Schematic algebra - Stratification

You're reading an English translation of the original Czech page.

Stratification means distribution of instances and classes into levels.

 Newton, Isaac

Partitioning into levels

Binomial theorem

Binomial theorem serves for computation (a+b)k. Coefficients of particular expressions as bk−s are caled binomial coefficients.

The theorem was formulated by I.Newton, computation of powers (a+b)k was known already to Italian mathematicians.

Powers of (a+b)k make system G(2,k) in the following sense. Nesting is marked by brackeds [].
(a + b)1= a + b
(a + b)²=[a]²+2ab +[b]2
(a + b)³=[a]³+3a²b+3ab²+[b]3
(a + b)4=[a]4+4a³b+4a²b²+2[ab]²+ 4ab³+ [b]4
(a + b)5=[a]5+5a4b+10a³b²+10a²b³+5ab4+ [b]5
(a + b)6=[a]6+6a5b+12a4b²+3[a²b]²+18a³b³+2[ab]³+12a²b4+3[ab²]²+ 6ab5+[b]6

Because product a∙b needn’t to be commutative, expression a² ∙ b² is not generally equal to [a∙b]² (similarly a4 ∙ b² not corresponds to [a² ∙ b]², ...). In such (non-commutative) case is also ( a + b )² = [a]²+ ab + ba + [b] ²... Leibnitz has derived a similar formula for n-th derivation of product (Leibnitz’s formula).

Binomial coefficients are used also for computation of differential sequences.

```        Powers (Newton)  Derivation (Leibnitz)  Difference
(a±b)k           (uv)(k)                u(k)(n)
───────────────────────────────────────────────────────────
k = 1   a±b              u'v+uv'                ut+1− ut
k = 2   a²±2ab+b²        u''v+2u'v'+uv''        ut+2− 2ut+1 + ut
```

Stratification in binary system

```    i gi  Binary schema  Distance schema   Level of class
──────────────────────────────────────────────────────────────
1    0        0000                             (0)        0
2    1        0001   0010   0100   1000        (4)        1
3    3        0011   0110   1100   1001        1(3)       2
4    5        0101   1010                      2(2)       2
5    7        0111   1110   1101   1011        1 1(2)     3
6   15        1111                             1 1 1 (1)  4
```
```    Level              0    1    2    3    4
────────────────────────────────────────────
All instances      1    4    6    4    1
Nested instances   1    0    2    0    1
Self instances     0    4    4    4    0
────────────────────────────────────────────
All classes        1    1    2    1    1
Nested classes     1    0    1    0    1
Self classes       0    1    1    1    0
```

Binomial coefficients determine distribution of all instances into particular levels.

We will be interested also for distribution of self and nested instances, resp. distribution of classes.

Let us compute how many instances a classes in system G(2,4) is on each level.

Nesting in layers

Binomial coefficients of powers (a+b)k arise from self instances and nested instances of powers (a+b)d ,d|k, according to schema:

```   k=1:    {1} + {1}
x     x
1     1

k=2:  1(1) + {2} + 1(1)
x
1
k=3:      1(1)+ {3} + {3} + 1(1)
x     x
1     1
2(2)   nested instances
+
k=4:  1(1)+ {4} + {4} + {4} + 1(1)  self instances
x     x     x
1     1     1       self classes
k=5:      1(1)+ {5} + {10} + {10} +  {5} + 1(1)
x      x      x      x
1      1      1      1
3(3)  2(2)  3(3)
+     +     +
k=6:    1(1)+ {6} + {6} + {6} + {6} + {6} +  1(1)
x     x     x     x     x
1     2     3     2     1
k=7:  1(1)+ {7} + {7} + {7} + {7} + {7} + {7} +  1(1)
x     x     x     x     x     x
1     3     5     5     3     1
2(2)
+
4(4)        4(4)        4(4)
+           +           +
k=8: 1(1)+ {8} + {8} + {8} + {8} + {8} + {8} + {8} + 1(1)
x     x     x     x     x     x     x
1     3     7     8     7     3     1
```

We observe nesting of systems G(2,d) into G(2,k ), d|k, on particular levels. Schema of all instances (Pascal’s triangle) is sum of self instances d(d), where (d) marks number of self classes in system G(2,d).

Extension for n>2

Total number of instances for G(2,k) makes Pascal’s triangle, generally for G(n,k) arithmetical triangles with base n.

``` G(2,k) k/L            0   1   2   3   4
──────────────────────────────────────
n=2    0           1
1         1   1
2       1   2   1
3     1   3   3   1
4   1   4   6   4   1
```

Numbers in Pascal’s triangle are called binomial coefficients. They determine number of ways from the root of the tree into given apex.

``` k/L                         0   1   2   3  ...
──────────────────────────────────────────
n=3  0                 1
1             1   1   1
2         1   2   3   2   1
3     1   3   6   7   6   3  1
4  1  4  10  16  19  16  10  4  1
```

The same holds also generally in arithmetical triangles.

