An identity related to the series $\sum_{n\geq 0}p(5n+4)x^n$ in Ramanujan's lost notebook

Question

While browsing through Ramanujan's original manuscript titled "The Lost Notebook" (the link is a PDF file with 379 scanned pages, so instead of a click it is preferable to download) I found this identity (numbered (4.5)) on page 139 $$x\cdot\frac{\left\{(1-x^5)(1-x^{10})(1-x^{15})\dots\right\} ^5} {(1-x)(1-x^2)(1-x^3)\dots} =\frac{x} {(1-x)^2}-\frac{x^2}{(1-x^2)^2}-\frac{x^3}{(1-x^3)^2}+\frac{x^4}{(1-x^4)^2}+\frac{x^6}{(1-x^6)^2}-\frac{x^7}{(1-x^7)^2}-\dots\tag{1}$$ which can be more compactly written as $$x\prod_{n\geq 1}\frac{(1-x^{5n})^5}{1-x^n}=\sum_{n\geq 1}\left(\frac{n}{5}\right)\frac{x^n}{(1-x^n)^2}$$ Next Ramanujan says "it follows from $(1)$ that $$\{(1-x)(1-x^2)(1-x^3)\dots \} ^5\{p(4)x+p(9)x^2+p(14)x^3+\dots\}=5\left\{\frac{x} {(1-x)^2}-\frac{x^2}{(1-x^2)^2}-\frac{x^3}{(1-x^3)^2}+\frac{x^4}{(1-x^4)^2}+\frac{x^6}{(1-x^6)^2}-\dots\right\}\tag{2}$$ and hence that $$p(4)+p(9)x+p(14)x^2+p(19)x^3+\dots=5\cdot\frac{\{(1-x^5)(1-x^{10})(1-x^{15})\dots\}^5 } {\{(1-x)(1-x^2)(1-x^3)\dots\}^6}\tag{3} $$ Here $p(n) $ denotes the number of unrestricted partitions of a positive integer $n $ and $\left(\dfrac{a}{p} \right) $ denotes the Legendre symbol.

The identity $(3)$ mentioned above is famous and is used to derive many partition congruences with $$p(5n+4)\equiv 0\pmod{5}$$ as an immediate consequence. A complete proof of $(3)$ based on different ideas is available in this answer.

It took me sometime to figure out as to how $(2)$ follows from $(1)$. The technique Ramanujan uses here is to expand both sides in powers of $x$ and then pick only the terms with $x^{5n}$ and replace $x^5$ by $x$. In this process we also make use of the fact that $$1+\sum_{n\geq 1}p(n)x^n=\frac{1}{(1-x)(1-x^2)(1-x^3)\dots}\tag{4}$$

What is really mysterious is the origin of the identity $(1)$ and I am trying to find a proof. I did check the nearby pages of the Lost Notebook and did not find any proofs or hint. However on the same page 139 of the notebook there is another identity (numbered (4.6)) $$\frac{\left\{(1-x)(1-x^2)(1-x^3)\dots\right\} ^5} {(1-x^5)(1-x^{10})(1-x^{15})\dots} =1-5\left(\frac{x} {1-x}-\frac{2x^2}{1-x^2}-\frac{3x^3}{1-x^3}+\frac{4x^4}{1-x^4}+\frac{6x^6}{1-x^6}-\frac{7x^7}{1-x^7}-\dots\right) \tag{5}$$ The identities $(1),(5)$ appear intimately tied to each other and seem to come out of nowhere.

Any proofs or references regarding proofs of the identities $(1),(5)$ are desired.

