Consider a finite Galois extension $L$ of $\mathbb Q$, of Galois group $G$. Let $k \geq 1$ be a fixed integer. Let $D$ be a subset of $G^k$ invariant by conjugation and by the natural action of the symmetric group $S\_k$ on $G^k$.


Let $A=A(L,D)$ be the set of all integers $n$ which are products of $k$ primes $n=p\_1\dots p\_k$ (variant: $k$ distinct primes) such that 
$(Frob\_{p\_1},\dots,Frob\_{p\_k}) \in D.$ 


(Here $Frob\_p$ is a Frobenius at $p$ in $G$, assuming $p$ is unramified in $L$. if one of the $p\_i$ is ramified, by convention say that $n$ is not in $A$.)


Let $A(x)$ be the number of integers $n<x$ in $A$. 



> 
> 
> > 
> > Is there a known equivalent for $A(x)$ when $x$ goes to infinity?
> > 
> > 
> > 
> 
> 
> 


Note that the case $k=1$ is Chebotarev's density theorem. In this case, $A(x) \sim \frac{|D|}{|G|} x/\log x $. In the general case, I would expect $$A(x) \sim c x/\log x \frac{(\log \log x)^{k-1}}{(k-1)!},$$ with a constant $c$ between $0$ and $1$ depending of $D$, $G$ (and perhaps only of $|D|$, $|G|$), based on the case $k=1$ and the case $D=G^k$, where $A$ is the set of all $k$-almost primes and the desired equivalent is a classical theorem of Landau. Because of these illustrious limit cases, I would not be surprised if this question was dealt with somewhere in the literature. But I couldn't find where.




---


I would already be interested in the case where $G$ is abelian. In this case the question takes a more classical form, which I explicit in case some readers are uncomfortable with Chebotarev. We can assume that $G=(\mathbb Z/a\mathbb Z)^\ast$. Then $D$ is a subset of $G^k$ invariant by permutation of the coordinates. The set $A$ is then the set of integers $n=p\_1\dots p\_k$ such that $(p\_1 \pmod{a},\dots,p\_k \pmod{a})$ is in $D$, and the question is still to find an equivalent for the number $A(x)$ of such integers $n$ which are $\leq x$.


**Edit**: I have made some computations in the simplest non trivial abelian case. This is the case $k=2$, and $L=Q(i)$ so that $G={1,-1}$ and $Frob\_p$ for an odd $p$ is just $p \mod{4}$.
You have $3$ basic subsets $D$ of $G^2$ invariant by $S\_2$ which are 
$D\_1 = \{(1,1)\}$, $D\_0=\{(1,-1),(-1,1)\}$, $D\_{-1}=\{(-1,-1)\}$.
The corresponding sets of integers are the sets $$A\_1=\{2-\text{almost primes } n=pq,\ \ p\equiv q \equiv 1 \pmod{4}\}$$ $$A\_0 =\{2-\text{almost primes } n,\ \ n\equiv -1 \pmod{4}\}$$ $$A\_{-1}=\{2-\text{almost primes } n=pq,\ \ p\equiv q \equiv -1 \pmod{4} \}.$$ I strongly expect in this case that the $A\_{-1},A\_1,A\_0$ take respectively a proportion $1/4,1/4,1/2$ (or $|D|/|G|^2$) of all $2$-almost primes. If true, this is probably not to hard to prove by the method indicated by Lucia in comment. However, I have not been able to "see this" in the computation. For $x=10^7$, for instance, those proportions are $0.300,0.200,0.500$; for $x=5 \cdot 10^7$, they are $0.292,0.500,0.208$. I think the domination of products of two primes congruent to $-1$ above the products of two primes congruent to $1$ is well explained by the "primes race": the advantage of primes congruent to $-1$ among small primes is felt even more for the product of two primes. But this has discouraged me of making more computations to guess the correct value of the constant $c$... 



Chebotarev density theorem for $k$-almost primes
I'll prove a more general result from which the Chebotarev question will follow. Let $P\_1$, $\ldots$, $P\_r$ be disjoint subsets of the primes such that the asymptotic formula 
$$ 
\sum\_{\substack{{p\le x}\\ {p\in P\_j}}} 1 \sim \alpha\_j \frac{x}{\log x} 
$$ 
holds with some $\alpha\_j >0$. Let $N\_k=N(k;a\_1,\ldots,a\_r)$ denote the set of integers that are products of $k$ primes with exactly $a\_j$ of these primes chosen from the set $P\_j$. Thus $a\_1+\ldots+a\_r=k$ and let's assume that all $a\_j \ge 1$.  

