Now I found the comment in A001519: Except for the initial term, positive values of x (or y) satisfying x^2 - 3xy + y^2 + 1 = 0. - Colin Barker , Feb 04 2014 and x^2 - 3xy + y^2 + 1 = 0 (1) is "my" eq (a (n+1) - a (n))^2 = a (n)*a(n + 1) - 1 (2) "Nothing new under the sun". But how I come across (2) - this another story. Thx to all.
Вторник, 5 июля 2016, 1:23 +03:00 от Gareth McCaughan < gareth.mccaughan@pobox.com >:
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun