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On partitions with bounded multiplicity restrictions on even parts and with the largest part being odd

Published Nov 03, 2026Fresh — added to the Hub within the last 30 daysv18 min read

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Abstract

Andrews and El Bachraoui (RACSAM 119, Article Number 4 (2025)) established identities relating $4$ -regular partitions to partitions with distinct even parts whose largest part is odd. As an extension to their work, in this paper we define partitions of a positive integer $n$ with bounded multiplicity restriction on even parts and whose largest part is odd. We prove some identities relating our partition functions and $2r$ -regular partition function, and show that identities due to Andrews and El Bachraoui follow as particular cases.

Introduction

Throughout this paper, let $a$ and $q$ be a complex numbers satisfying $|q| < 1$ . Define $q$ -series [1, 10] in standard notation as

(a;q)_0 = 1,\qquad(a;q)_n = \prod_{j=0}^{n-1}(1-aq^j)\qquad\text{and}\qquad(a;q)_\infty= \prod_{j=0}^{\infty}(1-aq^j). \tag*{(1)}

The following elementary consequences of (1.1) will be used in the paper without referring explicitely:

(a;q)_{m+n}=(a;q)_m(aq^m;q)_n,\qquad(a;q)_\infty=(a;q)_n(aq^n;q)_\infty,\qquad\text{and}\qquad(a;q)_\infty=(a;q^2)_\infty(aq;q^2)_\infty,

where $m$ and $n$ are positive integers. A non-increasing finite sequence of positive integers $\beta_1 \ge\beta_2 \ge\cdots\ge\beta_k > 0$ is said to be a partition of a positive integer $n$ if $n = \sum_{i=1}^{k}\beta_i$ . The integers $\beta_i$ are called the parts of the partition. If $p(n)$ denotes the number of partitions of $n$ , then its generating function [8] is given by

\sum_{n=0}^{\infty}p(n)q^n = \frac{1}{(q;q)_{\infty}};\qquad p(0)=1.

For example, the partitions of $p(7)=15$ and the relevant partitions are given by $7$ , $6+1$ , $5+2$ , $5+1+1$ , $4+3$ , $4+2+1$ , $4+1+1+1$ , $3+3+1$ , $3+2+2$ , $3+2+1+1$ , $3+1+1+1+1$ , $2+2+2+1$ , $2+2+1+1+1$ , $2+1+1+1+1+1$ , and $1+1+1+1+1+1+1$ .

For any integers $l \ge2$ , an $l$ -regular partition of a positive integer $n$ is a partition of $n$ wherein none of the parts is divisible by $l$ . Let $R_l(n)$ denote the number of $l$ -regular partitions of $n$ , then its generating function is given by

\sum_{n=0}^{\infty}R_l(n)q^n = \frac{(q^l;q^l)_{\infty}}{(q;q)_{\infty}}.

For example, the partitions of $R_4(7)=12$ and the relevant partitions are given by $7$ , $6+1$ , $5+2$ , $5+1+1$ , $3+3+1$ , $3+2+2$ , $3+2+1+1$ , $3+1+1+1+1$ , $2+2+2+1$ , $2+2+1+1+1$ , $2+1+1+1+1+1$ , and $1+1+1+1+1+1+1$ .

Also, for any nonnegative integer $n$ let $ped(n)$ denote the number of partitions of $n$ with distinct even parts. Then the generating function of $ped(n)$ [2] is given by

\sum_{n=0}^{\infty}ped(n)q^n = \frac{(-q^2;q^2)_{\infty}}{(q;q^2)_{\infty}} = \frac{(q^4;q^4)_{\infty}}{(q;q)_{\infty}}. \tag*{(2)}

For example, the partitions of $ped(7)=12$ and the relevant partitions are given by $7$ , $6+1$ , $5+2$ , $5+1+1$ , $4+3$ , $4+2+1$ , $4+1+1+1$ , $3+3+1$ , $3+2+1+1$ , $3+1+1+1+1$ , $2+1+1+1+1+1$ , and $1+1+1+1+1+1+1$ . The second equality of (1.2) follows from the first equality by elementary $q$ -series manipulations. Therefore, one clearly has, $R_4(n)=ped(n)$ . For the arithmetic properties and related result of $ped(n)$ see [2, 3, 6, 13, 14] and for $R_l(n)$ see [5, 7, 9, 11, 12].

Recently, Andrews and El Bachraoui [4] derived some relations relating certain restricted partitions into distinct even parts with restricted 4-regular partitions. They defined three partition functions, $\mathrm{DE1}(n)$ which counts the number of partitions of $n$ in which even parts are distinct and the largest part is odd; $\mathrm{DE2}(n)$ which counts the number of partitions of $n$ in which even parts are distinct, the largest part is odd and appears at least twice; and $\mathrm{DE3}(n)$ which counts the number of partitions of $n$ in which even parts are distinct, the largest part is odd and appears exactly once. For example, $\mathrm{DE1}(7)=7$ with the partitions given by $7$ , $5+2$ , $5+1+1$ , $3+3+1$ , $3+2+1+1$ , $3+1+1+1+1$ , $1+1+1+1+1+1+1$ ; $\mathrm{DE2}(7)=2$ with the partitions given by $3+3+1$ , $1+1+1+1+1+1+1$ ; and $\mathrm{DE3}(7)=5$ with the partitions given by $7$ , $5+2$ , $5+1+1$ , $3+2+1+1$ , $3+1+1+1+1$ . The corresponding generating functions [4] are defined as follows:

\sum_{n=0}^{\infty}\mathrm{DE1}(n)q^n=\sum_{n=0}^{\infty}\frac{(-q^2;q^2)_nq^{2n+1}}{(q;q^2)_{n+1}},

\sum_{n=0}^{\infty}\mathrm{DE2}(n)q^n=\sum_{n=0}^{\infty}\frac{(-q^2;q^2)_nq^{4n+2}}{(q;q^2)_{n+1}},

and

\sum_{n=0}^{\infty}\mathrm{DE3}(n)q^n=\sum_{n=0}^{\infty}\frac{(-q^2;q^2)_nq^{2n+1}}{(q;q^2)_n}.

