Representations found by the greedy algorithm

[Kimberling, Clark]

Andy,
 
Thanks for the response.  The representation I'm asking about is that
given by the greedy algorithm, which produces a unique representation
where the c_'s can be zero.  The problem remains:  does that
representation, for every n, consist of only finitely many terms?
 
Best wishes,
Clark
 
 
 


________________________________

From: A.N.W.Hone at kent.ac.uk [mailto:A.N.W.Hone at kent.ac.uk] 
Sent: Tuesday, December 19, 2006 9:10 AM
To: Kimberling, Clark
Cc: seqfans at seqfan.net; seqfan at ext.jussieu.fr
Subject: Re: Representations found by the greedy algorithm


Dear Clark, 
 
The problem as stated is ill-posed. 
 
You want to say that N = c_0/x + c_1/x^2 + c_2/x^3 + ... where c_j are
all positive integers, 
right? 
 
Then the solution is not unique, for the following reason. 
 
The golden mean x satisfies 
 
x^2=x+1, 
 
so that 
 
1=1/x +1/x^2,   ---(*) 
 
and therefore for any integer N we have 
 
N= N/x + N/x^2, 
 
which gives one solution (with only finitely many non-zero c_j), 
but there are infinitely many from (*), because note we can write 
 
N=N/x (1/x +1/x^2) + N/x^2 
   =N(1/x + 1/x^2)^2 
   =N(1/x +1/x^2)^M 
 
for any integer M, and so on. 

Merry Christmas to you and all the seqfans! 
Andy 

----- Original Message -----
From: "Kimberling, Clark" <ck6 at evansville.edu>
Date: Tuesday, December 19, 2006 2:37 pm
Subject: Representations found by the greedy algorithm
To: seqfans at seqfan.net, seqfan at ext.jussieu.fr

>  
> Suppose x=(1+sqrt(5))/2.  The greedy algorithm finds that 
> every positive
> integer N has a representation
> 
> N = c0/x + c1/x^2 + c2/x^3 + ...
> 
> Can someone prove that c1, c2, c3, ... are all 0 except for finitely
> many 1's?
> 
> Examples:
> 
> 4 = 6/x + 1/x^3 + 1/x^6
> 
> 5 = 8/x + 1/x^6
> 
> 6 = 9/x + 1/x^2 + 1/x^6
> 
> 22 = 35/x + 1/x^3 + 1/x^5 + 1/x^7 + 1/x^10.
> 
> Clark Kimberling
> 
> 
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