At risk of gilding the lily I have reworked my solution below, in order better to make the point that this kind of identity can often by wrangled in an elementary fashion by routine application of a small toolbox comprising lemma (4) -- (6). On 7/5/16, Zak Seidov <math-fun@mailman.xmission.com> wrote:
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.
This is a consequence of elementary identities in Fibonacci numbers F_n , proved via induction in both directions from basic recurrence F_{n+1} = F_n + F_{n-1} . (3) Lemma: For integer n , (4) -F_{-n} = (-1)^n F_{n} . Lemma: For integer m, n , (5) F_{n+m+1} = F_{n+1} F_{m+1} + F_n F_m . Lemma: For integer m,n, (6) F_{n+m} - ( F_{m+1} + F_{m-1} )F_{n} + (-1)^m F_{n-m} = 0 --- since via (5) with n -> n-1 , F_{n+m} = F_n F_{m+1} + F_{n-1} F_m ; and with n -> n-1 , m -> -m , (-1)^m F_{n-m} = (-1)^m ( F_n F_{1-m} + F_{n-1} F_{-m} ) , = F_n F_{m-1} - F_{n-1} F_m via (4). Adding the equations yields the result. QED Now for integer k denote b(k) == (F_{k+2})^2 - 3 F_{k+2} F_k + (F_k)^2 ; then b(0) = 1 , and b(k+1) - b(k) = F_{k+2}^2 - F_{k-2}^2 - 3 F_k F_{k+2} - 3 F_k F_{k-2}) = ( F_{k+2} - F_{k-2} ) ( F_{k+2} - 3 F_k + F_{k-2} ) , = 0 via (6) with m -> 2 , F_3 + F_1 = 3 . Hence for all k , b(k) = 1 . (7) [ Alternatively, umbral calculus dispenses with (4) -- (6) altogether: F_{k+2} - 3 F_k + F_{k-2} reduces to ( E^4 - 3 E^2 + 1 )F_k = ( E^2 + E - 1 ) ( E^2 - E - 1 )F_k = ( E^2 + E - 1 ) 0 = 0 , (8) where E F_k == F_{k+1} denotes the shift operator. ] Finally (2) is the special case of (7) with k -> 2n+1 , a(n) == F_{2n+1} . Fred Lunnon
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