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A347406
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Earliest sequence of distinct positive integers such that both gcd(a(n),a(n-k)) = 1 and gcd(a(n),a(n+k)) = 1, where k is each divisor of a(n) and n - k >= 1.
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3
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1, 2, 3, 5, 4, 7, 9, 8, 11, 13, 10, 17, 19, 14, 15, 23, 16, 29, 21, 26, 27, 25, 22, 31, 35, 32, 33, 37, 38, 41, 39, 34, 43, 47, 28, 53, 51, 20, 57, 59, 40, 61, 49, 44, 63, 67, 46, 71, 73, 52, 69, 79, 50, 83, 81, 55, 58, 77, 65, 82, 87, 85, 89, 74, 93, 95, 91, 86, 97, 101, 62, 103, 45, 64, 75
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OFFSET
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1,2
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COMMENTS
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The majority of terms are concentrated along two lines, the upper line has gradient of approximately 1.37 while the lower line has a gradient of approximately 1.02. Between these a third more random line also appear. See the linked image.
Small numbers with only 2 and 3 as prime divisors apparently take many terms to appear. For example a(210613) = 6, a(224221) = 18, while 12 and 24 have not appeared after 250000 terms.
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LINKS
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EXAMPLE
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a(3) = 3 as the divisors of 3 are 1 and 3, and a(3-1) = a(2) = 2, a(3+1) = a(4) = 5, and a(3+3) = a(6) = 7, and the gcd of 3 and each of these three numbers is 1. As a(3-3) = a(0) is not defined this term is ignored.
a(11) = 10 as the divisors of 10 are 1, 2, 5 and 10, and a(11-1) = a(10) = 13, a(11-2) = a(9) = 11, a(11-5) = a(6) = 7, a(11-10) = a(1) = 1, a(11+1) = a(12) = 17, a(11+2) = a(13) = 19, a(11+5) = a(16) = 23, and a(11+10) = a(21) = 27, and the gcd of 10 and each of these eight numbers is 1.
a(13) = 19 as the divisors of 19 are 1 and 19, and a(13-1) = a(12) = 17, a(13+1) = a(14) = 14, and a(13+19) = a(32) = 34, and the gcd of 19 and each of these three numbers is 1. Note that as a(11) = 10, and a(11+2) = a(13), where 2 is a divisor of 10, a(13) cannot equal 15 as gcd(10,15) > 1. This is the first term that differs from A347179.
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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