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Truncated versions of three identities of Euler and Gauss

Part of: Additive number theory; partitions Combinatorics

Published online by Cambridge University Press:  30 August 2022

Olivia X.M. Yao
Olivia X.M. Yao*
Affiliation:
School of Mathematical Sciences, Suzhou University of Science and Technology, Suzhou, Jiangsu Province 215009, P. R. China (yaoxiangmei@163.com)
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Abstract

In 2012, Andrews and Merca proved a truncated theorem on Euler's pentagonal number theorem. Motivated by the works of Andrews and Merca, Guo and Zeng deduced truncated versions for two other classical theta series identities of Gauss. Very recently, Xia et al. proved new truncated theorems of the three classical theta series identities by taking different truncated series than the ones chosen by Andrews–Merca and Guo–Zeng. In this paper, we provide a unified treatment to establish new truncated versions for the three identities of Euler and Gauss based on a Bailey pair due to Lovejoy. These new truncated identities imply the results proved by Andrews–Merca, Wang–Yee, and Xia–Yee–Zhao.

Keywords

partitionstheta functionstruncated sumspartition-theoretic interpretationsBailey pair

MSC classification

Primary: 11P81: Elementary theory of partitions
Secondary: 05A17: Partitions of integers

Information

Type
Research Article
Information
Proceedings of the Edinburgh Mathematical Society , Volume 65 , Issue 3 , August 2022 , pp. 775 - 798
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Copyright
Copyright © The Author(s), 2022. Published by Cambridge University Press on Behalf of The Edinburgh Mathematical Society

1. Introduction

Recall that a partition of a positive integer $n$ is an integer sequence $\lambda =(\lambda _1,\, \lambda _2,\, \ldots,\, \lambda _k)$ such that $\lambda _1\geq \lambda _2\geq \cdots \geq \lambda _k>0$ and $\lambda _1 +\lambda _2+\cdots +\lambda _k=n$. The numbers $\lambda _i$ are called parts of $\lambda$ [Reference Andrews2]. As usual, let $p(n)$ denote the number of partitions of $n$ and set $p(0)=1$ and $p(x)=0$ when $x$ is a negative integer. Euler found that the generating function for $p(n)$ is

(1.1)\begin{equation} \sum_{n=0}^{\infty} p(n) q^{n}=\frac{1}{(q;q)_\infty}. \end{equation}

Here and throughout the paper, we assume that $|q|<1$ and the $q$-shifted factorials are defined by

\begin{align*} (a;q)_\infty& :=\prod_{k=0}^{\infty} (1-aq^{k}), \\ (a;q)_n& :=\displaystyle\frac{(a;q)_\infty }{(aq^{n};q)_\infty},\\ (a_1,a_2,\ldots, a_k;q)_n & :=(a_1;q)_n(a_2;q)_n \cdots (a_k;q)_n. \end{align*}

In addition, the $q$-binomial coefficient is defined by

\[\left[\begin{array}{@{}c@{}} n \\ m \\ \end{array}\right] = \left[\begin{array}{@{}c@{}} n \\ m \\ \end{array}\right]_q:= \left\{ \begin{array}{cc} \displaystyle\dfrac{(q;q)_n}{(q;q)_m(q;q)_{n-m}}, & {\rm if\ } 0\leq m \leq n, \\ 0, & {\rm otherwise}. \end{array} \right.\]

The reciprocal of the right-hand side of (1.1) has an interesting series representation as follows:

(1.2)\begin{equation} (q;q)_\infty =\displaystyle\sum_{n=0}^{\infty} ({-}1)^{n} q^{n(3n+1)/2}(1-q^{2n+1}), \end{equation}

which is Euler's pentagonal number theorem.

In 2012, Andrews and Merca [Reference Andrews and Merca3] proved the following theorem which is a truncated version of Euler's pentagonal number theorem.

Theorem 1.1 Andrews and Merca [Reference Andrews and Merca3]

Let $k$ be a positive integer. Then,

(1.3)\begin{equation} \displaystyle\frac{1}{(q;q)_\infty}\displaystyle\sum_{n=0}^{k-1 }({-}1)^{n} q^{n(3n+1)/2}(1-q^{2n+1}) = 1+({-}1)^{k-1 } q^{k(k-1)/2}\sum_{n=k}^{\infty}\frac{q^{(k+1)n}}{(q;q)_n} \left[\begin{array}{@{}c@{}} n-1 \\ k -1 \\ \end{array}\right]. \end{equation}

In addition, Andrews and Merca [Reference Andrews and Merca3] presented a partition-theoretic interpretation of the truncated sum in (1.3), namely,

(1.4)\begin{equation} ({-}1)^{k-1} \displaystyle\sum_{j=1-k}^{k} ({-}1)^{j } p\left(n-\displaystyle\frac{j(3j-1)}{2}\right) =M_k(n), \end{equation}

where $M_k(n)$ is the number of partitions of $n$ in which $k$ is the least integer that is not a part and there are more parts $>k$ than there are $< k$.

Apart from Euler's pentagonal number theorem, there are two other classical theta identities due to Gauss:

(1.5)\begin{equation} \displaystyle\frac{( q;q)_\infty}{({-}q;q)_\infty} =1+2\sum_{n=1}^{\infty} ({-}1)^{n} q^{n^{2}} \end{equation}

and

(1.6)\begin{equation} \displaystyle\frac{(q^{2};q^{2})_\infty}{({-}q;q^{2})_\infty} =\displaystyle\sum_{n=0}^{\infty} ({-}q)^{n(n+1)/2}. \end{equation}

Motivated by Andrews and Merca's work, Guo and Zeng [Reference Guo and Zeng11] established the truncated versions of (1.5) and (1.6) in 2013. More precisely, they proved that for $k\geq 1$,

(1.7)\begin{equation} \displaystyle\frac{({-}q;q)_\infty}{ (q;q)_\infty} \left(1+2\displaystyle\sum_{j=1}^{k} ({-}1)^{j} q^{j^{2}}\right)=1+({-}1)^{k} \sum_{n=k+1}^{\infty} \frac{({-}q;q)_k({-}1;q)_{n-k} q^{n(k+1)}}{ (q;q)_n } \left[\begin{array}{@{}c@{}} n-1 \\ k \\ \end{array}\right] \end{equation}

and

(1.8)\begin{align} & \displaystyle\frac{({-}q;q^{2})_\infty}{ (q^{2};q^{2})_\infty} \displaystyle\sum_{j=0}^{ k-1} ({-}1)^{j} q^{2j^{2}+j}(1-q^{2j+1})\nonumber\\ & \quad=1+({-}1)^{k-1} \displaystyle\sum_{n=k }^{\infty} \displaystyle\frac{({-}q;q^{2})_k({-}q;q^{2})_{n-k} q^{ (2k+2)n-k}}{ (q^{2};q^{2})_n } \left[\begin{array}{@{}c@{}} n-1 \\ k-1 \\ \end{array}\right]_{q^{2}}. \end{align}

Recently, using the Rogers–Fine identity, Andrews and Merca [Reference Andrews and Merca4] proved the following two nice identities:

(1.9)\begin{align} & \displaystyle\frac{({-}q;q)_\infty}{ (q;q)_\infty} \left(1+2\displaystyle\sum_{j=1}^{k} ({-}1)^{j} q^{j^{2}}\right) =1+2({-}1)^{k}\frac{({-}q;q)_k}{ (q;q)_k }\sum_{j=0}^{\infty} \frac{q^{(k+1)(k+j+1)}({-}q^{k+j+2};q)_\infty }{ (1-q^{k+j+1})(q^{k+j+2};q)_\infty}, \end{align}
(1.10)\begin{align} & \frac{({-}q;q^{2})_\infty}{ (q^{2};q^{2})_\infty} \sum_{j=0}^{ k-1} ({-}1)^{j} q^{2j^{2}+j}(1-q^{2j+1}) \nonumber\\ & \quad = 1+({-}1)^{k-1}\frac{({-}q;q^{2})_k}{(q^{2};q^{2})_{k-1}}\sum_{j=0}^{\infty}\frac{q^{k(2j+2k+1)}({-}q^{2k+2j+3} ;q^{2})_\infty}{ (q^{2k+2j+2};q^{2})_\infty}, \end{align}

where $k$ is a positive integer. In addition, Wang and Yee [Reference Wang and Yee27] derived the following truncated identity of (1.5):

(1.11)\begin{align} & \displaystyle\frac{({-}q;q)_\infty}{ (q;q)_\infty} \displaystyle\sum_{n=0}^{k-1} ({-}1)^{n}(1-q^{2n+1}) q^{n^{2}} \nonumber\\ & \quad =1+({-}1)^{k-1}q^{{-}k} \sum_{n=k}^{\infty} \frac{({-}q;q)_k({-}1;q)_{n-k}q^{(k+1)n} }{(q;q)_n} \left[\begin{array}{@{}c@{}} n-1 \\ k-1 \\ \end{array}\right], \end{align}

where $k$ is a positive integer. Very recently, Xia et al. [Reference Xia, Yee and Zhao29] established new truncated theorems of the three classical theta series identities in (1.2), (1.5), and (1.6) by taking different truncated series than the ones chosen by Andrews–Merca and Guo–Zeng. More precisely, they proved that for $k\ge 1$,

(1.12)\begin{align} & \displaystyle\frac{1}{(q;q)_\infty}\displaystyle\sum_{n=0}^{k-1} (1-q^{n+1})(1-q^{2n+2}) ({-}1)^{n} q^{3n(n+1)/2}\nonumber\\ & \quad =1+({-}1)^{k-1} q^{k(k-1)/2} \sum_{n=k}^{\infty} \frac{q^{n(k+2)}}{(q;q)_n}{\left[\begin{matrix} {n-1} \\ {k-1} \end{matrix} \right]}, \end{align}
(1.13)\begin{align} & \displaystyle\frac{({-}q;q)_\infty }{ (q;q)_\infty} \displaystyle\sum_{n=0}^{k-1} (1-q^{ n+ 1 })^{2} ({-}1)^{n} q^{n^{2}+ n }\nonumber\\ & \quad =1+ ({-}1)^{k-1} \frac{({-}q;q)_{k}} {(q;q)_{k-1}} \sum_{j=0}^{\infty} \frac{q^{k(k+1+j) }({-}q^{k+2+j};q)_\infty}{ (q^{k+2+j};q)_\infty}, \end{align}
(1.14)\begin{align} & \frac{({-}q;q^{2})_\infty}{ (q^{2};q^{2})_\infty} \sum_{n=0}^{k-1} (1-q^{2n+2})(1-q^{4n+4}) ({-}1)^{n} q^{2n^{2}+n} \nonumber\\ & \quad = 1+q +({-}1)^{k-1}\frac{({-}q;q^{2})_{k+1}}{(q^{2};q^{2})_{k-1}} \sum_{j=0}^{\infty} \frac{q^{k(2k+2j+1)} ({-}q^{2k+2j+3};q^{2})_\infty}{ (q^{2k+2j+4};q^{2})_\infty}. \end{align}

Other recent investigations on the truncated theta series can be found in the papers of Ballantine et al. [Reference Ballantine, Merca, Passary and Yee5], Burnette and Kolitsch [Reference Burnette and Kolitsch6], Chan et al. [Reference Chan, Ho and Mao7], Chern [Reference Chern8], Kolitsch [Reference Kolitsch13], Mao [Reference Mao16, Reference Mao17], Merca [Reference Merca18–Reference Merca20, Reference Merca, Wang and Yee22], Merca and Yee [Reference Merca and Yee21], Seo and Yee [Reference Seo and Yee23, Reference Seo and Yee24], Wang and Yee [Reference Wang and Yee25, Reference Wang and Yee27], Xia [Reference Xia28], Xia and Zhao [Reference Xia and Zhao30], Yao [Reference Yao31], and Zhang [Reference Zhang33].

