A Variation On Continued Fractions

[Leroy Quet]

I should note that we can use this to transform sequences of positive 
integers into other sequences of positive integers.

If we have the positive integer sequence {b(k)},

we can get x = the continued fraction [b(0);b(1),b(2),...].

And then we can use the recursive algorithm below to get the sequence 
{a(k)},

where x =  /a(0),a(1),a(2),.../.

(And in some cases, each a(m) {for m >= 1} is b(m)b(m-1),  while in other 
cases the terms differ from b(m)b(m-1).)

thanks,
Leroy Quet



>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