For $\alpha \in (0,1)$, write $\alpha$ as a continued fraction like $\alpha=[a_1, a_2, \ldots]$ (note that the implicit $0$th coefficient $a_0=0$ has been omitted), and let $\frac{p_n}{q_n}$ be the $n$th convergent to $\alpha$. If you have $T$ the Gauss map, $T(x) = \frac{1}{x}-\left\lfloor\frac{1}{x}\right\rfloor$ (which acts on a continued fraction by $T([a_1,a_2,\ldots]) = [a_2, a_3, \ldots])$ then how can the value of $$\sum_{T^n x=x}\frac{1}{q_n^t},$$ be estimated for $t$ positive and fixed $n \in \mathbb{N}$?


  • $\begingroup$ I've taken the liberty of cleaning up some of the grammar of the question and adding a definition for the Gauss map; please let me know if I inadvertently changed the meaning of your question! $\endgroup$ Sep 25, 2012 at 18:01
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    $\begingroup$ Also, I suspect this question might do a bit better over on MathOverflow, since even the $n=1$ case seems to be remarkably difficult. $\endgroup$ Sep 25, 2012 at 18:08
  • $\begingroup$ i'm interested for $n \in \mathbb{N}$ large. and Thenk Steven for the correction in the grammar. $\endgroup$ Sep 25, 2012 at 18:20

1 Answer 1


I've decided to put my comments together in an answer so some readers can get a feel for what the question is about, and see the $n=1,2,3$ cases. I should also mention that this question's subject area seems relevant to the topic of dynamical zeta functions, though I don't know any details.

Setting the simple continued fraction expansion $x=[a_1,a_2,\cdots]$, we find that $T^nx=x$ means


and hence the SCFE is $a_1,\cdots,a_n$ repeated over and over again for such a fixed point; note the $a_i$s can be arbitrary positive integers for each $1\le i\le n$. Therefore we have


where the $q_n$'s are functions of the $x$'s and $c_n(k)$ counts the number of rationals in $(0,1)$ with denominator $k$ (in reduced form) expressible as $x=[a_1,\cdots,a_n]$. When $n=1$, the only rational number $x=[a_1]=1/a_1$ with denominator $k$ is $1/k$, so $c_n(k)=1$ and $\zeta_1(t)=\zeta(t)$ is the familiar Riemann zeta function. When $n=2$, look at the fractions of the form


Note $\gcd(ab+1,b)=\gcd(1,b)=1$ by the Euclidean algorithm, so the numerator and denominator as depicted share no common factor and the above is in reduced form. Hence

$$\zeta_2(t)=\sum_{k\ge2}\frac{1}{k^t}\sum_{ab+1=k}1=\sum_{k\ge2}\frac{\sigma_0(k-1)}{k^t}$$ where the divisor function $\sigma_(m)$ counts the number of positive integer divisors of $m$. For $n=3$ the situation becomes a bit more complicated; the fractions look like


Note again $\gcd(a(bc+1)+c,bc+1)=\gcd(c,bc+1)=\gcd(c,1)=1$ by the Euclidean algorithm and hence this is again in reduced form as depicted. Therefore we obtain the coefficients

$$f_3(k)=\#\{1\le a,b,c\in\Bbb N: a+abc+c=k\}.$$

So far I haven't had any luck obtaining a closed form for this; I've looked at a couple substitutions and at modular systems $k\equiv \bar{a}$ mod $\bar{c}$, $\bar{c}$ mod $\bar{a}$, to no avail. At any rate, while there's probably a closed-form for $q_n$ as a polynomial in $a_1,\cdots,a_n$ I haven't looked at, the $f_n$ functions will correspond to ever more complicated multivariable Diophantine equations involving more and more terms; I do not immediately see any clever way to tackle all of the $f_n$'s simultaneously.


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