For a given real number $x$, the continued fraction representation $x = [a_0; a_1, a_2, \cdots]$ where $(a_n)_{n \geq 0}$is defined by setting $x = \alpha_0$, then $a_i = \lfloor \alpha_i \rfloor$, and $\alpha_{i+1} = \frac{1}{\alpha_i - a_i}$ for $i \geq 0$. A convergent of $x$ is a rational number $p_n/q_n$ where $p_n/q_n = [a_0; a_1, \cdots, a_n]$, and $\gcd(p_n, q_n) = 1$.

My question is as follows. Is there a set $\mathcal{A} \subset \mathcal{P}$, where $\mathcal{P}$ denotes the set of prime numbers, and the set of $\mathfrak{C}_\mathcal{A}$ real numbers $x \in [0,1]$ with $x = [0; a_1, a_2, \cdots]$ such that $a_i \in \mathcal{A}$ for all $i \geq 1$, with the property that there exist infinitely many pairs of primes $p < q$ such that $p/q$ is a convergent for some element $x \in \mathfrak{C}_\mathcal{A}$?

The question is of interest because of the following 'additive' property of continued fractions:

$$\displaystyle \frac{1}{a + \frac{b}{d}} = \frac{d}{b + ad}$$

which implies that if $p/q$ is a convergent in $\mathfrak{C}_\mathcal{A}$ then so is $q/(p + aq)$. This is interesting because Zaremba's conjecture asserts that for a finite set of the form $\mathcal{B} = \{1, 2, \cdots, B\}$ with $B \geq 5$, the set of denominators $\mathfrak{D}_\mathcal{B}$ of those $d$ that appears as the denominator of a fraction $b/d$ which is the convergent of some number $z$ whose partial quotients all line $\mathcal{B}$ should be all but finitely many positive integers. If this hypothesis can be relaxed to a set of primes, then one can show that almost all positive integers can be written as the sum $p + aq$ where $a$ is from a finite family of primes which would greatly strengthen Chen's theorem and be ever so close to the vaunted Goldbach conjecture.


1 Answer 1


This is only half an answer. For almost all real $x$ we have an asymptotic formula for convergents with prime denominators, see Bykovskii, On the distribution of prime denominators of the approximants for almost all real numbers. Probably we should sieve one more time.


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