I claim that for fixed $k$ and as $x\to \infty$
$$ 
\sum\_{\substack {{n\le x}\\ {n\in N\_k}}} 1 
\sim k\prod\_{j=1}^{r} \frac{\alpha\_j^{a\_j}}{(a\_j)!} \frac{x}{(\log x)} (\log \log x)^{k-1}
$$ 
The argument that follows is standard. The case $k=1$ follows from our assumption, and now suppose that $k\ge 2$. 


Put $z=x^{1/\log \log x}$ and note that by partial summation 
$$ 
\sum\_{p\in P\_j, p\le z} \frac{1}{p} \sim \alpha\_j \log \log z \sim \alpha\_j \log \log x
$$ 
while 
$$ 
\sum\_{p \in P\_j, z< p \le x} \frac{1}{p} \sim \alpha\_j \log \log \log x. 
$$ 
Now suppose $n=p\_1\cdots p\_k$ is an element of $N(k;a\_1,\ldots,a\_r)$. We distinguish two cases: when the largest prime factor of $n$ is bigger than $z$ but all the other prime factors are smaller than $z$, and when the two largest prime factors of $n$ are both larger than $z$. (Of course there is also the case when all prime factors of $n$ are smaller than $z$, but this is clearly negligible.) The first case will be the main term and the second case will be a slightly smaller error. 


Let's look at the first case. Suppose the largest prime factor (say $p$) of $n$ lies 
in the set $P\_j$, so that the product of the smaller primes, say $m$, lies in $N(k-1;a\_1,\ldots,a\_{j}-1,\ldots,a\_r)$. These terms give, upon summing over the last prime $p$, (note that as $m\le z^{k-1}$ we have $\log x \sim \log (x/m)$)  

$$ 
\sim \sum\_{\substack{{m\in N(k-1;\ldots)} \\ {p|m\implies p\le z}} }
\frac{\alpha\_j x}{m \log (x/m)} 
\sim \frac{\alpha\_j x}{\log x} \prod\_{t=1, t\neq j}^{r} \frac{1}{a\_t!} \Big(\sum\_{p\le z, p\in P\_t} \frac 1p\Big)^{a\_t} \frac{1}{(a\_j-1)!} \Big(\sum\_{p\le z, p\in P\_j}\frac 1p\Big)^{a\_j-1}.
$$ 
This is 
$$
\sim a\_j \prod\_{t=1}^{r} \frac{\alpha\_t^{a\_t}}{a\_t!} \frac{x}{\log x} (\log \log x)^{k-1} 
$$
and summing it over $j$ from $1$ to $r$ gives the claimed asymptotic. 


Now for the second case. The sum over the largest prime $p$ gives a quantity bounded by (let $m$ be the product of the remaining primes and note that $m\le x^{(k-1)/k}$)
$$ 
\frac{x/m}{\log (x/m)} \ll \frac{x}{m \log x}. 
$$ 
Now sum this over the possible values of $m$, keeping in mind that $m$ has $k-1$ prime factors, and the largest prime factor of $m$ is assumed to lie in $[z,x]$.  

Thus these terms are bounded by 
$$ 
\ll \frac{x}{\log x} \Big(\sum\_{p\le x} \frac{1}{p} \Big)^{k-2} \Big(\sum\_{z< p\le x} \frac{1}{p}\Big) 
\ll \frac{x}{(\log x)} (\log \log x)^{k-2} (\log \log \log x).
$$ 


**Application to the Chebotarev question.** Suppose $C\_1$, $\ldots$, $C\_r$ are distinct conjugacy classes in $G$. Suppose each element of the set $D$ (which is a $k$-tuple) consists of $a\_1$ entries from $C\_1$, $a\_2$ entries from $C\_2$, $\ldots$, $a\_r$ entries from $C\_r$. A little combinatorics shows that the size of the set $D$ (assumed to be conjugation and permutation invariant) is 
$$
|D| = k!\prod\_{j=1}^{r} \frac{|C\_j|^{a\_j}}{a\_j!}.
$$ 
In our work above take $P\_j$ to be the primes $p$ for which Frobenius lies in the class $C\_j$. Then by Chebotarev our assumed asymptotic for the primes in $P\_j$ holds with $\alpha\_j=|C\_j|/|G|$. It follows that the number of integers in the original question is 
$$ 
\sim \frac{|D|}{|G|^k} \frac{x}{\log x} \frac{(\log \log x)^{k-1}}{(k-1)!}.
$$ 
This answers the Chebotarev question (assuming I got my combinatorics right) at least when $k$ is fixed. When $k$ grows, one has to do more work, but the Selberg-Delange method should work in this case with some effort; note here that one needs to use the structure of sets of Chebotarev primes, rather than just an arbitrary subset of the primes as above. 


