# 19. Calculation of t_value for dashed lines of arbitrary order

Prerequisites:

The function $\mathrm{t\_value}$, by definition, indicates which column of a dashed line a dash belongs to. For this reason, in principle, calculating this function is quite easy, because the definition itself shows us how to do it:

• A progressive number is assigned, starting from 1, to all the dashes, excluding those in column 0; the assignment must take place from top to bottom and from left to right;
• $\mathrm{t\_value}(x)$ is the column to which the $x$-th dash belongs.

Let’s take for example the dashed line $T_2$ up to column 10:

0 1 2 3 4 5 6 7 8 9 10
2
3

Let’s associate a progressive number to each dash, except for those in column 0:

0 1 2 3 4 5 6 7 8 9 10
2   1   3   4   6   8
3     2     5     7

We have 8 dashes in all, numbered from 1 to 8, in which $\mathrm{t\_value}$ is:
$\mathrm{t\_value}(1) = 2$;
$\mathrm{t\_value}(2) = 3$;
$\mathrm{t\_value}(3) = 4$;
$\mathrm{t\_value}(4) = \mathrm{t\_value}(5) = 6$;
$\mathrm{t\_value}(6) = 8$;
$\mathrm{t\_value}(7) = 9$;
$\mathrm{t\_value}(8) = 10$.

This method is simple enough to apply, but it has several disadvantages: it takes more and more time as the number of columns considered increases, and is not based on a formula but on an algorithm, so it is difficult to determine the behaviour of the function, its properties and so on, if not by trial and error.

For our purposes, however, we must try to get $\mathrm{t\_value}$ through a calculation. But it is not easy to find a formula that, given $x$, returns us $\mathrm{t\_value}(x)$ for a dashed line of any order. This is a still open problem of dashed line theory, but there are some partial results of some importance, which could become the basis for finding a universal algebraic expression.

## Calculation for the dashes of the first row

The formulas already known for the calculation of $\mathrm{t\_value}$ depend on the row containing the dash the value of which is to be calculated, and the row in turn can be calculated with other formulas. However, to simplify the search for a formula for $\mathrm{t\_value}$ that applies to all orders, we can think of treating initially only the case of the dashes of the first row (i.e. the calculation of $\mathrm{t\_value}(x)$ assuming that the $x$-th dash belongs to the first row). This assumption reduces the number of variables to consider, since we don’t have to worry about the row, but does not reduce the difficulty of the problem; in fact, as it can be seen from the already known formulas, the complexity of the formula does not depend on the particular row considered.

### Calculation for second and third order dashed lines

The formulas for calculating $\mathrm{t\_value}$ for dashes of first row of dashed lines of small orders are already known, as well as their properties. If the $x$-th dash belongs to the first row, for a second order dashed line $(n_1, n_2)$, due to Corollary of Theorem T.8 (Formula for calculation of linear second order function $\mathrm{t\_value}$ for first row) we have:

$$\mathrm{t\_value_2}(x) = n_1 \biggl \lceil \cfrac{n_2 \cdot x + 1}{n_1 + n_2} \biggr \rceil \tag{1}$$

For a third order dashed line $(n_1, n_2, n_3)$, instead, the formula for $\mathrm{t\_value}(x)$ can be obtained through the downcast from third order to first, applying the Theorem T.11 (Partial solution of the downcast characteristic equation of linear $\mathrm{t}$, from the third to the first order) starting by calculating $y$, setting $i = 1$, $j = 2$, $k = 3$, and, at the end, the Proposition T.1 (Linear first order $\mathrm{t}$ and $\mathrm{t\_value}$ functions):

$$\mathrm{t\_value_3}(x) = n_1 \biggl \lceil \cfrac{n_2 \cdot n_3 \cdot x + n_2 + n_3}{n_1 \cdot n_2 + n_1 \cdot n_3 + n_2 \cdot n_3} \biggr \rceil \tag{2}$$

These formulas are almost entirely composed of algebraic operations, apart from the symbol $\lceil \cdot \rceil$, which indicates rounding up.

Resuming the previous dashed line $(2, 3)$, of second order, let’s calculate $\mathrm{t\_value}(4)$ applying the formula (1). Let’s start by replacing the numeric values, which in our case are $n_1 = 2$, $n_2 = 3$ and $x = 4$:

$\mathrm{t\_value}(4) = 2 \biggl \lceil \cfrac{3 \cdot 4 + 1}{2 + 3} \biggr \rceil$

Let’s proceed with the calculation, until we obtain the final value:

$\mathrm{t\_value}(4) = 2 \biggl \lceil \cfrac{12 + 1}{5} \biggr \rceil = 2 \biggl \lceil \cfrac{13}{5} \biggr \rceil = 2 \cdot 3 = 6$

which is exactly the result we would have achieved by applying the definition.

