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Let $\mathbb{N}_+$ denote the set of positive integers. Consider the remainder function $\text{rem}:\mathbb{N}_+\times \mathbb{N}_+ \to \mathbb{N}\cup\{0\}$ defined by $$(n,d) \mapsto n - \Big(\Big\lfloor\frac{n}{d}\Big\rfloor\cdot d\Big) \in \{0,\ldots,d-1\}.$$

If $S\subseteq \mathbb{N}_+$ we define the upper density of $S$ to be $\mu^+(S) = \lim\sup_{n\to\infty}\frac{|S\cap \{1,\ldots,n+1\}|}{n+1}.$

For $n\in\mathbb{N}_+$ let $r_n:\mathbb{N}_+\to \mathbb{N}\cup\{0\}$ be defined by $x\mapsto \text{rem}(x,n).$ For $n\in\mathbb{N}_+$ and $S\subseteq \mathbb{N}_+$ we say that $S$ is remainder-balanced for $n$ if $$\mu^+\Big(r_n^{-1}(\{a\})\cap S\Big) = \mu^+\Big(r_n^{-1}(\{b\})\cap S\Big)$$ for all $a,b\in\{1,\ldots,n-1\}$.

Question. Is every prime remainder-balanced for $P$?

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    $\begingroup$ 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. $\endgroup$
    – Martin Sleziak
    Commented Jan 15, 2023 at 12:06
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    $\begingroup$ 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.) $\endgroup$
    – Martin Sleziak
    Commented Jan 15, 2023 at 12:12
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    $\begingroup$ 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 $\endgroup$
    – Wojowu
    Commented Jan 15, 2023 at 12:12
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    $\begingroup$ Will correct the mistakes -- apologies for this bad question. $\endgroup$
    – Dominic van der Zypen
    Commented Jan 15, 2023 at 12:18
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    $\begingroup$ @MartinSleziak Thank you - apologies for my sloppiness $\endgroup$
    – Dominic van der Zypen
    Commented Jan 15, 2023 at 12:34

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