Hopefully, nice(?) sequence relating to Fib, Luc.
Paul D Hanna
pauldhanna at juno.com
Tue Sep 21 15:47:12 CEST 2004
For your sequence, I get:
a(n) = ( 2*fibonacci(2*n+3) + fibonacci(2*n-2) )^2
and since
Lucas(n) = fibonacci(n+2) - fibonacci(n-2)
or
Lucas(2*n+2) = fibonacci(2*n+4) - fibonacci(2*n)
Then I believe that your result could be proved
using the form:
( 9*fibonacci(2n+2) + Lucas(2n+2))^2
although I do not have time right now to work it out.
Interesting Fibonacci relation, nonetheless!
-- creigh at o2online.de wrote:
Hi!
As neither
(a(n)) = (9, 100, 729, 5041, 34596, 237169, 1625625, 11142244, 76370121, 523448641, )
nor (sqrt(a(n))
appear to be listed at OEIS, I'm hoping the following relation will not
be a trivial consequence of some other formula:
81[Fib(2n+2)]^2 + [Luc(2n+2)]^2 = 4a(n) + 18[Fib(4n+4)]
[by the way, this was a found by slightly modifying the formula for "les,
ves," etc. at http://www.research.att.com/projects/OEIS?Anum=A097947 ]
Sincerely,
Creighton
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