# 9. Row computation in third order linear dashed lines

Prerequirements:

In the computation of the row of the $x$-th dash for third order linear dashed lines, as for the second order, there are particular moduli that, evaluating the appropriate conditions, convey information about the membership to a certain row $i$. However with respect to the second order there are also important differences:

• Complexity of calculations: manual row calculation may take some minutes;
• Generality: there exists a simple condition for row computation, but it’s valid only for dashed lines the components of which are two by two coprime. If you want to compute it generally, you have to treat some particular cases that sometimes can happen.

These problems, indeed, are not as serious as they may seem. The complexity of calculations is not a problem if you use a computer, neither they preclude the theoretical importance of the results. Also matrix calculus, for example, is hard to do manually even with small order matrixes, but this does not diminish the importance of matrixes in applications, neither makes the underlying theory of linear algebra loose its value. Concerning generality, as we’ll see in the section about the applications of dashed line theory, the dashed lines we are interested in for studying prime numbers are those having prime numbers themselves as components; being such, the components of these dashed lines are, even more so, two by two coprime, so the simple condition for row computation can be used.

We’ll see all this in detail below. The proofs are in Teoria dei tratteggi (Dashed line theory, in Italian), pages 163-172.

## Dashed lines with two by two coprime components

Let’s see what happens in dashed lines the components of which are two by two coprime, that are, as we said, the simpler case. As an example we consider the third order dashed line of this kind with the smallest possible components, that is $(2, 3, 5)$:

As for the second order, also for the third order there is a modulus that characterizes each row. For the first row, indicating the dashed line with $T = (n_1, n_2, n_3)$, it’s the following:

$$(n_2 + n_3) n_1 x \mathrm{\ mod\ } (n_1 n_2 + n_1 n_3 + n_2 n_3) \tag{1}$$

We can note a first similarity with the second order: the moduli that appear in Theorem T.2 (Computation of the $x$-th dash row in a second order linear dashed line) are computed with respect to $n_1 + n_2$, that by an algebraic point of view is a symmetrical polynomial with respect to both the valiables $n_1$ and $n_2$. Similarly, modulus (1) is computed with respect to the polynomial $n_1 n_2 + n_1 n_3 + n_2 n_3$ that is symmetrical with respect to $n_1$, $n_2$ and $n_3$. But we already recmet these polynomials talking about the periodicity of linear dashed lines. In particular, by the Corollary of Property T.4 (Number of dashes in a period of a linear dashed line with two by two coprime components), these are the polynomials that count the number of dashes in a period of a linear dashed line with two by two coprime components. According to this Corollary, the formula that computes the number of dashes in a period when the order is $k = 3$ is just $n_1 n_2 + n_1 n_3 + n_2 n_3$, and for $k = 2$ it’s $n_1 + n_2$.

Let’s return to modulus (1) now. Since the polynomial $n_1 n_2 + n_1 n_3 + n_2 n_3$ will often recur and, as we saw, it’s very meaningful, it’s worth giving it a name. We’ll call it $N$:

$$N := n_1 n_2 + n_1 n_3 + n_2 n_3 \tag{2}$$

With this notation, modulus (1) is rewritten as:

$$(n_2 + n_3) n_1 x \mathrm{\ mod\ } N \tag{3}$$

Doing the calculations for the dashed line $T = (n_1, n_2, n_3) = (2, 3, 5)$ of Figure 1, we have that $N = 2 \cdot 3 + 2 \cdot 5 + 3 \cdot 5 = 31$ and (3) becomes $(3 + 5) 2 x \mathrm{\ mod\ } N = 16 x \mathrm{\ mod\ } 31$. Now we compute this modulus for every ordinal $x$: Figure 2: Values of the function (n_2 + n_3) n_1 x mod N = 16x mod 31 for the dashed line (2, 3, 5), for various ordinals x

The numbers in the first row, highlighted in yellow, constitute a particular set, that we’ll call $R_T(1)$. It’s obtained, in the case of a dashed line $T = (n_1, n_2, n_3)$ with two by two coprime components, as follows:

$$R_T(1) := \{n_2 a + n_3 b \mid a \in \{1, \ldots, n_3\}, b \in \{1, \ldots, n_2\}\} \tag{4}$$

