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. I've made some progress in finding the patterns to their continued fractions. Can you find the next pattern $a_6(n)$? If it is like $a_5(n)$ it will be dependent on the moduli of some natural number $K$.

Beginning with $n=0$, the sequence is:
Edit [I made this to hard with an error. I should have said, Beginning with $n=1$, the sequence is:]

1, 2, 1, 3, 3, 9, 4, 1, 5, 2, 7, 9, 8, 1, 9, 2, 11, 2, 12, 1, 13, 2, 15, 16, 16, 1, 17, 2, 19, 1, 20, 1, 21, 2, 23, 1, 24, 1, 25, 2, 27, 1, 28, 1, 29, 2, 31, 1, 32, 1, 33, 2, 35, 2, 36, 1, 37, 2, 39, 2, 40, 1, 41, 2, 43, 2, 44, 1, 45, 2, 47, 2, 48, 1, 49, 2, 51, 3, 52, 1, 53, 2, 55, 3, 56, 1, 57, 2, 59, 3, 60, 1, 61, 2, 63, 3, 64, 1, 65, 2, 67, 4, 68 ,1, 69, 2, 71, 4, 72, 1, 73, 2, 75, 4, 76, 1, 77, 2, 79, 4, 80, 1, 81, 2, 83, 4, 84, 1, 85, 2, 87, 5, 88, 1, 89, 2, 91, 5, 92, 1, 93, 2, 95, 5, 96, 1, 97, 2, 99, 5, 100, 1, 101, 2, 103, 6, 104, 1, 105, 2, 107, 6, 108, 1, 109, 2, 111, 6, 112, 1, 113, 2, 115, 6, 116, 1, 117, 2, 119, 7, 120, 1, 121, 2, 123, 7, 124, 1, 125, 2, 127, 7, 128, 1, 129, 2, 131, 7, 132, 1.

The function for the pattern to the sequence could be piecewise defined, as the last known piece of $a_5(n)$ started at $n=24$. Whether the functions for $a_5(n)$ and $a_6(n)$ can be defined without piecewise functions has yet to be answered as far as I know.

Addendum June 10, 2011 [Now that I have $a_1(n)$ through $a_6(n)$, I find it difficult to find what they all have in common. Perhaps there is an easier pattern to find $a_1(n)$ from $a_0(n)$, $a_2(n)$ from $a_1(n)$, ...? Can you help me find it? I will ask something like this in the talk page soon.]

