33

Gosper showed how to produce the continued fraction for
$$\frac{axy+bx+cy+d}{exy+fx+gy+h}$$
and even more complicated expressions given the continued fractions for $x$ and $y$. As a special case, he covers fractional linear transformations of $x$, which of course includes adding $1/2$ or multiplying by $2$. I haven't digested this yet.
Here is a special ...

31

As you indicate, real algebraic numbers of degree $\leq 2$ have this property in view of Lagrange's classical result characterizing them by the eventual periodicicty of the continued fractions expansion. It may be useful to know (if you don't already) that $\alpha \in \mathbb{R}$ having bounded continued fractions coefficients is equivalent to the sharpness ...

answered Jul 31 '14 at 5:54

Vesselin Dimitrov

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23

The entry from May 24, 1796 is worked out in a more general form on February 16, 1797 [reproduced below from this scan]
$$1-a+a^3-a^6+a^{10}+\cdots=\frac{1}{\displaystyle 1+\frac{\strut a}{\displaystyle 1+\frac{\strut a^2-a}{\displaystyle 1+\frac{\strut a^3}{\displaystyle 1+\frac{\strut a^4-a^2}{\displaystyle 1+\frac{\strut a^5}{1+\cdots}}}}}}$$
so the ...

answered Jun 26 '19 at 20:19

Carlo Beenakker

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20

[Edited again to give a second identity relating $E$ to eta products]
Continued fraction or not, an expression
$q^{\frac{(r-s)^2}{8(r+s)}} f(\pm q^r, \pm q^s)$
is a modular form of weight $1/2$ for all integers $r,s$ with $r+s>0$,
because it is a sum $\sum_{n=-\infty}^\infty \pm q^{(cn+d)^2}$
with rational $c,d$ and periodic signs.
Therefore the quotient ...

answered Oct 6 '12 at 3:53

Noam D. Elkies

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16

The Gauss-Kuzmin Theorem says that if $x$ is chosen uniformly at random from (say) $[0,1)$ (thanks, John Bentin) then as $n\to\infty$ the probability that the $n$th partial quotient of $x$ is $k$ tends to $$\log_2{(k+1)^2\over k(k+2)}$$ It was widely believed in 1978, and is still widely believed today, that $\pi$ acts, in this regard, like a random number. ...

answered Apr 23 '14 at 6:20

Gerry Myerson

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15

My survey paper, Real numbers with bounded partial quotients, discusses this question a bit. It is in L'Enseignement Math. 38 (1992), 151-187. In particular it seems that the first person to raise the unboundedness of the partial quotients of an algebraic number of degree > 2 was perhaps Khintchine in 1949. Some weak results on the growth of the partial ...

15

To my knowledge, much more general question is open:
OPEN DIOPHANTINE PROBLEMS p. 15
Essentially nothing is known about the continued fraction expansion of a real algebraic number of degree ≥ 3; one does not know the answer to any of the following two questions.
Question 2.9. Does there exist a real algebraic number of degree ≥ 3 with bounded ...

14

No, there is not. Already $a_0$ is impossible to determine just by reading finitely many digits (namely, it is $1$ iff $\delta_i=1$ for all $i$). The same goes for the subsequent terms of the continued fraction, mutatis mutandis.

answered Aug 26 '15 at 11:19

Emil Jeřábek

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13

It's known that a quadratic irrational has a purely periodic continued fraction expansion if and only if it is greater than $1$ and its conjugate is between $-1$ and $0$. Your observation amounts to proving that $r+2 + (-3+\sqrt{D})/2$ has this property (note that adding $r + 2$ makes it start with $2r+1$). This amounts to checking that $-1 < (2r+1 - \...

