1 Introduction

A partition \(\lambda \) is a finite weakly decreasing sequence of positive integers \(\lambda =(\lambda _1,\lambda _2,\ldots ,\lambda _l)\). The terms \(\lambda _i\) are called the parts of \(\lambda \), and the number of parts of \(\lambda \) is called the length of \(\lambda \), denoted \(\ell (\lambda )\). The weight of \(\lambda \) is the sum of its parts, denoted \(|\lambda |\). We say \(\lambda \) is a partition of n if \(|\lambda |=n\).

An important part of the theory of partitions concerns partition identities, which have a long history starting with Euler’s celebrated partition theorem [8].

Theorem 1.1

The partitions of n into odd parts are equinumerous with the partitions of n into distinct parts.

For convenience, a partition \(\lambda \) is called odd if each part of \(\lambda \) is odd, and distinct if every pair of parts differs by at least one. Let \(\mathcal {O}\) and \(\mathcal {D}\) denote the set of odd and distinct partitions respectively. We next discuss several statistics on partitions and study their joint distribution on \(\mathcal {O}\) and \(\mathcal {D}\).

Given a partition \(\lambda =(\lambda _1,\lambda _2,\ldots ,\lambda _l)\), the alternating sum of \(\lambda \) is defined to be

$$\begin{aligned}\ell _a(\lambda ):=\lambda _1-\lambda _2+\lambda _{3}-\lambda _{4}+\cdots +(-1)^{l-1}\lambda _{l}.\end{aligned}$$

Denote by \(o(\lambda )\) the number of distinct odd parts occurring an odd number of times in \(\lambda \), and denote by \(n_o(\lambda )\) the number of odd parts in \(\lambda \). Define the minimal odd excludant of \(\lambda \), denoted \(\textrm{moe}(\lambda )\), to be the smallest positive odd integer that is not a part of \(\lambda \). This newer statistic was introduced by Andrews and Newman [2] in 2019, and has many fruitful applications [3, 12, 13]. We now give an example to illustrate these statistics. Let \(\lambda =(7,7,5,5,5,3,2,1,1)\). Then, \(\ell _a(\lambda )=4,o(\lambda )=2,n_o(\lambda )=8\), and \(\textrm{moe}(\lambda )=9\).

Let \(\mathcal {O}_{1,4}\) and \(\mathcal {O}_{3,4}\) be the set of odd partitions with the minimal odd excludant congruent to 1 and 3 modulo 4, respectively. Denote by \(\mathcal {D}_e\) and \(\mathcal {D}_o\) the set of distinct partitions of even and odd length, respectively. The first author of the present paper and Xu [15] presented a strong refinement of Euler’s identity, which reads as follows.

Theorem 1.2

The number of odd partitions of n with length l, m parts occurring an odd number of times and the minimal odd excludant congruent to 1 (respectively, 3) modulo 4 is the same as the number of distinct partitions of n into an even (respectively, odd) number of parts with m odd parts and alternating sum l. Equivalently,

$$\begin{aligned} \sum _{\lambda \in \mathcal {O}_{1,4}}x^{\ell (\lambda )}y^{o(\lambda )}q^{|\lambda |}&=\sum _{\pi \in \mathcal {D}_{e}}x^{\ell _a(\pi )}y^{n_{o}(\pi )}q^{|\pi |},\\ \sum _{\lambda \in \mathcal {O}_{3,4}}x^{\ell (\lambda )}y^{o(\lambda )}q^{|\lambda |}&=\sum _{\pi \in \mathcal {D}_{o}}x^{\ell _a(\pi )}y^{n_{o}(\pi )}q^{|\pi |}. \end{aligned}$$

In this paper, we strengthen further Theorem 1.2 by taking into account the relationship between the statistics \(o(\lambda )\) and \(n_o(\pi )\). Given a partition \(\lambda \), let \(o_{1,4}(\lambda )\) and \(o_{3,4}(\lambda )\) denote the number of distinct parts occurring an odd number of times in \(\lambda \) and congruent to 1 and 3 modulo 4, respectively. Clearly, \(o(\lambda )=o_{1,4}(\lambda )+o_{3,4}(\lambda )\). For a partition \(\pi =(\pi _1,\pi _2,\ldots ,\pi _l)\), we call each \(\pi _{2i-1}\) an odd indexed part, and each \(\pi _{2i}\) an even indexed part. Namely, whether \(\pi _i\) is an odd or even indexed part depends on the parity of the subscript i. Let \(n_{o,o}(\pi )\) and \(n_{o,e}(\pi )\) denote the number of odd and even indexed odd parts of \(\pi \), respectively. It is clear that \(n_o(\pi )=n_{o,o}(\pi )+n_{o,e}(\pi )\).

We come to the main result of this work.

Theorem 1.3

The number of odd partitions of n with length l, i parts congruent to 1 modulo 4 and occurring an odd number of times, j parts congruent to 3 modulo 4 and occurring an odd number of times, and the minimal odd excludant congruent to 1 (respectively, 3) modulo 4 is equal to the number of distinct partitions of n into an even (respectively, odd) number of parts with alternating sum l, i odd indexed odd parts and j even indexed odd parts. Equivalently,

$$\begin{aligned} \sum _{\lambda \in \mathcal {O}_{1,4}}x^{\ell (\lambda )}y^{o_{1,4}(\lambda )}z^{{o_{3,4}(\lambda )}}q^{|\lambda |}&=\sum _{\pi \in \mathcal {D}_{e}}x^{\ell _a(\pi )}y^{n_{o,o}(\pi )}z^{n_{o,e}(\pi )}q^{|\pi |},\\ \sum _{\lambda \in \mathcal {O}_{3,4}}x^{\ell (\lambda )}y^{o_{1,4}(\lambda )}z^{{o_{3,4}(\lambda )}}q^{|\lambda |}&=\sum _{\pi \in \mathcal {D}_{o}}x^{\ell _a(\pi )}y^{n_{o,o}(\pi )}z^{n_{o,e}(\pi )}q^{|\pi |}. \end{aligned}$$

We demonstrate Theorem 1.3 for \(n=10\) in Table 1.

Table 1 An illustration of Theorem 1.3
Full size table

Remark 1.4

Over the years, there have been many different refinements of Euler’s theorem; see [5,6,7, 9] and the references therein for more information. Theorem 1.3 is a new refinement of Euler’s result.

The rest of the paper is organized as follows. In Sect. 2, we present some preliminary results, which are useful to our later proofs. The goal of Sect. 3 is to establish the generating function to keep track of the statistics on odd partitions. In Sect. 4, we find an expression for the joint distribution of the statistics on distinct partitions of fixed length, and finish the proof of Theorem 1.3.

2 Preliminaries

Throughout this paper, we use the following standard notation

$$\begin{aligned} (a;q)_{0}&:=1,\\ (a;q)_{n}&:=\prod _{i=0}^{n-1}(1-aq^{i}),\ n\ge 1,\\ (a;q)_{\infty }&:=\lim _{n\rightarrow \infty }(a;q)_{n}, \end{aligned}$$

and always assume that \(|q|<1\).

In this work, we employ some fundamental tools in the theory of q-series. Among the most useful summation formulas is the q-binomial theorem, one special case of which [1, p. 7, Eq. (1.2.4)] is stated as follows.

Lemma 2.1

For \(|z|<\infty \),

$$\begin{aligned} \sum \limits _{n=0}^\infty \frac{z^nq^{n(n-1)/2}}{(q;q)_n}=(-z;q)_\infty . \end{aligned}$$
(2.1)

We also need the second Heine transformation [1, p. 9, Corollary 1.2.4].

Lemma 2.2

For \(|t|<1\) and \(|c/b|<1\),

$$\begin{aligned} _2\phi _1\left( {\begin{array}{c} {a,b} \\ {c} \end{array}};q,t\right) =\frac{(c/b;q)_\infty (bt;q)_{\infty }}{(c;q)_\infty (t;q)_{\infty }} {}_2\phi _1\left( {\begin{array}{c} {abt/c,b} \\ {bt} \end{array}};q,\frac{c}{b}\right) , \end{aligned}$$
(2.2)

where

$$\begin{aligned} _2\phi _1\left( {\begin{array}{c} {a,b} \\ {c} \end{array}};q,t\right) :=\sum \limits _{n=0}^\infty \frac{(a;q)_n(b;q)_n}{(c;q)_n(q;q)_n}t^n. \end{aligned}$$

Another important formula in q-series is the Rogers–Fine identity, which was first proved by Rogers [14] and rediscovered independently by Fine [10].

