A q-Series Congruence Inspired by Andrews and Ramanujan

Abstract

For each $s\in \{1,3,5\}$, we consider $R_s\left(n\right)$ to be the number of the partitions of n into parts not congruent to 0, $\pm s(mod12)$. In recent years, some relations for computing the value of $R_3\left(n\right)$ were studied. In this paper, we investigate the parity of $R_s\left(n\right)$ when $s\in \{1,5\}$ and derive the following congruence identity: ${\displaystyle \sum _{n=1}^{\infty }}\frac{{(-q;q)}_{n-1}^2(1+q^n)q^{n^2}}{{(q;q)}_{2n}}\equiv {\displaystyle \sum _{n=1}^{\infty }}\left(q^{n^2}+q^{3n^2}\right)(mod2).$ For each $s\in \{1,5\}$, the number of the partitions of n into parts not congruent to 0, $\pm s(mod12)$ is connected with two truncated theta series. Some open problems involving $R_1\left(n\right)$ and $R_5\left(n\right)$ are introduced in this context.

1. Introduction

The problem regarding the number of ways to represent a positive integer as a sum of positive integers can be dated back to Leibniz, but the theory of partitions began with Euler in the mid-eighteenth century [1]. Recall [2] that a partition of a positive integer n is a weakly decreasing sequence of positive integers whose sum is n. The Euler partition function $p\left(n\right)$ gives the number of partitions of n. In order to understand certain aspects of partitions, Euler introduced the idea of a generating function of a sequence. In particular, Euler showed that the generating function of the partition function $p\left(n\right)$, can be expressed as an elegant infinite product:

$$\sum _{n=0}^{\infty }p\left(n\right)q^n=\frac{1}{{(q;q)}_{\infty }}.$$

Here and throughout this paper, we use the following customary q-series notation:

$${(a;q)}_0=1,{(a;q)}_n=\prod _{k=0}^{n-1}(1-a{q}^k),{(a;q)}_{\infty }=\prod _{k=0}^{\infty }(1-a{q}^k).$$

Because the infinite product ${(a;q)}_{\infty }$ diverges when $a\ne 0$ and $\left|q\right|⩾1$, whenever ${(a;q)}_{\infty }$ appears in a formula, we assume $\left|q\right|<1$, and we use the compact notation

$${(a_1,a_2\dots ,a_n;q)}_{\infty }={(a_1;q)}_{\infty }{(a_2;q)}_{\infty }\cdots {(a_n;q)}_{\infty }.$$

Euler became interested in the reciprocal of the partition generating function, i.e., the infinite product ${(q;q)}_{\infty }$. He expanded this product by direct multiplication and found

$${(q;q)}_{\infty }=1-q-q^2+q^5+q^7-q^{12}-q^{15}+q^{22}+q^{26}-\cdots ,$$

where the exponents are generalized pentagonal numbers, i.e., $n(3n-1)/2$, $n\in \mathbb{Z}$. Then, Euler made the following conjecture

$$\begin{array}{cc}\hfill & \sum _{n=-\infty }^{\infty }{(-1)}^n{q}^{n(3n-1)/2}={(q;q)}_{\infty },\hfill \end{array}$$

(1)

for which he found a proof almost a decade later. As one can see in [2], Euler’s pentagonal numbers theorem plays an important role in the theory of partitions. For example, this theorem leads to an efficient method of computing the partition function $p\left(n\right)$, i.e.,

$$\begin{array}{c}\hfill \sum _{j=-\infty }^{\infty }{(-1)}^{j}p\left(n-j(3j-1)/2\right)={\delta }_{0,n},\end{array}$$

(2)

where $p\left(n\right)=0$ for any negative integer n, and ${\delta }_{i,j}$ is the Kronecker delta function. Jacobi and Franklin [3] reproved (1) by using theta functions and a combinatorial method, respectively, and a surprisingly simple algebraic proof of (1) was given by Shanks [4].

Euler’s pentagonal numbers theorem (1) has the following generalization which is known as the Jacobi triple product identity (see [5], Eq. (1.6.1)): for $z\ne 0$,

$$\begin{array}{cc}\hfill & \sum _{n=-\infty }^{\infty }{(-z)}^n{q}^{n(n-1)/2}={(z,q/z,q;q)}_{\infty }.\hfill \end{array}$$

(3)

