cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 19 results. Next

A279111 Number of non-equivalent ways to place 2 non-attacking kings on an n X n board.

Original entry on oeis.org

0, 0, 4, 13, 37, 75, 147, 246, 406, 610, 910, 1275, 1779, 2373, 3157, 4060, 5212, 6516, 8136, 9945, 12145, 14575, 17479, 20658, 24402, 28470, 33202, 38311, 44191, 50505, 57705, 65400, 74104, 83368, 93772, 104805, 117117, 130131, 144571, 159790, 176590, 194250, 213654
Offset: 1

Views

Author

Heinrich Ludwig, Dec 06 2016

Keywords

Comments

Rotations and reflections of placements are not counted. If they are to be counted, see A061995.

Examples

			There are 4 non-equivalent ways to place 2 non-attacking kings on a 3 X 3 board:
   K.K   K..   K..   .K.
   ...   ..K   ...   ...
   ...   ...   ..K   .K.

Links

Crossrefs

Cf. A061995, A279112 (3 kings), A279113 (4 kings), A279114 (5 kings), A279115 (6 kings), A279116 (7 kings), A279117, A236679.

Programs

  • Mathematica
    Table[(n^4 - 2 n^2 - 4 n + Boole[OddQ@ n] (2 n^2 - 4 n + 7))/16, {n, 43}] (* or *)
Rest@ CoefficientList[Series[x^3*(4 + 5 x + 3 x^2 - x^3 + x^4)/((1 - x)^5*(1 + x)^3), {x, 0, 43}], x] (* Michael De Vlieger, Dec 08 2016 *)
  • PARI
    concat(vector(2), Vec(x^3*(4 + 5*x + 3*x^2 - x^3 + x^4) / ((1 - x)^5*(1 + x)^3) + O(x^60))) \\ Colin Barker, Dec 07 2016

Formula

a(n) = (n^4 - 2*n^2 - 4*n + IF(MOD(n, 2) = 1, 2*n^2 - 4*n + 7))/16.
a(n) = (2*n^4 - 2*n^2 - 12*n + 7 - (2*n^2 - 4*n + 7)*(-1)^n)/32. - Bruno Berselli, Dec 07 2016
a(n) = 2*a(n-1) + 2*a(n-2) - 6*a(n-3) + 6*a(n-5) - 2*a(n-6) - 2*a(n-7) + a(n-8).
From Colin Barker, Dec 07 2016: (Start)
a(n) = n*(n - 2)*(n^2 + 2*n + 2)/16 for n even.
a(n) = (n - 1)*(n^3 + n^2 + n - 7)/16 for n odd.
G.f.: x^3*(4 + 5*x + 3*x^2 - x^3 + x^4) / ((1 - x)^5*(1 + x)^3).
(End)

A236560 Number T(n,k) of equivalence classes of ways of placing k 3 X 3 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=3, 0<=k<=floor(n/3)^2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 3, 6, 2, 1, 1, 6, 21, 29, 14, 1, 6, 53, 161, 174, 1, 10, 111, 665, 1713, 1549, 608, 107, 11, 1, 1, 10, 201, 1961, 9973, 24267, 29437, 17438, 4756, 459
Offset: 3

Views

Author

Christopher Hunt Gribble, Jan 30 2014

Keywords

Comments

The first 8 rows of T(n,k) are:
.\ k 0 1 2 3 4 5 6 7 8 9
n
3 1 1
4 1 1
5 1 3
6 1 3 6 2 1
7 1 6 21 29 14
8 1 6 53 161 174
9 1 10 111 665 1713 1549 608 107 11 1
10 1 10 201 1961 9973 24267 29437 17438 4756 459

Examples

			T(6,2) = 6 because the number of equivalence classes of ways of placing 2 3 X 3 square tiles in a 6 X 6 square under all symmetry operations of the square is 6. The portrayal of an example from each equivalence class is:
.___________      ___________      ___________
|     |     |    |     |_____|    |     |     |
|  .  |  .  |    |  .  |     |    |  .  |_____|
|_____|_____|    |_____|  .  |    |_____|     |
|           |    |     |_____|    |     |  .  |
|           |    |           |    |     |_____|
|___________|    |___________|    |_____|_____|
.
.___________      ___________      ___________
|     |     |    |_____ _____|    |_____      |
|  .  |     |    |     |     |    |     |_____|
|_____|_____|    |  .  |  .  |    |  .  |     |
|     |     |    |_____|_____|    |_____|  .  |
|     |  .  |    |           |    |     |_____|
|_____|_____|    |___________|    |_____|_____|

Links

Crossrefs

Cf. A054252, A236679, A236757, A236800, A236829, A236865, A236915, A236936, A236939.

