A pattern of periodic continued fraction

Question

I am interested in the continued fractions which $1$s are consecutive appears. For example, it is the following values.

$$ \sqrt{7} = [2;\overline{1,1,1,4}] \\ \sqrt{13} = [3;\overline{1,1,1,1,6}] $$ In this article, let us denote n consecutive $1$s as $1_n$. Applying this, the above numbers would be as follows.

$$ \sqrt{7} = [2;\overline{1_3,4}] \\ \sqrt{13} = [3;\overline{1_4,6}] $$

While investigating these numbers, the following pattern was found experimentally.

$$ \sqrt{F(n)^2m^2-(F(n)^2-L(n))m+\frac{(F(n)-1)(F(n)-3)}{4}-\frac{F(n-3)-1}{2}} =\left[F(n)m-\frac{F(n)-1}{2};\ \overline{1_{n-1},\ 2\left(F(n)m-\frac{F(n)-1}{2}\right)}\right] $$ ($m,n \in \mathbb{N},\ n\equiv\pm1\ (mod3),\ n > 3,\ $$F(n)$ is Fibonacci number, $L(n)$ is Lucas number)

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I confirm that it works correctly when $n$ and $m$ are single digits. If you find a proof or counterexample, please let me know.

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(2022/04/13 edit)

A general expression was derived. I think the expression I found is that special case. The condition is that the inside of the square root is always an integer.

Here are some concrete examples.

$$\begin{array}{|c|c|} \hline n & pattern \\ \hline 4 & \sqrt{9m^2-2m} = [3m-1;\overline{1_3,2(3m-1)}] \\ \hline 5 & \sqrt{25m^2-14m+2} = [5m-2;\overline{1_4,2(5m-2)}] \\ \hline 7 & \sqrt{169m^2-140m+29} = [13m-6;\overline{1_6,2(13m-6)}] \\ \hline 8 & \sqrt{441m^2-394m+88} = [21m-10;\overline{1_7,2(21m-10)}] \\ \hline \end{array}$$

Answer 1

I think your claims are correct but quite needlessly complicated. The theorem at the end of this answer shows a result which is both simpler to write out and more general.

Let $(F_n)_{n\geq 0}$ be the standard Fibonacci sequence, defined by $F_0=0,F_1=1$ and $F_{n+2}=F_n+F_{n+1}$ for $n\geq 1$.

Let $f(x)=\frac{1}{1+x}$, and $f^n=f\circ f \circ \ldots \circ f$ ($n$ times). It is easy to check by induction that

$$ f^n(x)=\frac{\big(F_{n+1}-F_n\big)+\big(2F_n-F_{n+1}\big)x}{F_n+(F_{n+1}-F_n)x} \tag{1} $$

Now let $a\geq 1$ be an integer. If we put $g(x)=f(\frac{1}{a+x})$,

$$ g(x)=\frac{\big(F_{n+1}-F_n\big)(x+a)+\big(2F_n-F_{n+1}\big)}{F_n(x+a)+F_{n+1}-F_n} \tag{2} $$

The roots of $g(x)=x$ are therefore defined by the equation

$$ x^2+ax-\bigg(\frac{F_{n+1}}{F_n}(a-1)+2-a\bigg)=0 \tag{3} $$

This is quadratic whose roots are $-\frac{a}{2}\pm \sqrt{\Delta}$ where

$$ \Delta = \frac{a^2}{4} + \bigg(\frac{F_{n+1}}{F_n}(a-1)+2-a\bigg) \tag{4} $$

For $n\geq 3$, we have $F_{n+1}=\frac{3}{2}F_{n}+\frac{1}{2}F_{n-3}$ and hence $F_{n+1}\geq \frac{3}{2}F_{n}$. It follows from (4) that $\Delta \geq \frac{a^2}{4} + 4\bigg(\frac{3}{2}(a-1)+2-a\bigg) \gt \frac{a^2}{4}$, so that the largest root $\alpha$ of $g(x)=x$ is positive. Thus :

Theorem. For any $n\geq 3$ and $a\geq 2$, there is a unique positive number whose continued fraction is $[\overline{1_{n},\ a}]$. This number is

$$ \alpha = -\frac{a}{2} + \sqrt{\frac{a^2}{4} + \bigg(\frac{F_{n+1}}{F_n}(a-1)+2-a\bigg)} \tag{5} $$

Update. When $a$ is of the form $a=F_n(2m+1)+1$, it is straightforward to compute that

$$ \begin{array}{lcl} \Delta &=& \frac{4F_n^3m^2 + (8F_nF_{n+1} + (4F_{n}^3 - 4F_n^2))m + (4F_nF_{n+1} + (F_n^3 - 2F_n^2 + 5F_n)}{4F_n} \\ &=& F_n^2 m^2 + (2F_{n+1} + F_{n}^2 - 1)m + F_{n+1} + \frac{F_n^2-2F_n+5}{4} \end{array} $$

So that $\Delta$ is an integer iff $F_n^2-2F_n+5$ is divisible by $4$. This is easily seen to be the case when $n\not\equiv 2$ modulo $3$.