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We need to provide a visual counting proof for the identity above.

At first, we count the number of pairs of $n$ tilings where at least one ends in a square. Then, conditioning on whether the first tiling ends in a square, we need prove the rest.

If we consider the R.H.S of the identity, we see that it is a strip with a length of $(2n-2)$ that we're tiling. How do we approach the problem then? How should one visualize the diagram, given that we need to find the tiling pairs where at least one ends in a square?

N.B what I'm asking for is a visual proof, not by any other means.

tex_mate
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  • We're using the combinatorial theorem, where it's stated that if there is a strip with length $n$, then there are $f_{(n)}$ ways to tile it with squares and dominoes, where $f_{(n)}$ is a fibonacci number. Here: $f_{(n)}$ = $F_{(n+1)}$ – tex_mate Feb 27 '18 at 12:45
  • I'm asking for a visual combinatorial proof. Not anything else. – tex_mate Feb 27 '18 at 12:51
  • Are you sure this is true? Plugging in $n = 3$ gives $f_9 - f_1 = f_4$, or equivalently $F_{10} - F_2 = F_5$, which is false. – mechanodroid Feb 27 '18 at 13:18
  • @mechanodroid Have a look at the formula from duplicate. The squares should not be in the index but for the $F_k$'s. This is a typo here in this question. – Dietrich Burde Feb 27 '18 at 13:23
  • @DietrichBurde True, but it still seems wrong. I believe it should be $f_n^2 - f_{n-2}^2 = f_{2n-1}$. – mechanodroid Feb 27 '18 at 13:50
  • @DietrichBurde hahah that's my question :) ..... ugh, looking at it now, I have just realised how much of a long way I have come... – Mr Pie Feb 27 '18 at 13:51
  • @mechanodroid This is exactly what the formula in the duplicate says, isn't it? – Dietrich Burde Feb 27 '18 at 13:53
  • @mechanodroid yes, what you wrote is true – tex_mate Feb 27 '18 at 13:53
  • @mechanodroid oh you're absolutely right. I'm still new to latex so I typed it wrong. That is, indeed what I meant to write – tex_mate Feb 27 '18 at 13:55

1 Answers1

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I believe your identity should in fact be:

$$f_n^2 - f_{n-2}^2 = f_{2n-1}$$

Consider the strip of length $2n$ which is breakable after the $n$-th place, in the sense that there is not a $2 \times 1$ tile at the positions $(n, n+1)$.

Clearly, such a strip can be tiled in $f_n^2$ ways since we are actually tiling two disjoint strips of length $n$.

On the other hand, consider the two possible cases:

  1. There are two $2 \times 1$ tiles at positions $(n-1, n)$ and $(n+1, n+2)$
  2. There is a $1 \times 1$ tile at the positions $(n)$ or $(n+1)$

In the first case, we are tiling two disjoint strips of length $n-2$, namely $(1, 2, \ldots, n-2)$ and $(n+3, n+4, \ldots, 2n)$. This can be done in $f_{n-2}^2$ ways.

In the second case, by removing a $1 \times 1$ tile from one of the positions $(n)$ or $(n+1)$ we obtain a tiling of a strip of length $2n - 1$ with no other restrictions. This can be done in $f_{2n-1}$ ways.

Therefore, $f_n^2 - f_{n-2}^2 = f_{2n-1}$.

mechanodroid
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    Yes, that's it. I did not consider the breakability of the tile at cell $n$. Thank you. And sorry for my vague question, I'm still fairly new to LaTex so I got the commands wrong. – tex_mate Feb 27 '18 at 14:00