Sum of prime remainders is greater than prime p

Question

This something like Waring meets the primes. Take some prime p(n) and divide it by all the primes less than one-half p(n) to find the sum of the remainders that are themselves also primes. Is there some sum of these remainders for p(s1) that is greater than p(n) for all subsequent primes greater than p(n)? Is there some p(s2) such than all primes greater than p(s2) will have a sum of remainders twice p(s2)? Some p(s3) for three times the sum of all its prime remainders for all primes greater than p(s3)? This could be continued for some p(s4), [p(s5)... Do you know if anyone has investigated this?

Answer 1

Here is a totally heuristic non-rigorous attempt.

For a prime $p$, there are about $n =\dfrac{p}{2 \ln p}$ primes less than $p/2$.

For each of these primes, say $q$, the probability of the remainder being prime is about $\dfrac{1}{\ln q}$.

Therefore the sum of these remainders is about $\sum_{q < p/2} \dfrac{q}{\ln q} \sim \left(\dfrac{p}{2\ln p}\right)^2 $.

This is larger than $p$ by a factor of $\dfrac{p}{(2\ln p)^2}$ which is greater than $1$, so the sum is usually much greater than $p$.