The aim of this paper is to deduce new truncated sums for the three classical theta series identities in (1.2), (1.5), and (1.6) by using a Bailey pair given by Lovejoy [Reference Lovejoy15], which imply the results proved by Andrews–Merca [Reference Andrews and Merca3], Wang–Yee [Reference Wang and Yee27], and Xia–Yee–Zhao [Reference Xia, Yee and Zhao29].

The main results of this paper can be stated as follows.

Theorem 1.2 Let $r$ and $k$ be positive integers. Then,

(1.15)\begin{align} & \displaystyle\frac{1}{(q;q)_\infty}\displaystyle\sum_{n=0}^{k-1} ( 1-q^{2n+r}) ({-}1)^{n} q^{n(3n+2r-1)/2} (q^{n+1};q)_{r-1}\nonumber\\ & \quad=1+({-}1)^{k-1} q^{k(k-1)/2}\displaystyle\sum_{n=k}^{\infty}\displaystyle\frac{q^{(k+r)n}}{(q;q)_n} \left[\begin{array}{@{}c@{}} n-1 \\ k-1 \\ \end{array}\right]. \end{align}

Theorem 1.3 Let $r$ and $k$ be positive integers. Then,

(1.16)\begin{align} & \displaystyle\frac{({-}q;q)_\infty }{ (q;q)_\infty}\displaystyle\sum_{n=0}^{k-1} (1-q^{2n+2r-1}) ({-}1)^{n} q^{n^{2}+(r-1)n} ( q^{n+1};q)_{2r-2}\nonumber\\ & \quad=({-}q;q)_{r-1} +\displaystyle\frac{({-}q;q)_{k+r-1}({-}1)^{k-1} }{(q;q)_{k-1}} \displaystyle\sum_{n=0}^{\infty} \frac{({-}q^{k+n+r};q)_\infty q^{k(k+r+n-1)}}{(q^{k+n+2r-1}; q)_\infty}, \end{align}
(1.17)\begin{align} & \frac{({-}q;q)_\infty }{ (q;q)_\infty} \sum_{n=0}^{k-1} (1-q^{ n+ r }) ({-}1)^{n} q^{n^{2}+ rn} ( q^{n+1};q)_{2r-1}\nonumber\\ & \quad=({-}q;q)_{r-1} +\frac{({-}q;q)_{k+r-1}({-}1)^{k-1} }{(q;q)_{k-1}} \sum_{n=0}^{\infty} \frac{({-}q^{k+n+r+1};q)_\infty q^{k(k+r+n)}}{(q^{k+n+2r}; q)_\infty}. \end{align}

Theorem 1.4 Let $r$ and $k$ be positive integers. Then,

(1.18)\begin{align} & \displaystyle\frac{({-}q;q^{2})_\infty}{(q^{2};q^{2})_\infty} \displaystyle\sum_{n=0}^{k-1} ({-}1)^{n} q^{2n^{2}+(2r-1)n}(1-q^{2n+2r-1}) (q^{2n+2};q^{2})_{2r-2}\nonumber\\ & \quad = ({-}q;q^{2})_{r-1} +\displaystyle\frac{({-}1)^{k-1} ({-}q;q^{2})_{k+r-1} }{ (q^{2};q^{2})_{k-1} } \displaystyle\sum_{n=0}^{\infty} \frac{ ({-}q^{2k+2n+2r+1};q^{2})_\infty q^{k(2k+2r+2n-1) }}{(q^{2k+4r+2n-2};q^{2})_\infty}, \end{align}
(1.19)\begin{align} & \frac{({-}q;q^{2})_\infty }{(q^{2};q^{2})_\infty}\sum_{n=0}^{k-1} ({-}1)^{n} q^{2n^{2}+(2r-1)n}(1-q^{4n+4r}) ( q^{2n+2};q^{2})_{2r-1}\nonumber\\ & \quad = ({-}q;q^{2})_{r} +\frac{({-}1)^{k-1} ({-}q;q^{2})_{k+r} }{ (q^{2};q^{2})_{k-1} } \sum_{n=0}^{\infty}\frac{ ({-}q^{2k+2n+2r+1};q^{2})_\infty q^{k(2k+2r+2n-1) }}{(q^{2k+4r+2n};q^{2})_\infty}. \end{align}

Remark Setting $r=1$ in (1.15) and (1.17)–(1.19), we arrive at (1.3), (1.13), (1.10), and (1.14), respectively. Putting $r=1$ in (1.16) and utilizing the $q$-binomial theorem stated in § 2, we can deduce (1.11). Taking $r=2$ in (1.15) yields (1.12).

Note that identities (1.5) and (1.6) are related to the overpartition function $\bar {p}(n)$ and the partitions with odd parts distinct function ${\rm pod}(n)$ (the pod function for short), which are two important partition functions. Recall that an overpartition of $n$ is a partition of $n$ where the first occurrence of each distinct part may be overlined. For instance, there are 14 overpartitions of 4:

\begin{align*} & 4,\ \bar{4},\ 3+1,\ \bar{3}+1,\ 3+\bar{1}, \ \bar{3}+\bar{1}, 2+2, \ \bar{2}+2, \\ & 2+1+1,\ \bar{2}+1+1,\ 2+\bar{1}+1, \ \bar{2}+\bar{1}+1, \ 1+1+1+1,\ \bar{1}+1+1+1. \end{align*}

For convenience, define $\bar {p}(0)$ to be $1$. The generating function for $\bar {p}(n)$ is

(1.20)\begin{equation} \displaystyle\sum_{n=0}^{\infty} \bar{p}(n)q^{n} =\displaystyle\frac{({-}q;q)_\infty}{ (q;q)_\infty}. \end{equation}

For more details on overpartition, see [Reference Corteel and Lovejoy9]. A partition with odd parts distinct is a partition in which odd parts are distinct but even parts are unrestricted. Let ${\rm pod}(n)$ denote the number of such partitions of $n$ with ${\rm pod}(0) = 1$. The generating function of ${\rm pod}(n)$ is

(1.21)\begin{equation} \displaystyle\sum_{n=0}^{\infty} {\rm pod}(n)q^{n} =\displaystyle\frac{({-}q;q^{2})_\infty}{ (q^{2};q^{2} )_\infty}. \end{equation}

For more details on partitions with odd parts distinct, see [Reference Hirschhorn and Sellers12]. Since the generating functions for $\bar {p}(n)$ and ${\rm pod}(n)$ are the reciprocals of the products in (1.5) and (1.6), Guo and Zeng [Reference Guo and Zeng11] established infinite families of linear inequalities for $\bar {p}(n)$ and $pod(n)$ based on (1.7) and (1.8). More precisely, they proved that, for $n\geq 1$ and $k\geq 1$,

(1.22)\begin{equation} ({-}1)^{k}\left(\bar{p}(n)+2\displaystyle\sum_{j=1}^{k} ({-}1)^{j }\bar{p}(n-j^{2})\right)\geq 0 \end{equation}

and

(1.23)\begin{equation} ({-}1)^{k-1}\displaystyle\sum_{j=0}^{k-1} ({-}1)^{j}( {\rm pod}(n-2j^{2}-j)-{\rm pod}(n-(j+1)(2j+1)))\geq 0. \end{equation}

Motivated by the fact that (1.2) has a partition-theoretic interpretation, namely $M_k(n)$, Guo and Zeng [Reference Guo and Zeng11] posed an open problem to give partition-theoretic interpretations for the two truncated sums in (1.7) and (1.8). Recently, based on (1.9) and (1.10), Andrews and Merca [Reference Andrews and Merca4] provided an affirmative answer to the question by introducing new partition functions $\overline {M}_k(n)$ and $MP_k(n)$. For more details, see the paper of Andrews and Merca [Reference Andrews and Merca4]. Note that identity (1.11) yields

(1.24)\begin{equation} \overline{p}(n-k^{2}) + ({-}1)^{k-1} \displaystyle\sum_{j={-}k}^{k} ({-}1)^{j} \overline{p}(n-j^{2}) \geq 0, \end{equation}

which was first conjectured by Guo and Zeng [Reference Guo and Zeng11], proved independently by Mao [Reference Mao16] and Yee [Reference Yee32]. Based on (1.12)–(1.14), Xia et al. [Reference Xia, Yee and Zhao29] derived infinite families of linear inequalities involving $p(n)$, $\bar {p}(n)$, and ${\rm pod}(n)$ which imply the positivity of the truncated sums in (1.4) and (1.24).

In this paper, we will also define several partition functions to provide partition-theoretic interpretations for the coefficients of the right-hand sides of (1.15)–(1.19). We defer their definitions to § 3–5 since the descriptions of those partitions are complex.

The rest of this paper is organized as follows. In § 2, we recall some basic facts on basic hypergeometric series and Bailey pairs and pose a lemma. In § 3–5, we prove Theorems 1.2–1.4 based on some formulas given in § 2 and present their corollaries.

