Consider the following sequence of rational numbers, starting with \( A_ 0 = 0 \) and defined recursively as

\[ A_n = \frac{ 1 } { 2 \lfloor A_{n-1} \rfloor - A_{n-1} + 1 }. \]

Show that every non-negative rational number appears in this sequence exactly once.

The sequence starts off as \( 0 , 1, \frac{1}{2}, 2, \frac{1}{3}, \frac{3}{2}, \frac{2}{3}, 3, \frac{1}{4}, \frac { 4}{3}, \ldots \).

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestClarification : What is \(R_n\)?

Log in to reply

Ooops, it should have been \(A_n\). Corrected.

Log in to reply

numberphile just posted a video on this

Log in to reply

The sequence that is generated is the same, which is the one obtained through the Stern-Brocot tree. There is a pretty nice proof by continued fractions to demonstrate that this sequenece hits all the rationals exactly once.

However, it is not immediately apparent why the above recurrence relation should lead to the Stern-Brocot tree. The relationship between consecutive terms of the Stern-Brocot tree that do not have the same parents is not clear, whereas we have a very simple description in the above recurrence relation.

So yes, one possible way of approach, is to show that the above recurrence relation leads to the Stern-Brocot tree, and hence conclude that it hits each rational number exactly once.

Log in to reply