Could you please help me to find the continued fraction of


Where $k$ is a positive integer.

All the best,

  • $\begingroup$ By that you mean $k = \sqrt{10 k + 3}$? If so, square both sides and use quadratic formula. $\endgroup$
    – Mark
    Commented Sep 14, 2018 at 16:01
  • $\begingroup$ No. I want to find continued fraction $\sqrt{10k+3}$ , where $k$ is positive integer. $\endgroup$
    – d.y
    Commented Sep 14, 2018 at 16:03
  • 1
    $\begingroup$ There is no general way to find this. It depends on the value of $k$. $\endgroup$ Commented Sep 14, 2018 at 16:22
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    $\begingroup$ There are some patterns for which the continued fraction really does follow a consistent pattern, such as $\sqrt{k^2 - 1}$ or $\sqrt {k^2 + 1}.$ They have a $k^2$ in them. The ones you ask about have no pattern at all. $\endgroup$
    – Will Jagy
    Commented Sep 14, 2018 at 18:12
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    $\begingroup$ Do all numerators need to be $1$? A simple continued fraction? $\endgroup$ Commented Sep 20, 2018 at 12:53

1 Answer 1


\begin{equation*} x+1=\sqrt{10k+3} \end{equation*} \begin{equation*} x^{2}+2x+1=10k+3 \end{equation*} \begin{equation*} x^{2}+2x=10k+2 \end{equation*} \begin{equation*} x\left(x+2\right)=10k+2 \end{equation*} \begin{equation*} x=\dfrac{10k+2}{2+x} \end{equation*} \begin{equation*} x=\dfrac{10k+2}{2+\dfrac{10k+2}{2+\dfrac{10k+2}{2+\dots}}} \end{equation*} Then, \begin{equation*} \sqrt{10k+3}=1+x \end{equation*} \begin{equation*} \sqrt{10k+3}=1+\dfrac{10k+2}{2+\dfrac{10k+2}{2+\dfrac{10k+2}{2+\dots}}} \end{equation*}

  • $\begingroup$ (+1) What was the motivation behind consider $x+1$ equal to the expression? $\endgroup$ Commented Sep 20, 2018 at 17:11
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    $\begingroup$ Although $k$ is a positive integer, the similarity to $\sqrt{3}=1+0.73205080\dots$ is undeniable. It's actually possible to add bigger values to $x$, but as the continued fraction appears from a recursive application, making the recursion $x\left(x+2\right)$ simple, i.e., tightly close, $x\left(x+2\right)+1$ is advisable. $\endgroup$ Commented Sep 20, 2018 at 17:50
  • $\begingroup$ the similarity to $\sqrt{3}=1+0.73205080\cdots$ is undeniable- Can you explain this statement? Why they need to be similar? $\endgroup$ Commented Sep 20, 2018 at 18:53

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