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

Vladimir Reshetnikov v.reshetnikov at gmail.com
Fri Nov 6 19:52:20 CET 2015


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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