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A084425
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Let b(0) = n, b(2*k+1) = c, where c > b(m) is the smallest number such that gcd(c,b(2*k))=1, b(2*k+2) = e where e < b(m) is the smallest number such that gcd(e,b(2*k+1))=1 for m = 0..2*k, until reaching e = 1. Then a(n) = the last c.
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1
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3, 4, 9, 6, 10, 10, 12, 12, 16, 25, 18, 18, 31, 26, 26, 28, 41, 34, 30, 30, 36, 40, 42, 42, 42, 42, 50, 48, 50, 56, 54, 58, 75, 64, 60, 60, 66, 68, 70, 72, 72, 93, 78, 80, 78, 86, 84, 84, 84, 90, 90, 98, 96, 100, 100, 102, 106, 127, 108, 108, 110, 116, 114, 118, 122
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OFFSET
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2,1
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LINKS
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EXAMPLE
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n = 6: 7, 5, 8, 3, 10, 1, hence a(6) = 10.
n = 7: 8, 5, 9, 4, 11, 6, 13, 3, 10, 1, hence a(7) = 10.
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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Edited by Frank Ellermann, Jun 07 2003
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STATUS
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approved
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