# Question #5bc5b

May 18, 2017

${F}_{5} = \left\{\frac{0}{1} , \frac{1}{5} , \frac{1}{4} , \frac{1}{3} , \frac{1}{2} , \frac{2}{3} , \frac{3}{4} , \frac{4}{5} , \frac{1}{1}\right\}$
${F}_{6} = \left\{\frac{0}{1} , \frac{1}{6} , \frac{1}{5} , \frac{1}{4} , \frac{1}{3} , \frac{1}{2} , \frac{2}{3} , \frac{3}{4} , \frac{4}{5} , \frac{5}{6} , \frac{1}{1}\right\}$
${F}_{7} = \left\{\frac{0}{1} , \frac{1}{7} , \frac{1}{6} , \frac{1}{5} , \frac{1}{4} , \frac{1}{3} , \frac{1}{2} , \frac{2}{3} , \frac{3}{4} , \frac{4}{5} , \frac{5}{6} , \frac{6}{7} , \frac{1}{1}\right\}$

#### Explanation:

The trick to recognizing these Farey sequences is to first look at the denominators.

For ${F}_{3}$, the denominators (in order) are:

$\left\{1 , 3 , 2 , 3 , 1\right\}$

The number 3 is the biggest denominator and there are 5 terms.

For ${F}_{4}$, the denominators are:

$\left\{1 , 4 , 3 , 2 , 3 , 4 , 1\right\}$

The number 4 is the biggest denominator and there are 7 terms.

The pattern for the denominators of all subsequent Farey sequences should be obvious.

The number of terms is always increased by 2. So, for ${F}_{5}$, the number of terms will be 9. And for ${F}_{6}$, the number of terms will be 11.

The first term is always $0 / 1$. The last term is always $1 / 1$. The middle term is always $1 / 2$. After 0, the numerators up to $1 / 2$ will always be $1$. The numerators after $1 / 2$ will simply count from 2 to the max number.

It may be easier to see if you let the last term $\frac{1}{1}$ be equal to the maximum denominator, divided by itself. So,

Last term of ${F}_{3}$ is $3 / 3$
Last term of ${F}_{4}$ is $4 / 4$
Last term of ${F}_{5}$ is $5 / 5$, etc.

There is always two more terms shoved in each subsequent Farey sequence. Let $m$ be the max denominator. Always, just after the zero term (the first term in the sequence) will be $1 / m$. And finally, right before the final 1 term (the next to last in the sequence) will be $\left(m - 1\right) / m$.

I'll let you explore the "questions to consider" on your own!