Problems & Puzzles:
Fibonacci dividing terms
Enoch Haga noticed that:
If you calculate the product & the sum of the first k
terms of the Fibonacci sequence (Pk & Sk), then Sk divides Pk for every
k=4m+2, m=1, 2, 3, ...
Fibonacci sequence is: 1,1,2,3,5,8,...
Q1. Can you probe or disapprove the Enoch note?
Contributions came from T. D. Noe, Anton Vrba & Alexander
Here is the nice proof from Noe:
Let F(k) and L(k) be the Fibonacci and Lucas numbers,
respectively. By induction, it is easy to show that S(n) = F(n+2)-1.
Hence, S(4n+2) = F(4n+4)-1.
Note that F(4n+4)-1 = F(2n+3)*L(2n+1) from the product expansion (19) in
the MathWorld entry for Lucas numbers. So we just have to show that
F(2n+3) and L(2n+1) divide P(4n+2), the product of the first 4n+2
Fibonacci numbers. When n>0, it is clear that F(2n+3) is one of the
factors of P(4n+2). So we just have to show that L(2n+1) divides
P(4n+2). Look at
F(4n+2), the last factor of the product, and note that F(4n+2) =
F(2n+1)*L(2n+1) by the addition formula of Fibonacci numbers (or the
product expansion again). Therefore, L(2n+1) also divides P(4n+2). QED.