Answer 1

Let us use the more standard variable $q$ instead of $x$. We start with the Rogers-Ramanujan continued fraction $$R(q) = \cfrac{q^{1/5}}{1 + \cfrac{q}{1 + \cfrac{q^{2}}{1 + \cfrac{q^{3}}{1 + \cdots}}}} = \frac{q^{1/5}}{1+}\frac{q}{1+}\frac{q^2}{1+}\frac{q^3}{1+}\cdots$$ which satisfies $$R(q) =q^{1/5}\prod_{n\geq 1}\frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}\tag{1}$$ (for a proof see this blog post). Taking logarithmic derivative with respect to $q$ we arrive at $$5q\frac{d} {dq} \log R(q) =1-5\sum_{n\geq 1}\left(\frac{n}{5}\right)\frac{nq^n}{1-q^n}$$ and the right hand side matches the right hand side of equation $(5)$ in question.

We will prove equation $(5)$ in question by showing that $$5q\frac{d} {dq} \log R(q) =\frac{\eta^5(q)}{\eta(q^5)}\tag{2}$$ where Dedekind eta function is defined by $$\eta(q) =q^{1/24}\prod_{n\geq 1}(1-q^n)$$ In fact we will show $(2)$ with $q$ replaced by $q^2$ ie $$5q\frac{d}{dq}\log R(q^2)=2\frac{\eta^5(q^2)}{\eta(q^{10})}\tag{3}$$ We will utilize modular equations of degree $5$ here and in that context let $k, l$ be elliptic moduli corresponding to nomes $q, q^5$ and $K, L$ denote the complete elliptic integrals of first kind with respect to these moduli. Also $k', l'$ are complementary moduli and $m=K/L$ is Ramanujan multiplier. The Dedekind eta function can be expressed in terms of elliptic integrals and moduli as $$\eta(q^2)=2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}\tag{4}$$ The right hand side of $(3)$ then equals $$2\cdot\frac{2^{-5/3}(2K/\pi)^{5/2}(kk')^{5/6}}{2^{-1/3}(2L/\pi)^{1/2}(ll')^{1/6}}=2^{-1/3}(2K/\pi)^2\sqrt{m}\frac{(kk') ^{5/6}}{(ll')^{1/6}}\tag{5}$$ On the other hand we have $$\frac{1}{R^5(q^2)}-11-R^5(q^2)=\frac{\eta^6(q^2)}{\eta^6(q^{10})}$$ (for a proof see my blog post 1 and post 2) and therefore solving a quadratic equation we get $$R^5(q^2)=\frac{\sqrt{(11+A(q))^2+4}-(11+A(q))}{2}\tag{6}$$ where $A(q) =\eta^6(q^2)/\eta^6(q^{10})$. Logarithmic differentiation of the above equation gives us $$5q\frac{d}{dq}\log R(q^2)=\dfrac{\dfrac{11+A(q)}{\sqrt {(11+A(q))^2+4}}-1}{\sqrt{(11+A(q))^2+4}-(11+A(q))}\cdot q\frac{dA}{dq}=-\frac{A(q)} {\sqrt{(11+A(q))^2+4}} \cdot q\frac{d} {dq} \log