What is the growth rate of the sum of powers of distinct primes closest to a given a integer?

Question

Let $n$ be a positive integer, and $$2 = p_1 < p_2 < \dots < p_m \le n$$ be the sequence of all primes less than or equal to $n$.

For each index $j$ let $p_j^{e_j}$ be the largest power of $p_j$ still less than or equal to $n$. Define

$$S_n = p_1^{e_1} + p_2^{e_2} + \dots + p_{m}^{e_m} $$

to be the sum of these prime powers. What is the growth rate of this series $S_n$?

We can get an upper bound of

$$S_n \le nm \sim \frac{n^2}{\ln n}$$

by the prime number theorem, and a lower bound of

$$S_n \ge \lfloor n/p_1\rfloor + \lfloor n/p_2\rfloor + \dots + \lfloor n/p_m\rfloor \sim n\ln \ln n $$

using the fact that each term $p_j^{e_j}$ is within a factor of $p_j$ of $n$ and asymptotics for the sum of the reciprocals of the primes.

However, there's still some gap between these two bounds. Is the precise asymptotic growth rate of $S_n$ known?

Answer 1

A better lower bound is $S(n)$, the sum of all primes below $n$, and this lower bound makes a good asymptotic value. One can tweak this by observing that for every term corresponding to a prime less than $\sqrt{n}$ that term is at least $n^{2/3}$, so a tighter lower bound like $S(n) - S(\sqrt{n}) + n^{7/6}/\log n$ is available. Since $S(n)$ is like $O(n^2/\log n)$, one wonders how good an asymptotic is desired.

Gerhard "Wonders What This Is For" Paseman, 2020.08.10.

Answer 2

Only a partial answer for now. Denote by $M$ the quantity $\max\{n-p_{j}^{e_{j}}\}$. Then $nm-m(m-1)/2\geq S_{n}\geq nm-M-(M-1)-(M-2)-\cdots\geq nm-mM+m(m-1)/2$. So determining $M$ would get us closer to the real order of growth of $S_{n}$.

Edited after Gerhard's contributions: let now $x_n$ be the solution of $x_n=\frac{\log n}{\log x_n}$. Then a lower bound for $S_{n}$ is $\sum_{p\leq x_{n}}p^{\lfloor\frac{\log n}{\log p}\rfloor}+S(n)-S(x_n)$ but this must be difficult to estimate.