Sum of $\frac{\Lambda(n)}{\log(n)}$ over divisors I am wondering if $\sum_{d\mid n}\frac{\Lambda(d)}{\log(d)}$ (which evaluates to $1/k$ for $d=p^k$ and $0$ otherwise) has any interesting significances or bounds.
For $n = p_1^{k_1}\dots p_s^{k_s}$, we would have
$$\sum_{d\mid n}\frac{\Lambda(d)}{\log(d)} = \sum_i H_{k_i}$$
where $H_k$ is the $k$-th harmonic number, but is there a good way to approximate this?
Here is the plot for $(n, \sum_{d\mid n}\frac{\Lambda(d)}{\log(d)})$ for $1\le n \le 50000$:

 A: You can for example derive the inequalities
$$
\sum\limits_{d|n} {\frac{{\Lambda (d)}}{{\log d}}} = \sum\limits_i {H_{k_i } }  \le \sum\limits_i {(\log (k_i  + 1) + \gamma )}  = \log d(n) + \omega (n)\gamma 
$$
and
$$
\sum\limits_{d|n} {\frac{{\Lambda (d)}}{{\log d}}} = \sum\limits_i {H_{k_i } }  \ge \sum\limits_i {\log (k_i  + 1)}  = \log d(n).
$$
Here $\gamma$ is the Euler$-$Mascheroni constant, $d(n)$ is the divisor function and $\omega(n)$ counts each distinct prime factor of $n$. In particular,
$$
\log 2 \le \mathop {\lim \sup }\limits_{n \to  + \infty } \frac{{\log \log n}}{{\log n}} \sum\limits_{d|n} {\frac{{\Lambda (d)}}{{\log d}}}  \le \log 2 + \gamma .
$$
A: We can begin by computing
\begin{align*}
\sum_{n<N}\sum_{d|n}\frac{\Lambda(d)}{\log(d)}&=\sum_{d<N}\sum_{n<N/d}\frac{\Lambda(d)}{\log(d)}\\
&=\sum_{d<N}\left[\frac{N}{d}\right]\frac{\Lambda(d)}{\log(d)}\\
&=\sum_{p<N}\left[\frac{N}{p}\right]+O\left(\sqrt{N}\right)\\
&=\sum_{p<N}\frac{N}{p}+O(\pi(N))\\
&=\sum_{p<N}\frac{N}{p}+O\left(\frac{N}{\log(N)}\right)\\
&=N\left(\log(\log(N))+M+O\left(\frac{1}{\log(N)}\right)\right)+O\left(\frac{N}{\log(N)}\right)
\end{align*}
Here, the last estimate is from Merten's third theorem with "simple" error bounds (i.e not dependent on PNT). Thus,
$$\frac{1}{N}\sum_{n<N}\sum_{d|n}\frac{\Lambda(d)}{\log(d)}=\log(\log(N))+M+O\left(\frac{1}{\log(N)}\right)$$
without PNT