Lemma 2.3

We have

$$\begin{aligned} \sum \limits _{n=0}^\infty \frac{(\alpha ;q)_n}{(\beta ;q)_n}\tau ^n =\sum \limits _{n=0}^\infty \frac{(\alpha ;q)_n(\alpha \tau q/\beta ;q)_n(\beta \tau )^nq^{n^2-n}(1-\alpha \tau q^{2n})}{(\beta ;q)_n(\tau ;q)_{n+1}}. \end{aligned}$$
(2.3)

We now establish some preliminary results, which play a central role in our later proofs.

Lemma 2.4

We have

$$\begin{aligned} \sum \limits _{k=0}^\infty \frac{(-xy^{-1}q;q^4)_k}{(-xzq^3;q^4)_k}(-xyq)^k&=\frac{1+xzq^{-1}}{1+xyq}\sum \limits _{k=0}^\infty \frac{(-xz^{-1}q^3;q^4)_k}{(-xyq^5;q^4)_k}(-xzq^{-1})^k,\end{aligned}$$
(2.4)
$$\begin{aligned} \sum \limits _{k=0}^\infty \frac{(-xz^{-1}q^3;q^4)_k}{(-xyq^5;q^4)_k}(-xzq^3)^k&=\frac{1+xyq}{1+xzq^3}\sum \limits _{k=0}^\infty \frac{(-xy^{-1}q^5;q^4)_k}{(-xzq^7;q^4)_k}(-xyq)^k. \end{aligned}$$
(2.5)

Proof

Replacing q by \(q^4\) and setting \(a=-xy^{-1}q,b=q^4,c=-xzq^3,t=-xyq\) in (2.2), we get

$$\begin{aligned}&\sum \limits _{k=0}^\infty \frac{(-xy^{-1}q;q^4)_k}{(-xzq^3;q^4)_k}(-xyq)^k \\&\quad =\frac{(-xzq^{-1};q^4)_\infty (-xyq^5;q^4)_\infty }{(-xzq^3;q^4)_\infty (-xyq;q^4)_\infty } \sum \limits _{k=0}^\infty \frac{(-xz^{-1}q^3;q^4)_k}{(-xyq^5;q^4)_k}(-xzq^{-1})^k\\&\quad =\frac{1+xzq^{-1}}{1+xyq}\sum \limits _{k=0}^\infty \frac{(-xz^{-1}q^3;q^4)_k}{(-xyq^5;q^4)_k}(-xzq^{-1})^k. \end{aligned}$$

Replacing x by \(xq^2\) and interchanging y with z in (2.4) produces (2.5). \(\square \)

Theorem 2.5

We have

$$\begin{aligned}&\sum \limits _{i=0}^\infty \frac{(-xzq^{-1};q^4)_ix^iy^iq^{2i^2+3i}}{(x^2q^2;q^4)_i(q^4;q^4)_i} \sum \limits _{k=0}^\infty \frac{(-xz^{-1}q^3;q^4)_kx^kz^kq^{2k^2+(4i+1)k}}{(x^2q^{4i+2};q^4)_k(q^4;q^4)_k}\\&\quad =\frac{(-xyq;q^4)_\infty (-xzq^3;q^4)_\infty }{(x^2q^2;q^4)_\infty } \sum \limits _{k=0}^\infty (-1)^k(xyq)^k\frac{(-xy^{-1}q;q^4)_k}{(-xzq^3;q^4)_k}. \end{aligned}$$

Proof

Replacing q by \(q^4\) and setting \(a=-q^2/\tau ,b=-xz^{-1}q^3,c=x^2q^{4i+2},t=xz\tau q^{4i+1}\) in (2.2), and then letting \(\tau \rightarrow 0\), we obtain

$$\begin{aligned}&\sum \limits _{k=0}^\infty \frac{(-xz^{-1}q^3;q^4)_kx^kz^kq^{2k^2+(4i+1)k}}{(x^2q^{4i+2};q^4)_k(q^4;q^4)_k}\\&\quad =\frac{(-xzq^{4i-1};q^4)_\infty }{(x^2q^{4i+2};q^4)_\infty } \sum \limits _{k=0}^\infty (-xz^{-1}q^3;q^4)_k(-xzq^{4i-1})^k. \end{aligned}$$

Employing the above identity, we can derive that

$$\begin{aligned}&\sum \limits _{i=0}^\infty \frac{(-xzq^{-1};q^4)_ix^iy^iq^{2i^2+3i}}{(x^2q^2;q^4)_i(q^4;q^4)_i} \sum \limits _{k=0}^\infty \frac{(-xz^{-1}q^3;q^4)_kx^kz^kq^{2k^2+(4i+1)k}}{(x^2q^{4i+2};q^4)_k(q^4;q^4)_k}\\&\quad =\sum \limits _{i=0}^\infty \frac{(-xzq^{-1};q^4)_ix^iy^iq^{2i^2+3i}}{(x^2q^2;q^4)_i(q^4;q^4)_i} \frac{(-xzq^{4i-1};q^4)_\infty }{(x^2q^{4i+2};q^4)_\infty }\sum \limits _{k=0}^\infty (-xz^{-1}q^3;q^4)_k(-xzq^{4i-1})^k\\&\quad =\frac{(-xzq^{-1};q^4)_\infty }{(x^2q^2;q^4)_\infty }\sum \limits _{i=0}^\infty \frac{x^iy^iq^{2i^2+3i}}{(q^4;q^4)_i} \sum \limits _{k=0}^\infty (-xz^{-1}q^3;q^4)_k(-xzq^{4i-1})^k\\&\quad =\frac{(-xzq^{-1};q^4)_\infty }{(x^2q^2;q^4)_\infty }\sum \limits _{k=0}^\infty (-xz^{-1}q^3;q^4)_k(-xzq^{-1})^k \sum \limits _{i=0}^\infty \frac{(xyq^{4k+5})^iq^{2i^2-2i}}{(q^4;q^4)_i}\\&\quad =\frac{(-xzq^{-1};q^4)_\infty }{(x^2q^2;q^4)_\infty }\sum \limits _{k=0}^\infty (-xz^{-1}q^3;q^4)_k(-xyq^{4k+5};q^4)_\infty (-xzq^{-1})^k\qquad \qquad (\hbox {by}\, (2.1))\\&\quad =\frac{(-xzq^{-1};q^4)_\infty (-xyq^5;q^4)_\infty }{(x^2q^2;q^4)_\infty }\sum \limits _{k=0}^\infty \frac{(-xz^{-1}q^3;q^4)_k}{(-xyq^5;q^4)_k}(-xzq^{-1})^k\\&\quad =\frac{(-xyq;q^4)_\infty (-xzq^{3};q^4)_\infty }{(x^2q^2;q^4)_\infty } \sum \limits _{k=0}^{\infty }\frac{(-xy^{-1}q;q^4)_k}{(-xzq^3;q^4)_k}(-xyq)^k.\qquad \qquad (\hbox {by}\, (2.4)) \end{aligned}$$

The proof is complete. \(\square \)

Corollary 2.6

We have

$$\begin{aligned}&\sum \limits _{i=0}^\infty \frac{(-xyq;q^4)_ix^iz^iq^{2i^2+5i}}{(x^2q^2;q^4)_{i+1}(q^4;q^4)_i} \sum \limits _{k=0}^\infty \frac{(-xy^{-1}q;q^4)_{k+1}x^{k+1}y^{k+1}q^{2k^2+(4i+3)k+1}}{(x^2q^{4i+6};q^4)_k(q^4;q^4)_k}\\&\quad =\frac{(-xyq;q^4)_\infty (-xzq^3;q^4)_\infty }{(x^2q^2;q^4)_\infty } \sum \limits _{k=1}^\infty (-1)^{k-1}(xyq)^k\frac{(-xy^{-1}q;q^4)_k}{(-xzq^3;q^4)_k}. \end{aligned}$$

