Extract a Pattern of Iterated continued fractions from convergents

Question

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.]

Answer 1

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

Answer 2

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.

Answer 3

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.