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A368231
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Lexicographically earliest infinite sequence of distinct positive numbers such that, for n>3, a(n) has a common factor with a(n-1) but not with a(n-2) or n.
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3
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1, 15, 35, 77, 143, 65, 30, 21, 91, 221, 85, 55, 33, 39, 182, 133, 95, 115, 69, 51, 170, 145, 203, 119, 102, 45, 155, 341, 154, 161, 207, 57, 190, 185, 407, 187, 153, 63, 217, 403, 130, 205, 123, 87, 319, 209, 247, 299, 138, 93, 589, 323, 238, 259, 111, 75, 70, 287, 451, 253, 230, 195, 377
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OFFSET
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1,2
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COMMENTS
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This is a variation of the Enots Wolley sequence A336957 and A360519, with an additional restriction that no term a(n) can have a common factor with n. For the sequence to be infinite a(n) must always have a prime factor that is not a factor of a(n-1)*(n+1). See the examples below.
Other than no term being a prime or prime power, see A336957, no term can be an even number with only two distinct prime factors. Clearly no term a(2*k) can be even, so if we assume that a(2*k+1) = 2^n*p^m, with n and m>=1, then a(2*k) must have p as a factor. But as a(2*k+2) must share a factor with a(2*k+1) and cannot have 2 as a factor, it must also have p as a factor. However that is not allowed as a(n) cannot share a factor with a(n-2), so no term can be even with only two distinct prime factors. Therefore the smallest even number is a(7) = 30.
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LINKS
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EXAMPLE
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a(2) = 15 as 15 is the smallest number that is not a prime power and does not have 2 as a factor.
a(3) = 35 as a(3) is chosen so it shares a factor with a(2) = 3*5 while not having 3 as a factor; it therefore must be a multiple of 5 while not being a power of 5. The smallest number meeting those criteria is 10, but a(2)*(3+1) = 15*4 = 60, and 10 has no prime factor not in 60, so choosing 10 would mean a(4) would not exist. The next smallest available number is 35.
a(4) = 77 as a(4) must be a multiple of 7 but not a power of 7, not a multiple of 2, 3 or 5, while having a prime factor not in 35*(4+1) = 165. The smallest number satisfying these criteria is 77.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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