Hubert Schaetzel

Pascal Trihedron Hubert Schaetzel

Collatz conjecture : Geography of the

come

Framework.

The Collatz conjecture is also called Syracuse conjecture, Ulam conjecture, Czech conjecture or 3x+1 problem. We studied this conjecture in another article [1] (writing also a simplified version [2]). The present work follows and completes the said study. The presentation of the conjecture and the results obtained at that time are not repeated here. The reader will refer to the said previous articles if necessary.

We have given at the said time an algorithm to evaluate the size of classes of identical altitude flight times (stopping times). Taking account of its structure, we called this routine "Pascal trihedron".

We contented ourselves of a mere enumeration of classes' populations (of identical altitude flight time) given in this text by theorem 15 page 77. The purpose of this article is to understand the geography of this trihedron and then catch the properties of the members of these classes which draw that geography. This study will in fact lead us towards two geographies, the second deriving from the first one.

Brief review and additional definitions.

We maintain all of our initial article notations. In particular:

-the number v of odd steps of altitude flight time (number of multiplications by 3 (plus 1), -the number w of even steps of altitude flight time (number of divisions by 2).

For an example, the number of odd and even steps in altitude flight time of 11 is given by the following There are 3 odd steps I and 5 even steps P before the first number resulting from the Collatz algorithm becomes smaller than 11.

Theorem 1

Except for finite loops, the number of w pairs and v odd steps are related by : w = int(Ln(3)/Ln (2).v)+1

(1)

Proof

For the infinite loops (if they exist), v and w are infinite and the relationship does not apply of course (to the nearest integer).

A multiplication of x by 3 and then adding 1 is equivalent to a multiplication by 3+ε, where ε is small in front of 3 when x is large. Performing the Collatz algorithm, we get y, the first number inferior to x, with x/2 < y < x (not including loops where y = x) and we have then also y = (3+ε 1 ).(3+ε 2 )…(3+ε v )/2 w .x. The ε i are all small in front of 3, since the exit of altitude flight by the Collatz algorithm is carried out for y > x/2 so for y remaining a large number (same order of magnitude as x). We have therefore (3+ε 1 ).(3+ε 2 )…(3+ε v ) = (3+ε) v with ε small in front of 3. We have then, according to previous observations, by substitution 1/2 < (3+ε) v /2 w < 1, which is also Ln(3+ε)/Ln (2).v < w < Ln(3+ε)/Ln (2).v+1, thus w = int(Ln(3+ε)/Ln(2)).v+1. This result is then tested in comparison with w = int(Ln(3)/Ln (2).v)+1 for x = 1 to x = 10 8 (for example). No exception is found to this rule for these x (in particular for pair number x, we have v = 0 and w = 1). However, as x becomes larger without exception, it is unlikely to see a sudden exception, as ε approximating 0 means that w = int(Ln(3+ε)/Ln (2).v)+1 approaches w = int(Ln(3)/Ln (2).v)+1.

Let us consider then x with w pair steps. The altitude flight cycle of x+k.2 w being the same as that of x, let us test then x k = x+k.2 w for any integer k with regard to x k /2 < y k ≤ x k , which is also Ln(3+ε k )/Ln (2).v ≤ w < Ln(3+ε k )/Ln(2).v+1. Number x k is therefore as large as wanted and hence the choice of an ε k as small as desired can be done. As Ln(3)/Ln(2) is transcendental and not in Q, Ln(3)/Ln (2).v is not an integer and there is necessarily an ε k , small enough among our tests, such as Ln(3+ε k )/Ln (2) is not in Q and Ln(3+ε k )/Ln (2).v is not an integer and close enough to the precedent one so that int(Ln(3+ε k )/Ln (2).v) = int(Ln(3)/Ln (2).v).

The result w = int(Ln(3)/Ln (2).v)+1 then applies for this case x k . Then, the result remains true for all x i = x+i.2 w , i a relative integer (i in Z) completing the proof.

Note that for x = 1 which loops on itself, the relationship still applies (because this number belongs to the family x = 1 mod 4 for which without exception v = 1, w = 2). For x small and negative, there are exceptions (by only one unit) found only for the well-known loops :

-For x = -1, we have v = 1, w = 1 and so w = int(Ln(3)/Ln (2).v) -For x = -5, we have v = 2, w = 3 and so w = int(Ln(3)/Ln (2).v) -For x = -17, we have v = 7, w = 11 and so w = int(Ln(3)/Ln (2).v)

We then have the populations of modulo 2 w families' associates (term that we adopted in the original article to refer to members of the same modulo 2 w family which all have same altitude flight time) given in the tables below :

We sum up the staffs at the last line (in red font and outside trihedron of staffs) : v = 1, w = 2, #(v)= 1, (2 w = 4) Family associates mod 4 5 (or 1)

1 1 v = 2, w = 4, #(v)=

Definition 1

We call these successive tables Pascal trihedron staffs' planes.

These planes take their complete form only when three identical columns occur at the end of the table (with a different column in front), that is for v ≥ 5 (w ≥ 8).

The number of columns of a plane is v-1.

Definition 2

We call the last line (in red) of each of them the simplified decomposition of the Pascal trihedron (of staffs) which we symbolize by #TPS(v). We have for example #TPS(7) = {1,3,5,7,7,7}.

Our first goal will be to put all of the set of associates, which we have given the list before each table above, at their proper places in the Pascal trihedron planes. Doing this, we get another set of tables that we also call Pascal trihedron planes. To distinguish them from the earlier set, we adopt the following denomination :

Definition 3

We call the new entities the Pascal trihedron associates' planes.

Here are a few examples which will ensue from our study, with underneath in red font and off trihedron, a reminder of the staffs : We identify a number a v (i,j) on the Pascal trihedron (of associates) by its plane number v and its position in line i and column j (where i = 1, j = 1 for the first row and column).

Important note.

We find effectively the simplified decomposition of the Pascal trihedron (of staffs). But, on the other hand, if we consider, for example the v = 7 plane, the fine structure of staffs in the last table, we would immediately choose a consolidation of staffs such as :

1 1 1 1 1 1 0 2 2 2 2 2 0 0 2 2 2 2 0 0 0 2 2 2 1 3 5 7 7 7
This is far from the one proposed earlier :

P 5/96 1 1 1 1 1 1 0 1 2 3 3 3 0 1 2 3 3 3 1 3 5 7 7 7
Our previous studies show that the staffs of the v plane v can be deducted additively from the fine structure of the staffs of the v-1 plane. A classification in the same mould would then probably allow finding some additive property among the associates' planes to generate more or less the v plane from the v-1plane. Yet this fine model seems set to wrong so far here. The 'geography' of the original staffing table has become totally unreadable. Nevertheless, we will see that anticipations at rank v find place from rank v-1 items.

In addition, we will highlight, in the last chapter of this article, a second geography that actually springs up the said additive property. Let us replace the odd steps I by integer 1 and even steps P by integer 0 (and les IP steps by 10). Previous IPIPPIPP becomes the binary number 10100100 that we convert to decimal 164. This one would be all the greater, while considering identical odd-numbered steps (and thus identical even-numbered steps), if the odd steps were more advanced (so IPIPIPPP will give 10101000 or 168).

Definition 5

We will call binary signature the binary number corresponding to the series IP... P... I... PP and decimal signature the result in decimal conversion.

The advantage of decimal signature lies mainly in the ease of a more concise reading.

A combination of bits 0 and 1 does not necessarily correspond to a series IP... P... I... PP generated by the algorithm of Collatz. For example, you can never have two consecutive odd steps I and thus 1 followed by 1 will never appear in a licit binary signature.

Definition 6

We call licit (or permissible, acceptable, legitimate…) signature that one that does not break any rule of the Collatz algorithm within an altitude flight routine.

Let us consider, for example, the binary signature 1000. It is not licit because the number of even steps w is greater than int(Ln(3)/Ln (2).v)+1, where v is the number of odd steps. The altitude flight time is exceeded in this writing. Similarly, writing 1000101010100 is not licit, even if we do have globally w = int(Ln(3)/Ln (2).v)+1, as the flight time was exceeded prematurely writing 1000 at the beginning of the sequence. For some given binary signature, in the same way, one has to check its validity at each new even intermediate step. For example, with signature 101010101010100100001000, the intermediate necessary checks for a licit signature are to be made at v 1 = 7, w 1 = 8, v 2 = 8, w 2 = 12 and v 3 = 9, w 3 = 15. We have w 1 = 8 < int(ln(3)/ln (2).v 1 )+1 = 12, w 2 = 12 < int(ln(3)/ln(2).v 2 )+1 = 13, w 3 = 15 = int(ln(3)/ln (2).v 3 )+1 thus corresponding effectively to a licit signature.

Theorem 2

The binary signatures of a given plane of the Pascal trihedron are all the same size.

Indeed, all the numbers of a plane v have an altitude flight time with v odd steps v and w = int(ln(3)/ln (2).v)+1 even steps and so the size of the signatures is identical and equal to v+w.

Theorem 4

Differences in binary signatures (where the two signatures exist) between columns are constants : s v (i,j)-s v (i,j-1) = cte v (j)

Proof Let us review this by an example as earlier : By construction, the facing doublets, from a column to the other, are characterised by the same permutation at the same place into the signature, the rest of it, left and right, being identical from one to the other. Their differences are thus identical (here 1000000000 = 2 9 ).

Note :

It is of course the same for decimal signatures.

Property 1

The difference between the first two columns is equal to 2 w-v and increases column after column by a factor of 4. v w 2 w-v 2 w-v+2 2 w-v+4 2 w-v+6 2 w-v+8 2 w-v+10 2 w-v+12

Proof

Let us go back first to the example of plane v = 7 (w = 12) signatures.

To go from the left to the right of this table consists simply to swap the 0 with the foregoing 1 at the position w-v-1+2c starting from the right, c being the number of the column in order to find the element of the next column. For example, for the line 5, swapping in 10101010(10)001001000 in column 3 is done at position 12-7-1+6 = 10 (and position 11 for 1) thus giving 10101010(01)001001000. Then for 101010(10)01001001000 in column 4 swapping is done at position 12-7-1+8 = 12 to give 101010(01)01001001000. To finish with, for 1010(10)0101001001000 in column 5 swapping is done in position 12-7-1+10 = 14 to give 1010(01)0101001001000. These permutations of (10) to (01) are applied in a systematic way for whatever v-plane starting always from a signature like 101010101010...100000...0, then gradually shifting along pairs (10) from those most to the right side towards those on the left side.

The decimal transcription of such writings is of course a gap of 2 w-v from column 1 to column 2 (the decimal values get lower), with a gradual increase of the said initial gap by a factor of 4 from a column to the next one.

Theorem 5

Differences in binary signatures (where signatures exist) between lines are constants : It is no more a simple permutation here in general, but by construction, identical entities that change at identical positions. The differences thus remain constant.

s v (i,j)-s v (i-1,j) = cte v (i) (3)

Note :

It is of course the same for decimal signatures.

Property 2

The difference between the first two lines is equal to 2 w-v-1 .

The reader can see this on the examples given in the table below. We have 2 8-5-1 , 8 = 2 10-6-1 , 16 = 2 12-7-1 and so on. We do not prove that property here (missing the general case anyway).

The anticipation of the values of these differences between lines would enable us to deduce the associates of any Pascal trihedron plane and is therefore of the utmost importance. We first give some examples of the decimal differences between lines observed before making an analysis, giving here the last column of the tables : This table offers prospects of anticipation. The principle is to infer the values at rank v from those at rank v-1.

v = 5 w = 8 Δw = 1 v = 6 w = 10 Δw = 2 v = 7 w = 12 Δw = 2 v = 8 w = 13 Δw = 1 v = 9 w = 15 Δw = 2 v =
To this end, we have placed in the header the values of

Δw v = Δw(v) = w(v) -w(v-1) = w v -w v-1
As w = int(Ln(3)/Ln(2).v)+1, Δw at rank v (noted as Δw(v) or Δw v ) is equal to 1 or 2. It is easy to see in first line, the differences of decimal signatures cte v (0) are multiplied by Δw(v) passing from v-1 to v. This is deducted immediately also from the ratio of the values previously announced at rank v-1 (that is 2 w(v-1)-(v-1)-1 ) and at rank v (that is 2 w(v)-v-1 ), hence a ratio of 2 Δw(v)-1 = if(Δw(v) = 1,1,2). This rule of multiplication at the first line is actually general up to a certain point that we will describe now.

The passage from rank v-1 to rank v is done by multiplications by Δw(v), by further splits and additions of new items : The example of the evolution from rank v = 9 to rank v = 10 shows the rapid complexification of the splitting for larger rankings. The ranks where Δw = 1 show a much stronger increase in complexity compared to the Δw = 2 ranks, as well for the splitting (previous table) that for new items (summary below). In this table, where we kept the order of appearance, we have grouped the values so that the sums are systematically equal in powers of 2, with the exception of the first or the last section which has to be added to the others to have the same powers of 2' property. When Δw(v-1) = 2, (hence at rank v-1,) the first line of each box and each subdivision is equal to 1. When Δw(v-1) = 1, then it is the last line of each box and subdivision which is equal to 4.

v = 5 w = 8 Δw = 1 v = 6 w =
v = 5 w = 8 Δw = 1 v = 6 w = 10 Δw = 2 v = 7 w = 12 Δw = 2 v = 8 w = 13 Δw = 1 v = 9 w = 15 Δw = 2 v = 10 w = 16 Δw = 1 v =
P 16/96

These few examples show the importance of the powers of 2 and what we will call the "history" (or background) of the process, namely here the previous ranks with the values of Δw(v) and Δw(v-1). We have compiled tables of differences cte v (i) up to rank v = 17 (the table including then 312455 associates). For a comprehensive study, the presentation of the tables in their entirety is necessary. For a complete locating, we will note the elements of the table at rank v by ddsl v (i,j), with line index i and column index j (i = 1 and j = 1 for the first element at the upper left).