A(q) $$ Next we observe that $$q\frac{d} {dq} \log A(q) =6q\frac{d}{dq}\log\eta(q^2)-6q\frac{d}{dq}\log \eta(q^{10})\tag{7}$$ Using the definition for another Ramanujan function $$P(q) =24q\frac{d}{dq}\log \eta(q) =1-24\sum_{n\geq 1}\frac{nq^n}{1-q^n}$$ we can see that the expression in $(7)$ equals $$\frac{P(q^2)-5P(q^{10})}{2}$$ and hence the left hand side of equation $(3)$ is $$\frac{A(q)(5P(q^{10})-P(q^2))} {2\sqrt{(11+A(q))^2+4}}\tag{8}$$ Now we know that $$A(q) =\frac{2^{-2}(2K/\pi)^3(kk')}{2^{-2}(2L/\pi)^3(ll')} =m^3\cdot\frac{kk'}{ll'}$$ and Ramanujan proved that $$5P(q^{10})-P(q^2)=\frac{4KL}{\pi^2}(3+kl+k'l')\sqrt{\frac{1+kl+k'l'}{2}}$$ and hence using expressions in $(5)$ and $(8)$ we have to prove that $$\frac{m^3(3+kl+k'l')^2(1+kl+k'l')}{(11+A(q))^2+4}=2^{7/3}\cdot\frac{(ll')^{5/3}}{(kk')^{1/3}}\tag{9}$$ Using modular equations of degree $5$ from my blog post we have \begin{align} \frac{m-1}{2^{4/3}}&=\frac{(ll') ^{5/12}}{(kk')^{1/12}}\tag{10a}\\ (k^3l)^{1/4}&=\frac{\rho+3m-5} {4m} \tag{10b}\\ (kl^3)^{1/4}&=\frac{\rho+m^2-3m}{4m}\tag{10c}\\ \frac{(ll') ^{1/2}}{(kk')^{1/2}}&=\frac{m(m-1)} {5-m}\tag{10d}\\ \rho&=\sqrt{m^3-2m^2+5m}\tag{10e} \end{align} Using $(10b),(10c)$ we get $$kl=\frac{\rho+3m-5}{4m}\cdot\frac{\rho+m^2-3m}{4m}$$ Replacing $k$ with $l'$, $l$ with $k'$ and $m$ with $5/m$ we see that $\rho$ is replaced by $5\rho/m^2$ and $$k'l'=\frac{\rho+3m-m^2} {4m} \cdot\frac{\rho+5-3m}{4m}$$ and hence $$kl+k'l'=\frac{\rho^2+(m^2-3m)(3m-5)} {8m^2}=\frac{m^2-4m+5}{2m}$$ and then $$(3+kl+k'l')^2(1+kl+k'l')=\frac{(m^2+2m+5)^2(m^2-2m+5)}{8m^3}$$ Next using $(10d)$ we have $$11+A(q) =11+m^3\frac{kk'}{ll'}=11+\frac{m(5-m)^2}{(m-1)^2}=\frac{m^3+m^2+3m+11}{(m-1)^2}$$ and $$(11+A(q))^2+4=\frac{(m^2+2m+5)^2(m^2-2m+5)}{(m-1)^4}$$ (verifying this needs a bit of patience, one can clear the denominator $(m-1)^4$ and check the equality of two polynomials of degree $6$ by verifying that their values are same for at least any $7$ values, say $0,\pm 1,\pm 2,\pm 3$, of the variable) and hence the left hand side of $(9)$ equals $(m-1)^4/8$. Using $(10a)$ we see that right hand side of $(9)$ also equals $(m-1)^4/8$. Thus the equation $(9)$ is true and our job is done.

