Proofs of some conjectures of Sun on representations by linear combinations of triangular numbers

Abstract

For positive integers a, b, c and d, let N(a, b, c, d; n) be the number of representations of n as $ax^2+by^2+cz^2+dw^2$ and T(a, b, c, d; n) be the number of representations of n as $a\frac{X(X+1)}{2}+b\frac{Y(Y+1)}{2}+c\frac{Z(Z+1)}{2}+d\frac{W(W+1)}{2}$, where x, y, z, w are integers, and n, X, Y, Z, W are nonnegative integers. Recently, Sun (J. Ramanujan Math. Soc., 35 (2020), 373–389) established a number of relations between T(a, b, c, d; n) and N(a, b, c, d; n), and posed some conjectures on such relations. In this paper, we confirm several Sun’s conjectures by employing Ramanujan’s theta function identities.

1 Introduction

Let $\mathbb {N}^{+}$, $\mathbb {N}$ and $\mathbb {Z}$ denote the set of positive integers, the set of nonnegative integers and the set of integers, respectively. Let $\mathbb {Z}^4=\mathbb {Z}\times \mathbb {Z}\times \mathbb {Z}\times \mathbb {Z}$ and $\mathbb {N}^4=\mathbb {N}\times \mathbb {N}\times \mathbb {N}\times \mathbb {N}$. For $a, b, c, d\in \mathbb {N}^{+}$ and $n\in \mathbb {N}$, we define

$$\begin{aligned} N(a,b,c,d;n)&=|\left\{ (x,y,z,w)\in \mathbb {Z}^4:ax^2+by^2+cz^2+dw^2=n\right\} | \end{aligned}$$

and

$$\begin{aligned} T(a,b,c,d;n)= & {} \bigg |\left\{ (x,y,z,w)\in \mathbb {N}^4:a\frac{x(x+1)}{2}+b\frac{y(y+1)}{2}\right. \nonumber \\{} & {} \left. +c\frac{z(z+1)}{2}+d\frac{w(w+1)}{2}=n\right\} \bigg |. \end{aligned}$$

We adopt the convention that $N(a,b,c,d;0)=T(a,b,c,d;0)=1$.

Attributed to Jacobi [1], we now know that

$$\begin{aligned} \textrm{N}(1,1,1,1;n)=8\sum _{d\mid n,4\not \mid d} d. \end{aligned}$$

(1.1)

Legendre [2] later discovered that

$$\begin{aligned} T(1,1,1,1;n)=\sum _{d\mid (2n+1)} d. \end{aligned}$$

(1.2)

By Ramanujan’s theta function identities, it is easy to establish the following relation:

$$\begin{aligned} T(1,1,1,1;n)=\frac{1}{8}N(1,1,1,1;2n+1), \end{aligned}$$

from which Legendre’s formula (1.2) follows immediately from Jacobi’s formula (1.1). It should be mentioned that Bateman and Knopp [3, Lemma 2.7] found the following relation between $r_k(n)$ and $t_k(n)$,

$$\begin{aligned} r_k(8n+k)=2^{k-1}\left( 2+\left( {\begin{array}{c}k\\ 4\end{array}}\right) \right) t_k(n),\quad \text {for } 1\le k\le 7, \end{aligned}$$

where $r_k(n)$ (resp. $t_k(n)$) be the number of representations of n as the sum of k squares (resp. triangular numbers). For $a, b, c, d\in \mathbb {N}^{+}$ with $1\le a+b+c+d\le 7$, Adiga, Cooper and Han [4] proved that

$$\begin{aligned} T(a,b,c,d;n)=\frac{1}{C(a,b,c,d)}N(a,b,c,d;8n+a+b+c+d), \end{aligned}$$

where

$$\begin{aligned} C(a,b,c,d)=16+8\left( \frac{\dot{\imath }_1 (\dot{\imath }_1-1)(\dot{\imath }_1-2)(\dot{\imath }_1-3)}{24} +\frac{\dot{\imath }_1(\dot{\imath }_1-1)\dot{\imath }_2}{2} +\dot{\imath }_1\dot{\imath }_3\right) \end{aligned}$$

and $\dot{\imath }_{\dot{\jmath }}$ denotes the number of elements in $\{a,b,c,d\}$ which are equal to $\dot{\jmath }$. Moreover, when $a+b+c+d=8$, Baruah, Cooper and Hirschhorn [5] found the relation between T(a, b, c, d; n) and $N(a,b,c,d;8n+8)-N(a,b,c,d;2n+2)$.

Seeking the relation between the number of representations as the sum of triangular numbers and the number of representations as the sum of squares is an interesting topic. In [6], Sun conjectured that for several values of a, b, c, d, T(a, b, c, d; n) is a linear combination of N(a, b, c, d; m) and $N(a,b,c,d;4\,m)$ with $m=8n+a+b+c+d$. It was proved by the present authors [7]. For more information on this issue, see [7] and references therein. Recently, Sun [8] established seventeen transformation formulae for the number of representing an integer as linear combinations of triangular numbers, and obtained many relations between T(a, b, c, d; n) and N(a, b, c, d; n). For example,

$$\begin{aligned} T(2,3,3,4;4n+3)&=\frac{1}{8}N(2,3,3,4;8n+9),\\ T(2,3,3,12;4n)&=\frac{1}{8}N(2,3,3,12;8n+5),\\ T(2,3,3,24;4n+2)&=\frac{1}{4}N(2,3,3,24;8n+12). \end{aligned}$$

At the end of paper [8], Sun posed some new conjectures about the relations between T(a, b, c, d; n) and N(a, b, c, d; n).

The aim of this paper is to confirm some of Sun’s conjectures in [8]. The main results are listed as follows.

Theorem 1

Let $n\in \mathbb {N}^{+}$. Then

$$\begin{aligned} T(1,2,3,10;4n+1)&=\frac{1}{12}N(1,2,3,10;8n+6), \end{aligned}$$

(1.3)

$$\begin{aligned} T(1,2,3,10;8n)&=\frac{1}{6}N(1,2,3,10;16n+4),\end{aligned}$$

(1.4)

$$\begin{aligned} T(1,2,3,10;4n+1)&=\frac{1}{36}N(1,2,3,10;32n+24),\end{aligned}$$

(1.5)

$$\begin{aligned} T(1,2,3,10;16n+10)&=\frac{1}{9}N(1,2,3,10;32n+24). \end{aligned}$$

(1.6)

Theorem 2

Let $n\in \mathbb {N}^{+}$. Then

$$\begin{aligned} T(1,3,6,30;4n)&=\frac{1}{12}N(1,3,6,30;8n+10),\end{aligned}$$

(1.7)

$$\begin{aligned} T(1,3,6,30;8n+1)&=\frac{1}{6}N(1,3,6,30;16n+12),\end{aligned}$$

(1.8)

$$\begin{aligned} T(1,3,6,30;4n)&=\frac{1}{36}N(1,3,6,30;32n+40),\end{aligned}$$

(1.9)

$$\begin{aligned} T(1,3,6,30;16n-1)&=\frac{1}{9}N(1,3,6,30;32n+8),\end{aligned}$$

(1.10)

$$\begin{aligned} T(1,3,6,30;8n+1)&=\frac{1}{30}N(1,3,6,30;64n+48),\end{aligned}$$

(1.11)

$$\begin{aligned} T(1,3,6,30;16n-1)&=\frac{1}{33}N(1,3,6,30;128n+32). \end{aligned}$$

(1.12)

Theorem 3

Let $n\in \mathbb {N}^{+}$. Then

$$\begin{aligned} T(2,5,10,15;4n+3)&=\frac{1}{12}N(2,5,10,15;8n+14),\end{aligned}$$

(1.13)

$$\begin{aligned} T(2,5,10,15;8n+6)&=\frac{1}{6}N(2,5,10,15;16n+20),\end{aligned}$$

(1.14)

$$\begin{aligned} T(2,5,10,15;4n+3)&=\frac{1}{36}N(2,5,10,15;32n+56),\end{aligned}$$

(1.15)

$$\begin{aligned} T(2,5,10,15;16n+8)&=\frac{1}{9}N(2,5,10,15;32n+24). \end{aligned}$$

(1.16)

Theorem 4

Let $n\in \mathbb {N}^{+}$. Then

$$\begin{aligned} T(5,6,15,30;4n+2)&=\frac{1}{12}N(5,6,15,30;8n+18),\end{aligned}$$

(1.17)

$$\begin{aligned} T(5,6,15,30;8n+7)&=\frac{1}{6}N(5,6,15,30;16n+28),\end{aligned}$$

(1.18)

$$\begin{aligned} T(5,6,15,30;16n+13)&=\frac{1}{9}N(5,6,15,30;32n+40). \end{aligned}$$

(1.19)

Remark 1

The proofs of Theorem 3 and Theorem 4 are similar to that of Theorem 1, which are omitted.