Here are the first tables (tables of origin) after divisions by 4.Δw(v) :

v = 5 1 1 1 v = 6 1 1 1 1 4 4 4 v = 7 2 2 2 2 2 1 1 1 1 1 7 7 7 7 1 1 1 1 31 31 31 1 1 1 v =
Σ i #ddsl v (i,j) = Σ i #a v (i,j) -1 (4)
The staffs of the table of associates are obtained by referring to the study made in our previous articles [1] and [2]. We thus have for Σ i #a v (i,j) and Σ i #ddsl v (i,j) respectively (first and second lines) : This rule allows studying only the last column and then coming back on the other elements at the end of the study. The second rule is so committed to this last column.

v = 5 1 2 2 2 0 1 1 1 P 17/96 v = 6 1 2 3 3 3 0 1 2 2 2 v =
Rule 2. Rule of the staffs of first generation.

The rule is to take over the second line of the above tables and perform the differences column to column. Due to the algorithm that gives the staffs of the Pascal trihedron, we have systematically three identical columns followed two voids in each of the seconds preceding lines (for v > 7).

v = 5 1 1 1 1 / / v = 6 1 2 2 2 1 1 / / v =
This rule allows to draw up tables of the last column following the said successive staffs and this process step can be done backwards. These tables are called as "first-generation": We can also call this rule, the rule of maximum element (of first generation) location, because by construction of the signatures, the difference of the signatures will be here always superior to all those preceding. It is actually the situation that would lead the reader, in absence of knowledge of the algorithm, to change column.

P 18/96 v = 5 1 v = 6 1 4 v = 7 2 7 31 1 1 1 v =
From how signatures are settled, all lines of the table that are thus obtained are identical items, except the voids at the top of lines and except the first line.

Rule 3. Rule of column sums.

Let us note by sddsl v (j) = Σ i ddsl v (i,j) the sum of column j. There are 5 final models in the maxima's location process for v < 18.

It turns out that the history must be analysed at least up to rank v-4 to tell about the origin of these models. Below, we give these models starting with the "history" which gives their rise :

Lists of v 8 13 (20) (25)

Property 3

The sum of the last column of a complete model is a power of 2.

The sum of all the elements of a complete model still is a power of 2.

In addition, the sum of all the elements and the sum of the last column of each generation are powers of 2.

Rule 6. Rule of the staffs of i-th generations.

The cuts on maximum elements correspond to the continuation of rule 2 by subtraction of staffs column to column.

For The construction of the model requires the last two generations up to v = 17. The last element of the latest generation (last line non-empty element on the right side) is equal 1 or 4 in consistency with the model at work. It will thus locate at first column for 1 to 3 models and in second for models 4 and 5.

Rule 9. Rule of extension of the model.

A model gives birth of the common extensions.

We will return to this extension notion later on with more details. For now, let give a few examples by adding at the end of tables the relevant ranks and their history Δw(v), Δw(v-1), Δw(v-2), Δw(v-3), …, Δw(6).

This history is stopped at the level of Δw(6) because this rank is effectively, after checking it, the rank suited to the proposed method.

v = 17 1 2 1 2 1 2 2 1 2 1 2 2 v = 12 2 2 1 2 1 2 2 v = 7 2 2 v = 5 1
The 1 of rank 5 in last line is crossed as its history is already out of fields. However, this value is recalled here for the perfect understanding of what will follow later on.

Extension of

v = 16 2 1 2 1 2 2 1 2 1 2 2 v = 11 2 1 2 1 2 2 v = 6 2
The reader will notice without any effort that any column with the same number at the end of chain is the identical over all its height regardless of the model.

Rule 10. Rule of the notches.

The histories settle the notches of 1 or 2 units up models.

The histories consist, as we already know, of isolated 1, isolated 2 or 2 in pair.

We have then :

Isolated 1 on column j Shift a notch on column j Isolated 2 on column j

No lag, no notch Pair of 2 on column j and column j + 1 Shift a notch for the pair on column j and j + 1 column Rule 11. Rule of linearity (affine rule).

The element r m(k) (i,j) of the review table in position (i, j) deduces from that in position (i+1, j) by :

r m(k) (i,j) = 2.r m(k) (i+1,j)-cte m(k) (j) (6)
where cte m(k) (j) are constants for a given model and column.

For each model, we give below the values of cte m(k) (j).

cte m(k) (1) cte m(k) (2) cte m(k) (3) cte m(k) (4) cte m(k) (5) cte m(k) (6) cte m(k) (7) By then the rapprochement with the history Δw(v-k) of models, we observe a common behaviour that makes us rewrite cte m(k) (j) as a form which is not dependent on a model but simply of v:

P 24/96 cte v (1) = 0 (7) cte v (2) = 1 (8) cte v (3) = 1+2 Δw(v-1)+1 (9) … cte v (j) = 1+2 Δw(v-1)+1 +2 Δw(v-2)+1 +…+2 Δw(v-j+2)+1
(10)

The element r m(k) (i,j) of the review table is thus freed in the recursive expression of the model :

r v (i,j) = 2.r v (i+1,j)-cte v (j) (11)
We have, to finish with, to find the column's endings values to control all aspects of the routine.

To do this, we will first write down two tables that we juxtapose :

m 2 m-1 2 m +1 (2 m +1)/2 m-1 2 m+2 2 m s1(m) s2(m) st(m) ∞ ∞ ∞ 2 ∞ ∞ ∞ ∞ 4 … … … … … … … … …
… … … … … … … … -∞ 0 1 ∞ 0 0 8 8 Let us have : sb(m) = (2 m +1)/2 m-1 (12) Then : s1(1) = 24 (13) s1(m) = s1(m-1)+2 m+2 (14) s2(1) = 5 (15) s2(m) = s2(m-1)+2 m (16) st(m) = s1(m)/s2(m) (17)
Note: The part m < 0 of the tables does not serve subsequently and is only here to show the harmony of the lower and upper boundaries : 8 = 4 * 2.

We then return to the models' extensions and do the differences column to column :

Differences column to column of model extension

v = 17 1 2 1 2 1 2 2 1 2 1 2 2 v = 12 2 2 1 2 1 2 2 v = 7 2 2 v = 5 1

Differences column to column of model extension

v = 16 1 2 1 2 2 1 2 1 2 2 v = 11 1 2 1 2 2 1 2 1 2 1 v = 6 1 2 1 2 1
Finally, let us do the ratio of each element in a column to the previous column (same line) :

Ratio of differences column to column of model extension

v = 17 1 2 1 2 1 2 2 1 2 2 2 v = 12 2 2 1 2 1 2 2 v = 7 2 2 v =
v = 16 2 1 2 1 2 2 1 2 1 2 2 v = 11 2 1 2 1 2 2 v = 6 2
Now, we can give the construction of the endings of column by proper links.

Rule 12. Rule of column endings.

Let us locate by dr m(k) (i,j) an element of a table of differences column to column of the extension of a given model and by dr m(k) (i_end,j), the last non-empty element of a column and by dr m(k) (i_end+i,j) on the same column the elements above offset by i-lines i. Then, we have :

P 27/96 Δw(v+2-j) = 1 isolated on column j dr m(k) (i_end+i, 1) = 2 i dr m(k) (i_end+i, 2) = sb(i+1)*dr m(k) (i_end+i, 1) = 2 i+1 +1 dr m(k) (i_end+i, j > 2) = 4*dr m(k) (i_end+i, j-1) Δw(v+2-j) = 2 isolated on column j dr m(k) (i_end+i, 1) = 2 i dr m(k) (i_end+i, 2) = sb(i+2)*dr m(k) (i_end+i+1, 1) = 2 i+2 +1 dr m(k) (i_end+i, j > 2) = st(i+1)*dr m(k) (i_end+i, j-1)
Δw(v+2-j) = 2 and Δw(v+2-j-1) = 2 in pair on column j and column j+1

dr m(k) (i_end+i, 1) = 2 i dr m(k) (i_end+i, 2) = sb(i+2)*dr m(k) (i_end+i+1, 1) = 2 i+2 +1 dr m(k) (i_end+i, j > 2) = st(i+2)*dr m(k) (i_end+i, j-1) dr m(k) (i_end+i, j+1 > 3) = st(i+1)*dr m(k) (i_end+i, j)
Note 1 : Δw(v) = 2 at the head of the list is considered to be isolated. Note 2 : When the model (it comes to model 3), displays the possibility at the head list to have either (Δw(v), Δw(v-1)) = (1,2) or (Δw(v), Δw(v-1)) = (1,2), then everything happens as if the first choice wins (as a result of the initial v = 5 certainly).

The endings of column's rule in fact give the evaluation of the whole table dr m(k) (i,j) and makes obsolete the use of rule 11.

We left it as it is interesting matter anyway.

Let us note also that the indexation of dr m(k) (i,j) is not essential which we express in the last rule.

Rule 13. Rule of limitation of model.

A model is unique and applicable only up to a certain rank v.

Due to the history Δw associated to w = int(ln(3)/ln(2).v)+1, as rank v increases, the events Δw = 1 or Δw = 2 arrive "randomly" and lead necessarily towards bifurcations featuring an infinity of more complex models. The models presented are however useful for priming of the routine.

All of the above rules allow going backwards up to the differences between the lines of the signatures for any v and thus to build a table of signatures. The table of signatures at rank v-1 and those in the previous ranks are not useful. The only necessity is the history Δw(v) to Δw(6).

Construction of the plane v of associates based on the signatures

Let us proceed by steps.

Property 4

An associate a, in column c (c = 1 identifying the first column) is of the form t.2 r -1, with r = v+1-c, r the number of systematic (IP) steps ahead of algorithm and t an odd integer. But, there is no integer t' such as a = t'.2 r+1 -1.

Proof

Indeed, this number a is odd and the first result of Collatz algorithm (a 3a+1 operation) is an even number 3.t.2 r -2 that can be divided by two providing 3.t.2 r-1 -1, thus returning to the original form. This number is necessarily odd as long as the exponent of 2 is strictly positive and must therefore undergo a step (IP) after that. We have to conduct r-1 other identical iterations thus leading to 3 r .t.2 0 -1 = 3 r .t-1. This time, this number is necessarily even since t is odd, showing indeed that we cannot have another r+1 th step (IP). Moreover, the formula r = v+1-c follows trivially from property 1.

Let us now address a fundamental point.

Theorem 6

For any licit signature, there is a number that generates this signature.

Proof

We start with an example for a good understanding of the algorithm that is involved. So let us look for the number m that generates the licit signature 101010101010100100001000. The Collatz algorithm imposes then the following three pairs of equations :

m = r 1 .2 v1 -1 r 1 .3 v1 -1 = m 1 .2 w1-v1 m 1 = r 2 .2 v2-v1 -1 P 28/96 r 2 .3 v2-v1 -1 = m 2 .2 w2-w1-(v2-v1) m 2 = r 3 .2 v3-v2 -1 r 3 .3 v3-v2 -1 = m 3 .2 w3-w2-(v3-v2)
Here the couples (v 1 = 7, w 1 = 8), (v 2 = 8, w 2 = 12), (v 3 = 9, w 3 = 15) summarize the signature and allow us to write the operations that take place according to the Collatz algorithm.

For more complex signatures, we will just have to add as many pairs of equations, that will be needed, based on the previous model.

The resolution of this system is then done by addressing in reverse the last equation up to the first one. It alternates diophantine equations of type ax+by = c and simple multiplication giving intermediate values m i before getting the final m. We have (a,b) = (2 i ,3 j ), so a and b are coprime, and the couple (x,y) = (cx 0 ,cy 0 ) is solution of equation ax 0 +by 0 = 1. By Bézout-Bachet theorem, we are sure of the existence of a solution (s 1 ,s 2 ) for this last equation that we complete with the arithmetic series (s 1 +k.y 0 ,s 2 -k.x 0 ). So, for our example : r 3 .3 9-8 -1 = m 3 .2 15-12-(9-8) , so that 3r 3 -1 = 4m 3 , so that r 3 = 3+4k 1 and m 3 =2+3k 1 m 2 = r 3 .2 9-8 -1, so that m 2 = 5+8k 1 r 2 .3 8-7 -1 = m 2 .2 12-8-(8-7) , so that 3r 2 -64k 1 = 41, so that r 2 = 35+64k 2 and k 1 = 1+3k 2 m 1 = r 2 .2 8-7 -1, so that m 1 = 69+128k 2 r 1 .3 7 -1 = m 1 .2 8-7 , so that 2187r 1 -256k 2 = 139, so that r 1 =1+256k 3 and k 2 = 8+2187k 3 m = r 1 .2 [START_REF]Each row has constant values (cf. theorem 9). We divide then the value by the factors set out in column "divisors" to get[END_REF] -1, so that m = 127+32768k 3

Thus we get finally a set of solution m = 127+2 15 .k = 127+2 w .k, with 127 the associate solution ranging between 0 and 2 w , where k is any integer (negative numbers are solutions also by adapting the flight altitude concept taking absolute values of numbers).

The pairs of equations' system, which we have written above, apply to any signature (including signatures which are not licit, knowing however that there is no need to resolve such systems). An intermediate pair of equations is written whenever appears more than one consecutive 0 in the signature. Each Bézout-Bachet equation leads in its following multiplicative equation to an intermediate m i of the form c i +2 wi .k j and the whole resolution results in a final m of the form c f +2 w .k where w is the number of 0 in the signature and k any relative integer (in Z), which completes our proof.

