A157791 Least number of lattice points on two adjacent sides from which every point of a square n X n lattice is visible.

1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5

Offset: 1

Author

T. D. Noe, Mar 06 2009

Keywords

• hard
• nonn

Comments

That is, the points are chosen from the 2n-1 points on two adjacent sides of the n X n lattice.

Examples

			a(11)= 4 because all 121 points are visible from (1,1), (1,2), (2,1), and (1,4).
a(25)= 4 because all 625 points are visible from (1,2), (4,1), (6,1), and (23,1).
		

Links

• Eric Weisstein's World of Mathematics, Visible Point

Crossrefs

Cf. A157639, A157720, A157790, A157792.

Programs

Mathematica

Join[{1}, Table[hidden=Table[{},{n^2}]; edgePts={}; Do[pt1=(c-1)*n+d; If[c==1||d==1, AppendTo[edgePts,pt1]; lst={}; Do[pt2=(a-1)*n+b; If[GCD[c-a,d-b]>1, AppendTo[lst,pt2]], {a,n}, {b,n}]; hidden[[pt1]]=lst], {c,n}, {d,n}]; edgePts=Sort[edgePts]; done=False; k=0; done=False; k=0; While[ !done, k++; len=Binomial[2n-1,k]; i=0; While[i
				

Extensions

More terms from Lars Blomberg, Nov 06 2014