The formula (2), as specified in the statement of Theorem T.11, is valid only if the components of the dashed line satisfy a particular inequality. To avoid additional hypotheses of this type, two downcasts can be made, first from the third order to the second one and then from the second to the first one, applying the Corollary of Theorem T.9 (Downcast function of linear $\mathrm{t}$, from the third to the second order) for first downcast, and the Corollary of Theorem T.8 (Formula for calculation of linear second order function $\mathrm{t\_value}$ for first row) for the other one; however this would result in a more complex final formula for $\mathrm{t\_value}$, with two rounding up operations. For the moment, for simplicity and to focus on the main ideas, we will limit ourselves to propose a generalization of the formula (2), neglecting the hypotheses regarding its applicability. Of course, these hypotheses, together with the alternative formula deriving from the approach with multiple downcasts, will have to be considered when everything has to be proved.

### Extension to higher orders

The fact that in the second and third order formulas there are, in the numerator and denominator of the fraction, only sums and products, allows us to rewrite them in a more compact form, using the elementary symmetric polynomials. If we indicate with $\mathrm{t\_value}_k$ the formula referred to the order $k$, we have:

$\mathrm{t\_value}_2(x) = n_1 \biggl \lceil \cfrac{\sigma_1(n_2) x + \sigma_0(n_2)}{\sigma_1(n_1, n_2)} \biggr \rceil$
$\mathrm{t\_value_3}(x) = n_1 \biggl \lceil \cfrac{\sigma_2(n_2, n_3) x + \sigma_1(n_2, n_3)}{\sigma_2(n_1, n_2, n_3)} \biggr \rceil$

Observing the formulas thus obtained, it is noted that they are characterized by a series of constants, which are all less than or equal to the order: for example, in the third order formula there are 2s, at the denominator there is a list of 3 elements, and so on. We can therefore rewrite these formulas in such a way that all these constants are calculated starting from the order, for example 2 can be rewritten as 3 – 1. This way, we’ll have:

$\mathrm{t\_value_2}(x) = n_1 \biggl \lceil \cfrac{\sigma_{2 - 1}(n_2) x + \sigma_{2 - 2}(n_2)}{\sigma_{2 - 1}(n_1, n_2)} \biggr \rceil$
$\mathrm{t\_value_3}(x) = n_1 \biggl \lceil \cfrac{\sigma_{3 - 1}(n_2, n_3) x + \sigma_{3 - 2}(n_2, n_3)}{\sigma_{3 - 1}(n_1, n_2, n_3)} \biggr \rceil$

But also the lists that are the subject of the function $\sigma$ can be reviewed with this in mind: $(n_1, n_2, n_3)$ is the list of components from $n_1$ to $n_3$, $(n_2, n_3)$ the list of components from $n_2$ to $n_3$, and so on. Applying this principle, if we indicate with “…” the intermediate elements (also including degenerate cases in which the resulting element is only one), the two formulas will become:

$\mathrm{t\_value_2}(x) = n_1 \biggl \lceil \cfrac{\sigma_{2 - 1}(n_2, \ldots, n_2) x + \sigma_{2 - 2}(n_2, \ldots, n_2)}{\sigma_{2 - 1}(n_1, \ldots, n_2)} \biggr \rceil$
$\mathrm{t\_value_3}(x) = n_1 \biggl \lceil \cfrac{\sigma_{3 - 1}(n_2, \ldots, n_3) x + \sigma_{3 - 2}(n_2, \ldots, n_3)}{\sigma_{3 - 1}(n_1, \ldots, n_3)} \biggr \rceil$

Written this way, the two formulas are very similar to each other, so much that, by setting $k \in \{2, 3\}$, both can be represented with a single formula:

$f_k(x) = n_1 \biggl \lceil \cfrac{\sigma_{k - 1}(n_2, \ldots, n_k) x + \sigma_{k - 2}(n_2, \ldots, n_k)}{\sigma_{k - 1}(n_1, \ldots, n_k)} \biggr \rceil$

This formula, which generalizes (1) and (2), is certainly valid for orders 2 and 3 (net of the hypotheses of Theorem T.11 for the third order, as we observed earlier). But we can try to take a step forward: if we could prove that it is also valid for any order, we would have found a universal formula to calculate $\mathrm{t\_value}$. We are therefore conducting a series of empirical checks to try to understand how this formula behaves, with the help of a special automated verification procedure, capable of determining, given an order $k$:

• If the formula is valid for the entire first period of the dashed line $T_k = (p_1, p_2,\cdots, p_k)$ (for simplicity, we are limiting ourselves to these particular dashed lines, used in our proof strategies, having the first prime numbers as components);
• In cases where the formula is not valid, how much is the difference between the calculated value and the real value. If there are more than one deviation, they are all indicated.