The paper is organized as follows. In Sect. 2, we present the necessary preliminaries. In Sects. 3 and 4, we prove Theorems 1 and 2, respectively.

2 Preliminaries

Ramanujan’s general theta function f(a, b) is defined by

$$\begin{aligned} f(a,b)=\sum _{n=-\infty }^\infty a^{n(n+1)/2}b^{n(n-1)/2},\quad |ab|<1. \end{aligned}$$

By [9, p. 34, Entry 18 (iv)], we see that for $n\in \mathbb {Z}$,

$$\begin{aligned} f(a,b) =a^{n(n+1)/2}b^{n(n-1)/2}f(a(ab)^n,b(ab)^{-n}). \end{aligned}$$

(2.1)

From the well-known Jacobi triple product identity in [9, p. 35, Entry 19],

$$\begin{aligned} f(a,b)=(-a;ab)_\infty (-b;ab)_\infty (ab;ab)_\infty , \end{aligned}$$

(2.2)

where

$$\begin{aligned} (a;q)_\infty :=\prod _{k=1}^\infty (1-aq^{k-1}). \end{aligned}$$

Setting $(a,b)=(q,q^2)$ and $(a,b)=(q,q^5)$ in (2.2), we readily obtain, respectively

$$\begin{aligned} f(q,q^2)=\frac{f_2f_3^2}{f_1f_6}, \end{aligned}$$

(2.3)

and

$$\begin{aligned} f(q,q^5)=\frac{f_2^2f_3f_{12}}{f_1f_4f_6}, \end{aligned}$$

(2.4)

where $f_k:=(q^k;q^k)_\infty$.

Two special cases of f(a, b) are

$$\begin{aligned} \varphi (q)&:=f(q,q)=\sum _{n=-\infty }^\infty q^{n^2}=\frac{f_2^5}{f_1^2f_4^2},\end{aligned}$$

(2.5)

$$\begin{aligned} \psi (q)&:=\frac{1}{2}f(1,q)=\sum _{n=0}^\infty q^{n(n+1)/2}=\frac{f_2^2}{f_1}. \end{aligned}$$

(2.6)

The generating functions for N(a, b, c, d; n) and T(a, b, c, d; n) are

$$\begin{aligned} \sum _{n=0}^\infty N(a,b,c,d;n)q^n&=\varphi (q^a)\varphi (q^b)\varphi (q^c)\varphi (q^d),\end{aligned}$$

(2.7)

$$\begin{aligned} \sum _{n=0}^\infty T(a,b,c,d;n)q^n&=\psi (q^a)\psi (q^b)\psi (q^c)\psi (q^d). \end{aligned}$$

(2.8)

Next, we require some theta function identities which will be used later.

Lemma 1

$$\begin{aligned} \varphi (q)&=\varphi (q^{4})+2q\psi (q^{8}),\end{aligned}$$

(2.9)

$$\begin{aligned} \psi (q)\psi (q^{3})&=\psi (q^{4})\varphi (q^{6})+q\varphi (q^{2})\psi (q^{12}),\end{aligned}$$

(2.10)

$$\begin{aligned} \varphi (q)\varphi (q^3)&=\varphi (q^{4})\varphi (q^{12})+4q^4\psi (q^{8})\psi (q^{24})+2q\psi (q^{2})\psi (q^{6}) . \end{aligned}$$

(2.11)

Proof

From [9, p. 40, Entry 25 (i), (ii)], we derive (2.9). Taking $(\mu ,\nu )=(2,1)$ in [9, p. 69, eq. (36.8)], we immediately obtain (2.10). Using (25.2.2) and (25.2.4) in [10, p. 219] to conclude (2.11). $\square$

The known ordinary methods used to deal with similar identities are not suitable for Sun’s conjectures considered here. To prove Sun’s statement, we devise the following two lemmas, which play an important role in our later proofs.

Lemma 2

$$\begin{aligned} \psi (q)\psi (q^{5})&=q^3\varphi (q^3)\psi (q^{30})+f(q,q^{5})f(q^{10},q^{20}),\end{aligned}$$

(2.12)

$$\begin{aligned}&=q^3\varphi (q^3)\psi (q^{30})+\left( f(q^8,q^{16})+qf(q^4,q^{20})\right) f(q^{10},q^{20}),\end{aligned}$$

(2.13)

$$\begin{aligned} \psi (q)\psi (q^{5})&=\psi (q^{6})\varphi (q^{15})+qf(q^2,q^4)f(q^{5},q^{25}),\end{aligned}$$

(2.14)

$$\begin{aligned}&=\psi (q^{6})\varphi (q^{15})+qf(q^2,q^4)\left( f(q^{40},q^{80})+q^5f(q^{20},q^{100})\right) . \end{aligned}$$

(2.15)

Proof

Putting $(\mu ,\nu )=(3,2)$ in (36.9) and (36.7) in [9, p. 69] to deduce (2.12) and (2.14), respectively. By [9, p. 46, Entry 30 (ii) and (iii)], we see that

$$\begin{aligned} f(a,b)&=f(a^3b,ab^3)+af(b/a,a^5b^3). \end{aligned}$$

(2.16)

Employing (2.16) in (2.12) and (2.14), we readily achieve (2.13) and (2.15), respectively.$\square$

Lemma 3

$$\begin{aligned} \varphi (q^2)\varphi (q^{10})+4q^3\psi (q^{4})\psi (q^{20})&=\varphi (q^3)\varphi (q^{15})+2q^2f(q,q^5)f(q^5,q^{25}),\end{aligned}$$

(2.17)

$$\begin{aligned} \psi (q^2)\varphi (q^{5})+q\varphi (q)\psi (q^{10})&=2q^2\psi (q^3)\psi (q^{15})+f(q,q^2)f(q^5,q^{10}). \end{aligned}$$

(2.18)

Proof

Putting $(\mu ,\nu )=(3,2)$ in [9, p. 68, eq. (36.3)], we obtain

$$\begin{aligned} \frac{1}{2}\left( \varphi (q)\varphi (q^{5})+\varphi (-q)\varphi (-q^{5})\right) =\sum _{m=0}^2q^{6m^2}f(q^{5(6+4m)},q^{5(6-4m)})\cdot f(q^{6+8m},q^{6-8m}). \end{aligned}$$

Employing (2.1) and (2.5), the above identity can be rewritten as follows,

$$\begin{aligned} \frac{1}{2}\left( \varphi (q)\varphi (q^{5})+\varphi (-q)\varphi (-q^{5})\right)&=\varphi (q^{6})\varphi (q^{30})+q^{6}f(q^{50},q^{10})\cdot q^{-2}f(q^{2},q^{10})\nonumber \\ {}&\quad +q^{24}\cdot q^{-10}f(q^{50},q^{10})\cdot q^{-10}f(q^{2},q^{10}) \nonumber \\&=\varphi (q^{6})\varphi (q^{30})+2q^{4}f(q^{2},q^{10})f(q^{10},q^{50}) . \end{aligned}$$