They further derived the following partition relations [4 Corollary 1, 2, 3 and 4], respectively:

(i) For $n>0$ , $\mathrm{DE1}(n)+\mathrm{DE1}(n-1)$ equals the number of the 4-regular partitions of $n$ .

(ii) For $n>0$ , $\mathrm{DE2}(n)+\mathrm{DE2}(n-3)$ equals the number of 4-regular partitions of $n$ into parts each $>1$ .

(iii) For $n>1$ , $\mathrm{DE3}(n+2)+\mathrm{DE3}(n-1)$ equals the number of 4-regular partitions of $n$ .

(iv) For $n>1$ , $\mathrm{DE3}(n+2)+\mathrm{DE3}(n-1)=\mathrm{DE1}(n)+\mathrm{DE1}(n-1)$ .

As an extension to the work done in [4], in this paper we define partitions of a positive integer $n$ with bounded multiplicity restrictions on even parts and with the largest part being odd. We establish their generating functions and prove some identities relating the newly defined partition functions and $2r$ -regular partition function. Moreover, we establish [4 Corollary 1, 2, 3 and 4] as particular cases of our results.

The paper is organized as follows. In § 2, we introduce our partition functions and establish their generating functions. In § 3, we prove some identities relating our partition functions and $2r$ -regular partition function. In Remark 3.12 we show that [4 Corollary 1, 2, 3 and 4] follow as particular cases of our results.

Partition functions and their generating functions

Definition 2.1. Let $h\geq0$ and $r\geq2$ be integers. Let $\mathcal{R}_{2r}(n,h)$ to be the number of partitions of nonnegative integer $n$ into parts not divisible by $2r$ and having exactly $h$ odd parts, counted with multiplicity.

For example, $\mathcal{R}_4(5,1)=3$ with the relevant partitions $5$ , $3+2$ , $2+2+1$ , and $\mathcal{R}_6(8,0)=4$ with the relevant partitions $8$ , $4+4$ , $4+2+2$ , $2+2+2+2$ .

Theorem 2.2. For any integers $h\geq0$ , $r\geq2$ , and $|xq|<1$ , we have

\sum_{n=0}^{\infty}\sum_{h=0}^{\infty}\mathcal{R}_{2r}(n,h)x^hq^n=\frac{(q^{2r};q^{2r})_{\infty}}{(xq;q^2)_{\infty}(q^2;q^2)_{\infty}}. \tag*{(3)}

Proof. Note that, no part is divisible by $2r$ in $2r$ -regular partitions of $n$ . Also, since every multiple of $2r$ is even, all odd parts $1,3,5,\ldots$ are allowed without any restriction. If $x$ counts the number of odd parts in the partitions, their contribution to $\mathcal{R}_{2r}(n,h)$ is given by

\prod_{j=0}^{\infty}\frac{1}{1-xq^{2j+1}}=\frac{1}{(xq;q^2)_{\infty}}. \tag*{(4)}

Also, since the allowed even parts are $2j$ with $r\nmid j$ , their contribution to $\mathcal{R}_{2r}(n,h)$ is given by

\prod_{\substack{j\geq1\\r\nmid j}}\frac{1}{1-q^{2j}}=\frac{\prod_{j=1}^{\infty}(1-q^{2rj})}{\prod_{j=1}^{\infty}(1-q^{2j})}=\frac{(q^{2r};q^{2r})_{\infty}}{(q^2;q^2)_{\infty}}. \tag*{(5)}

Multiplying (2.2) and (2.3), we obtain

\sum_{n=0}^{\infty}\sum_{h=0}^{\infty}\mathcal{R}_{2r}(n,h)x^hq^n =\frac{1}{(xq;q^2)_{\infty}}\frac{(q^{2r};q^{2r})_{\infty}}{(q^2;q^2)_{\infty}} =\frac{(q^{2r};q^{2r})_{\infty}}{(xq;q^2)_{\infty}(q^2;q^2)_{\infty}},

which completes the proof. ■

Remark 2.3. Putting $x=1$ in (2.1), and comparing the coefficients of $q^n$ , we obtain

\mathcal{R}_{2r}(n)=\sum_{h=0}^{\infty}\mathcal{R}_{2r}(n,h). \tag*{(6)}

Definition 2.4. Let $h\geq0$ , $r\geq2$ , and $s\geq1$ be integers. Let $\mathrm{BELO}_{r,s}(n,h)$ be the number of partitions of nonnegative integer $n$ in which every even part occurs at most $r-1$ times, the largest part is odd and occurs at least $s$ times, and the partition contains exactly $h$ odd parts, counted with multiplicity.

For example, $\mathrm{BELO}_{3,2}(8,2)=1$ with the partition $3+3+2$ and $\mathrm{BELO}_{2,1}(7,3)=3$ with the relevant partitions $5+1+1$ , $3+3+1$ , and $3+2+1+1$ .

Theorem 2.5. Let $r\geq2$ , $s\geq1$ , and $|xq|<1$ be integers. Then, we have

\sum_{n=0}^{\infty}\sum_{h=0}^{\infty}\mathrm{BELO}_{r,s}(n,h)x^hq^n =\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m(xq^{2m+1})^s}{(xq;q^2)_{m+1}(q^2;q^2)_m}. \tag*{(7)}

Proof. Suppose that the largest part of a partition counted by $\mathrm{BELO}_{r,s}(n,h)$ is $2m+1$ . Since the odd parts $1,3,\ldots,2m-1$ can occur without restriction and the largest odd part $2m+1$ occurs at least $s$ times, the contribution to $\mathrm{BELO}_{r,s}(n,h)$ from the odd parts is given by

\frac{(xq^{2m+1})^s}{(1-xq)(1-xq^3)\cdots(1-xq^{2m+1})} =\frac{(xq^{2m+1})^s}{(xq;q^2)_{m+1}}. \tag*{(8)}

where the power of $x$ records the number of odd parts.