For $n\in \mathbb{Z}$, the number $n(6n-1)$ is a generalized pentagonal number with an even index, while $n(6n-5)+1$ is a generalized pentagonal number with an odd index. The following bisection of

$${(q;q)}_{\infty }=\sum _{\begin{array}{c}n=-\infty \\ n\mathrm{even}\end{array}}^{\infty }{q}^{n(3n-1)/2}-\sum _{\begin{array}{c}n=-\infty \\ n\mathrm{odd}\end{array}}^{\infty }{q}^{n(3n-1)/2}$$

can be written as

$${(q;q)}_{\infty }={(-q^5,-q^7,q^{12};q^{12})}_{\infty }-q{(-q,-q^{11},q^{12};q^{12})}_{\infty },$$

where we have invoked the Jacobi triple product identity (3) with q replaced by $q^{12}$ and letting $z=-q$ or $z=-q^5$. Thus, we deduce the following identity

$$\begin{array}{c}\hfill 1-\frac{{(-q^5,-q^7,q^{12};q^{12})}_{\infty }}{{(q;q)}_{\infty }}+q\frac{{(-q,-q^{11},q^{12};q^{12})}_{\infty }}{{(q;q)}_{\infty }}=0.\end{array}$$

(4)

The last term of this identity appeared in a discussion of some of Ramanujan’s formulas from both his Notebooks and lost Notebook, where Andrews proved the following identity [6]:

$$\begin{array}{c}\hfill \sum _{n=1}^{\infty }\frac{{(-q;q)}_{n-1}^2(1+q^n)q^{n^2}}{{(q;q)}_{2n}}=\frac{q{(-q,-q^{11},q^{12};q^{12})}_{\infty }}{{(q;q)}_{\infty }}.\end{array}$$

(5)

Another proof of this identity can be seen in a paper published previously, in 1952, by Slater [7], Eq. 56.

In this paper, we provide the following q-series congruence.

Theorem 1.

For $\left|q\right|<1$,

$$\sum _{n=1}^{\infty }\frac{{(-q;q)}_{n-1}^2(1+q^n)q^{n^2}}{{(q;q)}_{2n}}\equiv \sum _{n=1}^{\infty }\left(q^{n^2}+q^{3n^2}\right)(mod2).$$

According to Euler’s recurrence relation (2), we can write

$$\sum _{k=-\infty }^{\infty }p\left(n-k(6k-1)\right)=1+\sum _{k=-\infty }^{\infty }p\left(n-1-k(6k-5)\right).$$

In addition, considering the equations

$$\sum _{k=-\infty }^{\infty }p\left(n-k(6k-1)\right)=\left[q^n\right]\frac{{(-q^5,-q^7,q^{12};q^{12})}_{\infty }}{{(q;q)}_{\infty }}$$

and

$$\sum _{k=-\infty }^{\infty }p\left(n-1-k(6k-5)\right)=\left[q^n\right]\frac{q{(-q,-q^{11},q^{12};q^{12})}_{\infty }}{{(q;q)}_{\infty }},$$

we observe that Theorem 1 can be easily derived by considering the following parity result involving sums of partition numbers $p\left(n\right)$ and Equations (4) and (5).

Theorem 2.

 (i)

For $n⩾0$,

$$\sum _{k=-\infty }^{\infty }p\left(n-k(6k-1)\right)\equiv 1(mod2),$$

if and only if n is square or thrice square.

 (ii)

For $n⩾0$,

$$\sum _{k=-\infty }^{\infty }p\left(n-k(6k-5)\right)\equiv 1(mod2),$$

if and only if $n+1$ is square or thrice square.

Very recently [8], we considered the partitions of n into parts not congruent to 0, $\pm 3(mod12)$ and obtained a similar parity result: for $n⩾0$,

$$\sum _{k=-\infty }^{\infty }p\left(n-k(6k-3)\right)\equiv 1(mod2),$$

if and only if $3n+1$ is a square. This result is equivalent to the the following q-series congruence:

$$\sum _{n=0}^{\infty }\frac{{(-q^2;q^2)}_n{q}^{n(n+1)}}{{(q;q)}_{2n+1}}\equiv \sum _{3n+1\mathrm{square}}{q}^n(mod2).$$

In order to prove Theorem 2, for each $s\in \{1,5\}$, we consider the function $R_s\left(n\right)$ counting the partitions of n into parts not congruent to 0, $\pm s(mod12)$. Elementary techniques in the theory of partitions give the following generating function for $R_s\left(n\right)$:

$$\begin{array}{c}\hfill \sum _{n=0}^{\infty }{R}_s\left(n\right)q^n=\frac{{(q^s,q^{12-s},q^{12};q^{12})}_{\infty }}{{(q;q)}_{\infty }}.\end{array}$$

(6)

For $s\in \{1,5\}$, it seems that these numbers $R_s\left(n\right)$ have not been investigated so far. Considering the Jacobi triple product identity (3) and the generating function (6), we easily derive the following results.

Corollary 1.