Formula

It appears that:
T(n,0) = 1, n>= 3
T(n,1) = (floor((n-3)/2)+1)*(floor((n-3)/2+2))/2, n >= 3
T(c+2*3,2) = A131474(c+1)*(3-1) + A000217(c+1)*floor(3^2/4) + A014409(c+2), 0 <= c < 3, c even
T(c+2*3,2) = A131474(c+1)*(3-1) + A000217(c+1)*floor((3-1)(3-3)/4) + A014409(c+2), 0 <= c < 3, c odd
T(c+2*3,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((3-c-1)/2) + A131941(c+1)*floor((3-c)/2)) + S(c+1,3c+2,3), 0 <= c < 3 where
S(c+1,3c+2,3) =
A054252(2,3), c = 0
A236679(5,3), c = 1
A236560(8,3), c = 2

A236757 Number T(n,k) of equivalence classes of ways of placing k 4 X 4 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=4, 0<=k<=floor(n/4)^2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 3, 1, 6, 9, 3, 1, 1, 6, 29, 35, 14, 1, 10, 75, 209, 174, 1, 10, 147, 765, 1234, 1, 15, 270, 2340, 7639, 6169, 1893, 242, 17, 1, 1, 15, 438, 5806, 34342, 79821, 80722, 36569, 7106, 459
Offset: 4

Views

Author

Christopher Hunt Gribble, Jan 30 2014

Keywords

Comments

The first 10 rows of T(n,k) are:
.\ k 0 1 2 3 4 5 6 7 8 9
n
4 1 1
5 1 1
6 1 3
7 1 3
8 1 6 9 3 1
9 1 6 29 35 14
10 1 10 75 209 174
11 1 10 147 765 1234
12 1 15 270 2340 7639 6169 1893 242 17 1
13 1 15 438 5806 34342 79821 80722 36569 7106 459

Examples

			T(8,3) = 3 because the number of equivalence classes of ways of placing 3 4 X 4 square tiles in an 8 X 8 square under all symmetry operations of the square is 3. The portrayal of an example from each equivalence class is:
._____________          _____________          _____________
|      |      |        |      |______|        |      |      |
|   .  |   .  |        |   .  |      |        |   .  |______|
|      |      |        |      |   .  |        |      |      |
|______|______|        |______|      |        |______|   .  |
|      |      |        |      |______|        |      |      |
|   .  |      |        |   .  |      |        |   .  |______|
|      |      |        |      |      |        |      |      |
|______|______|        |______|______|        |______|______|

Links

Crossrefs

Cf. A054252, A236679, A236560, A236800, A236829, A236865, A236915, A236936, A236939.

Formula

It appears that:
T(n,0) = 1, n>= 4
T(n,1) = (floor((n-4)/2)+1)*(floor((n-4)/2+2))/2, n >= 4
T(c+2*4,2) = A131474(c+1)*(4-1) + A000217(c+1)*floor(4^2/4) + A014409(c+2), 0 <= c < 4, c even
T(c+2*4,2) = A131474(c+1)*(4-1) + A000217(c+1)*floor((4-1)(4-3)/4) + A014409(c+2), 0 <= c < 4, c odd
T(c+2*4,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((4-c-1)/2) + A131941(c+1)*floor((4-c)/2)) + S(c+1,3c+2,3), 0 <= c < 4 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 Number T(n,k) of equivalence classes of ways of placing k 5 X 5 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=5, 0<=k<=floor(n/5)^2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 3, 1, 6, 1, 6, 12, 3, 1, 1, 10, 40, 44, 14, 1, 10, 97, 245, 174, 1, 15, 193, 925, 1234, 1, 15, 339, 2640, 6124, 1, 21, 555, 6617, 27074, 19336, 4785, 461, 23, 1
Offset: 5

Views

Author

Christopher Hunt Gribble, Jan 31 2014

Keywords

Comments

The first 11 rows of T(n,k) are:
.\ k 0 1 2 3 4 5 6 7 8 9
n
5 1 1
6 1 1
7 1 3
8 1 3
9 1 6
10 1 6 12 3 1
11 1 10 40 44 14
12 1 10 97 245 174
13 1 15 193 925 1234
14 1 15 339 2640 6124
15 1 21 555 6617 27074 19336 4785 461 23 1

