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A371218
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Greedy Cantor Dust Partition level 2.
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0
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3, 6, 7, 12, 15, 16, 19, 30, 33, 34, 39, 42, 43, 46, 55
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OFFSET
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1,1
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COMMENTS
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Start by removing the Greedy Cantor's Dust Partition (A348636) from the set of positive integers. We are left with: 3,6,7,8,9,12,15,16,17,... Now apply the A348636 algorithm on this set. Starting with the lowest term, consecutively partition the positive integers excluding the elements of Cantor's set (level 2) into sets s(1), s(2), s(3), ... so that no arithmetic progression of length 3 exists in a set. When choosing s(k), always choose k as small as possible. a(n) = smallest number in s(n).
This process can be applied iteratively to create an infinite sequence of Cantor-Dust-like sequences whose intersection is empty and whose union is the set of positive integers.
Define the original Cantor Dust sequence to be CD(1) and this CD(2). Is it always true for n < m that the number of elements less than k that are in CD(n) will always be equal to or more than the number of elements in CD(m)?
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LINKS
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EXAMPLE
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a(1) = 3, the first integer not in Greedy Cantor's Dust Partition (A348636).
a(2) = 6, the second integer not in Greedy Cantor's Dust Partition.
a(3) = 7, the third integer not in Greedy Cantor's Dust Partition and also not forming an arithmetic progression of length 3 with a(1) and a(2).
a(4) is not 8 because a(2), a(3), 8 would form an arithmetic progression of length 3.
a(4) is not 9 because a(1), a(2), 9 would form an arithmetic progression of length 3.
a(4) = 12, the sixth integer not in Greedy Cantor's Dust Partition and also not forming an arithmetic progression of length 3 with previously admitted integers.
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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STATUS
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approved
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