For $s\in \{1,5\}$, the partition function $R_s\left(n\right)$ is closely related to $p\left(n\right)$, i.e.,

$$R_s\left(n\right)=\sum _{k=-\infty }^{\infty }{(-1)}^{k+1}p\left(n-(k+1)(6k+s)\right).$$

Corollary 2.

For $s\in \{1,5\}$, in certain conditions, the number $R_s\left(n\right)$ satisfies Euler’s recurrence relation for the partition function $p\left(n\right)$,

$$\sum _{k=-\infty }^{\infty }{(-1)}^k{R}_s\left(n-k(3k-1)/2\right)={\omega }_s\left(n\right),$$

where

$${\omega }_s\left(n\right)=\left\{\begin{array}{cc}{(-1)}^{m+1},\hfill & \mathrm{if}n=(m+1)(6m+s),m\in \mathbb{Z},\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Related to $R_3\left(n\right)$, we remark on a similar result proved in [8]:

$$\sum _{k=-\infty }^{\infty }{(-1)}^k{R}_3\left(n-k(3k-1)/2\right)=\left\{\begin{array}{cc}{(-1)}^{m(m+1)/2},\hfill & \mathrm{if}n=3m(m+1)/2,m\in {\mathbb{N}}_0,\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right..$$

The organization of this paper is as follows. We first prove Theorem 2 in Section 2. The proof of Theorem 2 is based on Corollary 1 and a new linear recurrence relation for computing $R_s\left(n\right)$ when $s\in \{1,5\}$. This recurrence relation provides a simple criterion to establish the parity of $R_s\left(n\right)$. In Section 3, we show that the recurrence relation obtained in Section 2 is a specialization of a more general result. An infinite family of linear inequalities involving $R_5\left(n\right)$ is derived in this context. We propose as open problem a similar family of linear inequalities involving $R_1\left(n\right)$ (see Conjecture 1). In Section 4, we consider two truncated forms of Euler’s pentagonal number Theorem (1) in order to derive new open problems involving infinite families of linear inequalities for $R_s\left(n\right)$ when $s\in \{1,5\}$ (see Conjectures 3 and 5). We conclude the paper with a conjectured Ramanujan-type congruence involving $R_s\left(n\right)$ (see Conjectures 6).

2. Proof of Theorem 2

The following theta identity is often attributed to Gauss [2], p. 23, Equations (2.2.12):

$$1+2\sum _{n=1}^{\infty }{(-1)}^n{q}^{n^2}=\frac{{(q;q)}_{\infty }}{{(-q;q)}_{\infty }}.$$

(7)

By this identity, with q replaced by $q^2$, we obtain

$$1+2\sum _{n=1}^{\infty }{(-1)}^n{q}^{2n^2}=\frac{{(q^2;q^2)}_{\infty }}{{(-q^2;q^2)}_{\infty }}.$$

Multiplying both sides of (6) by (2), we obtain

$$\begin{array}{cc}\hfill \left(1+2\sum _{n=1}^{\infty }{(-1)}^n{q}^{2n^2}\right)\left(\sum _{n=0}^{\infty }{R}_s\left(n\right)q^n\right)& =\frac{{(q^2;q^2)}_{\infty }}{{(-q^2;q^2)}_{\infty }}·\frac{{(q^s,q^{12-s},q^{12};q^{12})}_{\infty }}{{(q;q)}_{\infty }}\hfill \\ \hfill & =\frac{{(q^2,q^6,q^{10},q^{12};q^{12})}_{\infty }}{{(q^{6-s},q^{6+s},q^3,q^9;q^{12})}_{\infty }}.\hfill \end{array}$$

(8)

Considering [9], Theorem 3, i.e.,

$$\begin{array}{cc}\hfill & 1+\sum _{n=1}^{\infty }\left(q^{n^2}+q^{3n^2}\right)=\frac{{(q^2,q^6,q^{10},q^{12};q^{12})}_{\infty }}{{(q,q^3,q^9,q^{11};q^{12})}_{\infty }},\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill & \sum _{n=1}^{\infty }\left(q^{n^2-1}-q^{3n^2-1}\right)=\frac{{(q^2,q^6,q^{10},q^{12};q^{12})}_{\infty }}{{(q^3,q^5,q^7,q^9;q^{12})}_{\infty }},\hfill \end{array}$$

(10)

we deduce:

$$\left(1+2\sum _{n=1}^{\infty }{(-1)}^n{q}^{2n^2}\right)\left(\sum _{n=0}^{\infty }{R}_5\left(n\right)q^n\right)=1+\sum _{n=1}^{\infty }\left(q^{n^2}+q^{3n^2}\right)$$

and

$$\left(1+2\sum _{n=1}^{\infty }{(-1)}^n{q}^{2n^2}\right)\left(\sum _{n=0}^{\infty }{R}_1\left(n\right)q^n\right)=\sum _{n=1}^{\infty }\left(q^{n^2-1}-q^{3n^2-1}\right).$$