(2.19)

Next, applying [9, p. 68, eq. (36.2)] with $A=B=1$, $\mu =3$, and $\nu =2$, we find that

$$\begin{aligned} \frac{1}{2}\left( f(q^{5},q^{5})f(q,q)-f(-q^{5},-q^{5})f(-q,-q)\right)&=q^{5}f(q^{40},q^{20})f(q^{16},q^{-4})\nonumber \\&\quad +q^{21}f(q^{60},1)f(q^{24},q^{-12})\nonumber \\ {}&\quad +q^{49}f(q^{80},q^{-20})f(q^{32},q^{-20}). \end{aligned}$$

(2.20)

Using (2.1) and (2.6) in (2.20),

$$\begin{aligned} \frac{1}{2}\left( \varphi (q)\varphi (q^{5})-\varphi (-q)\varphi (-q^{5})\right)&=qf(q^{40},q^{20})f(q^{4},q^{8})\nonumber \\&\quad +q^9f(q^{60},1)f(q^{12},1) +qf(q^{20},q^{40})f(q^{4},q^{8})\nonumber \\&=2qf(q^{4},q^{8})f(q^{20},q^{40})+4q^{9}\psi (q^{12})\psi (q^{60}). \end{aligned}$$

(2.21)

Combining (2.19) and (2.21) together to deduce that

$$\begin{aligned} \varphi (q)\varphi (q^5)= & {} \varphi (q^{6})\varphi (q^{30})+2q^{4}f(q^{2}, q^{10})f(q^{10},q^{50})\nonumber \\{} & {} + 2qf(q^{4},q^{8})f(q^{20},q^{40})+4q^{9}\psi (q^{12})\psi (q^{60}). \end{aligned}$$

From (2.9),

$$\begin{aligned} \varphi (q)\varphi (q^{5})&=\left( \varphi (q^{4})+2q\psi (q^{8})\right) \nonumber \\&\left( \varphi (q^{20})+2q^5\psi (q^{40})\right) \\&=\varphi (q^{4})\varphi (q^{20})+4q^6\psi (q^{8}) \psi (q^{40})+2q\psi (q^{8})\varphi (q^{20}) +2q^5\varphi (q^{4})\psi (q^{40}). \end{aligned}$$

From the above two identities for $\varphi (q)\varphi (q^5)$, we immediately see that

$$\begin{aligned} \varphi (q^{4})\varphi (q^{20})+4q^6\psi (q^{8})\psi (q^{40})&=\varphi (q^{6}) \varphi (q^{30})+2q^{4}f(q^{2},q^{10})f(q^{10},q^{50}),\\ 2q\psi (q^{8})\varphi (q^{20}) +2q^5\varphi (q^{4})\psi (q^{40})&=2qf(q^{4},q^{8})f(q^{20},q^{40})+4q^{9}\psi (q^{12})\psi (q^{60}), \end{aligned}$$

and the desired identities follow. $\square$

Lemma 4

$$\begin{aligned} \frac{f_3^2}{f_1^2}&=\frac{f_4^4f_6f_{12}^2}{f_2^5f_8f_{24}}+2q \frac{f_4f_6^2f_8f_{24}}{f_2^4f_{12}},\end{aligned}$$

(2.22)

$$\begin{aligned} f(q,q^2)&=\frac{f_1f_4^4f_{12}^2}{f_2^4f_8f_{24}}+2q \frac{f_1f_4f_6f_8f_{24}}{f_2^3f_{12}}. \end{aligned}$$

(2.23)

Proof

Identity (2.22) follows from [10, eq. (30.10.4)]. Multiplying both sides of (2.22) by $\frac{f_1f_2}{f_6}$ and applying (2.3), we immediately deduce (2.23). $\square$

To end this section, we present the following fact which will be used frequently and without be explicitly mentioned. Let $\{a(n)\}_0^\infty$ be the sequence defined by

$$\begin{aligned} \sum _{n=0}^\infty a(n)q^n=A_0(q^m)+qA_1(q^m)+\cdots +q^{m-1}A_{m-1}(q^m), \end{aligned}$$

(2.24)

where $A_i(q)$ is an arbitrary infinite series in q and $m\ge 2$. For $0\le i<m$, collecting those terms on each side of (2.24) where the powers of q are of the form $mn+i$, we have

$$\begin{aligned} \sum _{n=0}^\infty a(mn+i)q^n=A_i(q). \end{aligned}$$

(2.25)

3 Proof of Theorem 1

It follows from (2.7) and (2.9) that

$$\begin{aligned} \sum _{n=0}^\infty N(1,2,3,10;n)q^n&=\left( \varphi (q^{4})+2q\psi (q^{8})\right) \left( \varphi (q^{8})+2q^2\psi (q^{16})\right) \nonumber \\&\quad \times \left( \varphi (q^{12})+2q^3\psi (q^{24})\right) \left( \varphi (q^{40})+2q^{10}\psi (q^{80})\right) . \end{aligned}$$

(3.1)

Picking out the terms involving $q^{4n}$ and $q^{4n+2}$ in (3.1), we deduce that, respectively,

$$\begin{aligned} \sum _{n=0}^\infty N(1,2,3,10;4n)q^n&=\varphi (q)\varphi (q^3)\left( \varphi (q^2)\varphi (q^{10})\right. \nonumber \\&\left. \quad +4q^3\psi (q^{4})\psi (q^{20})\right) \nonumber \\&\quad +4q\psi (q^2)\psi (q^{6})\left( \varphi (q^{2}) \varphi (q^{10})+4q^3\psi (q^{4})\psi (q^{20})\right) \end{aligned}$$

(3.2)

and

$$\begin{aligned} \sum _{n=0}^\infty N(1,2,3,10;4n+2)q^n&=2\varphi (q)\varphi (q^3)\left( \psi (q^{4})\varphi (q^{10})+ q^2\varphi (q^2)\psi (q^{20})\right) \nonumber \\&\quad +8q\psi (q^{2})\psi (q^{6})\left( \psi (q^{4})\varphi (q^{10})+q^2 \varphi (q^2)\psi (q^{20})\right) . \end{aligned}$$

(3.3)

Substituting (2.11) into (3.3),

$$\begin{aligned} \sum _{n=0}^\infty N(1,2,3,10;4n+2)q^n&=2\left( \varphi (q^{4})\varphi (q^{12})\right. \quad +4q^4\psi (q^{8})\psi (q^{24})\\ {}&\quad \left. +2q\psi (q^{2})\psi (q^{6})\right) \left( \psi (q^{4})\varphi (q^{10})+q^2\varphi (q^2)\psi (q^{20})\right) \\&\quad +8q\psi (q^{2})\psi (q^{6})\left( \psi (q^{4})\varphi (q^{10})+q^2 \varphi (q^2)\psi (q^{20})\right) , \end{aligned}$$

from which we derive

$$\begin{aligned} \sum _{n=0}^\infty N(1,2,3,10;8n+6)q^n =12\psi (q)\psi (q^{3})\left( \psi (q^{2})\varphi (q^{5})+q\varphi (q) \psi (q^{10})\right) . \end{aligned}$$

(3.4)

Thanks to (2.8) and (2.10),

$$\begin{aligned} \sum _{n=0}^\infty T(1,2,3,10;n)q^n =\left( \psi (q^{4})\varphi (q^{6})+q\varphi (q^{2})\psi (q^{12})\right) \psi (q^{2})\psi (q^{10}), \end{aligned}$$

which implies that

$$\begin{aligned} \sum _{n=0}^\infty T(1,2,3,10;2n)q^n&=\psi (q^{2})\varphi (q^{3})\psi (q)\psi (q^{5}),\end{aligned}$$

