The $M_{2}$-rank of partitions without repeated odd parts modulo $6$ and $10$

Abstract

Lovejoy and Osburn (J Theor Nombres Bordeaux 21:313–334, 2009) proved formulas for the generating functions for $M_{2}$-rank differences modulo $3$ and $5$ for partitions without repeated odd parts. In this paper, we derive formulas for the generating functions for $M_{2}$-ranks modulo $6$ and $10$ for such partitions. With these generating functions, we find some inequalities between $M_{2}$-ranks modulo $3,5,6$, and $10$ of partitions without repeated odd parts.

Appendix: Proofs of Lemmas 3.2, 4.2 and 4.3

First, we derive some theta identities. Let

$$\begin{aligned} H(a,b,c,q):=\frac{\left( ab,q/(ab),bc,q/(bc),ca,q/(ca);q\right) _\infty (q;q)^2_\infty }{\left( a,q/a,b,q/b,c,q/c,abc,q/(abc);q\right) _\infty }. \end{aligned}$$

Then, we have the following equivalent version of Corollary 4.4 in [8]:

Lemma 6.1

$$\begin{aligned}&H(a,b,c,q)-H(a,b,d,q)=H(c,1/d,abd,q),\end{aligned}$$

(6.1)

$$\begin{aligned}&H(a,a,q/a,q^2)+H(b,b,q/b,q^2)=2H(a,q/a,b,q^2),\end{aligned}$$

(6.2)

$$\begin{aligned}&H(a,a,q/a,q^2)-H(b,b,q/b,q^2)=2H(a,q/a,b/q,q^2). \end{aligned}$$

(6.3)

We derive the following special cases. Replacing $q$ by $q^{50}$ and setting $(a,b,c,d)=(q^{10},q^{25},q^{10},-q^{-5})$ in (6.1), we get

$$\begin{aligned} \frac{J_{15,50}^2J_{20,50}}{J_{5,50}J_{10,50}^2J_{25,50}}- \frac{q^5J_{15,50}}{J_{10,50}J_{25,50}} =\frac{\overline{J}_{10,50}\overline{J}_{15,50}J_{15,50}}{J_{5,50}\overline{J}_{5,50}J_{10,50}\overline{J}_{20,50}}. \end{aligned}$$

Multiplying by $\frac{J_{5,50}J_{10,50}^2J_{25,50}}{J_{15,50}}$ throughout, the above equation becomes

$$\begin{aligned} J_{15,50}J_{20,50}- q^5J_{5,50}J_{10,50} =\frac{J_{5,50}J_{20,50}^2J_{25,50}}{J_{10,50}J_{15,50}}. \end{aligned}$$

(6.4)

Replacing $q$ by $q^{50}$ and setting $(a,b)=(q^{5},q^{15})$ in (6.3), we find

$$\begin{aligned} \frac{J_{10,100}}{J_{5,100}^2J_{45,100}^2}-\frac{J_{30,100}}{J_{15,100}^2J_{35,100}^2} =\frac{2q^5J_{10,100}J_{30,100}}{J_{5,100}J_{15,100}J_{35,100}J_{45,100}J_{50,100}}. \end{aligned}$$

(6.5)

Replacing $q$ by $q^{50}$ and setting $(a,b)=(q^{10},q^{20})$ in (6.3), we find that

$$\begin{aligned} \frac{J_{20,100}}{J_{10,100}^2J_{40,100}^2}-\frac{J_{40,100}}{J_{20,100}^2J_{30,100}^2} =\frac{2q^{10}}{J_{30,100}J_{40,100}J_{50,100}}. \end{aligned}$$

(6.6)

Next, we recall [9, Theorem 1.1]:

Lemma 6.2

For five complex parameters $A,b,c,d,e$ satisfying $A^2=bced,$ there holds the theta function identity

$$\begin{aligned}{}[A/b, A/c, A/d, A/e; q]_\infty - [b,c,d,e;q]_\infty = b[A, A/bc, A/bd, A/be; q]_\infty .\nonumber \\ \end{aligned}$$

(6.7)

We need the following special cases. Replacing $q$ by $q^{50}$ and setting $(A, b, c, d, e)=(-q^{25}, q^5, q^{10}, q^{15}, q^{20})$ in (6.7), we have

$$\begin{aligned} \overline{J}_{5,50}\overline{J}_{10,50}\overline{J}_{15,50}\overline{J}_{20,50}- J_{5,50}J_{10,50}J_{15,50}J_{20,50}= q^5\overline{J}_{0,50}\overline{J}_{5,50}\overline{J}_{10,50}\overline{J}_{25,50}.\nonumber \\ \end{aligned}$$

(6.8)

Now, we are ready to Lemma 3.2, 4.2, and 4.3, respectively.

