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Let $q_n$ be the $n$-th number in $\{\,p\in\mathbb P\,|\,p\equiv 3\pmod 4\,\}$ and define $r_n=\frac{\pi^2}{4}n\ln n$, then
$1\leq n\leq 1973\implies r_n\leq q_n$.

enter image description here

$\qquad\quad q_n$ is the blue curve and $r_n$ is the red curve.

But $r_{1974}=36958>36947=q_{1974},\;r_{10000}-q_{10000}=2989$ and $r_{100000}-q_{100000}=93029$. For $n>1800$ prime numbers of this form starts to come closer and the nice formula for $r_n$ loose all its accuracy.

 n   qn   rn
 1    3    0
 2    7    3
 3   11    8
 4   19   14
 5   23   20
 6   31   27
 7   43   34
 8   47   41
 9   59   49
10   67   57
11   71   65
12   79   74
13   83   82
14  103   91
15  107  100
16  127  109
17  131  119
18  139  128
19  151  138
20  163  148

Are there known formulas to estimate $q_n$?


Guided by a comment I managed to look up the formula $\pi(x;d,a)\sim\frac{x}{\varphi(d)\ln x}$, where $\varphi(d)$ is the same for primes of the form $nd+a$. In this case $\varphi(4)=2$ is the constant to replace $\frac{\pi^2}{4}$.

The formula above comes from A Computational Introduction to Number Theory and Algebra, Victor Shoup

Lehs
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    Half and half. So not particularly close to $\pi^2/4$.With the natural generalization to primes congruent to $a\pmod{m}$, where $\gcd(a,m)=1$. – André Nicolas Aug 07 '16 at 02:07
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    I have previously posted a question which might be of interest to you, regarding the distribution of primes that are equal $1\pmod4$ vs the distribution of primes that are equal $3\pmod4$, and regarding the distribution of primes that are equal $1\pmod6$ vs the distribution of primes that are equal $5\pmod6$. It's a little long, but in case you want to have a look, you can find it here. – barak manos Aug 07 '16 at 08:08
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    This paper may be of interest to you. –  Aug 07 '16 at 08:10
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    @barakmanos: I must have seen your post before because I upvoted it. But now, when my own interest is right on the same spot, I find your approach very interesting. Especially the asymmetry between different congruence classes, beyond the invarians in the extended Gauss formula in the end of my question. – Lehs Aug 07 '16 at 09:28
  • @user170039: Thanks for that interesting link! – Lehs Aug 07 '16 at 09:34

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