Let it be an arithmetic progression $A = \{a_1, a_1+k, a_1+2k,...,a_1+(n-1)k : k\in\mathbb N\}$, and let $S=\{x_1,x_2,...,x_r\}$ be some subset of $A$, such that the elements of $S$ are equi-distributed in the sense that the probability that a random element of $A$ belongs to $S$ is positive and defined by a function $P$.
If the above holds, it also seems to hold the following sum formula: $$\sum_{i=1}^{r}x_i = \frac{1}{P}\left(\frac{(a_1+a_r)r}{2}\right)+O(y)$$
Where I have not been able to determine properly a general magnitude of $y$ (although it seems pretty small with the examples I have checked).
I have not been able to prove the above, although I have checked it with some well-known series:
- If $x_i$ is some even number, and we consider the set of natural numbers, we have that $\sum_{i=1}^{r}x_i = 2\left(\frac{r^2+r}{2}\right)+O(y)$
- If $x_i$ is some prime number, and we consider the set of natural numbers, we have that $\sum_{i=1}^{r}x_i = \log(r)\left(\frac{r^2+r}{2}\right)+O(y)$
- If $x_i$ is some square-free integer, and we consider the set of natural numbers, we have that $\sum_{i=1}^{r}x_i = \frac{\pi^2}{6}\left(\frac{r^2+r}{2}\right)+O(y)$
I would appreciate
(i) If necessary (and true), a better definition of $P$ that implies that the conjectured generalization holds.
(ii) A theoretical explanation of why does this hold. I have not been able to prove it, although heuristically makes sense.
(iii) Which could be sharp general bounds for $y$.
Thanks in advance!
EDIT
I think I have at least the sketch of a proof, and it seems pretty simple. Just note that, according to the conditions, $x_i=a_i*\frac{1}{P}+O(t)$, and thus the sum formula follows. Therefore, the sharpness of the sum formula depends on the error of the probability function, as we have that $y\leq r*t$.
For instance, if $x_i=a_i*\frac{1}{P}+O(1)$, then $\sum_{i=1}^{r}x_i = \frac{1}{P}\left(\frac{(a_1+a_r)r}{2}\right)+O(r)$.
Just note that, according to the conditions, $x_i=a_i*\frac{1}{P}+O(t)$, and thus the sum formula follows. Therefore, the sharpness of the sum formula depends on the error of the probability function, as we have that $y\leq r*t$.
For instance, if $x_i=a_i*\frac{1}{P}+O(1)$, then $\sum_{i=1}^{r}x_i = \frac{1}{P}\left(\frac{(a_1+a_r)r}{2}\right)+O(r)$.