Advertisement

Annals of Combinatorics

, Volume 23, Issue 3–4, pp 489–509 | Cite as

Combinations of Ranks and Cranks of Partitions Moduli 6, 9 and 12 and Their Comparison with the Partition Function

  • Zafer Selcuk AyginEmail author
  • Song Heng Chan
Article
  • 42 Downloads

Abstract

Let \(L \in \{6,9,12 \} \). We determine the generating functions of certain combinations of three ranks and three cranks modulo L in terms of eta quotients. Then, using the periodicity of signs of these eta quotients, we compare their values with the values of \(\frac{p(n)}{L/3}\).

Keywords

Eisenstein series Dedekind eta function Eta quotients Modular forms Derivatives 

Mathematics Subject Classification

11A25 11E20 11F11 11F20 11F30 11Y35 

1 Introduction and Notation

Dyson [3] defined the rank of a partition to be the largest part minus the number of parts. Let p(n), N(an) and N(aLn) denote the number of partitions of n; the number of partitions of n with rank a; and the number of partitions of n with rank congruent to a modulo L, respectively. Dyson specifically conjectured that
$$\begin{aligned} N(a,5;5n+4)&=\frac{p(5n+4)}{5},\quad \mathrm{for }\,0 \le a \le 4 , \end{aligned}$$
(1.1)
$$\begin{aligned} N(a,7;7n+5)&=\frac{p(7n+5)}{7},\quad \mathrm{for }\,0 \le a \le 6, \end{aligned}$$
(1.2)
which provide combinatorial interpretations for Ramanujan’s partition congruences modulo 5 and 7. Equations (1.1) and (1.2) were proven by Atkin and Swinnerton-Dyer in [2]. The rank does not give a combinatorial interpretation for Ramanujan’s partition congruences modulo 11. Thus, Dyson also conjectured the existence of a statistics which can and called it the ‘crank’. Later Garvan defined cranks and showed that they give the desired combinatorial interpretation for congruences modulo 11 as well as 5 and 7. Let M(an) and M(aLn) denote the number of partitions of n with crank a and the number of partitions of n with crank congruent to a modulo L. We note the following elementary equations for future reference:
$$\begin{aligned}&\displaystyle N(a,L;n)=N(-a,L;n),~ M(a,L;n)=M(-a,L;n), \end{aligned}$$
(1.3)
$$\begin{aligned}&\displaystyle \sum _{n = 0} ^{\infty } \sum _{a=0}^{L-1} N(a,L;n) q^n =\sum _{n = 0} ^{\infty } p(n) q^n,~ \sum _{n = 0} ^{\infty } \sum _{a=0}^{L-1} M(a,L;n) q^n = \sum _{n = 0} ^{\infty } p(n) q^n,\nonumber \\ \end{aligned}$$
(1.4)
and for a shorthand notation, we use
$$\begin{aligned} N(a,b,c;L;n)&=N(a,L;n)+N(b,L;n)+N(c,L;n), \\ M(a,b,c;L;n)&=M(a,L;n)+M(b,L;n)+M(c,L;n). \end{aligned}$$
Let \(q=\mathrm{e}^{2 \pi i z}\) with \(z \in {\mathbb {H}}\), thus \(\vert q \vert <1\). We define the infinite product \((q;q)_{\infty }=\prod _{n=1}^{\infty } (1-q^n)\). A specialized version of this product, \(\eta (q)=q^{1/24}(q;q)_{\infty }\), is called the Dedekind eta function, whose quotients are called eta quotients.
In [5, Theorem 4.1], Kang proved the following relation between ranks and cranks:
$$\begin{aligned}&\sum _{m=-\infty }^{\infty } \sum _{n=0}^{\infty } \big ( N(3m-1;n) + N(3m;n) + N(3m+1;n)\big ) w^m q^n \nonumber \\&\quad = \frac{(q^3;q^3)_{\infty }}{(q;q)_{\infty }} \sum _{m=-\infty }^{\infty } \sum _{n=0}^{\infty } M(m;n) w^m q^{3n}. \end{aligned}$$
(1.5)
She then replaces w with \(-1\) to obtain
$$\begin{aligned}&\sum _{n=0}^{\infty } \big ( N(0,1,1;6;n) - N(2,2,3;6;n) \big ) q^n\nonumber \\&\quad =\frac{(q^3;q^3)_{\infty }}{(q;q)_{\infty }} \sum _{n=0}^{\infty } \big ( M(0,2;n) - M(1,2;n) \big ) q^{3n}= \frac{(q^3;q^3)^4_{\infty }}{(q;q)_{\infty }(q^6;q^6)^2_{\infty }}. \end{aligned}$$
(1.6)
Inspired by (1.5), we give generating functions of two variations of three combinations of the crank function. Then letting \(L \in \{6,9,12 \}\), we use her formula and our results to find the generating functions of
$$\begin{aligned}&N(3j-1,3j,3j+1;L;n) ,\quad \mathrm{for \,\,}0 \le j \le L/3-1, \end{aligned}$$
(1.7)
$$\begin{aligned}&M(3j-1,3j,3j+1;L;n),\quad \mathrm{for \,\,}0 \le j \le L/3-1, \end{aligned}$$
(1.8)
$$\begin{aligned}&M(3j-2,3j-1,3j;L;n) \mathrm{,\quad for \,}0 \le j \le L/3-1, \end{aligned}$$
(1.9)
in terms of the partition function and eta quotients. The signs of the coefficients of the eta quotients in the generating functions seem to be periodic (except a few cases). This lets us to compare (1.7)–(1.9) with \(\tfrac{p(n)}{L/3}\).

In the next section, we state the main results. In Sect. 3, we define the Jacobi theta function and give some preliminary results. In Sect. 4, we give the generating functions of two variations of combinations of three cranks. The combination in (1.8) is motivated by Kang’s results from [5], and the combination in (1.9) was chosen because among all other options, this one gives more elegant results. In Sect. 5, we prove Theorem 2.1. Fourier coefficients of all combinations of eta quotients that appear in Theorem 2.1 seem to be periodic in their signs. Some of these are proved in Sect. 6. The rest are left as conjectures, see Sect. 8. Section 7 is dedicated to the discussion on the connection of our results to recent results by Hickerson and Mortenson [4].

2 Main Results

Let
$$\begin{aligned} \displaystyle F(q)=\sum _{n =0}^{\infty } a_n q^n, \end{aligned}$$
then we denote \(\sum \nolimits _{n =0}^{\infty } \mathfrak {Re}(a_n) q^n\) by \( \mathfrak {Re}\big (F(q)\big )\) and \( \sum \nolimits _{n =0}^{\infty } \mathfrak {Im}(a_n) q^n\) by \( \mathfrak {Im}\big (F(q)\big )\). Below we state the main theorems.