In the case of the dashed line $(2, 3, 5)$, doing the calculations we have:

\begin{aligned} R_{(2, 3, 5)}(1) &= \{3 a + 5 b \mid a \in \{1, 2, 3, 4, 5\}, b \in \{1, 2, 3\}\} \\ &= \{8, 11, 14, 17, 20, 13, 16, 19, 22, 25, 18, 21, 24, 27, 30\} \end{aligned}

If you look closely at Figure 2, you can see that these are exactly the values of modulus (3) shown in the first row. In fact, generally speaking, it can be proved that:

$$\mathrm{t}_T(x) \in T \Leftrightarrow (n_2 + n_3) n_1 x \mathrm{\ mod\ } N \in R_T(1) \tag{5}$$

A similar argument is valid for the other lines. In order to evaluate the membership of the $x$-th dash to the second row, the modulus to be evaluated is the following:

$$(n_1 + n_3) n_2 x \mathrm{\ mod\ } N \tag{6}$$

Doing the calculations for $(n_1, n_2, n_3) = (2, 3, 5)$, this formula becomes $(2 + 5) 3 x \mathrm{\ mod\ } N = 21 x \mathrm{\ mod\ } 31$. We compute this modulus for every ordinal $x$ of the dashed line $(2, 3, 5)$: Figure 3: Values of the function (n_1 + n_3) n_2 x mod N = 21x mod 31 for the dashed line (2, 3, 5), for various ordinals x

The numbers shown in the second row, highlighted in yellow, constitute the following set, that we’ll call $R_T(2)$:

$$R_T(2) := \{n_1 a + n_3 b \mid a \in \{1, \ldots, n_3\}, b \in \{0, \ldots, n_1 - 1\}\} \tag{7}$$

In the case of the dashed line $(2, 3, 5)$, doing the calculations we’ll have:

\begin{aligned} R_{(2, 3, 5)}(2) &= \{2 a + 5 b \mid a \in \{1, 2, 3, 4, 5\}, b \in \{0, 1\}\} \\ &= \{2, 4, 6, 8, 10, 7, 9, 11, 13, 15\} \end{aligned}

And these are exactly the values in the second row of Figure 3. Generally speaking:

$$\mathrm{t}_T(x) \in T \Leftrightarrow (n_1 + n_3) n_2 x \mathrm{\ mod\ } N \in R_T(2) \tag{8}$$

Concerning the third row, the modulus to be computed is the following:

$$(n_1 + n_2) n_3 x \mathrm{\ mod\ } N$$

For the dashed line $(2, 3, 5)$ it becomes $(2 + 3) 5 x \mathrm{\ mod\ } N = 26 x \mathrm{\ mod\ } 31$ and assumes the following values: Figure 4: Values of the function (n_1 + n_2) n_3 x mod N = 26x mod 31 for the dashed line (2, 3, 5), for various ordinals x

The numbers in the third row, highlighted in yellow, constitute the set $R_T(3)$ defined as:

$$R_T(3) := \{n_1 a + n_2 b \mid a \in \{0, \ldots, n_2 - 1\}, b \in \{0, \ldots, n_1 - 1\}\} \tag{9}$$

that in the present case becomes:

\begin{aligned} R_{(2, 3, 5)}(3) &= \{2 a + 3 b \mid a \in \{0, 1, 2\}, b \in \{0, 1\}\} \\ &= \{0, 2, 4, 3, 5, 7\} \end{aligned}

This time the condition that characterizes the membership to the third row is the following:

$$\mathrm{t}_T(x) \in T \Leftrightarrow (n_1 + n_2) n_3 x \mathrm{\ mod\ } N \in R_T(3) \tag{10}$$

You can have a confirmation in Figure 4.

