OFFSET
9,6
LINKS
Christopher Hunt Gribble, C++ program
Christopher Hunt Gribble, Example graphics
FORMULA
It appears that:
T(n,0) = 1, n>= 9
T(n,1) = (floor((n-9)/2)+1)*(floor((n-9)/2+2))/2, n >= 9
T(c+2*9,2) = A131474(c+1)*(9-1) + A000217(c+1)*floor((9-1)(9-3)/4) + A014409(c+2), 0 <= c < 9, c odd
T(c+2*9,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((9-c-1)/2) + A131941(c+1)*floor((9-c)/2)) + S(c+1,3c+2,3), 0 <= c < 9 where
S(c+1,3c+2,3) =
A054252(2,3), c = 0
A236679(5,3), c = 1
A236560(8,3), c = 2
A236757(11,3), c = 3
A236800(14,3), c = 4
A236829(17,3), c = 5
A236865(20,3), c = 6
A236915(23,3), c = 7
A236936(26,3), c = 8
EXAMPLE
The first 17 rows of T(n,k) are:
.\ k 0 1 2 3 4
n
9 1 1
10 1 1
11 1 3
12 1 3
13 1 6
14 1 6
15 1 10
16 1 10
17 1 15
18 1 15 30 5 1
19 1 21 96 74 14
20 1 21 221 413 174
21 1 28 417 1525 1234
22 1 28 705 4290 6124
23 1 36 1107 10269 23259
24 1 36 1638 21630 73204
25 1 45 2334 41790 199436
.
T(18,3) = 5 because the number of equivalence classes of ways of placing 3 9 X 9 square tiles in an 18 X 18 square under all symmetry operations of the square is 5.
CROSSREFS
KEYWORD
tabf,nonn
AUTHOR
Christopher Hunt Gribble, Feb 01 2014
STATUS
approved