Proof

In Theorem 2.5, replacing x by \(xq^2\) and interchanging y with z yields

$$\begin{aligned}&\sum \limits _{i=0}^\infty \frac{(-xyq;q^4)_ix^iz^iq^{2i^2+5i}}{(x^2q^6;q^4)_i(q^4;q^4)_i} \sum \limits _{k=0}^\infty \frac{(-xy^{-1}q^5;q^4)_kx^ky^kq^{2k^2+(4i+3)k}}{(x^2q^{4i+6};q^4)_k(q^4;q^4)_k}\\&\quad =\frac{(-xyq^5;q^4)_\infty (-xzq^3;q^4)_\infty }{(x^2q^6;q^4)_\infty } \sum \limits _{k=0}^\infty (-xzq^3)^k\frac{(-xz^{-1}q^3;q^4)_k}{(-xyq^5;q^4)_k}\\&\quad =\frac{(-xyq;q^4)_\infty (-xzq^3;q^4)_\infty }{(x^2q^6;q^4)_\infty } \sum \limits _{k=0}^\infty (-xyq)^k\frac{(-xy^{-1}q^5;q^4)_k}{(-xzq^3;q^4)_{k+1}}, \end{aligned}$$

where we used (2.5) in the last step.

We then multiply both sides of the above identity by

$$\begin{aligned}\frac{xyq(1+xy^{-1}q)}{1-x^2q^2}\end{aligned}$$

to get the desired result after \(k+1\) replaced by k. \(\square \)

3 Odd Partitions

Let \(O_{1,4}(x,y,z;q)\) and \(O_{3,4}(x,y,z;q)\) be the trivariate generating function defined by

$$\begin{aligned} O_{1,4}(x,y,z;q)&:=\sum _{\lambda \in \mathcal {O}_{1,4}}x^{\ell (\lambda )}y^{o_{1,4}(\lambda )}z^{o_{3,4}(\lambda )}q^{|\lambda |},\\ O_{3,4}(x,y,z;q)&:=\sum _{\lambda \in \mathcal {O}_{3,4}}x^{\ell (\lambda )}y^{o_{1,4}(\lambda )}z^{o_{3,4}(\lambda )}q^{|\lambda |}. \end{aligned}$$

Theorem 3.1

We have

$$\begin{aligned} O_{1,4}(x,y,z;q)&=\frac{(-xyq;q^4)_\infty (-xzq^3;q^4)_\infty }{(x^2q^2;q^4)_\infty } \sum \limits _{k=0}^\infty (-1)^k(xyq)^k\frac{(-xy^{-1}q;q^4)_k}{(-xzq^3;q^4)_k},\\ O_{3,4}(x,y,z;q)&=\frac{(-xyq;q^4)_\infty (-xzq^3;q^4)_\infty }{(x^2q^2;q^4)_\infty } \sum \limits _{k=1}^\infty (-1)^{k-1}(xyq)^k\frac{(-xy^{-1}q;q^4)_k}{(-xzq^3;q^4)_k}. \end{aligned}$$

Proof

The generating function for odd partitions with the minimal odd excludant \(4k+1\) is

$$\begin{aligned}&\prod _{i=1}^k(xyq^{4i-3}+x^2q^{2(4i-3)}+x^3yq^{3(4i-3)}+x^4q^{4(4i-3)}+\cdots )\\&\qquad \times \prod _{i=1}^k(xzq^{4i-1}+x^2q^{2(4i-1)}+x^3zq^{3(4i-1)}+x^4q^{4(4i-1)}+\cdots )\\&\qquad \times \prod _{i=k+1}^\infty (1+xyq^{4i+1}+x^2q^{2(4i+1)}+x^3yq^{3(4i+1)}+x^4q^{4(4i+1)}+\cdots )\\&\qquad \times \prod _{i=k}^\infty (1+xzq^{4i+3}+x^2q^{2(4i+3)}+x^3zq^{3(4i+3)}+x^4q^{4(4i+3)}+\cdots )\\&\quad =\prod _{i=1}^k\frac{xyq^{4i-3}+x^2q^{8i-6}}{1-x^2q^{8i-6}} \prod _{i=1}^k\frac{xzq^{4i-1}+x^2q^{8i-2}}{1-x^2q^{8i-2}} \prod _{i=k+1}^\infty \frac{1+xyq^{4i+1}}{1-x^2q^{8i+2}} \prod _{i=k}^\infty \frac{1+xzq^{4i+3}}{1-x^2q^{8i+6}}\\&\quad =x^{2k}y^{k}z^kq^{4k^2}\frac{(-xy^{-1}q;q^4)_k(-xz^{-1}q^3;q^4)_k(-xyq^{4k+5};q^4)_\infty (-xzq^{4k+3};q^4)_\infty }{(x^2q^2;q^8)_k(x^2q^6;q^8)_k(x^2q^{8k+10};q^8)_\infty (x^2q^{8k+6};q^8)_\infty }\\&\quad =(x^2yz)^{k}q^{4k^2}\\&\quad \quad \times \frac{(-xy^{-1}q;q^4)_k(-xz^{-1}q^3;q^4)_k(-xyq^{4k+5};q^4)_\infty (-xzq^{4k+3};q^4)_\infty (1-x^2q^{8k+2})}{(x^2q^{2};q^4)_\infty }\\&\quad =\frac{(-xyq;q^4)_\infty (-xzq^3;q^4)_\infty }{(x^2q^2;q^4)_\infty } \frac{(-xy^{-1}q;q^4)_k(-xz^{-1}q^3;q^4)_k(1-x^2q^{8k+2})(x^{2}yz)^{k}q^{4k^2}}{(-xyq;q^4)_{k+1}(-xzq^3;q^4)_k}. \end{aligned}$$

Summing over all k, we obtain

$$\begin{aligned}&O_{1,4}(x,y,z;q)\\&\quad =\sum \limits _{k=0}^{\infty }\frac{(-xyq;q^4)_\infty (-xzq^3;q^4)_\infty }{(x^2q^2;q^4)_\infty }\\&\qquad \times \frac{(-xy^{-1}q;q^4)_k(-xz^{-1}q^3;q^4)_k(1-x^2q^{8k+2})(x^{2}yz)^{k}q^{4k^2}}{(-xyq;q^4)_{k+1}(-xzq^3;q^4)_k}\\&\quad =\frac{(-xyq;q^4)_\infty (-xzq^3;q^4)_\infty }{(x^2q^2;q^4)_\infty }\\ {}&\qquad \times \sum \limits _{k=0}^{\infty } \frac{(-xy^{-1}q;q^4)_k(-xz^{-1}q^3;q^4)_k(1-x^2q^{8k+2})(x^{2}yz)^{k}q^{4k^2}}{(-xyq;q^4)_{k+1}(-xzq^3;q^4)_k}. \end{aligned}$$

Replacing q by \(q^4\), and letting \(\alpha =-xy^{-1}q,\beta =-xzq^3,\tau =-xyq\) in the Rogers–Fine identity (2.3), we get

$$\begin{aligned}&\sum \limits _{k=0}^\infty \frac{(-xy^{-1}q;q^4)_k}{(-xzq^3;q^4)_k}(-xyq)^k \\ {}&\quad =\sum \limits _{k=0}^\infty \frac{(-xy^{-1}q;q^4)_k(-xz^{-1}q^3;q^4)_k(1-x^2q^{8k+2})x^{2k}y^kz^kq^{4k^2}}{(-xyq;q^4)_{k+1}(-xzq^3;q^4)_k}. \end{aligned}$$

We now arrive at

$$\begin{aligned} O_{1,4}(x,y,z;q)=\frac{(-xyq;q^4)_\infty (-xzq^3;q^4)_\infty }{(x^2q^2;q^4)_\infty } \sum \limits _{k=0}^\infty (-xyq)^k\frac{(-xy^{-1}q;q^4)_k}{(-xzq^3;q^4)_k}. \end{aligned}$$

