Squarefree products of a class of primes

Numbers which are the sum of two squares are the product of a square and a collection of distinct primes which are 1 or 2 mod 4.

Landau proved that there are $\sim kx/\sqrt{\log x}$ such numbers up to $x$ for a constant $k\approx0.76422.$ Restricting to the squarefree members reduces the incidence by a factor of $\zeta(2).$

So there are $\sim0.46459\ldots x/\sqrt{\log x}$ products of primes =1,2 mod 4 up to $x$. Does this generalize to other congruence classes of primes? That is, given $m$ and some set $S$ and numbers which are the product of distinct primes $\equiv s\pmod m$ for some $s\in S$ (with at least one $s$ coprime to $m$), is their density $kx/\sqrt{\log x}$ for some suitable constant $k$?

Bonus: Is there a good way to find the constant given $m$ and $S$?

• Sure. See proof method in volume 2 of LeVeque Topics in Number Theory. The connection with quadratic forms is that the set of "bad" primes is those for which $(\Delta|q) = -1.$ This reduces to representations by a single form with class number one, which happens just a few times for positive forms. – Will Jagy Apr 20 '15 at 18:48
• @WillJagy: Thanks -- post as an answer, maybe? – Charles Apr 20 '15 at 19:07
• Let $S\subset\mathcal{P}$ be some subset of primes with relative density $\delta$. Let $A$ be the set of integers which can be written as products of primes in $S$. Then we expect that $$\sum_{n\leq x}1_A(n)\sim\frac{kx}{(\log x)^{1-\delta}}$$ for some constant $k$ depending on $S$. We may need $S$ to be reasonably well behaved, such as the example of primes congruent to $s$ mod $m$, but in general this is the statement that we should be looking to prove. This is related to the singularity at $s=1$. – Eric Naslund Apr 27 '15 at 21:54