Note that the above also constitutes a proof for the derivative of Rogers-Ramanujan continued fraction $$R'(q) =\frac{R(q)} {5q}\cdot\frac{\eta^5(q)}{\eta(q^5)}$$ (rearranging equation $(2)$ here). The formula above and equivalent formulations are provided in this answer without proof.

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I found another interesting proof of identity $(5)$ in question in a paper titled A unified proof of Ramanujan's classical identities and Jacobi's four square theorem by Bhaskar Srivastava which appeared in Vietnam Journal of Mathematics Vol 38:3 (2010), pages 299-304.

This is based on Fourier series of elliptic functions and identities related to theta functions and is more in line with the methods of Ramanujan. We start with the identity $$\left(\frac {2K}{\pi}\right)^2\text{cs}^2\left(\frac{2Kx}{\pi},k\right)+\frac{4KE}{\pi^2}=\operatorname{cosec}^2x-8\sum_{n\geq 1}\frac{nq^{2n}}{1-q^{2n}}\cos 2nx\tag{11}$$ which is proved (as equation $(11)$) in this answer of mine.

To get rid of the term $4KE/\pi^2$ we subtract two instances of the above equation to get $$\left(\frac {2K}{\pi}\right)^2\left(\text{cs}^2\left(\frac{2Kx}{\pi},k\right)- \text{cs}^2\left(\frac{2Ky}{\pi},k\right)\right) =\operatorname{cosec} ^2x-\operatorname{cosec}^2y-8\sum_{n\geq 1}\frac{nq^{2n}}{1-q^{2n}} (\cos 2nx - \cos 2ny)$$ Rewriting the left hand side in terms of Jacobi theta functions we get $$\vartheta_3(q)^2\vartheta_4^2(q)\left(\frac{\vartheta_2^2(x,q)}{\vartheta_1^2(x,q)}-\frac{\vartheta_2^2(y,q)}{\vartheta_1^2(y,q)}\right)=\operatorname{cosec} ^2x-\operatorname{cosec}^2y-8\sum_{n\geq 1}\frac{nq^{2n}}{1-q^{2n}} (\cos 2nx - \cos 2ny)$$ Using the well known theta function identity $$\vartheta_1^2(x,q)\vartheta_2^2(y,q) - \vartheta_1^2(y,q)\vartheta_2^2(x,q)= \vartheta_2^2(q)\vartheta_1(x-y,q)\vartheta_1(x+y,q)\tag{12}$$ we now get our desired identity $$\vartheta_2^2(q)\vartheta_3^2(q)\vartheta_4^2(q) \cdot \frac{\vartheta_1(y-x,q) \vartheta_1(x+y,q)} {\vartheta_1^2(x,q)\vartheta_1^2(y,q)} = \operatorname{cosec} ^2x-\operatorname{cosec}^2y-8\sum_{n\geq 1}\frac{nq^{2n}}{1-q^{2n}} (\cos 2nx - \cos 2ny)\tag{13}$$ We also need here the product representations $$\vartheta_1(x,q)=2q^{1/4}\sin x\prod_{n\geq 1}(1-q^{2n})(1-e^{2ix}q^{2n}) (1-e^{-2ix}q^{2n})\tag{14}$$ and $$\vartheta_2^2(q)\vartheta_3^2(q)\vartheta_4^2(q)=4\sqrt{q}\prod_{n\geq 1}(1-q^{2n})^6\tag{15}$$ We now put $x=\pi/5,y=2\pi/5$ in equation $(13)$ and then the left hand side becomes $$4\sqrt{q}\prod_{n\geq 1}(1-q^{2n})^6\frac{\vartheta_1(\pi/5,q) \vartheta_1(3\pi/5,q)}{ \vartheta_1^2(\pi/5,q) \vartheta_1^2(2\pi/5,q)}$$ We have $$\sin(\pi/5)\sin(2\pi/5)=\frac{\sqrt{5}}{4}$$ and $$ \vartheta_1(3\pi/5,q)= \vartheta_1(2\pi/5,q)$$ and using $(14)$ $$\vartheta_1(\pi/5,q)\vartheta_1(2\pi/5,q)= 4\sqrt{q}\prod_{n\geq 1}(1-q^{2n})(1-q^{2n})(1-\zeta q^{2n}) (1-\zeta^2q^{2n}) (1-\zeta^3 q^{2n}) (1-\zeta^4q^{2n})=4\sqrt{q}\prod_{n\geq 1}(1-q^{2n})(1-q^{10n})$$ where $\zeta=e^{2\pi i/5}$. Thus we can see that the left hand side of $(13)$ simplifies to $$\frac{4}{\sqrt{5}}\prod_{n\geq 1}\frac{(1-q^{2n})^{5}}{1-q^{10n}}$$ On the other hand we observe that $$ \operatorname{cosec} ^2(\pi/5)-\operatorname{cosec}^2(2\pi/5)=\frac{4}{\sqrt{5}}$$ and $$\cos (2n\pi/5)-\cos(4n\pi/5)=\left(\frac{n}{5}\right)\frac{\sqrt{5}}{2}$$ and then the right hand side of $(13)$ becomes $$\frac{4}{\sqrt{5}}-4\sqrt{5}\sum_{n\geq 1}\left(\frac{n}{5}\right)\frac{nq^{2n}}{1-q^{2n}}$$ Dividing equation $(13)$ by $4/\sqrt{5}$ and replacing $q^2$ by $q$ we get identity $(5)$ in question.

