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Zaremba's conjecture asserts that if there exists some integer $A > 1$ such that for the set of numbers $x \in [0,1]$ whose partial quotients in the continued fraction representation are all bounded from above by $A$ contains all natural numbers as partial denominators. In other words, for each $d \in \mathbb{N}$ there exists $x$ with all partial quotients $0 = a_0, a_1, \cdots \leq A$ and an integer $n$ such that $[0; a_1, \cdots, a_n] = \frac{b}{d}$. Bourgain and Konotorvich recently proved that Zaremba's conjecture holds for a set of density 1, which may depend on the choice of $A$.

Is the following easier version of Zaremba's conjecture trivial? That is, if we take our partial quotients from a set $\mathcal{A}$ which is an infinite but sparse set, that one should immediately obtain all possible $d \in \mathbb{N}$ as partial denominators?

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    $\begingroup$ Are you sure you have Zaremba's conjecture right? What if $A=\{1\}$? What if $A$ is empty? $\endgroup$ May 8, 2014 at 22:51
  • $\begingroup$ I fixed the statement of the conjecture as to be more clear. $\endgroup$ May 9, 2014 at 0:41
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    $\begingroup$ There are obviously infinite sets so that not all denominators are possible, like $\lbrace 11, 12, 13, ... \rbrace$. So, are you trying to ask for a sparse set so that all denominators are possible? $\endgroup$ May 9, 2014 at 2:13
  • $\begingroup$ I wish to get an understanding why a weaker version of Zaremba's conjecture allowing for arbitrarily large partial quotients is not interesting, presumably because it is trivial. I guess I am looking for a set of the form $\mathcal{A} = \{1, 2, 3, \cdots, A\} \cup \{b_n\}$ for some sparse sequence $\{b_n\}$. $\endgroup$ May 9, 2014 at 2:18

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