Denote the (unsigned) [*Stirling numbers of the $1^{st}$-kind*](https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind) by ${n \brack k}$ and define
$$\mathbf{F}\_a(q)=\sum\_{m\geq1}\frac{q^{am}}{(1-q^m)^{2a}} \qquad \text{and} \qquad
\mathbf{G}\_b(q)=\sum\_{m\geq1}\frac{m^bq^m}{1-q^m}.$$


I have encountered the following property.



> 
> **QUESTION.** Is this true? Or, can you provide a reference to it.
> $$\mathbf{F}\_a(q)=\frac1{(2a-1)!}\sum\_{k=0}^{a-1}(-1)^k\left(\sum\_{j=-k}^k(-1)^j {a \brack a-k+j} \cdot {a\brack a-k-j}\right)\mathbf{G}\_{2a-1-2k}(q).$$
> 
> 
> 


**Example.** When $a=2$, the claim reads as
$$\sum\_{m\geq1}\frac{q^{2m}}{(1-q^m)^4}
=\frac16\sum\_{m\geq1}\frac{m^3q^m}{1-q^m}-\frac16\sum\_{m\geq1}\frac{mq^m}{1-q^m}.$$
Observe that the right-hand side enumerates **excess** in the sum of powers of divisors.


**Postscript.** I have found a better format.



> 
> **QUESTION.** Is this true? Or, can you provide a reference to it.
> $$\mathbf{F}\_a(q)=\frac1{(2a-1)!}\mathbf{G}(\mathbf{G}^2-1^2)(\mathbf{G}^2-2^2)(\mathbf{G}^2-3^2)\cdots(\mathbf{G}^2-(a-1)^2);$$
> where we adopt an umbral notation $\mathbf{G}^j$ to stand for $\mathbf{G}\_j$ and multiply accordingly.
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> 
> 



$q$-series and Stirling of the 1st kind

> 
> **QUESTION.** Is this true? Or, can you provide a reference to it.
> $$\mathbf{F}\_a(q)=\frac1{(2a-1)!}\mathbf{G}(\mathbf{G}^2-1^2)(\mathbf{G}^2-2^2)(\mathbf{G}^2-3^2)\cdots(\mathbf{G}^2-(a-1)^2);$$
> where we adopt an umbral notation $\mathbf{G}^j$ to stand for $\mathbf{G}\_j$ and multiply accordingly.
> 
> 
> 


Expanding the differences of two squares this is $$\mathbf{F}\_a(q) \stackrel?= \frac1{(2a-1)!}(\mathbf{G}+a-1)^{\underline{2a-1}}$$ and expanding the umbral definition of $\mathbf{G}$ this becomes $$\mathbf{F}\_a(q) \stackrel?= \frac1{(2a-1)!} 
\sum\_{m \geq 1}\frac{(m+a-1)^{\underline{2a-1}} q^m}{1-q^m} = \sum\_{m \geq a} \binom{m+a-1}{2a-1} \frac{q^m}{1-q^m}$$ where the range of $m$ in the final sum has been tweaked to the support of the binomial.


It's standard that $\frac{1}{(1-x)^k} = \sum\_{j \ge 0} \binom{j+k-1}{k-1} x^j$, so


$$\mathbf{F}\_a(q) = \sum\_{m \geq 1}\frac{q^{am}}{(1-q^m)^{2a}} = \sum\_{m \geq 1} \sum\_{j \ge 0} \binom{j+2a-1}{2a-1} q^{(j+a)m} $$


Subst $k = j+a$:


$$\mathbf{F}\_a(q) = \sum\_{m \geq 1} \sum\_{k \ge a} \binom{k+a-1}{2a-1} q^{km} = \sum\_{k \ge a} \binom{k+a-1}{2a-1} \sum\_{m \geq 1} q^{km} = \sum\_{k \ge a} \binom{k+a-1}{2a-1} \frac{q^k}{1 - q^k} \blacksquare$$