Examples

			T(10,3) = 3 because the number of equivalence classes of ways of placing 3 5 X 5 square tiles in an 10 X 10 square under all symmetry operations of the square is 3. The portrayal of an example from each equivalence class is:
._______________      _______________      _______________
|       |       |    |       |_______|    |       |       |
|       |       |    |       |       |    |       |_______|
|   .   |   .   |    |   .   |       |    |   .   |       |
|       |       |    |       |   .   |    |       |       |
|_______|_______|    |_______|       |    |_______|   .   |
|       |       |    |       |_______|    |       |       |
|       |       |    |       |       |    |       |_______|
|   .   |       |    |   .   |       |    |   .   |       |
|       |       |    |       |       |    |       |       |
|_______|_______|    |_______|_______|    |_______|_______|

Links

Crossrefs

Cf. A054252, A236679, A236560, A236757, A236829, A236865, A236915, A236936, A236939.

Formula

It appears that:
T(n,0) = 1, n>= 5
T(n,1) = (floor((n-5)/2)+1)*(floor((n-5)/2+2))/2, n >= 5
T(c+2*5,2) = A131474(c+1)*(5-1) + A000217(c+1)*floor(5^2/4) + A014409(c+2), 0 <= c < 5, c even
T(c+2*5,2) = A131474(c+1)*(5-1) + A000217(c+1)*floor((5-1)(5-3)/4) + A014409(c+2), 0 <= c < 5, c odd
T(c+2*5,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((5-c-1)/2) + A131941(c+1)*floor((5-c)/2)) + S(c+1,3c+2,3), 0 <= c < 5 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 Number T(n,k) of equivalence classes of ways of placing k 6 X 6 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=6, 0<=k<=floor(n/6)^2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 3, 1, 6, 1, 6, 1, 10, 16, 4, 1, 1, 10, 51, 50, 14, 1, 15, 125, 293, 174, 1, 15, 239, 1065, 1234, 1, 21, 423, 3075, 6124, 1, 21, 672, 7371, 23259, 1, 28, 1030, 16093, 81480, 51615, 10596, 808, 31, 1
Offset: 6

Views

Author

Christopher Hunt Gribble, Jan 31 2014

Keywords

Comments

The first 13 rows of T(n,k) are:
.\ k 0 1 2 3 4 5 6 7 8 9
n
6 1 1
7 1 1
8 1 3
9 1 3
10 1 6
11 1 6
12 1 10 16 4 1
13 1 10 51 50 14
14 1 15 125 293 174
15 1 15 239 1065 1234
16 1 21 423 3075 6124
17 1 21 672 7371 23259
18 1 28 1030 16093 81480 51615 10596 808 31 1

Examples

			T(12,3) = 4 because the number of equivalence classes of ways of placing 3 6 X 6 square tiles in a 12 X 12 square under all symmetry operations of the square is 4. The portrayal of an example from each equivalence class is:
._________________          _________________
|        |        |        |        |________|
|        |        |        |        |        |
|    .   |    .   |        |    .   |        |
|        |        |        |        |    .   |
|        |        |        |        |        |
|________|________|        |________|        |
|        |        |        |        |________|
|        |        |        |        |        |
|    .   |        |        |    .   |        |
|        |        |        |        |        |
|        |        |        |        |        |
|________|________|        |________|________|
.
._________________          _________________
|        |        |        |        |        |
|        |________|        |        |        |
|    .   |        |        |    .   |________|
|        |        |        |        |        |
|        |    .   |        |        |        |
|________|        |        |________|    .   |
|        |        |        |        |        |
|        |________|        |        |        |
|    .   |        |        |    .   |________|
|        |        |        |        |        |
|        |        |        |        |        |
|________|________|        |________|________|

Links

Crossrefs

Cf. A054252, A236679, A236560, A236757, A236800, A236865, A236915, A236936, A236939.