By equating the coefficient of $q^n$ on each side of these identities, for $s\in \{1,5\}$, we obtain the following linear recurrence relation:

$$R_s\left(n\right)+2\sum _{j=1}^{\infty }{(-1)}^j{R}_s(n-2j^2)={\rho }_s\left(n\right),$$

(11)

where

$${\rho }_5\left(n\right)=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}n=k^2\mathrm{or}n=3k^2,k\in {\mathbb{N}}_0,\hfill \\ 0,\hfill & \mathrm{otherwise}\hfill \end{array}\right.,$$

and

$${\rho }_1\left(n\right)=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}n=k^2-1,k\in {\mathbb{N}}_0,\hfill \\ -1,\hfill & \mathrm{if}n=3k^2-1,k\in {\mathbb{N}}_0,\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

We see that the numbers $R_s\left(n\right)$ and ${\rho }_s\left(n\right)$ have the same parity. The proof follows easily considering Corollary 1.

3. A Truncated Theta Identity of Gauss

Recall that an overpartition of a positive integer n is a partition of n in which the first occurrence of a part of each size may be overlined or not [10]. Let $\overline{p}\left(n\right)$ denote the number of overpartitions of an integer n. For example, $\overline{p}\left(4\right)=14$ because there are 14 possible overpartitions of 4:

$$\begin{array}{c}\left(4\right),\left(\overline{4}\right),(3,1),(3,\overline{1}),(\overline{3},1),(\overline{3},\overline{1}),(2,2),(\overline{2},2),(2,1,1),\hfill \\ (2,\overline{1},1),(\overline{2},1,1),(\overline{2},\overline{1},1),(1,1,1,1),(\overline{1},1,1,1).\hfill \end{array}$$

Since the overlined parts form a partition into distinct parts and the non-overlined parts form an ordinary partition, we have the following generating function for overpartitions,

$$\sum _{n=0}^{\infty }\overline{p}\left(n\right)q^n=\frac{{(-q;q)}_{\infty }}{{(q;q)}_{\infty }}.$$

For $s\in \{1,5\}$, the partition function $R_s\left(n\right)$ is closely related to $\overline{p}\left(n\right)$. We consider that $\overline{p}\left(x\right)=0$ when x is not a nonnegative integer.

Theorem 3.

 (i)

For $n>0$,

$$R_1\left(n\right)=\sum _{k=1}^{\infty }\left(\overline{p}\left(\frac{n+1-k^2}{2}\right)-\overline{p}\left(\frac{n+1-3k^2}{2}\right)\right).$$

 (ii)

For $n⩾0$,

$$R_5\left(n\right)=\overline{p}\left(\frac{n}{2}\right)+\sum _{k=1}^{\infty }\left(\overline{p}\left(\frac{n-k^2}{2}\right)+\overline{p}\left(\frac{n-3k^2}{2}\right)\right).$$

Proof. 

According to (8), we have

$$\sum _{n=0}^{\infty }{R}_s\left(n\right)q^n=\frac{{(-q^2;q^2)}_{\infty }}{{(q^2;q^2)}_{\infty }}·\frac{{(q^2,q^6,q^{10},q^{12};q^{12})}_{\infty }}{{(q^{6-s},q^{6+s},q^3,q^9;q^{12})}_{\infty }}.$$

By this relation, considering (9) and (10), we can write

$$\sum _{n=0}^{\infty }{R}_1\left(n\right)q^n=\left(\sum _{n=0}^{\infty }\overline{p}\left(n\right)q^n\right)\left(\sum _{n=1}^{\infty }\left(q^{n^2-1}-q^{3n^2-1}\right)\right),$$

and

$$\sum _{n=0}^{\infty }{R}_5\left(n\right)q^n=\left(\sum _{n=0}^{\infty }\overline{p}\left(n\right)q^n\right)\left(1+\sum _{n=1}^{\infty }\left(q^{n^2}+q^{3n^2}\right)\right).$$

The proof follows easily by equating the coefficient of $q^n$ on each side of these identities. □

