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Let $\mathbb{N}$ denote the set of positive integers, and let ${\bf P}\subseteq {\mathbb N}$ be the set of prime numbers. For $a\in \mathbb{N}$ let $$p(a) = \{p\in {\bf P}: (\exists k\in\mathbb{N})k\cdot p = a\}.$$ So we have $p(1)=0$, and $p(n) > 1$ for $n\in\mathbb{N}\setminus\{1\}$.

For $n\in \mathbb{N}$, let $\text{med}(n)$ be the median of the set $\{p(a): 1\leq a \leq n\}$.

Questions. Is the sequence $(\text{med}(n))_{n\in\mathbb{N}}$ bounded? If yes, does it have a limit, and is the (integer) value of the limit known?

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    $\begingroup$ The median is certainly unbounded, and grows like $\gg \log\log n$. This follows from the fact (Hardy-Ramanujan, reproved by Turan) that $p(a)=(1+o(1))\log\log a$ for almost all integers $a$. $\endgroup$
    – Thomas Bloom
    Commented Oct 27, 2021 at 8:25
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    $\begingroup$ Are you counting the number of distinct prime divisors? Is $p(2^{5})=1$? Is $p(2^{5})=5$? en.wikipedia.org/wiki/Prime_omega_function $\endgroup$
    – Joseph Van Name
    Commented Oct 27, 2021 at 13:21
  • $\begingroup$ That's right, Joseph, only distinct prime divisors, so $p(2^5) = 1$. Will edit the original question $\endgroup$
    – Dominic van der Zypen
    Commented Oct 27, 2021 at 18:21
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    $\begingroup$ @JosephVanName the result cited by Thomas works for both $\endgroup$
    – Fedor Petrov
    Commented Oct 27, 2021 at 18:52

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