# Prime race modulo $12$. When is the first sign change?

Define $\pi(x;q,a)$ as the number of primes less than or equal to $x$ which are congruent to $a$ modulo $q$.

Up to $x=10^{11}$ we have $\pi(x;12,1) \le \pi(x;12,-1)$.

What is the smallest $x$ for which $\pi(x;12,1) > \pi(x;12,-1)$?

Assuming GRH the sign changes infinitely often.

• That being said, Jason Sneed has apparently proven unconditionally that this does indeed change sign infinitely often (see math.uiuc.edu/~ford/wwwpapers/barriersIII.pdf). – Peter Humphries Aug 9 '17 at 13:23
• In any case, the first sign change is probably unknown. The usual way to find sign changes is well known and is based on the work of Ingham. It is the same method that Odlyzko and te Riele use to disprove Mertens conjecture. – Peter Humphries Aug 9 '17 at 13:25
• Greg Martin discusses mod 12 races in arxiv.org/pdf/math/0010086.pdf (although he doesn't specifically consider the current question). – Gerry Myerson Aug 9 '17 at 13:25
• @reuns, no, this is incorrect. Landau's theorem only shows that $\psi(x;q,a) - \psi(x;q,b)$ changes sign infinitely often, but this is not enough to conclude that $\pi(x;q,a) - \pi(x;q,b)$ does too. – Peter Humphries Aug 9 '17 at 14:04
• @reuns, yes, it is complicated; see Odlyzko and te Riele's paper for a similar proof. Basically you need to use many zeroes of these $L$-functions to overcome the constant $C$, which can be computationally quite demanding. – Peter Humphries Aug 9 '17 at 16:33

The comments above point to the paper Chebyshev's Bias by Rubinstein and Sarnak, and the paper of Martin. For context, let's consider first primes mod $3$.
Assuming you're familiar with the terminology of Rubinstein and Sarnak, they compute the bias of primes to be $2$ mod $3$ over $1$ mod $3$ to be $0.9990\ldots$ (This is $\delta(P_{3;N;R})$ in their notation). In 1978, Bays and Hudson computed that $608,981,813,029$ was the smallest integer $x$ such that $\pi_{3,2}(x)<\pi_{3,1}(x)$.
The paper of Martin shows the bias of primes to be $11$ mod $12$ over $1$ mod $12$ is much stronger: $\delta_{12;11,1}=0.999977$ (in Martin's slightly different notation.) Since this is a logarithmic scale, the 'extra' $0.000977$ means that the bias is almost two orders of magnitude stronger.
Meanwhile, Ford and Hudson showed in that your $x$ will be less than $10^{84}$.