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Define $l(n)$ to be the least prime factor of $n$ and, say, $l(1)=0$ for simplicity. Obviously we have $2\leq l(n)\leq n$ for $n\geq 2$. There appears to be very little information about the asymptotic behaviour of $l(n)$ available.

One may observe that

$$\sum_1^{\infty}\frac{l(n)}{n^s}=\zeta(s)\sum_p\frac{1}{p^{s-1}}\prod_{q<p}\left(1-\frac{1}{q^{s}}\right)$$

for $\sigma>2$, where $p,q$ are prime and the empty product is unity.

It seems a fair bet that the Dirichlet series on the left has a meromorphic continuation to the region $\sigma>1$ but I haven't proved this. It certainly is singular at $s=2$, thus so is the sum on the right. The type of singularity is not at all obvious, so without further investigation little more information is available from this naive approach.

Is $l(n)$ known to have some kind of mean value? If so, how is it calculated?

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Note that $l(n) \le \sqrt{n}$ unless $n$ is prime. This makes it easy to show that $\sum_{n \le x} l(n) \sim \sum_{p \le x} l(p) = \sum_{p \le x} p \sim \frac{1}{2} \frac{x^2}{\log x}$, as $x\to\infty$. (The last asymptotic formula comes from the prime number theorem and partial summation.) This was noted by Kalecki: On certain sums extended over primes or prime factors. (Polish. Russian, English summary) Prace Mat. 8 1963/1964 121–129.

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  • $\begingroup$ For "least prime factor other than n" you can problem come up with some estimate how many numbers are the product of two large primes, and that will overpower all the smaller "least prime factors" $\endgroup$ – gnasher729 Sep 13 '16 at 8:12

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