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Finally I also managed to find the proof of identity $(1)$ in question in a paper titled On the additive formulae of the theta functions and a collection of Lambert series pertaining to the modular equations of degree 5 by L.-C. Shen which appeared in the Transactions of the American Mathematical Society, Vol 345, Number 1, 1994, pages 323-345.

In this paper Shen discusses the technique used in Bhaskar Srivastava's paper referred above and uses it to prove identity $(1)$ of the question.

The relevant Fourier series needed is $$\left(\frac{2K}{\pi}\right)^2\text{dn}^2\left(\frac{2Kx}{\pi},k\right)=\frac{4KE}{\pi^2}+8\sum_{n\geq 1}\frac{nq^n}{1-q^{2n}}\cos 2nx\tag{16}$$ which is proved as equation $(7)$ in this answer of mine.

Using two instances of this identity we get $$\left(\frac{2K}{\pi}\right)^2\left(\text{dn}^2\left(\frac{2Kx}{\pi},k\right) - \text{dn}^2\left(\frac{2Ky}{\pi},k\right)\right) =8\sum_{n\geq 1}\frac{nq^n}{1-q^{2n}}(\cos 2nx-\cos 2ny)$$ Switching to theta functions on left hand side we get $$\vartheta_3^2(q) \vartheta_4^2(q)\left(\frac{ \vartheta_3^2(x, q) } { \vartheta_4^2(x, q) } - \frac{ \vartheta_3^2(y, q) } { \vartheta_4^2(y, q) } \right) =8\sum_{n\geq 1}\frac{nq^n}{1-q^{2n}}(\cos 2nx-\cos 2ny)$$ And using the theta function identity $$\vartheta_3^2(x,q)\vartheta_4^2(y,q)-\vartheta_3^2(y,q)\vartheta_4^2(x,q)=\vartheta_2^2(q)\vartheta_1(y-x,q)\vartheta_1(x+y,q)\tag{17}$$ we finally get $$\vartheta_2^2(q) \vartheta_3^2(q) \vartheta_4^2(q)\cdot\frac{\vartheta_1(y-x,q)\vartheta_1(x+y,q)}{\vartheta_4^2(x,q)\vartheta_4^2(y,q)}=8\sum_{n\geq 1}\frac{nq^n}{1-q^{2n}} (\cos 2nx - \cos 2ny)\tag{18}$$ Next we replace $q$ with $q^{5/2}$ in above equation and put $x=3\pi\tau/4,y=\pi\tau/4$ where $q=e^{\pi i\tau} $. The right hand side of $(18)$ now equals $$4\sum_{n\geq 1}\frac{nq^{5n/2}}{1-q^{5n}}(e^{3\pi in\tau/2} + e^{-3\pi in\tau/2}-e^{\pi in\tau/2}-e^{-\pi in\tau/2})=\sum_{n\geq 1}\frac{n(q^n-q^{2n}-q^{3n}+q^{4n})}{1-q^{5n}}=4\sum_{n\geq 1}\sum_{j\geq 1}\left(\frac{j}{5}\right)nq^{jn} $$ Interchanging the order of summation the above sum becomes $$4\sum_{j\geq 1}\left(\frac{j}{5}\right)\sum_{n\geq 1}nq^{jn}=4\sum_{n\geq 1}\left(\frac{n}{5}\right)\frac{q^n}{(1-q^n)^2}$$ The left hand side of $(18)$ is $$\vartheta_2^2(q^{5/2}) \vartheta_3^2(q^{5/2}) \vartheta_4^2(q^{5/2})\cdot\frac{\vartheta_1(-\pi\tau/2,q^{5/2})\vartheta_1(\pi\tau,q^{5/2})}{\vartheta_4^2(3\pi\tau/4,q^{5/2})\vartheta_4^2(\pi\tau/4,q^{5/2})}$$ Using the product representation $$\vartheta_4(x,q)=\prod_{n\geq 1}(1-q^{2n})(1-e^{2ix}q^{2n-1})(1-e^{-2ix}q^{2n-1})\tag{19}$$ we can see that the left hand side of equation $(18)$ becomes (after some effort) $$4q\prod_{n\geq 1}\frac{(1-q^{5n})^5}{1-q^n}$$ and thus the desired identity $(1)$ in question is established.