In the same time, we have proved also :

Theorem 7

Any odd integer of the arithmetic progression m modulo 2 w has same signature as m.

Definition 7

We call positive sorting the implementation of the classification of a plane v with such signatures as described above.

We can also write immediately (without developing more arguments).

Theorem 8

There is a positive sorting for each Pascal trihedron plane.

Pascal trihedron classification by positive sorting.

For plane v = 7, using positive sorting, we get the following associates : The differences from a line to another are as follow : The differences from a column to another are constants modulo 2 w . In the same way, the differences from a line to another : then table 2 evolves from either starting at the third line (for k = 1 for example), or at the fifth line (for k = 6, for example), ..., or at the second last line (for k = 2048 here), that is depending on the choice of k. In particular, varying k from 0 to 4095, we observe that the last line takes all values from 0 to 4095. Specifically, we have writing df the value of the last line (3 v = 2187) :

383
a v (i,j)-a v (i-1,j) = ct(i) mod 2 w (18) a v (i,j)-a v (i,j-1) = ct(j)
df = 1777 + k.3 v mod 2 w (20)
In the case under consideration here, as an example, table 2 will show respectively for k = 1293 (where the last line is identical to the first) and k = 3597 (where the last line is equal to 0) as follows :

df = i.3 v mod 2 w ( 21
)
where i is a relative integer (i = k-3597). We can give, still as an example, all expressions of the numbers written in table 3b :

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
As the initial values are obtained by multiplication by 3 (plus 1) and divisions by 2, the differences of these values are multiplications by 3 and divisions by 2 (sometimes corrected by 2 w before dividing by 2 in operating modulo 2 w ) and this as long as the signatures are in correspondence. This is not the case here for the first line in a red box, where the rule is different resulting in another type of evaluation (5+8t assessment is explained later).

P 32/96

Besides, we still can give, as an example, the values to add to the reference table values to pass from table i = 0 to i = 1, then for table i = 1 to i = 2... and then from table i to i + 1 : We observe this time multiplications by 3 or divisions by 2 (corrected sometimes by a 2 w addition before division) starting from the value x 0 = 0.

Let us now go back to the fundamentals. All what has been seen so far depends on what follows. The Collatz algorithm is based on equations like those :

m = r 1 .2 v1 -1 r 1 .3 v1 -1 = m 1 .2 w1-v1 m 1 = r 2 .2 v2-v1 -1 r 2 .3 v2-v1 -1 = m 2 .2 w2-w1-(v2-v1) m 2 = r 3 .2 v3-v2 -1 r 3 .3 v3-v2 -1 = m 3 .2 w3-w2-(v3-v2) … m i-1 = r i .2 v[i]-v[i-1] -1 r i .3 v[i]-v[i-1] -1 = m i .2 w[i]-w[i-1]-(v[i]-v[i-1]) … m j-1 = r j .2 v[j]-v[j-1] -1 r j .3 v[j]-v[j-1] -1 = m j .2 w[j]-w[j-1]-(v[j]-v[j-1]) (22)
The resolution is done starting at the bottom and working up the equations.

In the table below, we visualize changes in correspondence of signatures (in brackets) and differences in the red sidebars, recalling also the previous equations:

960 960+1024k [1] 238+1024k [1] 174+1024k 960 960+1024k [1] 86+1024k [1] 22+1024k 960 960+1024k [1] 882+1024k [1] 818+1024k 960 960+1024k [1] 540+1024k [1] 476+1024k 2208 160+1024k [0] m1 = 431+512k r2 = 27+32k [0] m1 = 79+512k r2 = 5+32k 2528 480+512k [0] 119+512k [0] 87+512k 2528 480+512k [0] 43+512k [0] 11+512k 2528 480+512k [0] 441+512k [0] 409+512k 2528 480+512k [0] 270+512k [0] 238+512k 2528 480+512k [1] 270+512k [1] 238+512k 3488 416+512k [1] 358+512k [1] 262+512k 3488 416+512k [1] 130+512k [1] 34+512k 3488 416+512k [1] 300+512k [1] 204+512k 1264 240+512k [0] m1 = 135+256k r2 = 17+32k [0] m1 = 119+256k r2 = 15+32k 1264 240+512k [0] 135+256k [0] 119+256k 3792 208+256k [0] 179+256k [0] 131+256k 3792 208+256k [0] 65+256k [0] 17+256k 3792 208+256k [0] 150+256k [0] 102+256k 3792 208+256k [1] 150+256k [1] 102+256k 3792 208+256k [1] 150+256k [1] 102+256k 3184 112+256k [1] 26+256k [1] 138+256k 3184 112+256k [1] 196+256k [1] 52+256k 3944 104+256k [0] m1 = 75+128k r2 = 19+32k [0] m1 = 51+128k r2=13+32k 3944 104+256k [0] 75+128k [0] 51+128k 3944 104+256k [0] 75+128k [0] 51+128k 3640 56+128k [0] 13+128k [0] 69+128k 3640 56+128k [0] 98+128k [0] 26+128k 3640 56+128k [1] 98+128k [1] 26+128k 3640 56+128k [1] 98+128k [1] 26+128k 3640 56+128k [1] 98+128k [1] 26+128k 2728 40+128k [1] 40+128k [1]
m2 = 5+16k r3 = 3+8k [0] 10+16k 2293 5+8k [1] 16k [0] m1 = 5+8k r2 = 3+4k 1149 5+8k [1] 16k [0] m2 = 5+8k r3 = 3+4k 3773 5+8k [1] 16k [0] m2 = 5+8k r3 = 3+4k 1077 5+8k [1] 16k [0] m2 = 5+8k r3 = 3+4k 3909 5+8k [1] 16k [0] m2 = 5+8k r3 = 3+4k 0 8k [0] 8k [1] 8k 0 8k [0] 8k [1] 8k 0 8k [0] 8k [1] 8k 0 8k [0] 8k [1] 8k 0 8k [0] 8k [1] 8k 0 4k [0] 4k [0] 4k 0 4k [0] 4k [0] 4k 0 4k [0] 4k [0] 4k 0 4k [0] 4k [0] 4k 0 4k [0] 4k [0] 4k 0 2k [0] 2k [0] 2k 0 2k [0] 2k [0] 2k 0 2k [0] 2k [0] 2k 0 2k [0] 2k [0] 2k 0 2k [0] 2k [0] 2k 0 k [0] k [0] k 0 k [0] k [0] k 0 k [0] k [0] k 0 k [0] k [0] k 0 k [0] k [0] k
In the first line of the previous table, we mention on the one hand the values v i and w i to be taken into account in the system of equations ( 8) and on the other hand the system of equations which thus results. The previous table is built, after writing the fundamental equations in their places, and then adding all the intermediate expressions that can be evaluated. We choose to this purpose the simplest possible terms. For example, if we have 5+8k = 5 mod 8 followed by signature [1] (reading from top to bottom), we perform the operation 3(5)+1 = 16 = 0 mod 8, or 8k (instead of 3(5+8k)+1 = 16+24 k = ..., -32, -8, 16, 40, ... list of numbers that is effectively included in ..., -32, -24, -16, -8, 0, 8, 16, 24, 32, 40, ...). For the divisions by 2, things are here simpler and 2x+2k.y systematically gives x+k.y.

We see in particular that the last line shows an arbitrary k : This means that there is effectively a choice for which the last line can display differences systematically equal to 0. Now, let us see the proof of the theorem (9).

Proof

It comes up to generalize the numerical observations made previously.

To switch from one signature to another, for a given v-plane, means resorting to a certain number of permutations of 10 and 01 or vice versa 10 to 01. (The red boxes in table 6 show that effectively).

Then we have to consider the effect of a first case of shifting, according to the following type, for the numbers on which the signatures act :

1 1 0 0 0 0 1 1
The operation is made generally up line (but down line gives exactly the same conclusion) :

(2x 1 -1)/3 (2x 2 -1)/3 2(y 1 -1)/3 2(y 2 -1)/3 1 2x 1 1 2x 2 0 (y 1 -1)/3 0 (y 2 -1)/3 0 x 1 0 x 2 1 y 1 1 y 2
Then, if the differences are identical x 2 -x 1 = r = y 2 -y 1 before permutation, we will find again identical differences 2(x 2 -1)/3-2(x 1 -1)/3 = 2r/3 = 2(y 2 -1)/3-2(y 1 -1)/3 after the swapping.

The second case that appeared above is the following :

1 0 1 0 0 1 0 1
The operation is performed here again up columns :

P 34/96 (2x 1 -1)/3 2(x 2 -1)/3 (2y 1 -1)/3 2(y 2 -1)/3 1 2x 1 0 (x 2 -1)/3 1 2y 1 0 (y 2 -1)/3 0 x 1 1 x 2 0 y 1 1 y 2
Then, if the differences are identical x 2 -x 1 = r = y 2 -y 1 before permutation, we find the same differences again 2(x 2 -1)/3-(2x 1 -1)/3 = (2r-1)/3 = 2(y 2 -1)/3-(2y 1 -1)/3 when swapping is completed.

The composition of these two cases provides a multitude of configurations that infer all identical differences after the area of permutations.

As for the establishment of a 0 difference at the end of the algorithm, it is just necessary to notice that starting from x+k.2 w , the application of v multiplications by 3 (plus 1) and w divisions by 2 leads to x'+k3 v , x being transformed into x' and k.2 w into k.3 v . Yet x'+k3 v mod 2 w , x' being fixed (as x is fixed), gives all the numbers from 0 to 2 w -1 (in a specific order), when k described 2 w -1. Indeed 3 v and 2 w are relative primes (for v > 0) and Bézout theorem allows asserting it. Hence it suffises to chosen a proper k for one of a pair of associates to obtain a null difference or any other given number modulo 2 w .

Consider then two pairs of signatures (although not the most general case) :

Line v+w [1] [1] [1] [1] [0] [0] [0] [0] … … … … Line j [0] [0] [1] [1] … … … … Line i [1] [1] [0] [0] … … … … Line q [1] [0] [1] [0] … … … … Line p [0] [1] [0] [1] … … … … Line 1 [0] [0] [0] [0]
Here, the unfilled lines have four identical elements (either all [0] or all [1]). By a suitable choice of the parameter k, the differences in the last line of each of the two pairs of signatures may be reduced to the same value (for example 0). When the process is then run up columns, line p creates not-identical differences that will be compensated after the q line and in the analogous way, line i will see compensation at line j. Finally, at line v+w, we will always have differences of equal values.

Therefore, all of the properties observed on the numerical application apply in general way. From a basic operation to another basic operation (permutation), differences turn out the same allowing concluding.

Note :

For a given plane, to a given modulo 2 w offset does correspond a priori a unique (binary or decimal) signatures difference, but the converse is false.

The following examples show exceptions to unity as early as the first decade of planes : We did not give completeness of the differences between lines (i.e. all of the redundancies) as in the previous table. We see exceptions to the bijection between values for the v = 9 and the v = 11 planes.

Difference between columns Difference between lines

Property 5

A ratio 1/2 for the differences in decimal signatures, resulting from a shift of the final 1, the rest of the signatures being identical, results in a ratio 2 for the differences between associates modulo 2 w .

Such lag being endemic in the tables of signatures, this ratio 2 (of the differences between associates modulo 2 w ) will regularly come out in our study. We give a simple example here without proof although it would be useful, example that we take in the v = 15 plane. The proof necessitates, a priori, the resolution of the following three systems (going up equations), resolution that we leave to the ambitious reader :

m = r 1 .2 v1 -1 r 1 .3 v1 -1 = m 1 .2 w1-v1 m 1 = r 2 .2 v2-v1 -1 r 2 .3 v2-v1 -1 = m 2 .2 w2-w1-(v2-v1) m 2 = r 3 .2 v3-v2 -1 r 3 .3 v3-v2 -1 = m 3 .2 w3-w2-(v3-v2) … m i-1 = r i .2 v[i]-v[i-1] -1 r i .3 v[i]-v[i-1] -1 = m i .2 w[i]-w[i-1]-(v[i]-v[i-1]) … m j-2 = r j .2 v[j-1]-v[j-2] -1 r j .3 v[j-1]-v[j-2] -1 = m j .2 w[j-1]-w[j-2]-(v[j-1]-v[j-2]) m j-1 = r j .2 v[j]-v[j-1] -1 r j .3 v[j]-v[j-1] -1 = m j .2 w[j]-w[j-1]-(v[j]-v[j-1]) m' = r 1 .2 v1 -1 r 1 .3 v1 -1 = m 1 .2 w1-v1 m 1 = r 2 .2 v2-v1 -1 r 2 .3 v2-v1 -1 = m 2 .2 w2-w1-(v2-v1) m 2 = r 3 .2 v3-v2 -1 r 3 .3 v3-v2 -1 = m 3 .2 w3-w2-(v3-v2) … m i-1 = r i .2 v[i]-v[i-1] -1 r i .3 v[i]-v[i-1] -1 = m i .2 w[i]-w[i-1]-(v[i]-v[i-1]) … m j-2 = r j .2 v[j-1]-v[j-2] -1 r j .3 v[j-1]-v[j-2] -1 = m j .2 w[j-1]-w[j-2]+1-(v[j-1]-v[j-2]) m j-1 = r j .2 v[j]-v[j-1] -1 r j .3 v[j]-v[j-1] -1 = m j .2 w[j]-w[j-1]-1-(v[j]-v[j-1]) m'' = r 1 .2 v1 -1 r 1 .3 v1 -1 = m 1 .2 w1-v1 m 1 = r 2 .2 v2-v1 -1 r 2 .3 v2-v1 -1 = m 2 .2 w2-w1-(v2-v1) m 2 = r 3 .2 v3-v2 -1 r 3 .3 v3-v2 -1 = m 3 .2 w3-w2-(v3-v2) … m i-1 = r i .2 v[i]-v[i-1] -1 r i .3 v[i]-v[i-1] -1 = m i .2 w[i]-w[i-1]-(v[i]-v[i-1]) … m j-2 = r j .2 v[j-1]-v[j-2] -1 r j .3 v[j-1]-v[j-2] -1 = m j .2 w[j-1]-w[j-2]+2-(v[j-1]-v[j-2]) m j-1 = r j .2 v[j]-v[j-1] -1 r j .3 v[j]-v[j-1] -1 = m j .2 w[j]-w[j-1]-2-(v[j]-v[j-1])
Now, let us say that positive sorting of a large staff's plane can be a long task. Thus, it may be interesting to find additional properties relating associates, hence what follows.

