The primes are equidistributed in the residue classes $1(\!\!\!\mod{4})$ and $3(\!\!\!\mod 4)$. We also know (for example, by Rubinstein-Sarnak) that the patterns cannot be eventually alternating, i.e. there exists no $n_0\in \mathbb N$ such that for all $n>n_0$ one has $$p_{n+1}\equiv -p_n \mod 4,$$ where $p_n$ is the $n$-th smallest prime. My question is whether there exists an easy proof of this, (maybe unconditional?)
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4$\begingroup$ See here for a simple proof, by MathOverflow's lucia: mathoverflow.net/questions/168378/… $\endgroup$– so-called friend DonCommented Feb 20, 2022 at 14:47
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$\begingroup$ it is great he foresaw my question! $\endgroup$– Dr. PiCommented Feb 21, 2022 at 11:34
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