﻿ Schematic algebra - Segmentation

# Schematic algebra - Segmentation

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### Segments of systems

The nesting (shown in the previous paragraphs) is a matter of G-systems with constant n. Now we are going to study G-systems with constant k.

```  G(2,2)    =>    G(3,2)
0                0
1   2            1   3
3                4
6   2
7   5
8
```

### Definition of segments

Each G-system can be divided to segments in the following way. New instances will be separated from instances of G-system of lower n.

E.g. let us observe the systems G(2,2) and G(3,2). The latter is only an extension of the former.

Every G(n,k) has n classes nested from G(n,1). Let us assume that these classes divide G-system into segments.

In segment s (s<n) has every such class number:

#### g(s) = (n−s) ∙ (nk−1) / (n−1)

To any segment existing in G(n,k) there exist a similar segment in G(n+1,k). This idea becomes clearer if we rewrite all numbers u with numbers (nk-1)-u.

At the same time we renumber segments to s'=(n−s).

E.g. segments in G(3,3):

 ``` u: 0 segment 2 1 3 9 2 6 18 4 12 10 5 15 19 7 21 11 8 24 20 ────────────────── 13 segment 1 14 16 22 17 25 23 ────────────────── 26 segment 0 ``` ``` u': 0 segment 0' ────────────────── 9 1 3 12 10 4 13 segment 1' ────────────────── 18 2 6 19 5 15 21 11 7 22 14 16 24 20 8 25 23 17 26 segment 2' ```

### Numeric systems

Let us write the numbers nk-1-u from the previous table as functions of n and look at quotients of the polynomials.

Numbers nk−1−u from the previous table as functions of n:

``` k=2                      k=3
segment 0                 segment 0
1                   1
segment 1                 segment 1
n       1           n2            1        n
n+1                 n2+n          n2+1    n +1
n2+n+1
```

Let us write coefficients of these functions (polynomials):

```
k=2                      k=3
segment 0                      segment 0
0 0                       0 0 0
segment 1                      segment 1
1 0    0 1                1 0 0      0 0 1      0 1 0
1 1                       1 1 0      1 0 1      0 1 1
1 1 1
```

Each G-system is a set of numbers from the n-th numeric system.

 The G-relation separates G(n,k) into n segments, where every segment s uses always just s symbols, s=0..n-1.

Sum of coefficients of polynomials is the same as for all instances of given class (instance of the same class have the same level).

Compositions of G-systems

 ``` Example of G(n,2) G(1,2) 00 00 00 ───────────────────────── 10 01 10 01 10 01 G(2,2) 11 11 11 ───────────────────────── 20 02 20 02 20 02 21 12 21 12 21 12 G(3,2) 22 22 22 ───────────────────────── 30 03 30 03 31 13 31 13 32 23 32 23 G(4,2) 33 33 ───────────────────────── 40 04 41 14 42 24 43 34 G(5,2) 44 ``` ``` Example of G(n,3) G(1,3) 000 000 000 ──────────────────────────────────────────── 100 001 010 100 001 010 100 001 010 110 101 011 110 101 011 110 101 011 G(2,3) 111 111 111 ──────────────────────────────────────────── 200 002 020 200 002 020 201 012 120 201 012 120 210 102 021 210 102 021 211 112 121 211 112 121 220 202 022 220 202 022 221 212 122 221 212 122 G(3,3) 222 222 ──────────────────────────────────────────── 300 003 030 301 013 031 302 023 032 310 103 130 311 113 131 312 123 132 320 203 230 321 213 231 322 223 232 330 303 330 331 313 331 332 323 332 G(4,3) 333 ```

### Electron peels in atoms

Number of electrons p in peel having auxiliary quantum number nl corresponds to number of instances in nl−th segment of G-systems of order k=2. (In the column z is usual denotation of peels).

