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A357985
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Counterclockwise square spiral constructed using the integers so that a(n) plus all other numbers currently visible from the current number equals n; start with a(0) = 0.
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2
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0, 1, 1, 1, 2, 1, 3, -1, 6, -2, -1, 0, 1, 9, -8, 15, -5, -7, -10, 14, -29, 58, -78, 101, -118, 150, -61, 309, -307, 553, -494, -186, -644, 315, -1177, 731, -1458, 3480, -5183, 7096, -8328, 9735, -10882, 7200, -29452, 31322, -52670, 51401, -65210, 61001, 11318, 135012, -109687, 259226, -221542
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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.
The magnitude of the numbers grow surprisingly quickly, e.g., a(150) = -4346232663618226. The only known terms that equal zero are a(0) and a(11); it is unknown whether more exist or if all integers eventually appear.
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LINKS
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EXAMPLE
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The spiral begins:
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-5....15...-8....9.....1 553
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-7 2....1.....1 0 -307
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-10 1 0.....1 -1 309
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14 3...-1.....6... -2 -61
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-29...58...-78...101...-118...150
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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 0,1), and 1 (at 1,0). These four numbers sum to 3, so a(6) = 3 so that 3 + 3 = 6.
a(7) = -1 as from square 7, at (0,-1) relative to the starting square, the numbers currently visible are 3 (at -1,-1), 1 (at -1,0), 2 (at -1,1), 0 (at 0,0), 1 (at 1,1), and 1 (at 1,0). These six numbers sum to 8, so a(7) = -1 so that -1 + 8 = 7.
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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