Period of Reciprocal of 53 is One Quarter of Maximal

Theorem
The decimal expansion of the reciprocal of $53$ has $\dfrac 1 4$ the maximum period, that is: $13$:
 * $\dfrac 1 {53} = 0 \cdotp \dot 01886 \, 79245 \, 28 \dot 3$

Proof
From Reciprocal of $53$:


 * $\dfrac 1 {53} = 0 \cdotp \dot 01886 \, 79245 \, 28 \dot 3$

Counting the digits, it is seen that this has a period of recurrence of $13$.

From Maximum Period of Reciprocal of Prime, the maximum period of recurrence of $\dfrac 1 p$ is $p - 1$.

We have that:
 * $13 = \dfrac {53 - 1} 4$

Hence the result.