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EXAMPLE
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For n = 6 the four regions of the last section of 6 are [2], [4, 2], [3], [6, 3, 2, 2, 1, 1, 1, 1, 1, 1, 1] therefore the "region numbers" are [8], [9, 9], [10], [11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11]. The sum of all region numbers is a(6) = 8+2*9+10+11^2 = 8+18+10+121 = 157, see below:
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. Last section Sum of
. of the set of Region region
k partitions of 6 numbers numbers
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11 6 11 11
10 3+3 10,11 21
9 4 +2 9, 11 20
8 2+2 +2 8,9, 11 28
7 1 11 11
6 1 11 11
5 1 11 11
4 1 11 11
3 1 11 11
2 1 11 11
1 1 11 11
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Total sum of region numbers is a(6) = 157
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