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Define the original Cantor Dust sequence to be CCD(1) and this CCD(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)?
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 sequence 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).
a(3) = 7, the third integer not in Greedy Cantor's Dust Partition and also not forming an arithmetic sequence 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 sequence progression of length 3.
a(4) is not 9 because a(1), a(2), 9 would form an arithmetic sequence progression of length 3.
a(4) = 12, the sixth integer not in Greedy Cantor's Dust Partition and also not forming an arithmetic sequence progression of length 3 with previously admitted integers.
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allocated for Gordon HamiltonGreedy Cantor Dust Partition level 2
3, 6, 7, 12, 15, 16, 19, 30, 33, 34, 39, 42, 43, 46, 55
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Start by removing the Greedy Cantor's Dust Partition (https://oeis.org/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 sequence 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 sequence C(1) and this C(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 or more than the number of elements in CD(m)?
a(1) = 3 (the first integer not in Greedy Cantor's Dust Partition (https://oeis.org/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 sequence of length 3 with a(1) and a(2).)
a(4) is not 8 because a(2),a(3),8 would form an arithmetic sequence of length 3.
a(4) is not 9 because a(1),a(2),9 would form an arithmetic sequence of length 3.
a(4) = 12 (the sixth integer not in Greedy Cantor's Dust Partition and also not forming an arithmetic sequence of length 3 with previously admitted integers.)
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Gordon Hamilton, Mar 31 2024
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allocated for Gordon Hamilton
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List the possible lengths of line segments achievable by connecting integral coordinates on a Cartesian grid. Starting from the n-th length, a(n) is the smallest number of consecutively greater lengths required to form a simple polygon with all vertices on integral Cartesian coordinates.
a(5) = 4 because i) the fifth, sixth, seventh and eighth lengths are sqrt(8), 3, sqrt(10), sqrt(13) and ii) a quadrilateral can be created using edges with these four lengths and iii) the fifth, sixth and seventh lengths alone cannot create a simple polygon with integral Cartesian vertices.
Starting with the n-th shortest Cartesian line segment, a(n) is the minimal number of consecutive line segments required to make a closed loopsimple polygon.
Gordon Hamilton, <a href="https://mathpickle.com/project/rootingfornasa/">"Rooting for NASA"</a>.
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