How to determine the values of $x$ such that $\frac{1}{x}=[1,1,1,\dots,x]$ with $n$ $1'$s?

Question

I'm not sure I understood the question correctly. In the hope of finding some pattern, I made a code in Mathematica that expands $[1,1,1,\dots,x]$ with $n$ $1$'s and solve that equation, for $n=2,3,4,\dots,20$ I obtained:

$$\begin{array}{cc}2&\begin{array}{cc}\begin{array}{c}x\to-\frac{1}{\sqrt{2}}\\\end{array}&\begin{array}{c}x\to\frac{1}{\sqrt{2}}\\\end{array}\\\end{array}\\3&\begin{array}{cc}\begin{array}{c}x\to-\frac{1}{\sqrt{3}}\\\end{array}&\begin{array}{c}x\to\frac{1}{\sqrt{3}}\\\end{array}\\\end{array}\\4&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{2}{5}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{2}{5}}\\\end{array}\\\end{array}\\5&\begin{array}{cc}\begin{array}{c}x\to-\frac{\sqrt{\frac{3}{2}}}{2}\\\end{array}&\begin{array}{c}x\to\frac{\sqrt{\frac{3}{2}}}{2}\\\end{array}\\\end{array}\\6&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{5}{13}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{5}{13}}\\\end{array}\\\end{array}\\7&\begin{array}{cc}\begin{array}{c}x\to-2\sqrt{\frac{2}{21}}\\\end{array}&\begin{array}{c}x\to2\sqrt{\frac{2}{21}}\\\end{array}\\\end{array}\\8&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{13}{34}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{13}{34}}\\\end{array}\\\end{array}\\9&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{21}{55}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{21}{55}}\\\end{array}\\\end{array}\\10&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{34}{89}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{34}{89}}\\\end{array}\\\end{array}\\11&\begin{array}{cc}\begin{array}{c}x\to-\frac{\sqrt{55}}{12}\\\end{array}&\begin{array}{c}x\to\frac{\sqrt{55}}{12}\\\end{array}\\\end{array}\\12&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{89}{233}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{89}{233}}\\\end{array}\\\end{array}\\13&\begin{array}{cc}\begin{array}{c}x\to-\frac{12}{\sqrt{377}}\\\end{array}&\begin{array}{c}x\to\frac{12}{\sqrt{377}}\\\end{array}\\\end{array}\\14&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{233}{610}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{233}{610}}\\\end{array}\\\end{array}\\15&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{377}{987}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{377}{987}}\\\end{array}\\\end{array}\\16&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{610}{1597}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{610}{1597}}\\\end{array}\\\end{array}\\17&\begin{array}{cc}\begin{array}{c}x\to-\frac{\sqrt{\frac{987}{646}}}{2}\\\end{array}&\begin{array}{c}x\to\frac{\sqrt{\frac{987}{646}}}{2}\\\end{array}\\\end{array}\\18&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{1597}{4181}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{1597}{4181}}\\\end{array}\\\end{array}\\19&\begin{array}{cc}\begin{array}{c}x\to-2\sqrt{\frac{646}{6765}}\\\end{array}&\begin{array}{c}x\to2\sqrt{\frac{646}{6765}}\\\end{array}\\\end{array}\\20&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{4181}{10946}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{4181}{10946}}\\\end{array}\\\end{array}\\\end{array}$$

But I don't think there is a visible pattern in there. Now I'm clueless again. Can you help?

Answer 1

Expanding the first few continued fractions:

$$[1,1,x]=1+\frac 1{1+\frac1x} = 1+\frac x{x+1} = \frac{2x+1}{x+1}$$

$$[1,1,1,x]=1+\frac1{[1,1,x]} = 1+\frac{x+1}{2x+1} = \frac{3x+2}{2x+1}$$

$$[1,1,1,1,x] = 1+\frac1{[1,1,1,x]} = 1+\frac{2x+1}{3x+2} = \frac{5x+3}{3x+2}$$

$$[1,1,1,1,1,x] = 1+\frac1{[1,1,1,1,x]} = 1+\frac{3x+2}{5x+3} = \frac{8x+5}{5x+3}$$

By following this pattern (or by induction), one can observe that

$$[\underbrace{1,1,\dots,1}_{n \text{ ones}},x] = \frac {F_{n+1}x+F_n}{F_nx+F_{n-1}}$$

hence we just need to solve

$$\frac1x=\frac {F_{n+1}x+F_n}{F_nx+F_{n-1}} \\ F_n x+F_{n-1}=F_{n+1}x^2+F_nx \\x = \sqrt{\frac{F_{n-1}}{F_{n+1}}}$$