On 04/07/2016 10:55, Zak Seidov via math-fun wrote:
Evident misprint (if only one) fixed:
If a(n)=3*a(n-1)-a(n-2), then (a (n+1) - a (n))^2 = a (n)*a(n + 1) - 1. How to prove/disprove it?
Yes, recurrence may be not sufficient. That is why Cf. A001519. And yes, (a (n+1) - a (n))^2 = a (n)*a(n + 1) - 1 can be easily checked (but I ask to prove/disprove).
I'm not sure what distinction you're making here; what's wrong with proving it the way I said? The solution to the recurrence is (as you can see e.g. in the OEIS entry) a(n) = 1/sqrt(5) [t^(2n-1)+t^(1-2n)] where t = (1+sqrt(5))/2. So now we just calculate: [a(n)-a(n+1)]^2 = 1/5 [t^(2n-1)(1-t^2)-t^-(2n+1)(1-t^2)]^2 = 1/5 t^2 [t^(2n-1)-t^-(2n+1)]^2 since t^2-1=t = 1/5 [t^2n-t^-2n]^2 = 1/5 [t^4n - 2 + t^-4n] a(n)a(n+1) = 1/5 [t^(2n+1)+t^-(2n+1)][t^(2n-1)+t^-(2n-1)] = 1/5 [t^4n+t^2+t^-2+t^-4n] and we will have LHS=RHS provided -2-(t^2+t^-2) = -5, which is in fact true. -- g