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# Talk:Table of convergents constants

## Code for table of decimal expansions

At first I used the Mathematica code

Quiet[Table[{l = x + Random[];
For[a = 1, a < 190, c = Convergents[l, 30];
l = FromContinuedFraction[c], a++];
Print["|-\[IndentingNewLine]| $\scriptstyle " , x, "\,< \,x\,< \," 1, x + 1, "\,$ ||", N[l, 36]]}, {x, 0,
64}]; // TableForm]

Then pasting the top of table on.

Then I saw that I could improve the accuracy by increasing the number of convergents used, like

Quiet[Table[{l = x + Random[];
For[a = 1, a < 190, c = Convergents[l, 50];
l = FromContinuedFraction[c], a++];
Print["|-\[IndentingNewLine]| $\scriptstyle " , x, "\,< \,x\,< \," 1, x + 1, "\,$ ||", N[l, 36]]}, {x, 0,
64}]; // TableForm]

There was still too much uncertianty in the 36 digits, so I think I should srttle for 10 digits from

Quiet[Table[{l = x + Random[];
For[a = 1, a < 80, c = Convergents[l, 80];
l = FromContinuedFraction[c], a++];
Print["|-\[IndentingNewLine]| $\scriptstyle " , x, "\,< \,x\,< \," 1, x + 1, "\,$ ||", N[l, 10]]}, {x, 0,
64}]; // TableForm]

## Table for Continued Fractions

Quiet[Table[{l = x + Random[];
For[a = 1, a < 90, c = Convergents[l, 90];
l = FromContinuedFraction[c], a++];
Print["|-\[IndentingNewLine]| $\scriptstyle " , x, "\,< \,x\,< \," 1, x + 1, "\,$ ||", N[l, 15]]}, {x, 0,
64}]; // TableForm]

## Italicizing variables

I previously italicized the ${\displaystyle \scriptstyle n\,}$ with

\text{Random} ~ n
\text{Convergents ~ constant} (n)

since it is the usual mathematical convention to italicize variables... — Daniel Forgues 14:21, 23 May 2011 (UTC)

## Open Problem

An open problem is to determine what "most" means: all irrational numbers?, all transcendental numbers?, ... all numbers having some unknown property that applies to "most" numbers? Are all rational numbers excluded? What about quadratic numbers? — Daniel Forgues 01:38, 20 May 2011 (UTC)

I think this is a better way of wording that, They posses an unknown quality.(Other numbers are "special cases.")— Marvin Ray Burns 21:09, 21 May 2011 (UTC)

I actually suspect that quadratic numbers ${\displaystyle \scriptstyle x\,}$ might not give the convergents constant of "most" numbers in ${\displaystyle \scriptstyle n\,<\,x\,<\,n+1\,}$ since the simple continued fraction of quadratic numbers is eventually periodic. Could it be that algebraic numbers of degree 3 or higher and transcendental numbers ${\displaystyle \scriptstyle x\,}$ give the convergents constant of "most" numbers in ${\displaystyle \scriptstyle n\,<\,x\,<\,n+1\,}$, since only for those ${\displaystyle \scriptstyle x\,}$ do you get an infinite not eventually periodic simple continued fraction. "Most" since the the transcendental numbers are uncountably infinite (cardinality ${\displaystyle \scriptstyle \aleph _{1}\,}$) while the algebraic numbers are countably infinite (cardinality ${\displaystyle \scriptstyle \aleph _{0}\,}$.)

Daniel Forgues 22:05, 22 May 2011 (UTC)

The computations of convergents constants thus far have been done by Mathematica, and a likely answer to this question is shown when Mathematica takes the continued fraction of 11/5; it gives both the integer part and the partial quotient for an answer of {2,5}. However, if you ask Mathematica to take the cf of 2.2 (a number known not to possess the above mentioned unknown quality), it gives only the integer part. So it seems like Mathematica returns a cf with less precision than desired when you start with one of the special cases, and you get something unexpected. That is why the special cases are so intransigent. —Marvin Ray Burns 03:02, 7 June 2011 (UTC)

### Integer part and partial denominators pattern

What is the best way to describe this pattern for odd rows and the pattern for even rows?

