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  • Divisor Matrix, Dinary Notation and Quantum Number Theory

    Abstract

    Abstract

    Introduce the concept of Dinary notation, use this to form a divisor matrix and explore its properties. Attempt to develop a Quantum Number Theory on the divisor notation as fermionic states.

    Introduction

    As mentioned in previous articles [Include Here] on this account, theere exist divisor functions that tell the number of divisors a number has, but also those that generate a map that tells which divisors they are notably the Cosine method described in [Include Here].

    This allows number to be written as strings with as many digits as that number. For example \[1=1 \\ 2=11 \\ 3=101 \\ 4=1101 \\ 5=10001 \\ 6=111001 \\\]

    The interpretation of this is that, 1 is dvisible by 1 (left hand digit), 2 is divisible by 1 and 2, three is divisible by 1, not 2, but 3, 4 by 1, and 2, not 3, but 4. etc.

    The numbers are generated by a cascading column sequence, in the left most column every number is 1, in the next across the sequence is 10101010... in the next 100100100100.. etc. Due to the triangular nature of this pattern, if one extrapolates the zeros back up the column from the diagonal row of 1’s, then the numbers are followed by 0’s such that it is fair to rewrite equation 1 as \[1=100000 \\ 2=110000 \\ 3=101000 \\ 4=110100 \\ 5=100010 \\ 6=111001 \\\]

    This can be interpreted as the truth that, no integer is divisible by an integer gereater than itself.

    We allow this grid to make a matrix of size \(n\) by \(n\) denoted by \(M_n\) such that \[M_6 = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{bmatrix}\]

    It can be seen the main diagonal is filled with 1’s as well as the left most column due to the fact any number is divisible by itself and 1. Thus prime numbers in this matrix will only feature these two 1’s.

    Then for any \(n\), \(M\) is a lower triangular matrix. If we sum across each row, i.e by operating on a length \(n\) vector filled with 1’s, the resulting vector will be filled with the number of divisors that each number has.

    Investigating the determinant of the matrix:

    \[det(M_2)=1 \\ det(M_3)=1 \\ det(M_4)=1 \\ det(M_5)=1 \\ det(M_6)=1\]

    So at least these matrices are invertible We have that each number is represented by a lower triangular matrix which contains sub matrices of all number below it. Taking directly from Wikipedia:

    The sum of two lower triangular matrices is lower triangular. The product of two lower triangular matrices is lower triangular. The inverse of a invertible lower triangular matrix is lower triangular. The product of a lower triangular matrix by a constant is a lower triangular matrix.

    ...in particular the lower triangular matrices also form a Lie algebra

    Investigating the product of two matrices:

    Take for example \(M_3\) \[M_3 \cdot M_3 = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 2 & 0 & 1 \end{bmatrix}\]

    However, what would make more sense is the Kronecker product to insert on matrix into each element of the other, this would yield a \(9x9\) matrix which may represent the number \(9\).

    \[M_3 \otimes M_3 = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1\cdot M_3 & 0\cdot M_3 & 0\cdot M_3 \\ 1\cdot M_3 & 1\cdot M_3 & 0\cdot M_3 \\ 1\cdot M_3 & 0\cdot M_3 & 1\cdot M_3 \end{bmatrix} \\= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 \end{bmatrix}\]

    This isn’t equal to \(M_9\) try taking the transpose of the second matrix as in the polynomial representation paper [Insert Here].

    \[M_3 \otimes M_{3}^{T} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1\cdot M_{3}^{T} & 0\cdot M_{3}^{T} & 0\cdot M_{3}^{T} \\ 1\cdot M_{3}^{T} & 1\cdot M_{3}^{T} & 0\cdot M_{3}^{T} \\ 1\cdot M_{3}^{T} & 0\cdot M_{3}^{T} & 1\cdot M_{3}^{T} \end{bmatrix} \\= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 \end{bmatrix}\]

    Inverses \[M_{2}^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} \\ M_{3}^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \\ M_{4}^{-1} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & -1 & 0 & 1 \end{bmatrix} \\ M_{5}^{-1} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 & 0 \\ 0 & -1 & 0 & 1 & 0 \\ -1 & 0 & 0 & 0 & 1 \end{bmatrix} \\ M_{6}^{-1} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 & 1 & 0 \\ 1 &-1 & -1 & 0 & 0 & 1 \end{bmatrix} \\ M_{7}^{-1} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0\\ -1 & 1 & 0 & 0 & 0 & 0 & 0\\ -1 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 1 & 0 & 0 & 0\\ -1 & 0 & 0 & 0 & 1 & 0 & 0\\ 1 &-1 & -1 & 0 & 0 & 1 & 0 \\ -1 & 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \\ M_{8}^{-1} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ -1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ -1 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 1 & 0 & 0 & 0 & 0\\ -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 1 &-1 & -1 & 0 & 0 & 1 & 0 & 0\\ -1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 1 \end{bmatrix} \\ M_{9}^{-1} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 1 &-1 & -1 & 0 & 0 & 1 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \\ M_{10}^{-1} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ -1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 1 &-1 & -1 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\ -1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ x & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \\\]

    Looking again for a column cascading pattern, on the left most column there is the pattern,
    1,-1,-1,0,-1,1,-1,0,... On the next colum we have the same pattern but spaced with 0’s
    0,1,0,-1,0,-1,0,0,... The next column across the same pattern with double zero padding
    0,0,1,0,0,-1,0,0,...