P 37/96 6 Intra-column properties.

Theorem 10

For v > 1, the number a v (1,1) in the first column of a Pascal trihedron associates' plane is the unique number of this plane, which has a cycle of length 2 w-v (modulo 2 w ), that is such as (a v (1,1)) 2^(w-v) = 1 mod 2 w (23) and (a v (1,1)) i ≠ 1 mod 2 w if 0 < i < 2 w-v (24)

For example in the preceding tables : 3 2^(4-2) = 1 mod 2 4 , 23 2^(5-3) = 1 mod 2 5 , 15 2^(7-4) = 1 mod 2 [START_REF]Each row has constant values (cf. theorem 9). We divide then the value by the factors set out in column "divisors" to get[END_REF] , 95 2^(8-5) = 1 mod 2 8 , 575 2^(10-6) = 1 mod 2 10 , 383 2^(12-7) = 1 mod 2 12 being the first occurrence of 1 mod 2 w when proceeding to successive iterations of m. We have also, for v = 1, 5 2^(2-1) = 1 mod 2 2 , but the length of the cycle is shorter than 2 (since 5 1 = 1 mod 2 2 ).

Proof

The first associate is the unique number of its ranked v family with multiplication steps executed prior to division steps in the Collatz algorithm : A multiplication step (I) is systematically followed by a phase of division (P) which is concatenated as an IP-step. The process is illustrated below and it is clear that there is only a single number that verifies this routine for a given v (since w is related to v in a unique way).

a v (l,1) = pr (3pr+1)/2 … … … … … res < pr IP … IP P … P P
From property 4 , the said number, at rank v, is necessarily of the form a v (1,1) = t.2 v -1, this corresponding to systematic IPtype sequences. Let us notice also that we have necessarily a v (1,1) ≠ t'.2 v+1 -1, which is ending the previous routine. The binomial development of (a v (1,1)) 2^(w-v) = (-1+t.2 v ) 2^(w-v) = 1+2 w-v .(t.2 v +x 1 .2 2v +…x i .2 i.v ) = 1+k.2 w is then equivalent to the first part of what we want to prove. As, per elsewhere, a v (1,1) ≠ t'.2 v+1 -1, there is no equality a v (1,1)) 2^(w-v) = 1+k.2 w-c with c > 0.

Note: All of the numbers (a v (1,1)+or(0,2))+k.2 v+1 , k an integer, have exactly same cycle length (modulo 2 w ) than a v (1,1) because of the form t.2 v -1. There are exactly 2 w-v such numbers in the range 1 to 2 w -1 (t = 1, 3,…, 2 w-v+1 -1) and they are the only ones to have this property. Among these numbers, a v (1,1) is the only one belonging to the Pascal trihedron plane v (as proved above). None of the numbers a v (1,1)+k. The cycles' length of the numbers a v (1,c) is doubled passing from one column to the next, that is (a v (1,c)) 2^(w-v+c-1) = 1 mod 2 w (25) and (a v (1,c)) i ≠ 1 mod 2 w if 0 < i < 2 w-v+c-1 (26) where c is the c th column.

We have thus as examples 11 2^(5-3+2-1) = 1 mod 2 5 and 7 2^(7-4+2-1) = 1 mod 2 [START_REF]Each row has constant values (cf. theorem 9). We divide then the value by the factors set out in column "divisors" to get[END_REF] .

Proof

We only have to repeat word for word the demonstration used previously and replace a v (1,1) by a v (1,c), t.2 v -1 by t.2 v+c -1 and t.2 v+1 -1 by t.2 v+c+1 -1.

The lengths of the cycles are illustrated below at the last line of each table : meaning i and v are always of the same order of magnitude.

v = 2, w = 4, #(v)= 1 Family mod 16, 3 4 v = 3, w = 5, #(v)=

Theorem 12

Associates a r and a s of a same column are linked by a relationship of the following type (i an odd integer) :

a s = a r i mod 2 w (31)

Proof

Let us consider column c. According to property 4, a r is of the form t.2 s -1 with t an odd integer. Integer s is the exact number of sequences of IP-type at the beginning of the Collatz algorithm for the integer a r , the last sequence being followed by at least one extra P sequence. Let us take the term to an odd exponent i, we get then by binomial development (t.2 s -1) i = x.t.2 s -1 where x is an odd number also. When i is incremented (= 0, 1, 2,…), the term (t.2 s -1) i mod 2 w takes (for any odd integer t) exactly 2 w-s different values. When i is now incremented by steps of 2 units (= 1, 3, 5, ...), the expression takes exactly half of these values, 2 w-s-1 different values. The odd integer t being fixed, the term x.t.2 s -1 mod 2 w takes then exactly 2 w-s /2 different values when x is incremented according to odd values, that is the same as x.2 s -1 mod 2 w to within a permutation. This are therefore the same numbers as those taken by (t.2 s -1) i mod 2 w . Thus (t.2 s -1) i mod 2 w describes all the integers set that can be taken by an expression such as x.t.2 s -1 mod 2 w . So a s is necessarily of the form a r i mod 2 w for some odd integer i.

Note: For even values of i and x, there is a difference of 2 between the terms of the set (t.2 s -1) i mod 2 w and those of the set x.2 s -1 mod 2 w classified in increasing values.

The reader can check by samples this relationship thanks to the values given below where the power i is given (the reference for each column being the first number of the previous subfamilies and recalling also here that the first non-trivial table starts at v = 5) : v = 5, w = 8, #(v)=7

Theorem 13

There is no iterative relationship between associates a r and a s of separate columns : P 40/96 a s ≠ a r i mod 2 w (32) Proof Otherwise, we would have a s = a r i mod 2 w with a r of the form t.2 u -1 for some odd integer t, a s of the form t'.2 u' -1 for some odd integer t' (u and u' permissible maximum values) and u ≠ u'. The role of a s and a r is interchangeable, namely that if a s = a r i mod 2 w , there are also some j such as a r = a s j mod 2 w . This also means that we can assume u' > u. By binomial development, we get (t.2 u -1) i = x.t.2 u -1 = t'.2 u' -1 mod 2 w when i is odd (and x will also be), which is impossible, because x.t.2 u-min(u,u') = t'.2 u'-min(u,u') mod 2 w-min(u,u') has a parity contradiction and we have (t.2 u -1) i = x.t.2 u+1 +1 = t'.2 u' -1 mod 2 w when i is even (and x will still be odd), that is also x.t.2 u +1 = t'.2 u'-1 mod 2 w-1 . To satisfy the parity, we have then u'-1 = 0, and as u' > u, we conclude u = 0, that means a r is even, number that does not interest us (all numbers in the Pascal trihedron are odd).

For example, for v = 7 (w = 12),

383

Intra-line properties.

There are no property specific to a single line based on the previous method as we shall see. The initial difficulty is, as we have seen, that there is no iterative link such as a s = a r i mod 2 w between two associates a r and a s of separate columns. Nevertheless, we can write links such as a s j = a r i mod 2 w , but these have a little pronounced differentiation character, only allowing the grouping of associates of identical staffs' lines.

Property 6

The strength of the relationship between two associates is expressed by : (a v (i,j+1)) 2^(w-k) = (a v (i,j)) 2^(w-k) mod 2 w (33

)
where k is the largest possible.

This is actually more a technique of classification than a property to be proven. For example, let us take the last column of the Pascal trihedron associates' plane for v = 7 (w = 12, modulo 4096) and let us focus on the two last columns. The differentiation is done this time thanks to the columns of 16, 8 and 4 iterations.

However, this method only links associates belonging to identical staff lines. We wrote, for example, pairs 2203/231 and 1435/3559, but 2203/3559 and 1435/231 matchings would give a table with the same characteristics when these associates are not on the same lines of the plane v = 7.

To go further, we adopt a different strategy.

Values of the first element of the Pascal trihedron planes.

We seek to deduce the value of a v+1 (1,1) from a v (1,1). To do this, we compare the current value to 6 times the precedent one plus 5 modulo 2 w and check the differences : The first associate at rank v is derived from the first associate at rank v-1 by multiplying by 6 and 5 with an additive correction td.2 v+3 modulo 2 w , td a relative integer.

v w ∆w a v (
We will come back to the values of td later on.

P 42/96

Theorem 14

Let us have a v (1,1) the first element of the Pascal trihedron v-plane. Let us have a 1 (1,1) = 1 (for v = 1, w = 2 and thus 5 > 2 w , and a 1 (1,1) = 5 is unsuitable to the present case where instead 1 is).

Let us have m(v) defied by the recursive relation : m(v+1) = (m(v)+r.3 v )/2 Δw(v+1)-1 (34) starting with m(1) = 1. Then : a v+1 (1,1) = (2.(a v (1,1)+r.2 w )-1)/3 (35)

for the smallest r such that m(v+1) and a v+1 (1,1) are simultaneous integers.

The integer m(v) is precisely the result of the Collatz algorithm on a v (1,1) when the altitude flight comes to an end (except for 1 which forms a loop with a result equal to the initial value). Moreover, the solution is necessarily reached for r in the set {0,1,2,3,4,5}.

For memory w = w(v) = int((Ln(3)/Ln( 2)).v)+1 and thus Δw(v+1) = w(v+1)-w(v) = int((Ln(3)/Ln(2)).( v+1)) -int((Ln(3)/Ln( 2)).v)

Proof

We focus first on the parameter t. Let us recall (property 4), that a v (1,1), ranked v, is necessarily of the form :

a v (1,1) = t.2 v -1 ( 36
)
where t is a positive integer.

It is the only acceptable form, otherwise the intermediate results would provide a premature even step. Then to finish Collatz algorithm with the objective to have only even steps, it is necessary and suffices to impose

3 v .t-1 = m.2 w-v (37)
where w-v is the remaining number of steps (which are even). The Bézout theorem assures us of the existence of a solution (t,m) of integers (here positive) to 3 v .t-m.2 w-v = 1, as 2 and 3 are relative primes, from which we derive then :

t = (1+m.2 w-v )/(3 v ) (38) Hence : a v (1,1) = (1+m.2 w-v ).(2/3) v -1 (39)
In practice, the unique number a v (1,1) is obtained by searching the smallest m such as (1+m.2 w-v ).( 2 The attentive reader noted a few values of m staying identical when we go from rank v-1 to rank v and other m values in a 1/2 ratio. This is related to the resolution of the Bézout equation for these two ranks (we note below w = w(v) at ranked v and w(v-1) at rank v -1 and ∆w their difference) : The anticipation of m seems simpler than that of a v (l,1) because of these cases.

3 v-1 .t 1 -m 1 .2 w(v-1)-v+1 = 1 (40) 3 v .t 2 -m 2 .2 w(v)-v = 1 (41) If w(v)-w(v-1) = ∆w = 1, one can have m 2 = m 1 (with t 2 = 3t 1 ), but you cannot have m 2 = m 1 /2. Case of v = 8,
More generally, parameter m is inferred from the v-plane to the v+1-plane, thanks to the twin equations ( 40) and (41), by the following relation :

m(v+1) = (m(v)+r(v).3 v )/2 Δw(v+1)-1 (42)
where m(v) is the value of m for the v-plane and m(v+1) and Δw(v+1) are respectively the values of m and Δw for the v+1plane and r(v) an integer.

A similar anticipation allows the evaluation of t :

t(v+1) = (t(v)+r(v).2 w-v )/3 (43)
where r (v) is the same parameter as that used previously.

These relationships enable completing the table above incrementing r until what t, m are integers.

Parameter m alternates integer and integer plus 1/2 values when r is incremented. Parameter t, meanwhile, alternates the integer, integer plus 1/3 and integer plus 2/3 when r is incremented. Thus, test r from 0 to 5 will provide certainly the adequate choice of r among them, knowing that we seek the minimum value of a v (1,1).

The frequency of the values r = 0, r = 1 and r = 2 giving the good a v (1,1) solution seems roughly the same order of magnitude. Similarly, the frequency of the values r = 3, r = 4 and r = 5 giving an a v (1,1) seems of the same order of magnitude, but this time half of the previous.

An alternative to the use of the relationship (39) is to take the recursive formula :

a v+1 (1,1) = (2.(a v (1,1)+r.2 w )-1)/3 (44)
As previously, the solution is reached when a v+1 (1,1) is an odd integer, r = r(v) taking one of the integer values between 0 and 5. However, and as previously, a single equation is not sufficient to identify the right solution. Indeed, when r describes 0 to 5, a v+1 (1,1) takes 2 odd integer's values and here the smallest of them is not necessarily the good solution. It is therefore always necessary to use parallel evaluations of m(v) in the absence of a simpler discriminative relation for r. If this relationship is related to the history of the Δw's, it seems relatively complex (little discrimination is observed on analysis of 6 generations of Δw).

