home *** CD-ROM | disk | FTP | other *** search
- Newsgroups: sci.math
- Path: sparky!uunet!gatech!rpi!batcomputer!munnari.oz.au!manuel!gerry@macadam.mpce.mq.edu.au
- From: gerry@macadam.mpce.mq.edu.au (Gerry Myerson)
- Subject: Re: Integer Sequence Problem
- Message-ID: <1992Jul30.030719.28573@newshost.anu.edu.au>
- Sender: news@newshost.anu.edu.au
- Organization: ceNTRe for Number Theory Research
- References: <1992Jul28.173452.24364@news.tu-graz.ac.at> <1992Jul28.225605.18476@pasteur.Berkeley.EDU>
- Date: Thu, 30 Jul 92 03:07:19 GMT
- Lines: 28
-
- In article <1992Jul28.225605.18476@pasteur.Berkeley.EDU>, luzeaux@bellini.berkeley.edu (Dominique Luzeaux) writes:
- >
- > In article <1992Jul28.173452.24364@news.tu-graz.ac.at>,
- > hhassler@iaik.tu-graz.ac.at (Hannes Hassler) writes:
- > |> For S an infinite, increasing sequence of positive integers (1=a_1 < a_2 <
- > |> a_3 < ... < a_i < a_{i+1} < ...), we define the VALUE val(S) of S
- > |>
- > |> val(S) := sup_{n>1} \sum_{i=1}^n a_i / a_{n-1}
-
- > |> PROBLEM: Find the minimum (infimum) of val(S) over all possible sequences S.
-
- [analysis deleted]
-
- > Furthermore, we can say that val(S) takes as only finite values q^2/(q-1)
- > where q is an integer >=2, hence 4 is the minimum of val(S) over all
- > sequences S.
-
- I won't argue with the analysis (which I've deleted), nor with min val(S)=4,
- but it's not true that val(S) can only be q^2/(q-1) with q an integer not
- less than 2. The Fibonacci sequence has val(S)=q^2/(q-1) with q the golden
- ratio ((1+sqrt5)/2). More generally, given real q>1, and any geometric
- series G with common ratio q, if S is a sequence of integers close enough to
- G (like, s_n=integer part of g_n), then val(S)=q^2/(q-1). Of course, the
- minimum value of q^2/(q-1) is 4, attained at q=2.
-
- > Dominique Luzeaux
-
- Gerry Myerson
-