Since $2m+1$ is the largest part in the partition of $n$ , the possible even parts are $2,4,\ldots,2m$ .

Since each even part $2j$ can occur $0,1,\ldots,r-1$ times, their contribution is given by

\prod_{j=1}^{m}\left(1+q^{2j}+q^{4j}+\cdots+q^{2(r-1)j}\right)=\prod_{j=1}^{m}\frac{1-q^{2rj}}{1-q^{2j}}=\frac{(q^{2r};q^{2r})_{m}}{(q^{2};q^{2})_{m}}. \tag*{(9)}

Multiplying (2.6) and (2.7), and then summing over $m\geq0$ , we arrive at the desired result. ■

Remark 2.6. Setting $x=1$ in (2.5) and

\sum_{h=0}^{\infty}\operatorname{BELO}_{r,s}(n,h)=\operatorname{BELO}_{r,s}(n), \tag*{(10)}

we obtain

\sum_{n=0}^{\infty}\operatorname{BELO}_{r,s}(n)q^{n} =\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_{m}q^{s(2m+1)}}{(q;q)_{m+1}(q^{2};q^{2})_{m}} =\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_{m}q^{s(2m+1)}}{(q;q)_{2m+1}}. \tag*{(11)}

Note that, $\operatorname{BELO}_{r,s}(n)$ counts the number of partitions of the nonnegative integer $n$ in which every even part occurs at most $r-1$ times and the largest part is odd and occurs at least $s$ times.

Definition 2.7. Let $h\geq0$ , $r\geq2$ , and $s\geq1$ be integers. For every integer $n\geq1$ , let $\operatorname{BELO}_{r,\leq s}(n,h)$ to be the number of partitions of $n$ in which every even part occurs at most $r-1$ times, the largest part is odd and occurs at most $s$ times, and the partition contains exactly $h$ odd parts, counted with multiplicity. For convenience, we assume that $\operatorname{BELO}_{r,\leq s}(0,h)=0$ .

For example, $\operatorname{BELO}_{3,\leq2}(8,2)=5$ with the relevant partitions $7+1$ , $5+3$ , $5+2+1$ , $3+3+2$ and $3+2+2+1$ , and $\operatorname{BELO}_{2,\leq1}(7,3)=2$ with the partitions $5+1+1$ and $3+2+1+1$ .

Theorem 2.8. Let $r\geq2$ , $s\geq1$ , and $|xq|<1$ be integers, then

\sum_{n=0}^{\infty}\sum_{h=0}^{\infty}\operatorname{BELO}_{r,\leq s}(n,h)x^{h}q^{n} =\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_{m}xq^{2m+1}\left(1-x^{s}q^{s(2m+1)}\right)}{(xq;q^{2})_{m+1}(q^{2};q^{2})_{m}}. \tag*{(12)}

Proof. The partitions counted by $\operatorname{BELO}_{r,s+1}(n,h)$ are precisely those partitions counted by $\operatorname{BELO}_{r,1}(n,h)$ in which the largest odd part occurs more than $s$ times. By Definitions 2.4 and 2.7, for $n\geq0$ , we obtain

\operatorname{BELO}_{r,\leq s}(n,h)=\operatorname{BELO}_{r,1}(n,h)-\operatorname{BELO}_{r,s+1}(n,h). \tag*{(13)}

Applying (2.5) with $s$ replaced by $1$ and $s+1$ i the right hand side of (2.11), we complete the proof. ■

Remark 2.9. Setting $x=1$ and

\sum_{h=0}^{\infty}\mathrm{BELO}_{r,\leq s}(n,h)=\mathrm{BELO}_{r,\leq s}(n) \tag*{(14)}

in (12), we obtain

\sum_{n=0}^{\infty}\mathrm{BELO}_{r,\leq s}(n)q^n=\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m q^{2m+1}\left(1-q^{s(2m+1)}\right)}{(q;q)_{2m+1}}.

Note that, $\mathrm{BELO}_{r,\leq s}(n)$ counts the partitions of $n$ in which every even part occurs at most $r-1$ times, the largest part is odd, and the largest part occurs at most $s$ times.

Definition 2.10. Let $r\geq2$ and $t\geq1$ be integers. For integer $n\geq0$ , let $\mathrm{BELO}_{r,=t}(n)$ denote the number of partitions counted by $\mathrm{BELO}_{r,t}(n)$ in which the largest odd part occurs exactly $t$ times.

For example, $\mathrm{BELO}_{3,=2}(8)=2$ with relevant partitions $3+3+2$ and $3+3+1+1$ , and $\mathrm{BELO}_{2,=1}(7)=5$ with the relevant partitions given by $7$ , $5+2$ , $5+1+1$ , $3+2+1+1$ , and $3+1+1+1+1+1$ .

Theorem 2.11. Let $r\geq2$ and $t\geq1$ be integers. Then, we have

\sum_{n=0}^{\infty}\mathrm{BELO}_{r,=t}(n)q^n=\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m q^{t(2m+1)}}{(q;q)_{2m}}.

Proof. It is easy to see that

\mathrm{BELO}_{r,=t}(n)=\mathrm{BELO}_{r,t}(n)-\mathrm{BELO}_{r,t+1}(n). \tag*{(15)}

Employing (11) with $s$ replaced by $t$ and $t+1$ in the right hand side of (15) and simplifying, we complete the proof.

Identities relating the partition functions and $2r$ -regular partition function

This section is devoted to proving identities relating our partition functions and the $2r$ -regular partition function.

Theorem 3.1. Let $h\geq0$ and $r\geq2$ be integers. Then for $n\geq1$ , we have

R_{2r}(n,h)=\mathrm{BELO}_{r,1}(n,h)+\sum_{j=1}^{r-1}\mathrm{BELO}_{r,j}(n-j,h+j),

where $\mathrm{BELO}_{r,s}(n,h)=0$ for any integer $n<0$ .

