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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.

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  • $\begingroup$ What are the $a_i$? $\endgroup$ – Roman Sep 25 at 17:17
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    $\begingroup$ 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. $\endgroup$ – user74846 Sep 25 at 17:36
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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[3]^2 l[2], 1/3 a[3] a[4] l[2], 1/3 (a[4]^2 + a[3] a[5]) l[2],
    1/3 (a[4] a[5] + a[3] a[6]) l[2] - 1/4 a[3]^3 l[3]}
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  • $\begingroup$ 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? $\endgroup$ – user74846 Sep 25 at 17:32
  • $\begingroup$ @J.M., would you mind addressing the above commented problem? $\endgroup$ – user74846 Sep 26 at 6:31
  • $\begingroup$ That's a different question altogether; since you already asked another question, I don't have to remind you to ask a new one. $\endgroup$ – J. M.'s discontentment Sep 26 at 10:03

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