[ Daniel Forgues 14:21, 23 May 2011 (UTC) ] (Start)

Tentative (since we have a small number of intervals) formulae for the pattern

${\displaystyle a_{0}(n)=a_{1}(n)=n,\quad n\geq 2,\,}$
${\displaystyle a_{2}(n)={\Bigg \lfloor }{\frac {n}{2}}{\Bigg \rfloor },\quad n\geq 2,\,}$
${\displaystyle a_{3}(n)=1+(2n-1)~{\Bigg (}{\frac {1+(-1)^{n}}{2}}{\Bigg )},\quad n\geq 6,\,}$
${\displaystyle a_{4}(n)=~?,\ldots \,}$

(End)

It is obvious that a_4(n), = 1 for odd n and 4*a_3(n) for even n. Marvin Ray Burns 23:57, 23 May 2011 (UTC)

What you are saying applies to a_3(n), the fourth term since the first term is a_0(n) (the integer part of a continued fraction being referred to as a_0,) a_4(n) is not obvious... — Daniel Forgues 17:32, 24 May 2011 (UTC)

It seems that

${\displaystyle a_{4}(n)=1\,}$

if ${\displaystyle \scriptstyle n\,}$ is odd, and

${\displaystyle a_{4}(n)={\Bigg \lceil }{\frac {n}{6}}{\Bigg \rceil },\quad n\geq 6,\,}$

if ${\displaystyle \scriptstyle n\,}$ is even, when you enter

Quiet[Table[{l = x + Random[];
For[a = 1, a < 30, c = Convergents[l, 30];
l = FromContinuedFraction[c], a++];, N[l, 10],
Print[x, "  ", ContinuedFraction[l, Floor[x/2] + 2] [ [ 5 ] ] ]}, {x,
6, 60}]; // TableForm]

into Mathematica.

It seems the following defines ${\displaystyle \scriptstyle a_{5}(n).\,}$

For ${\displaystyle \scriptstyle n{\bmod {6}}\,=\,0\,}$

${\displaystyle a_{5}(n)=\,}$
${\displaystyle 9n+4\,}$ for ${\displaystyle 1
${\displaystyle 9n+2\,}$ for ${\displaystyle 6
${\displaystyle 9n+1\,}$ for ${\displaystyle 12
${\displaystyle 9n+0\,}$ for ${\displaystyle 24

else

${\displaystyle a_{5}(n)=\,}$
${\displaystyle (n-1)/2\,}$ for ${\displaystyle n~{\bmod {~}}6=1\,}$
${\displaystyle 2\,}$ for ${\displaystyle n~{\bmod {~}}6=2\,}$
${\displaystyle (n-1)/2\,}$ for ${\displaystyle n~{\bmod {~}}6=3\,}$
${\displaystyle 1\,}$ for ${\displaystyle n~{\bmod {~}}6=4\,}$
${\displaystyle (n-1)/2\,}$ for ${\displaystyle n~{\bmod {~}}6=5.\,}$

To see it in Mathematica enter

Table[{n, Which[

n <= 6,
Which[Mod[n, 6] == 0, 9 n + 4, Mod[n, 6] == 1, (n - 1)/2,
Mod[n, 6] == 2, 2, Mod[n, 6] == 3, (n - 1)/2, Mod[n, 6] == 4, 1,
Mod[n, 6] == 5, (n - 1)/2],

n <= 12,
Which[Mod[n, 6] == 0, 9 n + 2, Mod[n, 6] == 1, (n - 1)/2,
Mod[n, 6] == 2, 2, Mod[n, 6] == 3, (n - 1)/2, Mod[n, 6] == 4, 1,
Mod[n, 6] == 5, (n - 1)/2],
n <= 24,
Which[Mod[n, 6] == 0, 9 n + 1, Mod[n, 6] == 1, (n - 1)/2,
Mod[n, 6] == 2, 2, Mod[n, 6] == 3, (n - 1)/2, Mod[n, 6] == 4, 1,
Mod[n, 6] == 5, (n - 1)/2],