(3.5)

$$\begin{aligned} \sum _{n=0}^\infty T(1,2,3,10;2n+1)q^n&=\varphi (q)\psi (q^{6})\psi (q)\psi (q^{5}). \end{aligned}$$

(3.6)

Substituting (2.13) into (3.6), and using (2.9) and (2.11) to arrive at

$$\begin{aligned}&\sum _{n=0}^\infty T(1,2,3,10;2n+1)q^n =\varphi (q)\psi (q^{6}) \left( q^3\varphi (q^3)\psi (q^{30})\right. \\&\left. \qquad +\left( f(q^8,q^{16})+qf(q^4,q^{20}) \right) f(q^{10},q^{20})\right) \\&\quad =q^3\psi (q^{6})\psi (q^{30}) \left( \varphi (q^{4})\varphi (q^{12})+4q^4\psi (q^{8})\psi (q^{24}) +2q\psi (q^{2})\psi (q^{6})\right) \\&\qquad +\psi (q^{6})f(q^{10},q^{20})\left( f(q^8,q^{16})+qf(q^4,q^{20}) \right) \left( \varphi (q^{4})+2q\psi (q^{8})\right) . \end{aligned}$$

Extracting the terms $q^{2n}$ for $n\ge 0$, replacing $q^2$ by q, and employing (2.3)–(2.6),

$$\begin{aligned} \sum _{n=0}^\infty T(1,2,3,10;4n+1)q^n&=2q^2\psi (q)\psi (q^{3})^2\psi (q^{15})\nonumber \\&\quad +\psi (q^{3})f(q^{5},q^{10})\left( f(q^4,q^{8})\varphi (q^{2}) +2qf(q^2,q^{10})\psi (q^{4})\right) \nonumber \\&=2q^2\psi (q)\psi (q^{3})^2\psi (q^{15}) +\psi (q)\psi (q^{3})f(q^{5},q^{10})\nonumber \\&\quad \left( \frac{f_1f_4^4f_{12}^2}{f_2^4f_8f_{24}} +2q\frac{f_1f_4f_6f_8f_{24}}{f_2^3f_{12}}\right) . \end{aligned}$$

(3.7)

Applying (2.23) and (2.18), (3.7) can be rewritten as follows,

$$\begin{aligned} \sum _{n=0}^\infty T(1,2,3,10;4n+1)q^n&=\psi (q)\psi (q^{3})\left( 2q^2\psi (q^{3})\psi (q^{15})+f(q,q^2)f(q^5,q^{10})\right) \nonumber \\&=\psi (q)\psi (q^{3})\left( \psi (q^{2})\varphi (q^{5})+q\varphi (q)\psi (q^{10})\right) . \end{aligned}$$

(3.8)

Combining (3.4) and above identity, (1.3) follows immediately.

In view of (2.11) and (3.2),

$$\begin{aligned} \sum _{n=0}^\infty N(1,2,3,10;4n)q^n&=\left( \varphi (q^{4})\varphi (q^{12})+4q^4\psi (q^{8})\psi (q^{24})+2q\psi (q^{2})\psi (q^{6})\right) \nonumber \\&\quad \quad \left( \varphi (q^2)\varphi (q^{10})+4q^3\psi (q^{4})\psi (q^{20})\right) \nonumber \\&\quad +4q\psi (q^{2})\varphi (q^2)\psi (q^{6})\varphi (q^{10}) +16q^4\psi (q^{2})\psi (q^{4})\psi (q^{6})\psi (q^{20}). \end{aligned}$$

(3.9)

It follows that

$$\begin{aligned} \sum _{n=0}^\infty N(1,2,3,10;8n+4)q^n&=6\psi (q)\psi (q^{3})\varphi (q)\varphi (q^{5}) +4q\psi (q^{2})\psi (q^{10})\\ {}&\quad \left( \varphi (q^{2})\varphi (q^{6})+4q^2\psi (q^{4})\psi (q^{12})\right) . \end{aligned}$$

Employing (2.9) and (2.10), we deduce that

$$\begin{aligned}&\sum _{n=0}^\infty N(1,2,3,10;8n+4)q^n-4q\psi (q^{2})\psi (q^{10}) \left( \varphi (q^{2})\varphi (q^{6})+4q^2\psi (q^{4})\psi (q^{12})\right) \\&\quad =6\left( \psi (q^{4})\varphi (q^{6})+q\varphi (q^{2})\psi (q^{12})\right) \quad \left( \varphi (q^{4})+2q\psi (q^{8})\right) \left( \varphi (q^{20})+2q^5\psi (q^{40})\right) , \end{aligned}$$

and

$$\begin{aligned} \sum _{n=0}^\infty N(1,2,3,10;16n+4)q^n&=6\psi (q^{2})\varphi (q^{3})\left( \varphi (q^{2}) \varphi (q^{10})+4q^3\psi (q^{4})\psi (q^{20})\right) \nonumber \\&\quad +12q\varphi (q)\psi (q^{6})\left( \psi (q^{4})\varphi (q^{10}) +q^2\varphi (q^{2})\psi (q^{20})\right) . \end{aligned}$$

(3.10)

Substituting (2.9) and (2.15) into (3.5), we find that

$$\begin{aligned} \sum _{n=0}^\infty T(1,2,3,10;2n)q^n&=\psi (q^{2})\varphi (q^{3}) \left( \psi (q^{6})\varphi (q^{15})+qf(q^2,q^4)\left( f(q^{40},q^{80}) +q^5f(q^{20},q^{100})\right) \right) \nonumber \\&=\psi (q^{2})\psi (q^{6}) \left( \varphi (q^{12})+2q^3\psi (q^{24})\right) \left( \varphi (q^{60})+2q^{15}\psi (q^{120})\right) \nonumber \\&\quad +q\psi (q^{2})f(q^2,q^4)\left( \varphi (q^{12})+2q^3\psi (q^{24})\right) \left( f(q^{40},q^{80})+q^5f(q^{20},q^{100})\right) , \end{aligned}$$

(3.11)

and

$$\begin{aligned} \sum _{n=0}^\infty T(1,2,3,10;4n)q^n&=\psi (q)\psi (q^{3}) \left( \varphi (q^{6})\varphi (q^{30})+4q^9\psi (q^{12})\psi (q^{60})\right) \\&\quad +\psi (q)f(q,q^2)\left( q^3\varphi (q^{6})f(q^{10},q^{50}) +2q^2\psi (q^{12})f(q^{20},q^{40})\right) . \end{aligned}$$

Using first (2.3) and (2.6), then (2.10) and (2.22), it follows that

$$\begin{aligned} \sum _{n=0}^\infty T(1,2,3,10;4n)q^n&=\psi (q)\psi (q^{3}) \left( \varphi (q^{6})\varphi (q^{30})+4q^9\psi (q^{12})\psi (q^{60})\right) \\&\quad +\frac{f_2^3}{f_6}\cdot \frac{f_3^2}{f_1^2} \left( q^3\varphi (q^{6})f(q^{10},q^{50})+2q^2\psi (q^{12})f(q^{20},q^{40})\right) \\&=\left( \psi (q^{4})\varphi (q^{6})+q\varphi (q^{2})\psi (q^{12})\right) \left( \varphi (q^{6})\varphi (q^{30})+4q^9\psi (q^{12})\psi (q^{60})\right) \\&\quad +\frac{f_2^3}{f_6}\cdot \left( \frac{f_4^4f_6f_{12}^2}{f_2^5f_8f_{24}}+2q\frac{f_4f_6^2f_8f_{24}}{f_2^4f_{12}}\right) \\&\quad \left( q^3\varphi (q^{6})f(q^{10},q^{50})+2q^2\psi (q^{12})f(q^{20},q^{40})\right) , \end{aligned}$$

from which we extract

$$\begin{aligned} \sum _{n=0}^\infty T(1,2,3,10;8n)q^n&=\psi (q^{2})\varphi (q^{3})^2\varphi (q^{15})+4q^5\varphi (q) \psi (q^{6})^2\psi (q^{30})\\&\quad +2q\psi (q^{6})f(q^{10},q^{20})\frac{f_2^4f_6^2}{f_1^2f_4f_{12}}\\&\quad +2q^2\varphi (q^{3})f(q^{5},q^{25})\frac{f_2f_3f_4f_{12}}{f_1f_6}. \end{aligned}$$