### Self classes

Number of self classes in segments is difference of numbers of self classes of two systems with adjacent bases, i.e.:

#### s(n,k) = v(n+1,k)−v(n,k)

Numbers of self classes in segments s(n,k) written as functions of variable n:

 ```k m(n,k) ──────────────────────────── 1 1 2 n 3 n∙(n+1) 4 (n/2)∙(2n²+3n+1) 5 n∙(n³+2n²+2n+1) 6 (n/6)∙(6n4+15n³+20n²+12n+1) 7 n∙(n5+3n4+5n³+5n²+3n+1) 8 (n/4)∙(4n6+14n5+28n4+35n³+26n²+11n+4) ``` ```0 1 2 3 4 5 6 7 ───────────────────────────────────────── 1 1 1 1 1 1 1 1 0 1 2 3 4 5 6 7 0 2 6 12 20 30 42 56 0 3 15 42 90 165 270 423 0 6 42 156 420 930 1806 3192 0 9 107 554 1910 5155 11809 24052 0 18 294 2028 8820 28830 77658 181944 0 30 780 7350 40590 161040 510510 1376340 ```

(Computation of s(n,4) is known as:  3= 1+2;  15= 4+5+6;   42= 9+10+11+12;   90= 16+17+18+19+20).

### All classes

Similarly as in the previous paragraph:

#### t(n,k) = m(n+1,k)−m(n,k)

Numbers of all classes in segments t(n,k) written as functions of variable n:

 ``` k t(n,k) ───────────── 1 1 2 n+1 3 n²+n+1 ``` ``` 0 1 2 3 4 5 6 7 ───────────────────────────────── 1 1 1 1 1 1 1 1 1 2 3 4 5 6 7 8 1 3 7 13 21 31 43 57 ```

### All instances

Similarly is also:

#### S(n,k) = M(n+1,k)−M(n,k) = (n+1)k−nk

 Kronecker Leopold, [] 1823-1891, German scholar and mathematician associated with the idea of arithmetization of mathematics, opponent of Cantor's concepts. He was interested in arithmetic of quadratic forms, theory of ideals and elliptic functions. Refused actual infinity. All mathematics have to be based on numbers and all the numbers on the natural numbers: "The natural numbers are from God, the other is the work of man."

Numbers of all instances in segments S(n,k) written as functions of variable n:

 ``` k S(n,k) ───────────────── 1 1 2 2n+1 3 3n²+3n+1 4 4n³+6n²+4n+1 ``` ``` 0 1 2 3 4 5 6 7 ───────────────────────────────────── 1 1 1 1 1 1 1 1 1 3 5 7 9 11 13 15 1 7 19 37 61 91 127 169 .... ```

Number S(n,k)  (called also Kronecker’s delta) – has relation to Bernoulli’s sequences (see Triangles) and to Last Fermat’s theorem.

## Periodic sequences

### Period of sequence

Let us consider sequence of numbers an. If there exist such number r, that for all n it holds an+r = an, then {an} is so called periodic sequence. Number r is period of sequence.

Least period of the given sequence is basic (natural) period.

### Members of differential sequences

Generally r-th member of s-th differential sequence

 as,r = ∑(−1)i ∙   ∙ ar+s−i

(sum for i=0..s)

In sequences of cubes is:

``` s\r  1  2  3  4  5   6  ...
────────────────────────────
0   0, 1, 8,27,64,125,...
1   1, 7,19,37,61,...
2   6,12,18,24,...
3   6, 6, 6,...
```

a2,3 = ∑(-1)i a3+2-i =  a5 -  a4 +  a3  = a5-2a4 +a3 = 64 - 2*27 + 8 = 18

In sequence of order s is s-th differential sequence constant, i.e. every number of this differential sequence equal to 0-th member:
as,0 = ∑(−1)i  as−i = Ks .

In case Kp= p! = 0 mod p (pεP) analogically it holds:

#### ai+p + (−1)p ai = 0 mod p

E.g. p-th power: (i + p)p + (−1)p ∙ ip = 0 mod p.

For p = 2: a0 = 0; a1 = 1; a2 = 4; a3 = 9; a4 = 16; a5 = 25; a2+a0= 4, a3+a1 = 10 is 0 mod 2.
For p = 3: a0 = 0; a1 = 1; a2 = 8; a3 = 27; a4 = 64; a5 = 125; a3−a0= 27, a4−a1 = 63 is 0 mod 3.

If pεP, tak it holds Wilson’s theorem: K(p)=(p−1)! ≡ −1 (mod p).

For every i = 1..p−1 is

= 0 mod p.

The first and last member rests:  a(p) + (−1)p ∙ a0 = −1 (mod p)

Because all members of p-th differential sequences equal, we can select other member, i-th (instead of 0-th):

### Natural period

Let qεQ has prime partition: q = ∏ (pjaj), pjεP, ajεZ. ¨

