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 A157639 Least number of lattice points from which every point of a square n X n lattice is visible. 5
 1, 1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 COMMENTS Adhikari and Granville give bounds on the size of a(n). From Jon E. Schoenfield, Aug 03 2009, Aug 09 2009, Sep 02 2009: (Start) Consider an arbitrarily large lattice. Define S1 as the square having both X and Y in the closed interval [1,3]. From a single viewpoint at (2,2), every lattice point on or inside S1 is visible. S1 (including its edges) covers 3 rows X 3 columns of lattice points, so a(3)=1. Also, since an n-row X n-column subset of S1 that includes the one viewpoint can be selected for n=1 and for n=2, this 1-viewpoint example is sufficient to show that a(n)=1 for n = 1..3. Define S2 as the square having both X and Y in [1,5]. Every lattice point in S2 is visible from at least one of the two viewpoints (3,1) and (3,4). S2 covers 5 rows X 5 columns of points, so a(5) <= 2. Since a 4-row X 4-column subset of S2 that includes the two viewpoints can also be selected, a(4) <= 2. Since no 1-viewpoint solution exists for n > 3, we have a(n)=2 for n=4 and n=5. Proceeding similarly, every lattice point where X and Y are both in [1,23] is visible from at least one of the three viewpoints (14,14), (15,14), and (14,15), so a(n) <= 3 for n = 2..23. Since no 2-viewpoint solution exists for n > 5, we have a(n)=3 for n = 6..23. Together, the three 4-viewpoint solutions {(20,20), (21,20), (20,21), (21,21)}, {(43,51), (72,65), (58,80), (57,66)}, and {(28,34), (105,105), (34,99), (99,40)} are sufficient to show that a(n) <= 4 for n = 24..132: in these solutions, every lattice point where X and Y are both in [1,m], where m = 40, 128, and 132, respectively, is visible from at least one viewpoint, and all four viewpoints would fit in a k-row X k-column subset of the m-row X m-column square, for k as small as 2, 30, and 78, respectively. Thus these three solutions demonstrate that a(n) <= 4 for the overlapping ranges n = 2..40, n = 30..128, and n = 78..132, respectively. Since (per an exhaustive search) no 3-viewpoint solution exists for n = 24..132, we have a(n)=4 for n = 24..132. Per exhaustive search, no 4-viewpoint solution exists for n=133, so a(133)=5. In summary: a(1..3)=1, a(4..5)=2, a(6..23)=3, a(24..132)=4, a(133)=5. (End) LINKS Sukumar Das Adhikari and Andrew Granville, Visibility in the Plane Eric Weisstein's World of Mathematics, Visible Point EXAMPLE a(3) = 1 because all 9 points are visible from the central (2,2) point. a(4) = 2 because all 16 points are visible from (1,2) or (2,1). a(6) = 3 because all 36 points are visible from (1,1), (1,2), or (2,1). a(24)= 4 because all 576 points are visible from (1,1), (1,2), (1,3), or (2,24). MATHEMATICA Table[sees=Table[{}, {n^2}]; Do[pt1=(c-1)*n+d; lst={}; Do[pt2=(a-1)*n+b; If[GCD[c-a, d-b]<2, AppendTo[lst, pt2]], {a, n}, {b, n}]; sees[[pt1]]=lst, {c, n}, {d, n}]; done=False; k=0; While[ !done, k++; len=Binomial[n^2, k]; i=0; While[i

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