Proof strategy based on spaces: method based on the approximate calculation of spaces

Prerequisites:

As we indicated in the previous article, our intent is to find conditions for a generic double dashed line $T^{(r_1, r_2, \ldots, r_k)}$ to have spaces in a given interval.
As a starting point, we can observe that we already know the interval in which we want to search for spaces, namely $(1, 2n)$ ; in particular, we know that the lower bound is always 1, and that this bound is excluded. So, given a generic double dashed line $D$ , the following possibilities can arise:

1 is a space. In this case it’s the first space, since 0 is certainly not. With the notations of dashed line theory, we have that $\mathrm{t\_space}_D(1) = 1$ . Then, if there are any spaces between 1 and $2n$ excluded, among them there will certainly be the second. So in this case it’s sufficient to prove that $\mathrm{t\_space}_D(2) \lt 2n$ .
1 is not a space. In this case, if there are any spaces between 1 and $2n$ excluded, the first one will certainly be among them, so in this case we must prove that $\mathrm{t\_space}_D(1) \lt 2n$ . But let’s re-examine the condition of the previous case: if the second space were smaller than $2n$ , a fortiori the first one would also be smaller. So, to prove that $\mathrm{t\_space}_D(1) \lt 2n$ , it’s sufficient to prove that $\mathrm{t\_space}_D(2) \lt 2n$ (even if this condition is not necessary).

As we have seen, in any case the condition $\mathrm{t\_space}_D(2) \lt 2n$ would guarantee the existence of at least one space in the interval $(1, 2n)$ . This, if $D$ is the double dashed line of the Hypothesis H.1 (second form), would guarantee the truth of the Hypothesis itself, which can therefore be further reformulated:

Hypothesis of existence of Goldbach pairs based on dashed lines (third form)

Let $2n \gt 4$ be an even number and let $k$ be the relative order of validity. Let $r_i := 2n \mathrm{\ mod\ } p_i$ , for every $i = 2, \ldots, k$ . Then the second space of the double dashed line $T_k^{(0, r_2, \ldots, r_k)}$ is less than $2n$ .

This reformulation brings us back to the initial problem of calculating the spaces, but this time in a double dashed line instead of a “single” one. But, as we said in the article Calculation of t_space for dashed lines of arbitrary order, explicitly calculating the spaces is very difficult even in single dashed lines, let alone in double ones! However, in our case it’s not necessary to calculate exactly what the second space is, but it is sufficient to prove that it is less than a certain fixed number; therefore even an approximate value would be fine, as long as we can guarantee that the real value does not deviate so much as to exceed $2n$ .

Reformulating the problem further, we can try to study spaces as a whole, rather than focusing on a single space (in this case, the second one). For example, given a double dashed line $D$ , we can reason like this:

Suppose we have found a function $B(m)$ such that, for every positive integer $m$ , the number of spaces od $D$ less than or equal to $m$ is at least $B(m)$ ;
Then, if $B(2n) \geq 2$ , it means that there exist at least two spaces smaller or equal to $2n$ or, equivalently, smaller than it (since $2n$ , being divisible by 2, is certainly not a space); this would prove Hypothesis H.1 (third form).

We have found some of these $B(m)$ functions for dashed lines up to the fourth order; for simplicity we have considered single linear dashed lines, but for double ones we can reason in a similar way.

In the following we’ll indicate with $s(T, m)$ the number of spaces less than or equal to $m$ of a dashed line $T$ .

