thanks for your comment. First, I don’t know why you say that the property you use is a fact. If you replace n by 2 and k by 1, the lefthand side becomes fib.(2*(1+1)) = fib.4 = 3 . On the other hand, the righthand side becomes fib.(1+1)*fib.(2*1) + fib.(1)*fib.(2*1-1) = fib.2 * fib.2 + fib.1 * fib.1 = 1*1 + 1*1 = 2 . Clearly, the property is not valid.

Second, the nice thing about formulating and using distributivity properties is that the induction is somehow hidden. If you read JFF0 (it’s in the Publications page), you see that the induction process is formulated in terms of loop invariance. Also, by formulating sufficient conditions for a function to distribute over the greatest common divisor, we avoid doing induction everytime.

Joao

Let’s assume for k, the proposition is true ie F(n*k) is a muliple of F(k)

Now we use the fact(I do not have a proof for it. This is based on observation.) that F(n*(k+1)) = [ F(k+1)]*F(n*k)+ [F(k)]*F(n*k-1)

Which clearly means that F(n*(k+1)) also is divisible by F(k).

-Manish

Th

If fib.k divides fib.(n*k), then the greatest common divisor of fib.k and fib.(n*k) is fib.k (it divides both, and is the largest divisor of itself).

Also, if fib.k is the greatest common divisor of itself and fib.(n*k), then it clearly divides fib.(n*k).

Do you agree?

Joao