Theorem 2.1

The generating functions of combinations of three ranks in (1.7) are given by
$$\begin{aligned} \sum _{n = 0} ^{\infty } N(0,1,1;6;n) q^n&= \frac{1}{2(q;q)_{\infty }} + \frac{(q^3;q^3)^4_{\infty }}{2(q;q)_{\infty }(q^6;q^6)^2_{\infty }} , \end{aligned}$$
(2.1)
$$\begin{aligned} \sum _{n = 0} ^{\infty } N(2,2,3;6;n) q^n&= \frac{1}{2(q;q)_{\infty }} - \frac{(q^3;q^3)^4_{\infty }}{2(q;q)_{\infty }(q^6;q^6)^2_{\infty }} ,\end{aligned}$$
(2.2)
$$\begin{aligned} \sum _{n = 0} ^{\infty } N(0,1,1;9;n) q^n&= \frac{1}{3(q;q)_{\infty }}+ \frac{2(q^3;q^3)^3_{\infty }}{3(q;q)_{\infty }(q^9;q^9)_{\infty }} ,\end{aligned}$$
(2.3)
$$\begin{aligned} \sum _{n = 0} ^{\infty } N(2,3,4;9;n) q^n&= \frac{1}{3(q;q)_{\infty }} - \frac{(q^3;q^3)^3_{\infty }}{3(q;q)_{\infty }(q^9;q^9)_{\infty }} ,\end{aligned}$$
(2.4)
$$\begin{aligned} \sum _{n = 0} ^{\infty } N(2,3,4;12;n) q^n&= \frac{1}{4(q;q)_{\infty }} - \frac{(q^3;q^3)^4_{\infty }}{4(q;q)_{\infty }(q^6;q^6)^2_{\infty }},\end{aligned}$$
(2.5)
$$\begin{aligned} \sum _{n = 0} ^{\infty } N(0,1,1;12;n) q^n&= \frac{1}{4(q;q)_{\infty }} + \frac{(q^3;q^3)^2_{\infty }(q^6;q^6)_{\infty }}{2(q;q)_{\infty }(q^{12};q^{12})_{\infty }}+\frac{(q^3;q^3)^4_{\infty }}{4(q;q)_{\infty }(q^6;q^6)^2_{\infty }} ,\end{aligned}$$
(2.6)
$$\begin{aligned} \sum _{n = 0} ^{\infty } N(5,5,6;12;n) q^n&= \frac{1}{4(q;q)_{\infty }} - \frac{(q^3;q^3)^2_{\infty }(q^6;q^6)_{\infty }}{2(q;q)_{\infty }(q^{12};q^{12})_{\infty }}+\frac{(q^3;q^3)^4_{\infty }}{4(q;q)_{\infty }(q^6;q^6)^2_{\infty }} . \end{aligned}$$
(2.7)
The generating functions of combinations of three cranks in (1.8) are given by
$$\begin{aligned} \sum _{n = 0} ^{\infty } M(0,1,1;6;n) q^n&= \frac{1}{2(q;q)_{\infty }}+\frac{(q^3;q^3)^4_{\infty }}{2(q;q)_{\infty }(q^6;q^6)^2_{\infty }} \nonumber \\&\quad -2q^2 \frac{(q^3;q^3)_\infty (q^4;q^4)_\infty (q^{36};q^{36})_\infty ^2}{(q^2;q^2)_\infty (q^{12};q^{12})_\infty (q^{18};q^{18})_\infty }, \end{aligned}$$
(2.8)
$$\begin{aligned} \sum _{n = 0} ^{\infty } M(2,2,3;6;n) q^n&= \frac{1}{2(q;q)_{\infty }}-\frac{(q^3;q^3)^4_{\infty }}{2(q;q)_{\infty }(q^6;q^6)^2_{\infty }} \nonumber \\&\quad +2q^2 \frac{(q^3;q^3)_\infty (q^4;q^4)_\infty (q^{36};q^{36})_\infty ^2}{(q^2;q^2)_\infty (q^{12};q^{12})_\infty (q^{18};q^{18})_\infty },\end{aligned}$$
(2.9)
$$\begin{aligned} \sum _{n = 0} ^{\infty } M(0,1,1;9;n) q^n&= \frac{1}{3(q;q)_{\infty }} + \frac{2(q^9;q^9)_{\infty }^3}{3(q;q)_{\infty } (q^{27};q^{27})_{\infty }} \nonumber \\&\quad -2 q^2 \frac{(q^3;q^3)_{\infty }(q^{27};q^{27})_{\infty }^2}{(q;q)_{\infty } (q^9;q^9)_{\infty }},\end{aligned}$$
(2.10)
$$\begin{aligned} \sum _{n = 0} ^{\infty } M(2,3,4;9;n) q^n&= \frac{1}{3(q;q)_{\infty }} - \frac{(q^9;q^9)_{\infty }^3}{3(q;q)_{\infty } (q^{27};q^{27})_{\infty }} \nonumber \\&\quad + q^2 \frac{(q^3;q^3)_{\infty }(q^{27};q^{27})_{\infty }^2}{(q;q)_{\infty } (q^9;q^9)_{\infty }} , \end{aligned}$$
(2.11)
$$\begin{aligned} \sum _{n = 0} ^{\infty } M(2,3,4;12;n) q^n&= \frac{1}{4(q;q)_{\infty }} -\frac{(q^3;q^3)^4_{\infty }}{4(q;q)_{\infty }(q^6;q^6)^2_{\infty }} \nonumber \\&\quad + q^2 \frac{(q^3;q^3)_\infty (q^4;q^4)_\infty (q^{36};q^{36})_\infty ^2}{(q^2;q^2)_\infty (q^{12};q^{12})_\infty (q^{18};q^{18})_\infty },\end{aligned}$$
(2.12)
$$\begin{aligned} \sum _{n = 0} ^{\infty } M(0,1,1;12;n) q^n&= \frac{1}{4(q;q)_{\infty }} +\frac{(q^3;q^3)^4_{\infty }}{4(q;q)_{\infty }(q^6;q^6)^2_{\infty }} \nonumber \\&\quad + \frac{(q^{2};q^{2})_\infty ^2 (q^{9};q^{9})_\infty ^2}{2(q;q)_\infty (q^{4};q^{4})_\infty (q^{18};q^{18})_\infty } \nonumber \\&\quad -q^2 \frac{(q^3;q^3)_\infty (q^4;q^4)_\infty (q^{36};q^{36})_\infty ^2}{(q^2;q^2)_\infty (q^{12};q^{12})_\infty (q^{18};q^{18})_\infty }, \end{aligned}$$
(2.13)
$$\begin{aligned} \sum _{n = 0} ^{\infty } M(5,5,6;12;n) q^n&= \frac{1}{4(q;q)_{\infty }} +\frac{(q^3;q^3)^4_{\infty }}{4(q;q)_{\infty }(q^6;q^6)^2_{\infty }} \nonumber \\&\quad -\frac{(q^{2};q^{2})_\infty ^2 (q^{9};q^{9})_\infty ^2}{2(q;q)_\infty (q^{4};q^{4})_\infty (q^{18};q^{18})_\infty } \nonumber \\&\quad -q^2 \frac{(q^3;q^3)_\infty (q^4;q^4)_\infty (q^{36};q^{36})_\infty ^2}{(q^2;q^2)_\infty (q^{12};q^{12})_\infty (q^{18};q^{18})_\infty }. \end{aligned}$$
(2.14)
The generating functions of combinations of three cranks in (1.9) are given by
$$\begin{aligned}&\sum _{n = 0} ^{\infty } M(0,1,2;6;n) q^n = \frac{1}{2(q;q)_{\infty }} + \frac{(q;q)_\infty (q^{9};q^{9})^2_\infty }{2(q^{2};q^{2})_\infty (q^{18};q^{18})_\infty }, \end{aligned}$$
(2.15)
$$\begin{aligned}&\sum _{n = 0} ^{\infty } M(3,4,5;6;n) q^n = \frac{1}{2(q;q)_{\infty }} - \frac{(q;q)_\infty (q^{9};q^{9})^2_\infty }{2(q^{2};q^{2})_\infty (q^{18};q^{18})_\infty },\end{aligned}$$
(2.16)
$$\begin{aligned}&\sum _{n = 0} ^{\infty } M(0,1,2;9;n) q^n = \frac{1}{3(q;q)_{\infty }} + \frac{2}{3} \mathfrak {Re} \left( \frac{(q^3;q^3)_{\infty } (\zeta _3 q^3; \zeta _3q^3)_{\infty }^2}{(q;q)_{\infty } (q^9;q^9)_{\infty }} \right. \nonumber \\&\quad \left. + (\zeta _3-1) q \frac{(q^3;q^3)_{\infty } (\zeta _3^2 q^3; \zeta _3^2 q^3)_{\infty } (q^{27};q^{27})_{\infty }}{(q;q)_{\infty } (q^9;q^9)_{\infty }} \right) , \end{aligned}$$
(2.17)
$$\begin{aligned}&\sum _{n = 0} ^{\infty } M(3,4,5;9;n) q^n = \frac{1}{3(q;q)_{\infty }} - \frac{1}{3} \mathfrak {Re} \left( \frac{(q^3;q^3)_{\infty } (\zeta _3 q^3; \zeta _3q^3)_{\infty }^2}{(q;q)_{\infty } (q^9;q^9)_{\infty }}\right. \nonumber \\&\quad \left. + (\zeta _3-1) q \frac{(q^3;q^3)_{\infty } (\zeta _3^2 q^3; \zeta _3^2 q^3)_{\infty } (q^{27};q^{27})_{\infty }}{(q;q)_{\infty } (q^9;q^9)_{\infty }} \right) \nonumber \\&\quad - \frac{1}{\sqrt{3}} \mathfrak {Im} \left( \frac{(q^3;q^3)_{\infty } (\zeta _3 q^3; \zeta _3q^3)_{\infty }^2}{(q;q)_{\infty } (q^9;q^9)_{\infty }} \right. \nonumber \\&\quad \left. + (\zeta _3-1) q \frac{(q^3;q^3)_{\infty } (\zeta _3^2 q^3; \zeta _3^2 q^3)_{\infty } (q^{27};q^{27})_{\infty }}{(q;q)_{\infty } (q^9;q^9)_{\infty }} \right) ,\end{aligned}$$
(2.18)
$$\begin{aligned}&\sum _{n = 0} ^{\infty } M(6,7,8;9;n) q^n = \frac{1}{3(q;q)_{\infty }} - \frac{1}{3} \mathfrak {Re} \left( \frac{(q^3;q^3)_{\infty } (\zeta _3 q^3; \zeta _3q^3)_{\infty }^2}{(q;q)_{\infty } (q^9;q^9)_{\infty }} \right. \nonumber \\&\quad \left. + (\zeta _3-1) q \frac{(q^3;q^3)_{\infty } (\zeta _3^2 q^3; \zeta _3^2 q^3)_{\infty } (q^{27};q^{27})_{\infty }}{(q;q)_{\infty } (q^9;q^9)_{\infty }} \right) \nonumber \\&\quad + \frac{1}{\sqrt{3}} \mathfrak {Im} \left( \frac{(q^3;q^3)_{\infty } (\zeta _3 q^3; \zeta _3q^3)_{\infty }^2}{(q;q)_{\infty } (q^9;q^9)_{\infty }} \right. \nonumber \\&\quad \left. + (\zeta _3-1) q \frac{(q^3;q^3)_{\infty } (\zeta _3^2 q^3; \zeta _3^2 q^3)_{\infty } (q^{27};q^{27})_{\infty }}{(q;q)_{\infty } (q^9;q^9)_{\infty }} \right) ,\end{aligned}$$
(2.19)
$$\begin{aligned}&\sum _{n = 0} ^{\infty } M(0,1,2;12;n) q^n = \frac{1}{4(q;q)_{\infty }} + \frac{(q;q)_\infty (q^{9};q^{9})^2_\infty }{4(q^{2};q^{2})_\infty (q^{18};q^{18})_\infty }\nonumber \\&\quad + \frac{(q^{4};q^{4})_\infty (q^{6};q^{6})_\infty ^4 }{2 (q^{2};q^{2})_\infty (q^{3};q^{3})_\infty (q^{12};q^{12})_\infty ^2},\end{aligned}$$
(2.20)
$$\begin{aligned}&\sum _{n = 0} ^{\infty } M(3,4,5;12;n) q^n = \frac{1}{4(q;q)_{\infty }} - \frac{(q;q)_\infty (q^{9};q^{9})^2_\infty }{4(q^{2};q^{2})_\infty (q^{18};q^{18})_\infty } \nonumber \\&\quad - q\frac{(q^{2};q^{2})^2_\infty (q^{3};q^{3})_\infty (q^{18};q^{18})^2_\infty }{2 (q;q)_\infty (q^{4};q^{4})_\infty (q^{6};q^{6})_\infty (q^{9};q^{9})_\infty },\end{aligned}$$
(2.21)
$$\begin{aligned}&\sum _{n = 0} ^{\infty } M(6,7,8;12;n) q^n = \frac{1}{4(q;q)_{\infty }} + \frac{(q;q)_\infty (q^{9};q^{9})^2_\infty }{4(q^{2};q^{2})_\infty (q^{18};q^{18})_\infty } \nonumber \\&\quad - \frac{(q^{4};q^{4})_\infty (q^{6};q^{6})_\infty ^4 }{2 (q^{2};q^{2})_\infty (q^{3};q^{3})_\infty (q^{12};q^{12})_\infty ^2}, \end{aligned}$$
(2.22)
$$\begin{aligned}&\sum _{n = 0} ^{\infty } M(9,10,11;12;n) q^n = \frac{1}{4(q;q)_{\infty }} - \frac{(q;q)_\infty (q^{9};q^{9})^2_\infty }{4(q^{2};q^{2})_\infty (q^{18};q^{18})_\infty } \nonumber \\&\quad + q \frac{(q^{2};q^{2})^2_\infty (q^{3};q^{3})_\infty (q^{18};q^{18})^2_\infty }{2 (q;q)_\infty (q^{4};q^{4})_\infty (q^{6};q^{6})_\infty (q^{9};q^{9})_\infty }. \end{aligned}$$
(2.23)

The following rank–crank difference formulae are direct consequences of Theorem 2.1.