Overestimation of the number of spaces in the first $m$ columns of a first order dashed line

Let $T = (n_1)$ be a first-order linear dash and let $m$ be a positive integer. Then, to obtain $s(T, m)$ , we must subtract from $m$ the number of dashes less than or equal to $m$ , the values of which are the multiples of $n_1$ less than or equal to $m$ . Their cardinality, by Proposition T.2, is given by $\left\lfloor \frac{m}{n_1} \right\rfloor$ , so:

s(T, m) = m - \left\lfloor \frac{m}{n_1} \right\rfloor \tag{1}

We can rewrite the expression as follows:

\begin{aligned} s(T, m) &= m - \left\lfloor \frac{m}{n_1} \right\rfloor \\ &= m - \frac{m - m \mathrm{\ mod\ } n_1}{n_1} \\ &= m - \frac{m}{n_1} + \frac{m \mathrm{\ mod\ } n_1}{n_1} \\ &= m \left(1 - \frac{1}{n_1} \right) + \frac{\mathrm{err\_s}(T, m)}{n_1} \end{aligned} \tag{2}

where we placed:

\mathrm{err\_s}(T, m) := m \mathrm{\ mod\ } n_1 \tag{3}

Looking at it this way, the value of $s(T, m)$ , or the number of spaces in the first $m$ columns of $T$ , is given by an easily computable formula, $m \left(1 - \frac{1}{n_1} \right)$ , plus an error term. Now, we are interested in the minimum number of spaces in the first $m$ columns (because we want to check that there are at least 2), so we are interested in the minimum value of the error term as $m$ varies. So let’s define:

\mathrm{min\_err\_s}(T) := \min_{1 \leq m \leq n_1} \mathrm{err\_s}(T, m) \tag{4}

where $m$ varies between 1 and $n_1$ so as to cover a whole period of the dashed line (which in this case has length $n_1$ because we are in the first order).

In this case the minimum value is very simple to calculate; in fact, by combining (3) and (4) we’ll obtain that:

\mathrm{min\_err\_s}(T) = \min_{1 \leq m \leq n_1} m \mathrm{\ mod\ } n_1 = 0

since moduli cannot be negative, and for $m = n_1$ we have that $m \mathrm{\ mod\ } n_1 = 0$ . So, by (2):

\begin{aligned} s(T, m) &= m \left(1 - \frac{1}{n_1} \right) + \frac{\mathrm{err\_s}(T, m)}{n_1} \\ & \geq m \left(1 - \frac{1}{n_1} \right) + \frac{\mathrm{min\_err\_s}(T)}{n_1} \\ &= m \left(1 - \frac{1}{n_1} \right) + \frac{0}{n_1} \\ &= m \left(1 - \frac{1}{n_1} \right) \end{aligned}

from which:

s(T, m) \geq m \left(1 - \frac{1}{n_1} \right)

Overestimation of the number of spaces in the first $m$ columns of a second order dashed line

Let $T = (n_1, n_2)$ be a second order linear dashed line and let $m$ be a positive integer. For simplicity, we’ll assume that the components are coprime.
To calculate $s(T, m)$ in this case, we can use the inclusion-exclusion principle, which we have already used to obtain the general characteristic equation of linear $\mathrm{t\_space}$ . According to this principle, we’ll have that:

s(T, m) = m - \left\lfloor \frac{m}{n_1} \right\rfloor - \left\lfloor \frac{m}{n_2} \right\rfloor + \left\lfloor \frac{m}{n_1 n_2} \right\rfloor

With simple algebraic steps, the above expression can be rewritten as follows:

s(T, m) = m \left(1 - \frac{1}{n_1} \right) \left(1 - \frac{1}{n_2} \right) + \frac{\mathrm{err\_s}(T, m)}{n_1 n_2}

where:

\mathrm{err\_s}(T, m) := n_2 (m \mathrm{\ mod\ } n_1) + n_1 (m \mathrm{\ mod\ } n_2) - m \mathrm{\ mod\ } n_1 n_2 \tag{5}

Even in this case we can define $\mathrm{min\_err\_s}(T)$ as the minimum value assumed by $\mathrm{err\_s}(T, m)$ as $m$ varies in a period of the dashed line:

\mathrm{min\_err\_s}(T) := \min_{1 \leq m \leq n_1 n_2} \mathrm{err\_s}(T, m)

from which:

s(T, m) \geq m \left(1 - \frac{1}{n_1} \right) \left(1 - \frac{1}{n_2} \right) + \frac{\mathrm{min\_err\_s}(T)}{n_1 n_2}

In this case, the calculation of $\mathrm{min\_err\_s}(T)$ is not immediate, because the expression (5), which must be minimized, is composed of three modular expressions that are not all of the same sign. As if that were not enough, $n_1$ and $n_2$ not only act as modules but also as multipliers, generating sudden variations in the value of $\mathrm{err\_s}(T, m)$ in the face of small variations in $m$ .