For n=3 each apex of the tree ramify into three directions.

Triangle G(3,k) determine number of possible ways of chess king from edge of chessboard to next fields [Vilenkin].

Similarly self instance split into levels:

For n=2: (a+b)^k
```k/L     0  1  2  3  4
──────────────────────────────
n=2 0   0
1   1   1
2   0   2   0
3   0   3   3   0
4   0   4   4   4   0
```

Numbers of nested instances are differences of numbers of all instances and numbers of self instances...

For n=3: (a+b+c)^k
```      k/L                   0   1   2   3 ..
───────────────────────────────────────
n=3  0                   0
1               1   1   1
2           0   2   2   2  0                        1  1  1
3        0  3   6   6   6  3  0                  1  2  2  2  1
4     0  4  8  16  16  16  8  4  0            1  2  4  4  4  2  1
5  0  5 15 30  45  50  45 30 15  5 0       1  3  6  9 10  9  6  3  1
```

There are 19 classes of triads in 12-tone musical system - see the following distance schemas:
11(10) 12(9) 13(8) 14(7) 15(6)
21(9) 22(8) 23(7) 24(6) 25(5)
31(8) 32(7) 33(6) 34(5)
41(7) 42(6) 43(5) 44(4)
51(6)

E.g. schema 21(9), makes 3-sound with whole tone and half tone (e.g. d-e-f), schema 34(5) makes minor triad (e.g. a-c-e), and schema43(5) major triad (e.g. c-e-g).

From these 19 classes, each of 18 classes have 12 transpositions. Only one class with schema 44(4) i.e. augmented triad (e.g. c-e-g#) is nested. This class has 4 instances. Therefore there is (in 12-tone system) 18∙12 = 216 self instances and 1∙4 nested instances, i.e. together 220 instances of triads and these triads makes 18 (self) + 1 (nested) = 19 classes.

Stratification instances and classes

Total number of nested instances as well as nested classes in system with prime k is equal to 2.

Number of self instances is always k-multiple of number of self classes:

V=k∙v

Method of nesting triads is depicted in the column (3) of the following table:

``` k (0)(1)(2)(3) (4) (5) (6) (7) (8) (9)(10)(11)(12)
─────────────────────────────────────────────────────────────────
12  1 12 66 220 495 792 924 792 495 220 66  12 1 všechny instance (M)
─────────────────────────────────────────────────────────────────
1  1                                          1
2  0                      2                   0
3  0              3               3           0 vnořené instance (W)
4  0          4           4           4       0
6  0      6      12      18      12      6    0
────────────────────────────────────────────────────────────────
12  0 12 60 216 480 792 900 792 480 216 60  12 0 vlastní instance (V)
- − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −
12  0  1  5  18  40  66  75  66  40  18  5   1 0  vlastní druhy (v)
─────────────────────────────────────────────────────────────────
1   1                                           1
2   0                      1                    0
3   0              1               1            0  vnořené druhy (w)
4   0          1           1           1        0
6   0      1       2       3       2      1     0
─────────────────────────────────────────────────────────────────
12  1  1  6  19  43  66  80  66  43  19  6   1  1  všechny druhy (m)
```

We get number of all instances (M) by return from numbers of self instances (V):

```     M(12,0) =  V(12,0) + V(12,1)  = 0 + 1 = 1
M(12,1) =  V(12,1) = 12
M(12,2) =  V(12,2) + V(6,1)  = 60 + 6 = 66
M(12,3) =  V(12,3) + V(4,1)  =216 + 4 = 220
```

Triangles

Triangles of numbers, that we will observe, arise from numbers of instances and classes in particular systems G(2,k). Triangles are symmetric, so it is enough to write half of them.

First column displays order k of given G-system, second column total number of elements (instances, classes). In the next columns are numbers of elements for particular levels L=0..κ(k).

Triangle of binomial coefficients (Pascal’s triangle) was known several thousand years ago, it was found in old Chinese records.

