Benedict Irwin edited Pascals.tex  over 9 years ago

Commit id: 350011b3cb00c4dce8a063cc038d9d294d37f8c7

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The third line is $(010;2);2$.\\  The n^{th} line is $010;2_1;2_2;2_3...;2_{n-1}$ \\  We can attempt similar triangles with other starting strings.  1  1;3= 111  111;3=  111  111  111 = 12321    12321;3=  12321  12321  12321 = 1367631    1367631;3  1367631  1367631  1367631    151807041     \begin{equation}  1 \\ 111 \\ 12321 \\ 13677631 \\ 151807041  \end{equation}