To check on an order, simply enter it in the box below and click on “Check”; the “Reset” button clears the box and the results, so that another check can be carried out, while the “Show every invalid value” box is used to show, in addition to the final result, also every single value that differs from the expected one. If the order is high, the operation may take a long time.

Order to be verified:
Show every invalid value

This way, varying from time to time the analyzed order, we have so far come to these conclusions:

• The formula is valid for $T_4 = (2, 3, 5, 7)$;
• For dashed lines from $T_5 = (2, 3, 5, 7, 11)$to $T_{10} = (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31)$, the formula is not always valid, and for the values of $x$ for which it isn’t, the relationship $f_k(x) = \mathrm{t\_value}(x) - 2$ applies;
• For $T_{11} = (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37)$, the formula is not always valid, and for the values of $x$ for which it isn’t, the relationship $f_k(x) = \mathrm{t\_value}(x) - w$, con $w \in \{2, 4\}$ applies.

In the light of these results, we can say that the value of the formula that we have found is not always equal to $\mathrm{t\_value}(x)$, but it would be if a certain amount were added to it, which we can assume it depends on $x$. We can therefore state the following hypothesis:

Universal formula for $\mathrm{t\_value}$

Let $T_k = (p_1, p_2, ... p_k)$ be the dashed line with the first $k \geq 1$ prime numbers as its components.Then, for each dash having ordinal $x$, belonging to the first line, we have:

$\mathrm{t\_value}(x) = p_1 \biggl \lceil \cfrac{\sigma_{k - 1}(p_2, \ldots, p_k) x + \sigma_{k - 2}(p_2, \ldots, p_k)}{\sigma_{k - 1}(p_1, \ldots, p_k)} \biggr \rceil + g(x)$

in which $g(x) ≥ 0$ is a “small” quantity depending on $x$, and which is zero up to $k = 4$.

This hypothesis is already proved in relation to the last part; in fact we have seen that $g(x) = 0$ for $2 \le k \le 4$, and the same thing also applies to first order dashed lines (not just $T_1 = (2)$, but all first order linear dashed lines), as it can be easily verified.

We know that the known formula for calculating $\mathrm{t\_value}$ for a first order dashed line $T = (n_1)$ is the following:

$$\mathrm{t\_value_1}(x) = n_1 \cdot x$$

So let’s start from the formula that we are assuming to be universal, and see if, by making intermediate calculations, we’ll come to the one known for the first order. Let’s start with the general expression:

$f_k(x) = n_1 \biggl \lceil \cfrac{\sigma_{k - 1}(n_2, \ldots, n_k) x + \sigma_{k - 2}(n_2, \ldots, n_k)}{\sigma_{k - 1}(n_1, \ldots, n_k)} \biggr \rceil + g(x)$

Let’s set $g(x) = 0$, which is what we want to verify, and $k = 1$, since we are referring to the first order:

$f_1(x) = n_1 \biggl \lceil \cfrac{\sigma_{1 - 1}(n_2, \ldots, n_1) x + \sigma_{1 - 2}(n_2, \ldots, n_1)}{\sigma_{1 - 1}(n_1, \ldots, n_1)} \biggr \rceil + 0$

Let’s continue with the intermediate calculations:

$f_1(x) = n_1 \biggl \lceil \cfrac{\sigma_0() x + \sigma_{-1}()}{\sigma_0()} \biggr \rceil$

Let’s replace $\sigma_0()$ with its value, which is 1:

$f_1(x) = n_1 \biggl \lceil \cfrac{1 \cdot x + \sigma_{-1}()}{1} \biggr \rceil$

We can set $\sigma_{-1}() = 0$, since the function is not defined for index -1. This seems reasonable to distinguish it from $\sigma_0() = 1$: if on one hand $\sigma_0()$is the neutral element of the product, which therefore preserves a product, $\sigma_{-1}()$ is the neutral element of the sum, which has the property of zeroing a product. With this assumption we get:

$f_1(x) = n_1 \biggl \lceil \cfrac{1 \cdot x + 0}{1} \biggr \rceil$

that is

$f_1(x) = n_1 \lceil x \rceil$

But $x$, rounded up, is equal to $x$ itself: then the formula becomes

$$f_1(x) = n_1 \cdot x$$

which is exactly what we expected it to be.

In total, therefore, Hypothesis H.3 is valid for all linear dashed lines from the first to the fourth order.

It remains to be seen whether Hypothesis H.3 is also valid for dashed lines of a higher order than the fourth; a further field of investigation is to understand if $g(x)$ can be calculated in turn by means of a formula with algebraic and integer operations.

## Calculation for dashes of any row

The ultimate goal is to determine a general formula for $\mathrm{t\_{value}}$ that applies to the dashes of any row of an arbitrary linear dashed line. For the moment, as we have specified above, we are focusing on the first row, however trying to find a formula that applies to all orders.