2. Preliminaries

In order to prove the main results of this paper, we recall some formulas on basic hypergeometric series and some facts on Bailey pairs, and then present a lemma.

First of all, the $q$-binomial theorem [Reference Gasper and Rahman10, p. 354, (II.3)] can be stated as follows:

(2.1)\begin{equation} \displaystyle\sum_{n=0}^{\infty} \displaystyle\frac{(a;q)_n}{(q;q)_n} z^{n}=\frac{(az;q)_\infty}{(z;q)_\infty}. \end{equation}

Setting $a=q^{-n}$ in (2.1) yields

(2.2)\begin{equation} \displaystyle\sum_{k=0}^{n} \displaystyle\frac{(q^{{-}n};q)_k }{( q;q)_k} z^{k}=(zq^{{-}n};q)_n. \end{equation}

Moreover, we require the following identity [Reference Gasper and Rahman10, p. 351, (I. 8)]

(2.3)\begin{equation} (zq^{{-}n};q)_n=(q/z;q)_n( -\displaystyle\frac{z}{q})^{n} q^{-(n^{2}-n)/2}.\end{equation}

and the fact that if $m\geq 1$ and $0< j\leq m$

(2.4)\begin{equation} (q^{j-m};q)_m=0. \end{equation}

Next, we recall some basic facts on Bailey pairs which will be used to prove an important lemma. Let $\{\alpha _n\}_{n\geq 0}$ and $\{\beta _n\}_{n\geq 0}$ be two sequences. Then the pair of sequence $(\alpha _n,\, \beta _n)$ is called a Bailey pair relative to $(a,\, q)$ if $\alpha _n$ and $\beta _n$ are related by the following identity [Reference Andrews1, Chapter 3]: for $n\geq 0$

(2.5)\begin{equation} \beta_n=\displaystyle\sum_{r=0}^{n} \displaystyle\frac{\alpha_r}{(q;q)_{n-r} (aq;q)_{n+r}}. \end{equation}

Finally, we prove a lemma, which will play a central role in our proofs.

Lemma 2.1 Let $m$ be a non-negative integer. Then

(2.6)\begin{align} & \displaystyle\frac{(aq;q)_m ({-}1)^{m} q^{m(m+1)/2}}{(q;q)_m}\displaystyle\sum_{n=0}^{m}\frac{(q^{{-}m},aq^{m+1},aq/bc;q)_n}{ (q,aq/b,aq/c;q)_n}\nonumber\\ & \quad= \displaystyle\sum_{n=0}^{m}\displaystyle\frac{(1-aq^{2n})(a,b,c;q)_n ({-}1)^{n} q^{n(n-1)/2}}{ (1-a)(q,aq/b,aq/c;q)_n}\bigg(\frac{aq}{bc}\bigg)^{n}. \end{align}

Proof. In 2012, Lovejoy [Reference Lovejoy15, (2.25) and (2.26)] found that $(\alpha _n,\, \beta _n)_{n\geq 0}$ is a Bailey pair relative to $(a,\,q)$, where

(2.7)\begin{equation} \alpha_n=\displaystyle\frac{(1-aq^{2n})(a;q)_n(b;q)_n(c;q)_n }{(1-a)(q;q)_n(aq/b;q)_n(aq/c;q)_n} ({-}1)^{n} q^{n(n-1)/2}\bigg(\frac{aq}{bc}\bigg)^{n} \end{equation}

and

(2.8)\begin{equation} \beta_n=\displaystyle\frac{(aq/bc;q)_n}{ (q;q)_n (aq/b;q)_n (aq/c;q)_n}. \end{equation}

Recently, based on an identity due to Liu [Reference Liu14], Wang and Yee [Reference Wang and Yee26] proved that if $(\alpha _n,\,\beta _n)$ is a Bailey pair relative to $(a,\, q)$ and $m$ is a non-negative integer, then

(2.9)\begin{equation} \displaystyle\frac{(aq;q)_m({-}1)^{m} q^{m(m+1)/2}}{ (q;q)_m}\displaystyle\sum_{n=0}^{m} (q^{{-}m};q)_n(aq^{m+1};q)_n\beta_n =\sum_{n=0}^{m} \alpha_n. \end{equation}

Combining (2.7)–(2.9), we arrive at (2.6). This completes the proof.

Setting $b \rightarrow \infty$ in (2.6), we get the following corollary:

Corollary 2.2 If $m$ is a non-negative integer, then

(2.10)\begin{equation} \displaystyle\frac{(aq;q)_m ({-}1)^{m} q^{m(m+1)/2}}{(q;q)_m} \displaystyle\sum_{n=0}^{m} \frac{(q^{{-}m},aq^{m+1} ;q)_n}{ (q, aq/c;q)_n} = \sum_{n=0}^{m} \frac{(1-aq^{2n})(a, c;q)_n q^{n^{2}}}{ (1-a)(q,aq/c;q)_n}\bigg(\frac{a }{ c}\bigg)^{n}. \end{equation}

3. Ordinary partitions

The aim of this section is to prove Theorem 1.2 and present its corollary.

3.1. Proof of Theorem 1.2

Taking $c \rightarrow \infty$ in (2.10) yields

(3.1)\begin{align} & \displaystyle\frac{(aq;q)_m ({-}1)^{m} q^{m(m+1)/2}}{(q;q)_m} \displaystyle\sum_{n=0}^{m} \frac{(q^{{-}m};q)_n(aq^{m+1} ;q)_n}{ (q ;q)_n}\nonumber\\ & \quad = \sum_{n=0}^{m} \frac{(1-aq^{2n})(a ;q)_n ({-}1)^{n} q^{n(3n-1)/2}}{ (1-a)(q;q)_n}a^{n}, \end{align}

where $m$ is a non-negative integer. Let $r$ be a positive integer. If we set $a=q^{r}$ in (3.1) and then multiply both sides by $(q;q)_r$, we arrive at

(3.2)\begin{align} & ({-}1)^{m} q^{m(m+1)/2}\displaystyle\sum_{n=0}^{m} \displaystyle\frac{(q^{{-}m};q)_n(q^{m+1};q)_{n+r}}{(q;q)_n}\nonumber\\ & \quad =\sum_{n=0}^{m} ({-}1)^{n} q^{n(3n+2r-1)/2} (1-q^{2n+r})(q^{n+1};q)_{r-1}. \end{align}

In view of (3.2),

(by (2.1))\begin{align} & \displaystyle\frac{1}{(q;q)_\infty} \displaystyle\sum_{n=0}^{m} ({-}1)^{n} q^{n(3n+2r-1)/2} (1-q^{2n+r})( q^{n+1};q)_{r-1}\nonumber\\ & \quad=\frac{1}{(q;q)_\infty} ({-}1)^{m} q^{m(m+1)/2}\sum_{n=0}^{m} \frac{(q^{{-}m};q)_n(q^{m+1};q)_{n+r}} {(q;q)_n}\nonumber\\ & \quad=\frac{({-}1)^{m} q^{m(m+1)/2}} {(q;q)_m} \sum_{n=0}^{m} \frac{(q^{{-}m};q)_n }{(q;q)_n (q^{m+n+r+1};q)_\infty }\nonumber\\ & \quad=\frac{({-}1)^{m} q^{m(m+1)/2}} {(q;q)_m} \sum_{n=0}^{m} \frac{(q^{{-}m};q)_n }{(q;q)_n }\sum_{j=0}^{\infty} \frac{q^{(m+n+r+1)j}}{(q;q)_j}\end{align}
(by (2.2))\begin{align} & \quad=\frac{({-}1)^{m} q^{m(m+1)/2}} {(q;q)_m} \sum_{j=0}^{\infty} \frac{q^{(m +r+1)j}}{(q;q)_j}\sum_{n=0}^{m}\frac{(q^{{-}m};q)_n }{(q;q)_n }q^{nj} \nonumber\\ & \quad=\frac{({-}1)^{m} q^{m(m+1)/2}} {(q;q)_m} \sum_{j=0}^{\infty} \frac{q^{(m +r+1)j}}{(q;q)_j} (q^{j-m};q)_m \end{align}
(by (2.4))\begin{align}& \quad=\frac{({-}1)^{m} q^{m(m+1)/2}} {(q;q)_m} ( (q^{ {-}m};q)_m+\sum_{j=m+1}^{\infty} \frac{q^{(m +r+1)j}}{(q;q)_j} (q^{j-m};q)_m )\end{align}
(by (2.3))\begin{align}& \quad=1+({-}1)^{m} q^{m(m+1)/2}\sum_{j=m+1}^{\infty}\frac{q^{(m+r+1)j}}{(q;q)_j}\left[\begin{array}{@{}c@{}} j-1 \\ m \\ \end{array}\right], \end{align}

which yields (1.15) after replacing $m$ by $k-1$. This completes the proof of Theorem 1.2.

3.2. A corollary of Theorem 1.2

From Theorem 1.2, we deduce the following corollary.

Corollary 3.1 Let $r$ be a positive integer and define

(3.3)\begin{equation} \displaystyle\sum_{j\geq 0} s_{r}(i,j) q^{j}:= ( 1-q^{2i+r}) ( q^{i+1};q)_{r-1} . \end{equation}

For $n\geq 1$ and $k\geq 1$,

(3.4)\begin{equation} ({-}1)^{k-1} \displaystyle\sum_{i=0}^{k-1} ({-}1)^{i} \sum_{j\geq 0} s_r(i,j) p\left(n-\displaystyle\frac{i(3i+2r-1)}{2}-j\right) =A(k,r,n), \end{equation}

where $A(k,\,r,\,n)$ is the number of partitions of $n$ in which $k$ is the least integer such that

  1. (1) every part $< k$ appears at least once;

  2. (2) for any integer $t$ with $0\leq t\leq r-1$,$k+t$ is not a part;

  3. (3) there are more parts $>k+r-1$ than there are $< k$.

