- Self-class triangle
- Central progression
- Systems G(2,k), k=2p-1
- Polynomials
- Differential sequences
- Symmetry
- Recurrent sequences

The triangle

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

The triangle of self-classes seems to be more general then Pascal triangle - the latter can be obtained by an integration of its members. Also some theorems about primes (Wilson theorems,...) are connected with its structure. Let us look on it more closely.

The progression {1,1,2,5,14,42,132,429,1430,...}, i.e. the sequence of greatest numbers in odd rows, is [2n+1|n]/(2n+1). *This progression is known also from the following diagram:*

1

1 5

1 4 14

1 3 9 28

1 2 5 14 42

*First column contains ones.**Members at first row are sum of theirs left neighbors.*

E.g. 2 = 1 + 1; 5 = 2 + 3; 14 = 5 + 9; 42 = 14 + 28.*Members on the following rows are sums of numbers from below and from left-up diagonal.*

E.g. 3 = 2 + 1; 9 = 5 + 4; 28 = 14 + 14.

When p is prime then for numbers c=[(2p-1)|i],i=1,2..p-1, it holds:

c = +-1 (mod p); c/(2p-1) = +-1 (mod p)

*E.g.:*

*[3|0] mod 2=+1; [3|1] mod 2=-1;*

*[5|0] mod 3=+1; [5|1] mod 3=-1; [5|2] mod 3=+1;*

*[9|0] mod 5=+1; [9|1] mod 5=-1; [9|2] mod 5=+1; [9|3] mod 5=-1; [9|4] mod 5=+1;*

In case k=2p-1 the number c determine the number of self-instances.

Therefore c/(2p-1) is the number of self-classes.

Self-class triangle for k =2p-1:

(p=2) 3: 0 1

(p=3) 5: 0 1 2

(p=5) 9: 0 1 4 9 14

(p=7) 13: 0 1 6 22 55 99 132

These rows mod p give always +1 or -1:

(p=2) 3: 0 +1

(p=3) 5: 0 +1 -1

(p=5) 9: 0 +1 -1 -1 -1

(p=7) 13: 0 +1 -1 +1 -1 +1 -1

Let us look on numbers in the self-class triangle as on polynomial coefficients:

k= 2: 1

k= 3: r + 1

k= 4: r^2+ 1r + 1

k= 5: r^3+ 2r^2+ 2r + 1

k= 6: r^4+ 2r^3+ 3r^2+ 2r + 1

k= 7: r^5+ 3r^4+ 5r^3+ 5r^2+ 3r + 1

k= 8: r^6+ 3r^5+ 7r^4+ 8r^3+ 7r^2+ 3r + 1

k= 9: r^7+ 4r^6+ 9r^5+ 14r^4+ 14r^3+ 9r^2+ 4r + 1

k=10: r^8+ 4r^7+ 12r^6+ 20r^5+ 25r^4+ 20r^3+ 12r^2+4r + 1

Written in another way:

k= 2: 1 (1 = 1; 0 = 0)

k= 3: (r+1) (2 = 2; 1 = 1)

k= 4: (r^2+r+1) (3 = 3; 2 = 2)

k= 5: (r+1)*(r^2+r+1) (2*3 = 6; 1+2 = 3)

k= 6: (r^2+r+1)^2 (3^2 = 9; 2*2 = 4)

k= 7: (r+1)*(r^2+r+1)^2 (2*3^2 = 18; 1+2*2= 5)

k= 8: (r^2+r+1)*(r^4+2r^3+4r^2+2r+1) (3*10 = 30; 2+4 = 6)

k= 9: (r+1)^3*(r^4+r^3+3r^2+r+1 ) (2^3*7 = 56; 3*1+4= 7)

k= 10: (r^2+r+1)^2*(r^4+2r^3+5r^2+2r+1) (3^2*11 = 99; 2*2+4= 8)

Number of all classes in system G(2,k) is in parentheses.

Let us look on differential sequences of numbers on diagonals.

d= 1: 0,1,1,1,1,1,1,...

d= 2: 0,1,1,2,2,3,3,...

1,0,1,0,1,0,1,0,

d= 3: 0,1,2,3,5,7,9,12,15,18,...

1,1,1,2,2,2,3,3,3,

0,0,1,0,0,1,0,0,

d= 4: 0,1,2,5,8,14,20,30,40,55,70,...