Corollary 2.2

We have
$$\begin{aligned}&\sum _{n = 0} ^{\infty } \big (N(0,1,1;6;n) - M(0,1,1;6;n) \big ) q^n \\&\quad = 2q^2 \frac{(q^3;q^3)_\infty (q^4;q^4)_\infty (q^{36};q^{36})_\infty ^2}{(q^2;q^2)_\infty (q^{12};q^{12})_\infty (q^{18};q^{18})_\infty } ,\\&\sum _{n = 0} ^{\infty } \big (N(0,1,1;9;n) - M(0,1,1;9;n) \big ) q^n \\&\quad = \frac{2(q^3;q^3)^3_{\infty }}{3(q;q)_{\infty }(q^9;q^9)_{\infty }}- \frac{2 (q^9;q^9)_{\infty }^3}{3(q;q)_{\infty } (q^{27};q^{27})_{\infty }} + 2 q^2 \frac{(q^3;q^3)_{\infty }(q^{27};q^{27})_{\infty }^2}{ (q;q)_{\infty } (q^9;q^9)_{\infty }},\\&\sum _{n = 0} ^{\infty } \big (N(2,3,4;12;n) - M(2,3,4;12;n) \big ) q^n \\&\quad = -q^2 \frac{(q^3;q^3)_\infty (q^4;q^4)_\infty (q^{36};q^{36})_\infty ^2}{(q^2;q^2)_\infty (q^{12};q^{12})_\infty (q^{18};q^{18})_\infty },\\&\sum _{n = 0} ^{\infty } \big (N(0,1,1;12;n) - M(0,1,1;12;n) \big ) q^n \\&\quad = \frac{(q^3;q^3)^2_{\infty }(q^6;q^6)_{\infty }}{2(q;q)_{\infty }(q^{12};q^{12})_{\infty }} - q\frac{(q^{2};q^{2})_\infty ^2 (q^{9};q^{9})_\infty ^2}{2(q;q)_\infty (q^{4};q^{4})_\infty (q^{18};q^{18})_\infty } \\&\qquad + q^2 \frac{(q^3;q^3)_\infty (q^4;q^4)_\infty (q^{36};q^{36})_\infty ^2}{(q^2;q^2)_\infty (q^{12};q^{12})_\infty (q^{18};q^{18})_\infty }.\\ \end{aligned}$$

The coefficients of the combinations of eta quotients in Theorem 2.1 seem to be alternating in sign (except a few cases), from which comparisons with 3p(n) / L can be deduced. In Sect. 6, we prove the following results. The rest of the comparisons are technically very challenging and thus left as conjectures, see Conjecture 8.1.

Theorem 2.3

We have
$$\begin{aligned}&N(0,1,1;6;2n) > \frac{p(2n)}{2} \mathrm{,\quad for \,all}~n \in {\mathbb {N}}_0, \end{aligned}$$
(2.24)
$$\begin{aligned}&N(0,1,1;6;2n+1) < \frac{p(2n+1)}{2} \mathrm{, \quad for \,all}~n \in {\mathbb {N}}, \end{aligned}$$
(2.25)
$$\begin{aligned}&N(2,2,3;6;2n) < \frac{p(2n)}{2} \mathrm{,\quad for \,all}~n \in {\mathbb {N}}_0, \end{aligned}$$
(2.26)
$$\begin{aligned}&N(2,2,3;6;2n+1) > \frac{p(2n+1)}{2}\mathrm{,\quad for \,all}~n \in {\mathbb {N}}, \end{aligned}$$
(2.27)
$$\begin{aligned}&N(0,1,1;9;n) > \frac{p(n)}{3}\mathrm{, \quad for \,all}~n \in {\mathbb {N}}_0 -\{3,7\}, \end{aligned}$$
(2.28)
$$\begin{aligned}&N(2,3,4;9;n) < \frac{p(n)}{3}\mathrm{,\quad for \, all}~n \in {\mathbb {N}}_0 -\{3,7\}, \end{aligned}$$
(2.29)
$$\begin{aligned}&N(2,3,4;12;2n) < \frac{p(2n)}{4}\mathrm{, \quad for \,all}~n \in {\mathbb {N}}_0, \end{aligned}$$
(2.30)
$$\begin{aligned}&N(2,3,4;12;2n+1) > \frac{p(2n+1)}{4}\mathrm{,\quad for \,all}~n \in {\mathbb {N}}, \end{aligned}$$
(2.31)
$$\begin{aligned}&N(0,1,1;12;n) > \frac{p(n)}{4} \mathrm{, \quad for \,all}~n \in {\mathbb {N}}_0, \end{aligned}$$
(2.32)
$$\begin{aligned}&N(5,5,6;12;n) < \frac{p(n)}{4} \mathrm{, \quad for \,all}~n \in {\mathbb {N}}_0, \end{aligned}$$
(2.33)
$$\begin{aligned}&M(0,1,2;6;2n) > \frac{p(2n)}{2} \mathrm{, \quad for \,all}~n \in {\mathbb {N}}_0-\{ 1\}, \end{aligned}$$
(2.34)
$$\begin{aligned}&M(0,1,2;6;2n+1) < \frac{p(2n+1)}{2} \mathrm{, \quad for \,all}~n \in {\mathbb {N}}_0, \end{aligned}$$
(2.35)
$$\begin{aligned}&M(3,4,5;6;2n) < \frac{p(2n)}{2} \mathrm{, \quad for \,all}~n \in {\mathbb {N}}_0-\{ 1\}, \end{aligned}$$
(2.36)
$$\begin{aligned}&M(3,4,5;6;2n+1) > \frac{p(2n+1)}{2} \mathrm{, \quad for \,all}~n \in {\mathbb {N}}_0. \end{aligned}$$
(2.37)

3 Further Notation and Preliminary Results

Let us define the Jacobi theta function by
$$\begin{aligned} j(w;q)=\sum _{n = - \infty }^{\infty } (-w)^n q^{\frac{n(n-1)}{2}}=\prod _{i = 0} ^{\infty } (1-wq^i) (1-q^{i+1}/w)(1-q^{i+1}), \end{aligned}$$
where w is a nonzero complex number. We have
$$\begin{aligned} j(q^k;q^{3k})=(q^k;q^k)_{\infty }. \end{aligned}$$
(3.1)
Let \(\zeta _N\) be an Nth root of unity, then from the definition we observe the following equalities which will come in handy in the remainder of the paper:
$$\begin{aligned}&\displaystyle j(w;q)j(\zeta _{3}w;q)j(\zeta _{3}^2w;q) =\frac{(q;q)_{\infty }^3}{(q^3;q^3)_{\infty }} j(w^3;q^3), \end{aligned}$$
(3.2)
$$\begin{aligned}&\displaystyle j(iq^9;q^9)=(1+i) \frac{(q^9;q^9)_{\infty } (q^{36};q^{36})_{\infty }}{(q^{18};q^{18})_{\infty }}, \end{aligned}$$
(3.3)
$$\begin{aligned}&\displaystyle j(i;q^3) =(1-i) \frac{(q^3;q^3)_{\infty }(q^{12};q^{12})_{\infty }}{(q^6;q^6)_{\infty }}, \end{aligned}$$
(3.4)
$$\begin{aligned}&\displaystyle j(iq^3;q^9)j(iq^6;q^9)= \frac{ (q^9;q^9)_{\infty }^2 (q^{12};q^{12})_{\infty }(q^{18};q^{18})_{\infty }}{(q^{6};q^{6})_{\infty }(q^{36};q^{36})_{\infty }}, \end{aligned}$$
(3.5)
$$\begin{aligned}&\displaystyle j(-1;q^3) = \frac{2 (q^6;q^6)_{\infty }^2}{(q^3;q^3)_{\infty }}, \end{aligned}$$
(3.6)
$$\begin{aligned}&\displaystyle j(\zeta _{3};q^3) = (1-\zeta _{3})(q^9;q^9)_{\infty }, \end{aligned}$$
(3.7)
$$\begin{aligned}&\displaystyle j(\zeta _{3} q^3; q^9)j(\zeta _{3} q^6; q^9) = \frac{(q^9;q^9)_{\infty }^4}{(q^3;q^3)_{\infty }(q^{27};q^{27})_{\infty }}, \end{aligned}$$
(3.8)
$$\begin{aligned}&\displaystyle j(\zeta _{3} q^9; q^9) = (1-\zeta _{3}^2)(q^{27};q^{27})_\infty , \end{aligned}$$
(3.9)
$$\begin{aligned}&\displaystyle j(- q^3; q^9) =j(- q^6; q^9) =\frac{(q^6;q^6)_{\infty }(q^9;q^9)_{\infty }^2}{(q^3;q^3)_{\infty } (q^{18};q^{18})_{\infty }}, \end{aligned}$$
(3.10)
$$\begin{aligned}&\displaystyle j(- q^9; q^9) = \frac{2(q^{18};q^{18})_{\infty }^2}{(q^{9};q^{9})_{\infty }},\end{aligned}$$
(3.11)
$$\begin{aligned}&\displaystyle j(q^{15};q^{36}) j(q^{33};q^{36})=\frac{ (q^3;q^3)_{\infty } (q^{18};q^{18})_{\infty }(q^{36};q^{36})_{\infty }^2}{(q^6;q^6)_{\infty }(q^9;q^9)_{\infty }} . \end{aligned}$$
(3.12)
Next, we state and prove Lemmas 3.1 and 3.2 which will be used in the next section to prove Theorems 4.1 and 4.2, respectively.

