A236679 Number T(n,k) of equivalence classes of ways of placing k 2 X 2 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=2, 0<=k<=floor(n/2)^2, read by rows.
1, 1, 1, 1, 1, 3, 4, 2, 1, 1, 3, 13, 20, 14, 1, 6, 37, 138, 277, 273, 143, 39, 7, 1, 1, 6, 75, 505, 2154, 5335, 7855, 6472, 2756, 459, 1, 10, 147, 1547, 10855, 50021, 153311, 311552, 416825, 361426, 200996, 71654, 16419, 2363, 211, 11, 1, 1, 10, 246, 3759, 39926, 291171
Offset: 2
Examples
T(4,2) = 4 because the number of equivalence classes of ways of placing 2 2 X 2 square tiles in a 4 X 4 square under all symmetry operations of the square is 4. The portrayal of an example from each equivalence class is: ._______ _______ _______ _______ | . | . | | . |___| | . | | |_______| |___|___| |___| . | |___|___| | . | . | | | | |___| | | . | |___|___| |_______| |_______| |___|___| |_______| The first 6 rows of T(n,k) are: .\ k 0 1 2 3 4 5 6 7 8 9 n 2 1 1 3 1 1 4 1 3 4 2 1 5 1 3 13 20 14 6 1 6 37 138 277 273 143 39 7 1 7 1 6 75 505 2154 5335 7855 6472 2756 459
Links
- Heinrich Ludwig, Table of n, a(n) for n = 2..107
- Christopher Hunt Gribble, C++ program
Crossrefs
Formula
It appears that:
T(n,0) = 1, n>= 2
T(n,1) = (floor((n-2)/2)+1)*(floor((n-2)/2+2))/2, n >= 2
T(c+2*2,2) = A131474(c+1)*(2-1) + A000217(c+1)*floor((2-1)(2-3)/4) + A014409(c+2), 0 <= c < 2, c odd
T(c+2*2,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((2-c-1)/2) + A131941(c+1)*floor((2-c)/2)) + S(c+1,3c+2,3), 0 <= c < 2 where
S(c+1,3c+2,3) =
A054252(2,3), c = 0
A236679(5,3), c = 1
Extensions
More terms from Heinrich Ludwig, Dec 11 2016 (The former entry A279118 from Heinrich Ludwig was merged into this entry by N. J. A. Sloane, Dec 21 2016)
Comments