Let
$$G(n)=\begin{cases}1 &\text{if }n \text{ is a prime }\equiv 3\bmod17\\0&\text{otherwise}\end{cases}$$
And let
$$P(n)=\begin{cases}1 &\text{if }n \text{ is a prime }\\0&\text{otherwise.}\end{cases}$$
How to prove that
$$\lim_{N\to\infty}\frac{\sum\limits_{n=1}^N G(n)}{\sum\limits_{n=1}^N P(n)}=\frac1{16}$$
And what is $$\lim_{N\to\infty} \frac{\sum\limits_{n=1}^N n\,G(n)}{\sum\limits_{n=1}^N n\,P(n)}?$$
And what is $O(f(n))$ of the fastest growing function $f(n)$ such that the following limit exists: $$\lim_{N\to\infty} \frac{\sum\limits_{n=1}^N f(n)\,G(n)}{\sum\limits_{n=1}^N f(n)\,P(n)}$$
And does this all follow directly from the asymptotic equidistribution of primes modulo most thing, if such a thing were known? And is it known?
The first sum follows from Siegel-Walfisz_theorem
Summation by parts on the second sum should yield for large $N$:
$$\frac{\sum\limits_{n=1}^N n\,G(n)}{\sum\limits_{n=1}^N n\,P(n)}=\frac{(N\sum\limits_{n=1}^N G(n))-\sum\limits_{n=1}^{N-1}\sum\limits_{k=0}^{n} G(k)}{\sum\limits_{n=1}^N n\,P(n)}=\frac{(N\sum\limits_{n=1}^N P(n)/16)-\sum\limits_{n=1}^{N-1}\sum\limits_{k=0}^{n} P(k)/16}{\sum\limits_{n=1}^N n\,P(n)}=\frac1{16}$$