For example, setting $r=3$ in Corollary 3.1 and simplifying yields

(3.5)\begin{align} & p(n-\displaystyle\frac{ k(3k+5)}{2})+p(n-\frac{3k^{2}+5k+2}{2})-p\left(n-\frac{3k^{2}+7k+2}{2}\right) \nonumber\\ & \quad +({-}1)^{k-1} \displaystyle\sum_{i={-}k}^{k} ({-}1)^{i} p\left(n-\displaystyle\frac{i(3i+1)}{2}\right)=A(k,3,n) . \end{align}

Taking $k=2$ in (3.5), we deduce that for $n\geq 1$,

(3.6)\begin{align} & -p(n)+p(n-1)+p(n-2) -p(n-5)-p(n-7)\nonumber\\ & \quad +p(n-11)+p(n-12)-p(n-14) =A(2,3,n). \end{align}

If $n=17$, then

\begin{align*} & -p(17)+p(16)+p(15)-p(12)-p(10)+p(6) +p(5)-p(3)\nonumber\\ & \quad={-}297+231+176-77-42+11+7-3\nonumber\\ & \quad=6=A(2,3,17) \end{align*}

since the six partitions in questions are

\[5+5+5+1+1,\ 6+5+5+1, \ 8+8+1,\ 9+7+1,\ 10+6+1,\ 11+5+1.\]

Proof of Corollary 3.1 First, we note that $\frac {q^{n(k+r)}}{(q;q)_n}$ is the generating function for partitions into $n$ parts $\geq k+r$. Furthermore,

\[q^{k(k-1)/2}\left[\begin{matrix} n-1 \\ k-1 \end{matrix} \right]= q^{1+2+\cdots +k-1}\left[\begin{matrix} n-1 \\ k-1\end{matrix} \right]\]

is the generating function for partitions into $k-1$ distinct parts $\leq n-1$, and by conjugation, this is the generating function for partitions into at most $n-1$ parts with every integer $< k$ appearing as a part. Thus,

(3.7)\begin{equation} \displaystyle\sum_{n=1}^{\infty} A(k,r,n)q^{n} =q^{k(k-1)/2} \sum_{n=k}^{\infty} \displaystyle\frac{q^{n(k+r)}}{ (q;q)_n}\left[ \begin{matrix} n-1 \\ k-1 \end{matrix}\right].\end{equation}

It follows from (1.1), (1.15), (3.3), and (3.7) that

(3.8)\begin{align} & ({-}1)^{k-1}\displaystyle\sum_{m=0}^{\infty} p(m)q^{m} \sum_{i=0}^{k-1} ({-}1)^{i} q^{i(3i+2r-1)/2}\sum_{j\geq 0} s_{r}(i,j)q^{j}\nonumber\\ & \quad=({-}1)^{k-1}+ \displaystyle\sum_{n=1}^{\infty} A(k,r,n) q^{n}. \end{align}

Equating the coefficients of $q^{n}$ on both sides of (3.8), we arrive at (3.4). This completes the proof.

4. Overpartitions

The objective of this section is to provide a proof of Theorem 1.3 and present its corollary.

4.1. Proof of Theorem 1.3

Taking $a=q^{2r-1}$ and $c=-q^{r}$ in (2.10) and then multiplying both sides by $(q;q)_{2r-1}$ yields

(4.1)\begin{align} & ({-}1)^{m} q^{m(m+1)/2}\displaystyle\sum_{n=0}^{m} \displaystyle\frac{(q^{{-}m};q)_n (q^{m+1};q)_{n+2r-1}}{ (q;q)_n({-}q^{r};q)_n}\nonumber\\ & \quad=\displaystyle\sum_{n=0}^{m} (1-q^{2n+2r-1}) ({-}1)^{n} q^{n^{2}+(r-1)n} (q^{n+1};q)_{2r-2}. \end{align}

In view of (4.1),

(by (2.1))\begin{align} & \displaystyle\frac{({-}q;q)_\infty} { (q;q)_\infty} \displaystyle\sum_{ n=0}^{m} (1-q^{2n+2r-1})({-}1)^{n} q^{n^{2}+(r-1)n} (q^{n+1};q)_{2r-2}\nonumber\\ & \quad=\frac{({-}q;q)_\infty} { (q;q)_\infty}({-}1)^{m} q^{m(m+1)/2}\sum_{n=0}^{m}\frac{(q^{{-}m};q)_n (q^{m+1};q)_{n+2r-1}}{ (q;q)_n({-}q^{r};q)_n}\nonumber\\ & \quad=\frac{({-}q;q)_{r-1}} { (q;q)_m}({-}1)^{m} q^{m(m+1)/2}\sum_{n=0}^{m}\frac{(q^{{-}m};q)_n ({-}q^{n+r};q)_\infty}{ (q;q)_n(q^{m+n+2r};q)_\infty}\nonumber\\ & \quad=\frac{({-}q;q)_{r-1}} { (q;q)_m}({-}1)^{m} q^{m(m+1)/2}\sum_{n=0}^{m}\frac{(q^{{-}m};q)_n }{ (q;q)_n } \sum_{j=0}^{\infty}\frac{({-}q^{{-}m-r};q)_j}{(q;q)_j} q^{(m+n+2r)j} \end{align}
(by (2.2))\begin{align}& \quad=\frac{({-}q;q)_{r-1}} { (q;q)_m}({-}1)^{m} q^{m(m+1)/2} \sum_{j=0}^{\infty}\frac{({-}q^{{-}m-r};q)_j }{(q;q)_j}q^{(m+2r)j}\sum_{n=0}^{m} \frac{(q^{{-}m};q)_n }{ (q;q)_n } q^{nj}\nonumber \\ & \quad=\frac{({-}q;q)_{r-1}} { (q;q)_m}({-}1)^{m} q^{m(m+1)/2} \sum_{j=0}^{\infty}\frac{({-}q^{{-}m-r};q)_j }{(q;q)_j}q^{(m+2r)j} (q^{j-m};q)_m \end{align}
(by (2.4))\begin{align}& \quad=\frac{({-}q;q)_{r-1}} { (q;q)_m}({-}1)^{m} q^{m(m+1)/2}((q^{ {-}m};q)_m + \sum_{j=m+1}^{\infty}\frac{({-}q^{{-}m-r};q)_j }{(q;q)_j}q^{(m+2r)j} (q^{j-m};q)_m )\end{align}
(by (2.3))\begin{align} & \quad=\frac{({-}q;q)_{r-1}} { (q;q)_m}({-}1)^{m} q^{m(m+1)/2}((q^{ {-}m};q)_m\nonumber\\ & \qquad + \sum_{j=0}^{\infty}\frac{({-}q^{{-}m-r};q)_{m+j+1} }{(q;q)_{m+j+1}}q^{(m+2r)(m+j+1)} (q^{j+1};q)_m )\nonumber\\ & \quad=({-}q;q)_{r-1} +\frac{({-}q;q)_{r-1}({-}1)^{m} q^{\frac{m(m+1)}{2}} ({-}q^{{-}m-r};q)_{m+1} }{(q;q)_m}\nonumber\\ & \qquad \times \sum_{j=0}^{\infty}\frac{({-}q^{1-r};q)_j q^{(m+2r)(m+j+1)} }{(q;q)_j(1-q^{m+j+1})}\nonumber\\ & \quad=({-}q;q)_{r-1} +\frac{({-}q;q)_{m+r}({-}1)^{m} q^{-(m+1)r} }{(q;q)_m}\sum_{j=0}^{\infty}\frac{({-}q^{1-r};q)_j q^{(m+2r)(m+j+1)} }{(q;q)_j(1-q^{m+j+1})} \end{align}
(by (2.1))\begin{align}& \quad=({-}q;q)_{r-1} +\frac{({-}q;q)_{m+r}({-}1)^{m} q^{-(m+1)r} }{(q;q)_m}\nonumber\\ & \qquad \times \sum_{j=0}^{\infty}\frac{({-}q^{1-r};q)_j q^{(m+2r)(m+j+1)} }{(q;q)_j }\sum_{n=0}^{\infty} q^{(m+j+1)n}\nonumber\\ & \quad=({-}q;q)_{r-1} +\frac{({-}q;q)_{m+r}({-}1)^{m} }{(q;q)_m} \sum_{n=0}^{\infty} q^{(m+1)(m+r+n)}\sum_{j=0}^{\infty} \frac{({-}q^{1-r};q)_j q^{(m+2r+n)j} }{(q;q)_j }\nonumber\\ & \quad=({-}q;q)_{r-1} +\frac{({-}q;q)_{m+r}({-}1)^{m} }{(q;q)_m} \sum_{n=0}^{\infty}\frac{({-}q^{m+n+r+1};q)_\infty q^{(m+1)(m+r+n)}}{(q^{m+n+2r};q)_\infty},\end{align}

which yields (1.16) after replacing $m$ by $k-1$ in the above identity.

Setting $a=q^{2r}$ and $c=-q^{r}$ in (2.10) then multiplying both sides by $\frac {(q;q)_{2r}}{1+q^{r}}$, we arrive at

(4.2)\begin{align} & ({-}1)^{m} q^{m(m+1)/2}\displaystyle\sum_{n=0}^{m}\displaystyle\frac{(q^{{-}m};q)_n (q^{m+1};q)_{n+2r }}{ (q;q)_n({-}q^{r };q)_{n+1}}\nonumber\\ & \quad = \displaystyle\sum_{n=0}^{m} (1-q^{ n+ r }) ({-}1)^{n} q^{n^{2}+ nr}(q^{n+1};q)_{2r-1} . \end{align}

