q^{2})_{\infty}^{2}\,(q^{20};q^{20})_{\infty}^{4}}\right),$

where

$\displaystyle Y_{1}$
$\displaystyle=244140625\,t-2539062500\,t^{2}-16835156250\,t^{3}$
$\displaystyle\quad+76030234375\,t^{4}-52238156250\,t^{5}-113476546875\,t^{6}$
$\displaystyle\quad+202929665625\,t^{7}-109146816250\,t^{8}-2326806875\,t^{9}$
$\displaystyle\quad+27353691750\,t^{10}-11209808800\,t^{11}+760841795\,t^{12}$
$\displaystyle\quad+444473492\,t^{13}+9297257\,t^{14}+8881\,t^{15},$
$\displaystyle Y_{2}$
$\displaystyle=-195312500\,t+2294921875\,t^{2}+2782421875\,t^{3}-31827109375\,t^{4}$
$\displaystyle\quad+48212765625\,t^{5}-7997218750\,t^{6}-37770181250\,t^{7}$
$\displaystyle\quad+34379741250\,t^{8}-8753861250\,t^{9}-2460581625\,t^{10}$
$\displaystyle\quad+1245062175\,t^{11}+88882605\,t^{12}+469293\,t^{13}+52\,t^{14}.$

Lemma 5.3 follows immediately. ∎

The RaduRK algorithm can also be used to introduce congruences modulo $64$ for
$b_{4}(n)$.

######  Theorem 5.4.

Let $\alpha\in\\{13,20,27,34,41,48\\}$. For all $n\geqslant 0$, we have

$b_{4}(49n+\alpha)\equiv 0\pmod{64}.$

Theorem 5.4 follows directly from the following lemma.

######  Lemma 5.5.

The following congruences hold.

  1. (i)

$\displaystyle{\prod\limits_{\alpha\in\\{13,20,34\\}}\sum\limits_{n=0}^{\infty}b_{4}(49n+\alpha)\,q^{n}\equiv
0\pmod{64^{3}}}$

  2. (ii)

$\displaystyle{\prod\limits_{\alpha\in\\{27,41,48\\}}\sum\limits_{n=0}^{\infty}b_{4}(49n+\alpha)\,q^{n}\equiv
0\pmod{64^{3}}}$

The proof of this lemma is quite similar to the proof of Lemmas 5.2 and 5.3,
so we omit the details. In the algorithm, we use

RK[28,4,{-1,0,1},49,13]

and

$\texttt{RK[28,4,\\{-1,0,1\\},49,27]}.$

We end this section by noting that it is likely that there exist infinite
families of congruences modulo $16$ or modulo $64$ for $b_{4}(n)$. However, we
were unable to find more congruences due to the running time of the RaduRK
program.

##  6 Open problems and concluding remarks

In this paper, inspired by the decompositions of $Q_{0}(n)$ and $Q_{2}(n)$ in
terms of the partition function $pod(n)$ given by Theorem 1.3, we derived new
infinite families of linear inequalities involving $pod(n)$. Here we show that
there is another way to decompose $Q_{0}(n)$ using $pod(n)$, and also
establish decompositions of $Q_{2}(n)$ and $b_{4}(n)$ in terms of the function
$\overline{p}(n)$ which enumerates the overpartitions of $n$ (see [15]).

######  Theorem 6.1.

For $n\geqslant 0$, we have

$\displaystyle
Q_{0}(n)=\sum_{k=0}^{\infty}pod\left(\frac{n-k(k+1)/2}{2}\right).$ (14)

###### Analytic proof.

By (3) we have

$\sum_{n=0}^{\infty}Q_{0}(n)\,q^{n}=(-q,-q^{2},-q^{3};q^{4})_{\infty}=(-q,-q^{3},q^{4};q^{4})_{\infty}\cdot\frac{(-q^{2};q^{4})_{\infty}}{(q^{4};q^{4})_{\infty}}.$

Next, using (2) and replacing $q$ by $q^{4}$ and $z$ by $-q$ in (7), we find

$\displaystyle\sum_{n=0}^{\infty}Q_{0}(n)\,q^{n}$
$\displaystyle=\left(\sum_{n=-\infty}^{\infty}q^{2n^{2}-n}\right)\left(\sum_{n=0}^{\infty}pod(n)\,q^{2n}\right)$
$\displaystyle=\left(\sum_{n=0}^{\infty}q^{n(n+1)/2}\right)\left(\sum_{n=0}^{\infty}pod(n)\,q^{2n}\right)$
$\displaystyle=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{\infty}pod\Big{(}\frac{n-k(k+1)/2}{2}\Big{)}\right)q^{n}.$

∎

###### Combinatorial proof.

We begin by recalling a result due to G. N. Watson [59]. If we denote by
$Q_{odd}(n)$ the number of partitions of $n$ into distinct odd parts, then

$Q_{odd}(n)=\sum_{k=0}^{\infty}p\left(\frac{n-k(k+1)/2}{4}\right).$ (15)

A combinatorial proof of (15) is given in [61] using abacus displays, which
were first introduced in [30]. For the convenience of the reader, we briefly
describe the construction here, using the language of [51, section 2].

