A054770

[Jud McCranie]

At 12:25 PM 6/8/00 -0400, John Conway wrote:
 > I understand what is meant by the zeroth term, but A54770 clearly states
> > that it is using 1,3,4 (A204), not 2,1,3,4 (A32):
>
>     In the original Handbook of Integer Sequences, EVERY sequence
>began with 1, whether it was logically there or not; and in the
>introduction to that book there was an explicit disclaimer making
>it clear that no official starts were being chosen.
>
>     In the Encyclopedia there's a slightly different disclaimer to
>the same effect, except that there there's an entry that tells you
>which number term is the first printed one (when there's an agreed
>numbering).  The fact that for the Lucas numbers this happens to be
>term number one doesn't make that into any kind of official start -
>for many sequences the first printed term is that for  n = 0, and
>a few even begin with the  n = 2  term (when there's some disagreement
>about the value at  n = 1).  It's a more or less random editorial
>decision, probably influenced by the convention of the original
>Handbook.

My point is that someone stated that the author of A54770 had "forgotten 
the zeroth Lucas number, 2, and if it was included you wouldn't get 
A54770."  My message was that it clearly states in A54770  that he is using 
Lucas numbers starting with 1, and that seems perfectly legitimate to me 
(to either include the zeroth term or not include it).  For instance, if 
you were doing something with sums of a fixed number distinct Fibonacci 
numbers, would you include the zeroth term, 0?  I don't think you would.

Also, I got the impression that the way the Lucas numbers are defined as 
they are is because they are the simplest such sequence that gives a 
sequence distinct from the Fibonacci numbers.  Starting with 1,2 just gives 
the Fibonacci numbers shifted by one place, so that isn't interesting.  The 
next case is 1,3; which gives a sequence distinct from the Fibonacci 
sequence.  That, to me, seems like a justification for starting the Lucas 
numbers at 1 - if that is clearly stated.



+--------------------------------------------------------+
|                  Jud McCranie                          |
|                                                        |
| 137*2^261147+1 is prime!  (78,616 digits, 5/2/00)      |
+--------------------------------------------------------+