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Consider $3 \ | \ k, \ k \ge 3$. If $p_R = k + 2$ is a prime, denote it a regular single prime, and if $p_L = k + 4$ is prime, an irregular single prime. Intuitively, with $n \to \infty$ the number of regular and irregular single primes in the range $p \le n$ should be equal: $(N_R(n) - N_L(n))/N_R(n) \to 0, \ n \to \infty$.

Is it a known statement and if so how is it proven?

Serge
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    This seems convoluted. Are you just looking at primes congruent to $2$ or to $1$ $\pmod 3$? If so, then Dirichlet tells us that the two have equal densities. – lulu Dec 27 '20 at 14:42
  • Despite of Chebycheff's biase (non-residues tend to be prefered), in the long run the ratio of both kind of primes will tend to $1/2$. For a similar prime race ($4k+1$ versus $4k+3$) it has been shown that the lead switches infinite many often and I guess this is the case here as well. – Peter Dec 27 '20 at 14:45
  • Nevertheless, both numbers will rarely be exactly equal. The difference between them does NOT converge to $0$. In fact, the numbers cannot be equal if we have an odd number of primes we have checked. So, you have to reformulate the conjecture. – Peter Dec 27 '20 at 14:48
  • Correct, there will be an infinite number of positions with $N_R \ne N_L$, the statement has been corrected. – Serge Dec 27 '20 at 20:09
  • Duplicate of https://math.stackexchange.com/questions/68981/primes-sum-ratio ? – Gerry Myerson Dec 28 '20 at 04:36
  • @GerryMyerson thanks. Probably I misunderstand something in: "If a,d are coprime, then the asymptotic proportion of primes congruent to a modulo d is equal to 1/φ(d), where φ is the totient function" but let's consider sequences: 1) S1: a = 1, d = 6; 2) S2: a = 5, d = 6. Clearly 1) a, d are coprime 2) S1, S2 are mutually exclusive and 3) contain all primes > 5. Then by the theorem the proportion of primes in a sufficiently large range $x$ would be $\sim \frac{2}{6φ(6)}$ and how would that agree with the PNT? – Serge Dec 28 '20 at 21:19
  • The statement you quoted is understood to be giving (asymptotic) proportions among the primes. That is, if you look only at the prime numbers, then the proportion congruent to $a$ modulo $d$ is $1/\phi(d)$. So, half the primes are $1\bmod6$, and half are $5\bmod6$ (asymptotically). – Gerry Myerson Dec 29 '20 at 06:30
  • @GerryMyerson, makes sense thanks. – Serge Dec 30 '20 at 14:25

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