About the sum $S(p_n)=\sum_{1\le k\lt n}\,p_n\mod\;p_k$

Question

For $\,p_n\gt2\,$ let's define the sum $\,S(p_n)=\sum_{1\le k\lt n}\,p_n\;mod\;p_k$, where $\,p_k\,$ represents the $\,k$-th prime.

The first terms of the sequence $\,S(p_n)\,$ (OEIS A033955 - sum of the remainders when the $\,n$-th prime is divided by primes up to the $\,(n-1)$-th prime) are:

$S(3)=1\;\;\;(p_2=3)$

$S(5)=3\;\;\;(p_3=5)$

$S(7)=4\;\;\;(p_4=7)$

$S(11)=8\;\;\;(p_5=11)$

$S(13)=13\;\;\;(p_6=13)$

$S(17)=18\;\;\;(p_7=17)$

$S(19)=27\;\;\;(p_8=19)$

$S(23)=29\;\;\;(p_9=23)$

$S(29)=46\;\;\;(p_{10}=29)$

The graphical representation of $\,S(p_n)\,$ up to $\,n=1229\,$ is the following:

The growing of $\,S(p_n)\,$ is well estimated by $$S(p_n)\sim n^2\cdot\log\log(n)\;\;\;\;\;\;\;\;(1)$$

Further, the modular equation $$S(p_n)\equiv0\mod p_n\;\;\;\;\;\;\;\;(2)$$ has, up to $\,n=78498$, only the following solutions:

$p_6=13\;\;\;S(13)=13$

$p_{39}=167\;\;\;S(167)=1002$

$p_{333}=2239\;\;\;S(2239)=123145$

$p_{36931}=439867\;\;\;S(439867)=2789196647$

I ask if the previous results, (1) and (2), can be motivated from a theoretical point of view.

Many thanks.

Answer 1

Others should provide sharper estimates of your $\ S(p_n)\ $ since I am not any n.th.pro. This is for the starters,

$$ S(p_n)\ \ge\ n*p_n - \sum_{p_k\le p_n} p_k \sim \frac 12\cdot n^2\cdot\log(n) $$

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One could do better,

$$ S(p_n)\ =\ n*p_n\ - \ \sum_m\,\sum_{p_k\le\frac{p_n}m} p_k $$

and now one needs a pro to handle this.