We can note that the definitions of the sets $R_T(1)$, $R_T(2)$ and $R_T(3)$ (formulas (4), (7) and (9)), are similar but have some differences about which components are involved and the possible values of the numbers $a$ and $b$ the components are multiplied by. Also formulas (5), (8) and (10), that characterize the membership to a row, have a strong similarity in structure, but the components are interchanged from one formula to the other one. Analyzing these differences it’s possible to obtain a single more general formula both for the sets $R_T(i)$, and for the characterization of the membership to the generic row $i$, as $i$ varies.

The general rules at the basis of the definitions of the sets $R_T(i)$, for $i = 1, 2, 3$, are the following:

• The involved components are those with an index different from $i$: in the definition of the set $R_T(1)$ there are the components different from $n_1$; in the definition of $R_T(2)$ there are those different from $n_2$; in the definition of $R_T(3)$ the ones different from $n_3$.
• Each of the involved components is multiplied by a number varying:
• Between 1 and the other involved component, if the latter has an index greater than $i$
• Between 0 and the other involved component minus one, if the latter has index less than $i$

In fact:

• Into $R_T(1)$ the component $n_2$ is multiplied by a number that varies between 1 and $n_3$, because the other component with index different from 1 is $n_3$ and the index 3 is greater than 1; a similar thing happens for the component $n_3$.
• Into $R_T(2)$ the component $n_1$ is multiplied by a number that varies between 1 and $n_3$, because the other component with index different from 2 is $n_3$ and the index 3 is greater than 2; instead the component $n_3$ is multiplied by a number that varies between 0 and $n_1 - 1$, because the other component with index different from 2 is $n_1$ and it has index less than 2.
• Into $R_T(3)$ the component $n_1$ is multiplied by a number that varies between 0 and $n_2 - 1$, because the other component with index different from 3 is $n_2$ and the index 2 in greater than 1; a similar thing happens for the component $n_2$.

Formally, definitions (4), (7) e (9) can be generalized by the following one, where it’s assumed that $\{i, j, k\} = \{1, 2, 3\}$:

$$R_T(i) := \left \{ n_j a + n_k b \vert \begin{matrix}a \in \begin{cases} \{1, \ldots, n_k\} & \text{if } k \gt i \\ \{0, \ldots, n_k - 1\} & \text{otherwise} \end{cases} \\ b \in \begin{cases} \{1, \ldots, n_j\} & \text{if } j \gt i \\ \{0, \ldots, n_j - 1\} & \text{otherwise} \end{cases} \end{matrix}\right\} \tag{4 + 7 + 9}$$

With the same assumption on $i$, $j$ and $k$, formulas (5), (8) e (10) can be also generalized as follows:

$$\mathrm{t}_T(x) \in T[i] \Leftrightarrow (n_j + n_k) n_i x \mathrm{\ mod\ } N \in R_T(i) \tag{5 + 8 + 10}$$

We can recap what we have seen so far with the following Theorem:

Computation of the row of the $x$-th dash for a third order linear dashed line with two by two coprime components

Let $T$ be a third order linear dashed line with indexes $I = \{1, 2, 3\}$, the components of which are two by two coprime. Let $\{i, j, k\} = \{1, 2, 3\}$. Then for all $x \gt 0$:

$$\mathrm{t}_T(x) \in T[i] \Leftrightarrow (n_j + n_k) n_i x \mathrm{\ mod\ } N \in R_T(i) \tag{5 + 8 + 10}$$

Where $N := n_1 n_2 + n_1 n_3 + n_2 n_3$ and $R_T(i)$ is defined by:

$$R_T(i) := \left \{ n_j a + n_k b \vert \begin{matrix}a \in \begin{cases} \{1, \ldots, n_k\} & \text{if } k \gt i \\ \{0, \ldots, n_k - 1\} & \text{otherwise} \end{cases} \\ b \in \begin{cases} \{1, \ldots, n_j\} & \text{if } j \gt i \\ \{0, \ldots, n_j - 1\} & \text{otherwise} \end{cases} \end{matrix}\right\} \tag{4 + 7 + 9}$$

