A210996 Number of free polyominoes with 2n cells.

1, 1, 5, 35, 369, 4655, 63600, 901971, 13079255, 192622052, 2870671950, 43191857688, 654999700403, 9999088822075, 153511100594603, 2368347037571252, 36695016991712879, 570694242129491412, 8905339105809603405, 139377733711832678648, 2187263896664830239467, 34408176607279501779592

Offset: 0

Author

Omar E. Pol, Sep 15 2012

Keywords

• nonn

Comments

It appears that we can write A216492(n) < A216583(n) < A056785(n) < A056786(n) < a(n) < A210988(n) < A210986(n), if n >= 3. - Omar E. Pol, Sep 16 2012

Examples

			For n = 1 there is only one free domino. For n = 2 there are 5 free tetrominoes. For n = 3 there are 35 free hexominoes. For n = 4 there are 369 free octominoes (see link section).
		

Links

• John Mason, Table of n, a(n) for n = 0..25
• Herman Tulleken, Polyominoes 2.2: How they fit together, (2019).
• Eric Weisstein's World of Mathematics, Polyomino
• Wikipedia, All 5 free tetrominoes, Illustration of a(2) = 5.
• Wikipedia, All 35 free hexominoes, Illustration of a(3) = 35.
• Wikipedia, All 369 free octominoes, Illustration of a(4) = 369.
• Wikipedia, Polyomino

Crossrefs

Bisection of A000105.

Cf. A000988, A056785, A056786, A210988, A210997, A216492, A216583.

Programs

Mathematica

A000105 = Cases[Import["https://oeis.org/A000105/b000105.txt", "Table"], {, }][[All, 2]];
Partition[A000105, 2][[All, 1]] (* Jean-François Alcover, Jan 03 2020 *)

Formula

a(n) = A000105(2n).

a(n) = A213376(n) + A056785(n). - R. J. Mathar, Feb 08 2023

Extensions

More terms from John Mason, Apr 15 2023