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A357991
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Lexicographically earliest counterclockwise square spiral constructed using the nonnegative integers so that a(n) plus all other numbers currently visible from the current number form a distinct sum; start with a(0) = 0.
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2
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0, 1, 1, 1, 2, 1, 3, 0, 4, 0, 0, 0, 1, 5, 0, 6, 0, 0, 1, 0, 2, 4, 0, 7, 0, 8, 0, 7, 0, 7, 0, 0, 0, 0, 0, 0, 0, 12, 0, 13, 0, 16, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 12, 0, 22, 0, 19, 0, 20, 1, 0, 0, 0, 0, 0, 0, 0, 0, 17, 0, 25, 0, 24, 0, 20, 1, 26, 0, 28, 0, 26, 0, 31, 0, 31, 0, 0, 0, 0
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OFFSET
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0,5
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COMMENTS
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A number is visible from the current number if, given that it has coordinates (x,y) relative to the current number, the greatest common divisor of |x| and |y| is 1.
In the first 50000 terms the smallest number that has not appeared is 9; it is unknown if all the positive numbers eventually appear.
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LINKS
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EXAMPLE
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The spiral begins:
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0---6---0---5---1 7
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0 2---1---1 0 0
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1 1 0---1 0 7
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0 3---0---4---0 0
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2---4---0---7---0---8
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a(6) = 3 as from square 6, at (-1,-1) relative to the starting square, the numbers currently visible are 1 (at -1,0), 0 (at 0,0), 1 (at 1,0), and 1 (at 0,1). These three numbers sum to 3, so a(6) = 3 so that 3 + 3 = 6, the smallest sum that has not previous occurred.
a(8) = 4 as from square 8, at (1,-1) relative to the starting square, the numbers currently visible are 0 (at 0,-1), 1 (at -1,0), 0 (at 0,0), 1 (at 1,0), and 1 (at 0,1). These five numbers sum to 3, so a(8) = 4 so that 3 + 4 = 7, the smallest sum that has not previous occurred. Note that a(7) = 0 and forms a sum of 8.
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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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