Instance triangles

```  k [M(2,k)] M(2,k,L)
0 [     1] 1
1 [     2] 1
2 [     4] 1  2
3 [     8] 1  3
4 [    16] 1  4   6
5 [    32] 1  5  10
6 [    64] 1  6  15  20
7 [   128] 1  7  21  35
8 [   256] 1  8  28  56  70
9 [   512] 1  9  36  84  126
10 [  1024] 1 10  45 120  210   252
11 [  2048] 1 11  55 165  330   462
12 [  4096] 1 12  66 220  495   792   924
13 [  8192] 1 13  78 286  715  1287  1716
14 [ 16384] 1 14  91 364 1001  2002  3003  3432
15 [ 32768] 1 15 105 455 1365  3003  5005  6435
16 [ 65536] 1 16 120 560 1820  4368  8008 11440 12870
17 [131072] 1 17 136 680 2380  6188 12376 19448 24310
18 [262144] 1 18 153 816 3060  8568 18564 31824 43758 48620
19 [524288] 1 19 171 969 3876 11628 27132 50388 75582 92378
```
```  k [W(2,k)] W(2,k,L)
0 [     1] 1
1 [     0] 0
2 [     2] 1 0
3 [     2] 1 0
4 [     4] 1 0 2
5 [     2] 1 0 0
6 [    10] 1 0 3 2
7 [     2] 1 0 0 0
8 [    16] 1 0 4 0  6
9 [     8] 1 0 0 3  0
10 [    44] 1 0 5 0  0 32
11 [     2] 1 0 0 0  0  0
12 [    76] 1 0 6 4 15  0 24
13 [     2] 1 0 0 0  0  0  0
```
```  k [V(2,k)] V(2,k,L)
0 [     1] 0
1 [     2] 1
2 [     2] 0  2
3 [     6] 0  3
4 [    12] 0  4   4
5 [    30] 0  5  10
6 [    54] 0  6  12  18
7 [   126] 0  7  21  35
8 [   240] 0  8  24  56   64
9 [   504] 0  9  36  81  126
10 [   990] 0 10  40 120  200   250
11 [  2046] 0 11  55 165  330   462
12 [  4020] 0 12  60 216  480   792   900
13 [  8190] 0 13  78 286  715  1287  1716
```

Class triangles

```  k [m(2,k)] m(2,k,L)
0 [    1] 1
1 [    2] 1
2 [    3] 1 1
3 [    4] 1 1
4 [    6] 1 1 2
5 [    8] 1 1 2
6 [   14] 1 1 3  4
7 [   20] 1 1 3  5
8 [   36] 1 1 4  7 10
9 [   60] 1 1 4 10 14
10 [  108] 1 1 5 12 21 28
11 [  188] 1 1 5 15 30 42
12 [  352] 1 1 6 19 43 66  80
13 [  632] 1 1 6 22 55 99 132
```
```  k [w(2,k)] w(2,k,L)
0 [    1] 1
1 [    0] 0
2 [    2] 1 0
3 [    2] 1 0
4 [    3] 1 0 1
5 [    2] 1 0 0
6 [    5] 1 0 1 1
7 [    2] 1 0 0 0
8 [    6] 1 0 1 0 2
9 [    4] 1 0 0 1 0
10 [    9] 1 0 1 0 2 1
11 [    2] 1 0 0 0 0 0
12 [   17] 1 0 1 1 3 0 5
13 [    2] 1 0 0 0 0 0 0
```
```  k [v(2,k)] v(2,k,L)
0 [    0] 0
1 [    2] 1
2 [    1] 0 1
3 [    2] 0 1
4 [    3] 0 1 1
5 [    6] 0 1 2
6 [    9] 0 1 2  3
7 [   18] 0 1 3  5
8 [   30] 0 1 3  7   8
9 [   56] 0 1 4  9  14
10 [   99] 0 1 4 12  20  25
11 [  186] 0 1 5 15  30  42
12 [  335] 0 1 5 18  40  66  75
13 [  630] 0 1 6 22  55  99 132
```

Triangle of self classes is analysed in separate chapter.

Substitution

Triangles can be mathematically contracted to half size – by substitution. Let us set (a+b)k a=x and b=1/x and yj = xj+ 1/xj, so y1 = x+1/x, y2 = x²+1/x²... Coefficients of new expressions cover just half of coefficients from Pascal’s triangle:
y1 = y1
y1² = (x+1/x)² = x²+2+1/x² = y2+ 2
y1³ = (x+1/x)³ = x³+3(x+1/x)+1/x³ = y3+ 3y1
y14 = (x+1/x)³ = .... = y4+ 4y2+ 6

Goniometrical functions

Let us compute expression (c−s)5 according to binomial theorem: The fifth row of Pascal’s triangle is
1 5 10 10 5 1

And therefore
(c−s)5 = c5 −5c4s +10c³s²−10c²s³+5cs4 −s5

Now let us mark s=sin(x), c= cos(x), t = tan(x). It holds:
cos(5x) = c5 −10c³s² +5cs4
sin(5x) = 5c4s −10c²s³ +s5
tan(5x) = (t5 +10t³ +5t)/(1−10t²+5t4)

Congruence of triangles

If we substitute numbers in triangle T by their remainders according to certain module r, we get new triangle T(r).

Let us observe properties of these triangles.

