Sequence of Smallest Numbers whose Reciprocal has Period n

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Theorem

Let $\left\langle{s_n}\right\rangle$ be the sequence defined as:

$s_n$ is the smallest positive integer the decimal expansion of whose reciprocal has a period of $n$

for $n = 0, 1, 2, \ldots$


Then $\left\langle{s_n}\right\rangle$ begins:

$1, 3, 11, 27, 101, 41, 7, 239, 73, 81, 451, \ldots$

This sequence is A003060 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


Proof

Demonstrated by inspection and calculation:

\(\displaystyle \frac 1 1\) \(=\) \(\displaystyle 1 \cdotp 0\) hence has a period of $0$
\(\displaystyle \frac 1 3\) \(=\) \(\displaystyle 0 \cdotp \dot 3\)
\(\displaystyle \frac 1 {11}\) \(=\) \(\displaystyle 0 \cdotp \dot 0 \dot 9\)
\(\displaystyle \frac 1 {27}\) \(=\) \(\displaystyle 0 \cdotp \dot 03 \dot 7\) Period of Reciprocal of 27 is Smallest with Length 3
\(\displaystyle \frac 1 {101}\) \(=\) \(\displaystyle 0 \cdotp \dot 009 \dot 9\)
\(\displaystyle \frac 1 {41}\) \(=\) \(\displaystyle 0 \cdotp \dot 0243 \dot 9\)
\(\displaystyle \frac 1 7\) \(=\) \(\displaystyle 0 \cdotp \dot 14285 \dot 7\) Period of Reciprocal of 7 is of Maximal Length
\(\displaystyle \frac 1 {239}\) \(=\) \(\displaystyle 0 \cdotp \dot 00418 \, 4 \dot 1\)
\(\displaystyle \frac 1 {73}\) \(=\) \(\displaystyle 0 \cdotp \dot 01369 \, 86 \dot 3\)
\(\displaystyle \frac 1 {81}\) \(=\) \(\displaystyle 0 \cdotp \dot 01234 \, 567 \dot 9\)


Sources