Therefore,

(by (2.1))\begin{align}& \displaystyle\frac{({-}q;q)_\infty }{(q;q)_\infty} \displaystyle\sum_{n=0}^{m} (1-q^{ n+ r }) ({-}1)^{n} q^{n^{2}+ nr}(q^{n+1};q)_{2r-1} \nonumber\\ & =\displaystyle\frac{({-}q;q)_\infty}{ (q;q)_\infty}({-}1)^{m} q^{m(m+1)/2}\displaystyle\sum_{n=0}^{m} \frac{(q^{{-}m};q)_n (q^{m+1};q)_{n+2r }}{ (q;q)_n({-}q^{r };q)_{n+1}}\nonumber\\ & =\frac{({-}q;q)_{r-1} ({-}1)^{m} q^{m(m+1)/2}}{(q;q)_m} \sum_{n=0}^{m} \frac{(q^{{-}m};q)_n ({-}q^{n+r+1};q)_\infty}{ (q;q)_n (q^{m+n+2r+1};q)_\infty} \nonumber\\ & =\frac{({-}q;q)_{r-1} ({-}1)^{m} q^{m(m+1)/2}}{(q;q)_m}\sum_{n=0}^{m} \frac{(q^{{-}m};q)_n }{ (q;q)_n } \sum_{j=0}^{\infty}\frac{({-}q^{{-}m-r};q)_j}{ (q;q)_j} q^{(m+n+2r+1)j} \end{align}
(by (2.2))\begin{align}& \quad=\frac{({-}q;q)_{r-1} ({-}1)^{m} q^{m(m+1)/2}}{(q;q)_m}\sum_{j=0}^{\infty} \frac{({-}q^{{-}m-r};q)_j}{ (q;q)_j} q^{(m+2r+1)j}\sum_{n=0}^{m} \frac{(q^{{-}m};q)_n }{ (q;q)_n }q^{nj}\nonumber\\ & \quad=\frac{({-}q;q)_{r-1} ({-}1)^{m} q^{m(m+1)/2}}{(q;q)_m} \sum_{j=0}^{\infty}\frac{({-}q^{{-}m-r};q)_j}{ (q;q)_j} q^{(m+2r+1)j} (q^{j-m};q)_m \end{align}
(by (2.4))\begin{align}& \quad=\frac{(\!{-}q;q)_{r-1} (\!{-}1)^{m} q^{m(m+1)/2}}{(q;q)_m}((q^{{-}m};q)_m\!+\!\sum_{j=m+1}^{\infty} \frac{({-}q^{{-}m-r};q)_j}{ (q;q)_j} q^{(m+2r+1)j} (\!q^{j-m};q)_m\!) \end{align}
(by (2.3))\begin{align}& \quad=\frac{({-}q;q)_{r-1} ({-}1)^{m} q^{m(m+1)/2}}{(q;q)_m}((q^{{-}m};q)_m\nonumber\\& \qquad +\sum_{j=0}^{\infty}\frac{({-}q^{{-}m-r};q)_{m+1+j}}{ (q;q)_{m+1+j}} q^{(m+2r+1)(m+1+j)} (q^{j+1};q)_m)\nonumber\\ & \quad= ({-}q;q)_{r-1}\left(1+\frac{({-}1)^{m} q^{m(m+1)/2} ({-}q^{{-}m-r};q)_{m+1} }{ (q;q)_m}\right.\nonumber\\& \qquad \left.\times \sum_{j=0}^{\infty}\frac{({-}q^{1-r};q)_{j}}{ (q;q)_{j}(1-q^{m+j+1})} q^{(m+2r+1)(m+1+j)}\right)\nonumber\\ & \quad= ({-}q;q)_{r-1} +\frac{({-}1)^{m} ({-}q;q)_{m+r} q^{-(m+1)r}}{(q;q)_m}\nonumber\\& \qquad \times\sum_{j=0}^{\infty}\frac{({-}q^{1-r};q)_{j}}{ (q;q)_{j}} q^{(m+2r+1)(m+1+j)}\sum_{n=0}^{\infty} q^{(m+j+1)n} \end{align}
(by (2.1))\begin{align}& \quad= ({-}q;q)_{r-1} +\frac{({-}1)^{m} ({-}q;q)_{m+r} }{ (q;q)_m}\sum_{n=0}^{\infty} q^{(m+1)(m+n+r+1)}\sum_{j=0}^{\infty}\frac{({-}q^{1-r};q)_{j}}{ (q;q)_{j}} q^{(m+2r+n+1)j}\nonumber\\ & \quad= ({-}q;q)_{r-1} +\frac{({-}1)^{m} ({-}q;q)_{m+r} }{ (q;q)_m}\sum_{n=0}^{\infty} q^{(m+1)(m+n+r+1)}\frac{({-}q^{m+r+n+2};q)_\infty }{ (q^{m+2r+n+1};q)_\infty} . \end{align}

If we set $m=k-1$ in the above identity, we arrive at the desired identity (1.17). This completes the proof of Theorem 1.3.

4.2. A corollary of Theorem 1.3

In the rest of this paper, we denote by $f_j$ the multiplicity of the part $j$ in a partition and by $f_{\overline {j}}$ the multiplicity of the part $\overline {j }$ in an overpartition, i.e., $f_{\overline {j}}=0,\,1$.

For example, from the following overpartition

\[\bar{5}+5+3+\bar{2}+2+2+2+1+1,\]

we have $f_{\bar {2}}=1$ and $f_2=3$.

We first define two partition functions $B(k,\,r,\,n)$ and $C(k,\,r,\,n)$. Let $B(k,\,r,\,n)$ count the number of overpartitions of $n$ satisfying the following conditions:

  1. (1) $f_{k}=k+r-1$;

  2. (2) $f_j=0$ if $k< j < k+2r-1$;

and let $C(k,\,r,\,n)$ count the number of overpartitions of $n$ satisfying the following conditions:

  1. (1) $f_{x}=k$, where $x$ is the smallest part $>k+r-1$ and $x$ exists;

  2. (2) $f_j=0$ if $k\leq j\leq x+r-1$ and $j\neq x$;

  3. (3) $f_{\overline {j}}=0$ if $k+r\leq j \leq x$.

For instance, $B(3,\,2,\,16)=11$ since there are 11 such partitions:

\begin{align*} & \overline{4}+3+3+3+3, \ \overline{3}+3+3+3+3+1, \ \overline{3}+3+3+3+3+ \overline{1}, \ 3+3+3+3+\overline{2}+2, \\ & 3+3+3+3+ 2 +2,\ 3+3+3+3+\overline{2}+1+1, \ 3+3+3+3+ 2 +2+\overline{1}+1,\\ & 3+3+3+3+\overline{2}+\overline{1}+1, \ 3+3+3+3+ 2 +2+1+1, \\ & 3+3+3+3+ \overline{1}+1+1+1, \ 3+3+3+3+ 1 +1+1+1. \end{align*}

Moreover, $C(3,\,2,\,16)=2$ since the two partitions in questions are

\[5+5+5+1,\quad 5+5+5+\overline{1}.\]

Note that the generating function for $B(k,\,r,\,n)$ is

(4.3)\begin{equation} \displaystyle\sum_{n=1}^{\infty} B(k,r,n)q^{n}=\displaystyle\frac{ ({-}q;q)_\infty q^{k(k+r-1)} }{(q;q)_{k-1}(q^{k+2r-1};q)_\infty}. \end{equation}

Let $x=k+r+n-1$ for some positive integer $n$. Then the generating function for $C(k,\,r,\,n)$ is

(4.4)\begin{equation} \displaystyle\sum_{n=1}^{\infty} C(k,r,n)q^{n}= \displaystyle\frac{({-}q;q)_{k+r-1} }{(q;q)_{k-1} }\sum_{n=1}^{\infty} \frac{({-}q^{k+n+r};q)_\infty q^{k(k+r+n-1)}}{(q^{k+n+2r-1};q)_\infty}. \end{equation}

By Theorem 1.3, we deduce the following corollary.

Corollary 4.1 Let $r$ be a positive integer and define

(4.5)\begin{align} \displaystyle\sum_{j\geq 0} t_{r}(i,j) q^{j}& := ( 1-q^{2i+2r-1}) ( q^{i+1};q)_{2r-2}, \end{align}
(4.6)\begin{align} \displaystyle\sum_{j\geq 0} u_{r}(i,j) q^{j}& := (1-q^{ i+ r }) ( q^{i+1};q)_{2r-1} . \end{align}

Then for $k\geq 1$ and $n> \frac {r(r-1)}{2}$,

(4.7)\begin{align} ({-}1)^{k-1} \displaystyle\sum_{i=0}^{k-1} ({-}1)^{i}\sum_{j\geq 0} t_r(i,j)\overline{p}(n-i^{2}-(r-1)i-j)& = B(k,r,n)+C(k,r,n), \end{align}
(4.8)\begin{align} ({-}1)^{k-1} \displaystyle\sum_{i=0}^{k-1} ({-}1)^{i} \sum_{j\geq 0} u_r(i,j)\overline{p}(n-i^{2}-ri-j) = & C(k,r,n). \end{align}

For instance, if we set $r=2$ and $k=3$ in Corollary 4.1, we find that for $n\geq 1$,

\begin{align*} & \overline{p}(n)-\overline{p}(n-1)-2\overline{p}(n-2) +2 \overline{p}(n-4) +2\overline{p}(n-5) -2\overline{p}(n-9)-2\overline{p}(n-10)\\ & \quad+\overline{p}(n-12)+\overline{p}(n-16)+\overline{p}(n-17) -\overline{p}(n-20)=B(3,2,n)+C(3,2,n),\\ & \overline{p}(n)\!-\!\overline{p}(n-1)\!-\!2\overline{p}(n-2) \!+\!2 \overline{p}(n-4) \!+\!2\overline{p}(n-5) \!-\!2\overline{p}(n-9)-2\overline{p}(n-10) + \overline{p}(n-15)\\ & \quad+2\overline{p}(n\!-\!16) \!+\!2\overline{p}(n\!-\!17) -\overline{p}(n\!-\!19)\!-\!2\overline{p}(n\!-\!20) -\overline{p}(n\!-\!21)\!+\!\overline{p}(n\!-\!24)= C(3,2,n). \end{align*}

If $n=16$, we have

\begin{align*} & \overline{p}(16) -\overline{p}(15)-2\overline{p}(14) +2 \overline{p}(12) +2\overline{p}(11) -2\overline{p}(7)-2\overline{p}(6) + \overline{p}(4)+\overline{p}(0)\\ & \quad=2062-1472-2\times 1040+2\times 504+2\times 344-2\times 64-2\times 40+14+1\\ & \quad=13=B(3,2,16)+C(3,2,16) \end{align*}

and

\begin{align*} & \overline{p}(16) -\overline{p}(15)-2\overline{p}(14) +2 \overline{p}(12) +2\overline{p}(11) -2\overline{p}(7)-2\overline{p}(6) +\overline{p}(1)+2\overline{p}(0)\\ & \quad=2062-1472-2\times 1040+2\times 504+2\times 344-2\times 64-2\times 40+2+2\times 1\\ & \quad=2= C(3,2,16). \end{align*}

Now, we turn to prove Corollary 4.1.

