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A280079
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Abscissa of points (x,y) of the square lattice such that x >= 0 and 0 <= y <= x, and ranked in order of increasing distance from the origin. Equidistant points are ranked in order of increasing ordinate.
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4
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0, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 3, 4, 5, 4, 5, 5, 4, 5, 6, 6, 6, 5, 6, 7, 7, 5, 6, 7, 7, 6, 8, 8, 7, 8, 6, 8, 7, 8, 9, 9, 9, 7, 8, 9, 9, 7, 10, 8, 10, 10, 9, 10, 8, 10, 9, 11, 11, 11, 10, 8, 11, 9, 10, 11, 12, 12, 9, 11, 12, 10, 12, 11, 12, 9, 10, 13, 12, 13, 11, 13, 13, 12, 10, 13, 11, 12, 13, 14, 14, 14, 10, 11, 14, 13, 12, 14, 13, 14, 11
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OFFSET
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1,4
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COMMENTS
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The points belong to the first half of the first quadrant, and in order are (0,0), (1,0), (1,1), (2,0), (2,1), (2,2), (3,0), (3,1), (3,2), (4,0), (4,1), (3,3), (4,2), (5,0), (4,3), (5,1), (5,2), etc.
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LINKS
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EXAMPLE
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a(12) = 3 since the twelfth point in distance from the origin is (3,3) at a distance of 3*sqrt(2) = 4.242640... whereas the eleventh is (4,1) at a distance of sqrt(17) = 4.12310... and the thirteenth is (4,2) at a distance of 2*sqrt(5) = 4.472113... .
The fourteenth and fifteenth points are respectively (5,0) and (4,3) and have the same distance 5 to the origin, but (5,0) has a smaller ordinate than (4,3), so a(14) = 5 and a(15) = 4.
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MATHEMATICA
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xmax = 20; (* Maximum explorative abscissa *)
(* t are points in the triangle of vertices (0, 0), (0, max) and (xmax, xmax) *)
t = Flatten[Table[{x, y}, {x, 0, xmax}, {y, 0, x}], 1];
nmax = Floor[xmax^2/4] (* Safe limit for correctly sorted sequence *)
Transpose[SortBy[t, {#[[1]]^2 + #[[2]]^2 &, #[[2]] &}]][[1]][[1 ;;
nmax]]
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CROSSREFS
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A280317 are the corresponding ordinates y.
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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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