"Egyptian Continued Fractions"

[Paul D Hanna]

Here are two related doubly-exponential sequences of partial quotients
involved in sums of some special simple continued fractions.  
Perhaps someone would like to extend these?  My lack of precision
prevents me from calculating more terms.
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(1) What are the smallest partial quotients of the continued fraction
whose fractional remainders add to unity?  

The continued fraction may begin like this (252238200 is approximate):
    0.4042503503074... = [0;2,2,9,91,14201,252238200,...]
so that
1 = [0;2,2,9,91,14201,...] + [0;2,9,91,14201,...] + [0;9,91,14201,...] + 
[0;91,14201,...] + [0;14201,...] + ...
  = .40425035 + .47371461 + .11097560 + .01098900 + .00007041 + ...

A more exact value of the real number defined by this property would also
be interesting.
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(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,...}, 
which is doubly-exponential in that  a(n+1)~a(n)^c 
where c seems to have a value near 1+sqrt(2).

Notice that each partial quotient is divisible by the prior one:
    if b(n) = a(n+1)/a(n), 
    then {b(k),k>0} = {1,5,26,2705,197033041,...}
Does this divisibility property continue to hold for higher terms?
 
Thanks,
        Paul