## New answers tagged continued-fractions

3
votes

### On four Ramanujan-type "Legendrian" sequences with a 3-term recurrence?

Stupid of me. As O. Gorodetsky mentions, these are classical:
$$F_1=(91\zeta(3)-2\pi^3\sqrt{3})/432$$
$$F_2=(28\zeta(3)-\pi^3)/64$$
$$F_3=(117\zeta(3)-2\pi^3\sqrt{3})/243$$
In addition, note that ...

1
vote

### On Zagier's missing continued fraction with multiple limits?

To complete the 12 cfracs in this post and the 4 in the next, all associated with 16 "sporadic sequences", then 13 of them have closed-forms, 1 has six limits (also with closed-forms but one ...

0
votes

Accepted

### On the continued fractions using Cooper's sequences $s_7,\, s_{10},\, s_{18}$ and the Zudilin-Cohen sequence

(This answers Question 2.)
Thanks to Cohen's 2022 paper, turns out there is a deg-$4$ and one can find polynomials $Q_k(n)$ for general deg-$k$ such that,
$$(n+1)^k s_{n+1} = Q_k (n)\, s_n - n^k s_{n-...

10
votes

Accepted

### On Zagier's missing continued fraction with multiple limits?

Set $Q=(1/2)L(\chi_{-3},2)$ (related to your Gieseking constant) and
$P=2\pi^2/81$. The limits are almost certainly (not proved),
\begin{align}
\lim_{m\to\infty}C_2(6m+0) &= -Q\\
\lim_{m\to\infty}...

2
votes

### On 12 cfracs: for Catalan's $K$, Gieseking's $\kappa$, and $\pi^2$, $\pi^3$, plus three for $\zeta(3)$ using Zagier's "six sporadic sequences"

Zagier answered your first question in his original paper. See the table in p. 11 here. He gives there evaluations of the continued fractions associated with his sequences A, C, D, E and F (note that ...

12
votes

Accepted

### On 12 cfracs: for Catalan's $K$, Gieseking's $\kappa$, and $\pi^2$, $\pi^3$, plus three for $\zeta(3)$ using Zagier's "six sporadic sequences"

We have $$C_2(-17,-6,-72)=-(5/8)L(\chi_{-3},2)$$ and
$$C_2(10,3,-9)=(1/2)L(\chi_{-3},2)$$ so both are proportional to what you
call Gieseking's constant but which is simply the value at 2 of the
L ...

0
votes

### Applications of finite continued fractions

Cornacchia's algorithm for solving the Diophantine equation
$x^{2}+dy^{2}=m$. See for details On Cornacchia’s algorithm
for solving the diophantine equation
$x^{2}+dy^{2}=m$ by F. Morain and J.-L. ...

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