I drew separately the gnomons for F r with r even and r odd, up to r = 12. I replaced the lengths of the sides with the terms in the Fibonacci sequence, and then I tried to find a rule for the gnomons corresponding to F 2n and F 2n+1 respectively.
For Fr with r even, I obtained the following table, with l j for j = 1, 2,..., 6, the lengths of the sides of the gnomon:
| l 1 | l 2 | l 3 | l 4 | l 5 | l 6 | |
| F 2 | F0 | F 0 | F 1 | F 1 | F 2 | F 2 |
| F 4 | F 4 | F 4 | F 2 | F 2 | F 3 | F 3 |
| F 6 | F 2 | F 2 | F 3 | F 3 | F 4 | F 4 |
| F 8 | F 3 | F 3 | F 4 | F 4 | F 5 | F 5 |
| F 10 | F 4 | F 4 | F 5 | F 5 | F 6 | F 6 |
| F 12 | F 5 | F 5 | F 6 | F 6 | F 7 | F 7 |
So, for the gnomon F2n I observed the following rule:
| F 2n | F n-1 | F n-1 | F n | F n | F n+1 | F n+1 |
This is an inductive process and at the end I have to verify it.
For F r with r odd, I obtained the following table, with lj for j = 1, 2,..., 6, the lengths of the sides of the gnomon:
| l 1 | l 2 | l 3 | l 4 | l 5 | l 6 | |
| F 5 | F 4 | F 3 | F 3 | F 2 | F 2 | F 1 |
| F 7 | F 5 | F 4 | F 4 | F 3 | F 3 | F 2 |
| F 9 | F 6 | F 5 | F 5 | F 4 | F 4 | F 3 |
| F 11 | F 7 | F 6 | F 6 | F 5 | F 5 | F 4 |
For the gnomon F2n+1 the following rule could be observed:
| F 2n+1 | F n+2 | F n+1 | F n+1 | F n | F n | F n-1 |
Using these rules, the gnomons for F20 and F21 have the following lengths of the sides:
| F 20 | F 9 | F 9 | F 10 | F 10 | F 11 | F 11 |
| F 21 | F 12 | F 11 | F 11 | F 10 | F 10 | F 9 |
Keeping into account that the area of the gnomon equals the corresponding number in the Fibonacci sequence, and also the shape of each kind of the gnomon, I found the following recursive relations:
F 2n = F n (F n-1 + F n+1 ) F 2n+1 = F n+2 F n + F n-1 F n+1
I hope that both relations are satisfied, but I don't know so much algebra to verify them.
As discovered in the solution to Gnomon I from May 2001 we also have the relations
F 2n = (F n+1 ) 2 - (F n-1 ) 2
and
F 2n+1 =(F n+1 ) 2 + (F n ) 2 .
These four relations are illustrated in the gnomon diagram above.