Is it practical to use infinite continued fraction to generate random numbers?

Question

I observed the pattern of this irrational number: $$\sqrt{1 + \sqrt{2}}$$ and realized that each element $a_i$ occurred very randomly. For the first 100 elements, this is the result:

[1,1,1,4,6,1,2,2,2,1,1,6,1,179,48,1,356,1,1,3,15,2,1,4,8,3,1,1,1,5,1,1,9,1,19,1,
2,13,2,1,1,4,2,1,1,3,2,1,1,4,15,1,4,5,1,7,6,1,6,6,2,3,38,1,4,1,9,3,1,2,1,2,1,2,1
,1,3,1,4,1,2,4,1,4,1,1,1,58,6,3,4,203,4,14,2,1,1,41,2,2]

As I increase the length of this sequence, the number were even more arbitrary. So I wonder is there any previous work or paper which relates to random number generator using continued fraction approach? Any idea? Thank you.

Answer 1

If one wants by some chance to model the Gauss–Kuzmin distribution then it might be feasible. Otherwise there seems to be little sense in it.

Answer 2

It is difficult to imagine sequences with a more brazenly skewed distribution than the above sequence. Such skewing is the norm with continued fractions.

There are various other problems with continued fractions as random number generators. One problem is that apart from the quadratic case, the quotients are not easy to compute. Another is that the quotients have some subtle dependencies.

Answer 3

Very little is known about the continued fraction expansions of numbers other than rationals and quadratic irrationals. In particular, it is widely believed but not proved that the continued fraction of an irrational of degree exceeding 2 has arbitrarily large partial quotients.

Almost all reals (all but a set of measure zero) have the same limiting density of 1s, 2s, 3s, etc., in their continued fraction expansion, but again it is believed-but-not-known that familiar irrationals such as $\pi$ and the one you cite are in the full-measure set.