Given an integer n, let Rn be the number of ways n can be reduced (by repeated use of factors 1 or 2) down to 1 or 0. Let Pn be the set of such paths from n to 1 or 0
The following tables lists the number of ways for n (in {2, 3, 4, 5}) to be reduced to 1 or 0:
For n=5
For n=4
For n=3
For n=2
P5
5 →-1 4 →-1 3 →-1 2 →-1 1
5 →-1 4 →-1 3 →-1 2 →-2 0
5 →-1 4 →-1 3 →-2 1
5 →-1 4 →-2 2 →-1 1
5 →-1 4 →-2 2 →-2 0
5 →-2 3 →-1 2 →-1 1
5 →-2 3 →-1 2 →-2 0
5 →-2 3 →-2 1
P4
4 →-1 3 →-1 2 →-1 1
4 →-1 3 →-1 2 →-2 0
4 →-1 3 →-2 1
4 →-2 2 →-1 1
4 →-2 2 →-2 0
P3
3 →-1 2 →-1 1
3 →-1 2 →-2 0
3 →-2 1
P2
2 →-1 1
2 →-2 0
Ways to reach 1 = 5
Ways to reach 0 = 3
Ways to reach 1 or 0 (R5) = 8
Ways to reach 1 = 3
Ways to reach 0 = 2
Ways to reach 1 or 0 (R4) = 5
Ways to reach 1 = 2
Ways to reach 0 = 1
Ways to reach 1 or 0 (R3 ) = 3
Ways to reach 1 = 1
Ways to reach 0 = 1
Ways to 1 or 0 (R2) = 2
Observation 3: Rn seems to be sum of Rn-1 and Rn-2 (with n ≥ 2 and R1 and R0 both taken to be 1).
Follows an informal reasoning to support the above observation. Suppose Rn is the number of ways 1 or 0 can be reached using reduction steps 1 or 2 and Rn-1 is the number of ways 1 or 0 can be reached similarly.
Then to calculate Rn+1, since n is reachable from n+1 by only one possible path (a reduction step of 1) and n-2 is reachable by only one possible (a reduction step of 2), Rn+1 must be the total of Rn-1 and Rn.
Observation 4: The set Pn has exactly Pn-1 paths to reach 1 and Pn-2 paths to reach 0 (for n ≥ 2).
Observation 5: ∀ n . Rn = Gn (Gn = Fn = Fn-1+ Fn-2 n ≥ 2 with F0 = F1 = 1)
Using the above observations, Fibn can be considered as the number of ways a given integer n can be reduced (by steps of 1 or 0) to 1.