
Chinese Leftovers II
Let $A_n$ be the smallest positive integer satisfying $A_n \bmod p_i = i$ for all $1 \le i \le n$, where $p_i$ is the $i$-th prime.
For example $A_2 = 5$, since this is the smallest positive solution of the system of equations
- $A_2 \bmod 2 = 1$
- $A_2 \bmod 3 = 2$
The system of equations for $A_3$ adds another constraint. That is, $A_3$ is the smallest positive solution of
- $A_3 \bmod 2 = 1$
- $A_3 \bmod 3 = 2$
- $A_3 \bmod 5 = 3$
and hence $A_3 = 23$. Similarly, one gets $A_4 = 53$ and $A_5 = 1523$.
Let $S(n)$ be the sum of all primes up to $n$ that divide at least one element in the sequence $A$.
For example, $S(50) = 69 = 5 + 23 + 41$, since $5$ divides $A_2$, $23$ divides $A_3$ and $41$ divides $A_{10} = 5765999453$. No other prime number up to $50$ divides an element in $A$.
Find $S(300000)$.