Graph of err_s(T, m) for the dashed line T = (4, 5) as m varies between 1 and 20

Starting from an in-depth study of the expression (5), we have proved the following Proposition:

Calculation of $\mathrm{min\_err\_s}$ for second order dashed lines

Let $T = (n_1, n_2)$ be a second order linear dashed line with coprime components. Then:

If $n_2 \geq (n_1 - 1)^2$ , then $\mathrm{min\_err\_s}(T) = -n_1(n_1 - 1)\left\lfloor \frac{n_2}{n_1} \right\rfloor$
Otherwise, $\mathrm{min\_err\_s}(T) \geq -n_1(n_1 - 1)\left\lfloor \frac{n_2}{n_1} \right\rfloor - (n_1 - 1)^2$

The proof is several pages long, and for brevity we’ll not report it here, but readers interested in it can contact us.

In dashed lines of orders greater than the second, we have generalized the first point of Proposition L.2, that is, we have found a formula for $\mathrm{min\_err\_s}$ valid when the last component of the dashed line is sufficiently large, or in general when the components are sufficiently large.

Overestimation of the number of spaces in the first $m$ columns of a third order dashed line

Let $T = (n_1, n_2, n_3)$ be a third order linear dashed line with pairwise coprime components, and let $m$ be a positive integer.
Applying the inclusion-exclusion principle, in this case we’ll obtain the following formula for $s(T, m)$ :

s(T, m) = m - \left\lfloor \frac{m}{n_1} \right\rfloor - \left\lfloor \frac{m}{n_2} \right\rfloor - \left\lfloor \frac{m}{n_3} \right\rfloor + \left\lfloor \frac{m}{n_1 n_2} \right\rfloor + \left\lfloor \frac{m}{n_1 n_3} \right\rfloor + \left\lfloor \frac{m}{n_2 n_3} \right\rfloor - \left\lfloor \frac{m}{n_1 n_2 n_3} \right\rfloor

With a little algebraic work, this can be rewritten as:

s(T, m) = m \left(1 - \frac{1}{n_1} \right) \left(1 - \frac{1}{n_2} \right) \left(1 - \frac{1}{n_3} \right) + \frac{\mathrm{err\_s}(T, m)}{n_1 n_2, n_3}

where:

\begin{aligned} \mathrm{err\_s}(T, m) := & n_2 n_3 (m \mathrm{\ mod\ } n_1) + n_1 n_3 (m \mathrm{\ mod\ } n_2) + n_1 n_2 (m \mathrm{\ mod\ } n_3) \\ & - n_3 (m \mathrm{\ mod\ } n_1 n_2) - n_1 (m \mathrm{\ mod\ } n_2 n_3) - n_2 (m \mathrm{\ mod\ } n_1 n_3) \\ & + m \mathrm{\ mod\ } n_1 n_2 n_3 \end{aligned} \tag{6}

Defining

\mathrm{min\_err\_s}(T) := \min_{1 \leq m \leq n_1 n_2 n_3} \mathrm{err\_s}(T, m)

we’ll have:

s(T, m) \geq m \left(1 - \frac{1}{n_1} \right) \left(1 - \frac{1}{n_2} \right) \left(1 - \frac{1}{n_3} \right) + \frac{\mathrm{min\_err\_s}(T)}{n_1 n_2 n_3} \tag{7}

