| The bigrade | ||
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The magic bigrade is a line of m numbers out of the numbers [1 .. mn] with k=0∑m-1Bk = m (mn + 1) / 2 and k=0∑m-1Bk2 = m (2 m2n + 3 mn + 1) / 6 |
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| centralized sum |
changing numbers Bk --> (mn + 1) / 2 + bk it follows: k=0∑m-1bk = 0 and k=0∑m-1bk2 = m (m2n - 1) / 12 |
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k=0∑m-1Bk =
k=0∑m-1(mn + 1) / 2 + bk =
m (mn + 1) / 2 + k=0∑m-1bk
m (mn + 1) / 2 ==> k=0∑m-1bk = 0 k=0∑m-1[(mn + 1) / 2 + bk]2 = k=0∑m-1[(mn + 1)2 / 4 + (mn + 1) * bk + bk2] = m (mn + 1)2 / 4 + k=0∑m-1 bk2 = m (2 m2n + 3 mn + 1) / 6 ==> k=0∑m-1 bk2 = m (2 m2n + 3 mn + 1) / 6 - m (mn + 1)2 / 4 = m (m2n - 1) / 12 |
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| bigrade move |
the move from one bigrade to another is given by B2k --> B1k + dk with 2 k=0∑m-1B1kdk = - 2 k=0∑m-1B2kdk = - k=0∑m-1dk2 |
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k=0∑m-1B2k2 =
k=0∑m-1(B1k + dk)2 =
k=0∑m-1B1k2 +
2 k=0∑m-1B1kdk +
k=0∑m-1dk2 ==> 2 k=0∑m-1B1kdk = - k=0∑m-1dk2 = - 2 k=0∑m-1B2kdk |
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The bigrade move make it possible to "hang" the square onto the first line with a set of differences the twisted condition these difffernces is curiously bound to the first lines numbers, the validity of these difference squares remain to be seen |
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