Formula

It appears that:
T(n,0) = 1, n>= 6
T(n,1) = (floor((n-6)/2)+1)*(floor((n-6)/2+2))/2, n >= 6
T(c+2*6,2) = A131474(c+1)*(6-1) + A000217(c+1)*floor(6^2/4) + A014409(c+2), 0 <= c < 6, c even
T(c+2*6,2) = A131474(c+1)*(6-1) + A000217(c+1)*floor((6-1)(6-3)/4) + A014409(c+2), 0 <= c < 6, c odd
T(c+2*6,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((6-c-1)/2) + A131941(c+1)*floor((6-c)/2)) + S(c+1,3c+2,3), 0 <= c < 6 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 Number T(n,k) of equivalence classes of ways of placing k 7 X 7 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=7, 0<=k<=floor(n/7)^2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 3, 1, 6, 1, 6, 1, 10, 1, 10, 20, 4, 1, 1, 15, 65, 59, 14, 1, 15, 153, 329, 174, 1, 21, 295, 1225, 1234, 1, 21, 507, 3465, 6124, 1, 28, 810, 8358, 23259, 1, 28, 1214, 17710, 73204
Offset: 7

Views

Author

Christopher Hunt Gribble, Jan 31 2014

Keywords

Comments

The first 13 rows of T(n,k) are:
.\ k 0 1 2 3 4 5 6 7 8 9
n
7 1 1
8 1 1
9 1 3
10 1 3
11 1 6
12 1 6
13 1 10
14 1 10 20 4 1
15 1 15 65 59 14
16 1 15 153 329 174
17 1 21 295 1225 1234
18 1 21 507 3465 6124
19 1 28 810 8358 23259
20 1 28 1214 17710 73204

Examples

			T(14,3) = 4 because the number of equivalent classes of ways of placing 3 7 X 7 square tiles in an 14 X 14 square under all symmetry operations of the square is 4. The portrayal of an example from each equivalence class is:
.___________________          ___________________
|         |         |        |         |_________|
|         |         |        |         |         |
|         |         |        |         |         |
|    .    |    .    |        |    .    |         |
|         |         |        |         |    .    |
|         |         |        |         |         |
|_________|_________|        |_________|         |
|         |         |        |         |_________|
|         |         |        |         |         |
|         |         |        |         |         |
|    .    |         |        |    .    |         |
|         |         |        |         |         |
|         |         |        |         |         |
|_________|_________|        |_________|_________|
.
.___________________          ___________________
|         |         |        |         |         |
|         |_________|        |         |         |
|         |         |        |         |_________|
|    .    |         |        |    .    |         |
|         |         |        |         |         |
|         |    .    |        |         |         |
|_________|         |        |_________|    .    |
|         |         |        |         |         |
|         |_________|        |         |         |
|         |         |        |         |_________|
|    .    |         |        |    .    |         |
|         |         |        |         |         |
|         |         |        |         |         |
|_________|_________|        |_________|_________|

Links

Crossrefs

Cf. A054252, A236679, A236560, A236757, A236800, A236829, A236915, A236936, A236939.

Formula

It appears that:
T(n,0) = 1, n>= 7
T(n,1) = (floor((n-7)/2)+1)*(floor((n-7)/2+2))/2, n >= 7
T(c+2*7,2) = A131474(c+1)*(7-1) + A000217(c+1)*floor(7^2/4) + A014409(c+2), 0 <= c < 7, c even
T(c+2*7,2) = A131474(c+1)*(7-1) + A000217(c+1)*floor((7-1)(7-3)/4) + A014409(c+2), 0 <= c < 7, c odd
T(c+2*7,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((7-c-1)/2) + A131941(c+1)*floor((7-c)/2)) + S(c+1,3c+2,3), 0 <= c < 7 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 Number T(n,k) of equivalence classes of ways of placing k 8 X 8 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=8, 0<=k<=floor(n/8)^2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 3, 1, 6, 1, 6, 1, 10, 1, 10, 1, 15, 25, 5, 1, 1, 15, 79, 65, 14, 1, 21, 187, 377, 174, 1, 21, 351, 1365, 1234, 1, 28, 606, 3900, 6124, 1, 28, 948, 9282, 23259, 1, 36, 1426, 19726, 73204, 1, 36, 2026, 38046, 199436
Offset: 8

Views

Author

Christopher Hunt Gribble, Feb 01 2014

Keywords

Comments

The first 16 rows of T(n,k) are:
.\ k 0 1 2 3 4
n
8 1 1
9 1 1
10 1 3
11 1 3
12 1 6
13 1 6
14 1 10
15 1 10
16 1 15 25 5 1
17 1 15 79 65 14
18 1 21 187 377 174
19 1 21 351 1365 1234
20 1 28 606 3900 6124
21 1 28 948 9282 23259
22 1 36 1426 19726 73204
23 1 36 2026 38046 199436

