A183869 a(n) = n + floor(sqrt(4*n + 5)); complement of A004116.

2, 4, 5, 7, 8, 10, 11, 12, 14, 15, 16, 18, 19, 20, 21, 23, 24, 25, 26, 28, 29, 30, 31, 32, 34, 35, 36, 37, 38, 40, 41, 42, 43, 44, 45, 47, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 59, 60, 62, 63, 64, 65, 66, 67, 68, 70, 71, 72, 73, 74, 75, 76, 77, 79, 80, 81, 82, 83, 84, 85, 86, 88, 89, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100, 101, 102, 103, 104, 105, 106, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 119, 120

Offset: 0

Author

Clark Kimberling, Jan 07 2011

Keywords

• nonn
• easy

Comments

a(n-2) appears to be the minimum number of rectangular tiles to place on an n X n grid of unit squares, possibly of different sizes, such that each side of every tile lies on a grid line, every unit square is covered by at most one tile, and each row and each column of the grid has exactly one unit square that is not covered by any tile. - Yifan Xie, Jul 19 2025 [The conjecture is proven. - Yifan Xie, Jul 24 2025]

Links

• Art of Problem Solving, 2025 IMO Problems/Problem 6
• Jinyuan Wang, Tiling constructions for a(n-2) on n X n grids, n = 2..16

Crossrefs

Cf. A000267, A004116.

Programs

Mathematica

a=4; b=5; Table[n+Floor[(a*n+b)^(1/2)],{n,0,100}]

PARI

a(n) = n + sqrtint(4*n+5); \\ Michel Marcus, Jul 19 2025