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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. 2
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 (list; graph; refs; listen; history; text; internal format)
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.

LINKS

Table of n, a(n) for n=1..100.

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

Sequence in context: A165360 A340542 A283303 * A116513 A122651 A343378

Adjacent sequences:  A280076 A280077 A280078 * A280080 A280081 A280082

KEYWORD

nonn

AUTHOR

Robert G. Wilson v and Andres Cicuttin, Dec 25 2016

STATUS

approved

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Last modified April 18 06:14 EDT 2021. Contains 343072 sequences. (Running on oeis4.)