What is the number of (positive) integer solutions of:
$$a_1+a_2+\ldots+a_i=n$$
where $a_1 \ge a_2 \ge \ldots\ge a_i\gt 0$ ?
Also, the order of summands does not matter.
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any limits on $n$ ? and number of variables a,b,c ? Also please include what you tried and where you are stuck. – advocateofnone May 31 '16 at 14:27
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2It sounds like you are asking for the number of restricted partitions of $n$ into $i$ parts. This would be notated as $p_i(n)$. Read more about it here. This happens to also equal the number of partitions of $n$ whose largest part is of size $i$, which is why the notation $p_i(n)$ does not distinguish between the two and just refers to it as a restricted partition of $n$. Convenient recurrence relations and generating functions are shown on the site, but closed form expressions are rather ugly. – JMoravitz May 31 '16 at 14:36
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@E.Girgin: In editing the Question you changed the final strong inequality $a_i \gt 0$ to the weak inequality $a_i \ge 0$. This has the effect of including partitions with fewer than $i$ parts in the counted number. I'm going to change this back, assuming you made this inadvertently. – hardmath May 31 '16 at 19:34
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Hint: You may deal with the sub-case $a_1>a_2>\ldots>a_m>0$ and $\sum a_k = n$ first.
By setting $a_1-a_2=b_1, a_2-a_3= b_2, \ldots a_m=b_m$, we want to count the solutions of: $$ b_1 + 2 b_2 +\ldots + m b_m = n $$ with $b_m\in\mathbb{N}^+$. That can be done through a classical argument. Then account for the cases in which some of the inequalities $a_1\geq a_2\geq a_3\ldots\geq a_m$ hold as equalities.
Jack D'Aurizio
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