Proof of Lemma 3.2

By (2.1), we know that

$$\begin{aligned} \frac{(q^2;q^2)_\infty }{(-q;q^2)_\infty }=\frac{J_{3,36}J_{15,36}J_{18,36}}{J_{6,36}J_{9,36}}-qJ_{9,36}. \end{aligned}$$

Expanding the right side and comparing both sides according to the powers of $q$ modulo $3$, we find that it suffices to prove the following three identities:

$$\begin{aligned} \frac{J_{3,36}^2J_{6,36}^3J_{12,36}^2J_{15,36}^2J_{18,36}^2}{2J_{36}^9}&= \frac{J_{18,36}^3J_{36}^3}{J_{9,36}^4}- \frac{J_{6,36}J_{18,36}^2J_{36}^3}{2J_{3,36}^2J_{15,36}^2},\end{aligned}$$

(6.9)

$$\begin{aligned} 0&= q\frac{J_{6,36}J_{18,36}^2J_{36}^3}{J_{3,36}J_{9,36}^2J_{15,36}}- q\frac{J_{6,36}J_{18,36}^2J_{36}^3}{J_{3,36}J_{9,36}^2J_{15,36}},\qquad \qquad \end{aligned}$$

(6.10)

$$\begin{aligned} \frac{J_{3,36}J_{6,36}^3J_{9,36}^2J_{12,36}^2J_{15,36}J_{18,36}^2}{2qJ_{36}^9}&= \frac{J_{18,36}^3J_{36}^3}{2qJ_{3,36}J_{9,36}^2J_{15,36}}- q^2\frac{J_{6,36}^2J_{18,36}J_{36}^3}{J_{3,36}^2J_{15,36}^2}. \end{aligned}$$

(6.11)

Simplifying, each of these three identities, we see that to prove (6.9), it suffices to show that

$$\begin{aligned} 2\frac{J_{18,36}}{J_{9,36}^4}-\frac{J_{6,36}}{J_{3,36}^2J_{15,36}^2}= \frac{J_{3,36}^2J_{6,36}^3J_{12,36}^2J_{15,36}^2}{J_{36}^{12}}. \end{aligned}$$

(6.12)

Noting that

$$\begin{aligned} \frac{J_{18,36}J_{36}^3}{J_{9,36}^4}&= \frac{\overline{J}_{9,18}}{J_{9,18}},\\ \frac{J_{6,36}J_{36}^3}{J_{3,36}^2J_{15,36}^2}&= \frac{\overline{J}_{3,18}}{J_{3,18}} \end{aligned}$$

and

$$\begin{aligned} \frac{J_{3,36}^2J_{6,36}^3J_{12,36}^2J_{15,36}^2}{J_{36}^{9}}= \frac{J_{3,18}J_{6,18}^2}{\overline{J}_{3,18}\overline{J}_{6,18}^2}, \end{aligned}$$

we find that (6.12) is equivalent to

$$\begin{aligned} \frac{2\overline{J}_{9,18}}{J_{9,18}}-\frac{\overline{J}_{3,18}}{J_{3,18}} =\frac{J_{3,18}J_{6,18}^2}{\overline{J}_{3,18}\overline{J}_{6,18}^2}. \end{aligned}$$

Multiplying by $\frac{J_{9,18}^2}{\overline{J}_{3,18}J_{3,18}J_{6,18}}$ throughout and rearranging, the above equation becomes

$$\begin{aligned} \frac{J_{9,18}^2J_{6,18}}{J_{3,18}^2J_{6,18}^2} +\frac{J_{9,18}^2J_{6,18}}{\overline{J}_{3,18}^2\overline{J}_{6,18}^2} =\frac{2J_{9,18}\overline{J}_{6,18}\overline{J} _{9,18}}{J_{3,18}J_{6,18}\overline{J}_{3,18}\overline{J}_{6,18}}. \end{aligned}$$

This follows from replacing $q,a$, and $b$ by $q^9,q^3$, and $-q^3$, respectively, in (6.2).

Equation (6.10) is trivial.

To prove (6.11), it suffices to show that

$$\begin{aligned} \frac{J_{18,36}^2}{J_{9,36}^2}-\frac{2q^3J_{6,36}^2}{J_{3,36}J_{15,36}}= \frac{J_{3,36}^2J_{6,36}^3J_{9,36}^2J_{12,36}^2J_{15,36}^2J_{18,36}}{J_{36}^{12}}, \end{aligned}$$

which is equivalent to

$$\begin{aligned} \frac{J_{6,12}^2}{J_{3,12}^2}-2q\frac{J_{2,12}^2}{J_{1,12}J_{5,12}}&=\frac{J_1^2 J_{2,12}}{J_{12}^3}. \end{aligned}$$

(6.13)

Equation (6.13) follows from two identities:

$$\begin{aligned} \frac{J_{6,12}^2}{J_{3,12}^2}+q\frac{J_{2,12}^2}{J_{1,12}J_{5,12}}&= \frac{J_{4,12}^2}{J_{1,12}J_{5,12}},\end{aligned}$$

(6.14)

$$\begin{aligned} \frac{J_{4,12}^2}{J_{1,12}J_{5,12}}-3q\frac{J_{2,12}^2}{J_{1,12}J_{5,12}}&= \frac{J_1^2 J_{2,12}}{J_{12}^3}. \end{aligned}$$

(6.15)

Equation (6.14) follows from replacing $q$ by $q^{12}$ and setting $(A, b, c, d, e)=(q^9, q, q^5, q^6, q^6)$ in (6.7).