1,1,3,3, 6, 6,10,10,15,15,

0,2,0, 3, 0, 4, 0, 5, 0

d= 5: 0,1,3,7,14 ,25,42,66,99,143,200,273,364,...

1,2,4,7 ,11,17,24,33, 44, 57, 73, 91,

1,2,3 , 4, 6,7, 9, 11, 13, 16, 18,

1,1, 1, 2, 1, 2, 2, 2, 3, 2,

0, 0, 1,-1, 1, 0, 0, 1, -1,

d= 6: 0,1,3,9,20,42,75,132,212,333,497,728,...

1,2,6,11,22,33,57,80,121,164,231,

1,4,5,11,11,24,23,41, 43, 67,

3,1,6, 0, 13,-1,18,2, 24

-2,5,-6,13,-14,19,-20,22,

7,-11,19,-27,33,-39,42, ...

d= 7: 0,1,4,12,30,66,132 , 245,429,715,1144,...

1,3,8,18,36,66 , 113,184,286,429,

2,5,10,18,30 , 47,71,102,143,

3,5,8,12 , 17, 24, 31, 41,

2,3,4 , 5, 7, 7, 10,

1 , 1 , 1, 2, 0, 3,

0 , 0, 1, -2, 3,

The sequences of diagonal r seem to have simpler form if number d is prime.

A kind of symmetry exists in differential sequences, if number n is prime.

E.g. in case n=7, the numbers 0,1,4,12,30,66,132 are on both edges of the figure:

n= 1 : 0 n= 7 : 0 ,1, 2, 3, 2, 1, 0

1, 3, 5, 5, 3, 1

n= 2 : 0,1 4, 8,10, 8, 4

1 12,18,18,12

30,36,30

n= 3 : 0,1,0 66,66

1,1 132

2

n= 5 : 0,1,1,1,0

1,2,2,1

3,4,3

7,7

14

Let us again see sequences a_{k}(n) on diagonals of self classes triangle:

**n=1:** 0,1,1, 1, 1, 1, 1, 1, 1, 1,...

**n=2:** 0,1,1, 2, 2, 3, 3, 4, 4, 5,...

**n=3:** 0,1,2, 3, 5, 7, 9,12, 15, 18,...

**n=4:** 0,1,2, 5, 8,14,20,30, 40, 55,...

**n=5:** 0,1,3, 7,14,25,42,66, 99,143,...

Let sequences r_{k}(n)} are defined recurrently:

- If k < 1 then r
_{k}(n) = 0; - If k = 1 then r
_{k}(n) = 1; - Else r
_{k}(n) = r_{k-c}(n) + z; where z= [n-2+k-1 | n-2 ] = [k-1 | n-2 ]

Sequences r_{k}(n) have form:

F_{n,c}= 1/ (1-x^{c}) (1-x)^{n-1}

Case c=n.

For n=1,2,3 recurrent sequences {r_{k}(n)} correspond to sequences of self classes {a_{k}(n)}.

For n=1,2..5 we get sequences:

**n=1**

r_{k}(1) = r_{k-1}(1) + [-2+k | -1] = r_{k-1} + 0

a(0)=1;

**F _{1,1}= 1/ (1-x)** = 1 + 1x

i.e. { r

**n=2**

r_{k}(2)= r_{k-2}(2)+ [ k-1 | 0] = r_{k-2}+ 1

a(1)=1 /(1-x);

**F _{2,2}=1/ [(1-x^{2}) (1-x)]** = 1 + 1x

i.e. { r

**n=3**

r_{k}(3)= r_{k-3}(3) + [ k | 1] = r_{k-3} + k

a(k)=1 /(1-x)^{2};

**F _{3,3}=1/[(1-x^{3}) (1-x)^{2}]** = 1 + 2x

i.e. { r

**n=4**

r_{k}(4) = r_{k-4}(4) + [k+1 | 2] = r_{k-4}+ k (k+1) / 2

a(k(k+1)/2)=1 /(1-x)^{3};

**F _{4,4}= 1/ [(1-x^{4}) (1-x)^{3}]** = 1 + 3x

i.e. { r

**n=5**

r_{k}(5) = r_{k-5}(5) + [ k+2 | 3] = r_{k-5}+ k (k+1)(k+2)/6

a(k (k+1)(k+2)/6)=1/(1-x)^{4};

**F _{5,5}= 1/ [(1-x^{5}) (1-x)^{4}]** = 1 + 4x

i.e. { r

Auxiliary expression a(z)= F_{n,c}(x)-x^{c}F_{n,c}(x).