Benedict Irwin edited Rules.tex  about 9 years ago

Commit id: d43af136e11f0196647d7b4ca86b36db2e7a93a5

deletions | additions      

       

Trial Rule Sets: \\  $N=1$ and $\{1,2,4,5,5,6\}a+\{4,1,2,3,6,3\}b \; mod\; 7=0$, \\  $N=2$ and $\{1,2,3,4,5,6\}a + \{6,5,4,3,2,1\}b \; mod \; 7 = \{3,6,2,5,1,4\}$,\\  then the family $f/7 \in \mathbb{Z}, \; \forall f \in F(a,b)$. Have for $a=0, b=4$ if $n - {2,4} \; mod \; 6 = 0$ number is divisible by $7$. \\  Have for $a=0, b=7$ if $n - {1,3} \; mod \; 6 = 0$ number is divisible by $7$. \\