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Let $R_n:=\frac{10^n-1}{10-1}$ (called a repunit) and $\mu$ be the Moebius function. Also $[n]:=\{1,2,3,\cdots, n\}, A_n:=\{m \in [n]| \mu (R_m)=0\}.$

What is the value of $\lim \limits_{n \rightarrow \infty} \frac{|A_n|}{n}?$

Find the values of that for $\mu(R_n)=1$ and $\mu(R_n)=-1.$

I computed several values using PIE.

$r_1=0.111111\cdots$, $r_2=0.151515\cdots$, $r_3=0.165945165945\cdots$, $r_4=0.1726517265\cdots$

Then $\lim \limits_{n \rightarrow \infty} \frac{|A_n|}{n}=\lim \limits_{n \rightarrow \infty} r_n. $

First minimal nonsquarefree repunit is $R_9$. So, $r_1= \frac{1}{9}$. Second minimal nonsquarefree repunit $R_n$ where $n$ is not multiple of 9 is $R_{22}$. So, $r_2= \frac{1}{9}+\frac{1}{22}-\frac{1}{198},$ etc.

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    For example, $1/11=0.09090909...$ which has period 2. Hence $p|R_2$.2012-07-26

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