On the number of integers whose prime factors are congruent to 1 modulo some given number. Let $m\ge2$ be an integer,
$$
A_m=\{k\in\mathbb{N}:\text{all prime factors of $k$ are}\equiv1\pmod m\},
$$
and $C_m(x)$ the counting function of $A_m$, that is, for $x>0$
$$
C_m(x)=\#\{k\in A_m:k\le x\}.
$$
For instance $A_2$ is the set of odd integers, and $C_2(x)\sim x/2$. As another example,
$$
A_3=\{1, 7, 13, 19, 31, 37, 43, 49, 61, 67, 73, 79, 91, 97, 103,\dots\}
$$
is sequence A004611.
By the prime number theorem on arithmetic progressions we have
$$
\frac{1}{\phi(m)}\,\frac{x}{\log x}\le C_m(x)\le\frac{x}{m}.
$$
Is anything known about the asymptotic behavior of $C_m(x)$ for $m>2$? 
 A: I find this particular problem interesting. So we are basically looking at the probability that a random integer $n$, all primes $p | n$ are congruent to $1\pmod m$. A well known mathematical result states that a random integer $n$, has on average, $\ln(\ln n)$ prime factors. So the probability all of these prime factors are congruent to $1 \pmod m$ is about $\left(\frac{1}{\phi(m)^{\ln(\ln n)}}\right)$ and we should expect
$$C_m(n) ≈ \sum_{k=1}^n \left(\frac{1}{\phi(m)^{\ln(\ln k)}}\right)$$
Note that the sum diverges for any particular $m$.
There is a better estimate: For all primes $q ≠ 1 \pmod m$, we want the probability such that $n$ is not divisible by $q$, which is $\left(\frac{q-1}{q}\right)$. In particular, for all primes $k<\sqrt n$:
$$K(n) = \prod_{k<\sqrt{n}} \left(\frac{k-1}{k}\right)$$
and for all primes $p=1\pmod m$
$$P_m(n) = \prod_{p<\sqrt{n}} \left(\frac{p-1}{p}\right)$$
The probability $n$ has all prime factors congruent to $1\pmod m$ is $\left(\frac{K(n)}{P_m(n)}\right)$.
Like with the previous estimate, we should be able to sum these probabilties to get an estimate for $C_m(n)$.