Examples

			T(16,3) = 5 because the number of equivalence classes of ways of placing 3 8 X 8 square tiles in an 16 X 16 square under all symmetry operations of the square is 5. The portrayal of an example from each equivalence class is:
._____________________        _____________________
|          |          |      |          |__________|
|          |          |      |          |          |
|          |          |      |          |          |
|     .    |     .    |      |     .    |          |
|          |          |      |          |     .    |
|          |          |      |          |          |
|          |          |      |          |          |
|__________|__________|      |__________|          |
|          |          |      |          |__________|
|          |          |      |          |          |
|          |          |      |          |          |
|    .     |          |      |     .    |          |
|          |          |      |          |          |
|          |          |      |          |          |
|          |          |      |          |          |
|__________|__________|      |__________|__________|
.
._____________________        _____________________
|          |          |      |          |          |
|          |__________|      |          |          |
|          |          |      |          |__________|
|     .    |          |      |     .    |          |
|          |          |      |          |          |
|          |     .    |      |          |          |
|          |          |      |          |     .    |
|__________|          |      |__________|          |
|          |          |      |          |          |
|          |__________|      |          |          |
|          |          |      |          |__________|
|     .    |          |      |     .    |          |
|          |          |      |          |          |
|          |          |      |          |          |
|          |          |      |          |          |
|__________|__________|      |__________|__________|
.
._____________________
|          |          |
|          |          |
|          |          |
|     .    |__________|
|          |          |
|          |          |
|          |          |
|__________|     .    |
|          |          |
|          |          |
|          |          |
|     .    |__________|
|          |          |
|          |          |
|          |          |
|__________|__________|

Links

Crossrefs

Cf. A054252, A236679, A236560, A236757, A236800, A236829, A236865, A236936, A236939.

Formula

It appears that:
T(n,0) = 1, n>= 8
T(n,1) = (floor((n-8)/2)+1)*(floor((n-8)/2+2))/2, n >= 8
T(c+2*8,2) = A131474(c+1)*(8-1) + A000217(c+1)*floor(8^2/4) + A014409(c+2), 0 <= c < 8, c even
T(c+2*8,2) = A131474(c+1)*(8-1) + A000217(c+1)*floor((8-1)(8-3)/4) + A014409(c+2), 0 <= c < 8, c odd
T(c+2*8,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((8-c-1)/2) + A131941(c+1)*floor((8-c)/2)) + S(c+1,3c+2,3), 0 <= c < 8 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 Number T(n,k) of equivalence classes of ways of placing k 9 X 9 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=9, 0<=k<=floor(n/9)^2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 3, 1, 6, 1, 6, 1, 10, 1, 10, 1, 15, 1, 15, 30, 5, 1, 1, 21, 96, 74, 14, 1, 21, 221, 413, 174, 1, 28, 417, 1525, 1234, 1, 28, 705, 4290, 6124, 1, 36, 1107, 10269, 23259, 1, 36, 1638, 21630, 73204, 1, 45, 2334, 41790, 199436
Offset: 9

Views

Author

Christopher Hunt Gribble, Feb 01 2014

Keywords

Examples

			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.

Links

Crossrefs

Cf. A054252, A236679, A236560, A236757, A236800, A236829, A236865, A236915, A236939.

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(9^2/4) + A014409(c+2), 0 <= c < 9, c even
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

A236939 Number T(n,k) of equivalence classes of ways of placing k 10 X 10 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=10, 0<=k<=floor(n/10)^2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 3, 1, 6, 1, 6, 1, 10, 1, 10, 1, 15, 1, 15, 1, 21, 36, 6, 1, 1, 21, 113, 80, 14, 1, 28, 261, 461, 174, 1, 28, 483, 1665, 1234, 1, 36, 819, 4725, 6124, 1, 36, 1266, 11193, 23259, 1, 45, 1878, 23646, 73204
Offset: 10

Views

Author

Christopher Hunt Gribble, Feb 01 2014

Keywords

Examples

			The first 17 rows of T(n,k) are:
.\ k  0     1     2     3     4
n
10    1     1
11    1     1
12    1     3
13    1     3
14    1     6
15    1     6
16    1    10
17    1    10
18    1    15
19    1    15
20    1    21    36     6     1
21    1    21   113    80    14
22    1    28   261   461   174
23    1    28   483  1665  1234
24    1    36   819  4725  6124
25    1    36  1266 11193 23259
26    1    45  1878 23646 73204
.
T(20,3) = 6 because the number of equivalence classes of ways of placing 3 10 X 10 square tiles in a 20 X 20 square under all symmetry operations of the square is 6.

