Fix an integer $a>1$. For $n \geq 1$ an integer, let $\pi_{n,1}(an)$ the number of primes $p \leq an$ such that $p \equiv 1 \pmod{n}$, and $\pi(an)$ the number of all primes $p \leq an$. Let $$Q_a(n) = \frac{\pi_{n,1}(an)}{\pi(an)} \phi(n),$$ where $\phi(n)$ is Euler's phi function.

If instead of fixing $a$ we fix $n$ and let $a$ goes to infinity, then by Dirichlet's theorem that $\lim_{a \rightarrow \infty} Q_a(n) = 1$. If we don't fix $n$ but let it goes to infinity fast enough relatively to $a$, for example $n=a^{1+\epsilon}$ with $\epsilon>0$ under (GRH), then one can prove that the limit is still 1 by some effective version of Dirichlet. But I am interested here in the case where $a$ is fixed. In this case, it is clear that $Q_a(n)$ varies to widely to have a limit when $n \rightarrow \infty$.

Hence let us tame $Q_a(n)$ by considering, following Cesaro, $C_a(n) = \frac{Q_a(1)+\dots+Q_a(n)}{n}$.

Does $C_a(n)$ have a limit when $a$ is fixed and $n$ goes to infinity ? If so what is this limit ?

I have made some sage computations for different values of $a$ ($a=2$ to $10$) and $C_a(n)$ seems to have a tendency to grow very slowly, though it is not clear if it is toward a finite limit or $+\infty$ -- or if the whole thing is just an artefact.

My motivation is trying to understand (if only conjecturally), in the simplest case I can think of, what happens to the effective Chebotarev density theorem beyond the version I can find in the literature. I'll appreciate any answer, be it unconditional, based on a conjecture like GRH, or even purely heuristic.

**Comments (added on October 16th):** I am interested in the question above for any integer $a$ but in my research a similar question arose with $a=8$. Actually I don't think the answer will really change of nature with $a$, so we can focus on the case $a=2$. In this case,
the only number congruent to $1$ modulo $n$ between $1$ and $2n$ which is susceptible to be prime is $n+1$, hence $Q_2(n)=0$ if $n+1$ is not prime, while when $n+1$ is a prime $p$,
$$Q_2(n)=\frac{\phi(p-1)}{\pi(2 (p-1))}.$$
By the way, this illustrates the fact that $Q_2(n)$ does not have a limit when $n \rightarrow \infty$: $0$ is obviously the inf.lim., but $+\infty$ is the sup.lim. : think on $n=p-1$
being for example the higher prime I'm a Sophie Germain's pair of prime, so that $\phi(n)=(p-1)/2-1$ and $Q_2(p-1) \sim \log p / 4$ which goes to inanity if one choses an infinite sequence of Germain's primes (which is widely expected to exist -- at this stage, I am perfectly happy to use any conjecture even if one can do otherwise).

Back to the question, one has: $$C_2(n) = \frac{1}{n} \sum_{1 < p \leq n,\ p \ \rm prime} \frac{\phi(p-1)}{\pi(2p-2)}.$$ Note that since $Q_2(n)$ is non-negative, one can replace it by an equivalent, so $$C_2(n) \sim D_2(n) := \frac{1}{n} \sum_{1 < p \leq n,\ p \ \rm prime} \frac{\phi(p-1) \log(p)}{2p},$$ and the question begins to look like a complicated version of a question already asked several times on mathoverflow about the Cesaro average of Euler's $\phi(n)$, which behaves much more smoothly than $\phi(n)$ itself. Experimentally, here is what I get for $C_2(n)$ for $n=2^k$, $k$ running from $1$ to $23$:

0.750000000000000 0.500000000000000 0.300000000000000 0.254482323232323 0.204751427085986 0.182394996041895 0.174044947095252 0.177096489596196 0.177412757367371 0.175004984083009 0.175280949354989 0.176774240882088 0.177008402332853 0.178414103595542 0.178516411591865 0.179091173423042 0.179809089385918 0.180252447106263 0.180775697751659 0.181112338150868 0.181529153981739 0.181858564625316 0.182136158910456

To me it looks like $C_2(n)$ has a limit or perhaps goes to infinity a little bit slower than $\log n$, but I am really not well-trained in the difficult art of divination of limits of sequences from their first terms... What do you think?

knownthat $n=a$ is fast enough? That would imply in particular a Linnik constant of at most 2. $\endgroup$ – Noam D. Elkies Oct 15 '12 at 1:05