Ramanujan proved that the average number of distinct divisors of $x$ for $x$ on $[1,n], ~\omega_{\text{av}}(n),$ and the average number of divisors including repetitions, $\Omega_{\text{av}}(n),$ are asymptotically equivalent,
$$\omega_{\text{av}}(n) \sim \Omega_{\text{av}}(n)\sim \log\log n.\hspace{10mm}(1)$$
Landau proved that
$$ \pi_k(n) \sim \frac{n}{\log n} \frac{(\log\log n)^{k-1}}{(k-1)!}. $$
It seems possible (not claiming it) that a sharper estimate of the average $\Omega_{\text{av}}(n)$ on a large interval $[1,n]$ would be
$$\Omega_2(n) = (1/n) \sum_{k=1}^r \frac{n}{\log n}\frac{k ( \log\log n)^{k-1}}{(k-1)!} .\hspace{10mm} (2)$$
for sufficiently large $r$, and it makes sense in any case that $\omega_{\text{av}}(n)\leq \Omega_{\text{av}}(n).$
Can we prove that
$$(1/n) \sum_{k=1}^r \frac{n}{\log n}\frac{k ( \log\log n)^{k-1}}{(k-1)!}-\log\log n \sim 1? $$
Numerically it seems to work. I removed my own attempt to prove this because it was apparently wrong.
We want to show that $(1/n) \sum_{k=1}^r \frac{n}{\log n}\frac{k ( \log\log n)^{k-1}}{(k-1)!}-\log\log n \sim 1 $ for large $r$.
If $f(x) = \sum_{k=1}^{\infty}\frac{k x^{k-1}}{(k-1)!} $, then
$\begin{array}\\ f(x) &= \sum_{k=1}^{\infty}\frac{k x^{k-1}}{(k-1)!}\\ &= \sum_{k=1}^{\infty}\frac{(k-1+1) x^{k-1}}{(k-1)!}\\ &= \sum_{k=1}^{\infty}\frac{(k-1) x^{k-1}}{(k-1)!}+\sum_{k=1}^{\infty}\frac{(1) x^{k-1}}{(k-1)!}\\ &= \sum_{k=2}^{\infty}\frac{(k-1) x^{k-1}}{(k-1)!}+\sum_{k=0}^{\infty}\frac{ x^{k}}{(k)!}\\ &= \sum_{k=2}^{\infty}\frac{ x^{k-1}}{(k-2)!}+e^x\\ &= \sum_{k=0}^{\infty}\frac{ x^{k+1}}{(k)!}+e^x\\ &=xe^x+e^x\\ &=e^x(x+1)\\ \end{array} $
Therefore $f(\ln \ln x) =e^{\ln \ln x}(\ln \ln x+1) =\ln x(\ln \ln x+1) $ so that $\frac1{\ln x}f(\ln \ln x) =e^{\ln \ln x}(\ln \ln x+1) =\ln \ln x+1 $.