Let us note that when we replace r(v) by r(v)+ 6, we have a' v+1 (1,1) = a v+1 (1,1)+2 w+2 which is indeed outside the domain [1,2 w+Δw ] where all of the numbers of the v+1 plane are supposed to be (knowing that a v+1 (1,1) is positive).

In addition, equation (37) means precisely that m = m(v) is the first number obtained after Collatz algorithm altitude flight on a v (1,1) is executed.

We have conducted this test until the value v = 17000 (with systematic surveys up to v = 3000).

A record for a short time.

It is easy to find numbers with flight time above an arbitrary value. For example, the Mersenne numbers a = 2 n -1 have n odd steps at the beginning of Collatz algorithm and hence an altitude flight time superior to n+int(Ln(3)/Ln(2).n)+1 (which is in the order of magnitude of 2,585 n) steps. Assuming a total flight time 5 times greater than the altitude flight time, the order of magnitude of the total flight time would be 13n. Thus a = 2 int(m/13) -1 would have a total flight time about m.

For our part, we are looking somewhat little different numbers, because of the factor t in front of 2 n , but the results are quite similar.

We stopped at a v=17000 (1,1), as computation time is being increasingly longer for a poor flight time gain. We have, restricting to this small sample, without assuming an absolute rule : We observe also that there are always two numbers t1 v and t2 v among t of the last table numbers such as : This regularity possibly hides some structure (modulo 2 i or other feature) which encompasses the rest of the numbers in each column. For example, we have 17^4=1, 29^4=17, 35^4=17 mod 2 6 for v = 7. But this type of relations very quickly becomes more complex for larger values of v (heterogeneity of the powers such that a number of a column produces either 1, or another number in the column). Thus, we did not find straight-forward rules at this stage. If such a structure is discovered, it will be time to review it on the third and following columns.

t v = 2, w = 4 1 v = 3, w = 5 3 3 v = 4, w = 7 1 1 15 v = 5,
t v (1,2) = t v (1,1)+or(0,1).2 w-v (45)
t1 v (1,2) + t2 v (1,2) = 2 w-v+1 (

An interesting difference

Let us go back to finish with td values by examining the differences t-td : The differences t-td are loosely correlated to 2 w-v-3 with sometimes straight returns to 0. If t(v)/t(v-1) = 1/3, then t-td at rank v is either equal to 0 or equal to 2 w-v-3 in the preceding table. For v = 26, we observe a factor 5, which contrasts with other differences. We can attribute, a priori, this characteristic to the fact that t(v)/t(v-1) < 1 without being equal to the 1/3 ratio. In all other cases, which are characterized by t(v) /t(v-1) > 1, we have here tdw = (t-td)/ 2 w-v-4 = 3, 6, 9 or 12.

P 51/96 v w w-v-3 t-td t-td t = t(v) t/3 (t-td)/2 w-v-4 3 5 -1 3
The overall situation is more complex. The % occurrences 14,6% 11,0% 9,6% 8,5% 7,9% 7,8% 7,0% 7,0% 5,9% 4,3% 4,0% 3,6% 3,4% 3,0% 2,5%

Returns to 0 are the most frequent, then masterfully 6 and 12.

If t(v)/t(v-1) = 1/3, then tdw at rank v is either 0, 2 or 4. Specifically, v tdw = 0 = 12, 19, 24, 29, 30, 37, 41, 67, 76, 77, 86, 87, 92, 93, 94, 97, 99, 122, 123, 146, 147, 152, 171, 172, 186, 189, 198, 199, 200, 217, 218, 221, 222, 243, 259, 268, 284, 2 85, 290, 292, 316, 317, 321, 322, 323, 349, 350, 401, 407, 408, 415, 418, 419, 421, 434, 435, 436, 437, 459, 460, 466, 471, 472, 473, 481, 490, 507, 514, 521, 524, v We find this data in the form of three points lined up at the bottom left of the chart below. For tdw = 3, 6, 9, 12, 15, the ratio t(v)/t(v-1) seems not possessing an upper boundary (when v tends to infinity). The tdw = 3 value is special because the points cross axis t(v)/t(v-1) = 1, which seems not the case for other values modulo 3. Values tdw = 5, 8, 10, 11, 14, 16 and 17 seem to have lower and upper boundaries, 5 and 10 being under the said axis, 8 crossing it, and 11, 14, 16 and 17 being above it.

P 53/96 9 Values of the second element of the Pascal trihedron planes.

We are talking about the number a v (1,2) lying on line 1 and column 2 of the Pascal trihedron.

We have 2.a v (1,2)+1 = a v (1,1) mod 2 w , or: The anticipation of the value of or(0,1) seems a difficult problem. In any case, it seems not related to ∆w, as ∆w never has three successive identical values while we observe this for or(0,1). It should be noted however that the term or(0,1) is directly related to the parity of m given sooner. If m is even then the value of or(0,1) is 0 otherwise it is 1 :

Conjecture 2 a v (1,2) = (a v (
or(0,1) = if(m = 0 mod 2,0,1) (48)
It is easy to find this result through an analysis of the Collatz algorithm as we have done in paragraph 8, knowing that the second element has just an odd step offset of one unit versus the first element.

All calculations performed, we find: a v (1,2) = (1+n.2 w-v+1 ).(2 v-1 /3 v )-1 (49) with n an integer n = (m+or(0,1).3 v )/2 (50) and if m is even, the value or(0,1) = 0 is required to get an integer n, and if m is odd, it is or(0,1) = 1 which we have to choose.

The reader will check that n = n(v) is the first resulting number of execution of the Collatz algorithm on a v (1,2) just after the P 54/96 altitude flight. We will address this point in a general framework later on.

10 Values of the elements of the first line of the Pascal trihedron planes.

The values of the first line a v (1,j+1), j ≥ 2, infer the preceding by recurrence :

Conjecture 3 a v (1,j+1) = a v (1,j)-d v (2,
2)+C(j-1,1).d v (3,3)-C(j-1,2).d v (4,4)+…+(-1) j .C(j-1,j-1).d v (j+1,j+1) mod 2 w (51)

where C(i,j) are the binomial coefficients C(i,j) = i!/(j!(i-j !)) and d v (i,j) meets d v (i+1,j+1) = (or(1,2 We proceed to differences modulo 2 w (here modulo 1024) column-by-column between the number left on the diagonal and the one above on the column, that is Having found these values, we go back to the elements of the first line with reverse operations :

d v (i+1,j+1) = d v (i,j)-d v (i,
d v (i,j+1) = d v (i,j)-d v (i+1,j+1) mod 2 w (57)
Let us have a v (1,1), d v (2,2), d v (3,3), d v (4,4), d v (5,5),… the numbers appearing on the main diagonal. The elements of the first line are then given by : (3,3)-C(j-1,2).d v (4,4)+…+(-1) j .C(j-1,j-1).d v (j+1,j+1) mod 2 w

a v (1,3) = a v (1,2)-d v (2,2)+d v (3,3) mod 2 w a v (1,4) = a v (1,3)-d v (2,2)+2d v (3,3)-d v (4,4) mod 2 w a v (1,5) = a v (1,4)-d v (2,2)+3d v (3,3)-3.d v (4,4)+d v (5,5) mod 2 w … a v (1,j+1) = a v (1,j)-d v (2,2)+C(j-1,1).d v
The coefficients in front of d v (i,i) are thus growing according to a Pascal triangle with alternating signs, C(i,j) being the binomial coefficients.

Alternative approach

Instead of making the differences according to the relation (12), let us proceed by the opposite : The elements of the main diagonal (excluding 575 and 736) infer each other by (or(0,1).2 w +d v (i,j))/d v (i+1,j+1) = 2 (59) meaning that the elements on the main diagonal are divided by two, with from time to time, the addition of w-1 .

d v (i+1,j+1) = -(d v (i,j)-d v (i,
The recurrence process is simple, but rests the problem around the anticipation of values or(1,2) (or else or(0,1) if we proceed by the alternative). Of course or(1,2) and or(0,1) are closely linked here.

This method applies similarly to all of the lines of the Pascal trihedron planes.

For example, for the tenth line of the plane corresponding to v = 12 (w = 20, 2 w = 1048576), we have the following The anticipation of the values or(1,2) (or or(0,1)) would solve largely, with conjectures on the first and second elements of a plane, the problem at hand.

The rest of the hereby text settles the problem a little differently but remains largely related to what has already been stated.

11 Pascal trihedron whole set anticipation. We proceed by an example (which is not yet conveniently sorted at this stage) for clarity.

We choose v = 9 (and w(at v) = 15), that is also v-1 = 8 (and w(at v-1) = 13).

The The values for v are the followings :

Meanwhile, let us rewrite the column "ratios" in a way that seems us most appropriate, taking into account the grouping of lines of identical staffs, thus highlighting the last three sets of data: Somehow, we find a kind of Pascal trihedron plane (with the last three identical columns). We will call these planes by the term (Pascal) secondary planes. However, everything remains to discover about their anticipation.

Working so far modulo 2 w(at v-1) (here 2 13 = 8192), the previous table cannot map the Pascal trihedron plane ranked v from that ranked v-1. We need resume the same work, but this time by making differences modulo 2 w(at v) (here 2 15 = 32768) between lines of equal lengths.

P 60/96

Dif2-Dif1 col 1 Dif2-Dif1 col 2 Dif2-Dif1 col 3 Dif2-Dif1 col 4 Dif2-Dif1 col 5 Dif2-Dif1 col 6 Dif2-Dif1 col 7 2 0 2 0 0 2 1 3 1 3 3 1 0 2 0 2 2 0 1 3 1 3 3 1 0 2 0 0 2 3 1 3 3 1 0 2 0 0 2 1 3 1 1 3 2 0 2 2 0 2 0 2 2 0 1 0 0 1 0 3 3 0 1 0 0 1 2 1 1 2 3 2 2 3 3 2 2 3 1 0 0 1 2 1 1 2 2 1 1 2 1 1 3 0 0 2 1 1 3 2 2 0 3 3 1 3 3 1 1 1 3 2 2 0 2 2 0 1 2 0 1 1 2 2 3 3 0 3 0 1 2 2 3 2 3
If the multiplicative factor between 2 w(at v-1) and 2 w(at v) is 2, we will get in these tables only 0 and 1, if it is instead 4, we will get 0, 1, 2 or 3. We will call these planes by the term tertiary planes.

If we can anticipate secondary and tertiary planes, we can build by successive iterations all values constituting the Pascal trihedron.

Conjecture 5

The values of secondary planes, ordered by positive sorting, can be deducted from one column to the next one by multiplication by 2 except the first column (trivially) on the one hand and the last two columns, which are identical to the previous one, on the other hand.

The reader can check this, for example, for the secondary plane of plane v = 9 given above, as well as the secondary plane of the planes v = 4 to v = 11 given in Chapter 12.

Knowledge of the first element in each row means knowing all the other objects.

Here, the rule applies from the first number (in parentheses), but this is not a general case. Indeed, let us note s v (i,j) an element of secondary plane v, i the line number line and j the column number, we have then:

Conjecture 6

For any secondary plane, the difference between the first element s v (

Conjecture 7

The associates of the first line at rank v are drawn from the associates next ranked v-1 by multiplying by 6 and adding 5 and after iterations of the result : As examples:

a v (1,j) = (6.a v-1 (1,j)+5) k mod
(5).6+5 = 3 mod 16 and 3 1 = 3 mod 16.

(3).6+5 = 23 mod 32 and 23 1 = 23 mod 32. but (0).6+5 = 5 mod 32 and 5 i ≠ 11 mod 32 for any i .

( 12 Sorted planes.

The reader will find below the first Pascal trihedron planes (of associates) classified by positive sorting and also related secondary and tertiary planes. The tertiary planes are :