Similarly, we have

$$\begin{aligned} O_{3,4}(x,y,z;q)&=\frac{(-xyq;q^4)_\infty (-xzq^3;q^4)_\infty }{(x^2q^2;q^4)_\infty }\\&\quad \times \sum \limits _{k=0}^{\infty } \frac{(-xy^{-1}q;q^4)_{k+1}(-xz^{-1}q^3;q^4)_k(1-x^2q^{8k+6})x^{2k+1}y^{k+1}z^kq^{(2k+1)^2}}{(-xyq;q^4)_{k+1}(-xzq^3;q^4)_{k+1}}\\&=\frac{(-xyq^5;q^4)_\infty (-xzq^3;q^4)_\infty (1+xy^{-1}q)xyq}{(x^2q^2;q^4)_\infty }\\&\quad \times \sum \limits _{k=0}^{\infty } \frac{(-xy^{-1}q^5;q^4)_{k}(-xz^{-1}q^3;q^4)_k(1-x^2q^{8k+6})x^{2k}y^kz^kq^{4k^2+4k}}{(-xyq^5;q^4)_{k}(-xzq^3;q^4)_{k+1}}\\&=\frac{(-xyq^5;q^4)_\infty (-xzq^3;q^4)_\infty (1+xy^{-1}q)xyq}{(x^2q^2;q^4)_\infty } \sum \limits _{k=0}^\infty \frac{(-xz^{-1}q^3;q^4)_k}{(-xyq^5;q^4)_k}(-xzq^3)^k, \end{aligned}$$

where the last equality was derived by replacing q by \(q^4\) and setting

$$\begin{aligned}\alpha =-xz^{-1}q^3,\beta =-xyq^5,\tau =-xzq^3\end{aligned}$$

in the Rogers–Fine identity (2.3).

Combining the above equation and (2.5) together, we get

$$\begin{aligned} O_{3,4}(x,y,z;q)&=\frac{(-xyq^5;q^4)_\infty (-xzq^3;q^4)_\infty (1+xy^{-1}q)xyq}{(x^2q^2;q^4)_\infty }\\&\quad \times \frac{1+xyq}{1+xzq^3}\sum \limits _{k=0}^\infty \frac{(-xy^{-1}q^5;q^4)_k}{(-xzq^7;q^4)_k}(-xyq)^k\\&=\frac{(-xyq;q^4)_\infty (-xzq^3;q^4)_\infty }{(x^2q^2;q^4)_\infty } \sum \limits _{k=0}^\infty \frac{(-xy^{-1}q;q^4)_{k+1}}{(-xzq^3;q^4)_{k+1}}(-1)^k(xyq)^{k+1}\\&=\frac{(-xyq;q^4)_\infty (-xzq^3;q^4)_\infty }{(x^2q^2;q^4)_\infty }\sum \limits _{k=1}^\infty (-1)^{k-1}(xyq)^k\frac{(-xy^{-1}q;q^4)_k}{(-xzq^3;q^4)_k}. \end{aligned}$$

This completes the proof. \(\square \)

4 Distinct Partitions

For a partition \(\lambda \), let \(n_{e,o}(\lambda )\) and \(n_{e,e}(\lambda )\) be the number of odd and even indexed even parts of \(\lambda \), respectively. Denote by \(\mathcal {D}_k\) the set of distinct partitions of length k, and define

$$\begin{aligned} D^o_k(x,y,z;q)&:=\sum \limits _{\lambda \in \mathcal {D}_k}x^{\ell _a(\lambda )}y^{n_{o,o}(\lambda )}z^{n_{o,e}(\lambda )}q^{|\lambda |},\\ D^e_k(x,y,z;q)&:=\sum \limits _{\lambda \in \mathcal {D}_k}x^{\ell _a(\lambda )}y^{n_{e,o}(\lambda )}z^{n_{e,e}(\lambda )}q^{|\lambda |}. \end{aligned}$$

It is straightforward to see that \(D^o_0(x,y,z;q)=1\) and \(D^e_0(x,y,z;q)=1\), and

$$\begin{aligned} D^o_1(x,y,z;q)&=\sum \limits _{i=1}^\infty x^{2i-1}yq^{2i-1}+\sum \limits _{i=1}^\infty x^{2i}q^{2i} =\frac{xyq}{1-x^2q^2}+\frac{x^2q^2}{1-x^2q^2}\\&=\frac{xyq(1+xy^{-1}q)}{1-x^2q^2},\\ D^e_1(x,y,z;q)&=\sum \limits _{i=1}^\infty x^{2i-1}q^{2i-1}+\sum \limits _{i=1}^\infty x^{2i}yq^{2i} =\frac{xq}{1-x^2q^2}+\frac{x^2yq^2}{1-x^2q^2}\\&=\frac{xq(1+xyq)}{1-x^2q^2}. \end{aligned}$$

We next deduce the recurrence for \(D^o_{k}(x,y,z;q)\) and \(D^e_{k}(x,y,z;q)\).

Lemma 4.1

For \(k\ge 1\), we have

$$\begin{aligned} D^o_{2k}(x,y,z;q)&=\frac{zq^{2k}}{1-q^{4k}}D^e_{2k-1}(x,y,z;q)+\frac{q^{4k}}{1-q^{4k}} D^o_{2k-1}(x,y,z;q),\end{aligned}$$
(4.1)
$$\begin{aligned} D^e_{2k}(x,y,z;q)&=\frac{q^{2k}}{1-q^{4k}}D^o_{2k-1}(x,y,z;q)+\frac{zq^{4k}}{1-q^{4k}} D^e_{2k-1}(x,y,z;q),\end{aligned}$$
(4.2)
$$\begin{aligned} D^o_{2k-1}(x,y,z;q)&=\frac{xyq^{2k-1}}{1-x^2q^{4k-2}}D^e_{2k-2}(x,y,z;q)\nonumber \\ {}&\quad +\frac{x^2q^{4k-2}}{1-x^2q^{4k-2}} D^o_{2k-2}(x,y,z;q),\end{aligned}$$
(4.3)
$$\begin{aligned} D^e_{2k-1}(x,y,z;q)&=\frac{xq^{2k-1}}{1-x^2q^{4k-2}}D^o_{2k-2}(x,y,z;q)\nonumber \\ {}&\quad +\frac{x^2yq^{4k-2}}{1-x^2q^{4k-2}} D^e_{2k-2}(x,y,z;q). \end{aligned}$$
(4.4)

Proof

Given a partition \(\lambda =(\lambda _1,\lambda _2,\ldots ,\lambda _{2k})\in \mathcal {D}_{2k}\), define \(\pi \) to be the partition

$$\begin{aligned}(\lambda _1-\lambda _{2k},\lambda _2-\lambda _{2k},\ldots ,\lambda _{2k-1}-\lambda _{2k}).\end{aligned}$$

Clearly, \(\pi \in \mathcal {D}_{2k-1}\).

It is easy to see that \(|\lambda |=|\pi |+2k\lambda _{2k}\) and \(\ell _a(\lambda )=\ell _a(\pi )\). Moreover, we have

$$\begin{aligned} n_{o,o}(\lambda )=\left\{ \begin{array}{ll} n_{e,o}(\pi ), &{} \quad \hbox {if }\lambda _{2k}\hbox { is odd;}\\ n_{o,o}(\pi ), &{} \quad \hbox {if }\lambda _{2k}\hbox { is even,} \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} n_{o,e}(\lambda )=\left\{ \begin{array}{ll} n_{e,e}(\pi )+1, &{} \quad \hbox {if }\lambda _{2k}\hbox { is odd;}\\ n_{o,e}(\pi ), &{} \quad \hbox {if }\lambda _{2k}\hbox { is even.} \end{array} \right. \end{aligned}$$

We now can conclude that

$$\begin{aligned} \sum \limits _{\lambda \in \mathcal {D}_{2k}}x^{\ell _a(\lambda )}y^{n_{o,o}(\lambda )}z^{n_{o,e}(\lambda )}q^{|\lambda |}&=\sum \limits _{i=1}^\infty zq^{2k(2i-1)} \sum \limits _{\pi \in \mathcal {D}_{2k-1}}x^{\ell _a(\pi )}y^{n_{e,o}(\pi )}z^{n_{e,e}(\pi )}q^{|\pi |}\\&\quad +\sum \limits _{i=1}^\infty q^{2k\times 2i} \sum \limits _{\pi \in \mathcal {D}_{2k-1}}x^{\ell _a(\pi )}y^{n_{o,o}(\pi )}z^{n_{o,e}(\pi )}q^{|\pi |}\\&=\frac{zq^{2k}}{1-q^{4k}}D^e_{2k-1}(x,y,z;q)\\ {}&\quad +\frac{q^{4k}}{1-q^{4k}} D^o_{2k-1}(x,y,z;q), \end{aligned}$$

which shows that (4.1) is true. Similarly, we can prove (4.2) and omit the details here.

