This method of Benjamin and Quinn uses a model of tiling a row of squares with stackable tiles and (non-stackable) dominoes. For example, in the diagram below, the first square may be tiled with 1-3 square tiles, or a domino may be placed on the first 2 squares. The second square may be tiled with 1-7 squares, or by using a domino on squares 0 & 1 or 1 & 2.

~~++++~~So, just tiling the first two squares, without dominoes there are ways of stacking squares; using a domino, there’s one way. There are 7 ways of tiling the second square alone – by using 1-7 tiles.

So we have: ~~++++~~ .

Let’s take the first 4 squares (0-3) of this diagram, which represent the continued fraction:

~~++++++++++++++~~

To tile these 4 squares, we could use:

~~++++~~• no dominoes. Then there are 3 ways of tiling the first square: with 1, 2 or 3 square tiles. So there will be ways of tiling without using dominoes.

~~++++~~• 2 dominoes. There’s only 1 way to do this.

~~++++~~• 1 domino. It could be placed on the first square, the 2nd or the 3rd.

~~++++~~~~++++~~a. If on the first, there are ways to tile the last two squares.

~~++++~~~~++++~~b. If on the 2nd, there are ways.

~~++++~~~~++++~~c. If on the 3rd, there are ways.

So adding these: ways. This gets the numerator of the fraction. To obtain the denominator, we remove the first square, leaving 7, 15, 1.

There are 3 possibilities re the placement of the domino:

~~++++~~• no domino. ways

~~++++~~• domino on squares 1 & 2. 1 way

~~++++~~• domino on squares 2 & 3. 7 ways

So the total is , and the continued fraction is .

**Why it works.**For positive numbers , define the

*continuant*to be the number of ways to tile a strip of length with dominoes (of length two) and stackable squares (of length one). For nonnegative integers and ,

since the first term counts tilings ending with a stack of tiles, and the second counts those ending with a domino. So for nonnegative ,

and for and ,

The first part counts tilings that don’t have a domino on cells and , while the second part counts those that do.~~++~~Also,

### Nonregular CFs

Nonregular continued fractions, i.e. of the form can be represented by a similar model, by using stackable dominoes also.

e.g.

The dominoes can be:

~~++++~~• not used. ways.

~~++++~~• on the first 2 cells, a stack of 1 or 2. ways.

~~++++~~• the the second 2 cells, a stack of 1 to 4. ways.

For the denominator, remove the first cell and find [3;4,5]:

~~++++~~• dominoes not used: ways.

~~++++~~• dominoes used. ways.

So the sought fraction is .

**references**

Benjamin, Quinn – Proofs that really count: the art of combinatorial proof, 2003

Benjamin, Zeilberger – Pythagorean primes and palindromic continued fractions