Benedict Irwin edited Rules.tex  almost 9 years ago

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Leaving Rules  \begin{equation}  \begin{array}  \; & [(a+3b) & mod \; 13 =0 & \wedge & n - 1 \; mod \; 6 = 0] \\ % 1 & 7 \vee & [(b-a-1) & mod \; 13 =0 & \wedge & n - 2 \; mod \; 6 = 0] \\ % 2 & 8  \vee & [(3a+b) & mod \; 13 =0 & \wedge & n - 3 \; mod \; 6 = 0] \\ % 3 & 9  \vee & [(4a+b+1)& mod \; 13 =0 & \wedge & n - 4 \; mod \; 6 = 0] \\ % 4 & 10 

\begin{bmatrix}  &8 & &5 & & &3 &7,10& & \\  1,2 & &3 & & & & & & & \\  & &1,5&2 &4 & &6  & & & & \\ 6  & & & &1 & &2 & & & \\ &5 & & & & &1 & & &2,4 \\  & & &4 & & & & &6  &1 &5 \\ & & &6  & & & & & & & \\ &2 & & & & & & & & \\  & &4 & &2 & & & & & &6  \\ & & & &3 & &3,6&  & &2 & & \\ \end{bmatrix}  \end{equation}  Leaving Rules Rules, if:  \begin{equation}  \begin{array}  \; & [(2a-b-2) & mod \; 17 =0 & \wedge & n - 1 \; mod \; 16 = 0] \\ % 1 

\vee & [(4a+b-6) & mod \; 17 =0 & \wedge & n - 3 \; mod \; 16 = 0] \\ % 3  \vee & [(3b+a+3) & mod \; 17 =0 & \wedge & n - 4 \; mod \; 16 = 0] \\ % 4  \vee & [(2b+a-6) & mod \; 17 =0 & \wedge & n - 5 \; mod \; 16 = 0] \\ % 5  \vee & [(5a+b+2) & mod \; 17 =0 & \wedge & n - 6 \; mod \; 16 = 0] \\ % 6  \end{array} \\  \frac{Z}{13} \frac{Z}{17}  \in \mathbb{Z} \end{equation} 

& & & & & &8 &6,7& & \\  & & & & & & & & & \\  & & & & & & & & & \\  6 & & & & & & & & & \\  & & & &6 & &10,15 & & & \\  & & & & & & & &6 & \\  7 & & & & & & & & & \\  & & & & & & & & & \\  &6  & & & & &&10,15  & & & \\ & & & & & & & & & \\  & & & & & & & & & \\  & & & & & & & & & \\  & & & & & & & & & \\  & & & & & &6  & & & & \\ \end{bmatrix}  \end{equation}  Leaving Rules, if:  \begin{equation}  \begin{array}  \; & [(4a-b+7) & mod \; 19 =0 & \wedge & n - 6 \; mod \; 18 = 0] \\ % 1  \end{array} \\  \frac{Z}{19} \in \mathbb{Z}  \end{equation}  \begin{equation}  \Gamma_{23}=  \begin{bmatrix} 

\\%9,9  \end{bmatrix}  \end{equation}  Leaving Rules, if:  \begin{equation}  \begin{array}  \; & [(9a-b-9) & mod \; 23 =0 & \wedge & n - 6 \; mod \; 22 = 0] \\ % 6  \end{array} \\  \frac{Z}{23} \in \mathbb{Z}  \end{equation}  \section{31}  Leaving Rules, if:  \begin{equation}  \begin{array}  \; & [(2a-3b+1) & mod \; 31 =0 & \wedge & n - 6 \; mod \; 30 = 0] \\ % 6  \end{array} \\  \frac{Z}{23} \in \mathbb{Z}  \end{equation}  \section{37}  For divisibility by $37$, (exceptionally common divisor) the general $\Gamma$ matrix can be constructed below 

& & & & & & & & & \\  \end{bmatrix}  \end{equation}  All entries have a $6$. So if $n-6 \; \mathrm{mod} \; 36 = 0$, $Z/37 \in \mathbb{Z}$.  \begin{equation}  \Gamma_{X}=