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.

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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

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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:
._____________          _____________          _____________
|      |      |        |      |______|        |      |      |
|   .  |   .  |        |   .  |      |        |   .  |______|
|      |      |        |      |   .  |        |      |      |
|______|______|        |______|      |        |______|   .  |
|      |      |        |      |______|        |      |      |
|   .  |      |        |   .  |      |        |   .  |______|
|      |      |        |      |      |        |      |      |
|______|______|        |______|______|        |______|______|
		

Crossrefs

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

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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:
._______________      _______________      _______________
|       |       |    |       |_______|    |       |       |
|       |       |    |       |       |    |       |_______|
|   .   |   .   |    |   .   |       |    |   .   |       |
|       |       |    |       |   .   |    |       |       |
|_______|_______|    |_______|       |    |_______|   .   |
|       |       |    |       |_______|    |       |       |
|       |       |    |       |       |    |       |_______|
|   .   |       |    |   .   |       |    |   .   |       |
|       |       |    |       |       |    |       |       |
|_______|_______|    |_______|_______|    |_______|_______|
		

Crossrefs

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

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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:
._________________          _________________
|        |        |        |        |________|
|        |        |        |        |        |
|    .   |    .   |        |    .   |        |
|        |        |        |        |    .   |
|        |        |        |        |        |
|________|________|        |________|        |
|        |        |        |        |________|
|        |        |        |        |        |
|    .   |        |        |    .   |        |
|        |        |        |        |        |
|        |        |        |        |        |
|________|________|        |________|________|
.
._________________          _________________
|        |        |        |        |        |
|        |________|        |        |        |
|    .   |        |        |    .   |________|
|        |        |        |        |        |
|        |    .   |        |        |        |
|________|        |        |________|    .   |
|        |        |        |        |        |
|        |________|        |        |        |
|    .   |        |        |    .   |________|
|        |        |        |        |        |
|        |        |        |        |        |
|________|________|        |________|________|
		

Crossrefs

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

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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:
.___________________          ___________________
|         |         |        |         |_________|
|         |         |        |         |         |
|         |         |        |         |         |
|    .    |    .    |        |    .    |         |
|         |         |        |         |    .    |
|         |         |        |         |         |
|_________|_________|        |_________|         |
|         |         |        |         |_________|
|         |         |        |         |         |
|         |         |        |         |         |
|    .    |         |        |    .    |         |
|         |         |        |         |         |
|         |         |        |         |         |
|_________|_________|        |_________|_________|
.
.___________________          ___________________
|         |         |        |         |         |
|         |_________|        |         |         |
|         |         |        |         |_________|
|    .    |         |        |    .    |         |
|         |         |        |         |         |
|         |    .    |        |         |         |
|_________|         |        |_________|    .    |
|         |         |        |         |         |
|         |_________|        |         |         |
|         |         |        |         |_________|
|    .    |         |        |    .    |         |
|         |         |        |         |         |
|         |         |        |         |         |
|_________|_________|        |_________|_________|
		

Crossrefs

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

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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:
._____________________        _____________________
|          |          |      |          |__________|
|          |          |      |          |          |
|          |          |      |          |          |
|     .    |     .    |      |     .    |          |
|          |          |      |          |     .    |
|          |          |      |          |          |
|          |          |      |          |          |
|__________|__________|      |__________|          |
|          |          |      |          |__________|
|          |          |      |          |          |
|          |          |      |          |          |
|    .     |          |      |     .    |          |
|          |          |      |          |          |
|          |          |      |          |          |
|          |          |      |          |          |
|__________|__________|      |__________|__________|
.
._____________________        _____________________
|          |          |      |          |          |
|          |__________|      |          |          |
|          |          |      |          |__________|
|     .    |          |      |     .    |          |
|          |          |      |          |          |
|          |     .    |      |          |          |
|          |          |      |          |     .    |
|__________|          |      |__________|          |
|          |          |      |          |          |
|          |__________|      |          |          |
|          |          |      |          |__________|
|     .    |          |      |     .    |          |
|          |          |      |          |          |
|          |          |      |          |          |
|          |          |      |          |          |
|__________|__________|      |__________|__________|
.
._____________________
|          |          |
|          |          |
|          |          |
|     .    |__________|
|          |          |
|          |          |
|          |          |
|__________|     .    |
|          |          |
|          |          |
|          |          |
|     .    |__________|
|          |          |
|          |          |
|          |          |
|__________|__________|
		

Crossrefs

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

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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.
		