Proof. From (2.1), we note that

\begin{aligned} \sum_{n=0}^{\infty}\sum_{h=0}^{\infty}\mathcal{R}_{2r}(n,h)x^h q^n-1 &=\frac{(q^{2r};q^{2r})_{\infty}}{(xq;q^2)_{\infty}(q^2;q^2)_{\infty}}-1\\ &=\sum_{m=0}^{\infty}\left(\frac{(q^{2r};q^{2r})_{m+1}}{(xq;q^2)_{m+1}(q^2;q^2)_{m+1}}-\frac{(q^{2r};q^{2r})_m}{(xq;q^2)_m(q^2;q^2)_m}\right)\\ &=\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m}{(xq;q^2)_m(q^2;q^2)_m}\left(\frac{1-q^{2r(m+1)}}{(1-xq^{2m+1})(1-q^{2m+2})}-1\right)\\ &=\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m}{(xq;q^2)_m(q^2;q^2)_m}\times\frac{xq^{2m+1}-xq^{4m+3}+q^{2m+2}-q^{2r(m+1)}}{(1-xq^{2m+1})(1-q^{2m+2})}\\ &=\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m}{(xq;q^2)_m(q^2;q^2)_m}\times\frac{(1-q^{2m+2})\left(xq^{2m+1}+\sum_{j=1}^{r-1}q^{j(2m+2)}\right)}{(1-xq^{2m+1})(1-q^{2m+2})}\\ &=\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m}{(xq;q^2)_{m+1}(q^2;q^2)_m}\left(xq^{2m+1}+\sum_{j=1}^{r-1}q^{j(2m+2)}\right)\\ &=\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m(xq^{2m+1})}{(xq;q^2)_{m+1}(q^2;q^2)_m} +\sum_{j=1}^{r-1}q^j x^{-j}\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m(xq^{2m+1})^j}{(xq;q^2)_{m+1}(q^2;q^2)_m}\\ &=\sum_{n=0}^{\infty}\sum_{h=0}^{\infty}BELO_{r,1}(n,h)x^h q^n +\sum_{j=1}^{r-1}q^j x^{-j}\sum_{n=0}^{\infty}\sum_{h=0}^{\infty}BELO_{r,j}(n,h)x^h q^n. \tag*{(16)} \end{aligned}

Equating the coefficients of $x^h q^n$ on both sides of (16), we complete the proof. ■

Corollary 3.2. Let $r \ge2$ be any integer. Then, for any integer $n \ge1$ , we have

\mathcal{R}_{2r}(n)=BELO_{r,1}(n)+\sum_{j=1}^{r-1}BELO_{r,j}(n-j). \tag*{(17)}

where $BELO_{r,s}(n)=0$ for any integer $n<0$ .

Proof. Putting $x=1$ in (3.1) and then employing (2.4) and (2.8), we obtain

\frac{(q^{2r};q^{2r})_{\infty}}{(q;q)_{\infty}}-1 =\sum_{n=0}^{\infty}BELO_{r,1}(n)q^n+\sum_{j=1}^{r-1}q^j\sum_{n=0}^{\infty}BELO_{r,j}(n)q^n. \tag*{(18)}

Equating the coefficients of $q^n$ on both sides of (3.3), we arroive at the desired result. ■

Theorem 3.3. Let $r \ge2$ and $t \ge2$ be integers. Then for $n \ge0$ , we have

\begin{aligned} \mathcal{R}_{2r}(n)={}&\mathrm{BELO}_{r,1}(n)+\sum_{j=1}^{t-1}\mathrm{BELO}_{r,j}(n-j)+\mathrm{BELO}_{r,t}(n+t)-\mathrm{BELO}_{r,t+1}(n+t)\\ &-\sum_{j=0}^{t-1}\mathrm{BELO}_{r,r+j}(n-r-j), \tag*{(19)} \end{aligned}

where $\mathrm{BELO}_{r,s}(n)=0$ for any integer $n<0$ .

Proof. From (11), we obtain

\begin{aligned} q^{-t}\sum_{n=0}^{\infty}\left(\mathrm{BELO}_{r,t}(n)-\mathrm{BELO}_{r,t+1}(n)\right)q^n \\ &=\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m q^{t(2m+1)-t}}{(q;q)_{2m+1}}-\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m q^{(t+1)(2m+1)-t}}{(q;q)_{2m+1}} \\ &=\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m q^{2mt}(1-q^{2m+1})}{(q;q)_{2m}(1-q^{2m+1})} \\ &=\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m q^{2mt}}{(q;q)_{2m}} \\ &=1+\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_{m+1}q^{2t(m+1)}}{(q;q)_{2m+2}} \\ &=1+\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m q^{2t(m+1)}(1-q^{2r(m+1)})}{(q;q)_{2m+1}(1-q^{2m+2})} \\ &=1+\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m q^{2t(m+1)}}{(q;q)_{2m+1}}\sum_{j=0}^{r-1}q^{2j(m+1)} \\ &=1+\sum_{j=0}^{r-1}q^{t+j}\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m q^{(t+j)(2m+1)}}{(q;q)_{2m+1}} \\ &=1+\sum_{j=0}^{r-1}q^{t+j}\sum_{n=0}^{\infty}\mathrm{BELO}_{r,t+j}(n)q^n. \tag*{(20)} \end{aligned}

which implies

\begin{aligned} &\sum_{n=0}^{\infty} BELO_{r,1}(n)q^n+\sum_{j=1}^{t-1}q^j\sum_{n=0}^{\infty}BELO_{r,j}(n)q^n+q^{-t}\sum_{n=0}^{\infty}\left(BELO_{r,t}(n)-BELO_{r,t+1}(n)\right)q^n\\ &\quad-\sum_{j=0}^{t-1}q^{r+j}\sum_{n=0}^{\infty}BELO_{r,r+j}(n)q^n =1+\sum_{n=0}^{\infty}BELO_{r,1}(n)q^n+\sum_{j=1}^{t-1}q^j\sum_{n=0}^{\infty}BELO_{r,j}(n)q^n\\ &\quad+\sum_{j=t}^{t+r-1}q^j\sum_{n=0}^{\infty}BELO_{r,j}(n)q^n-\sum_{j=r}^{r+t-1}q^j\sum_{n=0}^{\infty}BELO_{r,j}(n)q^n\\ &=1+\sum_{n=0}^{\infty}BELO_{r,1}(n)q^n+\sum_{j=1}^{r-1}q^j\sum_{n=0}^{\infty}BELO_{r,j}(n)q^n\\ &=\sum_{n=0}^{\infty}R_{2r}(n)q^n, \tag*{(21)} \end{aligned}

where the last equality follows from (18). Equating the coefficients of $q^n$ in (21), we complete the proof. ■