We have empirically verified that the following formula holds when $n_3$ is sufficiently large:

\begin{aligned} \mathrm{min\_err\_s}(T) = & -n_1(n_1 - 1)\left\lfloor \frac{n_2}{n_1} \right\rfloor(n_2 + 1)(n_2 + n_3 - 1) + \\ & + n_1 (n_1 - 1) n_2 (n_2 - 1) \left\lceil \frac{(n_2 - n_2 \mathrm{\ mod\ } n_1)(n_2 + n_3 + 1) - n_3}{n_1 n_2} \right\rceil \end{aligned}

Overestimation of the number of spaces in the first $m$ columns of a fourth order dashed line

Let $T = (n_1, n_2, n_3, n_4)$ be a fourth order linear dashed line with pairwise coprime components, and let $m$ be a positive integer.
Applying the inclusion-exclusion principle, in this case we obtain the following formula for $s(T, m)$ :

\begin{aligned} s(T, m) & = m \\ & - \left\lfloor \frac{m}{n_1} \right\rfloor - \left\lfloor \frac{m}{n_2} \right\rfloor - \left\lfloor \frac{m}{n_3} \right\rfloor - \left\lfloor \frac{m}{n_4} \right\rfloor \\ & + \left\lfloor \frac{m}{n_1 n_2} \right\rfloor + \left\lfloor \frac{m}{n_1 n_3} \right\rfloor + \left\lfloor \frac{m}{n_1 n_4} \right\rfloor + \left\lfloor \frac{m}{n_2 n_3} \right\rfloor + \left\lfloor \frac{m}{n_2 n_4} \right\rfloor + \left\lfloor \frac{m}{n_3 n_4} \right\rfloor \\ & - \left\lfloor \frac{m}{n_1 n_2 n_3} \right\rfloor - \left\lfloor \frac{m}{n_1 n_2 n_4} \right\rfloor - \left\lfloor \frac{m}{n_1 n_3 n_4} \right\rfloor - \left\lfloor \frac{m}{n_2 n_3 n_4} \right\rfloor \\ & + \left\lfloor \frac{m}{n_1 n_2 n_3 n_4} \right\rfloor \end{aligned}

Similarly to the formulas of the previous orders, we’ll obtain:

s(T, m) = m \left(1 - \frac{1}{n_1} \right) \left(1 - \frac{1}{n_2} \right) \left(1 - \frac{1}{n_3} \right) \left(1 - \frac{1}{n_4} \right) + \frac{\mathrm{err\_s}(T, m)}{n_1 n_2, n_3, n_4}

where:

\begin{aligned} \mathrm{err\_s}(T, m) & := \\ n_2 n_3 n_4 (m \mathrm{\ mod\ } n_1) + n_1 n_3 n_4 (m \mathrm{\ mod\ } n_2) + n_1 n_2 n_4 (m \mathrm{\ mod\ } n_3) + n_1 n_2 n_3 (m \mathrm{\ mod\ } n_4) \\ - n_3 n_4 (m \mathrm{\ mod\ } n_1 n_2) - n_1 n_4 (m \mathrm{\ mod\ } n_2 n_3) - n_2 n_4 (m \mathrm{\ mod\ } n_1 n_3) & \\ - n_2 n_3 (m \mathrm{\ mod\ } n_1 n_4) - n_1 n_3 (m \mathrm{\ mod\ } n_2 n_4) - n_1 n_2 (m \mathrm{\ mod\ } n_3 n_4) & \\ + n_4 (m \mathrm{\ mod\ } n_1 n_2 n_3) + n_3 (m \mathrm{\ mod\ } n_1 n_2 n_4) + n_2 (m \mathrm{\ mod\ } n_1 n_3 n_4) + n_1 (m \mathrm{\ mod\ } n_2 n_3 n_4) & \\ - m \mathrm{\ mod\ } n_1 n_2 n_3 n_4 \end{aligned} \tag{8}