Marvin Ray Burns
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  • Apart from typos and the fact that any understandable explanation should use simple examples, you cannot use "truly random" without explanation and you cannot used "it is believed" instead of saying whether this is your personal conjecture or someone else's conjecture. I will vote to close, but I will certainly vote to reopen if you edit your question, explain this conjecture and ask us if it makes sense / has relations to known conjectures and results. – Phira May 19 '11 at 10:26
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    Marvin: (+1) I'd vote to reopen, but don't yet have the "threshold rep" to do so. Also @Zev: could you delete, or modify, your comments given the effort Marvin has put into modifying the question, and the sincerity of the question? – amWhy May 19 '11 at 23:54
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    @Marvin: I've started a meta thread (http://meta.math.stackexchange.com/questions/2214/reopening-the-question-about-convergents) for gaining support for reopening the question. – Qiaochu Yuan May 20 '11 at 06:28
  • @Amy J.M.: I will remove my original comments, because Marvin has edited his wiki page to improve most of the aspects I touched on. However, I still stand by my vote to close. [Myself's post](http://meta.math.stackexchange.com/questions/2214/reopening-the-question-about-convergents/2218#2218) on the meta thread pretty much captures my thoughts on this question. – Zev Chonoles May 20 '11 at 22:43
  • @Zev: fair enough. I hope you didn't take my suggestion as an order? I didn't intend for my comment to come across that way. I just worry that new visitors, browsing questions for the first time (e.g. after they've been improved, etc) will read comments that don't necessarily apply anymore, and wonder "what the he...ck? Yikes!" – amWhy May 20 '11 at 23:04
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    This reminds me of Khinchin's constant http://en.wikipedia.org/wiki/Khinchin's_constant because it is also the limit value of a function on _almost_ _all_ continued fraction parameters. – Dan Brumleve May 21 '11 at 05:37
  • The link says "for most 10 < x < 11 the process returns 10.098057706624427..., not true for x = 10.1, 10.2, 10.5 and perhaps for other values." I wonder if it is true for sqrt(101)? According to the wiki article in my previous comment, solutions of quadratic equations do _not_ produce Khinchin's constant. – Dan Brumleve May 21 '11 at 05:57
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    @Dan Brumleve, The input of sqrt(101) into the function of iterated continued fractions from convergents gives 10.0980577066244279660274026688, when you use the Mathematica code, l = Sqrt[101]; Table[c = Convergents[l, 100]; l = FromContinuedFraction[c], {n, 1, 50}]; N[l, Floor[30]] . – Marvin Ray Burns May 21 '11 at 11:39
  • That value (which I guess is the same for most numbers between 10 and 11) to the precision it is given has three instances of "66" and three instances of three different other double digits ("77", "44", and "88"). I wonder why _that_ is true? – Dan Brumleve May 23 '11 at 03:58
  • Dan, I don't know if we are looking too hard. The pattern might lie in the denominators of the simple continued fraction. See https://oeis.org/wiki/Talk:Table_of_convergents_constants#Partial_denominators_pattern . – Marvin Ray Burns May 23 '11 at 23:58
  • I suggest that if you can find a way to extract this pattern (in base k) it may yield a proof that the attractors are irrational; I am thinking of the http://en.wikipedia.org/wiki/Proof_that_e_is_irrational – Dan Brumleve May 24 '11 at 04:29
  • @Dan Brumleve, I'm working a little harder to extract patterns to the continued fractions of these constants. – Marvin Ray Burns May 30 '11 at 21:40
  • @Dan: Re your numerological comment, note that even if the digits are selected at random, you expect to see a repeated digit at least once every $10$ digits (depending on the distribution of digits). – Yuval Filmus May 31 '11 at 06:29
  • I'm not sure it makes sense to calculate the continued fraction of the constant. It is a fixed point of some operation on continued fractions with non-integral coefficients, so you shouldn't expect much from its integral continued fraction. – Yuval Filmus May 31 '11 at 06:31
  • Yuval, indeed, it is numerological, but that is science. Marvin has now edited the question to render all comments out of context. – Dan Brumleve May 31 '11 at 06:35
  • I take my last comment back, given the new interpretation of the iteration. – Yuval Filmus May 31 '11 at 08:00
  • @Dan and etal, I'm sorry if my zeal for a more useful question made anyone's comment look out of place. I was afraid that if I started a new question about my work on that article, it would be closed as a duplicate, since this question was so general at first. P.S. It might take a while for me to digest all the comments and answers, but I think they will eventually help a lot! – Marvin Ray Burns Jun 01 '11 at 00:21
  • Marvin, don't sweat it. If you have other questions or puzzles about continued fractions (even if you already know the answers) you would be doing the site a service by posting them; currently there are only 17 questions tagged as such out of more than 13000 total. I've found that posting questions is a good way to practice engaging the community and to understand what other people find interesting and why. – Dan Brumleve Jun 01 '11 at 02:20

3 Answers3


Here is some theoretical calculation of the convergent constants, which also agrees with my numerical simulation. However, I get different constants than you! This is probably due to some small misunderstanding, but my method will probably work even using the "correct" formulation.

Edit: Here is the misunderstanding. My formula goes from $[x_0;x_1,x_2,\ldots]$ to $[p_0/q_0;p_1/q_1,p_2/q_2,\ldots]$. Your formula, on the other hand, has an extra step of expressing the latter as a usual integral continued fraction.

This misunderstanding shows that your pages aren't clear enough, please fix that. In more detail, you write $$x_1 = [a_0(1);a_1(1),a_2(1),\ldots] = \left[\frac{p_0(1)}{q_0(1)};\frac{p_1(1)}{q_1(1)},\frac{p_2(1)}{q_2(1)},\ldots\right],$$ but it's not really clear that the middle expression is the regular continued fraction for the value expressed by the continued fraction on the right.

