Fibonacci
numbers {F_n, n ≥ 0} satisfy the
recurrence relation

(1)

F_n+2 =
F_n+1 + F_n,

along
with the initial conditions F₁ = 1 and
F₀ = 0.

The
Fibonacci name has been attached to the sequence 0, 1, 1, 2, 3, 5, ... due to
the inclusion in his 1202 book Liber
Abaci of a rabbit reproduction puzzle: under certain
constraints the rabbit population at discrete times is given exactly by that
sequence. As naturally, the sequence is simulated by counting the tilings with
dominoes of a 2×n board:

A
tiling of a 2×n board may end with two horizontal dominoes or a single vertical
domino:

In
the former case, it‘s an extension of a tiling of a 2×(n-2) board; in the latter
case, it‘s an extension of a tiling of a 2×(n-1). If
T_n denotes the number of domino tilings of a 2×n
board, then clearly

T_n =
T_n-2 + T_n-1

which is the same recurrence relation that is satisfied
by the Fibonacci sequence. By a direct verification,
T₁ = 1, T₂ = 2,
T₃ = 3, T₄ = 5, etc.,
which shows that {T_n} is nothing but a shifted Fibonacci sequence. If
we define, T₀ =
1, as there is only 1 way to do nothing;
and T_-1 =
0, because there are no boards with negative side
lengths, then F_n =
T_n-1, for n ≥ 0.

The domino tilings
are extensively used in Graham,
Knuth, Patashnik and
by Zeitz. Benjamin
& Quinn economize
by considering only an upper 1×n portion of the board (and its tilings). This
means tiling a 1×n board with 1×1 and 1×2 pieces.

I‘ll
use Benjamin &
Quinn‘s
frugal tilings to prove Cassini‘s
Identity

F_n+1·F_n+1 -
F_n·F_n+2 = (-1)ⁿ

In terms of
the tilings, I want to prove that
T_n·T_n -
T_n-1·T_n+1 = (-1)ⁿ.

The meaning
of the term T_n·T_n is obvious: this is
the number of ways to tile two 1×n boards where the tilings of the two boards
are independent of each other. Similarly,
T_n-1T_n+1 is the number of ways to tile
two boards: one 1×(n-1) and one 1×(n+1). Now, the task is to retrieve the
relation between the two numbers annunciated by Cassini‘s identity.

Our setup
consists of two 1×n boards:

with the
bottom board shifted one square to the right:

The
tilings of the two boards may or may not have a fault line.
A fault line is a line on the two
boards at which the two tilings are breakable. For example, the tilings below
have three fault lines:

The
trick is now to swap tails: the pieces of the two tilings (along with the
boards) after the last fault line:

Since
the bottom board has been shifted just one square, the swap produces one tiling
of a 1×(n+1) - the top board in the diagram - and one tiling of a 1×(n-1) board
- the bottom board in the diagram. Note that the old faults have been preserved
and no new faults have been introduced.

Thus,
in the presence of faults, there is a 1-1 correspondence between two n-tilings
(T_n) and a pair of (n-1)- and (n+1)-tilings. The time is to account
for the faultless combinations, if any.

But
there are. Any 1×1 square induces a fault. This leaves exactly two faultless
tilings. If n is odd, both n-1 and n+1 are even, there is a unique pair of
(n-1)- and (n+1)-tilings:

If
n is even, there is a unique n-tiling that, when shifted, generates no fault
lines:

References

1. A. T. Benjamin, J. J.
   Quinn, Proofs
   That Really Count: The Art of Combinatorial
   Proof,
   MAA, 2003


2. R. Graham, D.
   Knuth, O. Patashnik, Concrete
   Mathematics,
   2nd edition, Addison-Wesley, 1994.


3. P.
   Zeitz, The
   Art and Craft of Problem Solving, John
   Wiley & Sons, 1999

|Contact| |Front
page| |Contents| |Algebra| |Store|

Copyright
? 1996-2011 Alexander
Bogomolny

本文转自：

http://www.cut-the-knot.org/arithmetic/combinatorics/FibonacciTilings.shtml

(转)Fibonacci Tilings,码迷,mamicode.com