Defining

\mathrm{min\_err\_s}(T) := \min_{1 \leq m \leq n_1 n_2 n_3 n_4} \mathrm{err\_s}(T, m)

we’ll have:

s(T, m) \geq m \left(1 - \frac{1}{n_1} \right) \left(1 - \frac{1}{n_2} \right) \left(1 - \frac{1}{n_3} \right) \left(1 - \frac{1}{n_4} \right) + \frac{\mathrm{min\_err\_s}(T)}{n_1 n_2 n_3 n_4} \tag{9}

We have empirically verified that the following formula holds when the dashed line components are sufficiently large:

\mathrm{min\_err\_s}(T) = n_4 P - (P - N) \left\lfloor \frac{F (n_4 + 1) - 1}{L}\right\rfloor + C

where:

\begin{aligned}L = & n_1 n_2 n_3 \\ P = & c_1 + c_2 n_3 + c_3 (n_3 + 1) \left(\left\lfloor \frac{c_4 (n_3 + 1)}{n_1 n_2} \right\rfloor - 1 \right) \\ N = & -c_3 n_3^2 + c_2 n_3 + c_5 + c_3 (n_3 + 1) \left\lfloor \frac{c_4 (n_3 + 1)}{n_1 n_2} \right\rfloor \\ F = & c_6 + n_1 n_2 \left\lfloor \frac{c_4 (n_3 + 1)}{n_1 n_2} \right\rfloor \\ C = & c_5 (n_3 - 1) + c_3 * (n_3 - 1) \left\lfloor \frac{c_4 (n_3 + 1)}{n_1 n_2} \right\rfloor \\ c_4 = & n_2 - 1 - n_2 \mathrm{\ mod\ } n_1 \\ c_6 = & n_2 (n_1 - 1) + 1 + n_2 \mathrm{\ mod\ } n_1 \\ c_3 = & n_1 (n_1 - 1) n_2 (n_2 - 1) \\ c_5 = & (n_2 (n_1 - 1) + n_2 \mathrm{\ mod\ } n_1) (n_2 - 1) (n_1 - 1) \\ c_2 = & n_1 (n_1 - 1) \left(((2 n_1 - 3) n_2 - (2 n_1 + 1)) \left\lfloor \frac{n_2}{n_1} \right\rfloor + 2 (n_2 \mathrm{\ mod\ } n_1) (n_2 - 1)\right) \\c_1 = & c_5 + c_3 \end{aligned}

Simplifying formulas

As it’s evident from the previous paragraphs, the complexity of the formulas of $\mathrm{min\_err\_s}$ grows quite rapidly as the order increases. We have therefore tried to simplify them, in order to understand something more.

A first simplification technique is based on the observation that the whole parts of fractions appearing in formulas are always multiplied by their denominator, so these expressions can be simplified using the property that states that, if $a$ and $b$ are two positive whole numbers, $b \left\lfloor \frac{a}{b} \right\rfloor = b \left\lfloor \frac{a - a \mathrm{\ mod\ } b}{b} \right\rfloor = a - a \mathrm{\ mod\ } b$ . By applying this technique, it’s possible to simplify the formulas as follows:

Second order:

\mathrm{min\_err\_s}(T) = - (n_1 - 1) (n_2 - n_2 \mathrm{\ mod\ } n_1)

Third order:

\begin{aligned} \mathrm{min\_err\_s}(T) = & (n_1 - 1) [n_1 (n_2)^2 - 3 n_2 n_3 - n_1 n_2 - n_2 + n_3 + 1 + 2 n_3 (n_2 \mathrm{\ mod\ } n_1) \\ - & (n_2 - 1) (((n_2 - n_2 \mathrm{\ mod\ } n_1) (n_3 + n_2 + 1) - n_3 - 1) \mathrm{\ mod\ } (n_1 n_2))] \end{aligned}