Letting \(\lambda =(\lambda _1,\lambda _2,\ldots ,\lambda _{2k-1})\in \mathcal {D}_{2k-1}\), define \(\pi \) to be the partition

$$\begin{aligned}(\lambda _1-\lambda _{2k-1},\lambda _2-\lambda _{2k-1},\ldots ,\lambda _{2k-2}-\lambda _{2k-1}).\end{aligned}$$

Obviously, \(\pi \in \mathcal {D}_{2k-2}\).

It is not hard to see that \(|\lambda |=|\pi |+(2k-1)\lambda _{2k-1}\) and \(\ell _a(\lambda )=\ell _a(\pi )+\lambda _{2k-1}\). Furthermore, we have

$$\begin{aligned} n_{o,o}(\lambda )=\left\{ \begin{array}{ll} n_{e,o}(\pi )+1, &{} \quad \hbox {if }\lambda _{2k-1}\hbox { is odd;}\\ n_{o,o}(\pi ), &{} \quad \hbox {if }\lambda _{2k-1}\hbox { is even,} \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} n_{o,e}(\lambda )=\left\{ \begin{array}{ll} n_{e,e}(\pi ), &{} \quad \hbox {if }\lambda _{2k-1}\hbox { is odd;}\\ n_{o,e}(\pi ), &{} \quad \hbox {if }\lambda _{2k-1}\hbox { is even.} \end{array} \right. \end{aligned}$$

We now can derive that

$$\begin{aligned} \sum \limits _{\lambda \in \mathcal {D}_{2k-1}}x^{\ell _a(\lambda )}y^{n_{o,o}(\lambda )}z^{n_{o,e}(\lambda )}q^{|\lambda |}&=\sum \limits _{i=1}^\infty x^{2i-1}yq^{(2k-1)(2i-1)}\\&\quad \times \sum \limits _{\pi \in \mathcal {D}_{2k-2}}x^{\ell _a(\pi )}y^{n_{e,o}(\pi )}z^{n_{e,e}(\pi )}q^{|\pi |}\\&\quad +\sum \limits _{i=1}^\infty x^{2i}q^{(2k-1)\times 2i}\\&\quad \times \sum \limits _{\pi \in \mathcal {D}_{2k-2}}x^{\ell _a(\pi )}y^{n_{o,o}(\pi )}z^{n_{o,e}(\pi )}q^{|\pi |}\\&=\frac{xyq^{2k-1}}{1-x^2q^{4k-2}}D^e_{2k-2}(x,y,z;q)\\&\quad +\frac{x^2q^{4k-2}}{1-x^2q^{4k-2}} D^o_{2k-2}(x,y,z;q), \end{aligned}$$

which proves (4.3). Similarly, we can show that (4.4) is true, and we omit the details here. \(\square \)

Theorem 4.2

For \(k\ge 0\), we have

$$\begin{aligned} D^o_{2k}(x,y,z;q)&=\frac{x^kq^{k(2k+1)}}{(x^2q^2;q^4)_k}\sum \limits _{i=0}^{k}\frac{(-xzq^{-1};q^4)_{i} (-xz^{-1}q^3;q^4)_{k-i}}{(q^4;q^4)_i(q^4;q^4)_{k-i}}y^iz^{k-i}q^{2i},\end{aligned}$$
(4.5)
$$\begin{aligned} D^e_{2k}(x,y,z;q)&=\frac{x^{k}q^{k(2k+1)}}{(x^2q^2;q^4)_{k}}\sum \limits _{i=0}^{k}\frac{(-xzq^3;q^4)_{i}(-xz^{-1}q^{-1};q^4)_{k-i} }{(q^4;q^4)_i(q^4;q^4)_{k-i}}y^{i}z^{k-i}q^{2(k-i)},\end{aligned}$$
(4.6)
$$\begin{aligned} D^o_{2k+1}(x,y,z;q)&=\frac{x^{k+1}q^{(k+1)(2k+1)}}{(x^2q^2;q^4)_{k+1}}\sum \limits _{i=0}^{k}\frac{(-xyq;q^4)_{i} (-xy^{-1}q;q^4)_{k-i+1}}{(q^4;q^4)_i(q^4;q^4)_{k-i}}y^{k-i+1}z^{i}q^{2i},\end{aligned}$$
(4.7)
$$\begin{aligned} D^e_{2k+1}(x,y,z;q)&=\frac{x^{k+1}q^{(k+1)(2k+1)}}{(x^2q^2;q^4)_{k+1}}\sum \limits _{i=0}^{k}\frac{(-xyq;q^4)_{i+1} (-xy^{-1}q;q^4)_{k-i}}{(q^4;q^4)_i(q^4;q^4)_{k-i}}y^{k-i}z^{i}q^{2(k-i)}. \end{aligned}$$
(4.8)

Proof

We proceed by induction on k. It is trivial for the case where \(k=0\).

We first prove (4.5) and (4.6). It follows from Lemma 4.1 that

$$\begin{aligned} \left[ \begin{array}{c} D^o_{2k}(x,y,z;q)\\ D^e_{2k}(x,y,z;q)\\ \end{array} \right] =\frac{q^{2k}}{1-q^{4k}} \left[ \begin{array}{cc} q^{2k} &{}z \\ 1 &{}zq^{2k} \\ \end{array} \right] \left[ \begin{array}{c} D^o_{2k-1}(x,y,z;q)\\ D^e_{2k-1}(x,y,z;q)\\ \end{array} \right] \end{aligned}$$
(4.9)

and

$$\begin{aligned} \left[ \begin{array}{c} D^o_{2k-1}(x,y,z;q)\\ D^e_{2k-1}(x,y,z;q)\\ \end{array} \right] =\frac{xq^{2k-1}}{1-x^2q^{4k-2}} \left[ \begin{array}{cc} xq^{2k-1} &{}y \\ 1 &{}xyq^{2k-1} \\ \end{array} \right] \left[ \begin{array}{c} D^o_{2k-2}(x,y,z;q)\\ D^e_{2k-2}(x,y,z;q)\\ \end{array} \right] , \end{aligned}$$

where we used the matrix notation and matrix product. We now conclude that

$$\begin{aligned} \left[ \begin{array}{c} D^o_{2k}(x,y,z;q)\\ D^e_{2k}(x,y,z;q)\\ \end{array} \right] = \left[ \begin{array}{cc} \frac{x^2q^{8k-2}+xzq^{4k-1}}{(1-q^{4k})(1-x^2q^{4k-2})} &{}\frac{xyq^{6k-1}+x^2yzq^{6k-2}}{(1-q^{4k})(1-x^2q^{4k-2})}\\ \frac{x^2q^{6k-2}+xzq^{6k-1}}{(1-q^{4k})(1-x^2q^{4k-2})} &{}\frac{xyq^{4k-1}+x^2yzq^{8k-2}}{(1-q^{4k})(1-x^2q^{4k-2})}\\ \end{array} \right] \left[ \begin{array}{c} D^o_{2k-2}(x,y,z;q)\\ D^e_{2k-2}(x,y,z;q)\\ \end{array} \right] . \end{aligned}$$

