Continued Fractions for Square Roots

In this section I'll consider continued fractions for quadratic irrationals of the form $\sqrt{d}$ , where d is a positive integer which is not a perfect square. Let's recall some earlier results we'll need.

We have an algorithm for constructing the continued fraction for a quadratic irrational. Suppose a quadratic irrational has been written in the form $x = \dfrac{m + \sqrt{d}}{s}$ , where:

(a) $m, s \in \integer$ and $s \ne 0$ .

(b) $s \mid d - m^2$ .

(c) d is a positive integer which is not a perfect square.

Set

$$m_0 = m, \quad s_0 = s, \quad x_0 = x, \quad a_0 = [x_0].$$

Then for $k \ge 0$ , define (in order)

$$\eqalign{ m_{k + 1} & = a_k s_k - m_k \cr \noalign{\vskip2pt} s_{k + 1} & = \dfrac{d - m_{k + 1}^2}{s_k} \cr \noalign{\vskip2pt} x_{k + 1} & = \dfrac{m_{k + 1} + \sqrt{d}}{s_{k + 1}} \cr \noalign{\vskip2pt} a_{k + 1} & = [x_{k + 1}] \cr}$$

The continued fraction for x is $[a_0, a_1, a_2, \ldots]$ , and it is known to be periodic.

In the case of $\sqrt{d}$ where d is a positive integer which is not a perfect square, I may take $m_0 = m = 0$ and $s_0 = s = 1$ .

The main result establishes the form of the continued fraction for $\sqrt{d}$ . To state it, I need a definition.

Definition. A sequence of numbers $b_1, \ldots b_k$ is palindromic if $b_i = b_{k - i + 1}$ for $i = 1, \ldots k$ .

The empty sequence is considered to be palindromic.

For example, the following sequences are palindromic:

$$\{10, 3, 4, 4, 3, 10\}, \quad \{17, 2, 7, 8, 7, 2, 17\}.$$

Theorem. Let d be a positive integer which is not a perfect square.

(a) The continued fraction for $\sqrt{d}$ has the form

$$\sqrt{d} = [a_0, \overline{a_1, \ldots a_{n - 1}, 2 a_0}].$$

(b) The sequence $\{a_1, \ldots a_{n - 1}\}$ is palindromic.

To illustrate, here is the continued fraction for $\sqrt{53}$ :

$$\sqrt{53} = [7, 3, 1, 1, 3, 14, 3, 1, 1, 3, 14, \ldots].$$

Note that $14 = 2 \cdot 7$ , and $\{3, 1, 1, 3\}$ is palindromic.

Likewise,

$$\sqrt{52} = [7, 4, 1, 2, 1, 4, 14, 4, 1, 2, 1, 4, 14, \ldots].$$

Note that $14 = 2 \cdot 7$ and $\{4, 1, 2, 1, 4\}$ is palindromic.

The proof of the theorem will use two theorems of Galois on purely periodic continued fractions which we proved earlier:

  1. Let $x = [a_0, a_1, \ldots]$ be a quadratic irrational. The continued fraction for x is purely periodic if and only if $x > 1$ and $-1 < \overline{x} < 0$ .
  1. Let $x = [\overline{a_0, a_1, \ldots a_{n - 1}, a_n}]$ be a purely periodic quadratic irrational. Then $-\dfrac{1}{\overline{x}}$ is purely periodic, and

$$-\dfrac{1}{\overline{x}} = [\overline{a_n, a_{n - 1}, \ldots a_1, a_0}].$$

Proof. (a) Step 1. We'll show: The continued fraction for $\sqrt{d} + [\sqrt{d}]$ is purely periodic.

Recall that $[\cdot]$ denotes the greatest integer function.

Since d is a positive integer and is not a perfect square, $d > 1$ and

$$\sqrt{d} + [\sqrt{d}] > 1.$$

The conjugate is $[\sqrt{d}] - \sqrt{d}$ . Since $\sqrt{d}$ is not an integer,

$$[\sqrt{d}] < \sqrt{d} < [\sqrt{d}] + 1.$$

The first of these two inequalities gives $[\sqrt{d}] - \sqrt{d} < 0$ . The second of these two inequalities gives

$$\eqalign{ \sqrt{d} - [\sqrt{d}] & < 1 \cr [\sqrt{d}] - \sqrt{d} & > -1 \cr}$$.