Crossrefs

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

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Author

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.
		

Crossrefs

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

A239473 Triangle read by rows: signed version of A059260: coefficients for expansion of partial sums of sequences a(n,x) in terms of their binomial transforms (1+a(.,x))^n ; Laguerre polynomial expansion of the truncated exponential.

Original entry on oeis.org

1, 0, 1, 1, -1, 1, 0, 2, -2, 1, 1, -2, 4, -3, 1, 0, 3, -6, 7, -4, 1, 1, -3, 9, -13, 11, -5, 1, 0, 4, -12, 22, -24, 16, -6, 1, 1, -4, 16, -34, 46, -40, 22, -7, 1, 0, 5, -20, 50, -80, 86, -62, 29, -8, 1, 1, -5, 25, -70, 130, -166, 148, -91, 37, -9, 1, 0, 6, -30, 95, -200, 296, -314, 239, -128, 46, -10, 1
Offset: 0

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Author

Tom Copeland, Mar 19 2014

Keywords

Comments

With T the lower triangular array above and the Laguerre polynomials L(k,x) = Sum_{j=0..k} (-1)^j binomial(k, j) x^j/j!, the following identities hold:
(A) Sum_{k=0..n} (-1)^k L(k,x) = Sum_{k=0..n} T(n,k) x^k/k!;
(B) Sum_{k=0..n} x^k/k! = Sum_{k=0..n} T(n,k) L(k,-x);
(C) Sum_{k=0..n} x^k = Sum_{k=0..n} T(n,k) (1+x)^k = (1-x^(n+1))/(1-x).
More generally, for polynomial sequences,
(D) Sum_{k=0..n} P(k,x) = Sum_{k=0..n} T(n,k) (1+P(.,x))^k,
where, e.g., for an Appell sequence, such as the Bernoulli polynomials, umbrally, (1+ Ber(.,x))^k = Ber(k,x+1).
Identity B follows from A through umbral substitution of j!L(j,-x) for x^j in A. Identity C, related to the cyclotomic polynomials for prime index, follows from B through the Laplace transform.
Integrating C gives Sum_{k=0..n} T(n,k) (2^(k+1)-1)/(k+1) = H(n+1), the harmonic numbers.
Identity A >= 0 for x >= 0 (see MathOverflow link for evaluation in terms of Hermite polynomials).
From identity C, W(m,n) = (-1)^n Sum_{k=0..n} T(n,k) (2-m)^k = number of walks of length n+1 between any two distinct vertices of the complete graph K_m for m > 2.
Equals A112468 with the first column of ones removed. - Georg Fischer, Jul 26 2023

Examples

			Triangle begins:
   1
   0    1
   1   -1    1
   0    2   -2    1
   1   -2    4   -3    1
   0    3   -6    7   -4    1
   1   -3    9  -13   11   -5    1
   0    4  -12   22  -24   16   -6    1
   1   -4   16  -34   46  -40   22   -7    1
   0    5  -20   50  -80   86  -62   29   -8    1
   1   -5   25  -70  130 -166  148  -91   37   -9    1
		

Crossrefs

For column 2: A001057, A004526, A008619, A140106.
Column 3: A002620, A087811.
Column 4: A002623, A173196.
Column 5: A001752.
Column 6: A001753.
Cf. Bottomley's cross-references in A059260.
Embedded in alternating antidiagonals of T are the reversals of arrays A071921 (A225010) and A210220.

Programs

  • Magma
    [[(&+[(-1)^(j+k)*Binomial(j,k): j in [0..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Feb 06 2018
    
  • Maple
    A239473 := proc(n,k)
        add(binomial(j,k)*(-1)^(j+k),j=k..n) ;
    end proc; # R. J. Mathar, Jul 21 2016
  • Mathematica
    Table[Sum[(-1)^(j+k)*Binomial[j,k], {j,0,n}], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 06 2018 *)
  • PARI
    for(n=0,10, for(k=0,n, print1(sum(j=0,n, (-1)^(j+k)*binomial(j, k)), ", "))) \\ G. C. Greubel, Feb 06 2018
    
