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Search: id:A111699
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| A111699 |
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a(1)=1. Skipping over integers occurring earlier in the sequence, count down b(n) (b(n)= number of earlier terms of {a(k)} which are coprime to n) from a(n-1) to get a(n). If this is <= 0, instead count up from a(n-1) b(n) positions (skipping already occurring integers) to get a(n). |
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+0
1
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| 1, 2, 4, 3, 8, 7, 13, 9, 16, 10, 22, 19, 31, 25, 11, 24, 42, 35, 6, 23, 40, 28, 53, 43, 5, 33, 58, 44, 73, 63, 20, 49, 75, 52, 14, 39, 86, 65, 21, 54, 100, 88, 30, 64, 17, 61, 115, 96, 38, 71, 113, 85, 142, 121, 78, 36, 98, 56, 132, 114, 177, 148, 105, 60, 125, 97, 174, 141, 83
(list; graph; listen)
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OFFSET
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1,2
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LINKS
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Diana Mecum, Table of n, a(n) for n = 1..900 [From Diana Mecum (diana.mecum(AT)gmail.com), Aug 04 2008]
Leroy Quet, Home Page (listed in lieu of email address)
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EXAMPLE
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The first 6 terms of the sequence can be plotted on the number line as:
1,2,3,4,*,*,7,8,*,*,*,*,*.
Now a(6) is 7. Counting b(7) = 5 (because all earlier terms but the 7 are coprime to 7) down from 7 gets a negative integer. So we instead count up 5 positions - skipping the 8 as we count - to arrive at 13 (which is at the right-most * of the number-line above).
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CROSSREFS
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Sequence in context: A101283 A125566 A166133 this_sequence A067179 A048767 A163511
Adjacent sequences: A111696 A111697 A111698 this_sequence A111700 A111701 A111702
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KEYWORD
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nonn
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AUTHOR
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Leroy Quet Nov 17 2005
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EXTENSIONS
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More terms from Diana Mecum (diana.mecum(AT)gmail.com), Aug 04 2008
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