All together, the conjugate satisfies

$$-1 < [\sqrt{d}] - \sqrt{d} < 0.$$

We showed above that $\sqrt{d} + [\sqrt{d}] > 1$ , so the hypotheses of the first of Galois's theorems is satisfied. Hence, $\sqrt{d} + [\sqrt{d}]$ is purely periodic. This completes the proof of Step 1.

Step 2. We'll show: The continued fraction for $\sqrt{d}$ has the form

$$\sqrt{d} = [a_0, \overline{a_1, \ldots a_{n - 1}, 2 a_0}].$$

Suppose the period of the continued fraction for $\sqrt{d} + [\sqrt{d}]$ is n. We write

$$\sqrt{d} + [\sqrt{d}] = [\overline{a_0, a_1, \ldots, a_{n - 1}}].$$

Then

$$a_0 = \left[\sqrt{d} + [\sqrt{d}]\right] = 2 [\sqrt{d}].$$

Thus,

$$\sqrt{d} + [\sqrt{d}] = [\overline{2 [\sqrt{d}], a_1, \ldots, a_{n - 1}}] = [2 [\sqrt{d}], a_1, \ldots, a_{n - 1}, \overline{2 [\sqrt{d}], a_1, \ldots, a_{n - 1}}].$$

Note that this is purely periodic, as we knew from Step 1. Moreover, it shows that if the period for the continued fraction for $\sqrt{d}$ is n, the same is true for the continued fraction for $[\sqrt{d}] + \sqrt{d}$ .

Subtract $[\sqrt{d}]$ from both sides to obtain

$$\sqrt{d} = [[\sqrt{d}], a_1, \ldots, a_{n - 1}, 2 [\sqrt{d}], \overline{a_1, \ldots, a_{n - 1}, 2 [\sqrt{d}]}].$$

That is, the repeating part is $\{a_1, \ldots, a_{n - 1}, 2 [\sqrt{d}]\}$ . The continued fraction has the form claimed.

Note that the infinite continued fraction for an irrational number is unique. So $a_1$ , ..., $a_{n - 1}$ are the same a's we'd have obtained if we'd computed the continued fraction for $\sqrt{d}$ using the standard algorithm.

This completes the proof of part (a).

(b) By the proof of (a), the continued fraction for $\sqrt{d} + [\sqrt{d}]$ is purely periodic. So suppose that

$$\sqrt{d} + [\sqrt{d}] = [\overline{a_0, a_1, \ldots, a_{n - 1}}].$$

I showed in the proof of (a) that

$$\sqrt{d} = \left[[\sqrt{d}], \overline{a_1, \ldots, a_{n - 1}, 2 [\sqrt{d}]}]\right].$$

Since $\sqrt{d} + [\sqrt{d}]$ is purely periodic, the second theorem of Galois shows that $\dfrac{-1}{\overline{\sqrt{d} + [\sqrt{d}]}} = \dfrac{1}{\sqrt{d} - [\sqrt{d}]}$ is purely periodic, and

$$\dfrac{1}{\sqrt{d} - [\sqrt{d}]} = [\overline{a_{n - 1}, \ldots, a_1, a_0}].$$

But

$$\sqrt{d} - [\sqrt{d}] = [0, \overline{a_1, \ldots, a_{n - 1}, 2 [\sqrt{d}]}] = \dfrac{1}{\overline{a_1, \ldots, a_{n - 1}, 2 [\sqrt{d}]}}.$$

Hence,

$$\dfrac{1}{\sqrt{d} - [\sqrt{d}]} = [\overline{a_1, \ldots, a_{n - 1}, 2 [\sqrt{d}]}].$$

Equating the terms in two expressions for $\dfrac{1}{\sqrt{d} - [\sqrt{d}]}$ , I obtain the equations

$$a_1 = a_{n - 1}, \ldots, a_{n - 1} = a_1.$$

This shows that the sequence $\{a_1, \ldots, a_{n - 1}\}$ is palindromic.

