cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A157790 Least number of lattice points on two opposite sides 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, 3, 4, 4, 4, 3, 4, 3, 4, 4, 4, 4, 6, 4, 5, 5, 4, 4, 7, 4, 5, 5, 6, 4, 8, 4, 6, 5, 6, 4, 8, 4, 6, 5, 7, 4, 8, 4, 6, 6, 6, 4, 8, 4, 8, 5, 6, 4, 8, 5, 7, 5, 6, 4, 8, 5, 6, 6, 6, 5, 8, 4, 6, 5
Offset: 1

Views

Author

T. D. Noe, Mar 06 2009

Keywords

Comments

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

Examples

			a(8) = 3 because all 64 points are visible from (1,1), (1,2), and (8,2).
a(9) = 4 because all 81 points are visible from (1,1), (1,2), (9,1), and (9,2).
		

Crossrefs

Programs

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

Extensions

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

Views

Author

T. D. Noe, Mar 06 2009

Keywords

Comments

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.

Examples

			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).
		

Crossrefs

Programs

  • Mathematica
    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
    				

Formula

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)

Extensions

More terms from Lars Blomberg, Nov 06 2014
Showing 1-2 of 2 results.