In particular, for the single possible values of $i$, $i = 1, 2, 3$, we have:

$$\mathrm{t}_T(x) \in T \Leftrightarrow (n_2 + n_3) n_1 x \mathrm{\ mod\ } N \in R_T(1) \tag{5}$$
$$\mathrm{t}_T(x) \in T \Leftrightarrow (n_1 + n_3) n_2 x \mathrm{\ mod\ } N \in R_T(2) \tag{8}$$
$$\mathrm{t}_T(x) \in T \Leftrightarrow (n_1 + n_2) n_3 x \mathrm{\ mod\ } N \in R_T(3) \tag{10}$$

Where

$$R_T(1) := \{n_2 a + n_3 b \mid a \in \{1, \ldots, n_3\}, b \in \{1, \ldots, n_2\}\} \tag{4}$$
$$R_T(2) := \{n_1 a + n_3 b \mid a \in \{1, \ldots, n_3\}, b \in \{0, \ldots, n_1 - 1\}\} \tag{7}$$
$$R_T(3) := \{n_1 a + n_2 b \mid a \in \{0, \ldots, n_2 - 1\}, b \in \{0, \ldots, n_1 - 1\}\} \tag{9}$$

By applying Theorem T.4, let’s establish which row does the sixth dash of the dashed line $T = (2, 3, 5)$, shown in Figure 1, belong to.

Let’s see if it belongs to the first row. So we have to compute the modulus $(n_2 + n_3) n_1 x \mathrm{\ mod\ } N$ of formula (3), that is $16 x \mathrm{\ mod\ } 31$, that for $x = 6$ assumes the value $96 \mathrm{\ mod\ } 31 = 3$. By formulas (5) and (4), the sixth dash belongs to the first row if and only if the number 3 belongs to $R_{(2,3,5)}(1)$, that is if it can be written in the form

$$3 = 3 a + 5 b,\ a \in \{1, 2, 3, 4, 5\},b \in \{1, 2, 3\} \tag{11}$$

But this is not possible. It can be written for example as $3 = 3 \cdot 1 + 5 \cdot 0$, but $0 \notin \{1, 2, 3\}$. Indeed, the smallest possible number that can be written as $3 a + 5 b$, where $a$ and $b$ are positive, is obtained by assigning to the two variables the smallest possible values, in this case $a = 1$ and $b = 1$. But doing so we obtain $3 \cdot 1 + 5 \cdot 1 = 3 + 5 = 8$, that is greater than 3, yet being the smallest possible number satisfying our constraints. So 3 is too small to be expressed in the form (11), and so by Theorem T.4 we conclude that the sixth dash does not belong to the second row.

Let’s see then if the sixth dash belongs to the second row. In this case we have to compute the modulus $(n_1 + n_3) n_2 x \mathrm{\ mod\ } N$ of formula (6), that is $21 x \mathrm{\ mod\ } 31$, that for $x = 6$ assumes the value $126 \mathrm{\ mod\ } 31 = 2$. By formulas (8) and (7), the sixth dash belongs to the second row if and only if the number 2 can be written in the form

$$2 = 2 a + 5 b,\ a \in \{1, 2, 3, 4, 5\},b \in \{0, 1\} \tag{12}$$

Also in this case 2 is a very small number, so the solution has to be searched among the smallest allowed values of $a$ and $b$. Indeed the solution is just $a = 1$ and $b = 0$:

$$2 = 2 \cdot 1 + 5 \cdot 0$$

So (12) has a solution, and by Theorem T.4 we can say that the sixth dash belongs to the second row, as we can see, by counting the positive valued dashes, in Figure 1.
Of course, if we tried to check the membership to the third row, we would obtain another equation without any solution, as happened for the first row.

