A054770

[karttu at megabaud.fi]

> At 02:27 AM 6/7/00 +0300, karttu at megabaud.fi wrote:
> 
>  > I think that basically you have forgotten the zeroth Lucas number 2,
> >present in the alternative entry of the Lucas sequence in EIS:
> >
> >ID Number: A000032 (Formerly M0155)
> >Sequence:  2,1,3,4,7,11,18,29,47,76,123,199,322,521,843,1364,2207,3571,
> 
> That is a different (generalized) Lucas sequence.  You can pick the first 
> two terms for a Fibonacci or Lucas sequence.  The standard Fibonacci starts 
> with 1,2 and the standard Lucas numbers start with 1,3.

Jud: "general" is often more useful than "particular".
Note that these sequences continue also to negative direction.
E.g. Fib(0) = 0, Fib(-1) = 1, Fib(-2) = -1, Fib(-3) = 2, Fib(-4) = -3, ...
thus Luc(n) = Fib(n-1)+Fib(n+1) keeps also with Luc(0) = Fib(+1)+Fib(-1).

Furthermore, I should have written "forgotten" in quotes,
as "forgotten". What I mean, is that as there is (*) a way
(* I'm guessing as long as I haven't read the article of Zeckendorf)
to write each integer as a sum of distinct Lucas numbers
(with L_0 = 2 included), (although not necessarily uniquely
even with "no consecutive terms" constraint present),
then to prove the conjecture of A054770 one just needs
to consider what is the form of those integers with "2" present
in Lucas number system when converted to Fibonacci number
system, i.e. Zeckendorf Expansion. Then consider the Wythoff
Array, and the answer should be quite easy.

Let's see:
A054770: 2,6,9,13,17,20,24,27,31,35,38,42

written in Zeckendorf Expansion
(remember, no adjacent 1's should occur):

   321
   41385321

 2       10 
 6     1001
 9    10001
13   100000
17   100101
20   101010
24  1000100
27  1010001
31  1010010
35 10000001
38 10000101
42 10010000
...

Now, what's the pattern here?


Terveisin,
   Cheers,

Antti