Newsgroups: sci.math
From: haoyuep@aol.com (Dan Hoey)
Date: 11 Jul 2002 04:20:06 GMT
Subject: Re: Rationals with Only 1's and 2's in Continued Fraction

pha...@ghs.org (Paul D. Hanna) wrote:

> Here is another sequence of rationals I cooked up (see sequence
> below).  These are the rationals >= 1 whose continued fractions
> consist of only 1's and 2's.

Yes, that's the _set_, but it took me a while to figure out what the
_sequence_ was--i.e, what order was being imposed on the set.  From
examination, it seems:

  q(F(n)+m) = 2 + 1/q(F(n-2)+m)    for      0 <= m < F(n-3), and
  q(F(n)+m) = 1 + 1/q(F(n-2)+m)    for F(n-3) <= m < F(n-1).

The first rule consists of prepending "2" to the terms of the continued
fraction, and the second consists of prepending "1".  This sequence
has the property that the term-sums of the continued fractions are
nondecreasing.  I'm not taking any stand on whether or not this is the
most natural ordering of the set.  But:

> Can anyone derive solutions to the following related questions?
> (1)What is the GEOMETRIC MEAN of this sequence of rationals?
> (2)What is the ARITHMETIC MEAN of this sequence of rationals?

It's easy to see that the arithmetic mean does not exist.  For note
that

  q(F(n)+m) = 1 + q(F(n)-F(n-3)+m)   for 0 <= m <= F(n-3)  (*).

Thus a tail of the sequence consists of terms one greater than in the
previous tail.  Since this tail is about 1/(phi^3) of the terms, it
will cause a repeated drift in the mean that does not approach
zero.

To be precise, let us define the finite sums

  s(n) = q(0)+q(1)+q(2)+...+q(n-1).

The difference (*) implies that

  s(F(n)) - s(F(n-1)+F(n-4)) = s(F(n)+F(n-3)) - s(F(n)) + F(n-3)

If the limits

         s(F(n))/F(n)          -> a,
  s(F(n)+F(n-3))/(F(n)+F(n-3)) -> b,

exist as n->oo, then this implies

  a.F(n) - b.(F(n-1)+F(n-4)) = b.(F(n)+F(n-3)) - a.F(n) + F(n-3)

  2.F(n).(a-b) = F(n-3)

  a-b = F(n-3)/(2 F(n)) -> 2 phi^-3 = 4 phi - 6 ~~ 0.472.

So if these limits exist, they are different.  The limits would have
to exist and be equal for the mean to exist.

For the geometric mean, we can do something similar with logarithms.

I'm interested in what the lim sup and lim inf of the means might be.

Dan Hoey <haoyuep@aol.com>           Posted and e-mailed