Lemma 3.1

We have
$$\begin{aligned}&w^2 j(\zeta _{3}w;q)j(\zeta _{3}^2w;q)+ \zeta _{3}^2 w^2 j(w;q)j(\zeta _{3}^2w;q)+ \zeta _{3}w^2 j(w;q)j(\zeta _{3}w;q)\\&\quad = 3 w^3 j(w^3q^3; q^9) j(w^3 q^6; q^9) + 3w^6 q^2 j(w^3q^9; q^9) j(w^3q^9;q^9). \end{aligned}$$

Proof

Noting that
$$\begin{aligned} \zeta _{3}^{m+1} +\zeta _{3}^{n+2m} + \zeta _{3}^{2n+2}= {\left\{ \begin{array}{ll} 3\zeta _{3}, &{}\mathrm{if \,\,}m \equiv 0 \pmod {3}~\mathrm{and}~n \equiv 1 \pmod {3},\\ 3\zeta _{3}^2, &{}\mathrm{if \,\,}m \equiv 1 \pmod {3}~\mathrm{and}~n \equiv 0 \pmod {3},\\ 3, &{}\mathrm{if \,\,}m \equiv 2 \pmod {3}~\mathrm{and}~n \equiv 2 \pmod {3},\\ 0, &{}\mathrm{otherwise,} \end{array}\right. } \end{aligned}$$
we have
$$\begin{aligned}&w^2 j(\zeta _{3}w;q)j(\zeta _{3}^2w;q)+ \zeta _{3}^2 w^2 j(w;q)j(\zeta _{3}^2w;q)+ \zeta _{3}w^2 j(w;q)j(\zeta _{3}w;q)\\&\quad = w^2 \sum _{n=-\infty }^{\infty } \sum _{m=-\infty }^{\infty } ( \zeta _{3}^{m+1} +\zeta _{3}^{n+2m} + \zeta _{3}^{2n+2} ) \\&\qquad \times (-1)^{n+m} w^{n+m} q^{n(n-1)/2+m(m-1)/2}\\&\quad = 3 \zeta _{3}^2 w^2 \sum _{n=-\infty }^{\infty } \sum _{m=-\infty }^{\infty } (-1)^{3n+3m+1} w^{3n+3m+1} q^{3n(3n-1)/2+(3m+1)(3m)/2} \\&\qquad + 3 \zeta _{3} w^2 \sum _{n=-\infty }^{\infty } \sum _{m=-\infty }^{\infty } (-1)^{3n+3m+1} w^{3n+3m+1} q^{(3m+1)3m/2+3n(3n-1)/2} \\&\qquad + 3 w^2 \sum _{n=-\infty }^{\infty } \sum _{m=-\infty }^{\infty } (-1)^{3n+3m+4} w^{3n+3m+4} \\&\qquad \times q^{(3n+2)(3n+1)/2+(3m+2)(3m+1)/2} \\&\quad = 3w^3 \sum _{n=-\infty }^{\infty } (-1)^{n} w^{3n} q^{3n} (q^9)^{(n^2 -n)/2} \sum _{m=-\infty }^{\infty } (-1)^{3m} w^{3m} q^{6m} (q^9)^{ (m^2-m)/2} \\&\qquad + 3w^6 q^2 \sum _{n=-\infty }^{\infty } (-1)^{n} w^{3n} q^{9n} (q^9)^{(n^2-n)/2} \sum _{m=-\infty }^{\infty } (-1)^{3m} w^{3m} q^{9m} (q^9)^{(m^2-m)/2} \\&\quad = 3w^3 j(w^3q^3; q^9) j(w^3 q^6; q^9) + 3w^6 q^2 j(w^3q^9; q^9) j(w^3q^9;q^9). \end{aligned}$$
\(\square \)

Lemma 3.2

We have
$$\begin{aligned}&j(\zeta _{3}w;q)j(\zeta _{3}^2w;q)+j(w;q)j(\zeta _{3}^2w;q)+j(w;q)j(\zeta _{3}w;q)\\&\quad =3 j(w^3 q^3; q^9)^2 +3w^3q j(w^3 q^9;q^9) j(w^3 q^6;q^9). \end{aligned}$$

Proof

Proceeding as in the proof of Lemma 3.1 with
$$\begin{aligned} \zeta _{3}^{m} +\zeta _{3}^{n+2m} + \zeta _{3}^{2n}= {\left\{ \begin{array}{ll} 3\zeta _{3}, &{}\mathrm{if}~m \equiv 1 \pmod {3}~\mathrm{and }~n \equiv 2 \pmod {3},\\ 3\zeta _{3}^2, &{}\mathrm{if}~m \equiv 2 \pmod {3}~\mathrm{and}~n \equiv 1 \pmod {3},\\ 3, &{} \mathrm{if}~m \equiv 0 \pmod {3}~\mathrm{and }~n \equiv 0 \pmod {3},\\ 0, &{}\mathrm{otherwise,} \end{array}\right. } \end{aligned}$$
we obtain
$$\begin{aligned}&j(\zeta _{3}w;q)j(\zeta _{3}^2w;q)+j(w;q)j(\zeta _{3}^2w;q)+j(w;q)j(\zeta _{3}w;q)\\&\quad =\sum _{n=-\infty }^{\infty } \sum _{m=-\infty }^{\infty } (\zeta _{3}^m + \zeta _{3}^{n+2m} + \zeta _{3}^{2n}) (-1)^{n+m} w^{n+m} q^{n(n-1)/2+m(m-1)/2} \\&\quad =3\zeta _{3} \sum _{n=-\infty }^{\infty } \sum _{m=-\infty }^{\infty } (-1)^{3n+3m+3} w^{3n+3m+3} q^{(3n+2)(3n+1)/2+(3m+1)3m/2} \\&\qquad +3\zeta _{3}^2 \sum _{n=-\infty }^{\infty } \sum _{m=-\infty }^{\infty } (-1)^{3n+3m+3} w^{3n+3m+3} q^{(3n+1)3n/2+(3m+2)(3m+1)/2} \\&\qquad +3 \sum _{n=-\infty }^{\infty } \sum _{m=-\infty }^{\infty } (-1)^{3n+3m} w^{3n+3m} q^{3n(3n-1)/2+3m(3m-1)/2} \\&\quad =3 w^3 q \sum _{n=-\infty }^{\infty } (-1)^{n} w^{3n} q^{9n} q^{(9n^2-9n)/2} \sum _{m=-\infty }^{\infty } (-1)^{m} w^{3m} q^{6m} q^{(9m^2-9m)/2} \\&\qquad +3 \sum _{n=-\infty }^{\infty } (-1)^{n} w^{3n} q^{3n} q^{(9n^2-9n)/2} \sum _{m=-\infty }^{\infty } (-1)^{m} w^{3m} q^{3m} q^{(9m^2-9m)/2} \\&\quad =3w^3 q j(w^3 q^9;q^9) j(w^3 q^6;q^9) + 3j(w^3q^3;q^9)^2. \end{aligned}$$
\(\square \)

The following two dissections of the Jacobi theta function will come in handy in the forthcoming arguments.

Lemma 3.3

For \(\vert A \vert , \vert B \vert < 1\), we have
$$\begin{aligned} j(-A^2 B;B^4) -A j(-A^2 B^3 ;B^4) =j(A;B). \end{aligned}$$

Proof

$$\begin{aligned}&j(-A^2 B;B^4) -A j(-A^2 B^3 ;B^4) \\&\quad = \sum _{n=-\infty }^{\infty } (A^2 B)^n (B^4)^{n(n-1)/2} -A \sum _{n=-\infty }^{\infty } (A^2 B^3)^n (B^4)^{n(n-1)/2}\\&\quad = \sum _{n=-\infty }^{\infty } A^{2n} B^{((2n)^2-(2n))/2} - \sum _{n=-\infty }^{\infty } A^{2n+1} B^{((2n+1)^2-(2n+1))/2}\\&\quad =j(A;B). \end{aligned}$$
\(\square \)

4 Generating Functions

In this section, we give the generating functions for two variations of combinations of three cranks, motivated by Kang’s results [5] for ranks. We use elementary manipulations to obtain our results. Noting that the generating function of cranks is given by
$$\begin{aligned} C(w;q) = \sum _{m=-\infty }^{\infty } \sum _{n = 0} ^{\infty } M(m;n) w^m q^n =\frac{(1-w)(q;q)_\infty ^2}{j(w;q)}, \end{aligned}$$
we state and prove Theorems 4.1 and 4.2.

Theorem 4.1

We have
$$\begin{aligned}&\sum _{m=-\infty }^{\infty } \sum _{n = 0} ^{\infty } \big (M(3m-1;n)+ M(3m;n)+M(3m+1;n)\big ) w^{m} q^n \\&\quad = \frac{(1-w) (q^3;q^3)_\infty }{(q;q)_\infty } \left( \frac{ j(w q^3; q^9) j(w q^6; q^9) + w q^2 j(w q^9; q^9)^2}{j(w;q^3)} \right) . \end{aligned}$$

Proof

Recall that \(\zeta _{3}\) is the third root of unity. Then, we have
$$\begin{aligned}&\frac{C(w;q)}{w(1-w)}+ \frac{C(\zeta _{3}w;q)}{\zeta _{3}w(1-\zeta _{3}w)} + \frac{C(\zeta _{3}^2w;q)}{\zeta _{3}^2w(1-\zeta _{3}^2w)} \nonumber \\&\quad =\frac{1}{w(1-w)} \left( \sum _{m=-\infty }^{\infty } \sum _{n = 0} ^{\infty } M(m;n) w^m q^n \right) \nonumber \\&\qquad + \frac{1}{\zeta _{3}w(1-\zeta _{3}w)} \left( \sum _{m=-\infty }^{\infty } \sum _{n = 0} ^{\infty } M(m;n) (\zeta _{3}w)^m q^n \right) \nonumber \\&\qquad + \frac{1}{\zeta _{3}^2w(1-\zeta _{3}^2w)} \left( \sum _{m=-\infty }^{\infty } \sum _{n = 0} ^{\infty } M(m;n) (\zeta _{3}^2w)^m q^n \right) \nonumber \\&\quad = \frac{3}{1-w^3} \left( \sum _{m=-\infty }^{\infty } \sum _{n = 0} ^{\infty } \big (M(3m-1;n) \nonumber \right. \\&\qquad \left. +\,\, M(3m;n)+M(3m+1;n)\big ) w^{3m} q^n \right) . \end{aligned}$$
(4.1)
On the other hand, using Lemma 3.1, we compute
$$\begin{aligned}&\frac{C(w;q)}{w(1-w)}+ \frac{C(\zeta _{3}w;q)}{\zeta _{3}w(1-\zeta _{3}w)} + \frac{C(\zeta _{3}^2w;q)}{\zeta _{3}^2w(1-\zeta _{3}^2w)} \nonumber \\&\quad =\frac{w^2 (q^3;q^3)_\infty \left( j(\zeta _{3}w;q)j(\zeta _{3}^2w;q)+ \zeta _{3}^2 j(w;q)j(\zeta _{3}^2w;q)+ \zeta _{3} j(w;q)j(\zeta _{3}w;q) \right) }{w^3 (q;q)_\infty j(w^3;q^3) } \nonumber \\&\quad = \frac{(q^3;q^3)_\infty }{ (q;q)_\infty j(w^3;q^3) } \left( 3 j(w^3q^3; q^9) j(w^3 q^6; q^9) + 3w^3 q^2 j(w^3q^9; q^9) j(w^3q^9;q^9) \right) . \end{aligned}$$
(4.2)
Combining (4.1) with (4.2) and replacing \(w^3\) by w, we obtain the desired result.