Next, we prove (6.15). Multiplying throughout by $\frac{J_{1,12}J_{4,12}J_{5,12}}{J_{12}}$, Eq. (6.15) becomes

$$\begin{aligned} \frac{J_{4,12}^3}{J_{12}}-3q\frac{J_{2,12}^2J_{4,12}}{J_{12}} =\frac{J_1^2 J_{1,12}J_{2,12}J_{4,12}J_{5,12}}{J_{12}^4}, \end{aligned}$$

which upon simplifying and rearranging gives,

$$\begin{aligned} \frac{J_1^3}{J_{3}}=\frac{J_{4}^3}{J_{12}} - 3q \frac{J_{12}^3J_2^2}{J_4 J_{6}^2}. \end{aligned}$$

Setting $b(q) = \frac{J_1^3}{J_3}$, we see that it suffices to show that

$$\begin{aligned} b(q) = b(q^4)-3q \frac{J_{12}^3J_2^2}{J_4 J_{6}^2}, \end{aligned}$$

which is proved in [13, Eq. (1.35)]. This completes the proof of Lemma 3.2. $\square$

Sketch of the proof of Lemma 4.2

By (2.2), we know that

$$\begin{aligned} \frac{(q^2;q^2)_\infty }{(-q;q^2)_{\infty }}\!=\!\frac{J_{15,100}J_{20,100}J_{35,100}J_{50,100}}{J_{10,100}J_{25,100}J_{40,100}} -q\frac{J_{5,100}J_{40,100}J_{45,100}J_{50,100}}{J_{20,100}J_{25,100}J_{30,100}} -q^3J_{25,100}. \end{aligned}$$

Expanding the right side of (4.5) and comparing both sides according to the powers of $q$ modulo $5$, we find that it suffices to prove the following five identities:

$$\begin{aligned}&\frac{J_{5,100}^2J_{10,100}J_{20,100}J_{30,100}^2J_{40,100}J_{45,100}^2J_{50,100}^2}{2J_{100}^9}\nonumber \\&- \frac{J_{10,100}^2J_{15,100}^2J_{20,100}J_{30,100}J_{35,100}^2J_{40,100}J_{50,100}^2}{2J_{100}^9}\nonumber \\&= \frac{J_{15,100}J_{20,100}J_{35,100}J_{50,100}}{J_{10,100}J_{25,100}J_{40,100}}\times \frac{2q^5J_{10,100}J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{15,100}^2J_{20,100}^2J_{25,100}^3J_{30,100}J_{35,100}^2J_{45,100}^3}\nonumber \\&-\frac{qJ_{5,100}J_{40,100}J_{45,100}J_{50,100}}{J_{20,100}J_{25,100}J_{30,100}}\times \frac{2q^4J_{30,100}J_{50,100}J_{100}^{15}}{J_{5,100}^2J_{10,100}J_{15,100}^3J_{25,100}^3J_{35,100}^3J_{40,100}^2J_{45,100}^2}\nonumber \\&-q^3J_{25,100}\times \frac{q^2J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{15,100}^3J_{20,100}J_{25,100}J_{35,100}^3J_{40,100}J_{45,100}^3},\end{aligned}$$

(6.16)

$$\begin{aligned}&\frac{qJ_{5,100}J_{10,100}^2J_{15,100}J_{30,100}^2J_{35,100}J_{40,100}^2J_{45,100}J_{50,100}}{J_{100}^9}\nonumber \\&= \frac{J_{15,100}J_{20,100}J_{35,100}J_{50,100}}{J_{10,100}J_{25,100}J_{40,100}}\times \frac{qJ_{20,100}J_{30,100}^2J_{50,100}J_{100}^{15}}{J_{5,100}^2J_{10,100}^2J_{15,100}^4J_{25,100}J_{35,100}^4J_{40,100}^3J_{45,100}^2}\nonumber \\&-q\frac{J_{5,100}J_{40,100}J_{45,100}J_{50,100}}{J_{20,100}J_{25,100}J_{30,100}}\times \frac{2q^5J_{10,100}J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{15,100}^2J_{20,100}^2J_{25,100}^3J_{30,100}J_{35,100}^2J_{45,100}^3}\nonumber \\&-q^3J_{25,100}\times \frac{q^3J_{10,100}^2J_{40,100}J_{50,100}J_{100}^{15}}{J_{5,100}^4J_{15,100}^2J_{20,100}^3J_{25,100}J_{30,100}^2J_{35,100}^2J_{45,100}^4},\end{aligned}$$