Theorem 3.4. Let $r \ge2$ and $s \ge1$ be integers. Then for every integer $n \ge0$ , we have

BELO_{r,s+1}(n)+BELO_{r,s+1}(n-1)+BELO_{r,s}(n-2s-1) = BELO_{r,s+2}(n)+BELO_{r,s}(n-1)+BELO_{r,s+r}(n-2s-r-1), \tag*{(22)}

where $BELO_{r,s}(n)=0$ for any integer $n<0$ .

Proof. For $m \ge1$ , we note that

(1-q^{2m})(1-q^{2m+1})\frac{(q^{2r};q^{2r})_m}{(q;q)_{2m+1}} = \frac{(q^{2r};q^{2r})_m}{(q;q)_{2m-1}} = (1-q^{2rm})\frac{(q^{2r};q^{2r})_{m-1}}{(q;q)_{2m-1}}.

Therefore,

\sum_{m=1}^{\infty}(1-q^{2m})(1-q^{2m+1})\frac{(q^{2r};q^{2r})_m q^{2sm}}{(q;q)_{2m+1}} = \sum_{m=1}^{\infty}(1-q^{2rm})\frac{(q^{2r};q^{2r})_{m-1}q^{2sm}}{(q;q)_{2m-1}} \tag*{(23)}

Simplifying (23), we obtain

\begin{aligned} \sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m q^{2sm}}{(q;q)_{2m+1}} -(1+q)\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m q^{2(s+1)m}}{(q;q)_{2m+1}} +q\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m q^{2(s+2)m}}{(q;q)_{2m+1}}\\ =q^{2s}\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m q^{2sm}}{(q;q)_{2m+1}} -q^{2s+2r}\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m q^{2(s+r)m}}{(q;q)_{2m+1}}. \tag*{(24)} \end{aligned}

Employing (11) in (24) and simplifying, we obtain

\begin{aligned} (1+q)\sum_{n=0}^{\infty}BELO_{r,s+1}(n)q^n+q^{2s+1}\sum_{n=0}^{\infty}BELO_{r,s}(n)q^n\\ =\sum_{n=0}^{\infty}BELO_{r,s+2}(n)q^n+q\sum_{n=0}^{\infty}BELO_{r,s}(n)q^n+q^{2s+r+1}\sum_{n=0}^{\infty}BELO_{r,s+r}(n)q^n. \tag*{(25)} \end{aligned}

(3.10) implies

\begin{aligned} \sum_{n=0}^{\infty}\left(\mathrm{BEL}_{O,r,s+1}(n)+\mathrm{BEL}_{O,r,s+1}(n-1)+\mathrm{BEL}_{O,r,s}(n-2s-1)\right)q^n &=\sum_{n=0}^{\infty}\left(\mathrm{BEL}_{O,r,s+2}(n)+\mathrm{BEL}_{O,r,s}(n-1)+\mathrm{BEL}_{O,r,s+r}(n-2s-r-1)\right)q^n. \tag*{(26)} \end{aligned}

Equating the coefficients of $q^n$ on both sides of (26), we arrive at the desired result. ■

Theorem 3.5. Let $h \ge0$ and $r \ge2$ be integers. Then for any integer $n \ge1$ , we have

\mathcal{R}_{2r}(n,h)=\mathrm{BEL}_{O,r,\le h}(n,h)+\sum_{j=1}^{r-1}\left(\mathrm{BEL}_{O,r,\le h+j}(n-j,h+j)-\mathrm{BEL}_{O,r,\le j-1}(n-j,h+j)\right).

where $\mathrm{BEL}_{O,r,\le0}(n,h)=0$ , and $\mathrm{BEL}_{O,r,\le s}(n,h)=0$ , for any integer $n<0$ .

Proof. For $t\ge1$ , subtracting the case $s=t-1$ from the case $s=t$ of (2.10) and noting $\mathrm{BEL}_{O,r,\le0}(n,h)=0$ , we obtain

\begin{aligned} \sum_{n=0}^{\infty}\sum_{h=0}^{\infty}\left(\mathrm{BEL}_{O,r,\le t}(n,h)-\mathrm{BEL}_{O,r,\le t-1}(n,h)\right)x^h q^n &=\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m(xq^{2m+1})^t(1-xq^{2m+1})}{(xq;q^2)_{m+1}(q^2;q^2)_m}\\ &=\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m(xq^{2m+1})^t}{(xq;q^2)_m(q^2;q^2)_m}. \tag*{(27)} \end{aligned}

Summing (3.12) over $t\ge k\ge1$ and using (2.5), we obtain

\begin{aligned} \sum_{t=k}^{\infty}\sum_{n=0}^{\infty}\sum_{h=0}^{\infty}\left(\mathrm{BEL}_{O,r,\le t}(n,h)-\mathrm{BEL}_{O,r,\le t-1}(n,h)\right)x^h q^n &=\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m}{(xq;q^2)_m(q^2;q^2)_m}\sum_{t=k}^{\infty}(xq^{2m+1})^t\\ &=\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m(xq^{2m+1})^k}{(1-xq^{2m+1})(xq;q^2)_m(q^2;q^2)_m}\\ &=\sum_{m=0}^{\infty}\frac{(q^{2r};q^{2r})_m(xq^{2m+1})^k}{(xq;q^2)_{m+1}(q^2;q^2)_m}\\ &=\sum_{n=0}^{\infty}\sum_{h=0}^{\infty}\mathrm{BEL}_{O,r,k}(n,h)x^h q^n. \tag*{(28)} \end{aligned}

