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 an analytic 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 in the absence of 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?

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?