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Timeline for Remainder-balancedness of primes

Current License: CC BY-SA 4.0

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Jan 21, 2023 at 11:03 history edited Dominic van der Zypen CC BY-SA 4.0
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Jan 21, 2023 at 11:03 comment added Dominic van der Zypen Thanks @WlodAA - will reformulate!
Jan 21, 2023 at 10:28 comment added Wlod AA The last sentence before Question seems to be messed up (just in case, you could fix it perhaps). Anyway, this sounds as the weak version of Dirichlet Theorem (if I managed to parse your formulation well enoiugh).
Jan 21, 2023 at 7:51 history edited Dominic van der Zypen CC BY-SA 4.0
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Jan 15, 2023 at 21:44 history edited Dominic van der Zypen CC BY-SA 4.0
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Jan 15, 2023 at 12:34 comment added Dominic van der Zypen @MartinSleziak Thank you - apologies for my sloppiness
Jan 15, 2023 at 12:21 history edited Dominic van der Zypen CC BY-SA 4.0
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Jan 15, 2023 at 12:18 comment added Dominic van der Zypen Will correct the mistakes -- apologies for this bad question.
Jan 15, 2023 at 12:12 comment added Wojowu I think you forgot to make your definition of remainder-balanced actually depend on $S$. If you meant $\mu^+(r_n^{-1}(\{a\})\cap S)$ in the definition, then the result is trivially true as primes, and all their subsets, have upper density $0$. If you meant to use relative density, the result is trivially false, as Martin points out. A corrected version is given by PNT in APs
Jan 15, 2023 at 12:12 comment added Martin Sleziak BTW is "$\mu^+\big(r_n^{-1}(\{a\}\big)$" (with the right bracket missing) suppose to be "$\mu^+\big(r_n^{-1}(\{a\}\cap S)\big)$"? (I see that Wojowu was faster than me - I was looking for appropriate link concerning the same result.)
Jan 15, 2023 at 12:06 comment added Martin Sleziak Maybe I misunderstood something, but it seems that $4\notin RB(P)$, since the only prime belonging to $4\mathbb N+2$ is two. And you can use basically the same argument for any integer which is not a prime.
Jan 15, 2023 at 11:54 history asked Dominic van der Zypen CC BY-SA 4.0