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I have found a number that satisfies this continued fraction:$$n=1+{1\over{2+{1\over{3+{1\over n}}}}}$$ With a value of about $1.4403$ after 9 layers of nesting. I've tried googling it and plugging it into Wolfram Alpha, but it doesn't seem to be used. Does it have any application or is some root of another number?

ADDENDUM

It has the exact value of $${13\pm(\sqrt{37}+\frac{1}{7}})\over{7}$$

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  • $\begingroup$ You have a $\frac{1}{n}$ in your fraction, should this be $\ldots$ instead? $\endgroup$
    – gd1035
    Jul 9, 2018 at 20:40
  • $\begingroup$ @gd1035 I prefer writing my continued fractions in a way that allows them to be solved algebraically, hence that notation. $\endgroup$
    – user189728
    Jul 9, 2018 at 20:42
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    $\begingroup$ @Blue ahhh the number was in an image format rather than text, so it wasn't indexed by my search engine. $\endgroup$
    – user189728
    Jul 9, 2018 at 20:42
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    $\begingroup$ @user189728 Regarding notation, you could write it $[1;2,3,\ldots,k,\ldots]$. Either way, note that you've assigned $n$ already on the left-hand side, so it would be best not to use it again for a different purpose on the right. $\endgroup$
    – Théophile
    Jul 9, 2018 at 20:52
  • $\begingroup$ @Théophile the $n$ on the right is equal to the one on the left $\endgroup$
    – user189728
    Jul 9, 2018 at 21:24

1 Answer 1

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The OEIS sequence A060997 is the decimal expansion of the continued fraction $\,1+1/(2+1/3(1+1/4+\dots)))\,$ with value $1.433127426722311758317183455775\dots$ and the exact formula is $\,I_0(2)/I_1(2)\,$ where $\,I_n(x)\,$ is the modified Bessel function of the first kind.

For the continued fraction $\,1+1/(2+1/(3+1/(1+1/(2+1/3+\dots))))) ,\,$ solving the equation $\, x =1+1/(2+1/(3+1/x)) \,$ simplifies to the quadratic equation $\, 0 = 7x^2 - 8x - 3 \,$ with solution $\, x = (4 + \sqrt{37})/7 \approx 1.440394647185.\,$ The decimal expansion is the sequence A177036 whose continued fraction is sequence A010882.

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    $\begingroup$ This result is for the continued fraction $[1,2,3,4,5,6...]$, not the continued fraction $[1,2,3,1,2,3,...]$ $\endgroup$ Jul 9, 2018 at 21:32
  • $\begingroup$ @MichaelSeifert The original title perhaps misleadingly wrote the continued fraction as $[1, 2, 3, \ldots]$ suggested the former (though the text specified the latter). $\endgroup$ Jul 9, 2018 at 22:00
  • $\begingroup$ Shouldn't the equation read $x=1+1/(2+1/(3+1/x))$ rather than $ x=1+1/(2+1/(3+x))$...? $\endgroup$
    – CiaPan
    Jul 9, 2018 at 22:35
  • $\begingroup$ @CiaPan Yes, my typo is fixed now. Thanks. $\endgroup$
    – Somos
    Jul 9, 2018 at 22:39

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