``` n\k 1  2  3  4  5  6  7  8  9 10 11 12
───────────────────────────────────────
0  0  0  0  0  0  0  0  0  0  0  0  0
1  0  1  1  1  1  1  1  1  1  1  1  1
2  0  0  2  0  2  4  2  0  8  4  2  4
3  0  1  0  1  3  3  3  1  0  9  3  9
4  0  0  1  0  4  4  4  0  1  6  4  4
5  0  1  2  1  0  1  5  1  8  5  5  1
6  0  0  0  0  1  0  6  0  0  6  6  0
7  0  1  1  1  2  1  0  1  1  9  7  1
8  0  0  2  0  3  4  1  0  8  4  8  4
9  0  1  0  1  4  3  2  1  0  1  9  9
10  0  0  1  0  0  4  3  0  1  0 10  4
11  0  1  2  1  1  1  4  1  8  1  0  1
12  0  0  0  0  2  0  5  0  0  4  1  0
───────────────────────────────────────
R  1  2  3  2  5  6  7  2  3 10 11  6
```

Number: R = ∏pj is called natural period.

Number of residual classes of expression nk mod k, n,kεN0, is equal to natural period R of number k.

## Natural classes

### Natural classes in G-system

Let us order instance of each class of G-system into ascending sequences. If all differences of instances {d1,d2,...} have form
B∙(nj), j=0..k−1,

we speak about natural class (N-class).
Differences {d1,d2,...} in G(n,k) are always divisible by n−1.

### Diference of instances

Construction of natural classes in G(2,4)

``` Instances          Ordering           Differences Type Nb
────────────────────────────────────────────────────────
0                 0                  x           −
1  2  4  8        1  2  4  8         1  2  4     N1
3  6 12  9        3  6  9 12         3  3  3     −
5 10        =>    5 10         =>    5           −
7 14 13 11        7 11 13 14         4  2  1     N1
15                15                  x           −
```

In system G(2,5), are all self classes natural.

``` Instances                      Differences      Type Nb
────────────────────────────────────────────────────────
0                             −                −
1   2   4   8   16            1   2   4   8    N1
5   9  10  18   20   =>       4   1   8   2    N1
11  13  21  22   26            2   8   1   4    N1
15  23  27  29   30            8   4   2   1    N1
31
```

Type natural class mark Nb, where b = B/(n−1). For n=2 is B=b=1 (examples above); for n=3, is B=2,4 a b=1,2; for n=4, is B=3,6,9 a b=1,2,3; and so on.

```  G(2,2): x       G(3,2): x        G(4,2): x
1 P             2 P              3  P
x               4 P              6  P
k=2; φ(2) = 1             x                9  P
2 P              x
x                3  P
6  P
x
3  P
x
G(2,3):          G(3,3):           G(4,3):
x               x                x           x
1  2 P          2  6 P           3 12 P      3 12 P
2  1 P          4 12 P           6 24 P      6 24 P
x               6  2 P           9 36 P     12  3 P
10  4            12  3 P     18  9
4 10            18  9        9 18
k=3; φ(3) = 2            12  4 P          21 21       24  6 P
x                9 18        x
2  6 P          24  6 P      3 12 P
6  2 P          33  6       12  3 P
x                6 33        x
21 21
36  9 P
G(2,4):          G(3,4):
x                 x              40
1  2  4 P         2   6  18 P    12   4  28
3  3  3           4  12  36 P    24  14   8
5                 8  16   8      10  30  10
4  2  1 P        10  30  10      36  12   4 P
x                14   8  34       x
16  32  16       2   6  18 P
k=4; φ(4) = 2              20               8  16   8
8  14  24      20
18   6   2 P    18   6   2 P
28   4  12       x
16  16  16
34   8  14
```

Differences of natural classes make some latin squares and similar figures:

```k=2:  1      k=3:  1 2      k=4: 1 2 4     k=5: 1 2 4 8
2 1           4 2 1          4 1 8 2
2 8 1 4
8 4 2 1
k=6:                 k=7:                      k=8:
1  2  4  8 16       1  2  4  8 16 32       1  2  4 8 16 32 64
16  8  4  2  1       8  1 16  2 32  4       4 32  1 8 64  2 16
16  4  1 32  8  2      16  2 64 8  1 32  4
2  8 32  1  4 16      64 32 16 8  4  2  1
4 32  2 16  1  8
32 16  8  4  2  1
```

### Identification of classes

For numbers N-classes g(j) v G(n,k) it holds:

#### (nj−1)∙g(j) mod (nk−1) = b∙(n−1)∙nj−1

where nsd(j, k) = 1.
Symbol b marks number typu N-classu, b=1..s.

For any N-class g from segment s  of system G(n,k) there exists:

·   contrast N-class g'; g+g'=nk-1.

·   corresponding N-class g' v G(n+1,k) (nesting)

·   corresponding N-class g' in segment s+1 (segmentation)

``` Instances  Differences  Type Nb
────────────────────────────────
0   -   -
1   3   9     2  6  N1
2   6  18     4 12  N2
4  12  10     6  2  N1
5  15  19    10  4  -
7  21  11     4 10  -
8  24  20    12  4  N2
13             -  -
14  16  22     2  6  N1
17  25  23     6  2  N1
26
```

For number of type N-class b=1, corresponding g-number is not divisible with nk−1.
E.g. in system G(2,4) classes g=1 a g=7 are relative prime to 15.

### Number of classes in G-systems

Each segment s has N-classes with number o type b=1..s.

In G(n,k) there exists (n−b) ∙ φ(k) N-classes g(j) with number of  type b.

Altogether is s∙φ(k) N-classes (i.e. k∙s∙φ(k) N-instances).

Total number N-classes v G(n,k) is:
n−1   n−1  n
∑(s∙φ(k))=φ(k) ∑(s)=φ(k) ∑(n−b)=φ(k)∙n∙(n−1)/2 N−classes
s=0   s=0  b=1

(Coefficient n(n−1)/2 reminds number of  bindings of n elements).

System G(3,3) has φ(3)∙3(3−1)/2=6 natural classes (with number of type b=1 or b=2):

### Number of classes in binary G-systems

There exist φ(k) N-classes g(j) in G(2,k).
(2j−1)∙g(j) mod (2k−1) = 2j−1 for nsd(j, k) = 1.

Sum of values g(j) (j=1..φ(k)) is:
∑(g) = 2k−2 ∙ φ(k)

E.g. for G(2,4) we get:

 ``` Instances Type Nb ────────────────────────── 0 − 1 2 4 8 N1 (j=1) 3 6 12 9 − 5 10 − 7 14 13 11 N1 (j=2) 15 − ``` ``` Solution: (mod 15) j = 1: 1∙g ≡ 1 g = 1 j = 2: 3∙g ≡ 2 g = 11 ```

Sum: 1+7= 8= 24−2 ∙ φ(4).

For G(2,5) it holds:

 ``` Instances Type Nb ─────────────────────────────── 0 − 1 2 4 8 16 N1 (j=1) 5 9 10 18 20 N1 (j=3) 11 13 21 22 26 N1 (j=2) 15 23 27 29 30 N1 (j=4) 31 − ``` ``` Solution: (mod 31) j = 1: 1g ≡ 1 g = 1 j = 2: 3g ≡ 2 g = 11 j = 3: 7g ≡ 4 g = 5 j = 4: 15g ≡ 8 g = 15 ```

Sum 1+11+5+15 = 32 = 25−2 ∙ φ(5).

Similarly for k = 7 (mod 127):
j= 1:  1∙g ≡ 1   g=  1   j= 4: 15∙g ≡  8   g=  9
j= 2:  3∙g ≡ 2   g= 43   j= 5: 31∙g ≡ 16   g= 21
j= 3:  7∙g ≡ 4   g= 55   j= 6: 63∙g ≡ 32   g= 63

Sum: 1+43+55+9+21+63 = 192 = 27−2 ∙ φ(7)

### Natural classes in M-systems

For natural classes (N-classes) g(j) it holds:

#### (nj−1)∙g(j) mod r = b∙nj−1

for nsd(j, k) = 1.

M(3,3) has all self classes natural, r = ( 33-1)/(3-1) = 26/2 = 13:

```   Instances  Differences  Type Nb
─────────────────────────────
0          -       -
1  3  9    2 6     N2
2  6  5    3 1     N1
4 12 10    6 2     N2
7  8 11    1 3     N1
13         -     -
```

M(3,4) has only 2 natural classes, r = ( 34−1)/(3−1) = 80/2 = 40:

 ``` Instances Differences Type Nb ──────────────────────────────────── 0 − − 1 3 9 27 2 6 18 N2(j=1) 2 6 18 14 4 8 4 − 4 12 36 28 8 16 8 − 5 15 10 − 7 21 23 29 14 2 6 − 8 24 32 16 8 8 8 − 10 30 20 − 11 33 19 17 6 2 14 − 13 39 37 31 18 6 2 N2(j=3) 20 − − 22 26 38 34 4 8 4 − 25 35 10 − 40 − − ``` ``` b = 1: (mod 40) j=1: (31−1)∙ g≡ 1 no solution j=2: (3²−1)∙ g≡ 3 no solution j=3: (3³−1)∙ g≡ 9 no solution b = 2: (mod 40) j=1: (31−1)∙ g≡ 2∙1 g = 1 j=2: (3²−1)∙ g≡ 2∙3 no solution j=3: (3³−1)∙ g≡ 2∙9 g = 13 ```

In M(3,5), r = (35−1)/(3−1) = 242/2 = 121 there exists for b = 1 natural classes:

 ``` Instances Type Nb ──────────────────────────── 0 − 1 3 9 27 81 − 2 6 18 54 41 − 4 12 36 108 82 − * 5 15 45 14 42 N1(j=3) 7 21 63 68 83 − 8 24 72 95 43 − 10 30 90 28 84 − 11 33 99 55 44 − 13 39 117 109 85 − 16 48 23 69 86 − 17 51 32 96 46 − 19 57 50 29 87 − * 20 60 59 56 47 N1(j=4) 22 66 77 110 88 − 25 75 104 70 89 − 26 78 113 97 49 − 31 93 37 111 91 − 34 102 64 71 92 − 35 105 73 98 52 − 38 114 100 58 53 − 40 120 118 112 94 − * 61 62 65 74 101 N1(j=1) 67 80 119 115 103 − * 76 107 79 116 106 N1(j=2) 121 − ``` ``` b = 1: (mod 121) j= 1: (31−1)∙ g ≡ 1 g = 61 j= 2: (3²−1)∙ g ≡ 3 g = 76 j= 3: (3³−1)∙ g ≡ 9 g = 5 j= 4: (34−1)∙ g ≡ 27 g = 20 ```

## Primes in R−systems

Number of primes lesser then given number r is denoted π(r). From the first Euler’s and Legendre’s estimations follows, that value π(r) grows approximately with the same speed as function r/ln(r). Complete prove of π(r) ~ r/ln(r) did J.Hadamard and Vallée−Poussin with help of methods of complex analysis. Values π(r) are tabulated, see e.g.[Narkiewicz]:

 Hadamard, Jacques , 1865-1963, French mathematician, proved (y.1896) prime theorem.
```       π(10²)  = 25
π(10³)  = 168
...
π(109)  = 50847534
```
 Vallée-Poussin, Charles-Jean de la , 1866-1962, Belgian mathematician, proved (y.1896) - independently of Hadamard - prime theorem.
```       …
...
π(1018) = 24739954287740860
π(1019) = 234057667276344607
π(1020) = 2220819602560918840
...
```
 Riemann, Georg Friedrich Bernhard , 1826-1866.

Better approximation of number of primes was made by G.Riemann.

P.L.Čebyšev (y.1852) has proved so called weak-form of prime-number theorem.
A.Selberg has proved (y.1949)  prime-number theorem without help of complex analysis.

 Dirichlet, Peter Lejeune [Dirichle], 1805-1859,

Dirichlet has proved, that any arithmetic sequence of the 1.order (a,a+d,a+2d,...) with characteristic [d,a], (a,d)=1, contains infinite number of primes.

### Primes in G-systems

We are interested if structure of G-system determines distribution of primes. Instances of classes relative to module r will contain all primes, that are not in prime partition of module r.
E.g. in G(3,3) is φ(26)=12 instances relative prime to 26:

Because 26 = 2*13, must be all primes less then 26, except 2 and 13, included among 12 instances: 1, 3, 5, 6, 7, 9, 15, 17, 19, 21, 23, 25. These are: 3,5,7,17,19 and 23.