(6.17)

$$\begin{aligned} 0&= \frac{J_{15,100}J_{20,100}J_{35,100}J_{50,100}}{J_{10,100}J_{25,100}J_{40,100}}\times \frac{q^2J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{15,100}^3J_{20,100}J_{25,100}J_{35,100}^3J_{40,100}J_{45,100}^3}\nonumber \\&-\frac{qJ_{5,100}J_{40,100}J_{45,100}J_{50,100}}{J_{20,100}J_{25,100}J_{30,100}}\times \frac{qJ_{20,100}J_{30,100}^2J_{50,100}J_{100}^{15}}{J_{5,100}^2J_{10,100}^2J_{15,100}^4J_{25,100}J_{35,100}^4J_{40,100}^3J_{45,100}^2}\nonumber \\&-q^3J_{25,100}\times \frac{2q^4J_{30,100}J_{50,100}J_{100}^{15}}{J_{5,100}^2J_{10,100}J_{15,100}^3J_{25,100}^3J_{35,100}^3J_{40,100}^2J_{45,100}^2}, \end{aligned}$$

(6.18)

$$\begin{aligned} 0&= \frac{J_{15,100}J_{20,100}J_{35,100}J_{50,100}}{J_{10,100}J_{25,100}J_{40,100}}\times \frac{q^3J_{10,100}^2J_{40,100}J_{50,100}J_{100}^{15}}{J_{5,100}^4J_{15,100}^2J_{20,100}^3J_{25,100}J_{30,100}^2J_{35,100}^2J_{45,100}^4}\nonumber \\&-\frac{qJ_{5,100}J_{40,100}J_{45,100}J_{50,100}}{J_{20,100}J_{25,100}J_{30,100}}\times \frac{q^2J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{15,100}^3J_{20,100}J_{25,100}J_{35,100}^3J_{40,100}J_{45,100}^3}\nonumber \\&-q^3J_{25,100}\times \frac{2q^5J_{10,100}J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{15,100}^2J_{20,100}^2J_{25,100}^3J_{30,100}J_{35,100}^2J_{45,100}^3},\end{aligned}$$

(6.19)

$$\begin{aligned}&-\frac{q^9J_{5,100}J_{10,100}^2J_{15,100}J_{20,100}^2J_{30,100}^2J_{35,100}J_{45,100}J_{50,100}}{J_{100}^9}\nonumber \\&= \frac{J_{15,100}J_{20,100}J_{35,100}J_{50,100}}{J_{10,100}J_{25,100}J_{40,100}}\times \frac{2q^4J_{30,100}J_{50,100}J_{100}^{15}}{J_{5,100}^2J_{10,100}J_{15,100}^3J_{25,100}^3J_{35,100}^3J_{40,100}^2J_{45,100}^2}\nonumber \\&-\frac{qJ_{5,100}J_{40,100}J_{45,100}J_{50,100}}{J_{20,100}J_{25,100}J_{30,100}}\times \frac{q^3J_{10,100}^2J_{40,100}J_{50,100}J_{100}^{15}}{J_{5,100}^4J_{15,100}^2J_{20,100}^3J_{25,100}J_{30,100}^2J_{35,100}^2J_{45,100}^4}\nonumber \\&-q^3J_{25,100}\times \frac{qJ_{20,100}J_{30,100}^2J_{50,100}J_{100}^{15}}{J_{5,100}^2J_{10,100}^2J_{15,100}^4J_{25,100}J_{35,100}^4J_{40,100}^3J_{45,100}^2}. \end{aligned}$$

(6.20)

The proof of the above five identities are quite similar to each other. We give the details of the proof of the first identity and omit the rest. After simplifying, we see that to prove (6.16), it suffices to show that

$$\begin{aligned}&\frac{J_{10,100}J_{20,100}J_{30,100}J_{40,100}J_{50,100}}{2J_{100}^{24}}\left\{ J_{5,100}^2J_{30,100}J_{45,100}^2-J_{10,100}J_{15,100}^2J_{35,100}^2\right\} \nonumber \\&\quad =\frac{2q^5J_{50,100}}{J_{5,100}J_{15,100}J_{20,100}J_{25,100}^4J_{35,100}J_{40,100}J_{45,100}}\nonumber \\&\qquad \times \left\{ \frac{1}{J_{5,100}^2J_{30,100}J_{45,100}^2}-\frac{1}{J_{10,100}J_{15,100}^2J_{35,100}^2}\right\} \nonumber \\&\qquad -\frac{q^5}{J_{5,100}^3J_{15,100}^3J_{20,100}J_{35,100}^3J_{40,100}J_{45,100}^3}. \end{aligned}$$

(6.21)

Multiplying by $J_{5,100}^2J_{15,100}^2J_{35,100}^2J_{45,100}^2$ throughout, (6.5) becomes

$$\begin{aligned} J_{5,100}^2J_{30,100}J_{45,100}^2&- J_{10,100}J_{15,100}^2J_{35,100}^2\nonumber \\&= -\frac{2q^5J_{5,100}J_{10,100}J_{15,100}J_{30,100}J_{35,100}J_{45,100}}{J_{50,100}}.\qquad \end{aligned}$$

(6.22)