Given a partition $\lambda$, fill the boxes of its Ferrers diagram with
alternating $0$s and $1$s starting with $0$ in the upper left corner. Draw
horizontal lines through rows ending in $0$ and vertical lines through columns
ending in $1$. The boxes at the intersection of the lines (pushed toward the
northwest) form the Ferrers diagram of a partition $\alpha$. Repeat the
process with horizontal lines through rows ending in $1$ and vertical lines
through columns ending in $0$ to obtain a partition $\beta$. The pair
$(\alpha,\beta)$ is referred to as the $2$-quotient of $\lambda$. The $2$-core
of $\lambda$ is the staircase partition $\delta$ whose Ferrers diagram is
obtained from the diagram of $\lambda$ by removing $2$-hooks (either two boxes
in a row or two boxes in a column) such that after every removal the obtained
diagram is the Ferrers diagram of a partition. Then

$|\lambda|=|\delta|+2|\alpha|+2|\beta|.$

It is shown in [30, Theorem 2.7.30] that the triple $(\delta,\alpha,\beta)$
completely determines $\lambda$. Clearly, if $\lambda$ is self-conjugate, then
$\alpha=\beta$ and $(\delta,\alpha)$ determines $\lambda$.

Given a partition $\mu\in\mathcal{Q}_{odd}(n)$, consider the partitions
$\mu^{sc}$ whose Ferrers diagram is obtained from that of $\mu$ by
straightening the hooks nested along the diagonal (this is the classical
bijection proving that the number of partitions of $n$ into distinct odd parts
equals the number of self-conjugate partitions of $n$). Define $\xi(\mu)$ to
be the partition $\alpha$ in the $2$-quotient $(\alpha,\alpha)$ of $\mu^{sc}$.
Then the transformation

$\xi:\mathcal{Q}_{odd}(n)\to\bigcup_{k=0}^{\infty}\mathcal{P}\left(\frac{n-k(k+1)/2}{4}\right)$

is a bijection.

To prove (14), start with a partition
$\lambda=(\lambda^{e},\lambda^{o})\in\mathcal{Q}_{0}(n)$. The parts of
$\lambda^{e}$ are congruent to $2$ modulo $4$. Let $\tilde{\lambda}^{e}$ be
the partitions whose parts are the parts of $\lambda^{e}$ divided by $2$.
Thus, $\tilde{\lambda}^{e}$ is a partition with distinct odd parts. We denote
by $\tilde{\lambda}^{o}$ the partition whose parts are the parts of
$\xi(\lambda^{o})$ multiplied by $2$. Let
$\psi(\lambda)=\tilde{\lambda}^{o}\cup\tilde{\lambda}^{e}$. Then

$\psi:\mathcal{Q}_{0}(n)\to\bigcup_{k=0}^{\infty}\mathcal{POD}\left(\frac{n-k(k+1)/2}{2}\right)$

is a bijection.

∎

In a similar way, $Q_{2}(n)$ and $b_{4}(n)$ can be expressed in terms of
$\overline{p}(n)$.

######  Theorem 6.2.

For $n\geqslant 0$, we have

$\displaystyle
Q_{2}(n)=\sum_{k=0}^{\infty}\overline{p}\left(\frac{n-k(k+1)/2}{4}\right),$

and

$\displaystyle
b_{4}(n)=\sum_{k=0}^{\infty}\overline{p}\left(\frac{n-k(k+1)/2}{2}\right).$

###### Analytic proof.

Arguing as in the proof of Theorem 6.1, we have

$\displaystyle\sum_{n=0}^{\infty}Q_{2}(n)\,q^{n}$
$\displaystyle=(-q,-q^{3},q^{4};q^{4})_{\infty}\cdot\frac{(-q^{4};q^{4})_{\infty}}{(q^{4};q^{4})_{\infty}}$
$\displaystyle=\left(\sum_{n=0}^{\infty}q^{n(n+1)/2}\right)\left(\sum_{n=0}^{\infty}\overline{p}(n)\,q^{4n}\right)$
$\displaystyle=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{\infty}\overline{p}\Big{(}\frac{n-k(k+1)/2}{4}\Big{)}\right)q^{n}$

and

$\displaystyle\sum_{n=0}^{\infty}b_{4}(n)\,q^{n}$
$\displaystyle=\frac{1}{(q;q^{2})_{\infty}(q^{2};q^{4})_{\infty}}=(-q;q)_{\infty}(-q^{2};q^{2})_{\infty}$
$\displaystyle=(-q,-q^{3};q^{4})_{\infty}\cdot\frac{(-q^{2};q^{2})_{\infty}}{(q^{2};q^{4})_{\infty}}$
$\displaystyle=(-q,-q^{3},q^{4};q^{4})_{\infty}\cdot\frac{(-q^{2};q^{2})_{\infty}}{(q^{2};q^{2})_{\infty}}$
$\displaystyle=\left(\sum_{n=0}^{\infty}q^{n(n+1)/2}\right)\left(\sum_{n=0}^{\infty}\overline{p}(n)\,q^{2n}\right)$
$\displaystyle=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{\infty}\overline{p}\Big{(}\frac{n-k(k+1)/2}{2}\Big{)}\right)q^{n}.$