    Noting that in the inverse matrices, the prime numbers so far all have a -1 in the left most column element.
    Both 1,6 have an element of 1, an observation is that these are perfect numbers
    4 and 8,9 have an element of zero, an observation is that these are numbers that contain a prime factor to a power higher than 1.

    Higher Patterns

    We can look for a 2x2 block matrix pattern in the factorisation. Recalling the definition of \(M_2\) and alongside the unit matrixa dn zero matrix defining a set of 2x2 matrices (based on order of appearence) \[a = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \;\; b = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \;\; c = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \;\; d = \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} =a+b\] Also note that \(I=b+c\) and \(M_2=a+b+c\).

    Using these, we can spit the divisor pattern into a 2x2 grib as below and identify the correct matrices to insert. It can be observed that a similar pattern to the previous one exists in this block matrix grid.

    \[\begin{array}{| c | c c | c c | c c | c c | c c | c c | c c | c c | c c | c c | c c | c c |} \hline 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 3 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 4 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 5 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 6 & 1 & 1 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 7 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 8 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 9 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 10 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 11 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 12 & 1 & 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 13 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 14 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 15 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 16 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ \hline 17 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 18 & 1 & 1 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ \hline 19 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 20 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \hline 21 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 22 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \hline 23 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 24 & 1 & 1 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \hline 25 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 26 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 27 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 28 & 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline \end{array}\]

    The pattern leaves the matrix \[\begin{bmatrix} 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & I & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & a & I & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & b & 0 & I & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & c & a & 0 & I & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & d & b & 0 & 0 & I & 0 & 0 & 0 & 0 & 0 \\ 2 & 0 & 0 & a & 0 & 0 & I & 0 & 0 & 0 & 0 \\ 2 & I & c & b & 0 & 0 & 0 & I & 0 & 0 & 0 \\ 2 & a & b & 0 & a & 0 & 0 & 0 & I & 0 & 0 \\ 2 & b & a & 0 & b & 0 & 0 & 0 & 0 & I & 0 \\ 2 & c & 0 & c & 0 & a & 0 & 0 & 0 & 0 & I \\ 2 & d & b & b & 0 & b & 0 & 0 & 0 & 0 & 0 & I \\ 2 & 0 & c & 0 & 0 & 0 & a & 0 & 0 & 0 & 0 & 0 & I \\ 2 & I & 0 & a & c & 0 & b & 0 & 0 & 0 & 0 & 0 & 0 & I \\ \end{bmatrix}\]

    However if one splits the 2 and d matrix it can be seen there are overlapping patterns. \[\begin{bmatrix} a+(I) & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ b+(a+c) & I & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & a & I & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & b & 0 & I & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & c & a & 0 & I & 0 & 0 & 0 & 0 & 0 \\ 2 & (a+b) & b & 0 & 0 & I & 0 & 0 & 0 & 0 \\ 2 & 0 & 0 & a & 0 & 0 & I & 0 & 0 & 0 \\ 2 & I & c & b & 0 & 0 & 0 & I & 0 & 0 \\ 2 & a & b & 0 & a & 0 & 0 & 0 & I & 0 \\ 2 & b & a & 0 & b & 0 & 0 & 0 & 0 & I \end{bmatrix}\]

    The block matrix in the top left corner \[\begin{bmatrix} 2 & 0 \\ 2 & I \end{bmatrix} =M_4 =\begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{bmatrix}\]

    The same grid can be used for a 3x3 system

    \[\begin{array}{| c | c c c | c c c | c c c | c c c | c c c | c c c | c c c | c c c |} \hline 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 3 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 4 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 5 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 6 & 1 & 1 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 7 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 8 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 9 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 10 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 11 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 12 & 1 & 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 13 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 14 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 15 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ \hline 16 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 17 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 18 & 1 & 1 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ \hline 19 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 20 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 21 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ \hline 22 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 23 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 24 & 1 & 1 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \hline 25 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 26 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 27 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 28 & 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 29 & 30 & \hline \end{array}\]

    One must define a large number of matrices to make sense of this pattern, there is a reccuring 3x3 matrix with a single element in the centre of the leftmost column which is the first non 0 or I matrix in a given column in the block matrix... A larger grid is needed to search for recurring patterns. A program will be made to use pixel map information to compress the number field further. A successful representation would lead to very quick checks of divisibility as one could home in on the exact block matrix and element needed to test. This is in effect a increase in the airity of the information field moving from a 2D number map to a 4D number map. A further abstraction into block matrices would create a 6D array which would be powerful but awkward to navigate.