Determine whether $$\sum_{t=2}^n\dfrac{\log t}{\Omega(t)}\sim n\log n$$
Let $r=n^{1/\Omega(n)}$, where $n$ is a positive integer and $\Omega(n)$ is the total number of prime factors of $n$. If $r$ is an integer, then it is prime and $n=r^{\Omega(n)}$, otherwise $r$ is strictly between the least and greatest prime factors of $n$.
It seems natural to compare the sum
$$s(n)=\sum_{t=2}^n\dfrac{\log t}{\Omega(t)}$$
to the second Chebyshev function,
$$\psi(n)=\sum_{p^k\leq n}\log p$$
and it appears that $\dfrac{s(n)}{\psi(n)}\sim\log n$. By $n=150000$ the L.H.S. is only greater by a factor of about $1.10129$.
Edit: I had forgotten that $\psi(n)\sim n$, so what I want to show is that
$$\sum_{t=2}^n\dfrac{\log t}{\Omega(t)}\sim n\log n$$
Using $$\sum_{t=2}^{n}\log\left(t\right)\sim n\log\left(n\right)$$ and$$\Omega\left(n\right)\sim\log\left(\log\left(n\right)\right)$$ as $n\rightarrow\infty$ we have, by partial summation,$$\sum_{t=2}^{n}\frac{\log\left(t\right)}{\Omega\left(t\right)}\sim\sum_{t=2}^{n}\frac{\log\left(t\right)}{\log\left(\log\left(t\right)\right)}\sim\frac{n\log\left(n\right)}{\log\left(\log\left(n\right)\right)}+\int_{3}^{n}\frac{1}{\log\left(\log\left(y\right)\right)^{2}}dy.$$ Now we can integrate by parts$$\int_{3}^{n}\frac{1}{\log\left(\log\left(y\right)\right)^{2}}dy=\left.\frac{y}{\log\left(\log\left(y\right)\right)^{2}}\right|_{3}^{n}+2\int_{3}^{n}\frac{y}{\log\left(\log\left(y\right)\right)^{3}y\log\left(y\right)}dy<\left.\frac{y}{\log\left(\log\left(y\right)\right)^{2}}\right|_{3}^{n}+2n\int_{3}^{n}\frac{1}{\log\left(\log\left(y\right)\right)y\log\left(y\right)}dy\sim\frac{n}{\log\left(\log\left(n\right)\right)^{2}}+2n\log\left(\log\left(\log\left(n\right)\right)\right)$$ so the integral gives a minor contribution than the first term. Hence$$\sum_{t=2}^{n}\frac{\log\left(t\right)}{\Omega\left(t\right)}\sim\frac{n\log\left(n\right)}{\log\left(\log\left(n\right)\right)}$$ as $n\rightarrow\infty.$