\(\square \)

Theorem 4.2

We have
$$\begin{aligned}&\sum _{m=-\infty }^{\infty }\sum _{n = 0} ^{\infty } (M(3m-2;n)+M(3m-1;n)+M(3m;n) ) w^{m} q^n \\&\quad =\frac{(1-w)(q^3;q^3)_\infty }{(q;q)_\infty } \left( \frac{ j(w q^3; q^9)^2 +w q j(w q^9;q^9) j(w q^6;q^9) }{ j(w;q^3)} \right) . \end{aligned}$$

Proof

Recalling \(\zeta _{3}\) is the third root of unity, then we have
$$\begin{aligned}&\frac{C(w;q)}{(1-w)}+ \frac{C(\zeta _{3}w;q)}{(1-\zeta _{3}w)} + \frac{C(\zeta _{3}^2w;q)}{(1-\zeta _{3}^2w)} \nonumber \\&\quad =\frac{1}{1-w} \left( \sum _{m=-\infty }^{\infty } \sum _{n = 0} ^{\infty } M(m;n) w^m q^n \right) \nonumber \\&\qquad + \frac{1}{1-\zeta _{3}w} \left( \sum _{m=-\infty }^{\infty } \sum _{n = 0} ^{\infty } M(m;n) (\zeta _{3}w)^m q^n \right) \nonumber \\&\qquad + \frac{1}{1-\zeta _{3}^2w} \left( \sum _{m=-\infty }^{\infty } \sum _{n = 0} ^{\infty } M(m;n) (\zeta _{3}^2w)^m q^n \right) \nonumber \\&\quad =\frac{3}{1-w^3} \left( \sum _{m=-\infty }^{\infty } \sum _{n = 0} ^{\infty } \big (M(3m-2;n)\right. \nonumber \\&\qquad \left. +\,M(3m-1;n)+M(3m;n) \big ) w^{3m} q^n \right) . \end{aligned}$$
(4.3)
On the other hand, we use Lemma 3.2 and obtain
$$\begin{aligned}&\frac{C(w;q)}{(1-w)}+ \frac{C(\zeta _{3}w;q)}{(1-\zeta _{3}w)} + \frac{C(\zeta _{3}^2w;q)}{(1-\zeta _{3}^2w)}=\frac{(q;q)^2}{j(w;q)}+\frac{(q;q)^2}{j(\zeta _{3}w;q)} + \frac{(q;q)^2}{j(\zeta _{3}^2w;q)} \nonumber \\&\quad =\frac{ (q^3;q^3)_\infty \left( j(\zeta _{3}w;q)j(\zeta _{3}^2w;q)+j(w;q)j(\zeta _{3}^2w;q)+j(w;q)j(\zeta _{3}w;q) \right) }{ (q;q)_\infty j(w^3;q^3) } \nonumber \\&\quad =\frac{(q^3;q^3)_\infty }{(q;q)_\infty j(w^3;q^3) } \left( 3 j(w^3 q^3; q^9)^2 +3 w^3 q j(w^3 q^9;q^9) j(w^3 q^6;q^9) \right) . \end{aligned}$$
(4.4)
Then, the theorem follows from combining (4.3) with (4.4) and replacing \(w^3\) by w. \(\square \)

5 Proof of Theorem 2.1

The proofs of (2.1) and (2.2) follow from (1.4) and (1.6). Next, we replace w by \(\zeta _3\) and by i in (1.5) to obtain
$$\begin{aligned}&\sum _{n=0}^{\infty } \big ( N(0,1,1;9;n) - N(2,3,4;9;n) \big ) q^n\nonumber \\&\quad = \frac{(q^3;q^3)_{\infty }}{(q;q)_{\infty }} \sum _{n=0}^{\infty } \big ( M(0,3;n)- M(1,3;n) \big ) q^{3n} \end{aligned}$$
and
$$\begin{aligned}&\sum _{n=0}^{\infty } \big ( N(0,1,1;12;n) - N(5,5,6;12;n) \big ) q^n \\&\quad =\frac{(q^3;q^3)_{\infty }}{(q;q)_{\infty }} \sum _{n=0}^{\infty } \big ( M(0,4;n) - M(2,4;n) \big ) q^{3n}, \end{aligned}$$
respectively. In [1], Andrews and Lewis proved that
$$\begin{aligned}&\sum _{n=0}^{\infty } \big ( M(0,3;n)- M(1,3;n) \big ) q^{n}= \frac{(q;q)^2_{\infty }}{(q^3;q^3)_{\infty }},\\&\sum _{n=0}^{\infty } \big ( M(0,4;n)- M(2,4;n) \big ) q^{n}= \frac{(q;q)_{\infty }(q^2;q^2)_{\infty }}{(q^4;q^4)_{\infty }}. \end{aligned}$$
Thus we have
$$\begin{aligned}&\sum _{n=0}^{\infty } \big ( N(0,1,1;9;n) - N(2,3,4;9;n) \big ) q^n = \frac{(q^3;q^3)^3_{\infty }}{(q;q)_{\infty }(q^9;q^9)_{\infty }}, \end{aligned}$$
(5.1)
$$\begin{aligned}&\sum _{n=0}^{\infty } \big ( N(0,1,1;12;n) - N(5,5,6;12;n) \big ) q^n = \frac{(q^3;q^3)^2_{\infty }(q^6;q^6)_{\infty }}{(q;q)_{\infty }(q^{12};q^{12})_{\infty }}. \end{aligned}$$
(5.2)
Hence (2.3) and (2.4) follow from (5.1) and (1.4). The relation (2.5) follows from the following observation and (2.2):
$$\begin{aligned}&\sum _{n=0}^{\infty } 2N(2,3,4;12;n) q^n = \sum _{n=0}^{\infty } N(2,2,3;6;n) q^n. \end{aligned}$$
We obtain (2.6) and (2.7) from (2.1), (5.2) and the following observation:
$$\begin{aligned}&\sum _{n=0}^{\infty } \big ( N(0,1,1;12;n) + N(5,5,6;12;n) \big ) q^n = \sum _{n=0}^{\infty } N(0,1,1;6;n) q^n. \end{aligned}$$
Below in Lemmas 5.15.3, we compute the expressions in Theorems 4.1 and 4.2 when w is replaced by second, third and fourth roots of unity. We use the following three lemmas to prove the rest of the equations similarly.

Lemma 5.1

We have
$$\begin{aligned}&\sum _{n=0}^{\infty } \big ( M(-1,0,1;6;n) - M(2,3,4;6;n) \big ) q^n \\&\quad =\frac{(q^3;q^3)^4_{\infty }}{(q;q)_{\infty }(q^6;q^6)^2_{\infty }} -4q^2 \frac{(q^3;q^3)_\infty (q^4;q^4)_\infty (q^{36};q^{36})_\infty ^2}{(q^2;q^2)_\infty (q^{12};q^{12})_\infty (q^{18};q^{18})_\infty },\\&\sum _{n=0}^{\infty } \big ( M(-2,-1,0;6;n) - M(1,2,3;6;n) \big ) q^n =\frac{(q;q)_\infty (q^{9};q^{9})^2_\infty }{(q^{2};q^{2})_\infty (q^{18};q^{18})_\infty }. \end{aligned}$$

Proof

We replace \(w=-1\) in Theorems 4.1 and 4.2. Then, we employ (3.6), (3.10) and (3.11) to get
$$\begin{aligned}&\sum _{n=0}^{\infty } \big ( M(-1,0,1;6;n) - M(2,2,3;6;n) \big ) q^n \\&\qquad = \frac{2 (q^3;q^3)_\infty }{(q;q)_\infty } \left( \frac{ j(- q^3; q^9) j(- q^6; q^9) - q^2 j(-q^9; q^9)^2}{j(-1;q^3)} \right) \\&\qquad =\frac{1}{(q;q)_\infty } \left( \frac{(q^9;q^9)_\infty ^4}{ (q^{18};q^{18})_{\infty }^2} - 4 q^2 \frac{(q^3;q^3)_{\infty }^2(q^{18};q^{18})_{\infty }^4}{(q^6;q^6)_{\infty }^2(q^{9};q^{9})_{\infty }^2} \right) ,\\&\sum _{n=0}^{\infty } \big ( M(-2,-1,0;6;n) - M(1,2,3;6;n) \big ) q^n\\&\qquad = \frac{2(q^3;q^3)_{\infty }}{(q;q)_{\infty } } \left( \frac{ j(- q^3; q^9)^2 - q j(- q^9;q^9) j(-q^6;q^9) }{ j(-1;q^3)} \right) \\&\qquad =\frac{1}{(q;q)_{\infty } } \left( \frac{ (q^9;q^9)_{\infty }^4}{ (q^{18};q^{18})_{\infty }^2} - 2 q \frac{(q^3;q^3)_{\infty }(q^9;q^9)_{\infty }(q^{18};q^{18})_{\infty }}{(q^6;q^6)_{\infty }} \right) . \end{aligned}$$
The eta quotients in brackets are holomorphic. Thus, the lemma follows from the Sturm Theorem ([6, Theorem 3.13]). \(\square \)

Lemma 5.2

We have
$$\begin{aligned}&\sum _{n=0}^{\infty } \big ( M(-1,0,1;9;n) - M(2,3,4;9;n) \big ) q^n\\&\quad = \frac{1}{(q;q)_{\infty }} \left( \frac{(q^9;q^9)_{\infty }^3}{(q^{27};q^{27})_{\infty }} \right. \left. -3 q^2 \frac{(q^3;q^3)_{\infty }(q^{27};q^{27})_{\infty }^2}{ (q^9;q^9)_{\infty }} \right) ,\\&\sum _{n=0}^{\infty } \big ( M(-2,-1,0;9;n) + \zeta _3 M(1,2,3;9;n) + \zeta _3^2 M(4,5,6;9;n) \big ) q^n\\&\quad =\frac{(q^3;q^3)_{\infty }}{(q;q)_{\infty } (q^9;q^9)_{\infty }} \left( (\zeta _3 q^3; \zeta _3q^3)_{\infty }^2 + (\zeta _3-1) q (q^{27};q^{27})_{\infty } (\zeta _3^2 q^3; \zeta _3^2 q^3)_{\infty } \right) . \end{aligned}$$

