The complete list of continued fractions like the Rogers-Ramanujan? I have two questions about q-continued fractions, but a little intro first. Given Ramanujan's theta function,
$$f(a,b) = \sum_{n=-\infty}^{\infty}a^{n(n+1)/2}b^{n(n-1)/2}$$
then the following,
$$A(q) = q^{1/8} \frac{f(-q,-q^3)}{f(-q^2,-q^2)}$$
$$B(q) = q^{1/5} \frac{f(-q,-q^4)}{f(-q^2,-q^3)}$$
$$C(q) = q^{1/3} \frac{f(-q,-q^5)}{f(-q^3,-q^3)}$$
$$D(q) = q^{1/2} \frac{f(-q,-q^7)}{f(-q^3,-q^5)}$$
$$E(q) = q^{1/1} \frac{f(-q,-q^{11})}{f(-q^5,-q^7)}$$
are q-continued fractions of degree $4,5,6,8,12$, respectively, namely,
$$A(q) = \cfrac{q^{1/8}}{1 + \cfrac{q}{1+q + \cfrac{q^2}{1+q^2 + \ddots}}},\;\;B(q) = \cfrac{q^{1/5}}{1 + \cfrac{q}{1 + \cfrac{q^2}{1 + \ddots}}}$$
$$C(q) = \cfrac{q^{1/3}}{1 + \cfrac{q+q^2}{1 + \cfrac{q^2+q^4}{1 + \ddots}}},\;\;\;\;D(q) = \cfrac{q^{1/2}}{1 + q +\cfrac{q^2}{1+q^3 + \cfrac{q^4}{1+q^5 + \ddots}}}$$
$$E(q) = \cfrac{q(1-q)}{1-q^3 + \cfrac{q^3(1-q^2)(1-q^4)}{(1-q^3)(1+q^6)+\cfrac{q^3(1-q^8)(1-q^{10})}{(1-q^3)(1+q^{12}) + \ddots}}}$$
The first three are by Ramanujan, the fourth is the Ramanujan-Gollnitz-Gordon cfrac, while the last is by Naika, et al (using an identity by Ramanujan). Let $q = e^{2\pi i \tau}$ where $\tau = \sqrt{-n}$ and these can be simply expressed in terms of the Dedekind eta function $\eta(\tau)$ as,
$$\tfrac{1}{A^4(q)}+16A^4(q) = \left(\tfrac{\eta(\tau/2)}{\eta(2\tau)}\right)^8+8$$
$$\tfrac{1}{B(q)}-B(q) = \left(\tfrac{\eta(\tau/5)}{\eta(5\tau)}\right)+1$$
$$\tfrac{1}{C(q)}+4C^2(q) = \left(\tfrac{\eta(\tau/3)}{\eta(3\tau)}\right)^3+3$$
$$\tfrac{1}{D(q)}-D(q) = \big(\tfrac{1}{A(q^2)}\big)^2$$
$$E(q) = \;???$$
Question 1: Does anybody know how to express $E(q)$ in terms of $\eta(\tau)$? (It's SO frustrating not to complete this list. I believe there might be a simple relationship between orders 6 and 12, just like there is between 4 and 8.) This cfrac can be found in "On Continued Fraction of Order 12", but the authors do not address this point.
Question 2: Excluding these five and the Heine cfrac which gives $\eta(\tau)/\eta(2\tau)$, are there any other q-continued fractions which yield an algebraic value at imaginary arguments?
 A: Based on Elkies' answer and an email by Michael Somos, we can give an alternative expression to my Question 1. If a sum is used, instead of a difference,
$$u=\frac{1}{E(q)}+E(q)$$
then,
$$\frac{u(u-4)^3}{(u-1)^3} = \left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^8$$
Or more simply, we can relate it to the cubic continued fraction $C(q)$ as,
$$u=\frac{1}{E(q)}+E(q) =\frac1{C(q)\,C(q^2)}$$
Since $\displaystyle C(q)=\frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}$, this implies,
$$u= \frac{\eta(4\tau)\,\eta^3(3\tau)}{\eta(\tau)\,\eta^3(12\tau)}$$
an eta quotient also mentioned by Elkies in his answer.
A: [Edited again to give a second identity relating $E$ to eta products]
Continued fraction or not, an expression
$q^{\frac{(r-s)^2}{8(r+s)}} f(\pm q^r, \pm q^s)$
is a modular form of weight $1/2$ for all integers $r,s$ with $r+s>0$,
because it is a sum $\sum_{n=-\infty}^\infty \pm q^{(cn+d)^2}$
with rational $c,d$ and periodic signs.
Therefore the quotient of two such expressions is a modular function,
and takes algebraic valus at quadratic imaginary values.
The quotient
$$
E(q) = q - q^2 + q^6 - q^7 + q^8 - q^9 + q^{11}
 - 2q^{12} + 2q^{13} - 2q^{14} + 2q^{15} \cdots
$$
looks like a modular unit $-$ its logarithmic derivative 
has small coefficients $-$ but not quite an eta product;
instead it seems to be a quotient of Klein forms:
$$
E(q) = q \prod_{n=1}^\infty (1-q^n)^{\chi(n)},
$$
where $\chi$ is the Dirichlet character of conductor $12$, given by
$$
\chi(n) = \cases{
  +1,& if $n \equiv \pm 1 \bmod 12$; \cr
  -1,& if $n \equiv \pm 5 \bmod 12$; \cr
  0,& otherwise.
  }
$$
Two identities relating $E$ to $\eta$ products, similar to but
somewhat more complicated than the ones you give for $A,B,C,D,$ are
$$
\frac1{E(q)} - E(q) =
 \frac{\eta(2\tau)^2 \eta(6\tau)^4}{\eta(\tau)\eta(3\tau)\eta(12\tau)^4},
$$
and (a bit simpler)
$$
\frac1{E(q)} + E(q) =
 \frac{\eta(4\tau)}{\eta(\tau)}
 \Bigl(\frac{\eta(3\tau)}{\eta(12\tau)}\Bigr)^3.
$$
