Let $$R(q) = \cfrac{q^{1/5}}{1 + \cfrac{q}{1 + \cfrac{q^{2}}{1 + \cfrac{q^{3}}{1 + \cdots}}}}$$


The following equality is famous:


$$\cfrac{q^{1/5}}{R(q)} = \prod\_{k>0} \cfrac{(1-q^{5k-2})(1-q^{5k-3})}{(1-q^{5k-1})(1-q^{5k-4})} ( = f(q))$$ 


The coefficients of $f(q)$ can be positive or negative. In fact,


$$f(q) = 1 + q - q^3 + q^5 + q^6 - q^7 - 2 q^8 + \cdots$$


Let


$$g(q) = \prod\_{k>0} f(q^k) = f(q)f(q^2)f(q^3) \cdots$$


$g(q)$


$= (1 + q - q^3 + q^5 + \cdots)(1 + q^2 - q^6 + q^{10} + \cdots)\cdots$


$= 1 + q + q^2 + q^3 + 2q^4 + 3q^5 + 3q^6 + 3q^7 + 4q^8 + 6q^9 + \cdots$


The coefficients of $g(q)$ seem [non-negative](https://oeis.org/A327716).
Are the coefficients of $g(q)$ non-negative? 



Are the coefficients of certain product of Rogers-Ramanujan Continued Fraction non-negative?
Notice that we can write
$$f(q)=\prod\_{n\geq 1} (1-q^n)^{-\left(\frac{n}{5}\right)}$$
therefore
$$g(q)=\prod\_{k\geq 1} f(q^k)=\prod\_{n\geq 1} (1-q^n)^{-a(n)}$$
where $a(n)=\sum\_{d|n}\left(\frac{d}{5}\right)$, where $\left(\frac{d}{5}\right)$ is the Legendre symbol. Now, $a(n)$ is easily seen to be multiplicative with $a(5^k)=1$, $a(p^k)=k+1$ when $p\equiv \pm 1\pmod{5}$, and $a(p^k)=\frac{1+(-1)^k}{2}$ when $p\equiv \pm 2\pmod{5}$. This means that $a(n)\geq 0$ for all $n$, so $g(q)$ is a product of series with nonnegative coefficients, and thus has nonnegative coefficients itself (or even nondecreasing ones as suspected in the comments).