13

Since the OP asked for other examples of this kind of numerology,I will give another one to support his observation
The function $\cos(\theta_{11})$ has the following closed form
$\cos(\theta_{11})=\frac{\sigma_{1}(11)}{22\sqrt{11}}-\frac{12\sum_{k=1}^{10} (2178k^2-572k^3+35k^4)\sigma_{1}(11-k)\sigma_{1}(k)}{161051\sqrt{11}}\tag1$
and continued fraction
...

13

For a "formula" for the continued fraction of algebraic numbers, in particular $2^{1/3}$, see Bombieri and van der Poorten. It's just not a simple pattern.
EDIT: Actually there's an error in the formula in the middle of page 152 there:
it should be
$$ \pmatrix{p_{h+1} & q_{h+1}\cr p_h & q_h\cr} = \pmatrix{c_{h+1} & 1\cr
1 & 0\cr} \pmatrix{...

13

It is known that $\ell(d)=O(\sqrt{d}\log d)$ and $\ell(d)=\Omega(\sqrt{d}/\log\log d)$.
See Cohn's paper (free access) for more details.
For numerical results and some further historical comments see Williams's paper.

12

Zaremba's Conjecture:
every integer appears as the denominator of a finite continued fraction whose partial quotients are bounded by an absolute constant.

12

Yes, there is. The algorithm is due to Bill Gosper - he is considering the more general problem of doing linear fractional transformations with continued fractions - adding $2^{-i}$ is a special case. See also Liardet and Stambul, 1998 for a fancier (and probably more readable) explanation.

answered Aug 26 '15 at 10:48

Igor Rivin

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11

This algorithm correspons to nearest integer continued fractions or centered continued fraction. The length of such fraction $l(a/b)$ can be expressed in terms of Gauss - Kuz'min statistics for classical continued fraction expansion, see The mean number of steps in the Euclidean algorithm with least absolute value remainders. It means that all results known ...

11

Notice that we can write
$$f(q)=\prod_{n\geq 1} (1-q^n)^{-\left(\frac{n}{5}\right)}$$
therefore
$$g(q)=\prod_{k\geq 1} f(q^k)=\prod_{n\geq 1} (1-q^n)^{-a(n)}$$
where $a(n)=\sum_{d|n}\left(\frac{d}{5}\right)$, where $\left(\frac{d}{5}\right)$ is the Legendre symbol. Now, $a(n)$ is easily seen to be multiplicative with $a(5^k)=1$, $a(p^k)=k+1$ when $p\equiv \...

answered Sep 23 '19 at 20:59

Gjergji Zaimi

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10

Lets call expansions
$$\langle
x_1,\ldots,x_m\rangle:=\cfrac{1}{x_1-{\atop\ddots\,\displaystyle{-\cfrac{1}{x_m}}}}$$
(as in Perron's book) reduced regular continued fractions (RRCF).
Probably they are older then Hirzebruch.
We'll prove more precise statement.
Theorem. If $(n,ab)=1$ and $n>ab$ then RRCF for all numbers
$$\left\{\frac{ab^{-1}\pmod{(n+kab)}...

10

Kind of a late response, but since no one else mentioned it I think it is worth pointing out that the procedure for computing homographies/linear fractional transformations on continued fractions (mentioned in Douglas Zare's answer) was also described explicitly in terms of finite state automata by George N. Raney, in "On continued fractions and finite ...

9

A copy of Hofstadter's paper is on OEIS here: Link to Eta-Lore paper
It is also referenced on OEIS at A001468.

9

(I have a comment regarding your Tree-Like Continued Fraction generalization, except I only have enough reputation to write answers.)
Given algebraic $x_0$, an instance of your generalization can be found by searching for some function who has a fixed point at $x_0$ and is the inverse of a polynomial.
Suppose a polynomial $p(x)$ has a root at $x_0$. ...

8

From M. Waldschmidt, "Open Diophantine Problems" (Moscow Mathematical Journal vol. 4, no. 1, 2004, pp. 245-305):
Does there exist a real algebraic number of degree $\geq 3$ with bounded partial quotients?
Does there exist a real algebraic number of degree $\geq 3$ with unbounded partial quotients?

