# Sequence of Smallest Numbers whose Reciprocal has Period n

## 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:

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

## 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 $\displaystyle \frac 1 {239}$ $=$ $\displaystyle 0 \cdotp \dot 00418 \, 4 \dot 1$ $\displaystyle \frac 1 {73}$ $=$ $\displaystyle 0 \cdotp \dot 01369 \, 86 \dot 3$ $\displaystyle \frac 1 {81}$ $=$ $\displaystyle 0 \cdotp \dot 01234 \, 567 \dot 9$