We now see that \(D^o_{2k}(x,y,z;q)\) satisfies

$$\begin{aligned} D^o_{2k}(x,y,z;q)&=\frac{x^2q^{8k-2}+xzq^{4k-1}}{(1-q^{4k})(1-x^2q^{4k-2})}D^o_{2k-2}(x,y,z;q)\\&\quad +\frac{xyq^{6k-1}+x^2yzq^{6k-2}}{(1-q^{4k})(1-x^2q^{4k-2})}D^e_{2k-2}(x,y,z;q)\\&=\frac{(xq^{4k-1}+z)xq^{4k-1}}{(1-q^{4k})(1-x^2q^{4k-2})}\frac{x^{k-1}q^{2k^2-3k+1}}{(x^2q^2;q^4)_{k-1}}\\&\quad \times \sum \limits _{i=0}^{k-1}\frac{(-xzq^{-1};q^4)_{i}(-xz^{-1}q^3;q^4)_{k-i-1}y^iz^{k-i-1}q^{2i}}{(q^4;q^4)_i(q^4;q^4)_{k-i-1}}\\&\quad +\frac{(yq^{2k}+xyzq^{2k-1})xq^{4k-1}}{(1-q^{4k})(1-x^2q^{4k-2})}\frac{x^{k-1}q^{2k^2-3k+1}}{(x^2q^2;q^4)_{k-1}}\\&\quad \times \sum \limits _{i=0}^{k-1}\frac{(-xzq^3;q^4)_{i}(-xz^{-1}q^{-1};q^4)_{k-i-1}y^{i}z^{k-i-1}q^{2(k-i-1)}}{(q^4;q^4)_i(q^4;q^4)_{k-i-1}}\\&=\frac{x^kq^{k(2k+1)}(1+xz^{-1}q^{4k-1})}{(x^2q^2;q^4)_k(1-q^{4k})}\\&\quad \times \sum \limits _{i=0}^{k-1}\frac{(-xzq^{-1};q^4)_{i}(-xz^{-1}q^3;q^4)_{k-i-1}y^iz^{k-i}q^{2i}}{(q^4;q^4)_i(q^4;q^4)_{k-i-1}}\\&\quad +\frac{x^kq^{k(2k+1)}}{(x^2q^2;q^4)_k(1-q^{4k})}\sum \limits _{i=0}^{k-1} \frac{(-xzq^{-1};q^4)_{i+1}(-xz^{-1}q^{-1};q^4)_{k-i-1}}{(q^4;q^4)_i(q^4;q^4)_{k-i-1}}\\&\quad \times y^{i+1}z^{k-i-1}q^{4k-2i-2}\\&=\frac{x^kq^{k(2k+1)}(1+xz^{-1}q^{4k-1})}{(x^2q^2;q^4)_k(1-q^{4k})}\\&\quad \times \sum \limits _{i=0}^{k-1}\frac{(-xzq^{-1};q^4)_{i}(-xz^{-1}q^3;q^4)_{k-i-1}y^iz^{k-i}q^{2i}}{(q^4;q^4)_i(q^4;q^4)_{k-i-1}}\\&\quad +\frac{x^kq^{k(2k+1)}}{(x^2q^2;q^4)_k(1-q^{4k})}\\ {}&\quad \times \sum \limits _{i=1}^{k}\frac{(-xzq^{-1};q^4)_{i}(-xz^{-1}q^{-1};q^4)_{k-i}y^{i}z^{k-i}q^{4k-2i}}{(q^4;q^4)_{i-1}(q^4;q^4)_{k-i}}\\&=\frac{x^kq^{k(2k+1)}}{(x^2q^2;q^4)_k(1-q^{4k})}\Bigg (\frac{(-xz^{-1}q^3;q^4)_{k-1}(1+xz^{-1}q^{4k-1})z^k}{(q^4;q^4)_{k-1}}\\&\quad +(1+xz^{-1}q^{4k-1})\\ {}&\quad \times \sum \limits _{i=1}^{k-1}\frac{(-xzq^{-1};q^4)_{i}(-xz^{-1}q^3;q^4)_{k-i-1}y^iz^{k-i}q^{2i}}{(q^4;q^4)_i(q^4;q^4)_{k-i-1}}\\&\quad +\sum \limits _{i=1}^{k-1}\frac{(-xzq^{-1};q^4)_{i}(-xz^{-1}q^{-1};q^4)_{k-i}y^{i}z^{k-i}q^{4k-2i}}{(q^4;q^4)_{i-1}(q^4;q^4)_{k-i}}\\ {}&\quad +\frac{(-xzq^{-1};q^4)_{k}y^{k}q^{2k}}{(q^4;q^4)_{k-1}}\Bigg )\\&=\frac{x^kq^{k(2k+1)}}{(x^2q^2;q^4)_k(1-q^{4k})}\Bigg (\frac{(-xz^{-1}q^3;q^4)_{k}z^k}{(q^4;q^4)_{k-1}}+\frac{(-xzq^{-1};q^4)_{k}y^{k}q^{2k}}{(q^4;q^4)_{k-1}}\\&\quad +(1+xz^{-1}q^{4k-1})\sum \limits _{i=1}^{k-1}\frac{(-xzq^{-1};q^4)_{i}(-xz^{-1}q^3;q^4)_{k-i-1}y^iz^{k-i}q^{2i}}{(q^4;q^4)_i(q^4;q^4)_{k-i-1}}\\&\quad +(1+xz^{-1}q^{-1})\sum \limits _{i=1}^{k-1}\frac{(-xzq^{-1};q^4)_{i}(-xz^{-1}q^{3};q^4)_{k-i-1}y^{i}z^{k-i}q^{4k-2i}}{(q^4;q^4)_{i-1}(q^4;q^4)_{k-i}}\Bigg )\\&=\frac{x^kq^{k(2k+1)}}{(x^2q^2;q^4)_k(1-q^{4k})}\Bigg (\frac{(-xz^{-1}q^3;q^4)_{k}z^k}{(q^4;q^4)_{k-1}} +\frac{(-xzq^{-1};q^4)_{k}y^{k}q^{2k}}{(q^4;q^4)_{k-1}}\\&\quad +\sum \limits _{i=1}^{k-1}\frac{(-xzq^{-1};q^4)_{i}(-xz^{-1}q^3;q^4)_{k-i-1}y^iz^{k-i}q^{2i}}{(q^4;q^4)_i(q^4;q^4)_{k-i}}\\&\quad \times \left( (1+xz^{-1}q^{4k-1})(1-q^{4k-4i})+(1+xz^{-1}q^{-1})(1-q^{4i})q^{4k-4i}\right) \Bigg )\\&=\frac{x^kq^{k(2k+1)}}{(x^2q^2;q^4)_k(1-q^{4k})}\Bigg (\frac{(-xz^{-1}q^3;q^4)_{k}z^k}{(q^4;q^4)_{k-1}}+\frac{(-xzq^{-1};q^4)_{k}y^{k}q^{2k}}{(q^4;q^4)_{k-1}}\\&\quad +\sum \limits _{i=1}^{k-1}\frac{(-xzq^{-1};q^4)_{i}(-xz^{-1}q^3;q^4)_{k-i-1}y^iz^{k-i}q^{2i}}{(q^4;q^4)_i(q^4;q^4)_{k-i}} \\ {}&\quad \times (1-q^{4k})(1+xz^{-1}q^{4k-4i-1})\Bigg )\\&=\frac{x^kq^{k(2k+1)}}{(x^2q^2;q^4)_k(1-q^{4k})}\Bigg (\frac{(-xz^{-1}q^3;q^4)_{k}z^k}{(q^4;q^4)_{k-1}}+\frac{(-xzq^{-1};q^4)_{k}y^{k}q^{2k}}{(q^4;q^4)_{k-1}}\\&\quad +\sum \limits _{i=1}^{k-1}\frac{(-xzq^{-1};q^4)_{i}(-xz^{-1}q^3;q^4)_{k-i}y^iz^{k-i}q^{2i}}{(q^4;q^4)_i(q^4;q^4)_{k-i}} (1-q^{4k})\Bigg )\\&=\frac{x^kq^{k(2k+1)}}{(x^2q^2;q^4)_k}\Bigg (\frac{(-xz^{-1}q^3;q^4)_{k}z^k}{(q^4;q^4)_{k}}+\frac{(-xzq^{-1};q^4)_{k}y^{k}q^{2k}}{(q^4;q^4)_{k}}\\&\quad +\sum \limits _{i=1}^{k-1}\frac{(-xzq^{-1};q^4)_{i}(-xz^{-1}q^3;q^4)_{k-i}y^iz^{k-i}q^{2i}}{(q^4;q^4)_i(q^4;q^4)_{k-i}}\Bigg )\\&=\frac{x^kq^{k(2k+1)}}{(x^2q^2;q^4)_k}\sum \limits _{i=0}^{k}\frac{(-xzq^{-1};q^4)_{i} (-xz^{-1}q^3;q^4)_{k-i}}{(q^4;q^4)_i(q^4;q^4)_{k-i}}y^iz^{k-i}q^{2i}. \end{aligned}$$

Thus, (4.5) is true. We can prove (4.6) in a similar manner, and omit the details here.

