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A280079
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.
4
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
OFFSET
1,4
COMMENTS
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.
EXAMPLE
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.
MATHEMATICA
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]]
CROSSREFS
A280317 are the corresponding ordinates y.
Sequence in context: A340542 A351167 A283303 * A116513 A122651 A343378
KEYWORD
nonn
AUTHOR
STATUS
approved