Some time ago, I already wrote about representation ideas of primes and
we saw that we run in troubles with this. Today, I want to present you a
similar approach.
Let’s start again with our equation
$$
x_{ij} = \left(2x_{i} + 1\right)x_{j} + x_{i} \\
y_{ij} = \left(2x_{ij} + 1\right)
$$
and the following represenation
$$
x_{1j} \ \ \ \ | 00010010010010010010 \dots \\
x_{2j} \ \ \ \ | 00000010000100001000 \dots \\
x_{\left(1,2\right),j} | 11101101101001100101 \dots
$$
The first line is given by \(x_{1j} = 3x_{j} + 1 = 4, 7, 10, 13, 16, 19\).
If we look a the numbers from 1 to 20 (from left to right), we represent
all numbers which are generated by \(x_{1j}\), by ‘1’ and the other
numbers by ‘0’. In the second line, we do the same for
\(x_{2j} = 5x_{j} + 2 = 7, 12, 17\).
In the third line we see \(x_{\left(1,2\right),j}\), which represents all
numbers which are not element of \(x_{1j}\) and not element of \(x_{2j}\) by
‘1’ and the others by ‘0’. So we can write
$$
x_{\left(1,2\right),j} = \overline{x_{1j}} \cdot \overline{x_{2j}}
$$
. Ok. What can we do with this, now?
At first, we look at \(x_{1j}\) and \(x_{2j}\). We will rewrite them to
matrices \(X_{\left(1\right)}^{n\times n}\) and
\(X_{\left(2\right)}^{n\times n}\). This matrices, all of the same size
\(n\times n\), have the numbers from the representation above as diagonal
entries. All other entries are ‘0’.
$$
X_{\left(1\right)}^{n\times n} :=
\begin{pmatrix}
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{pmatrix}
= \left(x_{\left(1\right),kj}\right)_{k=1,\dots,n \ , j=1,\dots, n} \ \delta_{kj}
$$
$$
X_{\left(2\right)}^{n\times n} :=
\begin{pmatrix}
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{pmatrix}
= \left(x_{\left(2\right),kj}\right)_{k=1,\dots,n \ , j=1,\dots, n} \ \delta_{kj}
$$
Here are
$$
x_{\left(i\right),kj} := \left\{
\begin{array}{l@{\quad \quad}l}
1 & \mathrm{if} \ k = \left(2x_{i} + 1\right)x_{l} + x_{i} \\
0 & \mathrm{else}
\end{array}
\right.
$$
and
$$
\delta_{kj} := \left\{
\begin{array}{l@{\quad \quad}l}
1 & \mathrm{if} \ k = j \\
0 & \mathrm{else}
\end{array}
\right.
$$
With this, we get \(\overline{X_{\left(i\right)}^{n\times n}}\) by
$$
\overline{X_{\left(i\right)}^{n\times n}} = \mathbb{1}_{n} - X_{\left(i\right)}^{n\times n}
$$
For an arbitrary number \(i=a,\dots,b\), \(a,b \in
\mathbb{N}\), \(a \le b\), of equations \(x_{ij}\) we receive
$$
X_{\left(a,\dots,b\right)}^{n \times n} = \prod_{i=a}^{b} \left(\mathbb{1}_{n} -
X_{\left(i\right)}^{n\times n}\right)
$$
and so
$$
x_{\left(a,\dots,b\right),kj} = \prod_{i=a}^{b} \left(1 - x_{\left(i\right),kj}\right)\delta_{kj} \\
$$
We received a matrix with ‘1’ entries at the places \(j=k\) which
represent primes and else ‘0’.