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A101543
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Triangle read by rows: First row = 2; n-th row (n>1) has n smallest positive integers not yet in the sequence such that each integer has a prime divisor in common with at least one element of the (n-1)st row.
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0
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2, 4, 6, 3, 8, 9, 10, 12, 14, 15, 5, 7, 16, 18, 20, 21, 22, 24, 25, 26, 27, 11, 13, 28, 30, 32, 33, 34, 17, 35, 36, 38, 39, 40, 42, 44, 19, 45, 46, 48, 49, 50, 51, 52, 54, 23, 55, 56, 57, 58, 60, 62, 63, 64, 65, 29, 31, 66, 68, 69, 70, 72, 74, 75, 76, 77, 37, 78, 80, 81, 82, 84
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OFFSET
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1,1
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COMMENTS
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Is this a permutation of the integers >= 2?
It appears that for n > 2, row n+1 always begins with the primes p such that 2p appears in row n and the rest of row n+1 consists of the smallest composite numbers not already used. The only way this pattern can break down is if we have to skip a composite number because it doesn't share a factor with any number in the previous row. Let f(n) be the last number in row n. To prove that this pattern continues, it suffices to show that f(n) < (f(n-1)-f(n-2)+1)^2, because the prime factors of row n-1 include all primes <= f(n-1)-f(n-2) and any composite number x has a prime factor <= sqrt(x). I have checked that f(n) < (f(n-1)-f(n-2)+1)^2 for all n up to 10000. In fact for 1000 < n <= 10000, f(n) < (f(n-1)-f(n-2)-300)^2. - David Wasserman, Mar 27 2008
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LINKS
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EXAMPLE
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7 is in the 5th row because it does not occur earlier and 14 is in the 4th row.
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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