Benedict Irwin edited Predicting the Series.tex  over 9 years ago

Commit id: 4584b88be889b9300b54f310b0e8a8f182d33892

deletions | additions      

       

\end{equation}  But we know, from the relationship, that \\  $D_1(3)=[D_1(19)-D_0(4\%4)]\;mod^*\;10=[7-6]\%10=1$ $D_1(03)=[D_1(19)-D_0(04\%4)]\;mod^*\;10=[7-6]\%10=1$  \\ $D_1(4)=[D_1(20)-D_0(5\%4)]\;mod^*\;10=[5-2]\%10=3$ $D_1(04)=[D_1(20)-D_0(05\%4)]\;mod^*\;10=[5-2]\%10=3$  \\ $D_1(5)=[D_1(1)-D_0(6\%4)]\;mod^*\;10=[0-4]\%10=6$ $D_1(05)=[D_1(01)-D_0(06\%4)]\;mod^*\;10=[0-4]\%10=6$  \\ $D_1(6)=[D_1(2)-D_0(7\%4)]\;mod^*\;10=[0-8]\%10=2$ $D_1(06)=[D_1(02)-D_0(07\%4)]\;mod^*\;10=[0-8]\%10=2$  \\ $D_1(13)=[D_1(9)-D_0(14\%4)]\;mod^*\;10=[2-4]\%10=8$ $D_1(13)=[D_1(09)-D_0(14\%4)]\;mod^*\;10=[2-4]\%10=8$  \\ $D_1(14)=[D_1(10)-D_0(15\%4)]\;mod^*\;10=[4-8]\%10=6$ \\  $D_1(15)=[D_1(11)-D_0(16\%4)]\;mod^*\;10=[9-6]\%10=3$ \\  $D_1(16)=[D_1(12)-D_0(17\%4)]\;mod^*\;10=[9-2]\%10=7$ \\