Any number with of a form $\frac{1}{n}$ has a decimal with a repetend of finite length that is never longer than $n$ (provable by Dirichlet principle). (Example: $\frac{92}{99}=0.929292\ldots$ in which case it is 92 that is repeating and the length of the series is 2.) Is there a way to find ALL numbers of the form $\frac{1}{n}$ which have repetends of length EXACTLY $n$? Are there infinitely many of them and does $\sum_{\text{$n$ with this property}}\frac{1}{n}$ converge or not?
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3$\begingroup$ Your question is about numbers of the form $\frac{1}{n}$, but your example is of the form $\frac{m}{n}$. Wouldn't $\frac{1}{99} = 0.010101\dots$ be better? $\endgroup$– Michael AlbaneseCommented May 2, 2022 at 12:10
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11$\begingroup$ The period of the decimal expression for $1/n$ is at most $n-1$, not $n$. Equality holds if and only if $n$ is prime and $10$ is a primitive root modulo $n$. $\endgroup$– Tom De MedtsCommented May 2, 2022 at 12:18
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This is a textbook example of a question for which one should turn to the OEIS for assistance. The first few elements of this set are
$$ 7, 17, 19, 23, 29, 47, 59, 61, 97, 109, 113, 131, 149, 167, 179, 181, 193, 223, 229, 233, 257, \dotsc$$
and the OEIS then tells you that these are the full reptend primes in base 10 (OEIS A001913). In fact just entering the first three elements 7, 17, 19 of this sequence into the OEIS will return the full reptend primes as the top search result.
Artin's primitive root conjecture predicts that this set has asymptotic density $$ \prod_p \left(1 - \frac{1}{p(p-1)}\right) = 0.373955\dots$$ in the set of all primes; since the sum of reciprocals of primes diverges, it thus predicts that your sum also diverges. This conjecture is known (Hooley - Artin's conjecture) under a sufficiently strong version of GRH, but remains open unconditionally; even the weaker statement that there are infinitely many such primes for a fixed base is unknown, though it is known for instance (see Heath-Brown - Artin's conjecture for primitive roots) that there are at most two prime bases for which the latter claim fails.
answered May 2, 2022 at 16:51
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1$\begingroup$ This is assuming that the OP wants their reptends to have length $n-1$ instead of $n$. $\endgroup$ Commented May 3, 2022 at 8:11
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1$\begingroup$ @GregMartin: That seems a reasonable assumption, since otherwise the question is trivial — the repetend always has length strictly less than $n$. $\endgroup$ Commented May 3, 2022 at 9:44