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A080634
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Start with a(1)=1. Then, for n>1, choose a(n)=1 or 2 so as to minimize the longest arithmetic progression in either S1(n) or S2(n), where S1(n)={k|a(k)=1,1<=k<=n} and S2(n)={k|a(k)=2,1<=k<=n}.
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0
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1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 2, 1, 2, 1, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 2, 2, 2, 2, 1, 2, 1, 1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 2, 2, 1, 2, 1, 2
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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EXAMPLE
| Given the first seven terms as {1,2,1,2,2,1,2}, we find that a(8)=1 gives S1(8)={1,3,6,8}, S2(8)={2,4,5,7}, both with maximum AP's of length 2, whereas a(8)=2 gives S1(8)={1,3,6}, S2(8)={2,4,5,7,8}, with max length AP's of 2 for S1 and 3 for S2. So a(8) must be assigned the value of 1.
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CROSSREFS
| Sequence in context: A049710 A025143 A174314 * A109925 A180227 A001468
Adjacent sequences: A080631 A080632 A080633 * A080635 A080636 A080637
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KEYWORD
| nonn
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AUTHOR
| John W. Layman (layman(AT)math.vt.edu), Feb 27 2003
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