5
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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?

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  • 2
    $\begingroup$ the stronger conjecture is that the coefficients form a nondecreasing series, which seems to be the case as far as I could check $\endgroup$ – Carlo Beenakker Sep 23 at 19:55
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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).

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