As we saw in the previous example, row computation for third order dashed lines is reduced to the resolution of equations of the kind

$$c_1 = c_2 a + c_3 b$$

in the unknowns $a$ and $b$, where $c_1$, $c_2$ and $c_3$ are constants and both the constants and the unknowns are integer numbers. They are called Diophantine equations, but in our case, in contrast to the classic theory of diophantine equations, we can’t accept all the solutions, but only those that satisfy particular constraints about $a$ and $b$. For example, taking the cue from formula (11), we may require $a$ and $b$ to be such that:

$$a \in {1, \ldots c_3}, b \in {1, \ldots c_2}$$

To our knowledge, there are no studies about diophantine equations subject to constraints of this kind. So, at the moment, the best resolution method is to reason about specific cases, as in the previous example; alternatively we may simply proceed by trial. In any case, we think it’s interesting to have found this link between two apparently far theories, dashed line theory and the one of diophantine equations. Maybe in the future the results of one of them will be applied to the other one.

## Generic dashed lines

Let’s see how row computation for third order linear dashed line changes when the components are not necessarily two by two coprime, but can be whatever.

In this case the implication towards right of Theorem T.4 is still valid, i.e. if the $x$-th dash belongs to the row with index $i$, then the modulus (5 + 8 + 10) belongs to the set $R_T(i)$. However the implication isn’t true in the opposite direction: if the modulus belongs to the set $R_T(i)$, there are some particular cases in which the dash belongs to a row with index different from $i$, as stated by the following Theorem.

Row computation of the $x$-th dash in any third order linear dashed line

Let $T$ be a third order linear dashed line with indexes $I = \{1, 2, 3\}$. Then for all $x \gt 0$:

$$\mathrm{t}_T(x) \in T[i] \Rightarrow (n_j + n_k) n_i x \mathrm{\ mod\ } N \in R_T(i) \tag{5+8+11 bis}$$

Where $N$ and $R_T(i)$ are defined as in Theorem T.4.

Conversely, if $(n_j + n_k) n_i x \mathrm{\ mod\ } N \in R_T(i)$ for some $i$, indicating with $t$ the dash $\mathrm{t}_T(x)$ and with $v$ its value, the following cases may occur:

• If $n_2 \nmid v$ or $i = 2$, then $t \in T[i]$
• Otherwise:
• If $i = 1$ and $n_2 n_3 = (n_2 + n_3)(n_1 - v \mathrm{\ mod\ } n_1)$, then:
$$\begin{cases} t \in T & \text{if } (n_2 + n_3) n_1 x \mathrm{\ mod\ } N = 2 n_2 n_3 \\ t \in T & \text{otherwise} \end{cases} \tag{14}$$
• If $i = 3$ and $n_1 n_2 = (n_1 + n_2)(n_3 - v \mathrm{\ mod\ }^{\star} n_3)$, then:
$$\begin{cases} t \in T & \text{if } (n_1 + n_2) n_3 x \mathrm{\ mod\ } N = 0 \\ t \in T & \text{otherwise} \end{cases} \tag{15}$$
• In all the other cases, $t \in T[i]$

The most important thing to remember about this Theorem is that the implication (5 + 8 + 10 bis) is valid also towards left:

• If $i = 2$, always
• If $i = 1$ or $i = 3$, at least when $n_2$ does not divide $v$, that is when there isn’t any dash of row 2 in the same column of $t$

The particular cases that occur when these conditions are not verified are rather rare in practice.

We can note that Theorem T.4 is indeed a corollary of Theorem T.6 (though for explanatory issues we preferred to present both as theorems). In fact, the only cases in which the implication (5 + 8 + 10 bis) is not valid also towards left, i.e. the cases in which $(n_j + n_k) n_i x \mathrm{\ mod\ } N \in R_T(i)$ but $t \notin T[i]$, occur when:

$$\begin{cases} n_2 n_3 = (n_2 + n_3)(n_1 - v \mathrm{\ mod\ } n_1) \text{ and } (n_2 + n_3) n_1 x \mathrm{\ mod\ } N = 2 n_2 n_3 & \text{if }i = 1 \\ n_1 n_2 = (n_1 + n_2)(n_3 - v \mathrm{\ mod\ }^{\star} n_3) \text{ and } (n_1 + n_2) n_3 x \mathrm{\ mod\ } N \neq 0 & \text{if } i = 3 \tag{16} \end{cases}$$

In particular, in order that these conditions are true, the following ones must necessarily be true too:

$$\begin{cases} (n_2 + n_3) \mid n_2 n_3 & \text{if }i = 1 \\ (n_1 + n_2) \mid n_1 n_2 & \text{if } i = 3 \tag{17} \end{cases}$$

In fact, for $i = 1$, if $n_2 + n_3$ does not divide $n_2 n_3$, the latter could not be written as $(n_2 + n_3)(n_1 - v \mathrm{\ mod\ } n_1)$, that is as a product of which one factor is just $n_2 + n_3$, so the first case of (16) cannot be true, being a logical conjunction of two conditions the first of which is false; a similar argument can be made for $i = 3$.