v = 1, w = 2, #(v)= 1 Family mod 4, 1 5 v = 2, w = 4, #(v)= 1 Family mod 32, 1 3 v = 3, w = 5, #(v)=
0 0 0 1 0 1 1 1 1 0 1 0 1 1 1 0 1 0 0 0 0 1 0 1 0 0 0 1 0 1 1 1 0 1 0 1 1 0 1 0 0 0 1 0 1 0 0 1 0 1 1 1 0 1 0 0 0 1 0 1 0 0 1 0 1 1 1 0 1 0 1 1 0 1 0 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 1 1 0 1 0 1 1 0 1 0 0 0 1 0 1 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 0 1 0 1 0 1 1 0 1 0 1 0 1 0 1 1 0 1 0 1 0 0 1 0 1 0 1 1 0 1 1 0 1 1 0 1 0 1 0 0 1 0 1 0 1 0 1 0 1 0 1 1 0 1 0 1 0 0 1 0 1 0 1 P 73/96 1 0 1 1 0 1 0 0 1 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 v = 11, w = 18, #(v)= 961 Family mod 262144 0 1 0 3 3 3 2 0 2 3 2 1 1 1 0 2 3 0 3 2 2 2 1 3 1 2 1 0 0 0 3 1 0 1 0 3 3 3 2 0 0 0 2 3 3 1 0 1 1 3 0 0 2 1 3 3 1 2 2 0 3 2 2 0 1 1 3 2 0 0 2 3 3 1 0 2 2 0 1 1 3 2 1 1 3 0 0 2 1 0 0 2 3 3 1 0 0 0 2 3 3 1 0 1 1 3 0 0 2 1 2 1 1 1 0 2 3 2 2 2 1 3 1 0 0 0 3 1 0 3 3 3 2 0 2 1 1 1 0 2 0 3 3 3 2 0 3 2 2 2 1 3 2 1 1 1 0 2 2 1 1 1 0 2 3 2 2 2 1 3 3 2 2 2 1 3 1 0 0 0 3 1 0 3 3 3 2 0 3 2 2 2 1 3 2 1 1 1 0 2 3 2 2 2 1 3 1 0 0 0 3 1 0 3 3 3 2 0 1 0 0 0 3 1 2 3 3 1 0 3 0 0 2 1 1 2 2 0 3 0 1 1 3 2 P 74/96 2 3 3 1 3 0 1 1 3 1 3 0 0 2 0 2 3 3 1 3 2 3 3 1 3 3 0 0 2 0 3 0 0 2 0 1 2 2 0 2 0 1 1 3 1 3 0 0 2 0 2 3 3 1 3 3 0 0 2 0 1 2 2 0 2 0 1 1 3 1 1 2 2 0 2 2 3 3 1 3 0 1 1 3 1 3 0 0 2 0 2 3 3 1 3 2 3 3 1 3 3 0 0 2 0 0 1 1 3 1 3 0 0 2 0 0 1 1 3 1 3 0 0 2 0 2 3 3 1 3 3 0 0 2 0 3 3 2 0 0 0 3 1 2 2 1 3 1 1 0 2 3 3 2 0 1 1 0 2 0 0 3 1 3 3 2 0 3 3 2 0 0 0 3 1 0 0 3 1 2 2 1 3 1 1 0 2 0 0 3 1 3 3 2 0 0 0 3 1 2 2 1 3 1 1 0 2 2 2 1 3 3 3 2 0 1 1 0 2 0 0 3 1 3 3 2 0 3 3 2 0 0 0 3 1 1 1 0 2 0 0 3 1 1 1 0 2 0 0 3 1 3 3 2 0 0 0 3 1 0 0 3 1 2 2 1 3 1 1 0 2 0 0 3 1 0 0 3 1 0 0 3 1 2 2 1 3 1 1 0 2 2 1 3 0 0 3 1 0 0 3 1 1 1 0 2 0 2 0 1 3 1 3 1 3 2 0 2 0 2 0 2 0 2 1 3 1 0 2 0 0 2 0 1 3 1 1 3 1 3 1 3 2 0 2 1 3 1 0 2 0 1 3 1 3 1 3 2 0 2 3 1 3 0 2 0 2 0 2 1 3 1 0 2 0 0 2 0 1 3 1 2 0 2 1 3 1 2 0 2 1 3 1 0 2 0 1 3 1 1 3 1 3 1 3 2 0 2 1 3 1 1 3 1 1 3 1 3 1 3 2 0 2 3 1 3 1 3 1 1 3 1 2 0 2 3 1 0 2 2 0 1 3 3 1 1 3 0 2 3 1 3 1 0 2 0 2 2 0 1 3 0 2 3 1 0 2 2 0 1 3 2 0 P 80/96
The scheme is totally reproducible for each of the two categories of matchings.

For the first category (red and green fonts), it is obviously a simple shift of the last 1, since signatures are stored with the constraint of an identical final block 10... 0 with a decreasing number of 0. By the method of signatures' construction with constraint, which is exhaustive to the left of the last 1, any signature of subblock n of type x... x0010…0 has necessarily a counterpart in the subblock n-1 (in his column) of type x...x0100...0 (the x…x being identical).

For the second category (blue and green fonts), where the offset of the 1's extends to the right without finding a counterpart in the subblock above it, there is an extra 01 after the last 1, then 00 or 000 at the end of signatures as Δw = 1 or Δw = 2. The offset being progressive and comprehensive in any subblock, a bijection results necessarily if there is an initial matching of the last 1 of the plan v-1 with the second last 1 of the plan v for subblocks in question. For this, it is necessary to check the matching between the first elements of the first subblock of the v-1 plans and v.

We start with the first element of the plan v-1 and the second component of the plan v : The last 1 examined of the plane v-1 item is effectively aligned with the second last of the other element to be considered in the plane v. When v is incremented, by construction, the signatures will grow up with simply 01 addition after the last 1 on the first list (green police components) and addition (if Δw = 2) or not (if Δw = 1) of a 0 at the signature end and 01 addition after the second last 1 of the second list (and same rule for the final 0's).

Having checked these first elements, it should be then check the setting of the other columns, noting the evolution from one to the other column. The case v-1 = 6 and v = 7 is representative : 1010101010100000 1010101010010000 1010101001010000 1010100101010000 1010010101010000 1010101010100100000 1010101010010100000 1010101001010100000 1010100101010100000 1010010101010100000 Indeed, starting from the first pair of elements, subject of the previous discussion, by construction, the other signatures will have the same final sequence of 0's (that is one less for the first line and the same number in second line because the first signature of the first subblock of a plan has an additional 0 as noted earlier). Thus, the last 1 of the elements of the first line remains aligned with the second last 1of the second line and by completeness (exhaustivity) all the preceding 1's (to the said last and second last) are aligned from one line to the other. By completeness also, the two lists in the presence will stop at the same time. For skeptics, as we still have to prove assertion (64), the underneath study will be reassuring as it is back on these final elements of lists. It suffices to have two 0 between the two last 1 of the last triangular subblock for this twinning to appear. Indeed, all items above or left have by construction a spacing between the latter two 1 equal or superior to this one, which allows to systematically switch 01 into 10 in the second last subblock. On the other hand, if a single 0 is there instead, then the twinning (in its entirety) is impossible.

Yet, this last element does increase as follows as v increases (we have not represented the cases v = 1 to v = 3 because this "last element" does not yet exist in these planes in a way): This arises from the fact that the last element is the one for which all of the 1's are most to the right. So, from v-1 to v, the latest addition which is 10 or 100 (depending on whether Δw = 1 or Δw = 2) will be done with an offset of 1 of a single notch to the left (because otherwise the signature would be the previous + the add and stop prematurely). Thus, the latter stores the whole history of the Δw in the 10 and 100 sequences of which it is formed, and this in order. In particular, the second last sequence is 10 if Δw prec1 = 1 and 100 if Δw prec1 = 2 which allows to conclude the proof.

v
On the other hand, the table also shows us that the elements of the last triangular subblock of a plan do end with 100 if Δw = 1 and 1000 if Δw = 2.

As the first subblock ends with w-v+1-1 sequences 0 and by construction the number of final 0's decrease from one subblock to another by one unit, we can easily count the number of subblocks at rank v, namely w-v-Δw.

We will finish our study by two properties and recall a third one. To facilitate the writing of these, we identify the constituting elements of subblocks by sb(v, n, i, j), where v is the plan, n the nth subblock (starting at 1), i the line index (starting at 1) in the subblock, j the index column (starting at 1) in the subblock. The first element of the plan v is spotted apart by sb(v,1,1,0).

For the first property, we focus on the first line of each subblock of a plane v.

Property 7

sb(v,n,1,j) = sb(v,n-1,1,j) + cte(n) mod 2 w j > 0 (65)

where cte(n) depends only on n (but not on j).

Property 8 generalizes property 7.

Property 8

There is a pair of two lists of length equal such as sb(v,n,1,j) = sb(v,n-1,1,j') + cte1(n) mod 2 w j et j' > 0 (66) and sb(v,n,1,j) = sb(v-1,n,1,j') + cte2(n) mod 2 w(v-1) j et j' > 0 (67)

where cte1(n) and cte2(n) depend only on n and w(v-1) means w at rank v-1.

P 82/96

We find in these formulas the additive properties that we had foreseen in the last paragraph of chapter 2 before tackling the notion of signature.

Proof

These properties are derived from theorem 9, which does indicate that equal differences of signatures means equal differences of associates modulo 2 w . The two pairs of lists are those discussed earlier as we compared signatures in red and green fonts and signatures in blue and green fonts. When signatures concern planes v-1 and v (blue and green fonts), the value modulo must obviously be taken on rank v-1, that is modulo 2 w(v-1) .

Example :

Numbering Finally, when ∆w prec1 (v) = 2, we have a twinning of the last two triangular subblocks and the differences between twin associates of these two blocks are thus equal.

The plan v = 7 (w = 12, 2 w = 4096) provides a synthesis of all these properties (768 = 2 * 384) : This construction applies to this second geography. The calculation efficiency stems from its simplicity.

Theorem 16

Let us have a v (r,s) the element on line r and column s of the Pascal trihedron v-plane.

Let us have m v (r,s) the result of the Collatz algorithm on a v (r,s) at the end of altitude flight.

We have then a v+1 (r,s) = (2.(a v (r,s)+r.2 w )-1)/3 (68) and m v+1 (r,s) = (m v (r,s)+r.3 v )/2 Δw(v+1)-1 (69)

for the smallest integer r such as a v+1 (r,s) and m v+1 (r,s) are simultaneous integers. Moreover, the solution is necessarily obtained for r in the set {0,1,2,3,4,5}.

Note

Passing from plane v to plane v + 1, the size of the plane increases and new integers a v+1 (r,s) where a v (r,s) is empty are to be evaluated directly.

  (10)0000 1010101010010(10)0000 1010101001010(10)0000 1010100101010(10)0000 1010010101010(10)0000 Line 2 1010101010100(01)0000 1010101010010(01)0000 1010101001010(01)0000 1010100101010(01)0000 1010010101010(0110001)000 10101010010(10001)000 10101001010(10001)000 10100101010(10001)000Line 10101010100(01010)000 10101010010(01010)000 10101001010(01010)000 10100101010(01010)000

  15, 22 and 29. If ∆w = 2, can have m 2 = m 1 /2 (with t 2 = 3t 1 ), but you cannot have m 2 = m 1 . Case of v = 4, 7, 12, 18, 19 and 30.

  The exponents are always inferior to 2 w-v+c-1 , c being the c th column. This is analogous to what has been said in the statement of theorem 11. The exponents have remarkably linked values, such 9 = 1+8, 25 = 9+16, 57 = 25+32, 89 = 25+64, 121 = 57+64, 185 = 121+64, 441 = 185+256, 825 = 185+128+512, 1657 = 121+512+1024, 1849 = 825+1024, 1977 = 441+512+1024, 2681 = 121+512+2048, 2073 = 25+2048. However, the plain anticipation of these values remains a mystery.

  When ∆w prec1 (v) = 2, we have a twinning of the last two triangular subblocks by moving the last 1 a step to the right (or what amounts to swapping 10 in 01) For example, for v = 7, Δw = 2, Δw prec1 = 2, the last two subblocks are as follows :

table : P 2

 :2

	11	34	17	52	26	13	40	20	10
		I	P	I	P	P	I	P	P
	/96

  The principle here is to continue the process of maxima's location on the last column of the table already obtained and to repeat it so long maxima are available, the only exception being the first element that is placed in the first column.

			31229	4	13	49	209	849
				2	2	2	2	2
				1	1	1	1	1
					7	7	7	7
					1	1	1	1
					4	4	4	4
						27	27	27
						1	1	1
						4	4	4
							123	123
							1	1
							4	4
	Then of third generation :
			849	2	7	27	123
				1	1	1	1
					4	4	4
	Rule 5. Rule of models.
	Then we have :
				sddsl v (j) = 4.Δw(v).sddsl v-1 (j-1)	(5)
	Examples :
	v	Δw(v)	sddsl v (1)	sddsl v (2)	sddsl v (3)	sddsl v (4)	sddsl v (5)	sddsl v (6)	sddsl v (7)
	6	2	1	4
	7	2	3	8	32
	8	1	7	24	64	256
	9	2	7	28	96	256	1024
	10	1	15	56	224	768	2048	8192
	11	2	15	60	224	896	3072	8192	32768
	This rule allows, knowing the elements of the last column of the previous table, to deduce all of the elements of the first
	line, from the knowledge of rank v-1.
	This then allows focusing only on the last column of the table.
	Rule 4. Rule of maxima.
	For v = 11, we will have a table of "second-generation":
	P 20/96

  Let us call review table the above table and, for us, the first column will now be the one after the numbers in parentheses.

	example v = 11
	Origin		4	14	33	64	107 150 193 193 193
	First generation 43 Going backwards generations, it may be possible to return to the original table as long as one controls the values of the 4 10 19 31 43 43
	elements of the first line.
	For v = 11, let us review these :
	First generation		8	25	93		381	1533	6653	31229
	Second generation	(31229)	4	13	49		209	849
	Third generation	(849)	2	7	27		123
	Fourth generation	(123)	1	4
	Rule 7. Rule of the powers of 2.
	The first column of the review table are decreasing powers of 2 resulting in 1
	Rule 8. Rule of model construction.

table 2

 2 Before we start the theorem proof and in order to easier understand the mechanisms involved, let us look at in details to signatures (Sig) and evolutions by Collatz algorithm (Alg) for a numerical example. We consider the second and third lines of the plane v = 7.So we check in the previous table equality of differences between associated two lines elements in correspondence. We see that this equality is maintained throughout the evolution by the algorithm of Collatz except in some places that we have bold (in black, green, and blue) with a return to identical values as soon as the first following line the so-called evolution.

	mod 2 w	(19)

In the table, on the right side, which focuses on the Collatz algorithm (Alg), the value of the first selected element 1855 has been replaced by 1855-4096 = -2241. Similarly for 2591-4096 = -1505, 1647-4096 = -2449 and 2203-4096 = -1893. On the other hand, we left unchanged 231. The goal here is to have coherence between two associated values, because the Collatz algorithm does not involve modulo operations, thus the differences of a value at head of column to his associated head value must be identical to ensure that the following data have some clarity. Thus, by doing as we do, we get a common first line's difference : 1087-(-2241) = 1823-(-1505) = 879-(-2449) = 3559-231 = 1435-(-1893) = 3328 and this without involving modulo operations. The differences column to column are then as follows (table of differences): will be unchanged except for the line in bold with the following result :

Table 3a

 3a

						et 3b
			k = 1293					k = 3597
	3328	3328	3328	3328	3328	3328	3328	3328	3328	3328
	1792	1792	1792	1792	1792	1792	1792	1792	1792	1792
	2944	2944	2944	2944	2944	2944	2944	2944	2944	2944
	640	640	640	640	640	640	640	640	640	640
	320	320	320	320	320	320	320	320	320	320
	960	960	960	960	2208	960	960	960	960	2208
	2528	2528	2528	2528	2528	2528	2528	2528	2528	2528
	3488	3488	3488	1264	1264	3488	3488	3488	1264	1264
	3792	3792	3792	3792	3792	3792	3792	3792	3792	3792
	3184	3184	3944	3944	3944	3184	3184	3944	3944	3944
	3640	3640	3640	3640	3640	3640	3640	3640	3640	3640
	2728	1820	1820	1820	1820	2728	1820	1820	1820	1820
	1364	1364	1364	1364	1364	1364	1364	1364	1364	1364
	2730	2730	2730	2730	2730	2730	2730	2730	2730	2730
	1365	1365	1365	1365	1365	1365	1365	1365	1365	1365
	245	3197	1725	3125	1861	2293	1149	3773	1077	3909
	2048	2048	2048	2048	2048	0	0	0	0	0
	1024	1024	1024	1024	1024	0	0	0	0	0
	2560	2560	2560	2560	2560	0	0	0	0	0
	3328	3328	3328	3328	3328	0	0	0	0	0

If we choose this last table as reference, the last line values are inferred by :

  This relationship allows case by case to find precisely the number of planes with associates of a given cycle length. However we look here simply for an approximate value when v is increasing.