```   0
1   3   9
2   6  18
4  12  10
5  15  19
7  21  11
8  24  20
13
14  16  22
17  25  23
26
```
```   R(17,12,210)
0
1   17   79   83  151   47  169  143  121  167  109  173
2   34  158  166   92   94  128   76   32  124    8  136
3   51   27   39   33  141   87    9  153   81  117   99
4   68  106  122  184  188   46  152   64   38   16   62
5   85  185  205  125   25
6  102   54   78   66   72  174   18   96  162   24  198
7  119  133  161
10  170  160  200   40   50
11  187   29   73  191   97  179  103   71  157  149   13
12  204  108  156  132  144  138   36  192  114   48  186
14   28   56  112
15   45  135  195  165   75
19  113   31  107  139   53   61  197  199   23  181  137
20  130  110  190   80  100
21  147  189   63
22  164   58  146  172  194  148  206  142  104   88   26
30   90   60  180  120  150
35  175
37  209  193  131  127   59  163   41   67   89   43  101
42   84  168  126
44  118  116   82  134  178   86  202   74  208  176   52
49  203   91   77
55   95  145  155  115   65
57  129   93  111  207  159  183  171  177   69  123  201
70  140
98  196  182  154
105
210
```
```Level:       1   2   3   4   5   6   7   8   9  10  11  12   * Total
──────────────────────────────────────────────────────────────────
Classes:     3   2   0   6   0   6   0   0   0   0   0  12   *   29
Instances:   3   4   0  24   0  36   0   0   0   0   0 144   *  211
```

Numbers n with higher ratio n/φ(n):

```  n      φ(n)    n/φ(n)
────────────────────
210     48     4,38
330     80     4,13
390     96     4,06
420     96     4,38
630    144     4,38
660    160     4,13
780    192     4,06
840    192     4,38
990    240     4,13
1050    240     4,38
1170    288     4,06
1260    288     4,38
```

## Sums of powers of instance numbers

### Summing in G-systems

Let us evaluate sum of numbers of instances in all classes of G(3,3):

```    0                0
1   3   9       13
2   6  18       26
4  12  10       26
5  15  19       39
7  21  11       39
8  24  20       52
13              13
14  16  22      52
17  25  23      65
26              26
```

Each sum is is divisible by number M(n,k)=(nk−1)/(n−1) = (3³−1)/(3−1) = 13.

V G(n,k), it holds for every class gi (sum through j):

### Computation of sum

It holds (v sum přes j):

#### ∑ ui (j) = L(gi) ∙ c(k,q) ∙ q/k)

where q is number of transpositions (i.e. order of initial nested system), L(gi) level of class gi and c(k,q) coefficient of nesting.

E.g. v G(3,3) it holds:

```    L(7) ∙c(3,1) ∙ 3/3 = 3∙ ((33−1)/(3−1))∙1 = 39
L(8) ∙c(3,1) ∙ 3/3 = 4∙ ((33−1)/(3−1))∙1 = 52
...
L(13)∙c(3,1) ∙ 1/3 = 3∙ ((33−1)/(3−1))/3 = 13
```
```    g│      │ L(g)  │k/q│  ∑ ui(j)
──┼──────┼───────┼───┼──────
0│ 000  │  0    │ 3 │    0
1│ 001  │  1    │ 1 │   13
2│ 002  │  2    │ 1 │   26
4│ 011  │  2    │ 1 │   26
5│ 012  │  3    │ 1 │   39
7│ 021  │  3    │ 1 │   39
8│ 022  │  4    │ 1 │   52
13│ 111  │  3    │ 3 │   13
14│ 112  │  4    │ 1 │   52
17│ 122  │  5    │ 1 │   65
26│ 222  │  6    │ 3 │   26
```

### Summing in G-systems

Let us mark se sum of all e-th powers of instances in given class. We are interested in what case r | se for all e=1..k−1. Quotient se/r is denoted by small letter, i.e. se.

In case, that se is not multiple of r, we write asterisk to this place.

```  G(2,2)    s1        G(2,3)          s1   s2
────────────         ───────────────────────
0                    0
1   2     1          1  2  4        1    3
3                    3  6  5        2   10
7

G(2,4)               s1   s2    s3
─────────────────────────────────
0
1   2   4   8        1    *    39
3   6  12   9        2   18   180
5  10                1    *    75
7  14  13  11        3    *   441
15

G(2,5)               s1   s2    s3    s4
───────────────────────────────────────
0
1   2   4   8  16    1   11   151  2255
3   6  12  24  17    2   34   668 14110
5  10  20   9  18    2   30   506 9102
7  14  28  25  19    3   65  1533 37949
11  22  13  26  21    3   61  1323 29965
15  30  29  27  23    4  104  2794 76748
31
```

### Summing in R-systems

We are interested, in what case all sums s1,s2,...,sk−1 are divisible by number r. The following systems have this property.

Let us separate systems according to Mersenne’s (M) or Fermat’s (F) systems.

M-systems:
R(2,2,3),    R(2,3,7),   R(2,5,31),   R(2,7,127),  R(3,3,13),   R(3,5,121),  R(4,2,5),   R(4,5,341),  R(5,3,31),   R(5,5,781),  R(6,2,7),    R(6,3,43),  R(7,5,2801), R(8,2,9),    R(8,3,73), R(8,5,4681), R(9,3,91),  R(10,2,11),  R(11,3,133), R(12,2,13), R(12,3,157),

F-Systems:
R(2,2,3),R(2,4,17)

All mentioned systems have prime order k or prime module r.

But only systems with prime module r does not have this property, see e.g. M-system R(9,3,91), where r = 91 = 7∙13:

```    R(9,3,91)            s1      s2
────────────────────────────────────
0
1     9    81         1      73
2    18    71         1      59
3    27    61         1      49
4    36    51         1      43
5    45    41         1      41
6    54    31         1      43
7    63    21         1      49
8    72    11         1      59
10    90    82         2     164
```

### Sums in R-systems of order 3

We are interested, for what systems of order 3 there exists sums of powers 1.a 2. degree (as e.g. v G(2,3)=R(2,3,7), M(3,3)=R(3,3,13),...). For class g we get:

```    Degree    Sum
───────────────────────────────────────────────────────
1.  g   + gn  + gn2   =  g(1+n+n²)  = g (n³−1)/(n−1)
2.  (g)² +(gn)²+(gn²)² =  g²(1+n²+n4) = g(n6−1)/(n²−1)
```

Sums of 1.degree are divisible by r:

• In G-systems G(n,3)=R(n,3,r), r=n³−1, if n = 2 (i.e.n−1=1).
• In all M-systems M(n,3)=R(n,3,r), r=(n³−1)/(n−1).

Sums of 2.degree are divisible by r:

• In G-systems G(n,3)=R(n,3,r), r=n³−1, if (n6−1)/(n²−1)/(n³−1) ε N, i.e. if (n−1) | (n²−n+1), what is possible in case n=2 only.
• In all M-systems M(n,3)=R(n,3,r), r=(n³−1)/(n−1), because ((n6−1)/(n²−1))/((n³−1)/(n−1))= (n³+1)/(n+1)=(n²−n+1) ε N.