A157790 -id:A157790 - cp's OEIS frontend

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

T. D. Noe, Mar 06 2009

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

			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).

Eric Weisstein's World of Mathematics, Visible Point

Cf. A157639, A157720, A157790, A157792.

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

More terms from Lars Blomberg, Nov 06 2014

A157792 Least number of lattice points on one side from which every point of a square n X n lattice is visible.

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 12, 12, 12, 13, 13, 13, 14, 14, 14, 15, 15, 15, 16, 16, 16, 17, 17, 17, 18, 18, 18, 19, 19, 19, 20, 20, 20, 21, 21, 21, 22, 22, 22, 23, 23, 23, 24, 24, 24, 25

Offset: 1

T. D. Noe, Mar 06 2009

That is, the points are chosen from the n points on one side of the n X n lattice. It appears that a(n) = ceiling((n+1)/3) for n > 8.

			a(3) = 2 because all 9 points are visible from (1,1) and (1,2).
a(5) = 3 because all 25 points are visible from (1,1), (1,2), and (1,4).
a(7) = 4 because all 49 points are visible from (1,1), (1,2), (1,3), and (1,6).
a(12)= 5 because all 144 points are visible from (1,1), (1,3), (1,6), (1,8), and (1,11).

Eric Weisstein's World of Mathematics, Visible Point

Cf. A157639, A157720, A157790, A157791.

Join[{1}, Table[hidden=Table[{},{n^2}]; edgePts={}; Do[pt1=(c-1)*n+d; If[c==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[n,k]; i=0; While[i

Conjectures from Chai Wah Wu, Aug 05 2022: (Start)
a(n) = a(n-1) + a(n-3) - a(n-4) for n > 12.
G.f.: x*(x^11 - x^9 - x^7 + x^6 - x^5 + x^4 - x^3 + x^2 + 1)/(x^4 - x^3 - x + 1). (End)

More terms from Lars Blomberg, Nov 06 2014

cp's OEIS Frontend

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

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