The following refinement of the Theorem will be used in our discussion of the Fermat-Pell equation. We refer to the algorithm for the continued fraction of a quadratic irrational --- "the quadratic irrational algorithm", for short --- described at the beginning of this section.

Proposition. Let d be a positive integer which is not a perfect square. Suppose the quadratic irrational algorithm is applied to:

(1) $\sqrt{d}$ , producing sequences $m_i$ , $s_i$ , $x_i$ , and $a_i$ .

(2) $[\sqrt{d}] + \sqrt{d}$ , producing sequences $m_i'$ , $s_i'$ , $x_i'$ , and $a_i'$ .

Then:

(a) $m_i = m_i'$ for $i \ge 1$ .

(b) $s_i = s_i'$ for $i \ge 0$ .

(c) $x_i = x_i'$ for $i \ge 1$ .

(d) $a_i = a_i'$ for $i \ge 1$ , and $2 a_0 = a_0'$ .

Proof. In the proof of part (a) of the preceding theorem, we showed that

$$\sqrt{d} = \left[[\sqrt{d}], \overline{a_1, \ldots, a_{n - 1}, 2 a_0} \right] = \left[[\sqrt{d}], a_1, \ldots, a_{n - 1}, 2 [\sqrt{d}], a_1, \ldots, a_{n - 1}, 2 [\sqrt{d}], \ldots \right].$$

$$[\sqrt{d}] + \sqrt{d} = \left[\overline{[2 \sqrt{d}], a_1, \ldots, a_{n - 1}} \right] = \left[2 [\sqrt{d}], a_1, \ldots, a_{n - 1}, 2 [\sqrt{d}], a_1, \ldots, a_{n - 1}, 2 [\sqrt{d}], \ldots \right].$$

Comparing terms, we see that (d) follows immediately.

In applying the quadratic irrational algorithm as in (1) and (2), we start with

$$m_0 = 0, \quad s_0 = 1, \quad x_0 = \sqrt{d}, \quad a_0 = [\sqrt{d}],$$

$$m_0' = [\sqrt{d}], \quad s_0' = 1, \quad x_0' = [\sqrt{d}] + \sqrt{d}, \quad a_0' = 2[\sqrt{d}].$$

In particular, we have $s_0 = s_0' = 1$ .

In addition,

$$m_1 = a_0 s_0 - m_0 = [\sqrt{d}] \cdot 1 - 0 = [\sqrt{d}],$$

$$s_1 = \dfrac{d - m_1^2}{s_0} = \dfrac{d - [\sqrt{d}]^2}{1} = d - [\sqrt{d}]^2$$,

$$x_1 = \dfrac{m_1 + \sqrt{d}}{s_1} = \dfrac{[\sqrt{d}] + \sqrt{d}}{d - [\sqrt{d}]^2},$$

$$m_1' = a_0' s_0' - m_0' = 2 [\sqrt{d}] \cdot 1 - [\sqrt{d}] = [\sqrt{d}],$$

$$s_1' = \dfrac{d - m_1'^2}{s_0'} = \dfrac{d - [\sqrt{d}]^2}{1} = d - [\sqrt{d}]^2$$,

$$x_1' = \dfrac{m_1' + \sqrt{d}}{s_1'} = \dfrac{[\sqrt{d}] + \sqrt{d}}{d - [\sqrt{d}]^2}.$$

Thus, $m_1 = m_1'$ , $s_1 = s_1'$ , and $x_1 = x_1'$ .

Suppose inductively that for some $i \ge 1$ ,

$$m_i = m_i', \quad s_i = s_i'.$$

Then

$$m_{i + 1} = a_i s_i - m_i = a_i' s_i' - m_i' = m_{i+1}',$$

$$s_{i + 1} = \dfrac{d - m_{i + 1}^2}{s_i} = \dfrac{d - m_{i + 1}'^2}{s_ii} = s_{i + 1}',$$

$$x_{i + 1} = \dfrac{m_{i + 1} + \sqrt{d}}{s_{i + 1}} = \dfrac{m_{i + 1}' + \sqrt{d}}{s_{i + 1}'} = x_{i + 1}'.$$

By induction, $m_i = m_i'$ for $i \ge 1$ , $s_i = s_i'$ for $i \ge 0$ , and $x_i = x_i'$ for $i \ge 1$ . This proves (a), (b), and (c).