But conditions (17) cannot occur if $n_1$, $n_2$ and $n_3$ are two by two coprime. The reason is that the sum of two coprime numbers is coprime to both, and so also to their product; but if their sum is coprime with their product, certainly cannot divide it. In our case, if $i = 1$ and $n_2$ and $n_3$ are coprime, then $n_2 + n_3$ is coprime to both $n_2$ and $n_3$, so also to $n_2 n_3$, so it cannot divide $n_2 n_3$, as it should be by the first case of (17). You can argument similarly for $i = 3$.

The statement of Theorem T.6 is difficult to apply in practice, because the necessary criteria for checking the membership to a row often involve $v$, that normally isn’t known a priori. However, resuming the previous remark, we think that the theorem may be simplified by removing references to $v$, for example the last part of the statement could become:

• If $i = 2$, then $t \in T[i]$
• Otherwise:
• If $i = 1$ and $(n_2 + n_3) \mid n_2 n_3$, then
$$\begin{cases} t \in T & \text{if } (n_2 + n_3) n_1 x \mathrm{\ mod\ } N = 2 n_2 n_3 \\ t \in T & \text{otherwise} \end{cases}$$
• If $i = 3$ and $(n_1 + n_2) \mid n_1 n_2$, then
$$\begin{cases} t \in T & \text{if } (n_1 + n_2) n_3 x \mathrm{\ mod\ } N = 0 \\ t \in T & \text{otherwise} \end{cases}$$
• In all the other cases, $t \in T[i]$

But this has to be verified and, possibly, proved.

We’ll complete this section with an example of when the implication of (5 + 8 + 10 bis) is not valid towards left. For this aim we consider the dashed line $(5, 3, 6)$ and its twelfth dash, that we’ll call $t$:

Let’s compute first of all the sets $R_T(i)$ for all row indexes: Figure 6: Sets R_T(i) for the dashed line T = (5,3,6), as i varies between 1 and 3.

Now, established that $N = n_1 n_2 + n_1 n_3 + n_2 n_3 = 5 \cdot 3 + 5 \cdot 6 + 3 \cdot 6 = 63$, by Theorem T.6 we have to ask ourselves: does $(n_j + n_k) n_i x \mathrm{\ mod\ } N$, for $x = 12$, belong to $R_T(i)$ for some $i$?
Let’s start from $i = 1$. In this case the modulus becomes:

$$(n_2 + n_3) n_1 x \mathrm{\ mod\ } N = (3 + 6)5 \cdot 12 \mathrm{\ mod\ } 63 = 45 \cdot 12 \mathrm{\ mod\ } 63 = 540 \mathrm{\ mod\ } 63 = 36 \tag{18}$$