Answer 2

Not entirely an answer yet

Theorem For $|q|<|ab|$, $$\frac{q(a^2,b^2,q^2/a^2,q^2/b^2;q^2)_{\infty}(q^2;q^2)_{\infty}^4}{ab(qab,qb/a,qa/b,q/ab;q^2)_{\infty}^2}=-\sum_{n=1}^{\infty}n\frac{(a^n-a^{-n})(b^n-b^{-n})}{q^n-q^{-n}} \tag{1}$$

The desired identity follows from setting $(q,a,b)\mapsto(q^5,\sqrt{q},q)$

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I saw this from "Ramanujan theta functions" from Shaun Cooper, Theorem 1.18. The proof in the book is very complicated (I haven't read it yet). Though $(1)$ seems reasonable to prove by setting $f(a,b)$ as the left-hand side and showing a functional formula. Expanding both sides as a double Laurent series and comparing coefficients

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Edit 16/3/2025 I will give proof $(1)$ by elementary means. Consider the following identity:

For $|q|<|x|,|y|,|z|$ and $|xyz|<\frac{1}{|q|}$ $$\frac{(qxy,qxz,qyz,q/xy,q/yz,q/xz,q,q;q)_{\infty}}{(qx,qy,qz,qxyz,q/x,q/y,q/z,q/xyz;q)_{\infty}}=1+A(x,y,z)\sum_{n=1}^{\infty}\frac{q^n}{1-q^n}\left(x^n+y^n+z^n-(xyz)^n-x^{-n}-y^{-n}-z^{-n}+(xyz)^{-n}\right)\tag{2}$$ Where $A(x,y,z)=\frac{(1-x)(1-y)(1-z)(1-xyz)}{(1-xy)(1-xz)(1-yz)}$.

Using $$\sum_{n=1}^{\infty}\frac{x^n q^n}{1-q^n}=\sum_{n=1}^{\infty}\frac{x q^n}{1-xq^n}$$ Dividing both sides by $A$ we get $$\frac{(xy,xz,yz,q/xy,q/xz,q/yz,q;q)_{\infty}}{(x,y,z,xyz,q/x,q/y,q/z,q/xyz;q)_{\infty}}=\frac{1}{A}+\sum_{k\geq 1}\left(\frac{xq^{k-1}}{1-xq^{k-1}}-\frac{q^k/x}{1-q^k/x}+\frac{yq^{k-1}}{1-yq^{k-1}}-\frac{q^k/y}{1-q^k/y}+\frac{zq^{k-1}}{1-zq^{k-1}}-\frac{q^k/z}{1-q^k/z}-\frac{xyzq^{k-1}}{1-xyzq^{k-1}}+\frac{q^k/xyz}{1-q^k/xyz}\right)\tag{3}$$ Consider the change of variables $(x,y,z,q)\mapsto(aq/bc,bq/ac,cq/ba,q^2)$ Write $$r(x,q)=\sum_{k\geq 1}\left(\frac{x q^{2k-1}}{1-x q^{2k-1}}-\frac{q^{2k-1}/x}{1-q^{2k-1}/x}\right)$$ so we write $(3)$ as $$\frac{q}{abc}\frac{(a^2,b^2,c^2,q^2/a^2,q^2/b^2,q^2/c^2,q^2,q^2;q^2)_{\infty}}{(qabc,qbc/a,qac/b,qab/c,qa/bc,qb/ac,qc/ab,q/abc;q^2)_{\infty}}$$ $$=r(ab/c,q)+r(ac/b,q)+r(bc/a;q)-r(abc,q) \tag{4}$$ If we write $$r(x,q)=\sum_{j=1}^{\infty}\frac{q^j(x^j-x^{-j})}{1-q^{2j}}$$ so we can also write (4) as $$\frac{q}{abc}\frac{(a^2,b^2,c^2,q^2/a^2,q^2/b^2,q^2/c^2,q^2,q^2;q^2)_{\infty}}{(qabc,qbc/a,qac/b,qab/c,qa/bc,qb/ac,qc/ab,q/abc;q^2)_{\infty}}=\sum_{n=1}^{\infty}\frac{(a^n-a^{-n})(c^n-c^{-n})(b^n-b^{-n})}{(q^n-q^{-n})}\tag{5}$$ Now we divide both sides by $c-c^{-1}$ and take $c\to 1$ to obtain $(1)$. Now the problem remain is a proof of $(2)$ which I will provide below.

