A298267 a(n) is the maximum number of heptiamonds in a hexagon of order n.

0, 3, 7, 13, 21, 30, 42, 54, 69, 85, 103, 123, 144, 168, 192, 219, 247, 277, 309, 342, 378, 414, 453, 493, 535, 579, 624, 672, 720, 771, 823, 877, 933, 990, 1050, 1110, 1173, 1237, 1303, 1371, 1440, 1512, 1584, 1659, 1735, 1813, 1893, 1974, 2058, 2142

Offset: 0

Author

Craig Knecht, Jan 15 2018

Keywords

• nonn

Comments

There are 24 heptiamonds.

It would be nice if this idea could be generalized to state that the hexagon can contain the maximum number of polyiamonds of any given size.

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..10000
• Craig Knecht, H2 Hexagon with 3 heptiamonds packed in.
• Craig Knecht, H3 hexagon with 7 heptiamonds packed in.
• Craig Knecht, H4 H5 H6 H7 heptiamond packing.
• Craig Knecht, Peripheral Buildouts.
• Craig Knecht, Proof notes.

Crossrefs

Cf. A033581 (The number of triangles in a hexagon), A291582 (hexiamond tiling).

Programs

Mathematica

Array[Floor[(6 #^2)/7] &, 50] (* Michael De Vlieger, Jan 20 2018 *)

Formula

a(n) = floor((6*n^2)/7).

Conjectures from Colin Barker, Jan 20 2018: (Start)

G.f.: x*(1 + x)*(3 - 2*x + 4*x^2 - 2*x^3 + 3*x^4) / ((1 - x)^3*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)).

a(n) = 2*a(n-1) - a(n-2) + a(n-7) - 2*a(n-8) + a(n-9) for n>8.

(End)