	Thus we have :
	((ln(3)/ln(2)-1).v ≈≤ i ≈≤ (ln(3)/ln(2)).v	(28)
	Hence
	(ln(2)/ln(3)).i ≈≤ v ≈≤ (ln(2)/(ln(3)-ln(2))).i (29)
	Considering the interval defined in this way, the number of planes v for large i is around :
	v ≈ i.ln²(2)/(ln(3).(ln(3)-ln(2))) ≈ 1,0786.i	(30)
	2
	Family mod 32,
			23		11
			4		8
	v = 4, w = 7, #(v)= 3
	Family mod 128
		15	7		59
		8		16	32
	v = 5, w = 8, #(v)=7
	Family mod 256
		95	175	39	219
			79		199	123
		8	16		32	64
	v = 6, w = 10, #(v)=12
	Family mod 1024
	575 287	367	999	923
		735	815	423	347
				975	583	507
	16	32	64	128 256
	v = 7, w = 12, #(v)=30
	Family mod 4096
	383 2239	2975	2031	615	2587
	1855	2591	1647	231	2203
	1087	1823	879	3559	1435
			4063	3119	1703	3675
			3295	2351	935	2907
					1231	3911	1787
					463	3143	1019
	32	64	128 256 512 1024

Evaluation of the number of planes including a cycle of given length

This evaluation is not crucial and may be skipped. From theorem 11, a plane v includes associates with different cycles of lengths 2 i where integer i ranges from w-v up to w-2. So, we get using the relationship between v and w : int((ln(3)/ln(2)).v)-v+1 ≤ i ≤ int((ln(3)/ln(2)).v)-1 (27) P 39/96

  By theorem 11, we have 2587 1024 = 1 mod 4096 and similarly 2203 1024 = 1 mod 4096, 1435 1024 = 1 mod 4096, 3675 1024 = 1 mod 4096, 2907 1024 = 1 mod 4096, 1787 1024 = 1 mod 4096 and 1019 1024 = 1 mod 4096. Let us divide then the exponent 1024 by 2, then by 4, then by 8 and so on. We get modulo 4096 a table which allows, thanks to the results on the columns of 64, 32 and 16 iterations, the following classification :

	Numbers of iterations of	1024	512	256	128	64	32	16	8	4	2	1
	615	1	1	2049 3073 3585 3841 1921 3009 3041		615
	231	1	1	2049 3073 3585 3841 3969 1985	481	113	231
	3559	1	1	2049 3073 3585 3841 3969 4033 3553 3559
	1703	1	1	2049 3073 3585 1793	897	2497	737	241	1703
	935	1	1	2049 3073 3585 1793	897	449	3809		935
	3911	1	1	2049 3073 1537	769	2433 3265 1121 3911
	3143	1	1	2049 3073 1537	769	2433 1217	97		3143
	Differentiation is still done by the columns of 64, 32 and 16 iterations.
	We can also realize differences modulo 4096 of these two tables and we get :
	Iterations Matchings	1024	512	256	128	64	32	16	8	4	2	1
	2587/615	0	2048 1024	512	2304 3200 3648 3360 2064 1972
	2203/231	0	2048 1024	512	2304 3200 3648 3360	16		1972
	1435/3559	0	2048 1024	512	2304 3200 3648 3360	16		1972
	3675/1703	0	2048 1024	512	2304 3200 3648 1312 1024	872	1972
	2907/935	0	2048 1024	512	2304 3200 3648 1312 1024 1972
	1787/3911	0	2048 1024	512	2304 3200 1600	288	528		1972
	1019/3143	0	2048 1024	512	2304 3200 1600	288	528		1972
	Numbers of iterations of	1024	512	256	128	64	32	16	8	4	2	1
	2587	1	2049 3073 3585 1793 2945 1473 2273 1009 3801 2587
	2203	1	2049 3073 3585 1793 2945 3521 1249	497	3545 2203
	1435	1	2049 3073 3585 1793 2945 3521 3297 3569 3033 1435
	3675	1	2049 3073 3585 1793	897	449	3809 1777 1113 3675
	2907	1	2049 3073 3585 1793	897	449	1761	753	601	2907
	1787	1	2049 3073 3585 3841 3969 4033 3553 1649 2585 1787
	1019	1	2049 3073 3585 3841 3969 4033 1505	625	2073 1019
	P 41/96

This table gives us lines' aggregate score {1,2,2,2} as expected by the simplified decomposition #TPS (7) = {1,3,5,7,7,7} of the studied plane. In particular, this technique allows always to find the correct first line, here through the column of 16 iterations.

Then let us do the same with the second last column :

  table below gives a wider range of values. To date, a general rule of anticipation of these values seems difficult to find.

	c	tdw	c	tdw	c	tdw	c	tdw	c	tdw	c	tdw	c	tdw	c	tdw	c	tdw	c	tdw
	55	15 105 12 155	8	205 16 255	3	305 14 355	6	405	2	455	8	505 15
	56	4	106	2	156	6	206 11 256 12 306 16 356 12 406	9	456	8	506	4
	57	6	107	6	157	8	207	6	257	2	307 17 357	4	407	0	457 12 507	0
	58	14 108 15 158 15 208	8	258	6	308	5	358 15 408	0	458	4	508	9
	59	14 109 16 159	8	209 15 259	0	309 12 359 11 409	3	459	0	509 14
	60	14 110 11 160	2	210 10 260 12 310 17 360 14 410 12 460	0	510 17
	61	4	111	8	161	9	211	6	261 16 311 16 361	5	411	5	461 12 511	4
	62	12 112 15 162	2	212	9	262 11 312 17 362 12 412	3	462 17 512 15
	63	16 113 11 163 12 213	2	263	2	313	2	363	8	413 12 463 10 513	5
	64	8	114	2	164	4	214 12 264	3	314	6	364	8	414	2	464 12 514	0
	65	9	115	9	165	3	215	4	265 12 315	3	365	6	415	0	465	5	515 12
	66	2	116	2	166 15 216	3	266 16 316	0	366	6	416 15 466	0	516 16
	67	0	117	9	167 16 217	0	267	5	317	0	367 14 417 10 467	9	517 14
	68	6	118	9	168 11 218	0	268	0	318 15 368	8	418	0	468 14 518	8
	69	6	119	2	169	2	219 15 269	6	319	4	369	2	419	0	469	8	519	2
	70	14 120	3	170	3	220	4	270	2	320	3	370 12 420	6	470	2	520	3
	71	8	121	6	171	0	221	0	271 15 321	0	371	8	421	0	471	0	521	0
	72	9	122	0	172	0	222	0	272	8	322	0	372	8	422 12 472	0	522	9
	73	8	123	0	173	3	223 12 273 14 323	0	373	3	423	2	473	0	523	2
	74	6	124 15 174 12 224 17 274 17 324	3	374	6	424 12 474	9	524	0
	75	2	125 11 175 11 225 10 275 10 325	3	375	9	425 16 475 14 525 12
	76	0	126	2	176	2	226	6	276	3	326 12 376	2	426 11 476 14 526 16
	77	0	127	3	177	3	227	2	277	9	327 14 377	9	427	2	477 11 527 11
	78	6	128	6	178	6	228 15 278	8	328 10 378 15 428	9	478	8	528	8
	79	3	129	6	179	2	229 16 279 12 329	6	379	4	429	8	479 12 529	9
	80	12 130	3	180	6	230	2	280 10 330 12 380 12 430 15 480 10 530	6
	81	11 131	6	181 14 231	6	281 15 331	4	381 16 431 14 481	0	531 14
	For the report tdw = (t-td)/2 w-v-4 , other values appear not present in the first list and we have stock of the occurrences
	(Nb_app) between v = 3 and v = 531 :
		tdw		0	6	12	8	2		3	9	15	14	4	16	11		10	5	17
	Nb_app		77	58	51	45	42		41	37	37	31	23	21	19		18	16	13
	c	tdw	c	tdw	c	tdw	c	tdw	c	tdw	c	tdw	c	tdw	c	tdw	c	tdw	c	tdw
	32	8	82	8	132	9	182	2	232	2	282 16 332	6	382	2	432	4	482 12
	33	6	83	6	133	8	183 15 233 15 283	8	333	8	383	3	433	3	483 11
	34	14	84	12 134 15 184	4	234	4	284	0	334	9	384 12 434	0	484 15
	35	17	85	4	135 10 185	3	235	9	285	0	335	8	385	5	435	0	485 10
	36	2	86	0	136	6	186	0	236	6	286	6	336 12 386	6	436	0	486	9
	37	0	87	0	137	9	187	9	237 14 287	2	337 17 387 15 437	0	487	8
	38	3	88	6	138	8	188	8	238 11 288	9	338	4	388	4	438	3	488 15
	39	12	89	9	139 15 189	0	239	2	289	3	339	3	389 15 439 12 489	2
	40	5	90	14 140 10 190	6	240 15 290	0	340	6	390 11 440 11 490	0
	41	0	91	5	141	3	191 14 241 10 291	3	341	6	391	2	441	2	491	9
	42	12	92	0	142	3	192 11 242	9	292	0	342 12 392	3	442	9	492	8
	43	14	93	0	143 12 193	8	243	0	293 12 343 16 393 12 443 15 493	6
	44	16	94	0	144	8	194	6	244	6	294	4	344	5	394	5	444 16 494	8
	45	17	95	15 145	8	195	3	245 15 295 12 345	6	395	3	445 14 495 12
	46	16	96	5	146	0	196 12 246 16 296 17 346 15 396 12 446 16 496	2
	47	5	97	0	147	0	197	5	247 11 297 10 347 10 397 11 447	5	497 12
	48	6	98	3	148	9	198	0	248	6	298	6	348	3	398 14 448 15 498	8
	49	14	99	0	149	9	199	0	249 14 299 14 349	0	399	8	449 10 499	2
	50	8	100	6	150	8	200	0	250 17 300	8	350	0	400	8	450	6	500	9
	51	8	101 15 151	3	201	3	251 10 301 12 351	6	401	0	451 14 501 15
	52	3	102	4	152	0	202	9	252	9	302 10 352 14 402 15 452 14 502 16
	53	15 103 12 153 12 203	2	253	2	303	9	353 17 403	4	453	4	503	8
	54	10 104	4	154	8	204 15 254	6	304 14 354	5	404	6	454 12 504	2
	P 52/96

  table :

	Line 10	713727 1028607 452351	112255	650687	934047	834799	161639	200475
		314880	472320	708480	538432	283360	949328	375416	38836
			157440	236160	878528	793504	665968	474664	711996
				78720	642368	963552	921040	857272	237332
					563648	321184 1006064 984808	428636
						806112	684880 1027320 492404
							927344	342440	513660
								463672	171220
									756124
	d 12 (i,i)	314880	157440	78720	563648	806112	927344	463672	756124
	d 12 (i,i)/d 12 (i+1,i+1)		2	2				2
	(2 w +d 12 (i,i))/d 12 (i+1,i+1)				2	2	2		2

  values of Pascal trihedron plane for v-1 are as follows:

					207	6983	4859
					7375	5959	3835
					7631	6215	4091
					6607	5191	3067
	255	4223	1983	6815	5871	4455
		3967	1727	6559	5615	4199
		3455	1215	6047	5103	3687
		2431	191	5023	4079	2663	539
			5439	2079	1135	7911
			4927	1567	623	7399
			3903	543	7791	6375
			4159	799	8047	6631
			3135	7967	7023	5607
				3551	2607	1191
				3039	2095	679
				2015	1071	7847
				2271	1327	8103
				1247	303	7079
					719	7495
	P 56/96

  As we perform offsets of the final 1's between subblocks, property 5 springs up with ratios of 2 from one to the other. Thus 21504 = 2*27136 mod 2 15 , 10240 = 2*21504 mod 2 15 and 5120 = 2*2560 mod 213 , 2048 = 2*5120 mod 2 13 , 4096 = 2*2048 mod 2 13 .