Triangle of all instances

Pascal’s triangle according to module 2 has fractal structure and corresponds to so called Sierpinski’s triangle. Binary numbers in particular rows makes products of odd Fermat’s numbers {3,5,15,17,51,85,255,257,...} (see Gauss' polygons).

```  T(2)                        k L     B  Partition  Indexes
─────────────────────────────────────────────────────────────
1                0 1     1  1   −
1 1               1 2     3  3   0
1 0 1              2 2     5  5   1
1 1 1 1             3 4    15  3∙5       0,1
1 0 0 0 1            4 2    17  17  2
1 1 0 0 1 1           5 4    51  3∙17      0,2
1 0 1 0 1 0 1          6 4    85  5∙17      1,2
1 1 1 1 1 1 1 1         7 8   255  3∙5∙17    0,1,2
1 0 0 0 0 0 0 0 1        8 2   257  257       3
1 1 0 0 0 0 0 0 1 1       9 4   771  3∙257     0,3
1 0 1 0 0 0 0 0 1 0 1     10 4  1285  3∙257     1,3
1 1 1 1 0 0 0 0 1 1 1 1    11 8  3855  3∙5∙257   0,1,3
1 0 0 0 1 0 0 0 1 0 0 0 1   12 4  4369  17∙257    2,3
1 1 0 0 1 1 0 0 1 1 0 0 1 1  13 8 13107  3∙17∙257  0,2,3
```

Level binary numbers (i.e. number of digits 1) in particular rows is given by expression L = 2m, where m is number of primes in partition.
For k=2t numbers gets form 2(2t)+1 (see F-systems).

```     T(2)             k L  Polynomial
───────────────────────────────────────────────────────────
1            0 1  1
1 1           1 2  x+1
1 0 1          2 2  x²+1
1 1 1 1         3 4  x³+x²+x+1
1 0 0 0 1        4 2  x4+1
1 1 0 0 1 1       5 4  x5+x4+x+1 =(x4+1)∙(x+1)
1 0 1 0 1 0 1      6 4  x6+x4+x²+1 =(x4+1)∙(x²+1)
1 1 1 1 1 1 1 1     7 8  x7+x6+x5+x4+x³+x²+x+1
1 0 0 0 0 0 0 0 1    8 2  x8+1
1 1 0 0 0 0 0 0 1 1   9 4  x&sup9;+x8+x+1 =(x8+1)∙(x+1)
```
For k=3 is: x³+x²+x+1=(x²+1)∙(x+1)= (x4−1)/(x−1), for k=7: x7+x6+x5+x4+x³+x²+x+1=(x+1)∙(x²+1)∙(x4+1)=(x8−1)/(x−1).

Derived triangles

Triangle of self classes arise by division of triangle of self instances. Odd number has always odd divisors only and for prime k triangles of self and of all elements (instances, classes) (excluding marginal instances) coincide. Therefore for odd prime k distribution of even and odd numbers in mentioned triangles is the same as in Pascal’s triangle.

Triangle of self instances has for odd k the same structure as triangle of self classes. Numbers B in rows of triangle of self instances are for even k all zero.

Triangle of self classes

```    T(2)                      k  L    B  Partition Correction
────────────────────────────────────────────────────────
0                         0  0    0   0
1 1                       1  2    3   3
1                         2  1    1   1
1 1                       3  2    3   3
1 1 1                     4  3    7   7
1 0 0 1                   5  2    9   3²
1 0 1 0 1                 6  3   21   3∙7       3∙7
1 1 1 1 1 1               7  6   63   3²∙7      3∙3∙7
1 1 1 0 1 1 1             8  6  119   7∙17
1 0 1 0 0 1 0 1           9  4  165   3∙5∙11    3∙5∙11
1 0 0 0 1 0 0 0 1        10  3  273   3∙7∙13    3∙7∙13
1 1 1 0 0 0 0 1 1 1      11  6  903   3∙7∙43    21∙43
1 1 0 0 0 1 0 0 0 1 1    12  5 1571   1571
1 0 0 1 1 0 0 1 1 0 0 1  13  6 2457   3³∙7∙13   7∙13∙27
```

There are pairs of numbers a,b that keep relation b=2a±1 in the last column. They exist for all prime k<=13, does it hold also for some higher values of k?

Triangle of all classes

```             T(2)                k   L     B  Partition
─────────────────────────────────────────────────────
1                 0   1     1   1
1 1                1   2     3   3
1 1 1               2   3     7   7
1 1 1 1              3   4    15   3∙5
1 1 0 1 1             4   4    27   3³
1 1 0 0 1 1            5   4    51   3∙17
1 1 1 0 1 1 1           6   6   119   7∙17
1 1 1 1 1 1 1 1          7   8   255   3∙5∙17
1 1 0 1 0 1 0 1 1         8   6   427   7∙61
1 1 0 0 0 0 0 0 1 1        9   4   771   3∙257
1 1 1 0 1 0 1 0 1 1 1      10   8  1879   1879
1 1 1 1 0 0 0 0 1 1 1 1     11   8  3855   3∙5∙257
1 1 0 1 1 0 0 0 1 1 0 1 1    12   8  6939   3∙2313
1 1 0 0 1 1 0 0 1 1 0 0 1 1   13   8 13107   3∙17∙257
```

Rows in triangle of nested instances and also in triangle of nested classes are for prime k of form xk+1.

