By continued fraction, I mean a simple (canonical) continued fraction.
By "tile the plane": I actually am interested in infinite sequences of tillable rectangles.
Continued fraction of $e$ can tile the plane
Continued fraction of Euler's number $e=2.7182\dots$ is nice and regular
$$[e]=[e_0;e_1,e_2,\dots]=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10\dots],$$
which is $[2; 1, 2,\dots]$ followed by blocks of three terms $[1,1,2k]$ for $k\ge 2$.
If we take a sequence of integer sided rectangles $r_0,r_1,r_2,\dots$ such that area of $r_i$ is equal to $e_i$, can we tile the "plane"? - where by "plane", I mean "one of the four quadrants".
We start in the origin $(0,0)$ and WLOG look at the quadrant $(x\ge 0,y\le 0)$. That is, we start at the top left corner, and continue our way down-and-right.
Then, to tile the "plane" (given quadrant), we can continue following pattern indefinitely:
That is, first observe $i=7$ where $r_0,r_1,\dots,r_i$ tile a $a_i\times b_i = 3\times 4$ rectangle. After this, every $6$th value of $i$ works by adding $[1,4k,1,1,4k+2,1]$ area rectangles, extending the sides of the tiled rectangle $a_i,b_i$ by $2$ (i.e. $|a_i-b_i|=1$ is maintained).
The above image uses the first $31$ terms $r_0,\dots,r_{30}$.
This was easy to find because the continued fraction of $e$ is nice and regular.
Can continued fraction of $\pi$ tile the plane?
Continued fraction of $\pi=3.1415\dots$ does not appear to have any obvious patterns
$$\pi=[\pi_0;\pi_1,\pi_2,\dots]=[3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2,\dots].$$
Notice that right off the bat, some larger values like $292=4\times 73$ start to appear.
Can we solve the same problem for $\pi$ as we did for $e$ above, and do it "regularly"?
That is,
Given $c\ge 0$, do there exist infinitely many $i$ such that integer sided rectangles $r_0,r_1,\dots,r_i$ with areas $\pi_0,\pi_1,\dots,\pi_i$, can tile a $a_i$ by $b_i$ rectangle, $|a_i-b_i|\le c$, for some $a_i,b_i$?
In the example of $e$, we see that $c=1$ works.
How small of a value of $c$ can we find, that works for $\pi$?
Can we even find any $c$ value that works?
There is some information about Pi Continued Fraction on mathworld, but I do not know if we can say enough about the terms of the continued fraction to draw conclusions about this problem.
Can we somehow utilize the known upper bounds on the irrationality measure of $\pi$, or any other known properties of this irrational transcendental number?
If there is no hope in solving the problem, can we do any better if the "integer sided" rectangles condition is relaxed to "rational sided", or removed (i.e. a side can be a real number)?
Remark
This question was inspired when I was trying to think of new visual representations of (approximations of) irrational numbers.
If I haven't made any mistakes, then,
for example, the following $81\times 8$ rectangle is tiled by integer rectangles of areas $\pi_0,\dots,\pi_{49}$:
As a bonus, the image was also made to have the following property:
- If you read the areas of rectangles in the image by starting in the upper left corner (red $1\times3$ rectangle) and continue visiting adjacent unvisited rectangles in the order of $8$ colors $\color{red}{R}\color{orange}{O}\color{yellow}{Y}\color{green}{G}\color{cyan}{C}\color{blue}{B}\color{purple}{P}\color{magenta}{M}$ (while preferring to visit the smaller area first if multiple choices are available), you can extract first $50$ terms of the continued fraction of $\pi$.
(If you have impaired color vision, you can use a tool like imagecolorpicker.com.)
That is, the above image represents (encodes) about first $56$ decimal digits of $\pi$.
Thought: If we generalize this from rectangles to polyominoes (and use something like the color rule above to guide the decoding of the image), we can get creative with our images.
It does not have to be $e$ or $\pi$ l. Any infinite continued fraction at all can tile the plane. This is illustrated here with $\sqrt{41}=[6,\overline{2,2,12}]$. The blocks below (which I had to get from a screenshot on my phone due to limited selections), shows the process.
Start with a row of six squares representing the 6 (blue). Now place two squares at the beginning of the second row for the first 2 and two immediately to the right of the first six squares (brown). You now have $[6,2,2]$. For the next set (yellow), start with the 12 to begin the third row, then the next two 2's in the first available slots of row 2 and row 1. You now have six elements $[6,2,2,12,2,2]$.
Continuing in this "Cantor-diagonal" pattern you will ultimately generate infinitely many rows and occupy infinitely many spaces in each for any infinite continued fraction. It's inelegant (and as noted above, so is my screenshot), but it proves a solution exists.