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Khinchin's Continued Fraction Theorem:

For almost all reals $r$ with continued fraction representation $[a_o; a_1, a_2, \dots ]$ the sequence $K_n(r) = \left(\prod_{i=1}^{n} a_i\right)^{1/n}$ converges to a constant $K$ (Khinchin's constant) independent of $r$.

Let $K(r)$ denote the limit of $K_n(r)$ and define a set $S=\{x\in \mathbb{R}: K(x) =K \}$. So $r\in S$ means roughly the digits in the continued fraction of $r$ are random. Let's say that $r\in S$ are "Khinchin random" numbers.

Questions:

Is $S^c$ closed under any operations? Is it the case that $a,b\notin S \implies a+b\notin S$ and/or $a\times b\notin S$?

For example, $\phi=[1;1,1,1 \dots ]\notin S$ but also $2\phi, \phi^3 \notin S$. It seems like polynomials in $\phi$ are not in $S$. Is it the case that the sums and products of non-Khinchin random numbers are not Khinchin random?

Can anything be said about whether $S$ and/or $S^c$ are dense in the reals?

The literature seems to talk about the measure of $S$ but it's not obvious to me whether there is always a member of $S$ between two distinct real numbers.

Where did this question come from? This question originally had two parts and we've separated it into two distinct posts. This post is the second part of this original question.

Martin Sleziak
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Mason
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  • It also seems $e\notin S$. So it might make sense to explore $\phi+e$ and $\phi \times e$ – Mason Dec 01 '19 at 04:12
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    A note, that I think suggests it isn't closed, is that if you take a Khinchin random continued fraction, and double the terms in it (or multiply by $n$), it is no longer Khinchin random. This gives a "large", although still measure zero, set of numbers. This makes it believable that you could generate all the reals with these numbers. Constructing explicit Khinchin random numbers is hard, so investigating specific numbers and combinations would be hard. Also both sets are dense. The rationals are not Khinchin random. Khinchin random are full measure on any open set so are dense. –  Dec 01 '19 at 05:08
  • It looks like $K(\phi^{2n-1})=A002878(n)$. You can evaluate these in Mathematica with f[x_]:=N[GeometricMean[Delete[ContinuedFraction[GoldenRatio^(x), 1000],1]],4] and Table[Round[f[2i-1]], {i, 30}] – Mason Dec 01 '19 at 05:14

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