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A257112
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Arrange numbers in a clockwise spiral with initial terms a(1)=1, a(2)=2, a(4)=4, a(6)=6, a(8)=8, a(11)=3, a(15)=5, a(19)=7, a(23)=9; thereafter each number is relatively prime to all of its four (N,S,E,W) neighbors, but shares a factor with each of its (N,S,E,W) neighbors at distance 2 and also satisfies an additional condition stated in the comments.
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9
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1, 2, 11, 4, 55, 6, 25, 8, 165, 14, 3, 16, 15, 26, 5, 12, 35, 18, 7, 22, 21, 32, 9, 28, 27, 10, 33, 20, 77, 34, 49, 38, 231, 46, 121, 24, 143, 36, 65, 44, 45, 52, 51, 58, 75, 56, 39, 40, 57, 50, 63, 62, 69, 64, 81, 68, 87, 17, 93, 136, 105, 74, 85, 42, 95, 48, 115, 54, 161
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OFFSET
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1,2
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COMMENTS
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To formulate the additional condition, let us call two numbers strictly connected if the set of prime divisors of one of them is a subset of the set of prime divisors of the other. Then the positions of two strictly connected terms should not be a knight's move apart.
Start with smallest number which has not yet appeared and satisfies the conditions: a(3)=11; thereafter always choose smallest number which has not yet appeared and satisfies the conditions.
This is a two-dimensional spiral analog of A098550.
In A098550 we have initial terms in the positions 1,2,3.
In the two-dimensional case we have 4 sides. So the initial TERMS are
9
8
7 6 1 2 3 (1)
4
5
But the POSITIONS in the spiral are indexed thus:
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7--8--9--10
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6 1--2
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5--4--3
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So the initial terms, by (1), are a(1)=1, a(2)=2, a(4)=4, a(6)=6, a(8)=8, ...
Conjecture: the sequence is a permutation of the positive integers. - Vladimir Shevelev, May 06 2015
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LINKS
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EXAMPLE
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The spiral begins
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21---32----9---28---27---10 etc.
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22 25----8--165---14
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7 6 1----2 3
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18 55----4---11 16
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35---12----5---26---15
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Formally the smallest a(12) is 10, but then 10 and 5 are strictly connected numbers on a knight move (and a(13) would not exist). So the smallest suitable a(12)=16.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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