Proof

These equations follow from employing (3.1), (3.7), (3.8) and (3.9) after replacing \(w=\zeta _3\) in Theorems 4.1 and 4.2, respectively. \(\square \)

Lemma 5.3

We have
$$\begin{aligned}&\sum _{n = 0} ^{\infty }\big (M(-1,0,1;12;n) -M(5,6,7;12;n) \big ) q^n =\frac{(q^{2};q^{2})_\infty ^2 (q^{9};q^{9})_\infty ^2}{(q;q)_\infty (q^{4};q^{4})_\infty (q^{18};q^{18})_\infty },\\&\sum _{n = 0} ^{\infty }\Big ( \big (M(-2,-1,0;12;n) -M(4,5,6;12;n) \big ) \\&\qquad + i \big (M(1,2,3;12;n)- M(7,8,9;12;n)\big ) \Big ) q^n \\&\quad = \frac{(q^{4};q^{4})_\infty (q^{6};q^{6})_\infty ^4 }{ (q^{2};q^{2})_\infty (q^{3};q^{3})_\infty (q^{12};q^{12})_\infty ^2} + i q \frac{(q^{2};q^{2})^2_\infty (q^{3};q^{3})_\infty (q^{18};q^{18})^2_\infty }{ (q;q)_\infty (q^{4};q^{4})_\infty (q^{6};q^{6})_\infty (q^{9};q^{9})_\infty }. \end{aligned}$$

Proof

We first substitute w by i in Theorem 4.1. Then by employing (3.3), (3.4) and (3.5) we have
$$\begin{aligned}&\frac{(1-i) (q^3;q^3)_{\infty }}{(q;q)_{\infty } } \left( \frac{ j(i q^3; q^9) j(i q^6; q^9) + i q^2 j(i q^9; q^9)^2}{j(i;q^3)} \right) \\&\quad = \frac{1}{(q;q)_{\infty }} \left( \frac{ (q^9;q^9)_{\infty }^2 (q^{18};q^{18})_{\infty }}{(q^{36};q^{36})_{\infty }} -2 q^2 \frac{ (q^6;q^6)_{\infty }(q^9;q^9)_{\infty }^2 (q^{36};q^{36})_{\infty }^2}{(q^{12};q^{12})_{\infty } (q^{18};q^{18})_{\infty }^2} \right) \\&\quad =\frac{(q^{2};q^{2})_\infty ^2 (q^{9};q^{9})_\infty ^2}{(q;q)_\infty (q^{4};q^{4})_\infty (q^{18};q^{18})_\infty }. \end{aligned}$$
The last equality follows from the Sturm Theorem ([6, Theorem 3.13]).
By Lemma 3.3, we have
$$\begin{aligned} j(iq^3;q^9)&=j(q^{15};q^{36}) -i q^3 j(q^{33};q^{36}),\\ j(iq^6;q^9)&=j(q^{21};q^{36}) -i q^6 j(q^{39};q^{36}), \end{aligned}$$
together with (3.3), (3.4) and w replaced by i in Theorem 4.2 we have
$$\begin{aligned}&\frac{(1-i)(q^3;q^3)_{\infty }}{(q;q)_{\infty } } \left( \frac{ j(i q^3; q^9)^2 +i q j(i q^9;q^9) j(i q^6;q^9) }{ j(i;q^3)} \right) \\&\quad = \frac{(q^6;q^6)_{\infty }}{(q;q)_{\infty } (q^{12};q^{12})_{\infty } } \left( j(q^{15};q^{36}) - q^3 j(q^{3};q^{36}) \right) \left( j(q^{15};q^{36}) + q^3 j(q^{3};q^{36}) \right) \\&\qquad + q \frac{(q^6;q^6)_{\infty }(q^9;q^9)_{\infty } (q^{36};q^{36})_{\infty }}{(q;q)_{\infty } (q^{12};q^{12})_{\infty } (q^{18};q^{18})_{\infty } } \left( q^6 j(q^{39};q^{36}) -j(q^{21};q^{36}) \right) \\&\qquad + i q \frac{(q^6;q^6)_{\infty }(q^9;q^9)_{\infty } (q^{36};q^{36})_{\infty } }{(q;q)_{\infty } (q^{12};q^{12})_{\infty } (q^{18};q^{18})_{\infty } } \left( q^6 j(q^{39};q^{36}) + j(q^{21};q^{36}) \right) \\&\qquad -2i q^3 \frac{(q^6;q^6)_{\infty }}{(q;q)_{\infty } (q^{12};q^{12})_{\infty } } \left( j(q^{15};q^{36}) j(q^{33};q^{36}) \right) \\&\quad = \frac{(q^6;q^6)_{\infty }}{(q;q)_{\infty } (q^{12};q^{12})_{\infty } } j(q^{3};-q^{9}) j(-q^3;-q^{9}) \\&\qquad + q \frac{(q^6;q^6)_{\infty }(q^9;q^9)_{\infty } (q^{36};q^{36})_{\infty }}{(q;q)_{\infty } (q^{12};q^{12})_{\infty } (q^{18};q^{18})_{\infty } } \left( -j(-q^{6};-q^{9}) \right) \\&\qquad + i q \frac{(q^6;q^6)_{\infty }(q^9;q^9)_{\infty } (q^{36};q^{36})_{\infty } }{(q;q)_{\infty } (q^{12};q^{12})_{\infty } (q^{18};q^{18})_{\infty } } \left( j(-q^{6};-q^{9}) \right) \\&\qquad -2i q^3 \frac{(q^6;q^6)_{\infty }}{(q;q)_{\infty } (q^{12};q^{12})_{\infty } } \left( j(q^{15};q^{36}) j(q^{33};q^{36}) \right) \\&\quad = \frac{(q^6;q^6)_{\infty }^2 (q^{18};q^{18})_{\infty }^5}{(q;q)_{\infty }(q^9;q^9)_{\infty }^2(q^{12};q^{12})_{\infty }(q^{36};q^{36})_{\infty }^2} \\&\qquad - q \frac{(q^6;q^6)_{\infty }^4(q^9;q^9)_{\infty } (q^{36};q^{36})_{\infty } }{(q;q)_{\infty }(q^3;q^3)_{\infty }(q^{12};q^{12})_{\infty }^2 (q^{18};q^{18})_{\infty }} \\&\qquad + i q \left( \frac{(q^3;q^3)_{\infty }(q^{18};q^{18})_{\infty }^4}{(q;q)_{\infty }(q^6;q^6)_{\infty }(q^9;q^9)_{\infty }(q^{36};q^{36})_{\infty }} \right. \\&\qquad \left. -2 q^2 \frac{ (q^3;q^3)_{\infty } (q^{18};q^{18})_{\infty } (q^{36};q^{36})_{\infty }^2}{(q;q)_{\infty }(q^9;q^9)_{\infty }(q^{12};q^{12})_{\infty }} \right) . \end{aligned}$$
In the second and third steps, we use Lemma 3.3 and (3.10)–(3.12), respectively. Then, the lemma follows from the Sturm Theorem ([6, Theorem 3.13]).

\(\square \)

6 Proof of Theorem 2.3

Let \(f(q)=\sum _{n} a_n q^n\) be the Fourier series expansion of f. We use [n]f(q) to denote \(a_n\).

6.1 Proofs of (2.24)–(2.27)

The following lemma from [7, Theorem 1.1] gives the desired result.

Lemma 6.1

[7, Theorem 1.1]. For all \(n \ge 1 \), we have
$$\begin{aligned}&[2n] \frac{(q^3;q^3)^4_{\infty }}{2(q;q)_{\infty }(q^6;q^6)^2_{\infty }} > 0,\\&[2n+1] \frac{(q^3;q^3)^4_{\infty }}{2(q;q)_{\infty }(q^6;q^6)^2_{\infty }} < 0. \end{aligned}$$

6.2 Proofs of (2.28)–(2.29)

Lemma 6.2

For all \(n \ge 0\), except \(n=3\), or 7, we have
$$\begin{aligned}{}[n]\frac{(q^3;q^3)^3_{\infty }}{(q;q)_{\infty }(q^9;q^9)_{\infty }} > 0. \end{aligned}$$

Proof

We have
$$\begin{aligned} \displaystyle \frac{q(q^9;q^9)^3_{\infty }}{(q^3;q^3)_{\infty }}=\frac{\eta ^3(q^9)}{\eta (q^3)} \in M_{1}(\Gamma _0(27),\chi _{-3}), \end{aligned}$$
from which we obtain
$$\begin{aligned} \frac{q(q^9;q^9)^3_{\infty }}{(q^3;q^3)_{\infty }}=\sum _{n = 0} ^{\infty } \sum _{d \mid {3n+1}} \displaystyle {\Bigl ({\frac{-3}{d}}\Bigr ) } q^{3n+1}. \end{aligned}$$
On the other hand, we have \(\sum _{d \mid {3n+1}} \displaystyle {\Bigl ({\frac{-3}{d}}\Bigr ) } \ge 0\), since
$$\begin{aligned} \sum _{\begin{array}{c} d \mid {3n+1},\\ d \equiv 1 \pmod {3} \end{array}} 1 \ge \sum _{\begin{array}{c} d \mid {3n+1},\\ d \equiv 2 \pmod {3} \end{array}} 1. \end{aligned}$$
Thus we have
$$\begin{aligned}{}[n]\frac{(q^3;q^3)^3_{\infty }}{(q;q)_{\infty }} \ge 0, \end{aligned}$$
(6.1)
with
$$\begin{aligned} \frac{(q^3;q^3)^3_{\infty }}{(q;q)_{\infty }}&=1+q+2q^2+2q^4+q^5+2q^6+q^8+2q^9+2q^{10}+2q^{12}\\&\quad +\,2q^{14}+3q^{16}+O(q^{17}). \end{aligned}$$
The lemma follows from observing that
$$\begin{aligned}{}[9n]\frac{1}{(q^9;q^9)_{\infty }} > 0. \end{aligned}$$
\(\square \)

Observe that (2.28) and (2.29) is a direct consequence of Lemma 6.2.