Proof. In view of (1.16), (1.20), and (4.3)–(4.5),

(4.9)\begin{align} & ({-}1)^{k-1}\displaystyle\sum_{m=0}^{\infty} \overline{p}(m)q^{m}\sum_{i=0}^{k-1} ({-}1)^{i} q^{i^{2}+(r-1)i}\sum_{j\geq 0}t_r(i,j)q^{j}\nonumber\\ & \quad=({-}1)^{k-1}({-}q;q)_{r-1} +\displaystyle\frac{({-}q;q)_{k+r-1} }{(q;q)_{k-1}} \displaystyle\sum_{n=0}^{\infty}\frac{({-}q^{k+n+r};q)_\infty q^{k(k+r+n-1)}}{(q^{k+n+2r-1}; q)_\infty}\nonumber\\ & \quad=({-}1)^{k-1}({-}q;q)_{r-1} +\frac{({-}q;q)_\infty q^{k(k+r-1)} }{(q;q)_{k-1}(q^{k+2r-1};q)_\infty}\nonumber\\ & \qquad + \frac{({-}q;q)_{k+r-1} }{(q;q)_{k-1} }\sum_{n=1}^{\infty}\frac{({-}q^{k+n+r};q)_\infty q^{k(k+r+n-1)}}{(q^{k+n+2r-1};q)_\infty}.\nonumber\\ & \quad=({-}1)^{k-1}({-}q;q)_{r-1}+\sum_{n=1}^{\infty} B(k,r,n)q^{n}+\sum_{n=1}^{\infty} C(k,r,n)q^{n}, \end{align}

which yields (4.7) after equating the coefficients of $q^{n}$ on both sides of (4.9).

Moreover, thanks to (1.17), (1.20), (4.4), and (4.6),

(4.10)\begin{align} & ({-}1)^{k-1}\displaystyle\sum_{m=0}^{\infty}\overline{p}(m)q^{m} \sum_{i=0}^{k-1} ({-}1)^{i} q^{i^{2}+ ri}\sum_{j\geq 0}u_r(i,j)q^{j}\nonumber\\ & \quad=({-}1)^{k-1}({-}q;q)_{r-1} +\displaystyle\frac{({-}q;q)_{k+r-1} }{(q;q)_{k-1}} \displaystyle\sum_{n=0}^{\infty}\frac{({-}q^{k+n+r+1};q)_\infty q^{k(k+r+n)}}{(q^{k+n+2r}; q)_\infty}\nonumber\\ & \quad=({-}1)^{k-1}({-}q;q)_{r-1} +\frac{({-}q;q)_{k+r-1} }{(q;q)_{k-1}}\nonumber\\ & \qquad \times \sum_{n=1}^{\infty}\frac{({-}q^{k+n+r};q)_\infty q^{k(k+r+n-1)}}{(q^{k+n+2r-1}; q)_\infty} \quad ({\rm replace\ }\ n\ {\rm by} \ n-1)\nonumber\\ & \quad=({-}1)^{k-1}({-}q;q)_{r-1} +\sum_{n=1}^{\infty} C(k,r,n)q^{n}. \end{align}

Equating the coefficients of $q^{n}$ on both sides of (4.10), we get the desired identity (4.8). The proof is complete.

5. Partitions with odd part distinct

The aim of this section is to prove Theorem 1.4 and present a corollary.

5.1. Proof of Theorem 1.4

Setting $(q,\,a,\,c) \rightarrow (q^{2},\,q^{4r-2},\,-q^{2r-1})$ in (2.10) and then multiplying both sides by $\frac {(q^{2};q^{2})_{2r-1}}{1+q^{2r-1}}$ yields

(5.1)\begin{align} & ({-}1)^{m} q^{m(m+1)}\displaystyle\sum_{n=0}^{m}\displaystyle\frac{(q^{{-}2m};q^{2})_n (q^{2m+2};q^{2})_{n+2r-1 } }{(q^{2};q^{2})_n({-}q^{2r-1};q^{2})_{n+1}}\nonumber\\ & \quad=\displaystyle\sum_{n=0}^{m} ({-}1)^{n} q^{2n^{2}+(2r-1)n}(1-q^{2n+2r-1}) (q^{2n+2};q^{2})_{2r-2}. \end{align}

Thanks to (5.1),

(by (2.1))\begin{align} & \displaystyle\frac{({-}q;q^{2})_\infty}{ (q^{2};q^{2})_\infty}\displaystyle\sum_{n=0}^{m} ({-}1)^{n} q^{2n^{2}+(2r-1)n}(1-q^{2n+2r-1}) (q^{2n+2};q^{2})_{2r-2}\nonumber\\ & \quad= \frac{({-}q;q^{2})_\infty}{ (q^{2};q^{2})_\infty} ({-}1)^{m} q^{m(m+1)}\sum_{n=0}^{m} \frac{(q^{{-}2m};q^{2})_n (q^{2m+2};q^{2})_{n+2r-1 } }{(q^{2};q^{2})_n({-}q^{2r-1};q^{2})_{n+1}}\nonumber\\ & \quad=\frac{({-}q;q^{2})_{r-1}({-}1)^{m} q^{m^{2}+m}}{ (q^{2};q^{2})_m}\sum_{n=0}^{m}\frac{(q^{{-}2m};q^{2})_n({-}q^{2n+2r+1};q^{2})_\infty }{(q^{2};q^{2})_n (q^{2m+2n+4r};q^{2})_\infty}\nonumber\\ & \quad=\frac{({-}q;q^{2})_{r-1}({-}1)^{m} q^{m^{2}+m}}{ (q^{2};q^{2})_m}\sum_{n=0}^{m} \frac{(q^{{-}2m};q^{2})_n }{(q^{2};q^{2})_n }\sum_{j=0}^{\infty} \frac{({-}q^{{-}2m-2r+1};q^{2})_j }{(q^{2};q^{2})_j} q^{(2m+2n+4r)j} \end{align}
(by (2.2))\begin{align}& \quad=\frac{({-}q;q^{2})_{r-1}({-}1)^{m} q^{m^{2}+m}}{ (q^{2};q^{2})_m }\sum_{j=0}^{\infty}\frac{({-}q^{{-}2m-2r+1};q^{2})_j }{(q^{2};q^{2})_j} q^{(2m+4r)j}\sum_{n=0}^{m}\frac{(q^{{-}2m};q^{2})_n }{(q^{2};q^{2})_n } q^{2nj}\nonumber\\ & \quad=\frac{({-}q;q^{2})_{r-1}({-}1)^{m} q^{m^{2}+m}}{ (q^{2};q^{2})_m }\sum_{j=0}^{\infty}\frac{({-}q^{{-}2m-2r+1};q^{2})_j }{(q^{2};q^{2})_j} q^{(2m+4r)j} (q^{2j-2m};q^{2})_m\end{align}
(by (2.4))\begin{align}& \quad=\frac{({-}q;q^{2})_{r-1}({-}1)^{m} q^{m^{2}+m}}{ (q^{2};q^{2})_m } ((q^{{-}2m};q^{2})_m \nonumber\\ & \qquad +\sum_{j=m+1}^{\infty} \frac{({-}q^{{-}2m-2r+1};q^{2})_j }{(q^{2};q^{2})_j} q^{(2m+4r)j} (q^{2j-2m};q^{2})_m)\end{align}
(by (2.3))\begin{align} & \quad= ({-}q;q^{2})_{r-1} +\frac{({-}q;q^{2})_{r-1} ({-}1)^{m} q^{m^{2}+m}}{ (q^{2};q^{2})_m }\nonumber\\ & \qquad \times \sum_{j=0}^{\infty} \frac{({-}q^{{-}2m-2r+1};q^{2})_{m+j+1} }{(q^{2};q^{2})_{m+j+1}} q^{(2m+4r)(m+j+1)} (q^{2j+2};q^{2})_m\nonumber\\ & \quad= ({-}q;q^{2})_{r-1} +\frac{({-}q;q^{2})_{r-1} ({-}1)^{m} q^{m^{2}+m} ({-}q^{{-}2m-2r+1};q^{2})_{m+1} }{ (q^{2};q^{2})_m }\nonumber\\ & \qquad \times \sum_{j=0}^{\infty} \frac{({-}q^{3-2r};q^{2})_{j} q^{(2m+4r)(m+j+1)}}{(q^{2};q^{2})_{j}(1-q^{2m+2j+2})}\nonumber\\ & \quad= ({-}q;q^{2})_{r-1} +\frac{({-}1)^{m} ({-}q;q^{2})_{m+r} }{ (q^{2};q^{2})_m }\sum_{j=0}^{\infty} \frac{({-}q^{3-2r};q^{2})_{j} q^{(2m+4r)j+(m+1)(2m+2r+1)}}{ (q^{2};q^{2})_{j} }\nonumber\\ & \qquad \times \sum_{n=0}^{\infty} q^{(2m+2j+2)n} \end{align}
(by (2.1))\begin{align}& \quad= ({-}q;q^{2})_{r-1} +\frac{({-}1)^{m} ({-}q;q^{2})_{m+r} }{ (q^{2};q^{2})_m }\sum_{n=0}^{\infty} q^{(m+1)(2m+2r+2n+1)}\nonumber\\ & \qquad \times \sum_{j=0}^{\infty}\frac{({-}q^{3-2r};q^{2})_{j} }{ (q^{2};q^{2})_{j} }q^{(2m+4r+2r)j }\nonumber\\ & \quad= ({-}q;q^{2})_{r-1} +\frac{({-}1)^{m} ({-}q;q^{2})_{m+r} }{ (q^{2};q^{2})_m }\sum_{n=0}^{\infty}\frac{ ({-}q^{2m+2r+2n+3};q^{2})_\infty q^{(m+1)(2m+2r+2n+1) }}{(q^{2m+4r+2n};q^{2})_\infty}, \end{align}

which yields (1.18) after replacing $m$ by $k-1$ in the above identity.