Substituting (3.13) with cases $k=1$ and $k=j$ in (3.1), we obtain

\begin{aligned} \sum_{n=0}^{\infty}\sum_{h=0}^{\infty} R_{2r}(n,h)x^h q^n-1 ={}&\sum_{t=1}^{\infty}\sum_{n=0}^{\infty}\sum_{h=0}^{\infty} \left(\mathrm{BELO}_{r,\leq t}(n,h)-\mathrm{BELO}_{r,\leq t-1}(n,h)\right)x^h q^n\\ &+\sum_{j=1}^{r-1}q^j x^{-j}\sum_{t=j}^{\infty}\sum_{n=0}^{\infty}\sum_{h=0}^{\infty} \left(\mathrm{BELO}_{r,\leq t}(n,h)-\mathrm{BELO}_{r,\leq t-1}(n,h)\right)x^h q^n\\ ={}&\sum_{n=0}^{\infty}\sum_{h=0}^{\infty}\left( \sum_{t=1}^{h}\left(\mathrm{BELO}_{r,\leq t}(n,h)-\mathrm{BELO}_{r,\leq t-1}(n,h)\right)\right.\\ &\left.\qquad+\sum_{j=1}^{r-1}\sum_{t=j}^{h+j} \left(\mathrm{BELO}_{r,\leq t}(n-j,h+j)-\mathrm{BELO}_{r,\leq t-1}(n-j,h+j)\right) \right)x^h q^n. \tag*{(29)} \end{aligned}

Comparing the coefficients of $x^h q^n$ on both sides of (3.14), we obtain

\begin{aligned} R_{2r}(n,h)={}&\sum_{t=1}^{h}\left(\mathrm{BELO}_{r,\leq t}(n,h)-\mathrm{BELO}_{r,\leq t-1}(n,h)\right)\\ &+\sum_{j=1}^{r-1}\sum_{t=j}^{h+j} \left(\mathrm{BELO}_{r,\leq t}(n-j,h+j)-\mathrm{BELO}_{r,\leq t-1}(n-j,h+j)\right)\\ ={}&\mathrm{BELO}_{r,\leq h}(n,h)+\sum_{j=1}^{r-1} \left(\mathrm{BELO}_{r,\leq h+j}(n-j,h+j)-\mathrm{BELO}_{r,\leq j-1}(n-j,h+j)\right). \end{aligned}

Hence, the proof is complete. ■

Corollary 3.6. Let $r\geq2$ be integer. Then for any integer $n\geq1$ , we have

R_{2r}(n)=\mathrm{BELO}_{r,\leq n}(n)+\sum_{j=1}^{\min\{r-1,n-1\}} \left(\mathrm{BELO}_{r,\leq n-j}(n-j)-\mathrm{BELO}_{r,\leq j-1}(n-j)\right).

where $\mathrm{BELO}_{r,\leq0}(n)=0$ , and $\mathrm{BELO}_{r,\leq s}(n)=0$ for any integer $n<0$ .

Proof. Setting $x=1$ in (3.14) and then employing (2.12), we obtain

\begin{aligned} \sum_{n=0}^{\infty}R_{2r}(n)q^n-1 ={}&\sum_{t=1}^{\infty}\sum_{n=0}^{\infty} \left(\mathrm{BELO}_{r,\leq t}(n)-\mathrm{BELO}_{r,\leq t-1}(n)\right)q^n\\ &+\sum_{j=1}^{r-1}q^j\sum_{t=j}^{\infty}\sum_{n=0}^{\infty} \left(\mathrm{BELO}_{r,\leq t}(n)-\mathrm{BELO}_{r,\leq t-1}(n)\right)q^n. \tag*{(30)} \end{aligned}

Equating the coefficients of $q^n$ on both sides of (3.15), we obtain

\begin{aligned} R_{2r}(n)={}&\sum_{t=1}^{\infty} \left(\mathrm{BELO}_{r,\leq t}(n)-\mathrm{BELO}_{r,\leq t-1}(n)\right)\\ &+\sum_{j=1}^{\min\{r-1,n-1\}}\sum_{t=j}^{\infty} \left(\mathrm{BELO}_{r,\leq t}(n-j)-\mathrm{BELO}_{r,\leq t-1}(n-j)\right). \tag*{(31)} \end{aligned}

Since $\mathrm{BELO}_{r,\leq t}(n)-\mathrm{BELO}_{r,\leq t-1}(n)$ counts the number of partitions of any positive integer $n$ in which every even part appears at most $r-1$ times and the largest odd part occurs exactly $t$ times, the multiplicity of the largest odd part in such partitions cannot exceed $n$ . Therefore,

\sum_{t=1}^{\infty}\left(\mathrm{BELO}_{r,\leq t}(n)-\mathrm{BELO}_{r,\leq t-1}(n)\right)=\mathrm{BELO}_{r,\leq n}(n) \tag*{(32)}

and

\sum_{t=j}^{\infty}\left(\mathrm{BELO}_{r,\leq t}(n-j)-\mathrm{BELO}_{r,\leq t-1}(n-j)\right)=\mathrm{BELO}_{r,\leq n-j}(n-j)-\mathrm{BELO}_{r,\leq j-1}(n-j). \tag*{(33)}

Employing (3.17) and (3.18) in (3.16), we complete the proof. ■

Theorem 3.7. Let $r\geq2$ and $s\geq1$ be integers. Then for any integer $n\geq0$ , we have

\begin{aligned} \mathrm{BELO}_{r,\leq s}(n)+\mathrm{BELO}_{r,\leq s}(n-1)+\mathrm{BELO}_{r,\leq s-1}(n-2s-1) ={}&\mathrm{BELO}_{r,\leq s+1}(n)+\mathrm{BELO}_{r,\leq s-1}(n-1)\\ &+\mathrm{BELO}_{r,\leq s+r-1}(n-2s-r-1)\\ &+\mathrm{BELO}_{r,\leq n-2s-1}(n-2s-1)\\ &-\mathrm{BELO}_{r,\leq n-2s-r-1}(n-2s-r-1), \end{aligned}

where $\mathrm{BELO}_{r,\leq0}(n)=0$ and $\mathrm{BELO}_{r,\leq s}(n)=0$ for any integer $n<0$ .

