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.

  • $\begingroup$ What are the $a_i$? $\endgroup$ – Roman Sep 25 at 17:17
  • 1
    $\begingroup$ 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

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]}
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.