Links

Crossrefs

Cf. A054252, A236679, A236560, A236757, A236800, A236829, A236865, A236915, A236936

Formula

It appears that:
T(n,0) = 1, n>= 10
T(n,1) = (floor((n-10)/2)+1)*(floor((n-10)/2+2))/2, n >= 10
T(c+2*10,2) = A131474(c+1)*(10-1) + A000217(c+1)*floor(10^2/4) + A014409(c+2), 0 <= c < 10, c even
T(c+2*10,2) = A131474(c+1)*(10-1) + A000217(c+1)*floor((10-1)(10-3)/4) + A014409(c+2), 0 <= c < 10, c odd
T(c+2*10,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((10-c-1)/2) + A131941(c+1)*floor((10-c)/2)) + S(c+1,3c+2,3), 0 <= c < 10 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
A236939(29,3), c = 9

A279112 Number of non-equivalent ways to place 3 non-attacking kings on an n X n board.

Original entry on oeis.org

0, 0, 2, 20, 138, 505, 1547, 3759, 8313, 16350, 30344, 52470, 87212, 138255, 212953, 317065, 461827, 655724, 915038, 1251720, 1688414, 2241365, 2941047, 3808915, 4884893, 6196650, 7795332, 9715914, 12022688, 14759115, 18004709, 21812685, 26280007, 31471000, 37502458
Offset: 1

Views

Author

Heinrich Ludwig, Dec 07 2016

Keywords

Comments

Rotations and reflections of placements are not counted. If they are to be counted, see A061996.

Examples

			There are 2 non-equivalent ways to place 3 non-attacking kings on a 3 X 3 board:
   K.K   K.K
   ...   ...
   ..K   .K.

Links

Crossrefs

Cf. A061996, A279111 (2 kings), A279113 (4 kings), A279114 (5 kings), A279115 (6 kings), A279116 (7 kings), A279117, A236679.

Programs

  • Mathematica
    Table[Boole[n > 1] (n^6 - 27 n^4 + 44 n^3 + 146 n^2 - 404 n + 240 + Boole[OddQ@ n] (8 n^3 - 21 n^2 + 16 n - 51))/48, {n, 35}] (* or *)
Rest@ CoefficientList[Series[x^3*(2 + 14 x + 76 x^2 + 93 x^3 + 102 x^4 - 17 x^5 - 36 x^6 - x^7 + 8 x^8 - x^9)/((1 - x)^7*(1 + x)^4), {x, 0, 35}], x] (* Michael De Vlieger, Dec 08 2016 *)
  • PARI
    concat(vector(2), Vec(x^3*(2 + 14*x + 76*x^2 + 93*x^3 + 102*x^4 - 17*x^5 - 36*x^6 - x^7 + 8*x^8 - x^9) / ((1 - x)^7*(1 + x)^4) + O(x^60))) \\ Colin Barker, Dec 07 2016

Formula

a(n) = (n^6 - 27*n^4 + 44*n^3 + 146*n^2 - 404*n + 240 + IF(MOD(n, 2) = 1, 8*n^3 - 21*n^2 + 16*n - 51))/48 for n >=2.
a(n) = 3*a(n-1) + a(n-2) - 11*a(n-3) + 6*a(n-4) + 14*a(n-5) - 14*a(n-6) - 6*a(n-7) + 11*a(n-8) - a(n-9) - 3*a(n-10) + a(n-11).
From Colin Barker, Dec 07 2016: (Start)
a(n) = (n^6 - 27*n^4 + 44*n^3 + 146*n^2 - 404*n + 240)/48 for n>1.
a(n) = (n^6 - 27*n^4 + 52*n^3 + 125*n^2 - 388*n + 189)/48 for n>1.
G.f.: x^3*(2 + 14*x + 76*x^2 + 93*x^3 + 102*x^4 - 17*x^5 - 36*x^6 - x^7 + 8*x^8 - x^9) / ((1 - x)^7*(1 + x)^4).
(End)
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