6.3 Proofs of (2.30)–(2.33)

We obtain (2.30) and (2.31) from (2.26) and (2.27), respectively; and (2.32) and (2.33) from the following lemma combined with (6.1).

Lemma 6.3

For all \(n \ge 0\) we have
$$\begin{aligned}&2[3n]\frac{(q^{6};q^{6})_{\infty }^{}}{(q^{3};q^{3})_{\infty }^{}(q^{12};q^{12})_{\infty }^{}}+[3n]\frac{(q^{3};q^{3})_{\infty }^{}}{(q^{6};q^{6})_{\infty }^{2}}> 0,\\&2[3n]\frac{(q^{6};q^{6})_{\infty }^{}}{(q^{3};q^{3})_{\infty }^{}(q^{12};q^{12})_{\infty }^{}}-[3n]\frac{(q^{3};q^{3})_{\infty }^{}}{(q^{6};q^{6})_{\infty }^{2}} > 0. \end{aligned}$$

Proof

Let
$$\begin{aligned} \displaystyle f_1(q)=\frac{(q^{6};q^{6})_{\infty }^{}}{(q^{3};q^{3})_{\infty }^{}(q^{12};q^{12})_{\infty }^{}}. \end{aligned}$$
Replacing q by \(-q\), we obtain
$$\begin{aligned} \displaystyle f_1(-q)=\frac{(q^{3};q^{3})_{\infty }^{}}{(q^{6};q^{6})_{\infty }^{2}}. \end{aligned}$$
That is, we have
$$\begin{aligned} 2f_1(q)+f_1(-q)&=2\frac{(q^{6};q^{6})_{\infty }^{}}{(q^{3};q^{3})_{\infty }^{}(q^{12};q^{12})_{\infty }^{}}+\frac{(q^{3};q^{3})_{\infty }^{}}{(q^{6};q^{6})_{\infty }^{2}}\\&=\sum _{n = 0} ^{\infty } \left( [n]f_1(q)+2 [2n] f_1(q) \right) q^n,\\ 2f_1(q)-f_1(-q)&=2\frac{(q^{6};q^{6})_{\infty }^{}}{(q^{3};q^{3})_{\infty }^{}(q^{12};q^{12})_{\infty }^{}}-\frac{(q^{3};q^{3})_{\infty }^{}}{(q^{6};q^{6})_{\infty }^{2}}\\&=\sum _{n = 0} ^{\infty } \left( [n]f_1(q)+2 [2n+1] f_1(q) \right) q^n. \end{aligned}$$
On the other hand, we observe that
$$\begin{aligned} \displaystyle f_1(q)=\frac{1}{(q^{3};q^{6})_{\infty }^{}(q^{12};q^{12})_{\infty }^{}}, \end{aligned}$$
that is, we have \([3n]f_1(q) > 0\) for all    \(n \ge 0\), from which the lemma follows.

\(\square \)

6.4 Proofs of (2.34)–(2.37)

These follow from Lemma 6.4 below.

Lemma 6.4

For all \(n > 1\) we have
$$\begin{aligned}{}[2n] \frac{(q;q)_\infty (q^{9};q^{9})^2_\infty }{(q^{2};q^{2})_\infty (q^{18};q^{18})_\infty }&> 0,\\ [2n-1] \frac{(q;q)_\infty (q^{9};q^{9})^2_\infty }{(q^{2};q^{2})_\infty (q^{18};q^{18})_\infty }&< 0.\\ \end{aligned}$$

Proof

Let
$$\begin{aligned} \displaystyle f_2(q)=\frac{(q;q)_\infty (q^{9};q^{9})^2_\infty }{(q^{2};q^{2})_\infty (q^{18};q^{18})_\infty }, \end{aligned}$$
then we have
$$\begin{aligned} f_2(-q)&=\frac{ (q^{2};q^{2})^2_\infty (q^{18};q^{18})^5_\infty }{(q;q)_\infty (q^{4};q^{4})_\infty (q^{9};q^{9})^2_\infty (q^{36};q^{36})^2_\infty } \\&= \frac{ (q^{18};q^{18})^5_\infty }{ (q^{9};q^{9})^2_\infty (q^{36};q^{36})^2_\infty } \frac{(q^{2};q^{2})^{2}_\infty (q^{3};q^{3})_\infty (q^{12};q^{12})_\infty }{(q;q)_\infty (q^{4};q^{4})_\infty (q^{6};q^{6})_\infty } \\&\quad \times \frac{(q^{6};q^{6})_\infty }{(q^{3};q^{3})_\infty (q^{12};q^{12})_\infty }. \end{aligned}$$
Let
$$\begin{aligned}&f_3(q)=\frac{ (q^{18};q^{18})^5_\infty }{ (q^{9};q^{9})^2_\infty (q^{36};q^{36})^2_\infty }=1+4q^9+4q^{36}+O(q^{243}),\\&f_4(q)= \frac{(q^{2};q^{2})^{2}_\infty (q^{3};q^{3})_\infty (q^{12};q^{12})_\infty }{(q;q)_\infty (q^{4};q^{4})_\infty (q^{6};q^{6})_\infty }=1+q+q^5+q^8+q^{16}+O(q^{21}),\\&f_5(q)= \frac{(q^{6};q^{6})_\infty }{(q^{3};q^{3})_\infty (q^{12};q^{12})_\infty }=1+q^3+q^6+2q^9+3q^{12}+O(q^{15}). \end{aligned}$$
It is well known that \(f_3(q)=1+ 4\sum _{n=1}^\infty q^{9n^2} \), thus \([n]f_3(q) \ge 0\) for all    \(n \in {\mathbb {N}}\). We observe that \(f_4(-q)\) is one of the Kac identities, whose coefficients are alternating in sign, see [8, Theorem 8.2 (2)]. Thus, \([n]f_4(q) \ge 0\) for all    \(n \in {\mathbb {N}}\). We also have
$$\begin{aligned} f_5(q) = \frac{(-q^{3};q^{3})_\infty }{(q^{12};q^{12})_\infty }, \end{aligned}$$
from which \([3n]f_5(q) > 0\) for all \(n \in {\mathbb {N}}_0\) follows. Thus \([n] f_4(q) f_5(q) > 0\) for all \(n \ge 3\). That is, we have \([n]f_3(q)f_4(q)f_5(q)=[n]f_2(-q) > 0\) for all \(n >2\). Then we replace q by \(-q\) in \(f_2(-q)\) to complete the proof of the lemma. \(\square \)

7 Further Discussion

In this section, we discuss the connection between Theorem 2.1 and recent results by Hickerson and Mortenson from [4]. Then, we discuss further possibilities in this direction. Let \(L \in {\mathbb {N}}\) be divisible by 3. Let
$$\begin{aligned} D(3j-1,3j,3j+1;L)= \sum _{n=0}^{\infty } \left( N(3j-1,3j,3j+1;L;n) -\frac{3p(n)}{L} \right) q^n. \end{aligned}$$
From [4, Theorem 4.1], one can observe that Appell–Lerch sums in \(D(3j-1,3j,3j+1;L)\) cancel out, leaving the generating function to be a theta function. This explains the beauty of the formulae in (2.1)–(2.7) is not a mere coincidence. Moreover, one can give the generating functions of combinations of ranks in (1.7) in terms of Jacobi theta functions. Below we illustrate this when \(j=0\) and \(L=3 \cdot 2^k\), where \(k \in {\mathbb {N}}_0\). By [4, Theorem 4.1] we have
$$\begin{aligned}&\sum _{n=0}^{\infty } N(0,1,1;L;n) q^n \nonumber \\&\quad = \frac{3}{L(q;q)_{\infty }} + \frac{(q;q)_{\infty }(q^3;q^3)_{\infty }^4}{2L (q^6;q^6)_{\infty }^2}\sum _{j=1}^{L-1}\frac{(1-\zeta _{L}^j)(2+\zeta _{L}^j)\zeta _{L}^{-2j} j(-\zeta _{L}^{2j};q)}{j(\zeta _{L}^j;q)j(-\zeta _{L}^{3j}q;q^3)j(-\zeta _{L}^{3j} q^2;q^3) }. \end{aligned}$$
(7.1)
Now let us consider
$$\begin{aligned}&\sum _{n=0}^{\infty } N(0,1,1;2L;n) q^n \\&\quad = \frac{3}{2L(q;q)_{\infty }} + \frac{(q;q)_{\infty }(q^3;q^3)_{\infty }^4}{4L (q^6;q^6)_{\infty }^2}\sum _{j=1}^{2L-1}\frac{(1-\zeta _{2L}^j)(2+\zeta _{2L}^j)\zeta _{2L}^{-2j} j(-\zeta _{2L}^{2j};q)}{j(\zeta _{2L}^j;q)j(-\zeta _{2L}^{3j}q;q^3)j(-\zeta _{2L}^{3j} q^2;q^3) }\\&\quad = \frac{1}{2} \sum _{n=0}^{\infty } N(0,1,1;L;n) q^n \\&\qquad + \frac{(q;q)_{\infty }(q^3;q^3)_{\infty }^4}{4L (q^6;q^6)_{\infty }^2}\sum _{j=0}^{L-1}\frac{(1-\zeta _{2L}^{2j+1})(2+\zeta _{2L}^{2j+1})\zeta _{2L}^{-4j-2} j(-\zeta _{2L}^{4j+2};q)}{j(\zeta _{2L}^{2j+1};q)j(-\zeta _{2L}^{6j+3}q;q^3)j(-\zeta _{2L}^{6j+3} q^2;q^3) }, \end{aligned}$$
where in the last line we use (7.1). This gives a recursive relation for N(0, 1, 1; L). In general if \(L=3 \cdot 2^k\), then we use this recursion to obtain
$$\begin{aligned}&\sum _{n=0}^{\infty } N(0,1,1;L;n) q^n \\&\quad = \frac{1}{2} \sum _{n=0}^{\infty } N(0,1,1;L/2;n) q^n \\&\qquad + \frac{(q;q)_{\infty }(q^3;q^3)_{\infty }^4}{2L(q^6;q^6)_{\infty }^2}\sum _{j=0}^{L/2-1}\frac{(1-\zeta _{L}^{2j+1})(2+\zeta _{L}^{2j+1})\zeta _{L}^{-4j-2} j(-\zeta _{L}^{4j+2};q)}{j(\zeta _{L}^{2j+1};q)j(-\zeta _{L}^{6j+3}q;q^3)j(-\zeta _{L}^{6j+3} q^2;q^3) }\\&\quad = \frac{1}{2^k} \sum _{n=0}^{\infty } N(0,1,1;3;n) q^n + \frac{(q;q)_{\infty }(q^3;q^3)_{\infty }^4}{ (q^6;q^6)_{\infty }^2} \\&\qquad \times \sum _{i=0}^{k-1} \frac{2^{i-1}}{L }\sum _{j=0}^{L/2^{i+1}-1}\frac{(1-\zeta _{L/2^{i}}^{2j+1})(2+\zeta _{L/2^{i}}^{2j+1})\zeta _{L/2^{i}}^{-4j-2} j(-\zeta _{L/2^{i}}^{4j+2};q)}{j(\zeta _{L/2^{i}}^{2j+1};q)j(-\zeta _{L/2^{i}}^{6j+3}q;q^3)j(-\zeta _{L/2^{i}}^{6j+3} q^2;q^3) }. \end{aligned}$$
That is by (1.4), we obtain the generating function of N(0, 1, 1; Ln) in terms of theta functions:
$$\begin{aligned}&\sum _{n = 0} ^{\infty } N(0,1,1;L;n) q^n \nonumber \\&\quad = \sum _{n = 0} ^{\infty } \frac{p(n)}{2^k} q^n + \frac{(q;q)_{\infty }(q^3;q^3)_{\infty }^4}{(q^6;q^6)_{\infty }^2}\nonumber \\&\qquad \times \sum _{i=0}^{k-1} \frac{2^{i-1}}{L} \sum _{j=0}^{L/2^{i+1}-1}\frac{(1-\zeta _{L/2^{i}}^{2j+1})(2+\zeta _{L/2^{i}}^{2j+1})\zeta _{L/2^{i}}^{-4j-2} j(-\zeta _{L/2^{i}}^{4j+2};q)}{j(\zeta _{L/2^{i}}^{2j+1};q)j(-\zeta _{L/2^{i}}^{6j+3}q;q^3)j(-\zeta _{L/2^{i}}^{6j+3} q^2;q^3) }. \end{aligned}$$
(7.2)
For a fixed \(L \in {\mathbb {N}}\), it might be possible to determine the signs of the theta function in (7.2), which will yield the relations between N(0, 1, 1; Ln) and \( \displaystyle \frac{p(n)}{2^k} \).

