TITLE: Prime races à la Mertens
QUESTION [8 upvotes]: I have just read the nice survey by Granville and Martin about prime races.
I wonder what happens if one changes the rules for the prime races as follows.
Fix $q$ a modulus (an integer $>1$). For $a$ an integer relatively prime to $q$, one has a Mertens-like formula 
$$\sum_{p<x, \atop p \equiv a \pmod{q}} \frac{1}{p} = \frac{1}{\phi(q)} \log\log x + M_{q,a} + O(1/\log x)$$ 
where $M_{q,a}$ is a constant. We can say that the "team $a$" wins the race mod $q$ if the constant $M_{q,a}$ is greater than the constants $M_{a,b}$ for others $b \pmod{q}$, and we can even make a complete ranking of the various teams (with possible ties). This is a reasonable notion, since stating that $M_{q,a} > M_{q,b}$ implies that the number of primes $p$ up to $x$ counted with harmonic density $1/p$ which are congruent to $a$ mod $q$ is greater than the same number for congruent to $b$ mod $q$, for every $x$ large enough. 

So what can be said about the ranking of the $M_{q,a}$'s for various $a$ mod $q$?

For example, with sage I have computed approximation of the values of $M_{8,a}$ for $a=1,3,5,7$, and I find numbers close to $-0.28,0.16,0.00,-0.11$ making the "team 3" the clear winner.

REPLY [10 votes]: I think this paper gives a pretty good answer:
http://projecteuclid.org/euclid.em/1317758092
In the appendix, Languasco and Zaccagnini give a proof of Norton that if $q \geq 2$ and $1 \leq a < q$ with $(q,a) = 1$, then as $q \to \infty$,
$$M_{q,a} = \begin{cases}
\displaystyle \frac{1}{a} + O\left(\frac{\log q}{\varphi(q)}\right) & \text{if $a$ is prime,} \\
\displaystyle O\left(\frac{\log q}{\varphi(q)}\right) & \text{otherwise.}
\end{cases}$$
This tells you that the race ought to be won by the smallest prime coprime to $q$, at least for $q$ sufficiently large. If you race two nonprimes against each other, however, then this says nothing.