  • Sage
    Trow = lambda n: sum((x-1)^j for j in (0..n)).list()
    for n in (0..10): print(Trow(n)) # Peter Luschny, Jul 09 2019

Formula

T(n, k) = Sum_{j=0..n} (-1)^(j+k) * binomial(j, k).
E.g.f: (exp(t) - (x-1)*exp((x-1)*t))/(2-x).
O.g.f. (n-th row): (1-(x-1)^(n+1))/(2-x).
Associated operator identities:
With D=d/dx, :xD:^n=x^n*D^n, and :Dx:^n=D^n*x^n, then bin(xD,n)= binomial(xD,n)=:xD:^n/n! and L(n,-:xD:)=:Dx:^n/n!=bin(xD+n,n)=(-1)^n bin(-xD-1,n),
A-o) Sum_{k=0..n} (-1)^k L(k,-:xD:) = Sum_{k=0..n} :-Dx:^k/k!
= Sum_{k=0..n} T(n,k) :-xD:^k/k! = Sum_{k=0..n} (-1)^k T(n,k)bin(xD,k)
B-o) Sum_{k=0..n} :xD:^k/k! = Sum_{k=0..n}, T(n,k) L(k,-:xD:)
= Sum_{k=0..n} T(n,k) :Dx:^k/k! = Sum_{k=0..n}, bin(xD,k).
Associated binomial identities:
A-b) Sum_{k=0..n} (-1)^k bin(s+k,k) = Sum_{k=0..n} (-1)^k T(n,k) bin(s,k)
= Sum_{k=0..n} bin(-s-1,k) = Sum{k=0..n} T(n,k) bin(-s-1+k,k)
B-b) Sum_{k=0..n} bin(s,k) = Sum_{k=0..n} T(n,k) bin(s+k,k)
= Sum_{k=0..n} (-1)^k bin(-s-1+k,k)
= Sum_{k=0..n} (-1)^k T(n,k) bin(-s-1,k).
In particular, from B-b with s=n, Sum_{k=0..n} T(n,k) bin(n+k,k) = 2^n. From B-b with s=0, row sums are all 1.
From identity C with x=-2, the unsigned row sums are the Jacobsthal sequence, i.e., Sum_{k=0..n} T(n,k) (1+(-2))^k = (-1)^n A001045(n+1); for x=2, the Mersenne numbers A000225; for x=-3, A014983 or signed A015518; for x=3, A003462; for x=-4, A014985 or signed A015521; for x=4, A002450; for x=-5, A014986 or signed A015531; and for x=5, A003463; for x=-6, A014987 or signed A015540; and for x=6, A003464.
With -s-1 = m = 0,1,2,..., B-b gives finite differences (recursions):
Sum_{k=0..n} (-1)^k T(n,k) bin(m,k) = Sum_{k=0..n} (-1)^k bin(m+k,k) = T(n+m,m), i.e., finite differences of the columns of T generate shifted columns of T. The columns of T are signed, shifted versions of sequences listed in the cross-references. Since the finite difference is an involution, T(n,k) = Sum_{j=0..k} (-1)^j T(n+j,j) bin(k,j)}. Gauss-Newton interpolation can be applied to give a generalized T(n,s) for s noninteger.
From identity C, S(n,m) = Sum_{k=0..n} T(n,k) bin(k,m) = 1 for m < n+1 and 0 otherwise, i.e., S = T*P, where S = A000012, as a lower triangular matrix and P = Pascal = A007318, so T = S*P^(-1), where P^(-1) = A130595, the signed Pascal array (see A132440), the inverse of P, and T^(-1) = P*S^(-1) = P*A167374 = A156644.
U(n,cos(x)) = e^(-n*i*x)*Sum_{k=0..n} T(n,k)*(1+e^(2*i*x))^k = sin((n+1)x)/sin(x), where U is the Chebyschev polynomial of the second kind A053117 and i^2 = -1. - Tom Copeland, Oct 18 2014
From Tom Copeland, Dec 26 2015: (Start)
With a(n,x) = e^(nx), the partial sums are 1+e^x+...+e^(nx) = Sum_{k=0..n} T(n,k) (1+e^x)^k = [ x / (e^x-1) ] [ e^((n+1)x) -1 ] / x = [ (x / (e^x-1)) e^((n+1)x) - (x / (e^x-1)) ] / x = Sum_{k>=0} [ (Ber(k+1,n+1) - Ber(k+1,0)) / (k+1) ] * x^k/k!, where Ber(n,x) are the Bernoulli polynomials (cf. Adams p. 140). Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of the m-th powers of the integers, their binomial transforms, and the Bernoulli polynomials.
With a(n,x) = (-1)^n e^(nx), the partial sums are 1-e^x+...+(-1)^n e^(nx) = Sum_{k=0..n} T(n,k) (1-e^x)^k = [ (-1)^n e^((n+1)x) + 1 ] / (e^x+1) = [ (-1)^n (2 / (e^x+1)) e^((n+1)x) + (2 / (e^x+1)) ] / 2 = (1/2) Sum_{k>=0} [ (-1)^n Eul(k,n+1) + Eul(k,0) ] * x^k/k!, where Eul(n,x) are the Euler polynomials. Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of signed m-th powers of the integers; their binomial transforms, related to the Stirling numbers of the second kind and face numbers of the permutahedra; and the Euler polynomials. (End)
As in A059260, a generator in terms of bivariate polynomials with the coefficients of this entry is given by (1/(1-y))*1/(1 + (y/(1-y))*x - (1/(1-y))*x^2) = 1 + y + (x^2 - x*y + y^2) + (2*x^2*y - 2*x*y^2 + y^3) + (x^4 - 2*x^3*y + 4*x^2*y^2 - 3*x*y^3 + y^4) + ... . This is of the form -h2 * 1 / (1 + h1*x + h2*x^2), related to the bivariate generator of A049310 with h1 = y/(1-y) and h2 = -1/(1-y) = -(1+h1). - Tom Copeland, Feb 16 2016
From Tom Copeland, Sep 05 2016: (Start)
Letting P(k,x) = x in D gives Sum_{k=0..n} T(n,k)*Sum_{j=0..k} binomial(k,j) = Sum_{k=0..n} T(n,k) 2^k = n + 1.
The quantum integers [n+1]q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)) = q^(-n)*(1 - q^(2*(n+1))) / (1 - q^2) = q^(-n)*Sum{k=0..n} q^(2k) = q^(-n)*Sum_{k=0..n} T(n,k)*(1 + q^2)^k. (End)
T(n, k) = [x^k] Sum_{j=0..n} (x-1)^j. - Peter Luschny, Jul 09 2019
a(n) = -n + Sum_{k=0..n} A341091(k). - Thomas Scheuerle, Jun 17 2022