Next, we divide by $J_{10,100}J_{30,100}$ on both sides of (6.5) and get

$$\begin{aligned} \frac{1}{J_{5,100}^2J_{30,100}J_{45,100}^2}-\frac{1}{J_{10,100}J_{15,100}^2J_{35,100}^2}= \frac{2q^5}{J_{5,100}J_{15,100}J_{35,100}J_{45,100}J_{50,100}}.\nonumber \\ \end{aligned}$$

(6.23)

Substituting (6.22) and (6.23) into (6.21), we find that (6.21) is equivalent to

$$\begin{aligned}&-q^5\frac{J_{5,100}J_{10,100}^2J_{15,100}J_{20,100}J_{30,100}^2J_{35,100}J_{40,100}J_{45,100}}{J_{100}^{24}}\nonumber \\&\quad =\frac{q^5}{J_{5,100}^2J_{15,100}^2J_{20,100}J_{35,100}^2J_{40,100}J_{45,100}^2}\nonumber \\&\qquad \times \left\{ \frac{4q^5}{J_{25,100}^4}-\frac{1}{J_{5,100}J_{15,100}J_{35,100}J_{45,100}}\right\} . \end{aligned}$$

(6.24)

Multiplying by $\frac{1}{q^5}J_{5,100}J_{15,100}J_{20,100} J_{35,100}J_{40,100}J_{45,100}J_{100}^{8}$ throughout and noting that,

$$\begin{aligned} -\frac{J_{5,100}^2J_{10,100}^2J_{15,100}^2J_{20,100}^2J_{30,100}^2J_{35,100}^2 J_{40,100}^2J_{45,100}^2}{J_{100}^{16}}&= -\frac{J_{5,25}^2J_{10,25}^2}{J_{25}^4},\\ \frac{4q^5J_{100}^8}{J_{5,100}J_{15,100}J_{25,100}^4J_{35,100}J_{45,100}}&= \frac{q^5\overline{J}_{0,25}^2\overline{J}_{5,25} \overline{J}_{10,25}}{J_{25}^4} \end{aligned}$$

and

$$\begin{aligned} -\frac{J_{100}^8}{J_{5,100}^2J_{15,100}^2J_{35,100}^2J_{45,100}^2} =-\frac{\overline{J}_{5,25}^2\overline{J}_{10,25}^2}{J_{25}^4}, \end{aligned}$$

we find that (6.24) is equivalent to

$$\begin{aligned} -J_{5,25}^2J_{10,25}^2=q^5\overline{J}_{0,25}^2\overline{J}_{5,25} \overline{J}_{10,25}-\overline{J}_{5,25}^2\overline{J}_{10,25}^2. \end{aligned}$$

This follows from (6.8). So, we complete the proof of (6.16). $\square$

Sketch of the proof of Lemma 4.3

Expanding the right side of (4.6) and comparing both sides according to the powers of $q$ modulo $5$, we find that it suffices to prove the five identities,

$$\begin{aligned}&\frac{J_{10,100}^2J_{15,100}^2J_{20,100}J_{30,100}J_{35,100}^2J_{40,100}J_{50,100}^2}{2J_{100}^9}\nonumber \\&\quad =\left\{ \frac{J_{30,100}J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{10,100}J_{15,100}^3J_{25,100}J_{35,100}^3J_{40,100}^2J_{45,100}^3}\right. \nonumber \\&\qquad \left. - 2q^5\frac{J_{10,100}J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{15,100}^2J_{20,100}^2J_{25,100}^3J_{30,100}J_{35,100}^2J_{45,100}^3}\right\} \nonumber \\&\qquad \times \frac{J_{15,100}J_{20,100}J_{35,100}J_{50,100}}{J_{10,100}J_{25,100}J_{40,100}} -q^3J_{25,100}\nonumber \\&\qquad \times \left\{ \frac{2q^7J_{10,100}^2J_{40,100}J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{15,100}^2J_{20,100}^3J_{25,100}^3J_{30,100}^2J_{35,100}^2J_{45,100}^3}\right. \nonumber \\&\qquad \left. +\frac{J_{10,100}^2J_{15,100}^2J_{20,100}J_{30,100}J_{35,100}^2J_{40,100}J_{50,100}^{2}}{2q^3J_{25,100}J_{100}^9}\right\} \nonumber \\&\qquad -\frac{qJ_{5,100}J_{40,100}J_{45,100}J_{50,100}}{J_{20,100}J_{25,100}J_{30,100}} \nonumber \\&\qquad \times \frac{q^4J_{10,100}J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{15,100}^3J_{20,100}^2J_{25,100}J_{30,100}J_{35,100}^3J_{45,100}^3}, \end{aligned}$$