n <= 100,
Which[Mod[n, 6] == 0, 9 n, Mod[n, 6] == 1, (n - 1)/2,
Mod[n, 6] == 2, 2, Mod[n, 6] == 3, (n - 1)/2, Mod[n, 6] == 4, 1,
Mod[n, 6] == 5, (n - 1)/2]
]}, {n, 1, 60}] // TableForm

and see it gives the same results as

Quiet[Table[{l = x + Random[];
For[a = 1, a < 30, c = Convergents[l, 30];
l = FromContinuedFraction[c], a++];, N[l, 10],
Print[x, "  ", ContinuedFraction[l, Floor[x/2] + 7] [ [ 6 ] ] ]}, {x,
1, 60}]; // TableForm]

Marvin Ray Burns 20:44, 29 May 2011 (UTC)

The pattern for ${\displaystyle \scriptstyle a_{6}(n)\,}$ is proven true in Mathematica for ${\displaystyle \scriptstyle n\,>\,24\,}$

In[189]:=p = Table[{n,
Which[Mod[n, 6] == 5, Floor[n*2/3], Mod[n, 6] == 4, 2,
Mod[n, 6] == 3, Floor[n*2/3] - 1, Mod[n, 6] == 2, 1,
Mod[n, 6] == 1, Floor[n*2/3], Mod[n, 6] == 0,
If[n > 24, Floor[(n - 1)/24], "Skip"]]}, {n, 25, 100}];

In[190]:= q = Quiet[Table[l = x + Random[];
Table[c = Convergents[l, 30];
l = FromContinuedFraction[c], {a, 1, 30}]; {x,
f = ContinuedFraction[l, a][[6 + 1]]}, {x, 25, 100}]];
In[191]:= FullSimplify[p == q]
Out[191]= True

Marvin Ray Burns 00:32, 9 June 2011 (UTC)

## Reference notes

Dan, I made a reference on Table of convergents constants‎ to a proof for cc([1,2]); the reference could use a little sprucing up. —Marvin Ray Burns 00:41, 22 June 2011 (UTC)

I have to debug the {{Cite web}} template, i.e. get rid of that spurious link that appears at the end (the link should only be wrapped under the title.) All the citation templates with web links (e.g. {{Cite web}}, {{Cite arXiv}}, ...) do that. Apart from that bug, all the citation templates work fine. — Daniel Forgues 01:43, 22 June 2011 (UTC)
Dan I made a small discovery on the cc(n) where 0<n<1: If an iterate (the result of FromContinuedFraction) is >1 then the cc will be the same as other cc's with the ame integral value of that iterate. Example, when computing cc(3/100)the iterates are
0.02997275204359673, 0.05968841510642278, 1.996553370372146,
1.4141268406012069, 1.4946426911647988, 1.4997306225168652,
1.4999865180117422, 1.4999993258613324, 1.4999999662930517,
1.4999999983146521, 1.4999999999157325, 1.4999999999957867,
1.4999999999997893, 1.4999999999999896, 1.4999999999999996,...

Once the iterate became 1.996553370372146, the cc was destined to become 3/2, as proven by Filmus, Yuval. I will put this note in the talk page.

## For 0<x<1

The following Mathematica experiments have indicated that for ${\displaystyle \scriptstyle 0\,<\,x\,<\,1\,}$ if an iterate>1 and is not an integer, no other iterate will be an integer and the convergents constant(cc) will be the same as other cc's with the same integral value of that iterate. If an iterate ever becomes an integer the cc will be that integer. Else the cc is 0.555753104278

For rational starting points

Table[{l = x/100,
TableForm[
Table[{c = Quiet[Convergents[Rationalize[l], 50]];
N[l = FromContinuedFraction[c]]}, {a, 6}],
TableHeadings -> Automatic]}, {x, 1, 400}] // TableForm

. For irrational l starting points
(These arn't exactly irrational, but are made to behave like irrationl numbers up to several digits.)

Table[{l = x/100 + Random[],
TableForm[
Table[{c = Quiet[Convergents[Rationalize[l], 50]];
N[l = FromContinuedFraction[c]]}, {a, 6}],
TableHeadings -> Automatic]}, {x, 1, 400}] // TableForm