Trigonometry related to Rogers–Ramanujan identities

Question

For integers $n\ge2$ and $k\ge2$, fix the notation $$ [m]=\sin\frac{\pi m}{nk+1} \quad\text{and}\quad [m]!=[1][2]\dots[m], \qquad m\in\mathbb Z_{>0}. $$ Consider the following trigonometric numbers: $$ a_i=\frac{[i+k-2]![n-i+k-2]!}{[k-2]![n+k-2]!}, \qquad i=1,2,\dots,n-1. $$ Is it possible for any $n$ to express the quantities $$ A_j=1-\frac{[k-1]\cdot [n+k-1]}{[j+k-1]\cdot [n-j+k-1]}, \qquad j=1,2,\dots,n-1, $$ as a product/quotient of terms of the form $(1-\text{product of some }a_i)$? If not (for $n\ge4$), is it possible to prove that?

The affirmative answer is known for $n=2$ and $n=3$. Namely, if $n=2$ so that we have only one $a_1$ and one $A_1$, then $$ A_1=1-a_1. $$ If $n=3$, then $$ A_j=(1-a_j)(1-a_1a_2) \quad\text{for } j=1,2. $$ (The last formula is a nice trigonometric identity, by the way.)

My question is motivated (in a very sophisticated way) by a recent question on Rogers–Ramanujan identities. The latter one reminded me about the problem of possible $\mathfrak{sl}_n$ generalizations of RRs in their classical form "a $q$-sum"="a $q$-product". Only the cases $n=2$ and $n=3$ are known; these are the Andrews–Gordon identities and the Andrews–Schilling–Warnaar identities (see [S. Ole Warnaar, Adv. in Math. 200 (2006) 403--434]). An indirect implication of such identities is the family of (highly nontrivial) numerical identities for the dilogarithm function; these come as the limit $q\to1$ specialisation and some multivariate asymptotics. The trigonometric identities above come into play from these considerations for $n=2$ and $n=3$; any answer for $n>3$ can shed some light on the existence of new RRs.

Answer 1

This is not a complete answer, but a strong computational evidence for $n=4$, $k=3$, $j=1$ being a counterexample even in the case of products/quotients of the form $1\pm a_{j_1}\cdots a_{j_k}$ for not necessarily distinct indices $j_1,\dots,j_k$, suggested by OP in the comments.

First, note that numbers $[m]\cdot I$, $a_i$, $A_j$ are naturally represented by elements of the cyclotomic field $\mathbb{Q}[\zeta]$, where $\zeta$ is a primitive $2(nk+1)$-th power root of unity. Namely, we have $[m] = (\zeta^m - \zeta^{-m})/(2I)$, while the imaginary unit $I$ cancels out in $a_i$ and $A_j$.

The element $A_1$ is an integer unit in $\mathbb{Q}[\zeta]$. The unit group here has 6 generators $(g_1, g_2, g_3, g_4, g_5, g_6) := (\zeta, -\zeta + 1, \zeta^2 + 1, \zeta^{10} - \zeta^5 + 1, \zeta^2 - \zeta + 1, \zeta^7 + \zeta)$, and their exponents in $A_1$ are $(19, 1, 1, 0, -1, 0)$. Note that the the order of $g_1$ is 26 (even!).

I've computed a basis of the units formed by products/quotients of numbers $1\pm a_1^u a_2^v a_3^w$ with $u+v+w\leq 30$. The basis contains four elements with the following exponents: $(0, 0, -1, -1, 1, -2)$, $(2, 0, 1, 1, 1, 1)$, $(-2, 0, 0, 0, 2, 0)$, $(0, 0, -2, -2, 0, 3)$. Since the exponents of $g_1$ are all even, they cannot deliver $A_1$, where the corresponding exponent is odd. Another reason why it's impossible is that the exponents of $g_2$ in the basis are all zero, while in $A_1$ the exponent of $g_2$ is 1.

PS. It may be well possible that for an expert it's a simple exercise to show that either of the two observations holds in general (i.e., without the $u+v+w\leq 30$ restriction), which will prove that $n=4$, $k=3$, $j=1$ is indeed a counterexample.