"Egyptian Continued Fraction" Expansions

[Paul D Hanna]

     The recent discussion of Egyptian fractions reminded me of something
that Hans Havermann and I worked on a while back.
I call these expansions, "Egyptian Continued Fractions" (if I may).

The "Egyptian continued fraction" (ECF) expansion of x is still a series
using a greedy algorithm, but instead of unit fractions, simple continued
fractions are used.  The smallest partial quotients {a(n)} are chosen so
that the following sum of continued fractions approach, but do not
exceed, x:
  x = a(0) + [0;a(1)] + [0;a(2),a(1)] + [0;a(3),a(2),a(1)] + ... 

For rational x=p/q, there are some nice properties for these terms.
For the several rational cases I tried, the following seems to hold.

  Let p and q be integers such that gcd(p,q)=1.
  Define a(n) by the ECF(p/q) where:
     p/q = sum_{n>=1} [0;a(n),a(n-1),...,a(1)].
  Let b(n) = a(n+1)/a(n); 
  then b(n) and c(n) are integers where
  b(n) = q*c(n)^2 + 1 (n>1) with c(1)=q
  Further, c(n)/c(n+1) = [0;a(n),a(n-1),...,a(1)].

I do not know if this holds for all integers p,q, where gcd(p,q)=1.
Maybe there is a simple proof?

Below I give two EXAMPLEs of ECF expansions of rational x.

I do not suppose that the same nice patterns apply to irrational x,
but I am sure there are exceptions. 

Perhaps there is a pattern to ECF(x) for x=exp(1)?

Regards,
    Paul
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EXAMPLE 1.
Egyptian Continued Fraction of 1/2, the terms of the expansion begin:
A1 = {3,6,114,82422,775518735906,49571011835891246676962576934,
1905670723258414192862600748187713148856128929433679181345327542068242,..
.}

These terms form the series of continued fractions:
1/2 = [0;3] + [0;6,3] + [0;114,6,3] + [0;82422,114,6,3] + 
      [0;775518735906,82422,114,6,3] + ...

Which corresponds to this series of c.f. convergents:
1/2 = 1/3 + 3/19 + 19/2169 + 2169/178773337 +
178773337/138642072323937340491 + ...
the numerators and denominators of these convergents are thus:
A2 = {1,3,19,2169,178773337,138642072323937340491,
6872627808122388145523700121830865817934119607931,...}

The ratios of the terms of A1 are integers:
A3 = {2,19,723,9409123,63919812044231139,
38443248436551744545237750536764136242163,...}

Now, apart from the initial term, there is a correspondence between
the sequences A3 and A2:
  A3(n) = 2*A2(n)^2 + 1, for n>1, with
  A3(1) = 2.
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EXAMPLE 2.
Egyptian Continued Fraction of 2/3, the terms of the expansion begin:
A1 = {2,6,78,39624,122706374856,596612992602856398979231968,
43676973734319055853289089522709656223485008972379594733782364384,...}

These terms form the series of continued fractions:
2/3 = [0;2] + [0;6,2] + [0;78,6,2] + [0;39624,78,6,2] + 
      [0;122706374856,39624,78,6,2] +...

Which corresponds to this series of c.f. convergents:
2/3 = 1/2 + 2/13 + 13/1016 + 1016/40257997 + 40257997/4939912870833724448
+...

the numerators and denominators of these convergents are thus:
A2 = {1,2,13,1016,40257997,4939912870833724448,
2947216201065475962006807395900856096025011661,...}

The ratios of the terms of A1 are integers:
A3 = {3,13,508,3096769,4862118967356028,
73208217514286067486338363925578714113,...}

Now, apart from the initial term, there is a correspondence between
the sequences A3 and A2:
  A3(n) = 3*A2(n)^2 + 1, for n>1, with
  A3(1) = 3.

The same pattern seems to hold for other rationals.
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