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Let $[n]:=\frac{1-q^n}{1-q}$. I wish to find a closed form of the following q-sum $$I(p):=\sum\limits_{n_1+\cdots+n_p=n\atop n_1,\ldots,n_p\ge 1}\ \ \ \frac{1}{[n_1]\cdots [n_p]}=？$$ For example, we have $$I(2)=\frac{1}{[n]}\sum\limits_{k=1}^{n-1}\frac{1+q^k}{[k]},$$ but how to obtain the general result for $p\ge 3$?

xuce1234
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    Before we begin, do you have a proof when $q \to1$? – Orat Mar 01 '19 at 15:11
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    Since for $q=1$ we have that $$ {{p!} \over {n!}}\left[ \matrix{ n \cr p \cr} \right] = \sum\limits_{\scriptstyle 1, \le ,k_{,j} \atop \scriptstyle ,k_{,1} , + ,k_{,2} , + , \cdots , + ,k_{,p} , = ,n} {{1 \over {k_{,1} k_{,2} \cdots k_{,p} }}} $$ where ${ n\brack p}$ indicates the un-signed Stirling N. 1st kind, then probably your sum is tied to the q-analog of those numbers. – G Cab Mar 01 '19 at 16:00

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