A030228 Number of chiral polyominoes with n cells.

0, 0, 0, 0, 2, 6, 25, 88, 335, 1215, 4534, 16823, 63159, 237679, 900341, 3423201, 13073163, 50095285, 192599091, 742576616, 2870584814, 11122879867, 43191525139, 168046317330, 654998425237, 2557224396342, 9999083912711, 39153000738695, 153511081627903

Offset: 0

Author

David W. Wilson

Keywords

• nonn

Comments

For n>0, A000105(n) + a(n) = A000988(n) because the number of free polyominoes plus the number of polyominoes lacking bilateral symmetry equals the number of one-sided polyominoes. - Graeme McRae, Jan 05 2006

For n>0, each chiral pair is counted as one. - Robert A. Russell, Feb 23 2022

Examples

			For a(4)=2, the two chiral tetrominoes are XXX and XX .
                                           X        XX
		

Links

• John Mason, Table of n, a(n) for n = 0..50 (terms 1..45 from Andrew Howroyd, 46..48 from Robert A. Russell)
• D. H. Redelmeier, Counting polyominoes: yet another attack, Discrete Math., 36 (1981), 191-203.

Crossrefs

Cf. A000988 (oriented), A000105 (unoriented), A030227 (achiral).

Cf. A006747, A006749, A144553 (subcategories).

Programs

Mathematica

A000105 = Cases[Import["https://oeis.org/A000105/b000105.txt", "Table"], {, }][[All, 2]];
A000988 = Cases[Import["https://oeis.org/A000988/b000988.txt", "Table"], {, }][[All, 2]];
a[n_] := A000988[[n]] - A000105[[n + 1]];
Array[a, 45] (* Jean-François Alcover, Sep 08 2019, after Graeme McRae *)

Formula

For n>0, a(n) = A000988(n) - A000105(n). - Graeme McRae, Jan 05 2006

a(n) = A006749(n) + A006747(n) + A144553(n). - Andrew Howroyd, Dec 04 2018

a(n) = A000105(n) - A030227(n). - Robert A. Russell, Feb 02 2019

For n>0, (A000988(n) - A030227(n)) / 2. - Robert A. Russell, Feb 23 2022

Extensions

Terms a(23) and beyond from Andrew Howroyd, Dec 04 2018

Name edited by Robert A. Russell, Feb 03 2019

a(0)=0 corrected by John Mason, Jan 12 2023