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A136348
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a(1)=1. Let m be the number of terms in the longest run of consecutive equal terms from among the first n terms of the sequence. a(n+1) = the number of terms, from among the first n terms of the sequence, that divide m.
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2
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1, 1, 2, 3, 3, 3, 5, 5, 5, 5, 3, 3, 3, 3, 3, 6, 6, 6, 6, 6, 6, 17, 17, 17, 17, 17, 17, 17, 2, 2, 2, 2, 2, 2, 2, 2, 11, 11, 11, 11, 11, 11, 11, 11, 11, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 11, 12, 12, 12, 12, 12
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OFFSET
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1,3
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LINKS
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EXAMPLE
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Among the first 20 terms of the sequence the run of five 3's (from a(11) to a(15)) is the longest run of consecutive equal terms. (So m = 5.) Among the first 20 terms of the sequence, there are six which divide 5. (a(1),a(2),a(7),a(8),a(9),a(10)) So a(21) = 6.
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PROG
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(PARI) See Links section.
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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