Find a fundamental unit in the ring of integers $\mathbb Z[\frac{1+\sqrt{141}}{2}]$ of $\mathbb Q(\sqrt{141})$ I have different corollaries for different numbers, the most appropriate for $141$ is the one below. I used an algorithm (don’t know if you know this, but $\beta_0=\sqrt{141}+\lfloor\sqrt{141}\rfloor, \quad\beta_{n+1}=\frac{1}{\beta_n-\lfloor\beta_n\rfloor}$ $a_n=\lfloor\beta_n\rfloor$ $p_n=p_{n-1}a_n+p_{n-2}, \quad q_n=q_{n-1}a_n+q_{n-2} $) to determine the continued fraction expansion of […]

Possible Duplicate: Why does this process, when iterated, tend towards a certain number? (the golden ratio?) Please post your favorit solution to the following Compute $x=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\ldots}}}$ Thank you

Every infinite continued fraction is irrational. But can every number, in particular those that are not the root of a polynomial with rational coefficients, be expressed as a continued fraction?

How can I find the smallest positive integer $N$, such that the continued fraction of $\sqrt{N}-\lfloor\sqrt{N}\rfloor$ begins with a given finite sequence containing a zero followed by positive integers ? For example, the sequence $[0,1,2,3,4]$ is given. We have to find the smallest number $N$, such that $\sqrt{N}-\lfloor\sqrt{N}\rfloor$ begins with $[0,1,2,3,4]$. The number $\sqrt{388}-\lfloor\sqrt{388}\rfloor$ has […]

Let $\dfrac{A_{n}}{B_{n}}$ be the $n^{th}$ convergent (approximant) $$ \frac{A_{n}}{B_{n}}=b_{0}+\dfrac{a_{1}}{b_{1}+\dfrac{a_{2}}{b_{2}+\dfrac{a_{3}}{\begin{array}{c} b_{3}+ \\ \\ \end{array} \begin{array}{cc} \ddots & \\ & \end{array} +\dfrac{a_{n-1}}{b_{n-1}+\dfrac{a_{n}}{b_{n}}}}}} $$ of a continued fraction. $A_{n}$, $A_{n-1}$, and $A_{n-2}$ satisfy the recurrence $$ \begin{eqnarray*} \begin{pmatrix} A_{n} \\ B_{n} \end{pmatrix} &=& \begin{pmatrix} A_{n-1} & A_{n-2} \\ B_{n-1} & B_{n-2} \end{pmatrix} \begin{pmatrix} b_{n} \\ a_{n} \end{pmatrix} \quad […]

First, how I arrived at a number having the property that the first $250$ digits after the decimal point contain $7777$ , $8888$ and $9999$ I wanted to construct a number which can be shown to be transcendental using the irrationality measure http://mathworld.wolfram.com/IrrationalityMeasure.html Consider the sequence $$a_1=0\ ,\ a_2=1\ ,\ a_n=a_{n-1}^2+a_{n-2}\ for \ n>2$$ Then, […]

I have been working on an article at https://oeis.org/wiki/Table_of_convergents_constants where I posted a table of “convergents constants” (defined at https://oeis.org/wiki/Convergents_constant) for a few numbers. It would be nice to support the article with some quality analysis. Before June 9, 2011, was starting to extract and clearly define a pattern to these constants cf the article. […]

Given a complex number $\begin{aligned}\frac{z}{n}=x+iy\end{aligned}$ and a gamma function $\Gamma(z)$ with $x\gt0$, it is conjectured that the following continued fraction for $\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)$ is true $$\begin{aligned}\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)=\frac{\displaystyle\Gamma\left(\frac{z}{4z+2n}\right)\Gamma\left(\frac{3z+2n}{4z+2n}\right)}{\displaystyle\Gamma\left(\frac{z+n}{4z+2n}\right)\Gamma\left(\frac{3z+n}{4z+2n}\right)}=\cfrac{2z+2n}{2z+n+\cfrac{(0z-n)(4z+3n)} {3(2z+n)+\cfrac{(2z+0n)(6z+4n)}{5(2z+n)+\cfrac{(4z+n)(8z+5n)}{7(2z+n)+\cfrac{(6z+2n)(10z+6n)}{9(2z+n)+\ddots}}}}}\end{aligned}$$ Or in gauss’s notation $$\begin{aligned}\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)=-\frac{1}{2z+2n}\underset{m=0}{\overset{\infty}{\mathbf K}}\frac{((2m-2)z+(m-2)n)((2m+2)z+(m+2)n)}{((2m+1)(2z+n)}\end{aligned}$$ Corollaries: 1):let $z=1$ and $n=2$,then we obtain a beautiful continued fraction for square root 2 $$\begin{aligned}{-1+\cfrac{3}{2+\cfrac{\frac{(-1)(5)}{(1)(3)}} {2+\cfrac{\frac{(1)(7)}{(3)(5)}}{2+\cfrac{\frac{(3)(9)}{(5)(7)}}{2+\cfrac{\frac{(5)(11)}{(7)(9)}}{2+\ddots}}}}}}=\sqrt{2}\end{aligned}$$ 2):However the most interesting case(for me […]

Does the classical cantor set have a nice description as a set of continued fractions? I made a (superficial) search and didn’t find anything, but I’m very tired right now, so please forgive me that I don’t put more effort in asking this.

Prove that for a natural number $n$, $$\prod_{k=1}^n \tan\left(\frac{k\pi}{2n+1}\right) = 2^n \prod_{k=1}^n \sin\left(\frac{k\pi}{2n+1}\right)=\sqrt{2n+1}.$$

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