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Search: id:A136348
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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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+0
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
(list; graph; listen)
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OFFSET
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1,3
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LINKS
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Leroy Quet, Home Page (listed in lieu of email address)
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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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CROSSREFS
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Cf. A136347.
Sequence in context: A081831 A111912 A096288 this_sequence A137847 A164023 A097224
Adjacent sequences: A136345 A136346 A136347 this_sequence A136349 A136350 A136351
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KEYWORD
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more,nonn
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AUTHOR
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Leroy Quet Dec 25 2007
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