We next turn to (4.7) and (4.8). It follows from (4.3) and (4.4) that

$$\begin{aligned} \left[ \begin{array}{c} D^o_{2k+1}(x,y,z;q)\\ D^e_{2k+1}(x,y,z;q)\\ \end{array} \right] =\frac{xq^{2k+1}}{1-x^2q^{4k+2}} \left[ \begin{array}{cc} xq^{2k+1} &{}y \\ 1 &{}xyq^{2k+1} \\ \end{array} \right] \left[ \begin{array}{c} D^o_{2k}(x,y,z;q)\\ D^e_{2k}(x,y,z;q)\\ \end{array} \right] . \end{aligned}$$

Combining the above equation and (4.9) together, we obtain

$$\begin{aligned} \left[ \begin{array}{c} D^o_{2k+1}(x,y,z;q)\\ D^e_{2k+1}(x,y,z;q)\\ \end{array} \right] = \left[ \begin{array}{cc} \frac{xyq^{4k+1}(1+xy^{-1}q^{4k+1})}{(1-q^{4k})(1-x^2q^{4k+2})} &{}\frac{xyzq^{6k+1}(1+xy^{-1}q)}{(1-q^{4k})(1-x^2q^{4k+2})}\\ \frac{xq^{6k+1}(1+xyq)}{(1-q^{4k})(1-x^2q^{4k+2})} &{}\frac{xzq^{4k+1}(1+xyq^{4k+1})}{(1-q^{4k})(1-x^2q^{4k+2})}\\ \end{array} \right] \left[ \begin{array}{c} D^o_{2k-1}(x,y,z;q)\\ D^e_{2k-1}(x,y,z;q)\\ \end{array} \right] . \end{aligned}$$

Now, we can verify that

$$\begin{aligned} D^o_{2k+1}(x,y,z;q)&=\frac{xyq^{4k+1}(1+xy^{-1}q^{4k+1})}{(1-q^{4k})(1-x^2q^{4k+2})}D^o_{2k-1}(x,y,z;q)\\&\quad +\frac{xyzq^{6k+1}(1+xy^{-1}q)}{(1-q^{4k})(1-x^2q^{4k+2})}D^e_{2k-1}(x,y,z;q)\\&=\frac{xyq^{4k+1}(1+xy^{-1}q^{4k+1})}{(1-q^{4k})(1-x^2q^{4k+2})} \frac{x^{k}q^{k(2k-1)}}{(x^2q^2;q^4)_{k}}\\&\quad \times \sum \limits _{i=0}^{k-1}\frac{(-xyq;q^4)_{i} (-xy^{-1}q;q^4)_{k-i}y^{k-i}z^{i}q^{2i}}{(q^4;q^4)_i(q^4;q^4)_{k-i-1}}\\&\quad +\frac{xyzq^{6k+1}(1+xy^{-1}q)}{(1-q^{4k})(1-x^2q^{4k+2})} \frac{x^{k}q^{k(2k-1)}}{(x^2q^2;q^4)_{k}}\\&\quad \times \sum \limits _{i=0}^{k-1}\frac{(-xyq;q^4)_{i+1} (-xy^{-1}q;q^4)_{k-i-1}y^{k-i-1}z^{i}q^{2(k-i-1)}}{(q^4;q^4)_i(q^4;q^4)_{k-i-1}}\\&=\frac{x^{k+1}q^{2k^2+3k+1}(1+xy^{-1}q^{4k+1})}{(1-q^{4k})(x^2q^2;q^4)_{k+1}}\\&\quad \times \sum \limits _{i=0}^{k-1}\frac{(-xyq;q^4)_{i} (-xy^{-1}q;q^4)_{k-i}y^{k-i+1}z^{i}q^{2i}}{(q^4;q^4)_i(q^4;q^4)_{k-i-1}}\\&\quad +\frac{x^{k+1}q^{2k^2+3k+1}(1+xy^{-1}q)}{(1-q^{4k})(x^2q^2;q^4)_{k+1}}\\&\quad \times \sum \limits _{i=0}^{k-1}\frac{(-xyq;q^4)_{i+1} (-xy^{-1}q;q^4)_{k-i-1}y^{k-i}z^{i+1}q^{4k-2i-2}}{(q^4;q^4)_i(q^4;q^4)_{k-i-1}}\\&=\frac{x^{k+1}q^{2k^2+3k+1}(1+xy^{-1}q^{4k+1})}{(1-q^{4k})(x^2q^2;q^4)_{k+1}}\\&\quad \times \sum \limits _{i=0}^{k-1}\frac{(-xyq;q^4)_{i} (-xy^{-1}q;q^4)_{k-i}y^{k-i+1}z^{i}q^{2i}}{(q^4;q^4)_i(q^4;q^4)_{k-i-1}}\\&\quad +\frac{x^{k+1}q^{2k^2+3k+1}(1+xy^{-1}q)}{(1-q^{4k})(x^2q^2;q^4)_{k+1}}\\&\quad \times \sum \limits _{i=1}^{k}\frac{(-xyq;q^4)_{i} (-xy^{-1}q;q^4)_{k-i} y^{k-i+1}z^{i}q^{4k-2i}}{(q^4;q^4)_{i-1}(q^4;q^4)_{k-i}}\\&=\frac{x^{k+1}q^{2k^2+3k+1}}{(1-q^{4k})(x^2q^2;q^4)_{k+1}}\Bigg (\frac{(1+xy^{-1}q^{4k+1})(-xy^{-1}q;q^4)_ky^{k+1}}{(q^4;q^4)_{k-1}}\\&\quad +(1+xy^{-1}q^{4k+1})\\ {}&\quad \times \sum \limits _{i=1}^{k-1}\frac{(-xyq;q^4)_{i} (-xy^{-1}q;q^4)_{k-i}y^{k-i+1}z^{i}q^{2i}}{(q^4;q^4)_i(q^4;q^4)_{k-i-1}}\\&\quad +(1+xy^{-1}q)\\ {}&\quad \times \sum \limits _{i=1}^{k-1}\frac{(-xyq;q^4)_{i} (-xy^{-1}q;q^4)_{k-i}y^{k-i+1}z^{i}q^{4k-2i}}{(q^4;q^4)_{i-1}(q^4;q^4)_{k-i}}\\&\quad +\frac{(1+xy^{-1}q)(-xyq;q^4)_kyz^kq^{2k}}{(q^4;q^4)_{k-1}}\Bigg )\\&=\frac{x^{k+1}q^{2k^2+3k+1}}{(1-q^{4k})(x^2q^2;q^4)_{k+1}}\Bigg (\frac{(1+xy^{-1}q^{4k+1})(-xy^{-1}q;q^4)_ky^{k+1}}{(q^4;q^4)_{k-1}}\\&\quad +\sum \limits _{i=1}^{k-1}\frac{(-xyq;q^4)_{i} (-xy^{-1}q;q^4)_{k-i}y^{k-i+1}z^{i}q^{2i}}{(q^4;q^4)_i(q^4;q^4)_{k-i}}\\&\quad \times \left( (1+xy^{-1}q^{4k+1})(1-q^{4k-4i})+(1+xy^{-1}q)(1-q^{4i})q^{4k-4i}\right) \\&\quad +\frac{(1+xy^{-1}q)(-xyq;q^4)_kyz^kq^{2k}}{(q^4;q^4)_{k-1}}\Bigg )\\&=\frac{x^{k+1}q^{2k^2+3k+1}}{(1-q^{4k})(x^2q^2;q^4)_{k+1}}\Bigg (\frac{(1+xy^{-1}q^{4k+1})(-xy^{-1}q;q^4)_ky^{k+1}}{(q^4;q^4)_{k-1}}\\&\quad +\sum \limits _{i=1}^{k-1}\frac{(-xyq;q^4)_{i} (-xy^{-1}q;q^4)_{k-i}y^{k-i+1}z^{i}q^{2i}}{(q^4;q^4)_i(q^4;q^4)_{k-i}}\\&\quad \times (1-q^{4k})(1+xy^{-1}q^{4k-4i+1})\\&\quad +\frac{(1+xy^{-1}q)(-xyq;q^4)_kyz^kq^{2k}}{(q^4;q^4)_{k-1}}\Bigg )\\&=\frac{x^{k+1}q^{2k^2+3k+1}}{(1-q^{4k})(x^2q^2;q^4)_{k+1}}\Bigg (\frac{(1+xy^{-1}q^{4k+1})(-xy^{-1}q;q^4)_ky^{k+1}}{(q^4;q^4)_{k-1}}\\&\quad +\sum \limits _{i=1}^{k-1}\frac{(-xyq;q^4)_{i} (-xy^{-1}q;q^4)_{k-i+1}y^{k-i+1}z^{i}q^{2i}}{(q^4;q^4)_i(q^4;q^4)_{k-i}}(1-q^{4k})\\&\quad +\frac{(1+xy^{-1}q)(-xyq;q^4)_kyz^kq^{2k}}{(q^4;q^4)_{k-1}}\Bigg )\\&=\frac{x^{k+1}q^{2k^2+3k+1}}{(x^2q^2;q^4)_{k+1}}\Bigg (\frac{(-xy^{-1}q;q^4)_{k+1}}{(q^4;q^4)_{k}}y^{k+1}\\&\quad +\sum \limits _{i=1}^{k-1}\frac{(-xyq;q^4)_{i} (-xy^{-1}q;q^4)_{k-i+1}}{(q^4;q^4)_i(q^4;q^4)_{k-i}}y^{k-i+1}z^{i}q^{2i}\\&\quad +\frac{(1+xy^{-1}q)(-xyq;q^4)_k}{(q^4;q^4)_{k}}yz^kq^{2k}\Bigg )\\&=\frac{x^{k+1}q^{(k+1)(2k+1)}}{(x^2q^2;q^4)_{k+1}}\sum \limits _{i=0}^{k}\frac{(-xyq;q^4)_{i} (-xy^{-1}q;q^4)_{k-i+1}}{(q^4;q^4)_i(q^4;q^4)_{k-i}}y^{k-i+1}z^{i}q^{2i}. \end{aligned}$$

