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Since the rationals are countable, you can list them in a sequence $(a_n)_{n\geq 0}$ such that each rational appears at least once in the sequence. Is there such a listing $(a_n)_{n \geq 0}$ for which $$\sum_{k = 0}^{\infty} a_kx^k =a_0 + a_1x + a_2x^2 + ...$$ has a closed form?

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    $\begingroup$ As a word of caution: if the answer turns out to be "no", a proof probably wouldn't be a good fit on this site. $\endgroup$ – Arthur Mar 7 '19 at 14:18
  • $\begingroup$ What do you mean by "nice"? $\endgroup$ – MathematicsStudent1122 Mar 7 '19 at 14:26
  • $\begingroup$ I don't have anything specific in mind; I guess that closed form is enough, so I'll delete the "nice". $\endgroup$ – Tanny Sieben Mar 7 '19 at 14:43
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The following are just some preliminary thougts on how we could start and approach the problem, which are anyway too long for a comment.

Along the track of the Farey sequence, we could build a 2 variables generating function for all the rationals in $[0,1]$ as $$ f(x,y) = \sum\limits_{1\, \le \,n} {\sum\limits_{1 \le m \le n} {\left[ {\gcd (n,m) = 1} \right]{m \over n}x^{\,n} y^{\,m} } } $$ where $[P]$ denotes the Iverson bracket.

We can also construct the o.g.f. above following instead the steps of the Stern-Brocot tree $$ \eqalign{ & f_0 (x,y) = 0x + xy \cr & f_1 (x,y) = 0x + {1 \over 2}x^{\,2} y + xy \cr & f_2 (x,y) = 0x + {1 \over 3}x^{\,3} y + {1 \over 2}x^{\,2} y + {2 \over 3}x^{\,3} y^{\,2} + xy \cr & \quad \vdots \cr} $$

We know the algorithm for constructing the single terms of the function, but (to my knowledge) not a closed expression for it,
even changing to exponential g.f. or others.

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