Applying first (2.5), (2.6), (2.3), (2.4) then (2.17) and (2.18) to achieve that

$$\begin{aligned} \sum _{n=0}^\infty T(1,2,3,10;8n)q^n&=\psi (q^{2})\varphi (q^{3}) \left( \varphi (q^{3})\varphi (q^{15})+2q^2\frac{f_2}{f_4^2}\cdot \frac{f_2f_3f_4f_{12}}{f_1f_6}f(q^{5},q^{25})\right) \nonumber \\&\quad +2q\varphi (q)\psi (q^{6}) \left( 2q^4\psi (q^{6})\psi (q^{30})+\frac{f_1^2f_4^2}{f_2^5}\cdot \frac{f_2^4f_6^2}{f_1^2f_4f_{12}}f(q^{10},q^{20})\right) \nonumber \\&=\psi (q^{2})\varphi (q^{3}) \left( \varphi (q^{3})\varphi (q^{15})+2q^2f(q,q^{5})f(q^{5},q^{25})\right) \nonumber \\&\quad +2q\varphi (q)\psi (q^{6}) \left( 2q^4\psi (q^{6})\psi (q^{30})+f(q^{2},q^{4})f(q^{10},q^{20})\right) \nonumber \\&=\psi (q^{2})\varphi (q^{3})\left( \varphi (q^{2})\varphi (q^{10})+4q^3\psi (q^{4})\psi (q^{20})\right) \nonumber \\&~~+2q\varphi (q)\psi (q^{6})\left( \psi (q^{4})\varphi (q^{10})+q^2\varphi (q^{2})\psi (q^{20})\right) . \end{aligned}$$

(3.12)

By (3.10) and (3.12), we get the desired relation (1.4).

Collecting those terms in which the power of q is even in (3.9) and then employing (2.8) and (2.9) yields

$$\begin{aligned}&\sum _{n=0}^\infty N(1,2,3,10;8n)q^n\\&\quad =\varphi (q)\varphi (q^{5})\left( \varphi (q^{2}) \varphi (q^{6})+4q^2\psi (q^{4})\psi (q^{12})\right) +24q^2\psi (q)\psi (q^{2})\psi (q^{3})\psi (q^{10})\\&\quad =\left( \varphi (q^{4})+2q\psi (q^{8})\right) \left( \varphi (q^{20})+2q^5\psi (q^{40})\right) \left( \varphi (q^{2})\varphi (q^{6})+4q^2\psi (q^{4})\psi (q^{12})\right) \\&\qquad +24q^2\sum _{n=0}^\infty T(1,2,3,10;n)q^n, \end{aligned}$$

and

$$\begin{aligned}&\sum _{n=0}^\infty N(1,2,3,10;16n+8)q^n\\&\quad =2\varphi (q)\varphi (q^{3})\psi (q^{4})\varphi (q^{10}) +2q^2\varphi (q)\varphi (q^{2})\varphi (q^{3})\psi (q^{20})\\&\qquad +8q\psi (q^{2})\psi (q^{4})\psi (q^{6})\varphi (q^{10}) +8q^3\psi (q^{2})\varphi (q^{2})\psi (q^{6})\psi (q^{20})\\&\qquad +24q\sum _{n=0}^\infty T(1,2,3,10;2n+1)q^n. \end{aligned}$$

Appealing to (3.3), we have

$$\begin{aligned} \sum _{n=0}^\infty N(1,2,3,10;16n+8)q^n&=\sum _{n=0}^\infty N(1,2,3,10;4n+2)q^n\\&\quad +24q\sum _{n=0}^\infty T(1,2,3,10;2n+1)q^n. \end{aligned}$$

Equating the coefficients of $q^{2n+1}$ to obtain

$$\begin{aligned} N(1,2,3,10;32n+24)=N(1,2,3,10;8n+6) +24T(1,2,3,10;4n+1). \end{aligned}$$

Hence, the relation (1.5) follows immediately from (1.3) and the above relation.

Employing first (2.6), (2.3) then (2.10) and (2.22) in (3.11),

$$\begin{aligned}&\sum _{n=0}^\infty T(1,2,3,10;4n+2)q^n\\&=2q\psi (q)\psi (q^{3})\left( \psi (q^{12})\varphi (q^{30})+q^{6}\varphi (q^{6})\psi (q^{60})\right) \\&\quad +\psi (q)f(q,q^2)\left( \varphi (q^{6})f(q^{20},q^{40})+2q^4\psi (q^{12})f(q^{10},q^{50})\right) \\&=2q\psi (q)\psi (q^{3}) \left( \psi (q^{12})\varphi (q^{30})+q^{6}\varphi (q^{6})\psi (q^{60})\right) \\&\quad +\frac{f_2^3}{f_6}\cdot \frac{f_3^2}{f_1^2}\left( \varphi (q^{6}) f(q^{20},q^{40})+2q^4\psi (q^{12})f(q^{10},q^{50})\right) \\&=2q\left( \psi (q^{4})\varphi (q^{6})+q\varphi (q^{2})\psi (q^{12})\right) \left( \psi (q^{12})\varphi (q^{30})+q^{6}\varphi (q^{6})\psi (q^{60})\right) \\&\quad +\frac{f_2^3}{f_6}\cdot \left( \frac{f_4^4f_6f_{12}^2}{f_2^5f_8f_{24}}+2q\frac{f_4f_6^2f_8f_{24}}{f_2^4f_{12}}\right) \left( \varphi (q^{6})f(q^{20},q^{40})+2q^4\psi (q^{12})f(q^{10},q^{50})\right) , \end{aligned}$$

from which we extract the terms involving $q^{2n}$, then using (2.5) and (2.3) yields

$$\begin{aligned} \sum _{n=0}^\infty T(1,2,3,10;8n+2)q^n&=2q\varphi (q)\psi (q^{6})\left( \psi (q^{6})\varphi (q^{15})+q^{3}\varphi (q^{3})\psi (q^{30})\right) \\&\quad +\frac{f_2^4f_6^2}{f_1^2f_4f_{12}} \left( \varphi (q^{3})f(q^{10},q^{20})+2q^2\psi (q^{6})f(q^{5},q^{25})\right) \\&=2q\varphi (q)\psi (q^{6})\left( \psi (q^{6})\varphi (q^{15})+q^{3}\varphi (q^{3})\psi (q^{30})\right) \\&\quad +\varphi (q)f(q^{2},q^{4}) \left( \varphi (q^{3})f(q^{10},q^{20})+2q^2\psi (q^{6})f(q^{5},q^{25})\right) \\&=2q\varphi (q)\psi (q^{6})\left( \psi (q^{6})\varphi (q^{15})+qf(q^{2},q^{4})f(q^{5},q^{25})\right) \\&\quad +\varphi (q)\varphi (q^{3}) \left( 2q^4\psi (q^{6})\psi (q^{30})+f(q^{2},q^{4})f(q^{10},q^{20})\right) . \end{aligned}$$

In light of (2.14), (2.11) and (3.6),

$$\begin{aligned} \sum _{n=0}^\infty T(1,2,3,10;8n+2)q^n&=2q\varphi (q)\psi (q^{6})\psi (q)\psi (q^{5})+\varphi (q)\varphi (q^{3})\\&\quad \left( 2q^4\psi (q^{6})\psi (q^{30})+f(q^{2},q^{4})f(q^{10},q^{20})\right) \\&=2q\sum _{n=0}^\infty T(1,2,3,10;2n+1)q^n +\left( \varphi (q^{4})\varphi (q^{12})\right. \\&\left. \quad +4q^4\psi (q^{8})\psi (q^{24}) +2q\psi (q^{2})\psi (q^{6})\right) \\&\quad \times \left( 2q^4\psi (q^{6})\psi (q^{30})+f(q^{2},q^{4})f(q^{10},q^{20})\right) . \end{aligned}$$