	237	101010101000101010010000	5599
	250	101010101000101010001000	15839	10240
	238	101010101000101001010000	21471
	251	101010101000101001001000	31711	10240
	239	101010101000100101010000	12511
	253	101010101000100101001000	22751	10240
	141	101010101001010000100	5023
	244	101010101001010000101000	25503		4096
	142	101010101001001000100	543
	247	101010101001001000101000	21023		4096
	143	101010101001000100100	7967
	249 144 111 252 211 145 254	Signatures 101010101001000100101000 101010101000101000100 101010101001010100000 101010101000101000101000 101010101001010101000000 101010101000100100100 101010101000100100101000	Associates 12063 2015 30687 6815 9375 1247 21727	Diff_ mod 2 w(v) !	4096 Diff_ mod 2 w(v-1) 4096 2560 4096
	211	101010101001010101000000	9375
	221	101010101001010100100000	3743	27136
	121	101010101001010010000	6559
	222	101010101001010010100000	11679		5120
	122	101010101001001010000	2079
	223	101010101001001010100000	7199		5120
	123	101010101000101010000	3551
	224	101010101000101010100000	16863		5120
	221	101010101001010100100000	3743
	231	101010101001010100010000	25247	21504
	222	101010101001010010100000	11679
	232	101010101001010010010000	415	21504
	223	101010101001001010100000	7199
	234	101010101001001010010000	28703	21504
	224	101010101000101010100000	16863
	237	101010101000101010010000	5599	21504
	131	101010101001010001000	6047
	233	101010101001010001010000	16287		2048
	132	101010101001001001000	1567
	235	101010101001001001010000	11807		2048
	133	101010101001000101000	799
	236	101010101001000101010000	2847		2048
	134	101010101000101001000	3039
	238	101010101000101001010000	21471		2048
	135	101010101000100101000	2271
	239	101010101000100101010000	12511		2048
	231	101010101001010100010000	25247
	241	101010101001010100001000	2719	10240
	232	101010101001010010010000	415
	242	101010101001010010001000	10655	10240
	233	101010101001010001010000	16287
	243	101010101001010001001000	26527	10240
	234	101010101001001010010000	28703
	245	101010101001001010001000	6175	10240
	235	101010101001001001010000	11807
	246	101010101001001001001000	22047	10240
	236	101010101001000101010000	2847
	248	101010101001000101001000	13087	10240
	P 83/96

14 Recursive construction of the Pascal trihedron set of values.

				Associates
	1	383	2239	2975	2031	615	2587
	2		1855	2591	1647	231	2203
	3			4063	3119	1703	3675
	4				1231	3911	1787
	5		1087	1823	879	3559	1435
	6			3295	2351	935	2907
	7				463	3143	1019
				Differences modulo 2 w
	1-2		384	384	384	384	384
	2-5		768	768	768	768	768
	3-6			768	768	768	768
	4-7				768	768	768

From results on t and m at rank v = 17000 given above and from the recursive algorithm equations ( 42), ( 43) and (36), the reader can find new a v (1,1) numbers with larger flight times (TFA and TF).

As for the value of a v (1,1), it increases haltingly with v in a mean ratio of 3. The graphs below give an outline of it expressing a v (1,1) 1/v according to v.

The increase in a v (1,1) is not regular also including reductions in values. Their frequency is 23% for the planes v from 1 to 3000. All these decreases correspond, a priori, to the general form a v+1 (1,1) = (2.a v (1,1)-1)/3. Increases from one plan to the next are sometimes significant to catch up "backlogs". So for v = 1945, we have a v (1,1) ≈ 2394.a v-1 (1,1).

The evolution with an average ratio of 3 is not an exclusive property to a v (1,1). It is a statistical property of the associates of a plane v. Indeed, let us suppose that we choose integers at random first among the numbers 1 to 2 w(v-1) , then among 1 to 2 w(v) , where w(v-1) and w(v) are large enough. For a large sample, the average of values in the first case would be of the order of magnitude of 2 w(v-1) /2 and for the second case of 2 w(v) /2, hence a ratio from the last to the first of 2 w(v) /2 w(v-1) = 2 Δw(v) = 2 Δw . We checked on a small sample the averages of the values of v-planes' associates. These do increase within an order of magnitude of 2 if Δw = 1 and 4 if Δw = 2. Due to the fact that w = int(ln(3)/ln (2).v)+1, the frequency of the events Δw = 1 is 2-ln(3)/ln (2) and that of the events Δw = 2 is ln(3)/ln(2)-1. The average increase of the associates' values on incrementing v is hence 2 2-ln(3)/ln [START_REF][END_REF] .4 ln(3)/ln(2)-1 = 2 ln(3)/ln(2) = 3.

More on parameter t

The study of t has allowed us to deepen the knowledge of the first element of a Pascal trihedron plane. Let us see this in a more general way. Therefore, let us give first the t-values for the first planes of the Pascal trihedron.

Conjecture 4

Let us take the divisors (in the column divisors) as follows :

-for the first set of lines (with v-2 columns): 2 v + 1 -for the second set of lines (with v-3 columns): 2 v (if v ≥ 6) -for the third set of lines (with v-4 columns): 2 v-1 (if v ≥ 7) -... -for the i-1 th set of lines (with v-i columns): 2 v -i + 3 (if v ≥ i+3) -... -for the last three sets of lines (with 5, 4 and 3 columns): 2 8

The last rule does not apply for v < 6, it does partially for v = 6 (the last column) and v = 7 (last two columns) and completely for v ≥ 8.

Then, the elements of the "ratios" column are all integers, except for one which is an integer + 1/2.

The half-integer corresponding line will be selected to take his place at the first line of the current plane (what we anticipated for our example). In each set of rows, we classify the values in the column "ratios" by increasing values (what we have also done from the beginning). This process gives a unique classification of the lines of each Pascal trihedron plane.

13 A classification for enumeration.

The objective is here to count the staffs of numbers of the Pascal trihedron planes. This has already been achieved in a previous article. However in the absence of a "geography" by signatures, we were limited to a conjecture on the said count instead of the theorem that we will establish now.

In fact, we will create a second geography. This new classification does not give Pascal trihedron planes in a compact "triangular" form, but in a form more indented on the left side. For an easy understanding, let us start by examples according to our regular practice. We choose two plans for reasons explained further, namely v = 8 (w = 13) and v = 9 (w = 15). We will not explain here again the notion of licit signatures. The tables are made up only of this kind. The principle of formation of this plane is the following :

We start from the first classification of the chosen plane and store items in the same columns as in this one.

The first line is identical to that of this first classification.

The first element of the plane is still at the top left and contains v sequences 10 at the beginning of signatures (reading from left to right) and w-v sequences 0 at the end, hence w-v+1 sequences 0 follow the most on the right 1 (called final 1 of signature subsequently).

The elements right on this line were then all w-v sequences 0 after the final 1.

The other elements are then classified according to a positive sorting of signatures with the additional constraint of a classification of the final part (which starts at the final 1), gradually decreasing the number of final 0.

P 77/96

The table of staffs obtained here is then : Each line in the previous table, offline summation, is matching with a "triangular subblock" (that we can number 1, 2, 3, ..., from the top down) of the Pascal trihedron plane v.

Each triangular subblock, so numbered, by construction, has the same number of final 0 off the first subblock whose first item has an extra 0 at the signature end (after the final 1) comparing to the other elements of the subblock.

We found previously (see ref [1] and [2]) and the relationships linking the staffing tables of rank v to those of rank v-1. Thus:

Theorem 15

Let us have #(v,i,j) an element of the staffing table (excluding the final red line). The number v is the number of the plane, the number i is the index of line starting at 1 (downwards), j is the index of the column starting at 1 (to the right).

Then we have :

The last line is doubled if ∆w prec1 (v) = 2, then incrementing the number of lines (64)

Proof :

It is recalled that any licit signature matches a number that generates it (see theorem 6) and that there is therefore exhaustivity in a list of licit signatures. Subsequently, we assimilate the #(v,i,j) count and the elements that it counts.

The first two lines of the theorem table are trivial (as a result of the construction procedure of the trihedron page 6).

Switch from one signature to another is done with a gradual shift of the 1 on the most right to the right, then the second, etc.

If there are matching signatures (bijection) on the first two subblocks to be treated, this correspondence will necessarily pass on a column until the final subblock.

Matching works on the first pair of subblocks by adopting #(v-1,1,j-1) and #(v,1,j) which means simply align the subblocks of the v-1 plane and the v plane on the right side, the first element of the plan v being thus isolated. We give the following example : The lines have been numbered for easier location. There are two types of matches, namely those of #(v-1,i,j-1) with #(v,i,j) in blue and green fonts and those of #(v,i-1,j) with #(v,i,j) in red and green fonts.

The first match is performed on the first two items (the first line of the table) 111 101010101001010100000 211 101010101001010101000000

Programming with Appendix 1

Construction of the v-plane.

The plane v = 1 includes a single item : the integer 1.

The plane v = 2 includes a single item : the integer 3. The choice of v, greater than or equal to 3, below enables to calculate the v-plan v line by line, each element being in its exact position in the columns, the empty positions being viewed by the number 0. The number of elements increasing exponentially with v, there is a rapid saturation of the system of calculation for comprehensive planes. Hence, the need for alternative code programming given afterwards.

)); i = 0; if(tabdpl == tabrech, break))); nbpl = nb-(2^w)*floor(nb/(2^w)); if(nbpl <> nb, print("Integer "nb" is represented by "nbpl" in the trihedron.")); print("The location of integer "nbpl" in the trihedron is plane "v", column "col", line "r-1".")}

Examples 2

Integer nb = 40160091701359 :

Integer 40160091701359 is represented by 111 in the trihedron. The location of integer 111 in the trihedron is plane 19, column 16, line 150386.

Integer nb = 110420130004991 :

The location of integer 110420130004991 in the trihedron is plane 30, column 10, line 5.

Appendix 3

Search for the value of an integer from its position in the Pascal trihedron.

The choice of the position (plane v, column, row) gives the modulo 2 w representative within the trihedron.

Program 3

{/* Give plan rank : v */ v = 7; /* Give column : column (<= v-1) */ column = 2; /* Give line : line (must be not void)*/ line = 2; infini = 10^10000;lm = log(3)/log(2); w = 1+floor(lm*v);alert = 0; sgorig = 0; rep = v+1-column; for(i = 1, rep, sgorig = concat(sgorig,1)); sgorig = concat(sgorig,0); for(i = 1, v-rep, sgorig = concat(sgorig,1)); for(i = 1, w-v-1, sgorig = concat(sgorig,0)); sgorig = sgorig[^1]; sgref = 0; for(t = 1,v, wp = 1+floor(lm*(t-1)); wc = 1+floor(lm*t); dww = wc-wp; if(dww == 1,sgref = concat(sgref,1), sgref = concat(sgref,1); sgref = concat(sgref,0)););sgref = sgref[^1]; invtaborig = 0;j = 1;

tabdpl[i]++; r++; if(tabdpl[column] == archive+1, if(column >=1, alert = 1;print("Wrong choice : Line position outside the trihedron."))); if(i > 1,for(j = 1, i-1, x = i-j; tabdpl

print("The element of the "v"-plane in line "line" and column "column" is equal to "nb2"."));} P 90/96

Exemple 3

The integer of plan v = 19 with coordinates (341651,16) is :

The element of the 19-plane in line 341651 and column 16 is equal to 1507847631

Integer of plane v = 14 with coordinates (3388,12) doesn't exist :

Wrong choice : Line position outside the trihedron.

Appendix 4

Getting a sequence of numbers answering successive additions of 1 at the beginning of the parity vector.

The choice of an integer aa, whose parity vector is vp, gives by the algorithm below a sequence of numbers having parity vectors [1, vp], then [1,1,vp,0], then [1,1,1,vp,0],... the number of final 0 being adjusted to the relationship linking v the number of multiplication steps to w the number of division steps, that is w = int(ln(3)/(2) .v)+1 (in altitude flight). The calculations are done without any recourse to the parity vectors (aiming a very high speed of execution). This program allows for example to recursively calculate the rest of the first elements of planes a v (1,1). It also allows to recursively calculate the series of the second elements of planes a v (1,2). Similarly, each element of a first line a v (1,i) gives an analog series a v+n (1,i). This continues throughout the second block of a Pascal trihedron plane : a v (i in second block, j) gives a v+n (i,j).

In addition, calculation gives jointly for a v (i,j) its flight altitude output value m v (i,j). The series of the 50 first elements of each plane (starting with 5) :

Integer 5 is represented by 1 in the trihedron. v m(v) a(v) Appendix 5

Search for the value of an integer from its parity vector.

The choice of licit parity vector "vectpar" gives the integer nb2 that generates it. The choice of a parity vector without precaution is often wrong. The number w of divisions by 2 and v of multiplications by 3 plus 1 are linked by w = int(ln(3)/ln(2).v)+1. Moreover, a bad choice can also come from the fact that too many divisions by 2 are selected at the beginning of the vector. A bad choice is detected (but not corrected).

Program 5

{/* Choose a valid parity vector */ vectpar = [1,0,1,1,1,0,1,1,1,0,0,0]; )-1;break)); print("The integer corresponding to the parity vector "vectpar" is "nb2".");)}

Examples 5

The integer corresponding to the parity vector

1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, Appendix 6

Getting the parity vector from the integer value.

The use of the Collatz algorithm gives the parity vector without difficulty. if(nbpl <> nb, print("Integer "nb" is represented by "nbpl" within the trihedron.")); print("Integer "nbpl" is located in plane "v" and as parity vector "vectpar".")}

Examples 6

Integer 576460752303423515 is represented by 27 within the trihedron. Integer 27 is located in plane 37 and has parity vector [1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0].

Integer 71 is located in plane 32 and as parity vector [1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0]