Suppose there is formal power series in infinite product form as follows: $$\prod_{d\geq 1} \left(1+\frac{u^d}{q^d-1}\right)^{a_d}$$, where $a_d$ are positive integers. Consider the expression $$\prod_{d\geq 1} \left(1+\frac{u^d}{q^d-1}\right)^{a_d}-1$$

Now, this expression has no 1 in it. So, when expressed as formal power series, the expression has the form $\sum_{n=1}^{\infty} b_nx^n$. Now, my question is can we write the above expression in an infinite product form as above by taking something out as common term. So, for example, observe that $b_1=\frac{a_1}{q-1}$. So, possibly can we write

$$\prod_{d\geq 1} \left(1+\frac{u^d}{q^d-1}\right)^{a_d}-1=\frac{a_1}{q-1}u\times (\text{an infinite product expression})$$ or, maybe in the form (Exponential generating function)$\times$ (an infinite product expression).

We note that the coefficients $b_n$ has the following form:

$$b_n=\sum_{\lambda \vdash n} \prod_{i=1}^{n} \frac{\binom{a_i}{m_i(\lambda)}}{(q^i-1)^{m_i(\lambda)}}$$

where, $\lambda\vdash n$ means $\lambda$ is a partiton of $n$ (hence the outer sum runs over all partitons of $n$), and $m_{i}(\lambda)$ denotes the number of times $i$ occur in the partition $\lambda$.

I know that my question probably seems a little confusing as it doesn't perfectly mention what exactly is it that I want to know, but anyway this is the best I could frame it. Thanks in advance for any kind of help.


The question is rather vague, and thus it may have multiple answers. Here is one.

Let $P$ denote the product of interest and let $Q:=\log(P)$, i.e. $$Q = \sum_{d\geq 1} a_d\log(1+\frac{u^d}{q^d-1}).$$ Then we have $$P-1=\exp(Q)-1=Q\frac{\exp(Q)-1}{Q}=Q\exp(\log(\frac{\exp(Q)-1}{Q})).$$ We have $$\log(\frac{\exp(Q)-1}{Q}) = \frac{1}{2}Q + \frac{1}{2}\sum_{k\geq 1} \frac{B_{2k}}{(2k)!k} Q^{2k},$$ where $B_{2k}$ are Bernoulli numbers, and thus $$P-1 = Q \sqrt{P} \prod_{k\geq 1} \sqrt{\exp( \frac{B_{2k}}{(2k)!k} Q^{2k} )}.$$ So we represented $P-1$ as some factor times the infinite product as it was requested by OP.

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