soft question - Magic trick based on deep mathematics

The following trick uses some relatively deep mathematics, namely cluster algebras. It will probably impress (some) mathematicians, but not very many laypeople.

Draw a triangular grid and place 1s in some two rows, like the following except you may vary the distance between the 1s:

1   1   1   1   1   1   1
  .   .   .   .   .   .
.   .   .   .   .   .   .
  .   .   .   .   .   .
.   .   .   .   .   .   .
  .   .   .   .   .   .
1   1   1   1   1   1   1

Now choose some path from the top row of 1s to the bottom row and fill it in with 1s also, like so:

1   1   1   1   1   1   1
  1   .   .   .   .   .
.   1   .   .   .   .   .
  1   .   .   .   .   .
.   1   .   .   .   .   .
  .   1   .   .   .   .
1   1   1   1   1   1   1

Finally, fill in all of the entries of the grid with a number such that for every 2 by 2 "subsquare"

  b
a   d
  c

the condition $ad-bc=1$ is satisfied, or equivalently, that $d=frac{bc+1}{a}$. You can easily do this locally, filling in one forced entry after another. For example, one might get the following:

1   1   1   1   1   1   1
  1   2   3   2   2   1
.   1   5   5   3   1   .
  1   2   8   7   1   .
.   1   3   11  2   1   .
  .   1   4   3   1   .
1   1   1   1   1   1   1

The "trick" is that every entry is an integer, and that the pattern of 1s quickly repeats, except upside-down. If you were to continue to the right (and left), then you would have an infinite repeating pattern.

This should seem at least a bit surprising at first because you sometimes divide some fairly large numbers, e.g. $frac{5cdot 11+1}{8} = 7$ or $frac{7cdot 3+1}{11} = 2$ in the above picture. Of course, the larger the grid you made initially, the larger the numbers will be, and the more surprising the exact division will be.

Incidentally, if anyone can provide a reference as to why this all works, I'd love to see it. I managed to prove that all of the entries are integers, and that they're bounded, and so there will eventually be repetition. However, the repetition distance is actually a simple function of the distance between the two rows of 1, which I can't prove.