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. 



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> Is $l(n)$ known to have some kind of mean value? If so, how is it calculated?
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Does the least prime factor have a mean of some sort?
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. 


