in reply to Re^3: Fibonacci numbers (again)
in thread Fibonacci numbers (again)
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.
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