Are the coefficients of certain product of Rogers-Ramanujan Continued Fraction non-negative?

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. Are the coefficients of $$g(q)$$ non-negative?

• the stronger conjecture is that the coefficients form a nondecreasing series, which seems to be the case as far as I could check – Carlo Beenakker Sep 23 at 19:55

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).