$\frac{(\sum_{i=0}^{n}ia_i)}{(\sum_{i=0}^{n}a_i)}=\frac{(\sum_{i=0}^{j}ib_i)}{(\sum_{i=0}^{j}b_i)}+\frac{(\sum_{i=0}^{k}ic_i)}{(\sum_{i=0}^{k}c_i)}$

Question

Let it be $A=\{a_0,...,a_n\}$, $B=\{b_0,...,b_j\}$ and $C=\{c_0,...,c_k\}$ three sets of equal or distinct integers, not necessarily ordered, each set having at least two values other than $0$. I am having troubles to find "interesting" conditions on the elements of these sets for this equality to hold:

$$\frac{\left(\sum_{i=0}^{n}ia_{i}\right)}{\left(\sum_{i=0}^{n}a_{i}\right)}=\frac{\left(\sum_{i=0}^{j}ib_{i}\right)}{\left(\sum_{i=0}^{j}b_{i}\right)}+\frac{\left(\sum_{i=0}^{k}ic_{i}\right)}{\left(\sum_{i=0}^{k}c_{i}\right)}$$

I derived that this equality is true if $$\sum_{i=0}^{n}a_{i}=\left(\sum_{i=0}^{j}b_{i}\right)\left(\sum_{i=0}^{k}c_{i}\right)$$ and $$\sum_{i=0}^{n}ia_{i}=\left(\sum_{i=0}^{k}c_{i}\right)\left(\sum_{i=0}^{j}ib_{i}\right)+\left(\sum_{i=0}^{k}ic_{i}\right)\left(\sum_{i=0}^{j}b_{i}\right)$$

An example of what I am looking for: If $\sum_{i=0}^{n}a_{i}$ is some prime number $\pm p$, then necessarily $\left(\sum_{i=0}^{j}b_{i}\right)=\pm 1$, $\left(\sum_{i=0}^{k}c_{i}\right)=\pm p$ and $\sum_{i=0}^{n}a_{i}$ divides $\left(\sum_{i=0}^{j}ib_{i}\right)$, or $\left(\sum_{i=0}^{k}c_{i}\right)=\pm 1$, $\left(\sum_{i=0}^{j}b_{i}\right)=\pm p$ and $\sum_{i=0}^{n}a_{i}$ divides $\left(\sum_{i=0}^{k}ic_{i}\right)$, which is only possible if some $b_i$ or $c_j$ is negative.

I have not been able to find any other meaningful insights. I would appreciate any hint or help.

Thanks in advance!

Edit

After @mathlove 's nice counterexample to the first equation, I want to clarify that it is derived from the second and the third ones; so ideally, I would want to find "interesting" conditions on the elements of the sets for the three equalities to hold.

Answer 1

If one generalizes your sum with $z$ as: $$ \sum_{i=0}^n a_i \: z^i = Q(n,z) \\ \sum_{i=0}^j b_i \: z^i = R(j,z) \\ \sum_{i=0}^k c_i \: z^i = S(k,z)$$ and $$ Q(n,z) = R(j,z) \cdot S(k,z) $$ then it is quite easy to see that taking the logarithmic derivative of each side and then multiplying by $z$ will give you: $$ z\frac{\frac{d}{dz}Q(n,z)}{Q(n,z)} = z\frac{\frac{d}{dz}R(j,z)}{R(j,z)} + z\frac{\frac{d}{dz}S(k,z)}{S(k,z)} $$ Setting $z = 1$ yields: $$ \frac{\sum_{i=0}^n i \cdot a_i}{\sum_{i=0}^n a_i } = \frac{\sum_{i=0}^j i \cdot b_i}{\sum_{i=0}^j b_i} + \frac{\sum_{i=0}^k i \cdot c_i}{\sum_{i=0}^k c_i} $$

Your second identity is just the derivative without taking the logarithm. Any "special" characteristics will come from the original equation. It may interest you to look into multiplicative functions, where $$ f(a\cdot b) = f(a)f(b)$$ when $a$ and $b$ are coprime.

Answer 2

Let $$\frac{\displaystyle\sum_{i=0}^{n}ia_{i}}{\displaystyle\sum_{i=0}^{n}a_{i}}=\frac{\displaystyle\sum_{i=0}^{j}ib_{i}}{\displaystyle\sum_{i=0}^{j}b_{i}}+\frac{\displaystyle\sum_{i=0}^{k}ic_{i}}{\displaystyle\sum_{i=0}^{k}c_{i}}\tag1$$

$$\sum_{i=0}^{n}a_{i}=\left(\sum_{i=0}^{j}b_{i}\right)\left(\sum_{i=0}^{k}c_{i}\right)\tag2$$

$$\sum_{i=0}^{n}ia_{i}=\left(\sum_{i=0}^{k}c_{i}\right)\left(\sum_{i=0}^{j}ib_{i}\right)+\left(\sum_{i=0}^{k}ic_{i}\right)\left(\sum_{i=0}^{j}b_{i}\right)\tag3$$

where $$\bigg(\displaystyle\sum_{i=0}^{n}a_{i}\bigg)\bigg(\displaystyle\sum_{i=0}^{n}b_{i}\bigg)\bigg(\displaystyle\sum_{i=0}^{n}c_{i}\bigg)\not=0\tag4$$

I would want to find "interesting" conditions on the elements of the sets for the three equalities to hold.

Since $(2)(3)\implies (1)$ under the condition that $(4)$, if I understand what you are saying correctly, then I think that what you want to find is "interesting" conditions on the elements of the sets for $(2)(3)$ to hold under the condition that $(4)$.

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I'm not sure if you find the following "interesting", but for any given values $$\sum_{i=0}^{j}b_{i}=P\ (\not=0),\sum_{i=0}^{j}ib_{i}=Q,\sum_{i=0}^{k}c_{i}=R\ (\not=0),\sum_{i=0}^{k}ic_{i}=S$$ and for any given $(b_2,b_3,\cdots, b_j),(c_2,c_3,\cdots,c_k)$, you can choose $b_0,b_1,c_0,c_1$ as follows : $$\begin{align}b_1&=Q-\bigg(2b_2+3b_3+\cdots +jb_j\bigg) \\\\b_0&=P-\bigg(b_1+b_2+\cdots +b_j\bigg) \\\\c_1&=S-\bigg(2c_2+3c_3+\cdots +kb_k\bigg) \\\\c_0&=R-\bigg(c_1+c_2+\cdots +c_k\bigg)\end{align}$$ Since it follows from $(2)(3)$ that $$\sum_{i=0}^{n}a_{i}=PR\ (\not=0)\qquad\text{and}\qquad\sum_{i=0}^{n}ia_{i}=RQ+SP$$ for any given $(a_2,a_3,\cdots, a_n)$, you can choose $a_0,a_1$ as follows : $$\begin{align}a_1&=RQ+SP-\bigg(2a_2+3a_3+\cdots +na_n\bigg) \\\\a_0&=PR-\bigg(a_1+a_2+\cdots +a_n\bigg)\end{align}$$

For these values, you can see that $(1),(2)$ and $(3)$ hold.