Proof. Note that,

\begin{aligned} \mathrm{BELO}_{r,k}(n)&=\sum_{h=0}^{\infty}\mathrm{BELO}_{r,k}(n,h)\\ &=\sum_{h=0}^{\infty}\left(\mathrm{BELO}_{r,\leq n}(n,h)-\mathrm{BELO}_{r,\leq k-1}(n,h)\right)\\ &=\mathrm{BELO}_{r,\leq n}(n)-\mathrm{BELO}_{r,\leq k-1}(n). \tag*{(34)} \end{aligned}

Using (3.19) in (3.7) with the corresponding values of $n$ and $k$ , we obtain

\begin{aligned} &\left(\mathrm{BELO}_{r,\leq n}(n)-\mathrm{BELO}_{r,\leq s}(n)\right)+\left(\mathrm{BELO}_{r,\leq n-1}(n-1)-\mathrm{BELO}_{r,\leq s}(n-1)\right)\\ &+\left(\mathrm{BELO}_{r,\leq n-2s-1}(n-2s-1)-\mathrm{BELO}_{r,\leq s-1}(n-2s-1)\right)\\ ={}&\left(\mathrm{BELO}_{r,\leq n}(n)-\mathrm{BELO}_{r,\leq s+1}(n)\right)+\left(\mathrm{BELO}_{r,\leq n-1}(n-1)-\mathrm{BELO}_{r,\leq s-1}(n-1)\right)\\ &+\left(\mathrm{BELO}_{r,\leq n-2s-r-1}(n-2s-r-1)-\mathrm{BELO}_{r,\leq s+r-1}(n-2s-r-1)\right). \tag*{(35)} \end{aligned}

Simplifying (3.20), we arrive at the desired result. ■

Theorem 3.8. Let $r \ge2$ and $t \ge2$ be integers. Then for any integer $n \ge0$ , we have

R_{2r}(n) = \mathrm{BELO}_{r,\le n}(n) + \sum_{j=1}^{t-1}\left(\mathrm{BELO}_{r,\le n-j}(n-j) - \mathrm{BELO}_{r,\le j-1}(n-j)\right) + \mathrm{BELO}_{r,\le t}(n+t) - \mathrm{BELO}_{r,\le t-1}(n+t) - \sum_{j=0}^{t-1}\left(\mathrm{BELO}_{r,\le n-r-j}(n-r-j) - \mathrm{BELO}_{r,\le r+j-1}(n-r-j)\right),

where $\mathrm{BELO}_{r,\le0}(n)=0$ , and $\mathrm{BELO}_{r,\le s}(n)=0$ for any integer $n<0$ .

Proof. Employing (34) in (19), we easiy arrive at the desired result. ■

Theorem 3.9. Let $r \ge2$ and $n \ge1$ be integers. Then

R_{2r}(n) = \sum_{t=1}^{\infty}\sum_{j=0}^{\min\{t,r-1\}}\mathrm{BELO}_{r,=t}(n-j),

where $\mathrm{BELO}_{r,=t}(n)=0$ for any integer $n<0$ .

Proof. From (17), we note that

\begin{aligned} R_{2r}(n) &= \mathrm{BELO}_{r,1}(n) + \sum_{j=1}^{r-1}\mathrm{BELO}_{r,j}(n-j) \\ &= \sum_{t=1}^{\infty}\mathrm{BELO}_{r,=t}(n) + \sum_{j=1}^{r-1}\sum_{t=j}^{\infty}\mathrm{BELO}_{r,=t}(n-j) \\ &= \sum_{t=1}^{\infty}\mathrm{BELO}_{r,=t}(n) + \sum_{t=1}^{\infty}\sum_{j=1}^{\min\{t,r-1\}}\mathrm{BELO}_{r,=t}(n-j) \\ &= \sum_{t=1}^{\infty}\left(\mathrm{BELO}_{r,=t}(n) + \sum_{j=1}^{\min\{t,r-1\}}\mathrm{BELO}_{r,=t}(n-j)\right) \\ &= \sum_{t=1}^{\infty}\sum_{j=0}^{\min\{t,r-1\}}\mathrm{BELO}_{r,=t}(n-j). \end{aligned}

Hence, the proof is complete. ■

Theorem 3.10. Let $r \ge2$ and $t \ge1$ be integers. Then for any integer $n \ge0$ , we have

\mathrm{BELO}_{r,=t+1}(n-1) + \mathrm{BELO}_{r,=t+1}(n-2) + \mathrm{BELO}_{r,=t}(n-2t-3) = \mathrm{BELO}_{r,=t+2}(n) + \mathrm{BELO}_{r,=t}(n-3) + \mathrm{BELO}_{r,=t+r}(n-2t-r-3),

where $\mathrm{BELO}_{r,=t}(k)=0$ , if $k$ is a negative integer.

Proof. Setting $s=t$ in (25), we obtain

\begin{aligned} (1+q)\sum_{n=0}^{\infty} BELO_{r,t+1}(n)q^n+q^{2t+1}\sum_{n=0}^{\infty} BELO_{r,t}(n)q^n &=\sum_{n=0}^{\infty} BELO_{r,t+2}(n)q^n+q\sum_{n=0}^{\infty} BELO_{r,t}(n)q^n\\ &\quad+q^{2t+r+1}\sum_{n=0}^{\infty} BELO_{r,t+r}(n)q^n. \tag*{(36)} \end{aligned}

Similarly, taking $s=t+1$ in (25), we obtain

\begin{aligned} (1+q)\sum_{n=0}^{\infty} BELO_{r,t+2}(n)q^n+q^{2t+3}\sum_{n=0}^{\infty} BELO_{r,t+1}(n)q^n &=\sum_{n=0}^{\infty} BELO_{r,t+3}(n)q^n+q\sum_{n=0}^{\infty} BELO_{r,t+1}(n)q^n\\ &\quad+q^{2t+r+3}\sum_{n=0}^{\infty} BELO_{r,t+r+1}(n)q^n. \tag*{(37)} \end{aligned}