In [11], Andrews and Merca defined ${\overline{M}}_k\left(n\right)$ to be the number of overpartitions of n in which the first part larger than k appears at least $k+1$ times. If $n=12$ and $k=2$, then we have ${\overline{M}}_2\left(12\right)=16$ because the overpartitions in question are

$$\begin{array}{c}(4,4,4),(\overline{4},4,4),(3,3,3,3),(\overline{3},3,3,3),(3,3,3,2,1),(3,3,3,\overline{2},1),(3,3,3,2,\overline{1}),\hfill \\ (3,3,3,\overline{2},\overline{1}),(\overline{3},3,3,2,1),(\overline{3},3,3,\overline{2},1),(\overline{3},3,3,2,\overline{1}),(\overline{3},3,3,\overline{2},\overline{1}),(3,3,3,1,1,1),\hfill \\ (3,3,3,\overline{1},1,1),(\overline{3},3,3,1,1,1),(\overline{3},3,3,\overline{1},1,1).\hfill \end{array}$$

We have the following truncated versions of the linear recurrence relation (11) obtained in the previous section.

Theorem 4.

Let k and n be positive integers. For $s\in \{1,5\}$,

$${(-1)}^k\left(R_s\left(n\right)+2\sum _{j=1}^k{(-1)}^k{R}_s(n-2j^2)-{\rho }_s\left(n\right)\right)=\sum _{j=0}^n{\rho }_s\left(j\right){\overline{M}}_k\left((n-j)/2\right),$$

where ${\overline{M}}_k\left(x\right)=0$ if x is not a nonnegative integer.

Proof. 

In [11], the authors considered the theta identity (2) and proved the following truncated form:

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

Replacing q by $q^2$ in this identity, we have:

$$\begin{array}{cc}\hfill & {(-1)}^k\left(\frac{{(-q^2;q^2)}_{\infty }}{{(q^2;q^2)}_{\infty }}\left(1+2\sum _{j=1}^k{(-1)}^j{q}^{2j^2}\right)-1\right)=\sum _{n=0}^{\infty }{\overline{M}}_k\left(n\right)q^{2n}.\hfill \end{array}$$

(12)

Multiplying both sides of (12) by the generating function of ${\rho }_s\left(n\right)$, we obtain:

$$\begin{array}{cc}\hfill & {(-1)}^k\left(\left(\sum _{n=0}^{\infty }{R}_s\left(n\right)q^n\right)\left(1+2\sum _{j=1}^k{(-1)}^j{q}^{2j^2}\right)-\sum _{n=0}^{\infty }{\rho }_s\left(n\right)q^n\right)\hfill \\ \hfill & =\left(\sum _{n=0}^{\infty }{\rho }_s\left(n\right)q^n\right)\left(\sum _{n=0}^{\infty }{\overline{M}}_k\left(n\right)q^{2n}\right),\hfill \end{array}$$

(13)

where we have invoked that

$$\frac{{(q^2;q^2)}_{\infty }}{{(-q^2;q^2)}_{\infty }}\sum _{n=0}^{\infty }{R}_s\left(n\right)q^n=\sum _{n=0}^{\infty }{\rho }_s\left(n\right)q^n.$$

The proof follows easily by equating the coefficient of $q^n$ on each side of the identity (13). □

Considering this theorem and the relation (9), we derive the following infinite family of linear inequalities.

Corollary 3.

Let k and n be positive integers. Then,

$${(-1)}^k\left(R_5\left(n\right)+2\sum _{j=1}^k{(-1)}^j{R}_5(n-2j^2)-{\rho }_5\left(n\right)\right)⩾0,$$

with strict inequality if $n⩾2k^2$.

There is a substantial amount of numerical evidence to conjecture the following inequality.

Conjecture 1.

Let k and n be positive integers. Then,

$${(-1)}^k\left(R_1\left(n\right)+2\sum _{j=1}^k{(-1)}^j{R}_1(n-2j^2)-{\rho }_1\left(n\right)\right)⩾0,$$

with strict inequality if $n=2k^2$ or $n⩾2k^2+2$.

Relevant to Theorem 4, it would be very appealing to have combinatorial interpretations of

$$\sum _{j=0}^n{\rho }_s\left(j\right){\overline{M}}_k\left((n-j)/2\right),$$

when $s\in \{1,5\}$.

4. Truncated Pentagonal Number Theorems

Andrews and Merca [12] defined $M_k\left(n\right)$ to be the number of partitions of n in which k is the least positive integer that is not a part, and there are more parts $>k$ than there are parts $<k$. If $n=18$ and $k=3$, then we have $M_3\left(18\right)=3$ because the three partitions in question are:

$$(5,5,5,2,1),(6,5,4,2,1),(7,4,4,2,1).$$

We have the following result.

Theorem 5.

