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<DIV dir=ltr align=left><FONT face=Arial color=#0000ff size=2><SPAN
class=304471815-19122006>Andy,</SPAN></FONT></DIV>
<DIV dir=ltr align=left><FONT face=Arial color=#0000ff size=2><SPAN
class=304471815-19122006></SPAN></FONT> </DIV>
<DIV dir=ltr align=left><FONT face=Arial color=#0000ff size=2><SPAN
class=304471815-19122006>Thanks for the
response. The representation I'm asking about is that
<STRONG>given by the greedy algorithm</STRONG>, which produces a unique
representation where the c_'s <STRONG>can be zero</STRONG>. The problem
remains: does <STRONG>that</STRONG> representation, for every n,
consist of only finitely many terms?</SPAN></FONT></DIV>
<DIV dir=ltr align=left><FONT face=Arial color=#0000ff size=2><SPAN
class=304471815-19122006></SPAN></FONT> </DIV>
<DIV dir=ltr align=left><FONT face=Arial color=#0000ff size=2><SPAN
class=304471815-19122006>Best wishes,</SPAN></FONT></DIV>
<DIV dir=ltr align=left><FONT face=Arial color=#0000ff size=2><SPAN
class=304471815-19122006>Clark</SPAN></FONT></DIV>
<DIV dir=ltr align=left><FONT face=Arial color=#0000ff size=2><SPAN
class=304471815-19122006></SPAN></FONT> </DIV>
<DIV dir=ltr align=left><FONT face=Arial color=#0000ff size=2><SPAN
class=304471815-19122006></SPAN></FONT> </DIV>
<DIV dir=ltr align=left><FONT face=Arial color=#0000ff size=2><SPAN
class=304471815-19122006></SPAN></FONT> </DIV><FONT face=Arial
color=#0000ff size=2></FONT><BR>
<DIV class=OutlookMessageHeader lang=en-us dir=ltr align=left>
<HR tabIndex=-1>
<FONT face=Tahoma size=2><B>From:</B> A.N.W.Hone@kent.ac.uk
[mailto:A.N.W.Hone@kent.ac.uk] <BR><B>Sent:</B> Tuesday, December 19, 2006 9:10
AM<BR><B>To:</B> Kimberling, Clark<BR><B>Cc:</B> seqfans@seqfan.net;
seqfan@ext.jussieu.fr<BR><B>Subject:</B> Re: Representations found by the greedy
algorithm<BR></FONT><BR></DIV>
<DIV></DIV>
<DIV>Dear Clark, </DIV>
<DIV><FONT face=Arial color=#0000ff size=2></FONT> </DIV>
<DIV>The problem as stated is ill-posed. </DIV>
<DIV><FONT face=Arial color=#0000ff size=2></FONT> </DIV>
<DIV>You want to say that N = c_0/x + c_1/x^2 + c_2/x^3 + ... where c_j are all
positive integers, </DIV>
<DIV>right? </DIV>
<DIV><FONT face=Arial color=#0000ff size=2></FONT> </DIV>
<DIV>Then the solution is not unique, for the following reason. </DIV>
<DIV><FONT face=Arial color=#0000ff size=2></FONT> </DIV>
<DIV>The golden mean x satisfies </DIV>
<DIV><FONT face=Arial color=#0000ff size=2></FONT> </DIV>
<DIV>x^2=x+1, </DIV>
<DIV><FONT face=Arial color=#0000ff size=2></FONT> </DIV>
<DIV>so that </DIV>
<DIV><FONT face=Arial color=#0000ff size=2></FONT> </DIV>
<DIV>1=1/x +1/x^2, ---(*) </DIV>
<DIV><FONT face=Arial color=#0000ff size=2></FONT> </DIV>
<DIV>and therefore for any integer N we have </DIV>
<DIV><FONT face=Arial color=#0000ff size=2></FONT> </DIV>
<DIV>N= N/x + N/x^2, </DIV>
<DIV><FONT face=Arial color=#0000ff size=2></FONT> </DIV>
<DIV>which gives one solution (with only finitely many non-zero c_j), </DIV>
<DIV>but there are infinitely many from (*), because note we can write </DIV>
<DIV><FONT face=Arial color=#0000ff size=2></FONT> </DIV>
<DIV>N=N/x (1/x +1/x^2) + N/x^2 </DIV>
<DIV> =N(1/x + 1/x^2)^2 </DIV>
<DIV> =N(1/x +1/x^2)^M </DIV>
<DIV><FONT face=Arial color=#0000ff size=2></FONT> </DIV>
<DIV>for any integer M, and so on. <BR></DIV>
<DIV>Merry Christmas to you and all the seqfans! </DIV>
<DIV>Andy </DIV>
<DIV><FONT face=Arial color=#0000ff size=2></FONT><FONT face=Arial color=#0000ff
size=2></FONT><BR>----- Original Message -----<BR>From: "Kimberling, Clark"
<ck6@evansville.edu><BR>Date: Tuesday, December 19, 2006 2:37
pm<BR>Subject: Representations found by the greedy algorithm<BR>To:
seqfans@seqfan.net, seqfan@ext.jussieu.fr<BR><BR>> <BR>> Suppose
x=(1+sqrt(5))/2. The greedy algorithm finds that <BR>> every
positive<BR>> integer N has a representation<BR>> <BR>> N = c0/x +
c1/x^2 + c2/x^3 + ...<BR>> <BR>> Can someone prove that c1, c2, c3, ...
are all 0 except for finitely<BR>> many 1's?<BR>> <BR>>
Examples:<BR>> <BR>> 4 = 6/x + 1/x^3 + 1/x^6<BR>> <BR>> 5 = 8/x +
1/x^6<BR>> <BR>> 6 = 9/x + 1/x^2 + 1/x^6<BR>> <BR>> 22 = 35/x +
1/x^3 + 1/x^5 + 1/x^7 + 1/x^10.<BR>> <BR>> Clark Kimberling<BR>>
<BR>> </DIV></BODY></HTML>