The rest is per the following (erroneous) interpretation. We start with some continued fraction. We compute the partial convergents, and use them as coefficients in a new continued fraction. We then compute the convergents of the latter, and use them as coefficients in another new continued fraction. And so on.

Let the convergents at some given point be $x_0,x_1,\ldots$. In order to calculate the convergents for the next iteration, we use the following formulae: $$ \begin{align*} P_0 &= x_0 & Q_0 &= 1 \\ P_1 &= x_0x_1 + 1 & Q_1 &= x_1 \\ P_n &= x_nP_{n-1}+P_{n-2} & Q_n &= x_nQ_{n-1}+Q_{n-2} \end{align*} $$ We immediately get that the values of $P_0,Q_0$ always stay the same, and so the limiting value $y_0$ of the first convergent is $x_0$.

Next, for the second convergent we have $$x'_1 = P_1/Q_1 = x_0 + 1/x_1.$$ Presumably, if this iteration is repeated, it will eventually reach a fixed point $y_1$ which satisfies $$y_1 = x_0 + 1/y_1.$$ It is easy to solve the quadratic to obtain $$y_1 = \frac{x_0 + \sqrt{x_0^2+4}}{2}.$$ We also obtain limiting values for $P_1,Q_1$, namely $\hat{P}_1 = y_0y_1+1$ and $\hat{Q}_1 = x_1$.

Continuing, for the third convergent we have $x'_2 = P_2/Q_2$. The fixed point $y_2$ satisfies $Q_2 y_2 = P_2$ or $$y_2 (y_2 \hat{Q}_1 + \hat{Q}_0) = y_2 \hat{P}_1 + \hat{P}_0.$$ We again get a quadratic for $y_2$ that we can solve (choosing the positive solution), and then deduce the limiting values $\hat{P}_2,\hat{Q}_2$.

We can continue this way to calculate all the limiting convergents. We naturally have $y_n \rightarrow y_\infty$, and so to derive a numerical approximation of the convergent constant, we can simply pick $n$ big.

So far I have been unable to find a closed formula for the convergent constant, but it's possible that one can come up with such a formula. Note that the constant only depends on $x_0$, which is the floor of the original number.

You can probably get an asymptotic series for the convergent constant this way. It will start $$C(x_0) = x_0 + \frac{1}{x_0} - \frac{3}{x_0^3} + \cdots.$$ We get one more term with each new $y_i$. Probably with some effort one can come up with a recurrence formula for the coefficients.

It's probably possible to prove that if you start with an irrational number, then you always converge to the convergent constant. You just have to prove that each of the convergents converges to the correct $y_n$, and you do that by induction.

The constants I get are different from yours. For example, for $x_0 = 10$ I get the constant $10.0980671369431$.

Here is some sage code that can be used to generate my constants. In order to get the constant of 10, use convergent(10).

def iteration(x, y):
   (A0,B0) = x
   (A1,B1) = y
   a = B1
   b = B0 - A1
   c = -A0
   z = (-b + sqrt(b*b-4*a*c))/(2*a)
   A2 = z*A1 + A0
   B2 = z*B1 + B0
   return (y, (A2,B2))