Let k and n be positive integers. For each $s\in \{1,5\}$, the partition functions $R_s\left(n\right)$ and $M_k\left(n\right)$ are related by

$$\begin{array}{c}{(-1)}^k\left({\omega }_s\left(n\right)-{\displaystyle \sum _{j=-(k-1)}^k}{(-1)}^j{R}_s\left(n-j(3j-1)/2\right)\right)\hfill \\ ={\displaystyle \sum _{j=-\infty }^{\infty }}{(-1)}^{j+1}{M}_k\left(n-(j+1)(6j+s)\right).\hfill \end{array}$$

Proof. 

In [12], Andrews and Merca considered Euler’s pentagonal number theorem (1) and proved the following truncated form:

$$\frac{{(-1)}^{k-1}}{{(q;q)}_{\infty }}\sum _{n=-(k-1)}^k{(-1)}^n{q}^{n(3n-1)/2}={(-1)}^{k-1}+\sum _{n=k}^{\infty }\frac{q^{\left(\genfrac{}{}{0pt}{}{k}{2}\right)+(k+1)n}}{{(q;q)}_n}\left[\begin{array}{c}n-1\\ k-1\end{array}\right],$$

(14)

where

$$\left[\begin{array}{c}n\\ k\end{array}\right]=\left\{\begin{array}{cc}{\displaystyle \frac{{(q;q)}_n}{{(q;q)}_k{(q;q)}_{n-k}}},\hfill & \mathrm{if}0⩽k⩽n,\hfill \\ 0,\hfill & \mathrm{otherwise}\hfill \end{array}\right.$$

is the q-binomial coefficient.

According to the Jacobi triple product identity (3), the generating function of ${\omega }_s\left(n\right)$ is given by

$$\sum _{n=0}^{\infty }{\omega }_s\left(n\right)q^n=\sum _{n=-\infty }^{\infty }{(-1)}^n{q}^{(n+1)(6n+s)}={(q^s,q^{12-s},q^{12};q^{12})}_{\infty }.$$

Multiplying both sides of (14) by ${(q^s,q^{12-s},q^{12};q^{12})}_{\infty }$, we obtain

$$\begin{array}{c}{(-1)}^{k-1}\left(\left({\displaystyle \sum _{n=1}^{\infty }}{R}_s\left(n\right)q^n\right)\left({\displaystyle \sum _{n=-(k-1)}^k}{(-1)}^n{q}^{n(3n-1)/2}\right)-{\displaystyle \sum _{n=0}^{\infty }}{\omega }_s\left(n\right)q^n\right)\hfill \\ =\left({\displaystyle \sum _{n=0}^{\infty }}{\omega }_s\left(n\right)q^n\right)\left({\displaystyle \sum _{n=0}^{\infty }}{M}_k\left(n\right)q^n\right),\hfill \end{array}$$

where we have invoked the generating function for $M_k\left(n\right)$ [12],

$$\sum _{n=0}^{\infty }{M}_k\left(n\right)q^n=\sum _{n=k}^{\infty }\frac{q^{\left(\genfrac{}{}{0pt}{}{k}{2}\right)+(k+1)n}}{{(q;q)}_n}\left[\begin{array}{c}n-1\\ k-1\end{array}\right].$$

The proof follows easily considering Cauchy’s multiplication of two power series. □

By the truncated form of Euler’s pentagonal number theorem (14), we see that the coefficients of $q^n$ in the series

$${(-1)}^k\left(1-\frac{1}{{(q;q)}_{\infty }}\sum _{n=-(k-1)}^k{(-1)}^n{q}^{n(3n-1)/2}\right)$$

are all nonegative. Related to this result on the truncated pentagonal number series, we remark that there is a substantial amount of numerical evidence to conjecture a stronger result.

Conjecture 2.

Let k and n be positive integers. If $s\in \{1,5\}$, then the coefficients of $q^n$ in the series

$${(-1)}^k\left(1-\frac{1}{{(q;q)}_{\infty }}\sum _{n=-(k-1)}^k{(-1)}^n{q}^{n(3n-1)/2}\right){(q^s,q^{12-s},q^{12};q^{12})}_{\infty }$$

are all nonnegative.

Assuming this conjecture, we immediately deduce that the partition function $R_s\left(n\right)$ satisfies for $s\in \{1,5\}$ the following infinite family of linear inequalities.

Conjecture 3.