Applying (3.8), we conclude that

$$\begin{aligned} \sum _{n=0}^\infty T(1,2,3,10;16n+10)q^n&=2\sum _{n=0}^\infty T(1,2,3,10;4n+1)q^n\\&\quad +2\psi (q)\psi (q^{3}) \left( 2q^2\psi (q^{3})\psi (q^{15})+f(q,q^{2})f(q^{5},q^{10})\right) \\&=4\sum _{n=0}^\infty T(1,2,3,10;4n+1)q^n. \end{aligned}$$

Equating the coefficient of $q^n$, and using (1.5), the final relation (1.6) follows immediately. $\square$

4 Proof of Theorem 2

From (2.7) and (2.9),

$$\begin{aligned} \sum _{n=0}^\infty N(1,3,6,30;n)q^n&=\left( \varphi (q^{4})+2q\psi (q^{8})\right) \left( \varphi (q^{12})+2q^3\psi (q^{24})\right) \\&\quad \times \left( \varphi (q^{24})+2q^6\psi (q^{48})\right) \left( \varphi (q^{120})+2q^{30}\psi (q^{240})\right) , \end{aligned}$$

which implies that

$$\begin{aligned} \sum _{n=0}^\infty N(1,3,6,30;4n)q^n&=\varphi (q)\varphi (q^3) \left( \varphi (q^6)\varphi (q^{30})+4q^9\psi (q^{12})\psi (q^{60})\right) \nonumber \\&\quad +4q\psi (q^{2})\psi (q^{6})\left( \varphi (q^6)\varphi (q^{30}) +4q^{9}\psi (q^{12})\psi (q^{60})\right) , \end{aligned}$$

(4.1)

and

$$\begin{aligned} \sum _{n=0}^\infty N(1,3,6,30;4n+2)q^n&=2q\varphi (q)\varphi (q^3) \left( \psi (q^{12})\varphi (q^{30})+q^6\varphi (q^6)\psi (q^{60})\right) \nonumber \\&\quad +8q^2\psi (q^{2})\psi (q^{6}) \left( \psi (q^{12})\varphi (q^{30})+q^6\varphi (q^6)\psi (q^{60})\right) . \end{aligned}$$

(4.2)

By (2.11) and (4.2),

$$\begin{aligned} \sum _{n=0}^\infty N(1,3,6,30;4n+2)q^n&=2q\left( \varphi (q^{4})\varphi (q^{12})+4q^4\psi (q^{8})\psi (q^{24})+2q\psi (q^{2})\psi (q^{6})\right) \\&\quad \left( \psi (q^{12})\varphi (q^{30})+q^6\varphi (q^6)\psi (q^{60})\right) \\&\quad +8q^2\psi (q^{2})\psi (q^{6}) \left( \psi (q^{12})\varphi (q^{30})+q^6\varphi (q^6)\psi (q^{60})\right) . \end{aligned}$$

This allows us to find that

$$\begin{aligned} \sum _{n=0}^\infty N(1,3,6,30;8n+2)q^n&=12q\psi (q)\psi (q^{3})\left( \psi (q^{6})\varphi (q^{15})\right. \nonumber \\&\left. \quad +q^3\varphi (q^3)\psi (q^{30})\right) . \end{aligned}$$

(4.3)

In light of (2.8) and (2.10),

$$\begin{aligned} \sum _{n=0}^\infty T(1,3,6,30;n)q^n&=\left( \psi (q^{4})\varphi (q^{6})+q\varphi (q^{2})\psi (q^{12})\right) \nonumber \\&\quad \psi (q^{6})\psi (q^{30}), \end{aligned}$$

from which we extract

$$\begin{aligned} \sum _{n=0}^\infty T(1,3,6,30;2n)q^n&=\psi (q^{2})\varphi (q^{3})\psi (q^3)\psi (q^{15}),\end{aligned}$$

(4.4)

$$\begin{aligned} \sum _{n=0}^\infty T(1,3,6,30;2n+1)q^n&=\varphi (q)\psi (q^3)\psi (q^{6})\psi (q^{15}). \end{aligned}$$

(4.5)

Substituting (2.9), (2.11) and (2.13) into (4.4), we see that

$$\begin{aligned} \sum _{n=0}^\infty T(1,3,6,30;2n)q^n&=\psi (q^{2})\varphi (q^{3}) \left( q^9\varphi (q^9)\psi (q^{90})\right. \\&\quad \left. +\left( f(q^{24},q^{48})+q^3f(q^{12},q^{60})\right) f(q^{30},q^{60})\right) \\&=q^9\psi (q^{2})\psi (q^{90}) \left( \varphi (q^{12})\varphi (q^{36})+4q^{12}\psi (q^{24})\psi (q^{72})\right. \nonumber \\&\left. \quad +2q^3\psi (q^{6})\psi (q^{18})\right) \\&\quad +\psi (q^{2})f(q^{30},q^{60})\left( \varphi (q^{12})+2q^3\psi (q^{24})\right) \nonumber \\&\quad \left( f(q^{24},q^{48})+q^3f(q^{12},q^{60})\right) . \end{aligned}$$

This implies that

$$\begin{aligned} \sum _{n=0}^\infty T(1,3,6,30;4n)q^n&=2q^6\psi (q)\psi (q^{3})\psi (q^{9})\psi (q^{45})\\&\quad +\psi (q)f(q^{15},q^{30})(\varphi (q^{6})f(q^{12},q^{24})\\&\quad +2q^3\psi (q^{12})f(q^6,q^{30})). \end{aligned}$$

Using first (2.3)–(2.6), then (2.23) and (2.18) to deduce that

$$\begin{aligned} \sum _{n=0}^\infty T(1,3,6,30;4n)q^n&=2q^6\psi (q)\psi (q^{3})\psi (q^{9})\psi (q^{45})\nonumber \\&\quad +\psi (q)\psi (q^{3})f(q^{15},q^{30})\cdot \frac{f_3}{f_6^2}\nonumber \\&\quad \cdot \left( \frac{f_{12}^5}{f_6^2f_{24}^2}\cdot \frac{f_{24}f_{36}^2}{f_{12}f_{72}} +2q^3\frac{f_{24}^2}{f_{12}} \cdot \frac{f_{12}^2f_{18}f_{72}}{f_6 f_{24}f_{36}}\right) \nonumber \\&=2q^6\psi (q)\psi (q^{3})\psi (q^{9})\psi (q^{45})\nonumber \\&\quad +\psi (q)\psi (q^{3})f(q^{15},q^{30}) \left( \frac{f_3f_{12}^4f_{36}^2}{f_6^4f_{24}f_{72}} +2q^3\frac{f_3f_{12}f_{18}f_{24}f_{72}}{f_6^3f_{36}}\right) \nonumber \\&=\psi (q)\psi (q^{3})\left( 2q^6\psi (q^{9})\psi (q^{45}) +f(q^3,q^6)f(q^{15},q^{30})\right) \nonumber \\&=\psi (q)\psi (q^{3})\left( \psi (q^{6})\varphi (q^{15}) +q^3\varphi (q^3)\psi (q^{30})\right) . \end{aligned}$$

(4.6)

This, with (4.3), yields the relation (1.7).