Here's an example which illustrates the result. Here are the first few values of m, s, x, and a for $\sqrt{42}$ :

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & m & & s & & x & & a & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 0 & & 1 & & $6.48074 \ldots$ & & 6 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 6 & & $2.08012 \ldots$ & & 2 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 1 & & $12.48074 \ldots$ & & 12 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 6 & & $2.08012 \ldots$ & & 2 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 1 & & $12.48074 \ldots$ & & 12 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 6 & & $2.08012 \ldots$ & & 2 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 1 & & $12.48074 \ldots$ & & 12 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 6 & & $2.08012 \ldots$ & & 2 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 1 & & $12.48074 \ldots$ & & 12 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 6 & & $2.08012 \ldots$ & & 2 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 1 & & $12.48074 \ldots$ & & 12 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

And here are the corresponding values of m, s, x, and a for $[\sqrt{42}] + \sqrt{42}$ :

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & m & & s & & x & & a & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 1 & & $12.48074 \ldots$ & & 12 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 6 & & $2.08012 \ldots$ & & 2 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 1 & & $12.48074 \ldots$ & & 12 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 6 & & $2.08012 \ldots$ & & 2 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 1 & & $12.48074 \ldots$ & & 12 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 6 & & $2.08012 \ldots$ & & 2 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 1 & & $12.48074 \ldots$ & & 12 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 6 & & $2.08012 \ldots$ & & 2 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 1 & & $12.48074 \ldots$ & & 12 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 6 & & $2.08012 \ldots$ & & 2 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 1 & & $12.48074 \ldots$ & & 12 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

You can see that the m's, x's, and a's agree beginning with the second line (index 1), and the s's agree from the start (index 0).

Proposition. Let d be a positive integer which is not a perfect square, and suppose $[\sqrt{d}] + \sqrt{d}$ has period n:

$$[\sqrt{d}] + \sqrt{d} = [\overline{a_0, a_1, \ldots, a_{n - 1}}].$$

Suppose the quadratic irrational algorithm is applied to $[\sqrt{d}] + \sqrt{d}$ , generating sequences $m_i$ , $s_i$ , and $x_i$ . Then:

(a) $x_i$ is purely periodic with period n.

(b) $x_0 \ne x_1, \ldots, x_{n - 1}$ .

Proof. (a) We saw in the derivation of the quadratic irrational algorithm that the $x_i$ 's produced by the algorithm are the same as the $x_i$ 's in the general continued fraction algorithm --- that is, $x_i$ in the quadratic irrational algorithm is the $i^{\rm th}$ partial quotient of $[\sqrt{d}] + \sqrt{d}$ . The $a_i$ 's are purely periodic with period n:

$$x = x_0 = [a_0, a_1, \ldots a_{n - 1}, a_0, a_1, \ldots a_{n - 1}, \ldots].$$

But

$$[a_0, a_1, \ldots a_{n - 1}, a_0, a_1, \ldots a_{n - 1}, \ldots] = [a_0, a_1, \ldots a_{n - 1}, x_n].$$

Comparing the two expansions, we see that

$$x_n = [a_0, a_1, \ldots a_{n - 1}, \cdots] = x_0.$$

On the other hand, if the $x_i$ 's repeat in n steps, then $a_i = [x_i]$ shows that the a's repeat in n steps as well. So the a's and x's have the same period, and $x_i$ is purely periodic with period n.

(b) Suppose on the contrary that $x_0 = x_i$ where $1 \le i \le n - 1$ . Then

$$[\overline{a_0, a_1, \ldots, a_{n - 1}}] = [\overline{a_i, a_{i + 1}, \ldots, a_{i + n - 1}}].$$

Thus,

$$\eqalign{ a_i & = a_0 \cr a_{i + 1} & = a_1 \cr & \vdots \cr}$$

Hence,

$$[\sqrt{d}] + \sqrt{d} = [\overline{a_0, a_1, \ldots, a_{i - 1}}].$$

But then $[\sqrt{d}] + \sqrt{d}$ has period $i < n$ , contradicting our assumption that the period is n.