Let k and n be positive integers. If $s\in \{1,5\}$, then

$${(-1)}^k\left({\omega }_s\left(n\right)-\sum _{j=-(k-1)}^k{(-1)}^j{R}_s\left(n-j(3j-1)/2\right)\right)⩾0.$$

Very recently, Xia and Zhao [13] defined ${\tilde{P}}_k\left(n\right)$ to be the number of partitions of n in which every part $⩽k$ appears at least once, and the first part larger than k appears at least $k+1$ times. For example, ${\tilde{P}}_2\left(17\right)=9$, and the partitions in question are:

$$\begin{array}{c}(5,3,3,3,2,1),(4,4,4,2,2,1),(4,4,4,2,1,1,1),(4,3,3,3,2,1,1),(3,3,3,3,2,2,1),\hfill \\ (3,3,3,3,2,1,1,1),(3,3,3,2,2,2,1,1),(3,3,3,2,2,1,1,1,1),(3,3,3,2,1,1,1,1,1,1).\hfill \end{array}$$

In analogy with Theorem 5, we have the following result.

Theorem 6.

Let k and n be positive integers. For each $s\in \{1,5\}$, the partition functions $R_s\left(n\right)$ and ${\tilde{P}}_k\left(n\right)$ are related by

$$\begin{array}{c}{(-1)}^{k-1}\left({\omega }_s\left(n\right)-{\displaystyle \sum _{j=-k}^k}{(-1)}^j{R}_s\left(n-j(3j-1)/2\right)\right)\hfill \\ ={\displaystyle \sum _{j=-\infty }^{\infty }}{(-1)}^{j+1}{\tilde{P}}_k\left(n-(j+1)(6j+s)\right).\hfill \end{array}$$

Proof. 

The proof of this identity is quite similar to the proof of Theorem 5. In [13], Xia and Zhao considered Euler’s pentagonal number Theorem (1) and proved the following truncated form:

$$\frac{1}{{(q;q)}_{\infty }}\sum _{n=-k}^k{(-1)}^n{q}^{n(3n-1)/2}=1+{(-1)}^k\frac{q^{k(k+1)/2}}{{(q;q)}_k}\sum _{n=0}^{\infty }\frac{q^{(n+k+1)(k+1)}}{{(q^{n+k+1};q)}_{\infty }}.$$

(15)

Multiplying both sides of (15) by ${(q^s,q^{12-s},q^{12};q^{12})}_{\infty }$, we obtain

$$\begin{array}{c}{(-1)}^k\left(\left({\displaystyle \sum _{n=1}^{\infty }}{R}_s\left(n\right)q^n\right)\left({\displaystyle \sum _{n=-k}^k}{(-1)}^n{q}^{n(3n-1)/2}\right)-{\displaystyle \sum _{n=0}^{\infty }}{\omega }_s\left(n\right)q^n\right)\hfill \\ =\left({\displaystyle \sum _{n=0}^{\infty }}{\omega }_s\left(n\right)q^n\right)\left({\displaystyle \sum _{n=0}^{\infty }}{\tilde{P}}_k\left(n\right)q^n\right),\hfill \end{array}$$

where we have invoked the generating function for ${\tilde{P}}_k\left(n\right)$ [13],

$$\sum _{n=0}^{\infty }{\tilde{P}}_k\left(n\right)q^n=\frac{q^{k(k+1)/2}}{{(q;q)}_k}\sum _{n=0}^{\infty }\frac{q^{(n+k+1)(k+1)}}{{(q^{n+k+1};q)}_{\infty }}.$$

The proof follows easily considering Cauchy’s multiplication of two power series. □

By the second truncated form of Euler’s pentagonal number theorem (15), we see that the coefficients of $q^n$ in the series

$${(-1)}^{k-1}\left(1-\frac{1}{{(q;q)}_{\infty }}\sum _{n=-k}^k{(-1)}^n{q}^{n(3n-1)/2}\right)$$

are all nonegative. Related to this result on the truncated pentagonal number series, we remark that there is a substantial amount of numerical evidence to conjecture a stronger result.

Conjecture 4.

Let k and n be positive integers. If $s\in \{1,5\}$, then the coefficients of $q^n$ in the series

$${(-1)}^{k-1}\left(1-\frac{1}{{(q;q)}_{\infty }}\sum _{n=-k}^k{(-1)}^n{q}^{n(3n-1)/2}\right){(q^s,q^{12-s},q^{12};q^{12})}_{\infty }$$

are all nonnegative.

Assuming this conjecture, we immediately deduce that the partition function $R_s\left(n\right)$ satisfies for $s\in \{1,5\}$ the following infinite family of linear inequalities.

Conjecture 5.

Let k and n be positive integers. If $s\in \{1,5\}$, then

$${(-1)}^{k-1}\left({\omega }_s\left(n\right)-\sum _{j=-k}^k{(-1)}^j{R}_s\left(n-j(3j-1)/2\right)\right)⩾0.$$

When $k\to \infty$, Theorem 5 or Theorem 6 yields Corollary 2. On the other hand, by Theorems 5 and 6, for each $s\in \{1,5\}$ we easily deduce an infinite family of decompositions of $R_s\left(n\right)$.