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A proof of $(2)$: Consider the function $f(x,y,z)$ defined as the LHS of $(2)$ $$f(x,y,z)=\frac{(qxy,qxz,qyz,q/xy,q/yz,q/xz,q,q;q)_{\infty}}{(qx,qy,qz,qxyz,q/x,q/y,q/z,q/xyz;q)_{\infty}}\tag{6}$$ Then it is easy to verify that $$f(xq,y,z)={\underbrace{\frac{(1-xy)(1-xz)(1-xq)(1-xyzq)}{(1-x)(1-xyz)(1-xyq)(1-xzq)}}_{F}}f(x,y,z)\tag{7}$$ Also the function $A$ defined before satisfies this functional equation as well. Define now the RHS as $g(x,y,z)$ $$g(x,y,z)=1+A(x,y,z)\sum_{n=1}^{\infty}\frac{q^n}{1-q^n}\left(x^n+y^n+z^n-(xyz)^n-x^{-n}-y^{-n}-z^{-n}+(xyz)^{-n}\right)$$ $$g(xq,y,z)=1+A(xq,y,z)\sum_{n=1}^{\infty}\frac{q^n}{1-q^n}\left(q^n(x^n-(xyz)^n)-q^{-n}(x^{-n}-(xyz)^{-n})+y^n+z^n-y^{-n}-z^{-n}\right)$$ $$=1+A(x,y,z)F\left\{\sum_{n=1}^{\infty}\frac{q^n}{1-q^n}\left(x^n-(xyz)^n+(xyz)^{-n}-x^{-n}+y^n-y^{-n}+z^n-z^{-n}\right)+\sum_{n=1}^{\infty}\left((xyzq)^n-(xq)^n+(xyzq)^{-n}-x^{-n}\right)\right\}$$ $$=F\left(\frac{1}{F}+A(x,y,z)\left(\frac{1}{1-xyzq}-\frac{1}{1-xq}+\frac{1}{xyz-1}-\frac{1}{x-1}\right)+A(x,y,z)\left\{\sum_{n=1}^{\infty}\frac{q^n}{1-q^n}\left(x^n-(xyz)^n+(xyz)^{-n}-x^{-n}+y^n-y^{-n}+z^n-z^{-n}\right)\right\}\right)$$ The functional formula is proved if we can show that $$\frac{1}{F}+A(x,y,z)\left(\frac{1}{1-xyzq}-\frac{1}{1-xq}+\frac{1}{xyz-1}-\frac{1}{x-1}\right)=1$$ Which is just expanding.

Now the function is symmetric in $x,y,z$ so the functional equation works for all three arguments. Verifying $f(1,y,z)=g(1,y,z)$ proves $(2)$.

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To be honest, I haven't worked out the full solution yet due to how cumbersome the calculation is (And currently I can't use WA).