Finally, similar arguments can be done for any \(3 | L\in {\mathbb {N}}\) other than \(3 \cdot 2^k\). So for any L divisible by 3, one can derive an equation similar to (7.2). Also, similar results can be derived for \(N(3j-1,3j,3j+1;L;n)\) for all \(0 \le j \le L/3-1\).

8 Remarks and Conjectures

Let
$$\begin{aligned} F(q)= \frac{(q^3;q^3)_{\infty } (\zeta _3 q^3; \zeta _3q^3)_{\infty }^2}{(q;q)_{\infty } (q^9;q^9)_{\infty }} + (\zeta _3-1) q \frac{(q^3;q^3)_{\infty } (\zeta _3^2 q^3; \zeta _3^2 q^3)_{\infty } (q^{27};q^{27})_{\infty }}{(q;q)_{\infty } (q^9;q^9)_{\infty }}. \end{aligned}$$
We tried linear combinations of all the eta quotients in \(M_1(\Gamma _1(162))\) to represent \((q,q)_{\infty } \mathfrak {Re}\big (F(q) \big )\) and \((q,q)_{\infty } \mathfrak {Im}\big ( F(q) \big )\) in terms of eta quotients to no avail. On the other hand, we have
$$\begin{aligned} j(\zeta _3 q^3; q^9)&= (\zeta _3 q^3;\zeta _3 q^3)_{\infty }= j(q^{36};q^{81}) - \zeta _3 q^3 j(q^{63};q^{81}) + \zeta _3^2 q^{15} j(q^{90};q^{81}), \\ j(\zeta _3q^6;q^9)&= (\zeta _3^2 q^3;\zeta _3^2 q^3)_{\infty } = j(q^{45};q^{81}) - \zeta _3 q^6 j(q^{72};q^{81}) + \zeta _3^2 q^{21} j(q^{99};q^{81}), \end{aligned}$$
from which the real and imaginary parts of F(q) can be worked out in terms of Jacobi theta function. This will give alternate formulae for (2.17)–(2.19). We prefer the formulae in (2.17)–(2.19).

We note that all the theta functions in (2.1)–(2.23), except (2.17)–(2.19), are weight 1 modular forms when multiplied by \((q,q)_{\infty }\).

Finally, we state the conjectures for the comparisons of crank combinations with partition function.

Conjecture 8.1

We have
$$\begin{aligned}&M(0,1,1;6;2n)< \frac{p(2n)}{2} \mathrm{,\quad for \,all \,\, }n \in {\mathbb {N}}, \\&M(0,1,1;6;2n+1)> \frac{p(2n+1)}{2} \mathrm{, \quad for\, all \,\, }n \in {\mathbb {N}}_0-\{ 1\}, \\&M(2,2,3;6;2n)> \frac{p(2n)}{2} \mathrm{, \quad for\, all \,\, }n \in {\mathbb {N}}, \\&M(2,2,3;6;2n+1)< \frac{p(2n+1)}{2} \mathrm{, \quad for \, all \,\, }n \in {\mathbb {N}}_0-\{ 1\}, \\&M(2,3,4;12;2n)> \frac{p(2n)}{4} \mathrm{, \quad for \,all \,\, }n \in {\mathbb {N}}, \\&M(2,3,4;12;2n+1)< \frac{p(2n+1)}{4} \mathrm{, \quad for \,all \,\, }n \in {\mathbb {N}}_0-\{ 1\}, \\&M(0,1,1;12;2n)< \frac{p(2n)}{4} \mathrm{, \quad for \,all \,\, }n \in {\mathbb {N}}_0-\{ 0,3,4\}, \\&M(0,1,1;12;2n+1)> \frac{p(2n+1)}{4} \mathrm{, \quad for \,all \,\, }n \in {\mathbb {N}}_0, \\&M(5,5,6;12;2n)< \frac{p(2n)}{4} \mathrm{, \quad for \,all \,\, }n \in {\mathbb {N}}_0, \\&M(5,5,6;12;2n+1)> \frac{p(2n+1)}{4} \mathrm{, \quad for \,all \,\, }n \in {\mathbb {N}}_0-\{ 0,1,5,7\}, \\&M(0,1,2;12;2n)> \frac{p(2n)}{4} \mathrm{, \quad for \,all \,\, }n \in {\mathbb {N}}_0, \\&M(0,1,2;12;2n+1)< \frac{p(2n+1)}{4} \mathrm{, \quad for \,all \,\, }n \in {\mathbb {N}}_0-\{ 1,2,3,5\}, \\&M(3,4,5;12;2n)< \frac{p(2n)}{4} \mathrm{, \quad for \,all \,\, }n \in {\mathbb {N}}_0, \\&M(3,4,5;12;2n+1)> \frac{p(2n+1)}{4} \mathrm{, \quad for \,all \,\, }n \in {\mathbb {N}}_0-\{ 0,5\}, \\&M(6,7,8;12;2n)> \frac{p(2n)}{4} \mathrm{, \quad for \,all \,\, }n \in {\mathbb {N}}_0-\{ 0,1,2\}, \\&M(6,7,8;12;2n+1)< \frac{p(2n+1)}{4} \mathrm{, \quad for \,all \,\, }n \in {\mathbb {N}}_0, \\&M(9,10,11;12;2n) < \frac{p(2n)}{4} \mathrm{, \quad for \,all \,\, }n \in {\mathbb {N}}_0-\{ 1,3\}, \\&M(9,10,11;12;2n+1) > \frac{p(2n+1)}{4} \mathrm{, \quad for \,all \,\, }n \in {\mathbb {N}}_0. \end{aligned}$$

Additionally, the referee pointed out that the signs of theta parts of M(0, 1, 1; 9; n), M(2, 3, 4; 9; n), M(0, 1, 2; 9; n), M(3, 4, 5; 9; n) and M(6, 7, 8; 9; n) are periodic modulo 9 when \(n\ge 467\). Thus, conjectures similar to Conjecture 8.1 can be stated for M(0, 1, 1; 9; n), M(2, 3, 4; 9; n), M(0, 1, 2; 9; n), M(3, 4, 5; 9; n) and M(6, 7, 8; 9; n).

Notes

Acknowledgements

We thank the referee for the editorial suggestions which improved the article and pointing out additional conjectures which appear in Sect. 8.

References

  1. 1.
    Andrews, G.E., Lewis, R.: The ranks and cranks of partitions moduli 2, 3, and 4. J. Number Theory 85(1), 74–84 (2000)MathSciNetCrossRefGoogle Scholar
  2. 2.
    Atkin, A.O.L., Swinnerton-Dyer, H.P.F.: Some properties of partitions. Proc. London Math. Soc. 4, 84–106 (1954)MathSciNetCrossRefGoogle Scholar
  3. 3.
    Dyson, F.J.: Some guesses in the theory of partitions. Eureka 8, 10–15 (1944)MathSciNetGoogle Scholar
  4. 4.
    Hickerson, D., Mortenson, E.: Dyson’s ranks and Appell-Lerch sums. Math. Ann. 367(1-2), 373–395 (2017)MathSciNetCrossRefGoogle Scholar
  5. 5.
    Kang, S.-Y: Mock Jacobi forms in basic hypergeometric series. Compos. Math. 145(3), 553-565 (2009)MathSciNetCrossRefGoogle Scholar
  6. 6.
    Kilford, L.J.P.: Modular Forms. A Classical and Computational Introduction. Imperial College Press, London (2008)CrossRefGoogle Scholar
  7. 7.
    Kim, B., Nam, H.: On a conjecture of Soon-Yi Kang on a certain partition rank difference. Ramanujan J. 35(3), 467–477 (2014)MathSciNetCrossRefGoogle Scholar
  8. 8.
    Köhler, G.: Eta Products and Theta Series Identities. Springer Monographs in Mathematics. Springer, Heidelberg (2011)CrossRefGoogle Scholar

Copyright information

© Springer Nature Switzerland AG 2019

Authors and Affiliations

  1. 1.Department of Mathematics and Computer ScienceUniversity of LethbridgeLethbridgeCanada
  2. 2.Division of Mathematical Sciences School of Physical and Mathematical SciencesNanyang Technological UniversitySingaporeSingapore

Personalised recommendations