Taking $(q,\,a,\,c)=(q^{2},\,q^{4r},\,-q^{2r+1})$ in (2.10) and then multiplying both sides by $(q^{2};q^{2})_{2r}$ yields

(5.2)\begin{align} & ({-}1)^{m} q^{m(m+1)}\displaystyle\sum_{n=0}^{m}\displaystyle\frac{(q^{{-}2m};q^{2})_n (q^{2m+2};q^{2})_{n+2r } }{(q^{2};q^{2})_n({-}q^{2r+1};q^{2})_n}\nonumber\\ & \quad=\displaystyle\sum_{n=0}^{m} ({-}1)^{n} q^{2n^{2}+(2r-1)n}(1-q^{4n+4r}) (q^{2n+2};q^{2})_{2r-1}. \end{align}

In view of (5.2),

(by (2.1))\begin{align} & \displaystyle\frac{({-}q;q^{2})_\infty}{ (q^{2};q^{2})_\infty}\displaystyle\sum_{n=0}^{m} ({-}1)^{n} q^{2n^{2}+(2r-1)n}(1-q^{4n+4r}) (q^{2n+2};q^{2})_{2r-1}\nonumber\\ & \quad= \displaystyle\frac{({-}q;q^{2})_\infty}{ (q^{2};q^{2})_\infty} ({-}1)^{m} q^{m(m+1)}\displaystyle\sum_{n=0}^{m} \frac{(q^{{-}2m};q^{2})_n (q^{2m+2};q^{2})_{n+2r } }{(q^{2};q^{2})_n({-}q^{2r+1};q^{2})_n}\nonumber\\ & \quad=\frac{({-}q;q^{2})_{r}({-}1)^{m} q^{m^{2}+m}}{ (q^{2};q^{2})_m}\sum_{n=0}^{m} \frac{(q^{{-}2m};q^{2})_n({-}q^{2n+2r+1}; q^{2})_\infty }{(q^{2};q^{2})_n (q^{2m+2n+4r+2};q^{2})_\infty}\nonumber\\ & \quad=\frac{({-}q;q^{2})_{r}({-}1)^{m} q^{m^{2}+m}}{ (q^{2};q^{2})_m}\sum_{n=0}^{m} \frac{(q^{{-}2m};q^{2})_n }{(q^{2};q^{2})_n }\sum_{j=0}^{\infty} \frac{({-}q^{{-}2m-2r-1};q^{2})_j }{(q^{2};q^{2})_j} q^{(2m+2n+4r+2)j} \end{align}
(by (2.2))\begin{align}& \quad=\frac{({-}q;q^{2})_{r}({-}1)^{m} q^{m^{2}+m}}{ (q^{2};q^{2})_m }\sum_{j=0}^{\infty}\frac{({-}q^{{-}2m-2r-1};q^{2})_j }{(q^{2};q^{2})_j} q^{(2m+4r+2)j}\sum_{n=0}^{m}\frac{(q^{{-}2m};q^{2})_n }{(q^{2};q^{2})_n } q^{2nj}\nonumber\\ & \quad=\frac{({-}q;q^{2})_{r}({-}1)^{m} q^{m^{2}+m}}{ (q^{2};q^{2})_m }\sum_{j=0}^{\infty}\frac{({-}q^{{-}2m-2r-1};q^{2})_j }{(q^{2};q^{2})_j} q^{(2m+4r+2)j} (q^{2j-2m};q^{2})_m \end{align}
(by (2.4))\begin{align} & \quad=\frac{({-}q;q^{2})_{r}({-}1)^{m} q^{m^{2}+m}}{ (q^{2};q^{2})_m } ((q^{{-}2m};q^{2})_m \nonumber\\ & \qquad +\sum_{j=m+1}^{\infty}\frac{({-}q^{{-}2m-2r-1};q^{2})_j }{(q^{2};q^{2})_j} q^{(2m+4r+2)j} (q^{2j-2m};q^{2})_m) \end{align}
(by (2.3))\begin{align}& \quad= ({-}q;q^{2})_{r} +\frac{({-}q;q^{2})_{r} ({-}1)^{m} q^{m^{2}+m}}{ (q^{2};q^{2})_m}\nonumber\\& \qquad \times \sum_{j=0}^{\infty}\frac{({-}q^{{-}2m-2r-1};q^{2})_{m+j+1} }{(q^{2};q^{2})_{m+j+1}} q^{(2m+4r+2)(m+j+1)} (q^{2j+2};q^{2})_m\nonumber\\ & \quad= ({-}q;q^{2})_{r} +\frac{({-}q;q^{2})_{r} ({-}1)^{m} q^{m^{2}+m} ({-}q^{{-}2m-2r-1};q^{2})_{m+1} }{ (q^{2};q^{2})_m }\nonumber\\& \qquad \times \sum_{j=0}^{\infty}\frac{({-}q^{1-2r};q^{2})_{j} q^{(2m+4r+2)(m+j+1)}}{(q^{2};q^{2})_{j}(1-q^{2m+2j+2})}\nonumber\\ & \quad= ({-}q;q^{2})_{r} +\frac{({-}1)^{m} ({-}q;q^{2})_{m+r+1} }{ (q^{2};q^{2})_m }\sum_{j=0}^{\infty}\frac{({-}q^{1-2r};q^{2})_{j} q^{(2m+4r+2)j+(m+1)(2m+2r+1)}}{ (q^{2};q^{2})_{j} }\nonumber\\& \qquad \times \sum_{n=0}^{\infty} q^{(2m+2j+2)n} \end{align}
(by (2.1))\begin{align}& \quad= ({-}q;q^{2})_{r} +\frac{({-}1)^{m} ({-}q;q^{2})_{m+r+1} }{ (q^{2};q^{2})_m }\sum_{n=0}^{\infty} q^{(m+1)(2m+2r+2n+1) }\nonumber\\ & \qquad \times \sum_{j=0}^{\infty}\frac{({-}q^{1-2r};q^{2})_{j} }{ (q^{2};q^{2})_{j} }q^{(2m+4r+2n+2)j }\nonumber\\ & \quad= ({-}q;q^{2})_{r} +\frac{({-}1)^{m} ({-}q;q^{2})_{m+r+1} }{ (q^{2};q^{2})_m }\sum_{n=0}^{\infty}\frac{ ({-}q^{2m+2n+2r+3};q^{2})_\infty q^{(m+1)(2m+2r+2n+1) }}{(q^{2m+4r+2n+2}; q^{2})_\infty} .\end{align}

If we replace $m$ by $k-1$ in the above identity, we get the desired identity (1.19). This completes the proof.

5.2. A corollary of Theorem 1.4

In order to state our corollary, we first define some partition functions.

Let $D(k,\,r,\,n)$ count the number of partitions of $n$ satisfying the following conditions:

  1. (1) $f_{x}=k$, where $x$ is the smallest part $\geq 2k+2r-1$, and $x$ is odd and exists;

  2. (2) $f_{2j+1}=0$ or $1$ if $1\leq 2j+1 \leq 2k+2r-3$ or $2j+1>x$;

  3. (3) $f_{2j}=0$ if $2k\leq 2j \leq x+2r-3$.

For example, $D(2,\,2,\,21)=5$ since the five partitions in questions are

\[9+9+3,\ 9+9+2+1, \ 7+7+5+2, \ 7+7+3+2+2,\ 7+7+2+2+2+1.\]

Let $x=2k+2r+2n-1$ for some integer $n\geq 0$. Then the generating function for $D(k,\,r,\,n)$ is

(5.3)\begin{equation} \displaystyle\sum_{n=1}^{\infty} D(k,r,n)q^{n}=\displaystyle\frac{ ({-}q;q^{2})_{k+r-1} }{ (q^{2};q^{2})_{k-1} }\sum_{n=0}^{\infty}\frac{ ({-}q^{2k+2n+2r+1};q^{2})_\infty q^{k(2k+2r+2n-1) }}{(q^{2k+4r+2n-2};q^{2})_\infty}. \end{equation}

Next, define $E(k,\,r,\,n)$ to be the number of partitions of $n$ satisfying the following conditions:

  1. (1) $f_{2k+2r-1}=k$ or $\ k+1$;

  2. (2) $f_{2j+1}=0$ or $1$ if $j\neq k+r-1$;

  3. (3) $f_{2j}=0$ if $k\leq j\leq k+2r-1$.

For instance, $E(2,\,2,\,21)=4$ because the four partitions in questions are

\[7+7+7,\ 7+7+5+2,\ 7+7+3+2+2, \ 7+7+2+2+2+1.\]

It is easy to see that the generating function for $E(k,\,r,\,n)$ is

(5.4)\begin{equation} \displaystyle\sum_{n=0}^{\infty} E(k,r,n)q^{n}=\displaystyle\frac{({-}q;q^{2})_\infty q^{k(2k+2r-1)}}{ (q^{2};q^{2})_{k-1}(q^{2k+4r};q^{2})_\infty}. \end{equation}

Finally, let $F(k,\,r,\,n)$ count the number of partitions of $n$ satisfying the following conditions:

  1. (1) $f_{x}=k$, where $x$ is the smallest part $>2k+2r-1$, and $x$ is odd and exists;

  2. (2) $f_{2j+1}=0$ or $1$ if $1\leq 2j+1 \leq 2k+2r-1$ or $2j+1>x$;

  3. (3) $f_{2j+1}=0$ if $2k+2r+1\leq 2j+1 < x$;

  4. (4) $f_{2j}=0$ if $2k\leq 2j \leq x+2r-1$.

For instance, $F(2,\,2,\,21)=2$ since the two partitions in questions are

\[9+9+3,\quad 9+9+2+1.\]

Let $x=2k+2r+2n-1$ for some positive integer $n$. Then the generating function for $F(k,\,r,\,n)$ is

(5.5)\begin{equation} \displaystyle\sum_{n=1}^{\infty} F(k,r,n) q^{n} = \displaystyle\frac{ ({-}q;q^{2})_{k+r} }{ (q^{2};q^{2})_{k-1} } \sum_{n=1}^{\infty} \frac{ ({-}q^{2k+2n+2r+1};q^{2})_\infty q^{k(2k+2r+2n-1) }}{(q^{2k+4r+2n};q^{2})_\infty}. \end{equation}

Corollary 5.1 Let $r$ be a positive integer and define

(5.6)\begin{align} \displaystyle\sum_{j\geq 0} v_{ r}(i,j) q^{j}& :=(1-q^{2i+2r-1}) (q^{2i+2 };q^{2})_{2r-2}, \end{align}
(5.7)\begin{align} \displaystyle\sum_{j\geq 0} w_{r}(i,j) q^{j}& := (1-q^{4i+4r}) (q^{2i+2 };q^{2})_{2r-1}. \end{align}

Then for $k\geq 1$ and $n> r^{2}$,

(5.8)\begin{align} ({-}1)^{k-1} \displaystyle\sum_{i=0}^{k-1} ({-}1)^{i} \sum_{j\geq 0} v_r(i,j) {\rm pod} (n-2i^{2}-(2r-1)i-j)& = D(k,r,n), \end{align}
(5.9)\begin{align} ({-}1)^{k-1} \displaystyle\sum_{i=0}^{k-1} ({-}1)^{i} \sum_{j\geq 0} w_r(i,j) {\rm pod} (n-2i^{2}-(2r-1)i-j) & = E(k,r,n)+F(k,r,n). \end{align}

