we know that the probability that a given $n\in\mathbb{N}$ is a prime is $\frac{1}{\log n}$ and all primes except 2 and 3 are of form $6n\mp 1$. We can deduce that the probability that $6n-1$ is $\frac{1}{2\log n}$? Thank you.
2026-02-23 01:05:18.1771808718
Primes of form $6n-1$
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The primes are almost evenly distributed between $6n+1$ and $6n-1$ (and this goes for any such division, like $4n+1$ vs $4n-1$, or $8n+1$ vs $8n+3$ vs $8n+5$ vs $8n+7$). More rigorously, it is known that for a natural number $k$, the ratio between primes below $k$ of one form and primes below $k$ of the other form tends to $1$ as $k$ goes to $\infty$. So as far as probability estimates for large $k$ are concerned, it's relatively safe to assume that they are the same.
However, it has been observed that primes which are quadratic residues (i.e. $6n+1$) do seem to be slightly less common if you look at the difference rather than the ratio. This phenomenon is known as Chebyshev's bias. As pointed out by Joriki in the comments below, this bias has been proven assuming the generalized Riemann hypothesis and the grand simplicity hypothesis, but we don't know yet whether it's actually true.