Tiling a square with similar non-congruent rectangles. What is the aspect ratio of the rectangles as n grows large? I recently saw a question here on mathoverflow: «For what n and t can a square be partitioned into n similar rectangles in t congruence classes?», where Joseph Gordon gave a proof that, indeed, a square can be partitioned into n non-congruent similar rectangles for any $n\ge3$. His method involves the use of Fibonacci-numbers. For n=3, the aspect ratio,r, of the similar rectangles is the square of the plastic number (also known as the Padovan constant).
Using his method I did some calclulations for a few n (stopped at n=16, where I found r approximately 1.6180355) and the results seem to suggest that as n grows larger, the aspect ratio of the similar rectangles tends towards the golden ratio. Can this be proved/disproved in any way?
 A: You are correct.
Let  $\varphi$ be the golden ratio and $r_n$ be the ratio you want (long side to short).
According to the answer you mention, for a certain positive constant $A_n:$ $$r_n=\frac{F_{n-1} A_n + F_{n-2}}{F_{n-2} A_n + F_{n-3}}. \tag{*}$$  It follows that (for odd $n$) $$\frac{F_{n-1}}{F_{n-2}} <r_n < \frac{F_{n-2}}{F_{n-3}}.  $$ The same is true for even $n$ except the inequalities are reversed. Furthermore, the two rationals differ by exactly  $\frac{1}{F_{n-2}F_{n-3}}.$
It is also well known that (again, up to reversing the inequalities) $$\frac{F_{n-1}}{F_{n-2}} < \varphi < \frac{F_{n-2}}{F_{n-3}}. \tag{**}$$ This establishes the result and shows that the convergence is rapid.

The exact values of $A_n$ were not needed here, but in fact they converge  quickly to $\frac{\sqrt 5}{\varphi}$ the root of the linear equation $$(A-1)\varphi^2-1.$$
$A_n$ is the real root of the cubic polynomial $$f_n(A)=(A-1)(F_{n-1} A + F_{n-2})^2 - (F_{n-2} A + F_{n-3})^2.$$ We see from $(**)$ that $f_n$ is almost exactly $$(A-1)(F_{n-2}\varphi A + F_{n-3}\varphi)^2 - (F_{n-2} A + F_{n-3})^2\\ =\left((A-1)\varphi^2-1\right)(F_{n-2} A + F_{n-3})^2.$$
