

A119611


Number of free polyominoes in {4,5} tessellation of the hyperbolic plane.


0




OFFSET

0,4


COMMENTS

Each tessellation in the hyperbolic plane is represented by a SchlĂ¤fli symbol of the form {p,q}, which means that q regular pgons surround each vertex. There exists a hyperbolic tessellation {p,q} for every p,q such that (p2)*(q2) > 4. I have, in a paper not referenced here explicitly, described polyiamonds and enumerated in the Klein curve, topologically derived from the {3,7} or dually the {7,3} hyperbolic tessellation, as a sideeffect of defining polyheptagons in the Klein quartic.


LINKS

Table of n, a(n) for n=0..6.
Don Hatch, Hyperbolic Planar Tesselations: {4,5}.
Eric Weisstein's World of Mathematics, Polyomino.


EXAMPLE

For n = 0,1,2,3 the polyominoes in Euclidean Golombics A000105 are essentially same as in the {4,5} tessellation of the hyperbolic plane, with redefinition of "straight line" and angular deficiency at a vertex.
For n = 4, the square tetromino does not exist. In its place is the cutsquare, a pentagonal pentomino with one cell removed; see n = 5.
For n = 5, we have modified versions of 11 of the 12 Euclidean pentominoes, but not the Ppentomino, as that has the square tetromino as a subpolyomino, with one adjacent monomino. In place of the P we have 4 unique hyperbolic pentominoes. First, the aforementioned pentagonal pentomino, with 5fold symmetry, embedded in the space where 5 right angles define a full rotation. Next, the cutsquare tetromino can have an adjacent monomino in 4 nonisomorphic positions. 12  1 + 4 = 15 hyperbolic pentominoes.
For n = 6 we lose the 8 Euclidean hexominoes that have the square tetromino as a subpolyomino. In their place, we have the pentagonal pentomino with an adjacent monomino; the cutsquare with an adjacent domino in 12 nonisomorphic positions; and the cutsquare with two separate adjacent monominoes in 16 nonisomorphic positions. 35  8 + 29 = 56 hyperbolic hexominoes.
Comment about the example "For = 5, etc.": What? You have described five replacements for the P. So I count 12  1 + 5 = 16.  Don Knuth, Oct 14 2016


CROSSREFS

Cf. A000105.
Sequence in context: A137533 A121392 A216388 * A005976 A187981 A048192
Adjacent sequences: A119608 A119609 A119610 * A119612 A119613 A119614


KEYWORD

nonn,uned,more


AUTHOR

Jonathan Vos Post, Jun 04 2006


STATUS

approved



