# "Egyptian Continued Fractions"

Hans Havermann hahaj at rogers.com
Tue Mar 4 07:33:09 CET 2003

```> (2) Determine the smallest partial quotients, {a(n)}, of the finite
> continued fractions that satisfy the following "Egyptian Continued
> Fraction" rule.
>
> Define a sequence of rationals such that r(n+1) = a(n+1) + 1/r(n), with
> r(1)=2,
> such that the partial sums defined by s(n+1) = s(n) + 1/r(n+1)
> approach a
> limit of unity, s(n) --> 1, as n grows.
>
> That is, what are the smallest partial quotients whose recursive finite
> continued fractions, as described below, have a sum of unity?
> 1 = [0;2] + [0;2,2] + [0;10,2,2] + [0;260,10,2,2] +
>     [0;703300,260,10,2,2] + [0;138573337735300,703300,260,10,2,2] +...
>   = .5 + .4 + .0961538461 + .0038447319 + .0000014218 + ...
>
> The sequence begins (offset=1):
>     {2,2,10,260,703300,138573337735300,...}

If your sixth term is correct, then I made a mistake. I get:

{2, 2, 10, 260, 703300, 128651592765800,
11640481755119007104771565805489000,
174323957340151900503011819340136847884611255011003423918589496240629570
05321114000,
353740762056033689486577658309972790809155987659540786695130928586606919
371344399898763107921518297972974011484694661477109061192634800647984428
5957315700814399797273046063160266969362962117033370000,
218135972532862908642707228617722324508953118045357496975224454171691188
092875176630601741077556050438402604522263729836224568408659746907919254
757283451163642261637041557107538399975730150618359367141417342645627172
194454424076003579611205630735299281167690337339421022113787131221265472
554590579778974109013743868611138902136326168754188853623524596324601215
438488074711997431539439036009315189850642110213730002416031332040362543
404826316910024517197528959543587966870939620000, ...}

```