What is the sum of the amount powers of primes $p$ between $n-2p$ and $n$, with $n>p$, and$2 < p < \frac{n}{2}$?
By powers of primes, I mean an integer of the form $p^k$, not an integer of the form $p^a q^b \cdots$. So far, I have tried this, by letting $\psi$ represent the quantity above:
$$\psi = \sum_{q} (\log_q(n)-\log_q(n-2q)) = \sum_{q} \frac{ \ln( \frac{n}{n-2q})}{\ln(q)},$$ where $q$ ranges across the odd primes less than or equal to $\frac{n}{2}$. I'm not sure how to simplify further, or whether the sum above ever gets very large. In case my sum is correct and you see some sort of simplification, please also discuss behavior of $\psi$ as $n \rightarrow \infty$.
Partial Answer :
the number of primes power less or equal to $n$.
you have $ \sum \limits_{ q \leq n} \frac{1}{2}\lfloor \ln_q (n) \rfloor (\lfloor \ln_q (n) \rfloor +1) = \sum \limits_{t=1}^{\ln_2(q)} \sum \limits_{n^{\frac{1}{t+1} }< q \leq n^{\frac{1}{t}}} \frac{1}{2}t(t+1) = \sum \limits_{t=1}^{\ln_2(n) } \frac{1}{2} t (t+1)( 1+\pi(n^{\frac{1}{t}})-\pi(n^{\frac{1}{t+1}})) = 1+\pi(n)-\pi(\sqrt{n})+3 \pi(\sqrt{n}) -3 \pi(\sqrt[3]{n}) + \cdots = 1+\sum \limits_{j=1}^{\ln_2(n)} j \pi(n^{\frac{1}{j}})$
$1+\pi(n)+2\pi(\sqrt{n}) < 1+\sum \limits_{j=1}^{\ln_2(n)} j \pi(n^{\frac{1}{j}}) < 1+\pi(n)+\frac{\log (n) \log (2 n)}{2 \log ^2(2)} \pi(\sqrt{n})$
So $ 1+\sum \limits_{j=1}^{\ln_2(n)} j \pi(n^{\frac{1}{j}}) \approx \frac{n}{\ln n}$.
I am too lazy to complete the answer but you need to subtract the same approach as above but now for $n-2p$.
Hope it helped.