Distribution of prime divisors. What is the distribution of the product of the distinct prime divisors?

Question

Let $N_n$ be an integer randomly chosen from among the first $n$ integers: $P[N_n = m] = 1/n$ for $ 1\le m \le n.$ Let $Q_{ni}$ be the distinct prime divisors of $N_n$, ordered by size: $Q_{n1} > Q_{n2} > \cdots,$ where $Q_{nv}=1$ if $N_n$ has fewer than $v$ distinct prime divisors, and let $T_n = \Pi_v Q_{nv}$ be the product of the distinct prime divisors. In this case, why is $$E[\log T_n] = \sum_{p \le n} \frac{1}{n} \Big \lfloor \frac{n}{p}\Big \rfloor \log p?$$

Answer 1

We have $$E[\log T_n]=\frac 1n\times \sum_{i=1}^n \sum_{p\,|\,i}\log p$$

But the double sum can be rewritten as $$\sum_{p≤n}a(n)_p\log p$$

Where $a(n)_p$ is defined to be the number of integers $i\in \{1,\cdots, n\}$ which are divisible by $p$. It is easy to see that $$a(n)_p=\Big \lfloor \frac np\Big \rfloor$$ and we are done.