Fourth order:

\begin{aligned} \mathrm{min\_err\_s}(T) = & (n_1 - 1) [2 n_1 n_2 (n_2 - 1) n_4 - 7 n_2 n_3 n_4 - n_2 n_4 + 3 n_3 n_4 + n_4 \\ + & (n_2 - 1) (n_3 - 1) B - 2 n_4 (n_2 - 1) A + (4 n_3 n_4) (n_2 \mathrm{\ mod\ } n_1)] \end{aligned}

where:

\begin{aligned}A := & ((n_2 - 1 - (n_2 \mathrm{\ mod\ } n_1)) (n_3 + 1)) \mathrm{\ mod\ } (n_1 n_2) \\ B := & (((n_2 - 1 - (n_2 \mathrm{\ mod\ } n_1)) n_3 + n_1 n_2 - A) (n_4 + 1) - 1) \mathrm{\ mod\ } (n_1 n_2 n_3) \end{aligned} \tag{10}

We also used another technique to further simplify the formulas, introducing an approximation. We observed that, by keeping the components of the dashed line fixed up to the penultimate one and making the last one vary, the graph of $\mathrm{min\_err\_s}$ that is obtained can be approximated with a straight line, which is always below it and which passes through the origin. For example, for the dashed lines $(2, 5, n_3)$ (where the first two components are fixed and the third varies) the following graph is obtained:

Graph of min_err_s(T) for third order dashed lines T of the type (2, 5, n_3) with n_3 up to 50, compared with the graph of the straight line that best approximates the graph by defect.

To calculate the angular coefficient of the line, we observed that $\mathrm{min\_err\_s}(T)$ always decreases by the same amount when the product of the other components is added to the last component of the dashed line. So, the angular coefficient can be calculated as the ratio between the variation of the function value and the variation of the last component (which is the product of the previous components). For example, in the dashed lines $(2, 5, n_3)$ we found that $\mathrm{min\_err\_s}$ decreases by 120 if $n_3$ is increased by $2 \cdot 5 = 10$ , so the angular coefficient of the line is given by:

\frac{\Delta \mathrm{min\_err\_s}}{\Delta n_3} = \frac{-120}{10} = -12

Since we have the general formula for $\mathrm{min\_err\_s}$ up to the fourth order, using this principle we can find the general formula of the angular coefficient of the line, which for example for the third order will be given by:

\frac{\mathrm{min\_err\_s}(n_1, n_2, n_3) - \mathrm{min\_err\_s}(n_1, n_2, n_3 + n_1 n_2)}{n_1 n_2}

With this technique you can find the following formulas for the slope of the line:

for the second order:
$- (n_1 - 1) n_1$
for the third order:

$- (n_1 - 1)(3 n_2 - 1 - 2 (n_2 \mathrm{\ mod\ } n_1))$
for the fourth order:

$(n_1 - 1) (2 n_1 n_2 (n_2 - 1) - 7 n_2 n_3 - n_2 + 3 n_3 + 1 - 2 (n_2 - 1) A + 4 n_3 (n_2 \mathrm{\ mod\ } n_1))$

where $A$ is given by (10).

Remembering that the line is a function of the last component of the dashed line and that it passes through the origin, the equation of the line for each order is therefore the following:

for the second order:
$y(n_2) = - (n_1 - 1) n_1 n_2$
for the third order:

$y(n_3) = - (n_1 - 1)(3 n_2 - 1 - 2 (n_2 \mathrm{\ mod\ } n_1)) n_3$
for the fourth order:
$y(n_4) = (n_1 - 1) (2 n_1 n_2 (n_2 - 1) - 7 n_2 n_3 - n_2 + 3 n_3 + 1 - 2 (n_2 - 1) A + 4 n_3 (n_2 \mathrm{\ mod\ } n_1)) n_4$

