# Double summation involving partition of a number n as a condition

I am dealing with the expression: $$b_n=\sum_{m=2}^{\lfloor\frac{n}{3}\rfloor}\left\{(-1)^m\frac{l_m}{m+1}\sum_{i_1+i_2+...+i_m=n}a_{i_1}a_{i_2}...a_{i_m}\right\}$$ The $$i_k\geq3$$, for any $$k$$. Perhaps integer partition of $$n$$ into $$m$$ parts can be considered as $$m$$ varies. All I need to do is, to simplify the term to analyse how $$b_n$$ behaves. So how do I put this in Mathematica? Thank you for your efforts.

• What are the $a_i$? – Roman Sep 25 at 17:17
• Those are again some complicated summations, but I'm not going deep into it.. All I need to know is whether b_n assumes any zero value for some n. That's why this question is asked. – user74846 Sep 25 at 17:36

The $$i_k\ge 3$$ restriction means $$b_n$$ is $$0$$ for $$n<6$$ (why?).

With that,

Table[Sum[(-1)^m l[m]/(m + 1) Sum[Product[a[i], {i, id}],
{id, IntegerPartitions[n, {m}, Range[3, n]]}],
{m, 2, Quotient[n, 3]}],
{n, 6, 9}]
{1/3 a^2 l, 1/3 a a l, 1/3 (a^2 + a a) l,
1/3 (a a + a a) l - 1/4 a^3 l}

• b_n is 0 for n<6 according to the requirement of a set of calculations. Also, how do I write this expression if we do not ignore that the partition is same if the order is changed? For example: [1,1,2] is not same as [1,2,1] is not same as [2,1,1]. Then how do I count all such possibilities in the summation involving a_{i_k}s? – user74846 Sep 25 at 17:32
• @J.M., would you mind addressing the above commented problem? – user74846 Sep 26 at 6:31
• That's a different question altogether; since you already asked another question, I don't have to remind you to ask a new one. – J. M.'s discontentment Sep 26 at 10:03