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Maximum degree of the polyomino graph PG(n) defined in A367435.
+20
5
0, 0, 1, 4, 10, 28, 38, 67, 80, 115, 139, 186, 203
Number of Hamiltonian cycles in the polyomino graph PG(n) defined in A367435.
+20
4
COMMENTS
A cycle and its reverse are not both counted.
We follow the convention in A003216 that the complete graphs on 1 and 2 nodes have 1 and 0 Hamiltonian cycles, respectively, so that a(1) = a(2) = 1 and a(3) = 0, but it could also be argued that a(1) = a(2) = 0 and/or a(3) = 1.
FORMULA
a(n) > 0 for 4 <= n <= 13.
Number of polyominoes with n cells that have the maximum degree ( A367437(n)) in the polyomino graph PG(n) defined in A367435.
+20
4
1, 1, 2, 2, 3, 1, 1, 1, 4, 2, 2, 1, 9
a(n) is the degree of the polyomino with binary code A246521(n+1) in the polyomino graph PG(n) defined in A367435.
+20
4
0, 0, 1, 1, 4, 3, 4, 3, 2, 10, 8, 3, 9, 10, 9, 8, 9, 10, 8, 4, 2, 15, 28, 15, 12, 12, 10, 17, 14, 19, 20, 15, 14, 15, 13, 18, 20, 9, 14, 13, 17, 4, 12, 16, 18, 11, 9, 10, 15, 22, 19, 10, 19, 14, 16, 3, 36, 36, 35, 31, 28, 30, 36, 22, 29, 37, 16, 11, 28, 13, 24
COMMENTS
Number of free polyominoes that can be made from the polyomino with binary code A246521(n+1) by moving one of its cells (not counting itself), where the intermediate (the set of cells remaining when the cell to be moved is detached) is required to be a (connected) polyomino. Can be read as an irregular triangle, whose m-th row contains A000105(m) terms, m >= 1.
EXAMPLE
As an irregular triangle:
0;
0;
1, 1;
4, 3, 4, 3, 2;
10, 8, 3, 9, 10, 9, 8, 9, 10, 8, 4, 2;
...
For n = 8, A246521(8+1) = 30 is the binary code of the S-tetromino. By moving one cell of the S-tetromino, we can obtain the L, O, and T tetrominoes (but not the I tetromino), so a(8) = 3.
Independence number of the polyomino graph PG(n) defined in A367435.
+20
4
COMMENTS
Maximum size of a set of free polyominoes with n cells such that no polyomino in the set can be obtained from another by moving one of its cells, where the intermediate (the set of cells remaining when the cell to be moved is detached) is required to be a (connected) polyomino.
Define the n-omino graph to be the graph whose vertices are each of the n-ominoes, two of which are joined by an edge if one can be obtained from the other by cutting out one of the latter's component squares (thus obtaining an (n-1)-omino for most cases) and gluing it elsewhere. The sequence counts the edges in these graphs.
+10
8
0, 0, 1, 8, 47, 266, 1339, 6544, 29837, 133495, 585002, 2542563, 10959656
COMMENTS
In some cases the act of removing a component square (temporarily) disconnects the polyomino before the component is reattached elsewhere. - Sean A. Irvine, Apr 13 2020 See A367435 for the case where the cells remaining after detaching the square to be moved must be a connected polyomino. - Pontus von Brömssen, Nov 18 2023
Minimum size of a set of polyominoes with n-1 cells such that all free polyominoes with n cells can be obtained by adding one cell to one of the polyominoes in the set.
+10
3
COMMENTS
a(8) <= 54, a(9) <= 160.
Apparently, a(n) is close to A365621(n+1) for n <= 8. Is this just a coincidence?
EXAMPLE
For n <= 4, all polyominoes with n-1 cells are needed to obtain all polyominoes with n cells by adding one cell, so a(n) = A000105(n-1). For n = 5, all but the square tetromino are needed to obtain all pentominoes, so a(5) = A000105(4)-1 = 4. For n = 6, there are 5 different sets of a(6) = 8 pentominoes that are sufficient to obtain all hexominoes. One of these sets consists of the I, L, N, P, U, V, W, and Y pentominoes. The X pentomino is the only pentomino that does not appear in any of these sets. The I, L, N, and W pentominoes are needed in all such sets.
For n = 7, there are 8 different sets of a(7) = 19 hexominoes that are sufficient to obtain all heptominoes. 14 hexominoes appear in all these sets, 10 appear in none of them.
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