In triangle of nested instances have the same form also for k=2t (as well as in Pascal’s triangle).

Zeroising of triangles

Clearing sequences

Let us mark numbers in k−th row of triangle c(k,s) and let us look for such sequence of numbers d(s), i.e.{d0,d1,d2,...}, to keep:

∑c(k,s)∙d(s) = 0 (for s=0,..,k)

```  d0+d1= 1-1= 0
d0+2d1+d2= 1-2+1= 0
d0+3d1+3d2+d3= 1-3+3-1= 0
```

Let us constrain to sequence with d0=1. To nullify Pascal’s triangle

sequence {1,−1,1,−1,...} is needed:

```  d0+d1=1-1= 0
d0+2d1+d2=1-2+1= 0
d0+2d1+2d2+d3=1-2+2-1= 0
d0+3d1+4d2+3d3+d4=1-3+4-3+1= 0
```

This sequence is nullified also by triangle

of all classes without marginal members:

``` d0+d1= 1 -1= 0
d0+d1+d2= 1-1+0= 0
d0+2d1+2d2+d3= 1-2+0+1= 0
d0+2d1+3d2+2d3+d4= 1-2+0+2-1= 0
```

Triangle of self classes is nullified by sequence

{1,−1,0,1,−1,0,...}:

Stirling’s triangle of 1.druhu is nullified by unitary sequence {1,1,1,1,...}, Stirling’s triangle of 2.druhu by alternating sequence of factorials {1,−1,2,−6,24,−120,...}, see Stirling’s triangles.

Bernoulli’s numbers

Now let us make the same without the last member in each row of triangle of self classes. We will sum:

∑c(k,s)∙d(s) = 0 (for s=0,..,k−1)

(We could also skip the first member -instead of the last- and consider expression:

(b+1)k−bk

i.e. so called Kronecker’s δ (delta)).
In Pascal’s triangle we get system of equations:
``` d0+2d1 = 0
d0+3d1+3d2 = 0
d0+4d1+6d2+4d3 = 0
d0+5d1+10d2+10d3+5d4 = 0
†裰ۅ梄B﷩裰ۅ쉸mŏ2š裰ۅҰ```
裰ۅ쉸m챈ŏ~ ɤ�ۅҰItem23裰ۅ쉸m챈ŏ{ɚ裰ۅҰ캑ۅҰItem8…�ۅҰItem22�ۅҰItem21†�ۅҰItem20†�ۅҰItem19ㅲ�ۅҰItem18†�ۅҰItem17†�ۅҰItem16†�ۅҰ

Thereof (after substitution d0=1) we get sequence {1,−1/2,1/6,0,−1/30,0,1/42,0,−1/30,...}. Members of this sequence are so called Bernoulli’s numbers. All Bernouli’s numbers with odd indexes except B1=−1/2 are zero.

Numbers was introduced by Jacob Bernoulli in the formula for sum of integer powers. They are used for summing of progressions (e.g. sum of Lerch’s progressions ∑nk∙f(n,k), where f(n,k) are Fermat’s coefficients), in theory of differential sequences, in Kummer’s theory of regular primes, in mathematical analysis,...
In progression x/(1−e−x) Bernouli’s numbers are the coefficients at xn/n!:

``` d0+d1 = 0
d0+2d1+2d2 = 0
d0+2d1+3d2+2d3 = 0
d0+3d1+5d2+5d3+3d4 = 0
d0+3d1+7d2+8d3+7d4+3d5 = 0
```
x/(1−e−x) = 1 +(1/2)x+(1/6)x²/2!−(1/30)x4/4!+(1/42)x6/6!−...

Analogy of Bernoulli‘s numbers

From triangle of self classes we get equation:

Z nich (for d0=1) we get sequence

{1,−1,1/2,−1/4,+1/4,−5/12,..}.

```  d0+2d1 = 0
d0+2d1+2d2 = 0
d0+3d1+4d2+3d3 = 0
d0+3d1+5d2+5d3+3d4 = 0
```
From triangle of all classes (without marginal members) is:

Thereof sequence {1,−1/2,1/6,−1/9,...}.

Möbius‘s function

 Möbius, August Ferdinand [], 1790-1868

So called Möbius' function μ(k) was defined by A.F.Möbius (1832), but before him the same function was used by C.F.Gauss (1801) . Gauss has proved that sum of primitive roots for given module depends on this function and similar computation used also later.

Parity of prime factors

Definition of Möbius' function is simple, but it is is a bit peculiar and difficult to understand.

Value μ(k) depends on number of prime factors of number k, it determine their parity..

• In case of even numbers μ(k)=+1, in opposite case μ(k)=−1.

E.g. μ(15)=μ(3∙5) = +1, μ(30)=μ(2∙3∙5) = −1.

• For number 1 is defined μ(1)=1. All primes p have odd parity, i.e. μ(p)=−1.
• If n is divisible by some square (i.e. by second power, excluding number 1) then μ(n)=0.

μ(k)=−1: 2,3,5,7,11,13,17,19,23,29,30,31,37,41,42,43,47,...

μ(k)= 0: 4,8,9,12,16,18,20,24,25,27,28,32,36,40,44,45,48,49,50,...

μ(k)=+1: 1,6,10,14,15,21,22,26,33,34,35,38,39,46,...

Sum of primitive roots

Sum of primitive roots according to module r is determined by Möbius' function μ(r−1):