def convergent(x0, n = 100):
   x0 = RR(x0)
   A0 = x0
   B0 = 1
   B1 = (x0 + sqrt(x0*x0+4))/2
   A1 = x0*B1 + 1
   x = (A0,B0)
   y = (A1,B1)
   for i in range(n):
      x, y = iteration(x, y)
   return y
Yuval Filmus
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  • Marvin gives this code: l = Sqrt[101]; Table[c = Convergents[l, 100]; l = FromContinuedFraction[c], {n, 1, 50}]; N[l, Floor[30]] which wolframalpha.com won't evaluate. Is it a problem with the precision? – Dan Brumleve May 31 '11 at 06:41
  • @Dan: Now I see what the problem is. This code makes it clear what Marvin intended. – Yuval Filmus May 31 '11 at 07:56
  • @Yuval Filmus, I've been looking at oeis.org/wiki/Convergents_constant, and I'm somewhat at a loss to see the ambiguity of which you pointed out. Instead of the middle expression being the regular continued fraction for the value expressed by the continued fraction on the right. The writer meant that we make a generalized continued fraction from the convergents of the cf on the left. I would like the page to be clear without people resorting to my code. What changes do you think I can make to clarify things? – Marvin Ray Burns Jun 02 '11 at 23:24
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    @Marvin: I would suggest giving an *algorithm* to generate the sequence, consisting of three steps: (1) express the current number as a (usual) continued fraction, (2) compute the sequence of partial convergents $r_1,r_2,\ldots$, (3) calculate the new number to be $[r_1;r_2,r_3,\ldots]$. I missed step (1). I also suggest removing the huge displays and using the standard notation $[\cdot;\cdot,\cdot,\cdot,\ldots]$ instead. – Yuval Filmus Jun 03 '11 at 02:44
  • @Yuval Filmus, I am looking into your suggestions, and am considering putting your 3 steps at the top of the section titled "Iterated continued fractions from convergents." Would you mind rewriting them exactly as you think they should be written if they were put in that place? As for the large cf's, they seem to be rather common in the OEIS Wiki. Perhaps I could re-emphasis them in the standard notation somehow. – Marvin Ray Burns Jun 04 '11 at 01:25
  • @Marvin: In order to get from the current iterate $x_i$ to the next one $x_{i+1}$, (1) Express $x_i$ as a continued fraction $[a_0;a_1,a_2,\ldots]$, (2) Compute the partial convergents $r_0 = a_0, r_1 = a_0 + 1/a_1, \ldots$, (3) The next iterate is $x_{i+1} = [r_0;r_1,r_2,\ldots]$. – Yuval Filmus Jun 04 '11 at 16:14
  • @Yuval Filmus, I implemented a couple of your suggestions at https://oeis.org/wiki/Convergents_constant#Iterated_continued_fractions_from_convergents . – Marvin Ray Burns Jun 05 '11 at 00:19
  • @Marvin: Now all that remains is to follow my other answer and try to prove rigorously some facts. At least the first few values of the eventual continued fractions should be approachable. – Yuval Filmus Jun 05 '11 at 00:29
  • @Yuval Filmus, I never answered your inquire,"I get the constant 10.0980671369431. If you understand why, please tell!" I got 10.0980577066244279660274026688 from l = Sqrt[101]; Table[c = Convergents[l, 100]; l = FromContinuedFraction[c], {n, 1, 60}]; N[l, Floor[30]]. I stopped at that many iterations and convergents because when I added more the value no longer changed How many of each did you use? I would be interested if you used more than I did and still had that differing answer. – Marvin Ray Burns Jun 11 '11 at 23:16
  • @Marvin: See my edit, the iteration I consider here is different than yours (and easier to analyze). I calculate the partial convergents and then put them back as coefficients in a new continued fraction, and then *directly* calculate the new partial convergents. In contrast, you first re-express the continued fraction as an integral one, and only then calculate the new partial convergents. – Yuval Filmus Jun 11 '11 at 23:38
  • @Yuval Filmus, It is very interesting how close my convergents constant(101), cc(101), is to your cc(101). I wonder what values do you get for cc(x) for x from 2 to 65, like the ones I posted at https://oeis.org/wiki/Table_of_convergents_constants? I wonder if perhaps they differ by a small constant multiple, especially as n->infinity. If they differ by any predictable factor then your analysis would describe my convergents constants, except for that factor, which might even be transient! I could be hoping for too much, though. – Marvin Ray Burns Jun 12 '11 at 22:49
  • @Marvin: It is true (quite surprisingly) that the continued fraction is of the form $[n;n,\lfloor n/2\rfloor, \cdots]$, and that explains why the two numbers are close to each other. The rest of the continued fraction is probably unrelated, in particular the next coefficient is not well-behaved for even $n$. – Yuval Filmus Jun 13 '11 at 01:40
  • @Yuval Filmus, I was wondering, if you get a chance to do so, would you write a short transition from the analysis you did, to one that supports the cc's that I have talked about? I have a little trouble really understanding your analysis, (for example, like why you solved a quadratic equation). I know that when I rewrite your code in Mathematica I will understand your analysis a little better, but to write your analysis rigorously enough to post is a pretty big step for me at this time. Thank you for having helped me so far. – – Marvin Ray Burns Jun 14 '11 at 00:03
  • @Marvin: The analysis for the two sets of constants is quite different. There is no reason that the analysis for my constants would apply to yours. What do you mean by "a short transition from the analysis [I] did, to one that supports the cc's that [you] have talked about"? – Yuval Filmus Jun 14 '11 at 04:12
  • @Yuval Filmus, I was operating under the mistaken assumption the the analysis of the two were similar. – Marvin Ray Burns Jun 15 '11 at 18:18

