I'm not sure what's the accepted terminology is in this regard, but --- following M. Kojima --- let us call a $p$-regular infinite continued fraction $\big[a_0, a_1, a_2, \dots \big]$ an expansion of the form

\begin{equation} a_0 + {p \over {\displaystyle a_1 + {\displaystyle p \over a_2 + {\displaystyle p \over {\displaystyle \ddots } }}}} \end{equation}

where $p$ is a prime number, $a_0$ is an integer, and $a_1, a_2, a_3 \dots$ are positive integers. The standard recursive techniques for analysing simple (finite) continued fractions --- where $p$ is replaced by $1$ --- can be easily modified to verify that any $p$-regular infinite continued fraction converges. However, unlike the simple infinite continued fractions, $p$-regular infinite continued fractions may have rational (indeed integer) limits, for example:

\begin{equation} p \ = \ p-1 \ + \ {p \over {\displaystyle p-1 \ + \ {\displaystyle p \over {\displaystyle p-1 \ + \ {\displaystyle p \over {\displaystyle p-1 \ + \ {p \over \ddots}} }} }}} \end{equation}

${ \bf\text{Question:}}$ If two $p$-regular infinite continued fractions $\big[a_0, a_1, a_2, \dots \big]$ and $\big[b_0, b_1, b_2, \dots \big]$ are equal must it be the case that $a_i = b_i$ for all indices $i \geq 0$ ?

thanks, Ines

  • 3
    $\begingroup$ I think you've practically answered this yourself. Take for instance $p=3$. Then $3 = [2, 2, 2, \ldots]$ is one representation, but for another take $[1, 1, 4, 1, 4, 1, 4, 1, 4, \ldots]$ with $1, 4$ repeating. $\endgroup$ – Todd Trimble Oct 31 '15 at 23:25
  • 1
    $\begingroup$ $p+\frac{p}{p}=1+\frac{p}{1}$ $\endgroup$ – Alexey Ustinov Nov 1 '15 at 6:09

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