In our experimental tests, this line has always been a good estimate of $\mathrm{min\_err\_s}$ , so we concluded that the following relations may hold, but they still need to be proven:

for the second order:
$\mathrm{min\_err\_s}(n_1, n_2) \geq - (n_1 - 1) n_1 n_2$
for the third order:
$\mathrm{min\_err\_s}(n_1, n_2, n_3) \geq - (n_1 - 1)(3 n_2 - 1 - 2 (n_2 \mathrm{\ mod\ } n_1)) n_3 \tag{11}$
for the fourth order:

$\begin{aligned}\mathrm{min\_err\_s}(n_1, n_2, n_3, n_4) \geq & (n_1 - 1) [2 n_1 n_2 (n_2 - 1) - 7 n_2 n_3 - n_2 + 3 n_3 \\ & + 1 - 2 (n_2 - 1) A + 4 n_3 (n_2 \mathrm{\ mod\ } n_1)] n_4 \end{aligned} \tag{12}$

Many of the formulas presented in this article, precisely those that calculate or approximate

\mathrm{min\_err\_s}

for the third and fourth order, are the result of a generalization process, started from the analysis of concrete cases and carried out using specific computer programs developed by us. As meticulous as it has been, this work cannot replace a real proof; demonstrating the formulas obtained is therefore an open point.

Final considerations and further developments

Formulas (11) and (12), which underestimate (according to experimental evidence, with good approximation) $\mathrm{min\_err\_s}$ for the third and fourth orders respectively, allow us to intuit what the behavior of the function may be for higher orders. Developing the products, we can note that the terms with the greatest weight are $-3 n_1 n_2 n_3$ in the third-order formula and $-7 n_1 n_2 n_3 n_4$ in the fourth-order one. Recalling that, in formulas (7) and (9), $s(T, m)$ has been rewritten so that $\mathrm{min\_err\_s}$ is divided by $n_1 n_2 n_3$ in the third order and by $n_1 n_2 n_3 n_4$ in the fourth, the two terms with greater weight contribute to the increase of $s(T, m)$ by respectively $\frac{-3 n_1 n_2 n_3}{n_1 n_2 n_3} = -3$ in the third order and $\frac{-7 n_1 n_2 n_3 n_4}{n_1 n_2 n_3 n_4} = -7$ in the fourth. Now, we can observe that $3 = 2^2 - 1$ and $7 = 2^3 - 1$ . In particular, 3 and 7 represent exactly the number of negative terms that appear in the formulas of $\mathrm{err\_s}(T, m)$ (6) and (8), respectively for the third and fourth order. Considering for example the formula (6) of the third order, we have:

Positive terms: $n_2 n_3 (m \mathrm{\ mod\ } n_1)$ , $n_1 n_3 (m \mathrm{\ mod\ } n_2)$ , $n_1 n_2 (m \mathrm{\ mod\ } n_3)$ , $m \mathrm{\ mod\ } n_1 n_2 n_3$
Negative terms: $- n_3 (m \mathrm{\ mod\ } n_1 n_2)$ , $- n_1 (m \mathrm{\ mod\ } n_2 n_3)$ , $- n_2 (m \mathrm{\ mod\ } n_1 n_3)$

Trying to calculate $\mathrm{min\_err\_s}$ , we tried to find the minimum value that the sum of all these terms, positive and negative, can assume when $m$ varies. Without going into too much detail, it’s clear that each positive term can be 0 (for example, if $m \mathrm{\ mod\ } n_1 = 0$ the first positive term is 0) and obviously not less than 0, while each negative term cannot exceed $n_1 n_2 n_3$ in absolute value (for example, $|- n_3 (m \mathrm{\ mod\ } n_1 n_2)| = n_3 (m \mathrm{\ mod\ } n_1 n_2) \leq n_3 (n_1 n_2 - 1) \leq n_1 n_2 n_3$ ). So, the minimum value we are looking for cannot be less than the value that would be obtained in the hypothetical case where all four positive terms are 0 and all three negative terms are $- n_1 n_2 n_3$ :

\mathrm{min\_err\_s}(n_1, n_2, n_3) \geq - 3 n_1 n_2 n_3

Similarly, for the fourth order we can get that:

\mathrm{min\_err\_s}(n_1, n_2, n_3, n_4) \geq - 7 n_1 n_2 n_3 n_4

since there are 7 negative terms and each cannot be greater than $n_1 n_2 n_3 n_4$ in absolute value.