```  r   Primitive roots        Sum  mod r  Partition of r−1   μ(r−1)
────────────────────────────────────────────────────────────────
11  2,6,7, 8                  23    1     10 = 2∙5         1
13  2,6,7,11                  26    0     12 = 3∙2²        0
31  3,11,12,13,17,21,22,24   123   −1     30 = 2∙3∙5      −1
```

Sums through divisors

The following sums go through all divisors d of number k, i.e. d|k (k>1).

Sum Möbius' functions of divisors is zero:

∑(μ(d)) = 0

For number of divisors τ(d) it holds:

∑(μ(d)∙τ(k/d))=∑(μ(k/d)∙τ(d)) = 1

For sum of divisors σ(d) it holds:

∑(μ(d)∙σ(k/d))=∑(μ(k/d)∙σ(d)) = k

Euler’s function is determined by relation:

φ(k)=∑(μ(d)∙k/d)=∑(μ(k/d)∙d)

E.g. for k=12 (φ(12) = 4):

``` d   μ(d) k/d  μ(d)∙k/d   Děl.k/d      τ(k/d) μ(d)∙τ(k/d)  σ(k/d)  μ(d)∙σ(k/d)
──────────────────────────────────────────────────────────────────────────────
1    1    12    12       1,2,3,4,6,12   6        6         28        28
2   −1     6    −6       1,2,3,6        4       −4         12       −12
3   −1     4    −4       1,2,4          3       −3          7        −7
4    0     3     0       1,3            2        0          4         0
6    1     2     2       1,2            2        2          3         3
12    0     1     0       1              1        0          1         0
─────────────────────────────────────────────────────────────────────────────
x   ∑0     x     4                      x        1          x        12
```
 Landau, Edmund [], 1877-1938

Mangoldt‘s relation

E.Landau has proved:

∑(μ(n)/n) = 0

for n=1..∞

I.e.: 1−1/2−1/3−1/5+1/6−1/7+1/10−1/11−1/13+1/14+1/15−1/17−1/19+... = 0

Binomial self instances

Let us consider case k=6 of binomial equation xk−1=0 (see). System M6(x) contains systems M2(x) and M3(x), and these contains M1(x).

Let us write in short composition of systems. We will compare composition of expression to values of Möbius' function for particular numbers d|k.

```   d  k/d  μ(d)
──────────────────────────────
1   6    1    M1 = V1
2   3   −1    M2 = V1∙V2
3   2   −1    M3 = V1∙V3
6   1    1    M6 = V1∙V2∙V3∙V6
```

Let us compute V6; expressions Mk/d for that μ(d)=1 are in numerator, expressions with μ(d)=−1 are in denominator: V6 = M6/(V1∙V2∙V3) = M6∙M1/(M2∙M3)

Generally it holds:

Aplication in G-systems

Number of self classes

To determine number self classes Gauss uvádí the following rule [Gauss II]:

 Now it is easy to derive from this theorem meaning of expression (n);... If n= aa bb cc..., where a,b,c,... are distinct primes, then n(n)=pn − ∑pn/a+ ∑pn/ab−∑pn/abc+..., where ∑pn/abc.. denote set of all expressions of type pn/abc.., where quantities a,b,c,... anyhow rearrange. So e.g. for n=36: 36(36)=p36−p18−p12+p6

Let us clear this example. Number 36 has partition into product of primes 36 = 2²∙3². So particular members of expression for 36∙v(p,36) have exponents:

36 = 36, 36/2 = 18, 36/3 = 12 a 36/2/3 = 6

Let us determine number of self classes v(n,36) according to Gauss' rule from relation:

v(n,36) = (n36−n18−n12+n6)/36

Let us note, that the sign is determined by values of Möbius' function for particular divisors of numbers 36.

Stratification of instances and classes

Numbers in table of stratification of instances and classes can be derived with help of Möbius' function [Read].

From nk instances we get value V(2,12) by count out of nested instances, i.e. these, that have d transpositions, where d|k (d≤k):

V(n,k) = ∑(μ(k/d)∙nd), for d|k

For n=2, k=12:

```   d   k/d   nd  μ(k/d)  μ(k/d)∙nd  M(n,k)−V(n,d)
────────────────────────────────────────────────────────
12    1     4096      1   4096     4096
6    2       64     −1    −64     −(6+12+18+12+6)=−54
4    3       16     −1    −16     −(4+4+4)=−12
3    4        8      0      0     −(3+3)=−6
2    6        4     +1     +4     −2
1   12        2      0      0     −(1+1)=−2
────────────────────────────────────────────────────────
4020     4020
```

It holds V(2,12) = 4020 = 12∙335=12∙v(12,2), i.e. relation gives correct number of self instances V(n,k).

But counts out (e.g.−64,−16) resp. counts in (+4) of instances does not express numbers of nested instances from particular G(n,d).

Gauss‘s theorem

Because number of self instances must be always divisible by order k, it holds:

V(n,k) ≡ 0 (mod k)

Therefore for n,kεN it holds [Narkiewicz] (sum for d|k):

∑(μ(k/d) nd) ≡ 0 (mod k)

The relation was proved by Gauss (1846) for kεP, demonstration for kεN was completed later by J.A.Serret.

E.g. for n=2

for k=3: 21μ(3)+ 2³μ(1) = 2∙(−1)+8 = 6 ≡ 0 (mod 3)

for k=9: 21μ(9)+ 2³μ(3)+ 2&sup9; μ(1)= 2∙0 +8∙(−1)+ 512= 504 ≡ 0 (mod 9)

for k=6: 21μ(6)+ 2²μ(3)+ 2³ μ(2) + 26 μ(1)= 2∙1 +4∙(−1)+ 8∙(−1) + 64 = 54 ≡ 0 (mod 6)

Sternek’s relation

Let F(n) = ∏f(n/d)μ(d). Then each prime, that divide F(n), divide also f(n), but not divide f(i) for i=1,2,...,n−1 (R.D.von Sternek,1896).

 Dedekind, Richard [], 1831-1916,

Law of inversion of numeric functions

The following two relations are equivalent: (Dedekind (1857), Liouville (1857)),[Vinogradov]

∑ f(d) = g(k), for d|k, k≥1

∑ μ(d) g(k/d) = f(k) for d|k, k≥1

Modifiction of binomial theorem

Differences of the form u'(t) = ut+1− ut can in some cases appear to be unnatural. We will try to show, that it is more suitable to use the form

u'(t) = ut+1−a∙ut

where a is some number.

Powers of algebraic numbers

Let us consider powers of algebraic numbers (a+b√d)t = vt+ ut√d. We begin with cases a=b=1 for d=2 and d=3:

```  (1+√2)1 =  1 +  1√2  (1+√3)1 =  1 +  1√3
(1+√2)2 =  3 +  2√2  (1+√3)2 =  4 +  2√3
(1+√2)3 =  7 +  5√2  (1+√3)3 = 10 +  6√3
(1+√2)4 = 17 + 12√2  (1+√3)4 = 28 +  16√3
(1+√2)5 = 41 + 29√2  (1+√3)5 = 76 +  44√3
(1+√2)6 = 99 + 70√2  (1+√3)6 =208 + 120√3
```

Numbers vt makes here the first differential sequence of numbers ut:

```Coefficients (1+√2)k                Coefficients (1+√3)k
─────────────────────────────────────────────────────────
1  2  5  12  29  70 169  s(0,t)=ut  1 2  6  16  44  120 328..
1  3  7   17  41  99 . s(1,t)=vt    1 4  10  28  76  208 .
2  4  10   24  58 .. s(2,t)        3  6   18  48 132 ..
2  6  14   34  ... s(3,t)          3  12  30  84 ..
4  8   20    ... s(4,t)            9   18  54 ..
4  12  ...      s(5,t)              9   36 ...
8   ..         s(6,t)                27 ...
```

Pisot‘s numbers

Real numbers that powered to high exponent approach natural numbers are called Pisot’s numbers.

E.g. number [(√ 5+1)/2]99 has dot followed by 20 zeros, number [(√ 5+1)/2]100 has dot followed by 20 digits 9.

Let us notice sequence of ratios in the previous schemas:
{1/1,3/2,7/5,17/12,41/29,99/70,239/169,... }. Values of fractions approach value √2.
Therefore 5∙√2=7.07107, 29∙√2 = 41.01219, 169∙√2 = 239.00209. It is why powers of Pisot’s algebraic numbers go near to natural numbers.

Differential sequences

Let us mark h-th differential sequence s(h,t). Every second row is 2-times multiple of original row for d=2, e.g. {2,6,14,34,..} = 2∙ {1,3,7,17,..}, 3-times multiple for d=3, e.g. {3,12,30,84,..} = 3∙ {1,4,10,28,..} and so on. In case a=b=1 it holds:

s(h+2,t) = d ∙ s(h,t)

For negative d are sequences less regural, but relation of differential sequences keeps:

```Coefficients  (1+√−2)k
──────────────────────────────────────────────────
+1  +2  +1  −4  −11 −10  +13   +56  ... s(0,t)=ut
+1  −1  −5  −7  +1    +23  +43 ...    s(1,t)=vt
−2  −4  −2  +8   +22  +20 ...       s(2,t)
−2  +2  +10 +14  −2 ...           s(3,t)
4   8   4  −16  ...            s(4,t)
4  −4  −20 ...               s(5,t)
−8 −16   ...                s(6,t)
```

Recurrent relations

Let use rewrite relation s(h+2,t) = d ∙ s(h,t) to form of differential equation u''(t) − u(t)∙d = 0.

If we define:

u'(t) = ut+1−ut
u''(t) = ut+2−2ut+1+ ut

we get:

u''(t) − u(t)∙d = 0
ut+2−2ut+1+ ut− ut∙d = 0
ut+2 = 2ut+1+(d−1)ut

resp.

ut = 2ut−1+(d−1)∙ut−2

In case a=b=1 it is not necessary to compute powers of algebraic numbers. Coefficients ut follows from given recurrent relations, coefficients vt are their first differential sequences.
E.g. for d=2, ut = 2ut−1+ ut−2, i.e. ut = {1, 2, 2∙2+1=5, 2∙5+2=12, 2∙12+5 = 29, ...}

General case of powers

In general case of powers (a+b√d)t the previous relations fail. We have to change computation of differential sequences.

E.g. in cases a=2 b=1 for d=2 a d=3 we get:

```(2+√2)1 =  2 +  1√2          (2+√3)1 =  2 + 1√3
(2+√2)2 =  6 +  4√2          (2+√3)2 =  7 + 4√3
(2+√2)3 = 20 + 14√2          (2+√3)3 = 26 + 15√3
(2+√2)4 = 68 + 48√2          (2+√3)4 = 97 + 56√3
(2+√2)5 =232 +164√2          (2+√3)5 =362 +209√3
```

Sequence vt does not make differential sequence of ut and we can’t get relations analogical to relations in the previous paragraphs.

Let us correct definition of differential sequences:

u'(t) = ut+1−a∙ut

u''(t)= (ut+2−a∙ut+1)− (ut+1−a∙ut)= ut+2−(a+1)ut+1+ a∙ut

We get:

``` Coefficients (2+√2)k  Coefficients (2+√3)k
───────────────────────────────────────────────────────────
1  4  14  48  164 ...    s(0,t)=ut   1  4  15  56  209 ...
2  6   20  68  ...      s(1,t)=vt     2  7  26  97 ...
2  8   28   ...       s(2,t)         3  12 45 ...
4  12  ...          s(3,t)           6  21 ...
4   ...           s(4,t)             9  ...
```

Now sequence vt are differential sequences of ut and relation (derived in the previous paragraphs) holds:

Symbolic notation

Powers (a+b√d)t:

```(a+b√d)1=  a   + b√d
(a+b√d)2= a2+  b2d     + b(2a)√d
(a+b√d)3= a3+ 3ab2d   + b(3a2+b2d)√d
(a+b√d)4= a4+ 6a2b2d+  b4d2    + b(4a3+4ab2d)√d
(a+b√d)5= a5+10a3b2d+  5ab4d2  +  b(5a4+10a2b2d+b4d2)√d
(a+b√d)6= a6+15a4b2d+15a2b4d2+b6d3     + b(6a5+20a3b2d+6ab4d)√d
```

Differential sequences:

```1  2a  3a²+b²d  4a³+4ab²d  5a4+10a²b²d+b4d²  a5+6ab4d²+20a³b²d
a   a²+b²d a³+3ab²d  a4+6a²b²d+b4d² a5+10a³b²d+5ab4d²
b²d  2ab²d  3a²b²d+b4d²  4a³b²d+4ab4d²
ab²d  a²b²d+b4d²  a³b²d+ 3ab4d²
b4d²  2ab4d²
ab4d²
```

In case, when differential sequences respect relation u'(t) = ut+1−a∙ut are coefficients of powers of algebraic numbers determined by recurrent sequences according to relation:

Golden number

Number (√5−1)/2 is called golden number (in connection to Fibonacci’s sequence and golden section) . Similarly silver number ((√8−2)/2 and bronze number (√13−3)/2 were derived. For such numbers xk it holds xk = 1/(k+xk), so xk is (positive) solution of equation x²+kx−1 =0.

E.g. for k=2 gets equation x²+2x−1=0 solution (−2+√8)/2 [= √2−1], i.e. silver number.

Euler‘s number

From sequences of factorials and super-factorials (see Factorials) come these algebraic numbers:

```   2 +  1√d     326 + 120√d
5 +  2√d    1957 + 720√d
16 +  6√d    ....
65 + 24√d
```

Values of fractions { 2/1,5/2,16/6,65/24,326/120,1957/720,... } get near to the Euler’s number e.

Is it possible to set operation (to select value d,..) in a such way, that algebraic numbers satisfy (power) relation?