(6.25)

$$\begin{aligned}&-\frac{qJ_{5,100}J_{10,100}^2J_{15,100}J_{30,100}^2J_{35,100}J_{40,100}^2J_{45,100}J_{50,100}}{J_{100}^9}\nonumber \\&\quad =\frac{J_{15,100}J_{20,100}J_{35,100}J_{50,100}}{J_{10,100}J_{25,100}J_{40,100}} \times \frac{2q^6J_{50,100}J_{100}^{15}}{J_{5,100}^2J_{15,100}^3J_{20,100}J_{25,100}^3J_{35,100}^3J_{40,100}J_{45,100}^2}\nonumber \\&\qquad -\left\{ \frac{J_{30,100}J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{10,100}J_{15,100}^3J_{25,100}J_{35,100}^3J_{40,100}^2J_{45,100}^3}\nonumber \right. \\&\qquad \left. - \frac{2q^5J_{10,100}J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{15,100}^2J_{20,100}^2J_{25,100}^3J_{30,100}J_{35,100}^2J_{45,100}^3}\right\} \nonumber \\&\qquad \times \frac{qJ_{5,100}J_{40,100}J_{45,100}J_{50,100}}{J_{20,100}J_{25,100}J_{30,100}} -q^3J_{25,100}\nonumber \\&\qquad \times \frac{q^3J_{30,100}J_{50,100}J_{100}^{15}}{J_{5,100}^2J_{10,100}J_{15,100}^4J_{25,100}J_{35,100}^4J_{40,100}^2J_{45,100}^2}, \end{aligned}$$

(6.26)

$$\begin{aligned}&\frac{J_{10,100}^3J_{15,100}J_{25,100}^2J_{35,100}J_{40,100}^3J_{45,100}^3J_{50,100}^2}{2q^3J_{20,100}J_{100}^9}\nonumber \\&\quad =\left\{ \frac{2q^7J_{10,100}^2J_{40,100}J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{15,100}^2J_{20,100}^3J_{25,100}^3J_{30,100}^2J_{35,100}^2J_{45,100}^3}\nonumber \right. \\&\qquad \left. +\frac{J_{10,100}^2J_{15,100}^2J_{20,100}J_{30,100}J_{35,100}^2J_{40,100}J_{50,100}^{2}}{2q^3J_{25,100}J_{100}^9}\right\} \nonumber \\&\qquad \times \frac{J_{15,100}J_{20,100}J_{35,100}J_{50,100}}{J_{10,100}J_{25,100}J_{40,100}} -q^3J_{25,100}\nonumber \\&\qquad \times \frac{q^4J_{10,100}J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{15,100}^3J_{20,100}^2J_{25,100}J_{30,100}J_{35,100}^3J_{45,100}^3}\nonumber \\&\qquad -\frac{qJ_{5,100}J_{40,100}J_{45,100}J_{50,100}}{J_{20,100}J_{25,100}J_{30,100}} \!\times \!\frac{2q^6J_{50,100}J_{100}^{15}}{J_{5,100}^2J_{15,100}^3J_{20,100}J_{25,100}^3J_{35,100}^3J_{40,100}J_{45,100}^2},\nonumber \\ \end{aligned}$$

(6.27)

$$\begin{aligned}&-\frac{J_{5,100}J_{20,100}^3J_{25,100}^2J_{30,100}^3J_{45,100}J_{50,100}^2}{2q^2J_{40,100}J_{100}^9}\nonumber \\&\qquad =\frac{J_{15,100}J_{20,100}J_{35,100}J_{50,100}}{J_{10,100}J_{25,100}J_{40,100}} \times \frac{q^3J_{30,100}J_{50,100}J_{100}^{15}}{J_{5,100}^2J_{10,100}J_{15,100}^4J_{25,100}J_{35,100}^4J_{40,100}^2J_{45,100}^2}\nonumber \\&\qquad -\left\{ \frac{2q^7J_{10,100}^2J_{40,100}J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{15,100}^2J_{20,100}^3J_{25,100}^3J_{30,100}^2J_{35,100}^2J_{45,100}^3}\nonumber \right. \\&\qquad \left. +\frac{J_{10,100}^2J_{15,100}^2J_{20,100}J_{30,100}J_{35,100}^2J_{40,100}J_{50,100}^{2}}{2q^3J_{25,100}J_{100}^9}\right\} \nonumber \\&\qquad \times q\frac{J_{5,100}J_{40,100}J_{45,100}J_{50,100}}{J_{20,100}J_{25,100}J_{30,100}} -q^3J_{25,100}\nonumber \\&\qquad \times \left\{ \frac{J_{30,100}J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{10,100}J_{15,100}^3J_{25,100}J_{35,100}^3J_{40,100}^2J_{45,100}^3}\nonumber \right. \\&\qquad \left. - \frac{2q^5J_{10,100}J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{15,100}^2J_{20,100}^2J_{25,100}^3J_{30,100}J_{35,100}^2J_{45,100}^3}\right\} , \end{aligned}$$

