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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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    You have not defined these $r_i$.2012-07-11
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    You still haven't defined the $r_i$, at any rate I can't extract a definition from your calculation of the first two values.2012-07-11
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    Next one is $R_{42}$. So $r_3 = \frac{1}{9} + \frac{1}{22} + \frac{1}{42} - \frac{1}{(9,22)} - \frac{1}{(9,42)} - \frac{1}{(22,42)} + \frac{1}{(9,22,42)},$ where $(a,b)$ in the least multiple of $a$ and $b.$2012-07-11
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    Please, write up a *definition* of $r_i$ and edit it into the body of the question.2012-07-11
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    Note that $R_n | R_m $ if $n|m$.2012-07-11
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    You might be interested in http://oeis.org/A046412 which lists the first 55 nonsquarefree repunits. It also has links to large tables of factors of repunits.2012-07-12
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    Another edit, another failure to put in a definition of $r_i$. I can only conclude that you don't know what $r_i$ means, yourself.2012-07-13
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    I mean $(a,b)$ is the least commom multiple of $a$ and $b$.2012-07-13
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    For a prime $p$, $ p|R_n$ if $1/p$ has a period $n$. Exception : $p$ is not 2,3,5!!!2012-07-26
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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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