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Search: id:A111118
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| A111118 |
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a(1) = 1; skipping over integers occurring earlier in the sequence, count down c(n) (c(n) = n-th composite) from a(n) to get a(n+1). If this is <= 0, instead count up from a(n) c(n) positions (skipping already occurring integers) to get a(n+1). |
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+0
3
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| 1, 5, 11, 2, 13, 23, 9, 26, 7, 28, 3, 31, 52, 29, 55, 25, 57, 22, 59, 19, 62, 17, 64, 15, 66, 10, 68, 6, 71, 115, 69, 117, 63, 119, 60, 121, 56, 124, 53, 126, 50, 128, 47, 131, 45, 133, 43, 135, 40, 137, 38, 140, 35, 142, 33, 144, 30, 147, 24, 149, 18, 151, 14, 153, 8, 156
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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If we did not skip earlier occurring integers when counting, we would instead have sequence A100298.
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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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The first 4 terms of the sequence can be plotted on the number line as:
1,2,*,*,5,*,*,*,*,*,11,*,*.
Now a(4) is 2. Counting c(4) = 9 down from 2 gets a negative integer. So we
instead count up 9 positions -- skipping the 5 and 11 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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Cf. A100298, A110080, A002808.
Sequence in context: A087463 A160366 A160923 this_sequence A159199 A125683 A125685
Adjacent sequences: A111115 A111116 A111117 this_sequence A111119 A111120 A111121
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KEYWORD
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nonn
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AUTHOR
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Leroy Quet Oct 15 2005
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EXTENSIONS
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More terms from Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), Oct 17 2005
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