(6.28)

$$\begin{aligned} 0&= \frac{J_{15,100}J_{20,100}J_{35,100}J_{50,100}}{J_{10,100}J_{25,100}J_{40,100}} \times \frac{q^4J_{10,100}J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{15,100}^3J_{20,100}^2J_{25,100}J_{30,100}J_{35,100}^3J_{45,100}^3}\nonumber \\&-q\frac{J_{5,100}J_{40,100}J_{45,100}J_{50,100}}{J_{20,100}J_{25,100}J_{30,100}} \times \frac{q^3J_{30,100}J_{50,100}J_{100}^{15}}{J_{5,100}^2J_{10,100}J_{15,100}^4J_{25,100}J_{35,100}^4J_{40,100}^2J_{45,100}^2}\nonumber \\&-q^3J_{25,100}\times \frac{2q^6J_{50,100}J_{100}^{15}}{J_{5,100}^2J_{15,100}^3J_{20,100}J_{25,100}^3J_{35,100}^3J_{40,100}J_{45,100}^2}. \end{aligned}$$

(6.29)

The proof of the above five identities are quite similar to each other. We give the details of the proof of the first identity and omit the rest. After simplifying, we find that, to prove (6.25), it suffices to show that

$$\begin{aligned}&\frac{J_{10,100}^2J_{15,100}^2J_{20,100}J_{30,100}J_{35,100}^2J_{40,100}J_{50,100}^2}{J_{100}^9}\nonumber \\&\quad =\left\{ \frac{J_{20,100}J_{30,100}J_{50,100}^2J_{100}^{15}}{J_{5,100}^3J_{10,100}^2J_{15,100}^2J_{25,100}^2J_{35,100}^2J_{40,100}^3J_{45,100}^3}\nonumber \right. \\&\qquad \left. -\frac{2q^5J_{50,100}^2J_{100}^{15}}{J_{5,100}^3J_{15,100}J_{20,100}J_{25,100}^4J_{30,100}J_{35,100}J_{40,100}J_{45,100}^3}\right\} \nonumber \\&\qquad -\left\{ \frac{q^5J_{10,100}J_{40,100}J_{50,100}^2J_{100}^{15}}{J_{5,100}^2J_{15,100}^3J_{20,100}^3J_{25,100}^2J_{30,100}^2J_{35,100}^3J_{45,100}^2}\nonumber \right. \\&\qquad \left. +\frac{2q^{10}J_{10,100}^2J_{40,100}J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{15,100}^2J_{20,100}^3J_{25,100}^2J_{30,100}^2J_{35,100}^2J_{45,100}^3}\right\} . \end{aligned}$$

(6.30)

Replacing $q$ by $q^{1/2}$ in (6.6), we have

$$\begin{aligned} \frac{J_{10,50}}{J_{5,50}^2J_{20,50}^2}-\frac{J_{20,50}}{J_{10,50}^2J_{15,50}^2} =\frac{2q^{5}}{J_{15,50}J_{20,50}J_{25,50}}. \end{aligned}$$

Noting that $\frac{J_{k,50}}{J_{50}}=\frac{J_{k,100}J_{50-k,100}}{J_{100}^2}$ and rearranging, the above equation gives

$$\begin{aligned} \frac{J_{10,100}J_{40,100}}{J_{5,100}^2J_{20,100}^2J_{30,100}^2J_{45,100}^2}&= \frac{J_{20,100}J_{30,100}}{J_{10,100}^2J_{15,100}^2J_{35,100}^2J_{40,100}^2}\nonumber \\&+\frac{2q^5}{J_{15,100}J_{20,100}J_{25,100}^2J_{30,100}J_{35,100}}. \end{aligned}$$

(6.31)

Multiplying by $\frac{J_{50,100}^2J_{100}^{15}}{J_{5,100}^3J_{25,100}^2J_{40,100}J_{45,100}^3}$ throughout and subtracting

$$\begin{aligned} \frac{4q^5J_{50,100}^2J_{100}^{15}}{J_{5,100}^3J_{15,100}J_{20,100}J_{25,100}^4J_{30,100}J_{35,100}J_{40,100}J_{45,100}^3} \end{aligned}$$

on both sides, (6.31) becomes

$$\begin{aligned}&\frac{J_{20,100}J_{30,100}J_{50,100}^2J_{100}^{15}}{J_{5,100}^3J_{10,100}^2J_{15,100}^2J_{25,100}^2J_{35,100}^2J_{40,100}^3J_{45,100}^3}\nonumber \\&\qquad -\frac{2q^5J_{50,100}^2J_{100}^{15}}{J_{5,100}^3J_{15,100}J_{20,100}J_{25,100}^4J_{30,100}J_{35,100}J_{40,100}J_{45,100}^3}\nonumber \\&\quad =\frac{J_{10,100}J_{50,100}^2J_{100}^{15}}{J_{5,100}^5J_{20,100}^2J_{25,100}^2J_{30,100}^2J_{45,100}^5}\nonumber \\&\qquad -\frac{4q^5J_{50,100}^2J_{100}^{15}}{J_{5,100}^3J_{15,100}J_{20,100}J_{25,100}^4J_{30,100}J_{35,100}J_{40,100}J_{45,100}^3}. \end{aligned}$$

(6.32)

