Self-class triangle

Self-class triangle

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.

Central progression

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

Systems G(2,k), k=2p-1

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

Polynomials

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.

Differential sequences

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.

Symmetry

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

Recurrent sequences

Let us again see sequences ak(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 rk(n)} are defined recurrently:

Sequences rk(n) have form:
Fn,c= 1/ (1-xc) (1-x)n-1

Case c=n.
For n=1,2,3 recurrent sequences {rk(n)} correspond to sequences of self classes {ak(n)}.

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

n=1

rk(1) = rk-1(1) + [-2+k | -1] = rk-1 + 0
a(0)=1;
F1,1= 1/ (1-x) = 1 + 1x1+ 1x2+ 1x3+ 1x4+ 1x5+ 1x6+...
i.e. { rk(1)} = { 0,1,1,1,1,1,1,1,1,1...}

n=2

rk(2)= rk-2(2)+ [ k-1 | 0] = rk-2+ 1
a(1)=1 /(1-x);
F2,2=1/ [(1-x2) (1-x)] = 1 + 1x1+ 2x2+ 2x3+ 3x4+ 3x5+...
i.e. { rk(2)} = { 0,1,1,2,2,3,3,4,4,5...}

n=3

rk(3)= rk-3(3) + [ k | 1] = rk-3 + k
a(k)=1 /(1-x)2;
F3,3=1/[(1-x3) (1-x)2] = 1 + 2x1+ 3x2+ 5x3+ 7x4+ 9x5+ ...
i.e. { rk(3)} = { 0,1,2,3,5,7,9,12,15,18,...}

n=4

rk(4) = rk-4(4) + [k+1 | 2] = rk-4+ k (k+1) / 2
a(k(k+1)/2)=1 /(1-x)3;
F4,4= 1/ [(1-x4) (1-x)3] = 1 + 3x1+ 6x2+ 10x3+ 16x4+ 24x5+ ...
i.e. { rk(4)} = { 0,1,3,6,10,16,24,34,46,61,...}

n=5

rk(5) = rk-5(5) + [ k+2 | 3] = rk-5+ k (k+1)(k+2)/6
a(k (k+1)(k+2)/6)=1/(1-x)4;
F5,5= 1/ [(1-x5) (1-x)4] = 1 + 4x1+ 10x2+ 20x3+ 35x4+ ...
i.e. { rk(5)} = { 0,1,4,10,20,35,57,88,130...}

Auxiliary expression a(z)= Fn,c(x)-xcFn,c(x).


Schematic algebra