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A111698
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a(1)=1. Skipping over integers occurring earlier in the sequence, count down a composite from a(n) to get a(n+1) so that a(n+1) is the smallest possible positive integer arrived at this way. If there are no positive integers at a distance of a composite number of yet unused integers, instead count up from a(n) 4 (the lowest composite positive integer) positions (skipping already occurring integers) to get a(n+1).
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1
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1, 5, 9, 2, 7, 12, 3, 10, 15, 4, 13, 18, 6, 16, 21, 8, 19, 24, 11, 22, 27, 14, 25, 30, 17, 28, 33, 20, 31, 36, 23, 34, 39, 26, 37, 42, 29, 40, 45, 32, 43, 48, 35, 46, 51, 38, 49, 54, 41, 52, 57, 44, 55, 60, 47, 58, 63, 50, 61, 66, 53, 64, 69, 56, 67, 72, 59, 70, 75, 62, 73, 78
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OFFSET
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1,2
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COMMENTS
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I have found two patterns for this sequence. The first is that there is a pattern 0,3,6,0,3,6,0,3,6,... which states the lengths of the "LessThanList" for each term. In other words, a(6) = 12. There are six integers less than 12 which are not already listed in the sequence at this point, {3,4,6,8,10,11}. a(7) = 3. There are no integers not already on the list which are less than 3 at this point. a(8) = 10. There are three integers less than 10 which are not already on the list at this point, {4,6,8}. Also, after the 14th term, the sequence becomes regular in the following way. The difference between successive terms is as follows: 5,-13,11,5,-13,11,... . - Diana L. Mecum, Aug 15 2008
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LINKS
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EXAMPLE
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The first 5 terms of the sequence can be plotted on the number line as:
1,2,*,*,5,*,7,*,9,*,*,*.
Now a(5) is 7. Counting down from 7 gets a noncomposite (1,2, or 3) number of steps to arrive at each yet unused positive integer. So we instead count up 4 positions, skipping the 9 as we count, to arrive at 12 (which is at the rightmost * of the number line above).
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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