These two overestimations have been obtained with a simplistic reasoning, considering a hypothetical worst case that might never occur; in fact, when $m$ varies, usually the positive and negative terms always balance each other in part, and at first sight it’s not possible to establish whether it’s actually possible to find a value of $m$ that produces a “perfect imbalance” like the one hypothesized (or almost). However, our detailed analysis, at least up to the fourth order, gave an affirmative answer to this question: it’s indeed always possible to find a value of $m$ that generates a case in which $\mathrm{min\_err\_s}$ approaches $-3 n_1 n_2 n_3$ for the third order and $-7 n_1 n_2 n_3 n_4$ for the fourth, thus assuming a value close to that of the worst case hypothesised.

From the point of view of the proof strategy we are following, this is a big problem. In fact, if the multiplicative coefficient (-3 for the third order and -7 for the fourth) were to grow as the number of negative terms in the expression of $\mathrm{min\_err\_s}$ , it would grow exponentially in absolute value with respect to the order of the dashed line (in particular, as $-(2^k - 1)$ for dashed lines of order $k$ ). But, as we noted before, the multiplicative coefficient represents the contribution of the error term to the bound of $s(T, m)$ . We would therefore have an error that grows exponentially, causing the right-hand side of the higher-order formulas similar to (7) and (9) to soon become a negative number, since the effect of the exponentially decreasing right-hand side would heavily prevail over the modestly growing term on the left. The inequalities would therefore attest that $s(T, m)$ is greater than or equal to a negative number: information that is trivially true by definition, and therefore of no use. Have we therefore made a mistake in our proof strategy?

To clarify, we need to remember the context, starting from the validity interval. If it’s true that it is possible to find a value of $m$ that generates exponential errors in the increase of $s(T, m)$ , it is also true that we have searched for this $m$ throughout a period of the dashed line, given by the product of the components, which grows at least exponentially (if the order is $k$ , we have $k$ components; if all the components were 2, their product would be $2^k$ , which is already an exponential, but the true product is even bigger, since the components are increasing and only the first can be 2). Therefore, the search for the minimum in a set of (at least) exponential size generates an exponential error term.
But in reality we do not need to search in the entire period of the dashed line; we are only interested in the validity interval, because it’s in that portion of the hatching that we want to look for spaces. The question we must ask ourselves is therefore the following: if in the search for $\mathrm{min\_err\_s}$ we limited ourselves to considering the values of $m$ within the validity interval, would the error continue to grow exponentially with respect to the order, or would there be a different type of trend? The answer is in the following article:

Overestimation of the number of spaces up to a column included in the validity interval

Proof strategy based on spaces: method based on the approximate calculation of spaces

Overestimation of the number of spaces in the first $m$ columns of a first order dashed line

Overestimation of the number of spaces in the first $m$ columns of a second order dashed line

Overestimation of the number of spaces in the first $m$ columns of a third order dashed line

Overestimation of the number of spaces in the first $m$ columns of a fourth order dashed line

Simplifying formulas

Final considerations and further developments

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Overestimation of the number of spaces in the first m columns of a first order dashed line

Overestimation of the number of spaces in the first m columns of a second order dashed line

Overestimation of the number of spaces in the first m columns of a third order dashed line

Overestimation of the number of spaces in the first m columns of a fourth order dashed line

Simplifying formulas

Final considerations and further developments

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Overestimation of the number of spaces in the first $m$ columns of a first order dashed line

Overestimation of the number of spaces in the first $m$ columns of a second order dashed line

Overestimation of the number of spaces in the first $m$ columns of a third order dashed line

Overestimation of the number of spaces in the first $m$ columns of a fourth order dashed line