[seqfan] Re: Conjecture: floor(L(n)/sqrt(5)) = F(n) - (1 - (-1)^n)/2

Frank Adams-Watters franktaw at netscape.net
Fri Nov 6 21:28:05 CET 2015


L(n) = phi^n + (1-phi)^n
F(n) = (phi^n - (1-phi)^n) / sqrt(5)

Those two formulas are all you need for this, noting that |1-phi| < 1.

Franklin T. Adams-Watters


-----Original Message-----
From: Vladimir Reshetnikov <v.reshetnikov at gmail.com>
To: seqfan <seqfan at seqfan.eu>
Sent: Fri, Nov 6, 2015 2:08 pm
Subject: [seqfan] Conjecture: floor(L(n)/sqrt(5)) = F(n) - (1 - (-1)^n)/2


Dear Seqfans,

Take a look at these sequences:

http://oeis.org/A000032 - Lucas
numbers L(n).
http://oeis.org/A000045 - Fibonacci numbers
F(n).
http://oeis.org/A052952 - F(n+2) - (1 - (-1)^n)/2.

It appears that
floor(A000032(n+2)/sqrt(5)) = A052952(n), or in other
words, floor(L(n)/sqrt(5))
= F(n) - (1 - (-1)^n)/2. I checked it for n =
0..50000.

Do you have any ideas
how to prove
it?

--
Thanks
Vladimir

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