8

Before you can formulate your question precisely, you need a better notion of distance between two pictures of a set $S \subset \mathbb{C}$ (where in this case, $S$ is the union of Ford circles). If $\iota_1, \iota_2$ are pictures (bijections from square subsets $A_1, A_2$ of $\mathbb{C}$ to the unit square), then we can define distance as the Hausdorff ...

answered Nov 30 '15 at 15:50

Adam P. Goucher

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7

This graph is looks like a graf of the function which replaces partial quotient (in nearest integer continued fraction) in the following way: $$a_i+~\leftrightarrow~a_i+1-.$$
For example
$$0+\cfrac{1}{3-\cfrac{1}{4+\cfrac{1}{5+0}}}\leftrightarrow
1-\cfrac{1}{2+\cfrac{1}{5-\cfrac{1}{6-0}}}.
$$
\begin{align*}
\frac{1}{2}=\frac{1}{2+0}&\leftrightarrow1-\...

7

The following remarks do not answer your questions completely, but they may nonetheless be helpful.
Note first that computing the base-$b$ expansion of $1/n$ is essentially the same thing as computing the powers of $b$ modulo $n$. Here is one way to write the steps of the base-$b$-expansion algorithm, which makes the connection clear:
\begin{align*}
1 &...

7

Guy, Unsolved Problems In Number Theory, F21, attributes to Bohuslav Divis the conjecture that in each real quadratic field there is an irrational with all partial quotients 1 or 2; more generally, same question but with 1 and 2 replaced by any pair of distinct positive integers.

7

Define the parent of a Ford circle to be the smaller of its two larger neighbors; then this parent relation defines the Stern–Brocot tree on the points of tangency of the circles. The path in this tree to a given real number is closely related to its continued fraction expansion. A path in the tree can be represented combinatorially by a binary sequence that ...

7

One can prove by induction that some of given ratios have nice continued fraction expansions:
\begin{gather*}
\frac{\alpha_n}{\beta_n }=[2;1^n,2,1^{n-1},\ldots,2,1,1,2,1]\to 1+\varphi=\varphi^2;\\
\frac{\gamma_n}{\delta_n }=[2;1^n,2,1^{n-1},\ldots,2,1,1,2]\to 1+\varphi=\varphi^2;\\
\frac{\alpha_n}{\gamma_n}=[1;2,1,1,2,1,1,1,2,\ldots,1^{n-1},2,1^n,2]\to \psi;\...

7

Let $P_{n}(z)=\gamma_{n}z^{n}+\cdots$ be a sequence of orthonormal polynomials with respect to some weight $w$ on $\mathbb{R}$. Given an $n\geq0$, consider the following Riemann-Hilbert problem for the $2\times2$ matrix-valued function $Y_{n}(z)$ of the complex variable $z$,
1) $z\to Y_{n}(z)$ is analytic on $\mathbb{C}\setminus\mathbb{R}$,
2) $Y_{n,+}(z)=...

6

Just in case, there is a connection with Hankel determinants. Dividing both sides by $z^2$, one has $$\sum_k\mu_k z^k= \frac{\tan z}{z} = \frac{1}{1 - \cfrac{z^2}{3- \cfrac{z^2}{5 - \ldots}}} $$
Then one defines for all $m,n\geq 0$ the Hankel determinant $H^n_m$ is
$$H^n_m=\det (\mu_{i+j})_{n\leq i,j\leq n+m-1}.$$ Let $z_1, z_2,\dots$ be the poles of $\...

6

Here are some results which suggest that perhaps what happens is neither very simple nor very random. But see the end for a better result.
These are some of the simplest continued fractions and what they lead to. This also tells you what results in simple continued fractions because $r-\frac12$ and $r+\frac12$ have the same continued fraction after the ...

answered Apr 28 '13 at 21:06

Aaron Meyerowitz

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