Extensions

Inverse array added by Tom Copeland, Mar 26 2014
Formula re Euler polynomials corrected by Tom Copeland, Mar 08 2024

A052264 Number of 5 X n binary matrices up to row and column permutations.

Original entry on oeis.org

1, 6, 34, 190, 1053, 5624, 28576, 136758, 613894, 2583164, 10208743, 38013716, 133872584, 447620002, 1426354541, 4346885204, 12710830673, 35768703586, 97125981825, 255111287298, 649598148384, 1606754306778, 3867515638005, 9074220508038, 20784247213232
Offset: 0

Views

Author

Vladeta Jovovic, Feb 04 2000

Keywords

Crossrefs

A diagonal of the array A(m,n) described in A028657. - N. J. A. Sloane, Sep 01 2013

Programs

Formula

G.f.: (x^68 - 2*x^67 + 10*x^66 + 32*x^65 + 175*x^64 + 794*x^63 + 3441*x^62 + 13186*x^61 + 46027*x^60 + 146118*x^59 + 427347*x^58 + 1155432*x^57 + 2912873*x^56 + 6875608*x^55 + 15281029*x^54 + 32094658*x^53 + 63945531*x^52 + 121210914*x^51 + 219194198*x^50 + 378998758*x^49 + 627863648*x^48 + 998282344*x^47 + 1525746624*x^46 + 2244502676*x^45 + 3181886869*x^44 + 4351201210*x^43 + 5744918381*x^42 + 7328807372*x^41 + 9039504349*x^40 + 10785767638*x^39 + 12455264802*x^38 + 13925287384*x^37 + 15077477135*x^36 + 15812782150*x^35 + 16065602576*x^34 + 15812782150*x^33 + 15077477135*x^32 + 13925287384*x^31 + 12455264802*x^30 + 10785767638*x^29 + 9039504349*x^28 + 7328807372*x^27 + 5744918381*x^26 + 4351201210*x^25 + 3181886869*x^24 + 2244502676*x^23 + 1525746624*x^22 + 998282344*x^21 + 627863648*x^20 + 378998758*x^19 + 219194198*x^18 + 121210914*x^17 + 63945531*x^16 + 32094658*x^15 + 15281029*x^14 + 6875608*x^13 + 2912873*x^12 + 1155432*x^11 + 427347*x^10 + 146118*x^9 + 46027*x^8 + 13186*x^7 + 3441*x^6 + 794*x^5 + 175*x^4 + 32*x^3 + 10*x^2 - 2*x + 1)/((x^6 - 1)^2*(x^4 + x^3 + x^2 + x + 1)^6*(x^3 - x^2 + x - 1)^6 * (x^2 + x + 1)^6*(x + 1)^10*(x - 1)^24).