Here is some analysis for the actual definition.

Suppose that the original continued fraction is $[a;b,c,d,\ldots]$. The first few convergents are $$a \quad a + 1/b \quad a + 1/(b + 1/c) \quad \ldots$$ Therefore, the continued fraction with convergents as coefficients is equal to $$a + 1/(a + 1/b + 1/(a + 1/(b + 1/c) + \cdots)).$$ In general, we would expect that $1/b + 1/(a + \cdots) < 1$; this will happen eventually. In that case, we can recover the second coefficient of the continued fraction as $a$.

Now we're at the case $[a;a,c,d,\ldots]$. Substituting $b = a$ above, the next iteration is equal to $$a + 1/(a + 1/a + 1/(a + 1/(a + 1/c) + \cdots)).$$ Let's express that as an integral continued fraction. After peeling off the first two coefficients, we are left with $$ \frac{1}{1/a + 1/(a + 1/(a + 1/c))} \approx \frac{1}{1/a + 1/a} = a/2. $$ Therefore in general, the next coefficient should be $\lfloor a/2 \rfloor$.

Now the analysis splits into two cases, whether $a$ is even or odd. You can get $a_3,a_4$ this way. Since $a_4$ involves division by $6$, we know have $6$ cases. And so on.

In order to prove that the process almost always converges to the constant, one needs to be more careful and show that the estimates above are mostly true. Probably one can get some conditions on the original continued fraction, and deduce from them that convergence happens "for most values", with some precise meaning.

This analysis will also help explain why you get different behavior for small $n$. However, the heuristic estimates I use should give you the value of all coefficients "for large $n$" - how large depends on the actual coefficient.

Yuval Filmus
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  • Now that I see from your explanation why $a_1(n)=a_0(n)=n, a_2(n)=floor(n/2)$, how should I argue that the special cases are exempt from the same analysis? For example let n=2.2; then the "convergents constant" of 2.2 = 2.0, where $a_0(2.2)=2$ and $a_1(2.2)$ !=2. – Marvin Ray Burns Jun 06 '11 at 01:50
  • @Marvin: My guess is that the analysis works for all irrationals. This should be relatively easy to show (if true). It should also work for some rationals, but that's a bit more delicate. Experimentation may help. – Yuval Filmus Jun 06 '11 at 02:47
  • Did you happen to notice that 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 it to take the cf of 2.2, it gives only the integer part! That is what happens when you start with any one of most special cases in the algorithm for convergents constants, and why you get something else instead of the answer for most x such that 2 – Marvin Ray Burns Jun 06 '11 at 23:17
  • @Marvin: I don't actually use Mathematica. If decimals confuse Mathematica, don't use them. Can you find any rational counterexamples? If not, the proof should be much easier. – Yuval Filmus Jun 07 '11 at 00:45
  • I quoted you at https://oeis.org/wiki/Table_of_convergents_constants. If there is anything I need to change, let me know. – Marvin Ray Burns Jun 08 '11 at 18:03

With a little help from "Phil" at Math2.org, in the thread found at http://math2.org/mmb/thread/43656, I conjectured a partial pattern for a6(n). Daniel Forgues converted it into latex at https://oeis.org/wiki/Table_of_convergents_constants. Thanks for all your help! I still need to work harder on the proof, though.

Marvin Ray Burns
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