Applying (2.11) in (4.1) gives

$$\begin{aligned} \sum _{n=0}^\infty N(1,3,6,30;4n)q^n&=\left( \varphi (q^{4})\varphi (q^{12})+4q^4\psi (q^{8})\psi (q^{24})+2q\psi (q^{2})\psi (q^{6})\right) \nonumber \\&\quad \left( \varphi (q^6)\varphi (q^{30})+4q^9\psi (q^{12})\psi (q^{60})\right) \nonumber \\&\quad +4q\psi (q^{2})\psi (q^{6})\varphi (q^6)\varphi (q^{30})\nonumber \\&\quad +16q^{10}\psi (q^{2})\psi (q^{6})\psi (q^{12})\psi (q^{60}) . \end{aligned}$$

(4.7)

Hence we have

$$\begin{aligned} \sum _{n=0}^\infty N(1,3,6,30;8n+4)q^n&=6\psi (q)\psi (q^{3})\varphi (q^3)\varphi (q^{15}) +4q^4\psi (q^{6})\psi (q^{30})\nonumber \\&\quad \left( \varphi (q^{2})\varphi (q^{6})+4q^2\psi (q^{4})\psi (q^{12})\right) \nonumber \\&=6\left( \psi (q^{4})\varphi (q^{6})+q\varphi (q^{2})\psi (q^{12})\right) \nonumber \\&\quad \left( \varphi (q^{12})+2q^3\psi (q^{24})\right) \left( \varphi (q^{60})+2q^{15}\psi (q^{120})\right) \nonumber \\&\quad +4q^4\psi (q^{6})\psi (q^{30}) (\varphi (q^{2})\varphi (q^{6})\nonumber \\&\quad +4q^2\psi (q^{4})\psi (q^{12})), \end{aligned}$$

(4.8)

and

$$\begin{aligned} \sum _{n=0}^\infty N(1,3,6,30;16n+12)q^n&=12q\psi (q^{2})\varphi (q^{3})\left( \psi (q^{12})\varphi (q^{30}) +q^6\varphi (q^{6})\psi (q^{60})\right) \nonumber \\&\quad +6\varphi (q)\psi (q^{6})\left( \varphi (q^{6})\varphi (q^{30})+4q^9\psi (q^{12})\psi (q^{60})\right) . \end{aligned}$$

(4.9)

By (2.15) and (2.9), (4.5) can be rewritten as follows,

$$\begin{aligned} \sum _{n=0}^\infty T(1,3,6,30;2n+1)q^n&=\varphi (q)\psi (q^{6}) (\psi (q^{18})\varphi (q^{45})+q^3f(q^6,q^{12})\nonumber \\&\quad \left( f(q^{120},q^{240}) +q^{15}f(q^{60},q^{300})\right) )\nonumber \\&=\psi (q^{6})\psi (q^{18}) \left( \varphi (q^{4})+2q\psi (q^{8})\right) \nonumber \\&\quad \left( \varphi (q^{180})+2q^{45}\psi (q^{360})\right) \nonumber \\&\quad +q^3\psi (q^{6})f(q^6,q^{12}))\left( \varphi (q^{4})+2q\psi (q^{8})\right) \nonumber \\&\quad \left( f(q^{120},q^{240})+q^{15}f(q^{60},q^{300})\right) . \end{aligned}$$

(4.10)

This implies that

$$\begin{aligned} \sum _{n=0}^\infty T(1,3,6,30;4n+1)q^n&=\psi (q^3)\psi (q^{9}) (\varphi (q^{2})\varphi (q^{90})\\&\quad +4q^{23}\psi (q^{4})\psi (q^{180}))\\&\quad +\psi (q^3)f(q^3,q^6)(q^9\varphi (q^{2})f(q^{30},q^{150})\nonumber \\&\quad +2q^2\psi (q^{4})f(q^{60},q^{120})). \end{aligned}$$

If we utilize first (2.6), (2.3), then (2.10) and (2.22), we achieve that

$$\begin{aligned} \sum _{n=0}^\infty T(1,3,6,30;4n+1)q^n&=\psi (q^3)\psi (q^{9}) \left( \varphi (q^{2})\varphi (q^{90})\right. \nonumber \\&\left. \quad +4q^{23}\psi (q^{4})\psi (q^{180})\right) \\&\quad +\frac{f_6^3}{f_{18}}\cdot \frac{f_9^2}{f_3^2} \left( q^9\varphi (q^{2})f(q^{30},q^{150})\right. \nonumber \\&\left. \quad +2q^2\psi (q^{4})f(q^{60},q^{120})\right) \\&=\left( \psi (q^{12})\varphi (q^{18})+q^3\varphi (q^{6})\psi (q^{36})\right) \\&\quad \left( \varphi (q^{2})\varphi (q^{90})+4q^{23}\psi (q^{4})\psi (q^{180})\right) \\&\quad +\frac{f_6^3}{f_{18}}\cdot \left( \frac{f_{12}^4f_{18}f_{36}^2}{f_6^5f_{24}f_{72}} +2q^3\frac{f_{12}f_{18}^2f_{24}f_{72}}{f_6^4f_{36}}\right) \\&\quad \left( q^9\varphi (q^{2})f(q^{30},q^{150})+2q^2\psi (q^{4})f(q^{60},q^{120})\right) . \end{aligned}$$

Extracting those terms in which the power of q is even, then using (2.3)–(2.6), (2.17) and (2.18), we deduce that

$$\begin{aligned} \sum _{n=0}^\infty T(1,3,6,30;8n+1)q^n&=\varphi (q)\psi (q^{6})\varphi (q^{9})\varphi (q^{45}) +4q^{13}\psi (q^{2})\varphi (q^3)\psi (q^{18})\psi (q^{90})\nonumber \\&\quad +2q\psi (q^{2})f(q^{30},q^{60})\frac{f_6^4f_{18}^2}{f_3^2f_{12}f_{36}}\nonumber \\&\quad +2q^6\varphi (q)f(q^{15},q^{75})\frac{f_6f_9f_{12}f_{36}}{f_3f_{18}}\nonumber \\&=\varphi (q)\psi (q^{6}) \left( \varphi (q^{9})\varphi (q^{45})+2q^6f(q^3,q^{15})f(q^{15},q^{75})\right) \nonumber \\&\quad +2q\psi (q^{2})\varphi (q^3) (2q^{12}\psi (q^{18})\psi (q^{90})\nonumber \\&\quad +f(q^{6},q^{12})f(q^{30},q^{60})) \nonumber \\&=\varphi (q)\psi (q^{6})\left( \varphi (q^{6})\varphi (q^{30})+4q^9\psi (q^{12})\psi (q^{60})\right) \nonumber \\&\quad +2q\psi (q^{2})\varphi (q^3)\left( \psi (q^{12})\varphi (q^{30})+q^6\varphi (q^{6})\psi (q^{60})\right) . \end{aligned}$$

(4.11)

Combining (4.9) and (4.11) gives the desired relation (1.8).

Selecting terms whose the power of q is even in (4.7), using (2.8) and (2.9), we find that

$$\begin{aligned} \sum _{n=0}^\infty N(1,3,6,30;8n)q^n&=\varphi (q^3)\varphi (q^{15})\left( \varphi (q^{2})\varphi (q^{6})+4q^2\psi (q^{4})\psi (q^{12})\right) \nonumber \\&\quad +24q^5\psi (q)\psi (q^{3})\psi (q^{6})\psi (q^{30})\end{aligned}$$

(4.12)

$$\begin{aligned}&=\left( \varphi (q^{12})+2q^3\psi (q^{24})\right) \left( \varphi (q^{60})+2q^{15}\psi (q^{120})\right) \nonumber \\&\quad \left( \varphi (q^{2})\varphi (q^{6})+4q^2\psi (q^{4})\psi (q^{12})\right) \nonumber \\&\quad +24q^5\sum _{n=0}^\infty T(1,3,6,30;n)q^n, \end{aligned}$$

(4.13)

and

$$\begin{aligned} \sum _{n=0}^\infty N(1,3,6,30;16n+8)q^n&=\left( 2q\psi (q^{12})\varphi (q^{30})+2q^7\varphi (q^6)\psi (q^{60})\right) \nonumber \\&\quad \left( \varphi (q)\varphi (q^3)+4q\psi (q^2)\psi (q^6)\right) \nonumber \\&\quad +24q^2\sum _{n=0}^\infty T(1,3,6,30;2n)q^n. \end{aligned}$$

(4.14)

Substituting (4.2) into (4.14), and comparing the coefficients of $q^{2n+2}$ to deduce that

$$\begin{aligned} N(1,3,6,30;32n+40)= N(1,3,6,30;8n+10)+24 T(1,3,6,30;4n). \end{aligned}$$

Hence, relation (1.9) follows immediately from the above relation and (1.7).