For instance, if $k=2$ and $r=2$, we find that for $n\geq 4$,

\begin{align*} & -{\rm pod}(n) + {\rm pod}(n-2) +{\rm pod}(n-3) +{\rm pod}(n-4)\nonumber\\ & -{\rm pod}(n-6) -{\rm pod}(n-7) -{\rm pod}(n-10)-{\rm pod}(n-11) +{\rm pod}(n-14)\nonumber\\ & +{\rm pod}(n-15)+{\rm pod}(n-16) -{\rm pod}(n-20)=D(2,2,n) \end{align*}

and

\begin{align*} & -{\rm pod}(n)+{\rm pod}(n-2) + {\rm pod}(n-4) +{\rm pod}(n-5) -{\rm pod}(n-9)-2{\rm pod}(n-10)\\ & -{\rm pod}(n-11) -{\rm pod}(n-13)+{\rm pod}(n-15) +{\rm pod}(n-16) +{\rm pod}(n-18)\\ & +{\rm pod}(n-19) -{\rm pod}(n-20)+{\rm pod}(n-21) +{\rm pod}(n-25) -{\rm pod}(n-27)\\ & -{\rm pod}(n-29) -{\rm pod}(n-31) +{\rm pod}(n-35)=E(2,2,n)+F(2,2,n). \end{align*}

If $n=21$, then

\begin{align*} & -{\rm pod}(21) + {\rm pod}(19) +{\rm pod}(18) +{\rm pod}(17) -{\rm pod}(15) -{\rm pod}(14)\\ & -{\rm pod}(11)-{\rm pod}(10) +{\rm pod}(7) +{\rm pod}(6)+{\rm pod}(5) -{\rm pod}(1)=5=D(2,2,21) \end{align*}

and

\begin{align*} & -{\rm pod}(21)+{\rm pod}(19) + {\rm pod}(17) +{\rm pod}(16) -{\rm pod}(12)-2{\rm pod}(11) -{\rm pod}(10) -{\rm pod}(8) \nonumber\\ & +{\rm pod}(6) +{\rm pod}(5) +{\rm pod}(3) +{\rm pod}(2) -{\rm pod}(1)+{\rm pod}(0) =6=E(2,2,21)+F(2,2,21). \end{align*}

To conclude this section, we present a proof of Corollary 5.1.

Proof of Corollary 5.1 It follows from (1.18), (1.21), (5.3), and (5.6) that

\begin{align*} & ({-}1)^{k-1} \displaystyle\sum_{m=0}^{\infty} {\rm pod}(m) q^{m}\sum_{i=0}^{k-1} ({-}1)^{i} q^{2i^{2}+(2r-1)i}v_r(i,j)q^{j}\nonumber\\ & \quad= ({-}1)^{k-1} ({-}q;q^{2})_{r-1} +\displaystyle\frac{ ({-}q;q^{2})_{k+r-1} }{ (q^{2};q^{2})_{k-1} }\sum_{n=0}^{\infty}\frac{ ({-}q^{2k+2n+2r+1};q^{2})_\infty q^{k(2k+2r+2n-1) }}{(q^{2k+4r+2n-2};q^{2})_\infty}\nonumber\\ & \quad= ({-}1)^{k-1} ({-}q;q^{2})_{r-1} +\sum_{n=1}^{\infty} D(k,r,n) q^{n}, \end{align*}

which yields (5.8) after equating the coefficients of $q^{n}$ on both sides of the above identity.

Similarly, in light of (1.19), (1.21), (5.4), (5.5), and (5.7) that

(5.9)\begin{align} & ({-}1)^{k-1} \displaystyle\sum_{m=0}^{\infty} {\rm pod}(m) q^{m}\sum_{i=0}^{k-1} ({-}1)^{i} q^{2i^{2}+(2r-1)i}w_r(i,j)q^{j}\nonumber\\ & \quad= ({-}1)^{k-1}({-}q;q^{2})_{r} + \displaystyle\frac{({-}q;q^{2})_{k+r} }{ (q^{2};q^{2})_{k-1} }\displaystyle\sum_{n=0}^{\infty} \frac{ ({-}q^{2k+2n+2r+1};q^{2})_\infty q^{k(2k+2r+2n-1) }}{(q^{2k+4r+2n};q^{2})_\infty}\nonumber\\ & \quad= ({-}1)^{k-1}({-}q;q^{2})_{r} +\sum_{n=1}^{\infty} E(k,r,n)q^{n} +\sum_{n=1}^{\infty} F(k,r,n)q^{n}. \end{align}

Equating the coefficients of $q^{n}$ on both sides of (5.10), we arrive at (5.9). This completes the proof.

6. Concluding Remarks

In this paper, based on a Bailey pair given by Lovejoy [Reference Lovejoy15], we present a unified treatment to establish new truncated theorems on the three classic theta function identities due to Euler and Gauss. From those theorems, we can present partition-theoretic interpretations for the alternating sums in (3.4), (4.7), (4.8), (5.8), (5.9) and deduce infinite families of linear inequalities on the three classical partition functions $p(n)$, $\overline {p}(n)$ and $pod(n)$. We also remark that (1.9) can be deduced from (2.10). If we take $a\rightarrow 1$ in (2.10), we have

(6.1)\begin{equation} ({-}1)^{m} q^{m(m+1)/2}\displaystyle\sum_{n=0}^{m} \displaystyle\frac{(q^{{-}m},q^{m+1};q)_n }{(q,q/c;q)_n}=1+\sum_{n=1}^{m} \frac{(c;q)_n}{(q/c;q)_n}(1+q^{n})q^{n^{2}}c^{{-}n}. \end{equation}

Setting $c=-1$ in (6.1) yields

(6.2)\begin{equation} ({-}1)^{m} q^{m(m+1)/2}\displaystyle\sum_{n=0}^{m} \displaystyle\frac{(q^{{-}m},q^{m+1};q)_n }{(q,-q;q)_n}=1+2\sum_{n=1}^{m} ({-}1)^{n} q^{n^{2}}. \end{equation}

Utilizing (6.2) and the method for proving (1.16), we can deduce (1.9). In addition, setting $a\rightarrow 1$ in (3.1) yields

(6.3)\begin{equation} ({-}1)^{m} q^{m(m+1)/2}\displaystyle\sum_{n=0}^{m}\displaystyle\frac{(q^{{-}m};q)_n(q^{m+1};q)_n}{(q;q)_n} =1+\sum_{n=1}^{m} ({-}1)^{n} q^{n(3n-1)/2}(1+q^{n}).\end{equation}

Employing (6.3) and the method for proving (1.15), we prove that for $k\geq 1$,

\[\frac{1}{(q;q)_\infty}\left(1+\sum_{j=1}^{k-1} ({-}1)^{j} q^{j(3j-1)/2}(1+q^{j})\right)=1+({-}1)^{k-1}q^{k(k-1)/2}\sum_{n=k}^{\infty} \frac{q^{nk}}{(q;q)_k}\left[ n-1 \atop k-1\right].\]

The above identity implies that for $k\geq 1$

\[({-}1)^{k-1}\sum_{j=1-k}^{k-1} ({-}1)^{j} p(n-j(3j+1)/2) =\widehat{P}_k(n),\]

where $\widehat {P}_k(n)$ is the number of partitions of $n$ in which $k$ is the least integer such that every part $< k$ appears at least once and there are more parts $>k-1$ than there are $< k$. This solves an open problem given by Merca [Reference Merca18].

Andrews–Merca [Reference Andrews and Merca3] and Guo–Zeng [Reference Guo and Zeng11] posed the following conjecture: -

conjecture Let $m,\, r$ be positive integer with $r \leq m/2$ and define

\[\sum_{n=0}^{\infty} J_{m,r}(n) q^{n}:=\frac{1}{(q^{r},q^{m-r},q^{m};q^{m})_\infty}.\]

Then for $n\geq 1$ and $k\geq 1$,

(6.4)\begin{equation} ({-}1)^{k-1}\displaystyle\sum_{j=0}^{k-1} ({-}1)^{j} (J_{m,r}(n-j(mj+m-2r)/2) -J_{m,r}(n-(j+1)(mj+2r)/2))\geq 0 \end{equation}

with strict inequality if $n \geq k(mk +m-2r)/2$.

This conjecture was settled by Mao [Reference Mao16] and Yee [Reference Yee32] independently, and then it was reproved by Wang and Yee [Reference Wang and Yee26] later.

Setting $(m,\,r)=(3,\,1)$, and $(m,\,r)=(2,\,1)$ in (6.4), respectively, we get the following inequalities on $p(n)$ and $\overline {p}(n)$:

(6.5)\begin{equation} ({-}1)^{k-1}\displaystyle\sum_{j=0}^{k-1} ({-}1)^{j} (p(n-\displaystyle\frac{j(3j+1)}{2}) - p(n-\frac{(j+1)(3j+2)}{2}) )\geq 0, \end{equation}

and

(6.6)\begin{equation} \overline{p}(n-k^{2}) + ({-}1)^{k-1}\displaystyle\sum_{j={-}k}^{k} ({-}1)^{j}\overline{p}(n-j^{2}) \geq 0. \end{equation}

In fact, inequality (6.5) is a consequence of (1.3) which was first proved by Andrews and Merca [Reference Andrews and Merca3]. In [Reference Xia, Yee and Zhao29], Xia et al. proved that for $n\geq 1$ and $k\geq 1$,

(6.7)\begin{align} & p(n-\displaystyle\frac{3k(k+1)}{2}) +({-}1)^{k-1} \displaystyle\sum_{j={-}k}^{k} ({-}1)^{j} p(n-\frac{j(3j+1)}{2}) \ge 0, \end{align}
(6.8)\begin{align} & \overline{p}(n-k(k+1)) + ({-}1)^{k-1} \sum_{j={-}k}^{k} ({-}1)^{j} \overline{p}(n-j^{2}) \ge 0, \end{align}

which are stronger than (6.5) and (6.6). So it would be interesting if one can find an inequality which is stronger than (6.4) for general positive integers $m$ and $r$ with $1\leq r \leq m/2$.

Acknowledgements

The author cordially thanks the anonymous referee for his/her helpful comments. This work was supported by the Natural Science Foundation of Jiangsu Province of China (no. BK20221383) and the National Natural Science Foundation of China (no. 11971203).

Declaration of competing interest

The author declares no conflicts of interest to this work.

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