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We have three sequences of positive integers $l$, $p$ and $q$ such that: $$ p_1 \geq p_2 \geq \cdots \geq p_k\text{ and } q_1 \geq q_2 \geq \cdots \geq q_k \geq \cdots \geq q_h \text{ where: } k < h $$

and

$$ 0\leq l_1 \leq l_2 \cdots \leq l_k\leq \cdots \leq l_h $$

and $$ p_1+2p_2+\cdots+kp_k < q_1+2q_2+\cdots+kq_k+\cdots+hq_h $$

and

$$ p_1+p_2+\cdots+p_k = q_1+q_2+\cdots+q_k+\cdots+q_h=n $$

I search an upper bound for this formula:

$$ l_1(p_1-q_1)+l_2(p_2-q_2)+\cdots+l_k(p_k-q_k) $$

If we put $r_i = p_i-q_i$, we can re-write it as:

$$ l_1r_1+l_2r_2+\cdots+l_kr_k $$

We distinguish 3 cases:

$$ r_i=r^{+}_i\text{ if } r_i>0 $$

$$ r_i=r^{-}_i\text{ if } r_i<0 $$

$$ r_i=r^0_i\text{ if } r_i=0 $$

So, we can find a first upper bound like that:

$$ l_1r_1+l_2r_2+\cdots+l_kr_k \leq \sum_{i\in I^{+}} l_ir^{+}_i $$

such that $I^{+}$ is the set of indices where $r_i>0$.

For my work, i need a better upper bound using the presented parameters.

Additional informations may be useful:

$$ \sum_{i=1}^h r_i = 0 $$

$$ \sum_{i=1}^k r_i = -\sum_{i=k+1}^h r_i $$

The upper bound is needed to prove another result that uses it in its proof.

Best regards

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    $\begingroup$ You should explain why the upper bound that you have is not good enough, and what would be good enough for you. $\endgroup$ Commented Feb 14 at 22:44
  • $\begingroup$ @IosifPinelis A good upper bound that include the $r_{i}^{-}$, because the mentioned upper bound is very far from the real value. If there is a transformation of the formula like Abel transformation that can be used is also a good think to use is also welcome. $\endgroup$ Commented Feb 15 at 12:16

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