Extracting the terms involving $q^{2n-1}$ in (4.10), using (2.10) and (2.22), gives

$$\begin{aligned}&\sum _{n=1}^\infty T(1,3,6,30;4n-1)q^n\nonumber \\&=\psi (q^{3})\psi (q^{9}) \left( 2q\psi (q^{4})\varphi (q^{90})+2q^{23}\varphi (q^{2})\psi (q^{180})\right) \\&\quad +\psi (q^{3})f(q^3,q^{6}))\left( q^2\varphi (q^{2})f(q^{60},q^{120})+2 q^{10}\psi (q^{4})f(q^{30},q^{150})\right) \\&=\left( \psi (q^{12})\varphi (q^{18})+q^3\varphi (q^{6})\psi (q^{36})\right) \left( 2q\psi (q^{4})\varphi (q^{90})+2q^{23}\varphi (q^{2})\psi (q^{180})\right) \\&\quad +\frac{f_6^3}{f_{18}} \left( \frac{f_{12}^4f_{18}f_{36}^2}{f_6^5f_{24}f_{72}} +2q^3\frac{f_{12}f_{18}^2f_{24}f_{72}}{f_6^4f_{36}}\right) \nonumber \\&\quad \left( q^2\varphi (q^{2})f(q^{60},q^{120})+2 q^{10} \psi (q^{4})f(q^{30},q^{150})\right) , \end{aligned}$$

and

$$\begin{aligned}&\sum _{n=1}^\infty T(1,3,6,30;8n-1)q^n\nonumber \\&\quad =2q^2\varphi (q^{3})\psi (q^{18}) \left( \psi (q^{2})\varphi (q^{45})+q^{11}\varphi (q)\psi (q^{90})\right) \\&\quad +\frac{f_6^5}{f_3^2f_{12}^2}\cdot \frac{f_{12}f_{18}^2}{f_6f_{36}} \left( q\varphi (q)f(q^{30},q^{60})+2q^{5}\psi (q^{2})f(q^{15},q^{75})\right) \\&=2q^2\varphi (q^{3})\psi (q^{18}) \left( \psi (q^{2})\varphi (q^{45})+q^{11}\varphi (q)\psi (q^{90})\right) \\&\quad +\varphi (q^{3})f(q^6,q^{12}) \left( q\varphi (q)f(q^{30},q^{60})+2q^{5}\psi (q^{2})f(q^{15},q^{75})\right) \\&=2q^2\psi (q^{2})\varphi (q^{3})\left( \psi (q^{18})\varphi (q^{45}) +q^{3}f(q^6,q^{12})f(q^{15},q^{75})\right) \\&\quad +q\varphi (q)\varphi (q^{3}) \left( 2q^{12}\psi (q^{18})\psi (q^{90})+f(q^6,q^{12})f(q^{30},q^{60})\right) . \end{aligned}$$

Invoking (2.14), (2.11) and (4.4),

$$\begin{aligned} \sum _{n=1}^\infty T(1,3,6,30;8n-1)q^n&=2q^2\sum _{n=0}^\infty T(1,3,6,30;2n)q^n+ q(\varphi (q^4)\varphi (q^{12})\nonumber \\&\quad +4q^4\psi (q^{8})\psi (q^{24})+2q\psi (q^{2})\psi (q^{6}))\\&\quad \times \left( 2q^{12}\psi (q^{18})\psi (q^{90})+f(q^6,q^{12})f(q^{30},q^{60})\right) . \end{aligned}$$

Utilizing (4.6) to deduce that

$$\begin{aligned} \sum _{n=1}^\infty T(1,3,6,30;16n-1)q^n&=2q\sum _{n=0}^\infty T(1,3,6,30;4n)q^n +2q\psi (q)\psi (q^{3})\\&\quad \left( 2q^{6}\psi (q^{9})\psi (q^{45})+f(q^3,q^{6})f(q^{15},q^{30})\right) \\&=4q\sum _{n=0}^\infty T(1,3,6,30;4n)q^n. \end{aligned}$$

This and (1.9) implies (1.10).

Applying (2.11) and (4.13) to deduce that

$$\begin{aligned} \sum _{n=0}^\infty N(1,3,6,30;16n)q^n&=\left( \varphi (q)\varphi (q^{3})+4q\psi (q^{2})\psi (q^{6})\right) \nonumber \\&\quad \left( \varphi (q^{6})\varphi (q^{30})+4q^{9}\psi (q^{12})\psi (q^{60})\right) \nonumber \\&\quad +24q^3\sum _{n=0}^\infty T(1,3,6,30;2n+1)q^n\nonumber \\&=\left( \varphi (q^4)\varphi (q^{12})+4q^4\psi (q^{8})\psi (q^{24})+6q\psi (q^{2})\psi (q^{6})\right) \nonumber \\&\quad \left( \varphi (q^{6})\varphi (q^{30})+4q^{9}\psi (q^{12})\psi (q^{60})\right) \nonumber \\&\quad +24q^3\sum _{n=0}^\infty T(1,3,6,30;2n+1)q^n, \end{aligned}$$

(4.15)

and

$$\begin{aligned}&\sum _{n=0}^\infty N(1,3,6,30;32n+16)q^n-24q\sum _{n=0}^\infty T(1,3,6,30;4n+1)q^n\\&\quad =6\psi (q)\psi (q^{3})\varphi (q^{3})\varphi (q^{15})+4q^{4}\psi (q^{6})\psi (q^{30})\nonumber \\&\quad \left( \varphi (q^2)\varphi (q^{6})+4q^2\psi (q^{4})\psi (q^{12})\right) . \end{aligned}$$

Combining (4.8) and the above identity yields,

$$\begin{aligned} \sum _{n=0}^\infty N(1,3,6,30;32n+16)q^n= & {} \sum _{n=0}^\infty N(1,3,6,30;8n+4)q^n\\{} & {} +24q\sum _{n=0}^\infty T(1,3,6,30;4n+1)q^n, \end{aligned}$$

which gives

$$\begin{aligned} N(1,3,6,30;64n+48)=N(1,3,6,30;16n+12) +24 T(1,3,6,30;8n+1). \end{aligned}$$

Hence, combine the above relation and (1.8) to derive (1.11).

From (4.15) and (4.12), we have

$$\begin{aligned} \sum _{n=0}^\infty N(1,3,6,30;32n)q^n&=\left( \varphi (q^2)\varphi (q^{6})+4q^2\psi (q^{4})\psi (q^{12})\right) \varphi (q^{3})\varphi (q^{15})\\&\quad +24q^5\psi (q)\psi (q^{3})\psi (q^{6})\psi (q^{30})\\&\quad +24q\sum _{n=1}^\infty T(1,3,6,30;2(2n-1)+1)q^n\\&=\sum _{n=0}^\infty N(1,3,6,30;8n)q^n\\&\quad +24q\sum _{n=1}^\infty T(1,3,6,30;4n-1)q^n. \end{aligned}$$

Comparing the coefficients of $q^{4n+1}$ on both sides gives

$$\begin{aligned} N(1,3,6,30;128n+32)=N(1,3,6,30;32n+8)+24T(1,3,6,30;16n-1). \end{aligned}$$

Employing (1.10) and the above relation, we get the final relation (1.12). $\square$