Therefore, (4.7) is true. A similar argument yields (4.8). \(\square \)

As a consequence of Theorem 4.2, we have the following result.

Corollary 4.3

We have

$$\begin{aligned} \sum \limits _{k=0}^\infty D_{2k}^o(x,y,z;q)&=\frac{(-xyq;q^4)_\infty (-xzq^3;q^4)_\infty }{(x^2q^2;q^4)_\infty } \sum \limits _{k=0}^\infty (-1)^k(xyq)^k\frac{(-xy^{-1}q;q^4)_k}{(-xzq^3;q^4)_k},\\ \sum \limits _{k=0}^\infty D_{2k+1}^o(x,y,z;q)&=\frac{(-xyq;q^4)_\infty (-xzq^3;q^4)_\infty }{(x^2q^2;q^4)_\infty } \sum \limits _{k=1}^\infty (-1)^{k-1}(xyq)^k\frac{(-xy^{-1}q;q^4)_k}{(-xzq^3;q^4)_k}. \end{aligned}$$

Proof

Employing (4.5), it is routine to check that

$$\begin{aligned} \sum \limits _{k=0}^\infty D^o_{2k}(x,y,z;q)&=\sum \limits _{k=0}^\infty \frac{x^kq^{k(2k+1)}}{(x^2q^2;q^4)_k}\sum \limits _{i=0}^{k}\frac{(-xzq^{-1};q^4)_{i} (-xz^{-1}q^3;q^4)_{k-i}y^iz^{k-i}q^{2i}}{(q^4;q^4)_i(q^4;q^4)_{k-i}}\\&=\sum \limits _{i=0}^\infty \frac{(-xzq^{-1};q^4)_iy^iz^{-i}q^{2i}}{(q^4;q^4)_i}\\ {}&\quad \times \sum \limits _{k=i}^\infty \frac{(-xz^{-1}q^3;q^4)_{k-i}x^kz^kq^{k(2k+1)}}{(x^2q^2;q^4)_k(q^4;q^4)_{k-i}}\\&=\sum \limits _{i=0}^\infty \frac{(-xzq^{-1};q^4)_iy^iz^{-i}q^{2i}}{(q^4;q^4)_i}\\ {}&\quad \times \sum \limits _{k=0}^\infty \frac{(-xz^{-1}q^3;q^4)_{k}x^{k+i}z^{k+i}q^{(k+i)(2k+2i+1)}}{(x^2q^2;q^4)_{k+i}(q^4;q^4)_{k}}\\&=\sum \limits _{i=0}^\infty \frac{(-xzq^{-1};q^4)_ix^iy^iq^{2i^2+3i}}{(x^2q^2;q^4)_i(q^4;q^4)_i}\\ {}&\quad \times \sum \limits _{k=0}^\infty \frac{(-xz^{-1}q^3;q^4)_{k}x^{k}z^{k}q^{2k^2+(4i+1)k}}{(x^2q^{4i+2};q^4)_{k}(q^4;q^4)_{k}}\\&=\frac{(-xyq;q^4)_\infty (-xzq^3;q^4)_\infty }{(x^2q^2;q^4)_\infty }\sum \limits _{k=0}^\infty (-1)^k(xyq)^k\frac{(-xy^{-1}q;q^4)_k}{(-xzq^3;q^4)_k}, \end{aligned}$$

where the last equality follows from Theorem 2.5.

Similarly, we can show that

$$\begin{aligned} \sum \limits _{k=0}^\infty D^o_{2k+1}(x,y,z;q)&=\sum \limits _{i=0}^\infty \frac{(-xyq;q^4)_ix^iz^iq^{2i^2+5i}}{(x^2q^2;q^4)_{i+1}(q^4;q^4)_i}\\&\quad \times \sum \limits _{k=0}^\infty \frac{(-xy^{-1}q;q^4)_{k+1}(xy)^{k+1}q^{2k^2+(4i+3)k+1}}{(x^2q^{4i+6};q^4)_k(q^4;q^4)_k}. \end{aligned}$$

Applying Corollary 2.6 to the above equality yields the second equation. \(\square \)

We conclude this section with the following result, which shows Theorem 1.3 is true.

Corollary 4.4

We have

$$\begin{aligned} O_{1,4}(x,y,z;q)&=\sum \limits _{k=0}^\infty D^o_{2k}(x,y,z;q),\\ O_{3,4}(x,y,z;q)&=\sum \limits _{k=0}^\infty D^o_{2k+1}(x,y,z;q). \end{aligned}$$

Proof

The desired result follows immediately from Theorem 3.1 and Corollary 4.3. \(\square \)

5 Concluding Remarks

It was pointed out by one anonymous referee that Berkovich and Uncu [4] have discussed the location of odd parts in distinct partitions. They established the following elegant result.

Theorem 5.1

The number of distinct partitions of n with i odd indexed odd parts and j even indexed odd parts is the same as the number of distinct partitions of n with i parts congruent to 1 modulo 4 and j parts congruent to 3 modulo 4.

As an immediate consequence of Theorem 1.3 and Theorem 5.1, we get a simple refinement of Euler’s partition theorem.

Corollary 5.2

The set of odd partitions of n with i different parts congruent to 1 modulo 4 and occurring an odd number of times, and j different parts congruent to 3 modulo 4 and occurring an odd number of times is equinumerous with the set of distinct partitions of n with i parts that are congruent to 1 modulo 4, and j parts that are congruent to 3 modulo 4.

In fact, Corollary 5.2 can be easily shown by applying Glaisher’s merging/splitting proof of Theorem 1.1. See [11] for details. If we take other pairs of statistics into account, such as the length of odd partitions and the alternating sum of distinct partitions or the minimal odd excludant of odd partitions and the length of distinct partitions, Corollary 5.2 is no longer true. Namely, Corollary 5.2 cannot be strengthened to be similar in flavor to Theorem 1.3.