∎

###### Combinatorial proof.

We start with a partition
$\lambda=(\lambda^{e},\lambda^{o})\in\mathcal{Q}_{2}(n)$. The parts of
$\lambda^{e}$ are congruent to $0$ modulo $4$. Let $\bar{\lambda}^{e}$ be the
overpartition whose parts are the parts of $\lambda^{e}$ divided by $4$ and
each part is overlined. Let
$\rho(\lambda)=\xi(\lambda^{o})\cup\bar{\lambda}^{e}$. Then

$\rho:\mathcal{Q}_{2}(n)\to\bigcup_{k=0}^{\infty}\overline{\mathcal{P}}\left(\frac{n-k(k+1)/2}{4}\right)$

is a bijection.

To prove the second identity we make use of the Fu-Tang combinatorial proof
[21] of another identity due to G. N. Watson [59]. If we denote by $Q(n)$ the
number of partitions of $n$ into distinct parts, then the Fu-Tang bijection

$\zeta:\mathcal{Q}(n)\to\bigcup_{k=0}^{\infty}\mathcal{P}\left(\frac{n-k(k+1)/2}{2}\right)$

(see Section 2 of [10] for succinct description of $\zeta$) yields

$Q(n)=\sum_{k=0}^{\infty}p\left(\frac{n-k(k+1)/2}{2}\right).$ (16)

Start with $\lambda=(\lambda^{e},\lambda^{o})\in\mathcal{B}_{4}(n)$. The parts
of $\lambda^{e}$ are congruent to $2$ modulo $4$. We divide each part of
$\lambda^{e}$ by $2$ to obtain a partition $\tilde{\lambda}^{e}$ into odd
parts. We use Glaisher’s bijection to transform $\tilde{\lambda}^{e}$ into a
partition with distinct parts whose parts we overline to obtain an
overpartition $\bar{\lambda}^{e}$. Next, we apply Glaisher’s bijection to
$\lambda^{o}$ to obtain a partition $\tilde{\lambda}^{o}$ into distinct parts
and let $\chi(\lambda)=\zeta(\tilde{\lambda}^{o})\cup\bar{\lambda}^{e}$. Then
the function

$\chi:\mathcal{B}_{4}(n)\to\bigcup_{k=0}^{\infty}\overline{\mathcal{P}}\left(\frac{n-k(k+1)/2}{2}\right)$

is a bijection. ∎

We note that Theorems 6.1 and 6.2 provide new identities of Watson type (more
on such identities can be found in [10]).

Numerical evidence suggests that $Q_{0}(n)$ and $Q_{2}(n)$ satisfy the
following linear homogeneous inequalities analogous to those given by
Corollaries 4.2 and 4.4.

######  Conjecture 1.

For $n,k\geqslant 0$, we have

$\displaystyle(-1)^{k-1}\left(\sum_{j=0}^{2k-1}(-1)^{j(j+1)/2}\,Q_{0}\big{(}n-j(j+1)/2\big{)}-\xi_{n}\right)\geqslant
0,$

with strict inequality if and only if $n\geqslant k(2k+1)$.

######  Remark.

For example,

$\displaystyle Q_{0}(n)-Q_{0}(n-1)\geqslant\xi_{n},$ $\displaystyle
Q_{0}(n)-Q_{0}(n-1)-Q_{0}(n-3)+Q_{0}(n-6)\leqslant\xi_{n},$ $\displaystyle
Q_{0}(n)-Q_{0}(n-1)-Q_{0}(n-3)+Q_{0}(n-6)+Q_{0}(n-10)-Q_{0}(n-15)\geqslant\xi_{n}.$

######  Conjecture 2.

For $n,k\geqslant 0$, we have

$\displaystyle(-1)^{k-1}\left(\sum_{j=0}^{2k-1}(-1)^{j(j+1)/2}\,Q_{2}\big{(}n-j(j+1)/2\big{)}-\chi_{n}\right)\geqslant
0,$

with strict inequality if and only if $n\geqslant k(2k+1)$.

######  Remark.

For example,

$\displaystyle Q_{2}(n)-Q_{2}(n-1)\geqslant\chi_{n},$ $\displaystyle
Q_{2}(n)-Q_{2}(n-1)-Q_{2}(n-3)+Q_{2}(n-6)\leqslant\chi_{n},$ $\displaystyle
Q_{2}(n)-Q_{2}(n-1)-Q_{2}(n-3)+Q_{2}(n-6)+Q_{2}(n-10)-Q_{2}(n-15)\geqslant\chi_{n}.$

## Acknowledgements

The authors thank an anonymous referee for many useful suggestions that helped
improve the presentation of the article.

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