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Definition: Let $\omega(n)$ be number of the distinct prime factors of a $n$. Arrange the distinct prime factors in ascending order. The median prime factor $p_m(n)$ of $n$ is defined as the middle prime factor in this ordered list and is given by the prime factor in the position $\left \lfloor \frac{\omega(n)+1}{2} \right \rfloor$. Clearly if $n$ is prime then $p_m(n) = n$. If $\omega(n)$ is even then although there are two primes in the middle, our definition implies that we consider the smaller of these two middle most primes as our median prime.

Question: Is it true that

$$ \sum_{k=2}^n p_m(k) \sim \frac{n^2}{2 \log n} $$

Experimental data for $n \le 1.5 \times10^{9}$ give the constant in the RHS as $0.51244$ and the this value is decreasing at a very slow rate provoding evidence to the conjecture that it approach $1/2$?

Nilotpal Sinha
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    Relevant: https://citeseerx.ist.psu.edu/document?repid=rep1&type=pdf&doi=87327d56b51dd26029c9de0e677ff6eaec2e124f – Zubin Mukerjee Nov 05 '24 at 19:48
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    Also relevant: https://projecteuclid.org/journals/illinois-journal-of-mathematics/volume-68/issue-3/The-distribution-of-intermediate-prime-factors/10.1215/00192082-11417186.short – Zubin Mukerjee Nov 05 '24 at 19:54
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    Limit distribution: https://www.jeanmariedekoninck.mat.ulaval.ca/fileadmin/Documents/Publications/2019_the_limit_distribution_of_the_middle_prime_factors_of_an_integer.pdf – Zubin Mukerjee Nov 05 '24 at 19:55
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    A more precise asymptotics is $$ \operatorname{Li} (n^2 ) + \mathcal{O}!\left( \frac{n^2}{\log n}{\rm e}^{ - c\sqrt{\log n} } \right) $$ with some $c>0$. – Gary Nov 06 '24 at 03:09

1 Answers1

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It is well-known that the sum of just the primes up to $n$ is asymptotic to $\frac{n^2}{2 \log n}$. So what you're asking is to show that the contribution from the non-prime values is negligible.

It's easy to see that if $k$ has at least $2$ distinct prime factors, then $p_m(k) < \sqrt{k}$: that's because $k$ must have at least one prime factor that's strictly greater than $p_m(k)$ (as observed by OP, in the edge case of exactly $2$ distinct primes, we round towards the lower one).

Likewise, if $k$ is not prime but $\omega(k)=1$, then $k$ is a power (i.e. a square or higher) of $p_m(k)$ and so $p_m(k) \le \sqrt{k}$.

Therefore the contribution from non-primes up to $n$ is at most $\sum_{k=2}^n \sqrt{k} = O(n^{3/2})$, which is certainly negligible compared to $n^2/\log n$.

So the answer to the question is YES, without needing to know anything about the fine distribution of middle prime factors.

Erick Wong
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