Theorem 7.

Let k and n be positive integers. If $s\in \{1,5\}$, then

$$\begin{array}{c}{R}_s\left(n\right)={\displaystyle \sum _{j=-\infty }^{\infty }}{(-1)}^{j+1}(M_k\left(n+k(3k+1)/2-(j+1)(6j+s)\right)\hfill \\ +{\tilde{P}}_k\left(n+k(3k+1)/2-(j+1)(6j+s)\right))\hfill \end{array}.$$

Proof. 

We have

$$\begin{array}{c}{\displaystyle \sum _{j=-\infty }^{\infty }}{(-1)}^{j+1}{M}_k\left(n-(j+1)(6j+s)\right)+{\displaystyle \sum _{j=-\infty }^{\infty }}{(-1)}^{j+1}{\tilde{P}}_k\left(n-(j+1)(6j+s)\right)\hfill \\ ={(-1)}^k\left({\omega }_s\left(n\right)-{\displaystyle \sum _{j=-(k-1)}^k}{(-1)}^j{R}_s\left(n-j(3j-1)/2\right)\right)\hfill \\ +{(-1)}^{k-1}\left({\omega }_s\left(n\right)-{\displaystyle \sum _{j=-(1-k)}^k}{(-1)}^j{R}_s\left(n-j(3j-1)/2\right)\right)\hfill \\ =R_s\left(n-k(3k+1)/2\right).\hfill \end{array}$$

The proof follows easily replacing n by $n+k(3k+1)/2$. □

In the context given by Theorem 7, it would be very appealing to have combinatorial interpretations of

$$\sum _{j=-\infty }^{\infty }{(-1)}^{j+1}{M}_k\left(n-(j+1)(6j+s)\right)$$

and

$$\sum _{j=-\infty }^{\infty }{(-1)}^{j+1}{\tilde{P}}_k\left(n-(j+1)(6j+s)\right)$$

when $s\in \{1,5\}$.

5. Concluding Remarks

A parity result involving sums of partition numbers $p\left(n\right)$ has been introduced in this paper considering the partition function $R_s\left(n\right)$, which counts the partitions of n into parts not congruent to 0, $\pm s(mod12)$ with $s\in \{1,5\}$. Our result is based on the fact that for each $s\in \{1,5\}$, almost all values of $R_s\left(n\right)$ are divisible by 2.

Identities of the form (9) or (10) were investigated by Cooper and Hirschhorn [9] in a paper published in 2001. Can other their identities be used to derive similar results to Theorem 1?

Ramanujan discovered various surprising congruences for $p\left(n\right)$ when n is in certain special arithmetic progressions. For example:

• $p(5n+4)\equiv 0(mod5)$,
• $p(7n+5)\equiv 0(mod7)$,
• $p(11n+6)\equiv 0(mod11)$.

Related to our partition function $R_s\left(n\right)$, we remark on the following congruence modulo 4.

Theorem 8.

For all $n,\alpha ⩾0$,

 (i)

$R_1\left(4^{\alpha }(8n+6)\right)\equiv 0(mod4)$;

 (ii)

$R_5\left(4^{\alpha }(8n+7)\right)\equiv 0(mod4)$.

Proof. 

By (11), we see that $R_5\left(n\right)$ is even when n is not a square or thrice square. In addition, since $4^{\alpha }(8n+7)$ is never a square or thrice square, by (11) we deduce that

$$R_5\left(4^{\alpha }(8n+7)\right)+2\sum _{j=1}^{\infty }{(-1)}^j{R}_5\left(4^{\alpha }(8n+7)-2j^2\right)=0.$$

(16)

On the other hand, it is not difficult two show that the Diophantine equations

$$a^2+2b^2=4^{\alpha }(8n+7)\mathrm{and}3a^2+2b^2=4^{\alpha }(8n+7)$$

do not have solutions. This means that $4^{\alpha }(8n+7-2j^2)$ is never a square or thrice square. Thus, we deduce that $R_5\left(4^{\alpha }(8n+7)-2j^2\right)\equiv 0(mod2)$. Considering (16), we obtain $R_5\left(4^{\alpha }(8n+7)\right)\equiv 0(mod4)$. In a similar way, we deduce the first congruence.

□

There is a substantial amount of numerical evidence to propose as an open problem the following Ramanujan-type congruence for $R_s\left(n\right)$ when $s\in \{1,5\}$.

Conjecture 6.

For all $n\neg \equiv 0(mod3)$, we have

 (i)

$R_1(32n+3)\equiv 0(mod3),$

 (ii)

$R_5(32n+4)\equiv 0(mod3).$

It would be very appealing to have solutions and generalizations of this open problem.