Re^4: Fibonacci numbers (again)

[ambrus]

It's all just convention, but I opt for 0,1,1,2,3,5,..., and I have a reason.

If you define F_0 = 0, F_1 = 1, than the nice equation gcd(F_n, F_k) = F_{gcd(n,k)} is true, and as a special case F_k divides F_n iff k divides n. This nice equation would be much more ugly if F_0 != 0.

Update: in case anyone's interested, any integer sequence defined by a second order homogenous linear recursion and starting element 0 has this property. That is, if s_0 = 0, s_1 is and integer, and s_{k+2} = A s_{k+1} + B s_k for given A, B integers; then gcd(s_n, s_k) = s_{gcd(n,k)}.

Update 2006 nov 4: the above paragraph is false. Sorry.