Multiplying (36) by $q^2$ , we obtain

\begin{aligned} (q^2+q^3)\sum_{n=0}^{\infty} BELO_{r,t+1}(n)q^n+q^{2t+3}\sum_{n=0}^{\infty} BELO_{r,t}(n)q^n &=q^2\sum_{n=0}^{\infty} BELO_{r,t+2}(n)q^n+q^3\sum_{n=0}^{\infty} BELO_{r,t}(n)q^n\\ &\quad+q^{2t+r+3}\sum_{n=0}^{\infty} BELO_{r,t+r}(n)q^n. \tag*{(38)} \end{aligned}

Subtracting (37) from (38), we obtain

\begin{aligned} &(q^2+q^3)\sum_{n=0}^{\infty} BELO_{r,t+1}(n)q^n+q^{2t+3}\sum_{n=0}^{\infty} BELO_{r,t}(n)q^n-(1+q)\sum_{n=0}^{\infty} BELO_{r,t+2}(n)q^n\\ &\quad-q^{2t+3}\sum_{n=0}^{\infty} BELO_{r,t+1}(n)q^n\\ &=q^2\sum_{n=0}^{\infty} BELO_{r,t+2}(n)q^n+q^3\sum_{n=0}^{\infty} BELO_{r,t}(n)q^n+q^{2t+r+3}\sum_{n=0}^{\infty} BELO_{r,t+r}(n)q^n\\ &\quad-\sum_{n=0}^{\infty} BELO_{r,t+3}(n)q^n-q\sum_{n=0}^{\infty} BELO_{r,t+1}(n)q^n-q^{2t+r+3}\sum_{n=0}^{\infty} BELO_{r,t+r+1}(n)q^n. \tag*{(39)} \end{aligned}

Rearranging the terms in (39) and using (15), we obtain

\begin{aligned} (q+q^2)\sum_{n=0}^{\infty} BELO_{r,t+1}(n)q^n+q^{2t+3}\sum_{n=0}^{\infty} BELO_{r,t}(n)q^n &=\sum_{n=0}^{\infty} BELO_{r,t+2}(n)q^n+q^3\sum_{n=0}^{\infty} BELO_{r,t}(n)q^n\\ &\quad+q^{2t+r+3}\sum_{n=0}^{\infty} BELO_{r,t+r}(n)q^n\\ &=\sum_{n=0}^{\infty}\bigl(BELO_{r,t+2}(n)+BELO_{r,t}(n-3)+BELO_{r,t+r}(n-2t-r-3)\bigr)q^n. \tag*{(40)} \end{aligned}

Equating the coefficients of $q^n$ on both sides (40), we arrive at the desired result. ■

Theorem 3.11. Let $r \ge2$ and $s,t \ge1$ be integers. Then for integer $n \ge0$ , we have

BELO_{r,=t}(n)=\delta_{n,t}+\sum_{j=0}^{r-1}\left(BELO_{r,s}(n-2t-j)-BELO_{r,\le t+j-1}(n-2t-j)+BELO_{r,\le s-1}(n-2t-j)\right).

where $\delta_{n,t}$ is defined by

\delta_{n,t}=\begin{cases}1, & \text{if } n=t,\\0, & \text{if } n\ne t.\end{cases}

Furthermore, $BELO_{r,\le0}(n)=0$ , and $BELO_{r,\le s}(n)=0$ for any integer $n<0$ .

Proof. Employing (15) in (20), we obtain

q^{-t}\sum_{n=0}^{\infty}BELO_{r,=t}(n)q^n=1+\sum_{j=0}^{r-1}q^{t+j}\sum_{n=0}^{\infty}BELO_{r,t+j}(n)q^n. \tag*{(41)}

Substracting the case $k=t+j$ from the case $k=s$ of (34) and simplifying, we obtain

BELO_{r,t+j}(n)=BELO_{r,s}(n)-BELO_{r,\le t+j-1}(n)+BELO_{r,\le s-1}(n). \tag*{(42)}

Employing (42) in (41) and then multiplying by $q^t$ , we obtain

\sum_{n=0}^{\infty}BELO_{r,=t}(n)q^n=q^t+\sum_{j=0}^{r-1}q^{2t+j}\sum_{n=0}^{\infty}\left(BELO_{r,s}(n)-BELO_{r,\le t+j-1}(n)+BELO_{r,\le s-1}(n)\right)q^n. \tag*{(43)}

Equating the coefficients of $q^n$ on both sides of (43), we arrive at the desired result. ■

Remark 3.12. Noting that, $\mathrm{DE1}(n)=BELO_{2,1}(n)$ , $\mathrm{DE2}(n)=BELO_{2,2}(n)$ , and $\mathrm{DE3}(n)=BELO_{2,=1}(n)$ , Andrews and El Bachraoui [4] Corollary 1, 2, 3 and 4 stated in the introduction section follow from Corollary 3.2 and Theorem 3.11 as particular cases. Precisely, Corollary 1, 2 and 3 of [4] follow from Corollary 3.2 and Corollary 4 of [4] follows from Theorem 3.11 proved in Section 3 of this paper.

Declarations

Funding: This research did not receive funding.

Author Contributions. Both authors contributed equally to this work.

Conflict of Interest. The authors declare that there is no conflict of interest regarding the publication of this paper.

Human and Animal Rights. The authors declare that there is no research involving human participants or animals in the context of this paper.

Data Availability Statement. Data sharing is not applicable to this paper as no datasets were generated or analyzed during the current study.

References

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On partitions with bounded multiplicity restrictions on even parts and with the largest part being odd

Abstract

Introduction

Partition functions and their generating functions

Identities relating the partition functions and $2r$ -regular partition function

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References

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AI contributions

More in Mathematics

More in Mathematics

Abstract

Introduction

Partition functions and their generating functions

Identities relating the partition functions and 2r2r2r-regular partition function

Declarations

References

More in Mathematics

Identities relating the partition functions and $2r$ -regular partition function