Look first at the
Gnomon 1 problem.
To investigate further the representation of Fibonacci numbers by
gnomons it will help to use a little algebra. We denote by $F_r$
the $r^{th}$ Fibonacci number in the sequence: $$0 , 1 , 1 , 2 ,
3 , 5 , 8, 13 , 21 , 34 , \cdots$$so that $$F_0 =0, F_1 = 1, F_2
= 1, F_3=2, F_4 = 3, \cdots$$Now the Fibonacci rule can be
written as $$F_{r+1} = F_r + F_{r-1}$$ Here are the gnomons for
$F_4 , F_5, F_6$ and $F_7$ . Note that each gnomon has six sides
(hexagonal) and that when $r$ is even the $F_r$-gnomon is a
square with a square cut out of the corner and when $r$ is odd
the $F_r$-gnomon is a rectangle (not a square) with a rectangle
cut out of the corner.
Draw some gnomons for yourself and mark the six lengths of the
edges of the gnomons with the corresponding Fibonacci numbers.
Draw dotted lines dividing each gnomon into two parts
illustrating the Fibonacci rule. [You may like to cut out a set
of gnomons for yourself so that you can fit them together, two
by two, like pieces of a jigsaw to demonstrate the Fibonacci
rule.]
To give the solution to this question, draw the gnomons for
$F_{2n}$ (note this will be a square with a square cut out of
one corner) and $F_{2n+1}$ ( two squares attached to each other
making a rectangle with a rectangle cut out of the corner). In
each case label the lengths of the six sides with the Fibonacci
numbers as $F_{n-1} , F_n , F_{n+1} ,$ etc. thus giving a
general specification for the gnomon corresponding to any
Fibonacci number. Now check that your specification works for
the Fibonacci numbers $F_{20}$ and $F_{21}$ .