As you can see in Figure 6, modulus 36 appears in $R_T(i) = R_T(1)$, but also in $R_T(2)$. This fact alone is sufficient to tell that the implication of (5 + 8 + 10 bis) is not valid towards left, because the modulus belongs to $R_T(i)$ for several $i$s: if the implication was valid also towards left, the dash $t$ would stay in the row with index 1 and in the one with index 2 at the same time, that would be absurdum. So let’s see how, by applying Theorem T.6, it’s possible to solve this ambiguity about row.
First of all, looking at Figure 5 we know that the value of $t$ is $v = 18$, that is a multiple of $n_2 = 3$, so neither of the conditions $n_2 \nmid v \Rightarrow 3 \nmid 18$ or $i = 2$, that would guarantee the membership to row $T[i] = T$, is true. Moreover we are just in the case when $n_2 n_3 = (n_2 + n_3)(n_1 - v \mathrm{\ mod\ } n_1)$: in fact $n_2 n_3 = 18$ and $(n_2 + n_3)(n_1 - v \mathrm{\ mod\ } n_1) = (3 + 6)(5 - 18 \mathrm{\ mod\ } 5) = 9(5 - 3) = 18$. Then for establishing which row $t$ belongs to, we have to apply formula (14), i.e. to check if modulus $(n_2 + n_3) n_1 x \mathrm{\ mod\ } N$ is equal to $2 n_2 n_3 = 2 \cdot 3 \cdot 6 = 36$. It’s indeed so (see formula (18)), thus by (14) we have that $t \in T$, that is the correct result, as you can see in Figure 5.

As we noted earlier, in order to apply Theorem T.6 it was necessary to know $v$, and we knew it because we already drew the dashed line table. But for the same reason we also saw that the dash belongs to the third row, without having applied Theorem T.6. However, if the simplified version we saw in the previous remark was true, we would be able to establish the dash row without drawing the table first, by simply evaluating the condition $(n_2 + n_3) \mid n_2 n_3$, that is true because $n_2 + n_3 = 3 + 6 = 9$ divides $n_2 n_3 = 3 \cdot 6 = 18$.

We arrived to say that $t \in T$ starting from $i = 1$, but this choice of $i$ was arbitrary: what would have happened if we had chosen other $i$s?
Let’s see for $i = 2$. In this case we have to evaluate the modulus of formula (8):

$$(n_1 + n_3) n_2 x \mathrm{\ mod\ } N = (5 + 6)3 \cdot 12 \mathrm{\ mod\ } 63 = 33 \cdot 12 \mathrm{\ mod\ } 63 = 396 \mathrm{\ mod\ } 63 = 18$$

This number stays only in $R_T(1)$. Should we then conclude that $t \in T$? No, because the condition to be evaluated is

$$(n_j + n_k) n_i x \mathrm{\ mod\ } N \in R_T(i)$$

While in this case $i = 2$ and $(n_1 + n_3) n_2 x \mathrm{\ mod\ } N \notin R_T(2)$: if this does not happen, we cannot go ahead, no matter if the modulus belongs to some other $R_T(\cdot)$. In this case, the only information we are able to get about the row is that $t \notin T$. We can deduct that by applying the negation of (5 + 8 + 10 bis): in fact, if $t$ was in $T$, the modulus $(n_1 + n_3) n_2 x \mathrm{\ mod\ } N$ would belong to $R_T(2)$; but if it’s not so, it means that $t \notin T$.

If instead we had started from $i = 3$, we would have obtained:

$$(n_1 + n_2) n_3 x \mathrm{\ mod\ } N = (5 + 3)6 \cdot 12 \mathrm{\ mod\ } 63 = 48 \cdot 12 \mathrm{\ mod\ } 63 = 576 \mathrm{\ mod\ } 63 = 9$$

that stays in $R_T(i) = R_T(3)$, but also in $R_T(1)$. But in this case it’s not true that $n_1 n_2 = (n_1 + n_2)(n_3 - v \mathrm{\ mod\ }^{\star} n_3)$, in fact $n_1 n_2 = 5 \cdot 3 = 15$ and $n_3 - v \mathrm{\ mod\ }^{\star} n_3 = 6 - 18 \mathrm{\ mod\ }^{\star} 6 = 6 - 6 = 0$, so their product is 0. Then we have to stop, because there are not the preconditions for evaluating condition (15). Differently from the case of $i = 2$, in this case we have no information about the row of $t$, even about the not-membership to a row.

Concluding, the only $i$ that lets us compute the row of $t$ is $i = 1$. It’s interesting to note that this happens though the dash does not belong to the row with index 1, but this is one of the “oddities” of Theorem T.6 with respect to the much simpler Theorem T.4. Though cases like this are rare in practice, thanks to this example we can have an idea of the complications we are avoiding when we suppose a dashed line to have two by two coprime components.