Note that by periodicity, $x_0 \ne x_i$ if $i = 1, \ldots, n - 1 \mod{n}$ .

The next result will be used in our discussion of the Fermat-Pell equation.

Theorem. Let d be a positive integer which is not a perfect square. Suppose the period of $\sqrt{d}$ is n, so

$$\sqrt{d} = [a_0, \overline{a_1. \ldots, a_{n - 1}, 2 a_0}].$$

If the quadratic irrational algorithm is applied to $\sqrt{d}$ , generating sequences $m_i$ , $s_i$ , $x_i$ , then:

(a) $s_i = 1$ if and only if $i = k n$ for $k \ge 0$ .

(b) $s_i \ne -1$ for all i.

Proof. (a) The continued fraction for $[\sqrt{d}] + \sqrt{d}$ is purely periodic with period n. Consequently, if the quadratic irrational algorithm is applied for $[\sqrt{d}] + \sqrt{d}$ generating sequences $m_i'$ , $s_i'$ , $x_i'$ , then $x_0' = x_{k n}'$ for $k \ge 0$ .

In our discussion of the quadratic irrational algorithm, we showed that $x_i' = x_j'$ if and only if $(m_i', s_i') = (m_j', s_j')$ . Hence, $s_0' = s_{k n}'$ for $k \ge 0$ . But the second-to-the-last proposition above shows that $s_i = s_i'$ for $i \ge 0$ . Therefore, $s_{k n} = s_0 = 1$ for $k \ge 0$ .

Conversely, suppose $s_i = 1$ . Then

$$x_i = \dfrac{m_i + \sqrt{d}}{s_i} = m_i + \sqrt{d}.$$

If $i = 0$ , we're done. Otherwise, $i \ge 1$ , so again by the second-to-the-last proposition above,

$$x_i' = x_i = m_i + \sqrt{d}.$$

Since $x_i'$ is generated from $[\sqrt{d}] + \sqrt{d}$ , an earlier result shows that the sequence $x_i'$ is purely periodic. By Galois's theorem, the conjugate of $x_i' = m_i + \sqrt{d}$ satisfies

$$-1 < m_i - \sqrt{d} < 0, \quad\hbox{or}\quad \sqrt{d} - 1 < m_i < \sqrt{d}.$$

Thus, $m_i = [\sqrt{d}]$ . It follows that

$$x_i' = [\sqrt{d}] + \sqrt{d} = x_0'.$$

Therefore, $i = k n$ by the preceding result.

(b) Suppose on the contrary that $s_i = -1$ . Then $i \ge 1$ (since $s_0 = 1$ ) and

$$x_i = \dfrac{m_i + \sqrt{d}}{s_i} = -m_i - \sqrt{d}.$$

Since $i \ge 1$ , I have

$$x_i' = x_i = -m_i - \sqrt{d}.$$

But $x_i'$ is purely periodic, so our results on purely periodic continued fractions give

$$-m_i - \sqrt{d} > 1 \quad\hbox{and}\quad -1 < -m_i + \sqrt{d} < 0.$$

The first inequality gives $-\sqrt{d} - 1 > m_i$ , while the second inequality gives $\sqrt{d} < m_i$ . Together, these give $\sqrt{d} < -\sqrt{d} - 1$ , a contradiction. Therefore, $s_i \ne -1$ for all i.

It isn't true in general that for period n we have $s_0 = s_{k n}$ . For instance, here's a quadratic irrational with period 3:

$$\dfrac{-11 + \sqrt{37}}{-4} = [1, 4, \overline{2, 1, 3}].$$

We have:

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & m & & s & & x & & a & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & -11 & & -4 & & 1.2293093674254452 & & 1 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 7 & & 3 & & 4.36092084343274 & & 4 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 5 & & 4 & & 2.7706906325745546 & & 2 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 3 & & 7 & & 1.2975375043283168 & & 1 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 4 & & 3 & & 3.3609208434327393 & & 3 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 5 & & 4 & & 2.7706906325745546 & & 2 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & 3 & & 7 & & 1.2975375043283168 & & 1 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

We don't have $s_0 = 1$ and we don't have $s_0 = s_3$ , $s_0 = s_6$ , and so on.

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