A Variation On Continued Fractions

[Leroy Quet]

[cross-posted to sci.math.]

Let x, a positive real, be represented as:
/a(0),a(1),a(2),a(3),.../,

which is short-hand for:

x = a(0)*(1 + 1/(a(1)*(1 + 1/(a(2)*(1 +...))))).

Now this may seem arbitrary at first.
But if we calculate a possible set of positive integer a()'s using a 
Greedy-like    algorithm, this representation may seem more "natural".

For any x >= 1,

Let a(0) = floor(x).

Let, with x(0) = x;

x(m) =  1/(x(m-1)/a(m-1) -1)

and

a(m) = floor(x(m))

for all positive integers m (until, if ever, a(m) = x(m)).


Now, with some x's, anyway, there is more than one representation (set of 
a's) which gets a particular x.

For one thing, letting a(0) =b(0), a(m) = b(m)b(m-1), for m = positive 
integers,
always works, where b(k) is the k_th term in the simple continued 
fraction of x.

But with x = pi, using the term-calculation algorith above, we get:

pi = /3,21,111,.../,

but 7*15 = b(1)*b(2) = 105, not 111.

But sometimes, a given x has the same representation, whether the a's are 
calculated using the above algorithm or we have a(m) = b(m)b(m-1).

An example (probably the "best" example) of this is
x = (sqrt(5)+1)/2, which is
/1,1,1,1,1,1,.../.

In any case, which x's have different representations and which x's have 
the same representation, whether calculated using the algorithm or by 
having
a(m) = b(m)b(m-1)?

And what are such expansions, using the algorithm, of some well-known 
constants?

thanks,
Leroy Quet