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Apr 20, 2021 at 12:51 answer added username timeline score: 1
S Apr 22, 2015 at 20:30 history bounty ended CommunityBot
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S Apr 14, 2015 at 19:16 history bounty started Twi
S Apr 14, 2015 at 19:16 history notice added Twi Draw attention
Apr 12, 2015 at 16:21 history edited Twi CC BY-SA 3.0
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Apr 12, 2015 at 16:09 comment added tituf If you slightly modify your function as $$F_n(q,i):= \sum_{i_1,\dots,i_n} \frac{q^{i_1+\dots+i_n}}{(1-q^{i_1+i_2})\dots(1-q^{i_{n-1}+i_n})(1-q^{i_n+i})} \,,$$ then you get a recurrence: $$ F_n(q,i) = \sum_j \frac{q^j}{1-q^{j+i}}\, F_{n-1}(q,j) \,.$$ Hope this can help you.
Apr 12, 2015 at 13:59 history asked Twi CC BY-SA 3.0