# Number of primes with $-1\pmod 6$ vs. Number of primes with $+1\pmod 6$

Have not been able to get an answer to this on http://math.stackexchange.com, so trying here too...

Given the following two sets:

• $P^-(n) = \{p \leq n : p \equiv -1\pmod 6\}$
• $P^+(n) = \{p \leq n : p \equiv +1\pmod 6\}$

For example:

• $P^-(40) = \{5,11,17,23,29\}$
• $P^+(40) = \{7,13,19,31,37\}$

Given the following two functions:

• $C^-(n)=|P^-(n)|$
• $C^+(n)=|P^+(n)|$

For example:

• $C^-(40) = 5$
• $C^+(40) = 5$

Questions:

1. Has it been proved that $\forall k \exists n : k=|C^-(n)-C^+(n)|$?

2. Has any bound been proved for $|C^-(n)-C^+(n)|$ relatively to $n$ (e.g., $\ln \ln n$)?

3. What is the largest known value of $|C^-(n)-C^+(n)|$, and for what value of $n$ does it hold?

• You might find this paper interesting dms.umontreal.ca/~andrew/PDF/PrimeRace.pdf – Jonah Sinick May 12 '14 at 7:28
• @Jonah Sinick: Thanks. I have already looked into the Prime Races subject, and found out that it's mostly about prime numbers of the form $4n\pm1$ (not $6n\pm1$ as in my question). Have I missed anything? – barak manos May 12 '14 at 7:50
• I was going to link to the same paper. It talks about primes which are $3n+1$ vs $3n+2$ as well. – Douglas Zare May 12 '14 at 8:08
• You got a very good answer on m.se, you just don't know one when you see it. – Gerry Myerson May 12 '14 at 9:20
• The m.se question: math.stackexchange.com/questions/787641/… – Janne Kokkala May 12 '14 at 10:17

Assume that the Riemann hypothesis for the non-principal $L$-series $\pmod{3}$ is false, say, this series has a zero $\rho=\sigma+i\gamma$ with $\sigma>1/2$. Then Turan and Knapowski have shown that both $C^-(n)-C^+(n)>n^{\sigma-\epsilon}$ and $C^-(n)-C^+(n)<-n^{\sigma-\epsilon}$ happen infinitely often.
If the Riemann hypothesis holds true, then $\frac{C^-(e^t)-C^+(e^t)}{e^{t/2}}$ is (up to a small correction) almost periodic, hence if there exists a sufficiently large value for $n$ such that $C^-(n)-C^+(n)>0.01\sqrt{n}$, say, then there are infinitely many $n$ such that $C^-(n)-C^+(n)>0.0099\sqrt{n}$, and similarly in the other direction.
Hence to prove that $C^-(n)-C^+(n)$ is unbounded in both directions one has to compute the corrections to make the difference almost periodic, and look for $n$ which make the difference large. You can model the proof on Kaczoroswki's article "On the Shanks-Rényi Race Problem mod 5", J. Number Theory 50, 106-118.
If you assume the Riemann hypothesis together with linear independence of the imaginary parts of the zeros of the $L$-series, then the various summands in the explicit formula behave like independent random variables, and you immediately get that $\limsup\frac{C^-(n)-C^+(n)}{\sqrt{n}}=\infty$, $\liminf\frac{C^-(n)-C^+(n)}{\sqrt{n}}=-\infty$.