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I am trying to figure out the coefficients of q-Pochhammer function for special case $(q^n;q)_{\infty}$. I was trying to calculate this using Jacoby's identities but still no success.

EDITED: In order to clarify the question. We have the $$(q^n;q)_{\infty}=\prod_{i \ge 1}{(1-q^{n+i})}=\sum{a_{n,i}q^i}$$

I am trying to calculate the $a_{n,i}$.

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  • $\begingroup$ some more details please $\endgroup$ – tired Feb 14 '17 at 16:50
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I cannot give you a closed expression for the coefficients; however you can use the fact that $$(z;q)_n=\frac{(z;q)_\infty}{(q^n z;q)_\infty}$$ to write $(q^n;q)_\infty= \frac{(q;q)_\infty}{(q;q)_{n-1}}$. Now use Euler's pentagonal number theorem $$ (q;q)_\infty = \sum_{k \in \mathbb{Z}} (-1)^k q^{\frac 1 2 k (3k-1)} $$ as well as the expansion $$ \frac{1}{(z;q)_n}=\sum_{k=0}^\infty \binom{n+k-1}{k}_q z^k, $$ where $\binom{n}{m}_q$ denotes the $q$-binomial coefficients, which are polynomials in $q$, to obtain

$$(q^n;q)_\infty = \bigg(\sum_{k \in \mathbb{Z}} (-1)^k q^{\frac 1 2 k (3k-1)} \bigg) \bigg( \sum_{r=0}^\infty \binom{n+r-2}{r}_q q^r\bigg) $$.

This is a series in $q$ with only positive powers. Hope this helps.

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