Next, multiplying by $\frac{q^5J_{10,100}J_{40,100}J_{50,100}^2J_{100}^{15}}{J_{5,100}^2J_{15,100}J_{20,100}^3J_{25,100}^2J_{30,100}^3J_{35,100}J_{45,100}^2}$ on the both sides of (6.5) and rearranging, we get

$$\begin{aligned}&\frac{q^5J_{10,100}J_{40,100}J_{50,100}^2J_{100}^{15}}{J_{5,100}^2J_{15,100}^3J_{20,100}^3J_{25,100}^2J_{30,100}^2J_{35,100}^3J_{45,100}^2}\nonumber \\&\qquad +\frac{2q^{10}J_{10,100}^2J_{40,100}J_{50,100}J_{100}^{15}}{J_{5,100}^3J_{15,100}^2J_{20,100}^3J_{25,100}^2J_{30,100}^2J_{35,100}^2J_{45,100}^3}\nonumber \\&\quad =\frac{q^5J_{10,100}^2J_{40,100}J_{50,100}^2J_{100}^{15}}{J_{5,100}^4J_{15,100}J_{20,100}^3J_{25,100}^2J_{30,100}^3J_{35,100}J_{45,100}^4}. \end{aligned}$$

(6.33)

Substituting (6.32) and (6.33) into (6.30), we find that (6.30) is equivalent to

$$\begin{aligned}&\frac{J_{10,100}^2J_{15,100}^2J_{20,100}J_{30,100}J_{35,100}^2J_{40,100}J_{50,100}^2}{J_{100}^9}\nonumber \\&\quad =\frac{J_{10,100}J_{50,100}^2J_{100}^{15}}{J_{5,100}^5J_{20,100}^2J_{25,100}^2J_{30,100}^2J_{45,100}^5}\nonumber \\&\qquad -\frac{4q^5J_{50,100}^2J_{100}^{15}}{J_{5,100}^3J_{15,100}J_{20,100}J_{25,100}^4J_{30,100}J_{35,100}J_{40,100}J_{45,100}^3}\nonumber \\&\qquad -\frac{q^5J_{10,100}^2J_{40,100}J_{50,100}^2J_{100}^{15}}{J_{5,100}^4J_{15,100}J_{20,100}^3J_{25,100}^2J_{30,100}^3J_{35,100}J_{45,100}^4}. \end{aligned}$$

(6.34)

This follows from the two identities,

$$\begin{aligned}&\frac{J_{50,100}^2J_{100}^{15}}{J_{5,100}^4J_{15,100}^2J_{20,100}J_{30,100}J_{35,100}^2J_{40,100}J_{45,100}^4}\nonumber \\&\qquad -\frac{4q^5J_{50,100}^2J_{100}^{15}}{J_{5,100}^3J_{15,100}J_{20,100}J_{25,100}^4J_{30,100}J_{35,100}J_{40,100}J_{45,100}^3}\nonumber \\&\quad =\frac{J_{10,100}^2J_{15,100}^2J_{20,100}J_{30,100}J_{35,100}^2J_{40,100}J_{50,100}^2}{J_{100}^9} \end{aligned}$$

(6.35)

and

$$\begin{aligned}&\frac{J_{10,100}J_{50,100}^2J_{100}^{15}}{J_{5,100}^5J_{20,100}^2J_{25,100}^2J_{30,100}^2J_{45,100}^5} -\frac{q^5J_{10,100}^2J_{40,100}J_{50,100}^2J_{100}^{15}}{J_{5,100}^4J_{15,100}J_{20,100}^3J_{25,100}^2J_{30,100}^3J_{35,100}J_{45,100}^4}\nonumber \\&\quad =\frac{J_{50,100}^2J_{100}^{15}}{J_{5,100}^4J_{15,100}^2J_{20,100}J_{30,100}J_{35,100}^2J_{40,100}J_{45,100}^4}. \end{aligned}$$

(6.36)

We obtain (6.35) after multiplying by $\frac{J_{15,100}J_{35,100}J_{50,100}^2J_{100}^{15}}{q^5J_{5,100}J_{30,100}J_{45,100}}$ on both sides of (6.24).

Next, we prove (6.36). Noting that $\frac{J_{k,50}}{J_{50}}=\frac{J_{k,100}J_{50-k,100}}{J_{100}^2},$ we find that (6.4) is equivalent to

$$\begin{aligned}&J_{15,100}J_{20,100}J_{30,100}J_{35,100}- q^5J_{5,100}J_{10,100}J_{40,100}J_{45,100}\nonumber \\&\quad =\frac{J_{5,100}J_{20,100}^2J_{25,100}^2J_{30,100}^2J_{45,100}}{J_{10,100}J_{15,100}J_{35,100}J_{40,100}}. \end{aligned}$$

(6.37)

Multiplying by $\frac{J_{10,100}J_{50,100}^2J_{100}^{15}}{J_{5,100}^5J_{15,100}J_{20,100}^3J_{25,100}^2J_{30,100}^3J_{35,100}J_{45,100}^5}$ on both sides of the above equation, we recover (6.36). This completes our proof of (6.25). $\square$