Extensions

Name clarified by Ching Pong Siu, Aug 30 2022

A112465 Riordan array (1/(1+x), x/(1-x)).

Original entry on oeis.org

1, -1, 1, 1, 0, 1, -1, 1, 1, 1, 1, 0, 2, 2, 1, -1, 1, 2, 4, 3, 1, 1, 0, 3, 6, 7, 4, 1, -1, 1, 3, 9, 13, 11, 5, 1, 1, 0, 4, 12, 22, 24, 16, 6, 1, -1, 1, 4, 16, 34, 46, 40, 22, 7, 1, 1, 0, 5, 20, 50, 80, 86, 62, 29, 8, 1, -1, 1, 5, 25, 70, 130, 166, 148, 91, 37, 9, 1, 1, 0, 6, 30, 95, 200, 296, 314, 239, 128, 46, 10, 1
Offset: 0

Views

Author

Paul Barry, Sep 06 2005

Keywords

Comments

Inverse is A112466. Note that C(n,k) = Sum_{j = 0..n-k} C(j+k-1, j).

Examples

			Triangle starts
   1;
  -1, 1;
   1, 0, 1;
  -1, 1, 1,  1;
   1, 0, 2,  2,  1;
  -1, 1, 2,  4,  3,  1;
   1, 0, 3,  6,  7,  4,  1;
  -1, 1, 3,  9, 13, 11,  5, 1;
   1, 0, 4, 12, 22, 24, 16, 6, 1;
Production matrix begins
  -1, 1;
   0, 1, 1;
   0, 0, 1, 1;
   0, 0, 0, 1, 1;
   0, 0, 0, 0, 1, 1;
   0, 0, 0, 0, 0, 1, 1;
   0, 0, 0, 0, 0, 0, 1, 1;
   0, 0, 0, 0, 0, 0, 0, 1, 1; - _Paul Barry_, Apr 08 2011
		

Crossrefs

Columns: A033999(n) (k=0), A000035(n) (k=1), A004526(n) (k=2), A002620(n-1) (k=3), A002623(n-4) (k=4), A001752(n-5) (k=5), A001753(n-6) (k=6), A001769(n-7) (k=7), A001779(n-8) (k=8), A001780(n-9) (k=9), A001781(n-10) (k=10), A001786(n-11) (k=11), A001808(n-12) (k=12).
Diagonals: A000012(n) (k=n), A023443(n) (k=n-1), A152947(n-1) (k=n-2), A283551(n-3) (k=n-3).
Main diagonal: A072547.
Sums: A078008 (row), A078024 (diagonal), A092220 (signed diagonal), A280560 (signed row).

Programs

  • Haskell
    a112465 n k = a112465_tabl !! n !! k
    a112465_row n = a112465_tabl !! n
    a112465_tabl = iterate f [1] where
       f xs'@(x:xs) = zipWith (+) ([-x] ++ xs ++ [0]) ([0] ++ xs')
    -- Reinhard Zumkeller, Jan 03 2014
    
  • Magma
    A112465:= func< n,k | (-1)^(n+k)*(&+[(-1)^j*Binomial(j+k-1,j): j in [0..n-k]]) >;
    [A112465(n,k): k in [0..n], n in [0..13]]; // G. C. Greubel, Apr 18 2025
    
  • Mathematica
    T[n_, k_]:= Sum[Binomial[j+k-1, j]*(-1)^(n-k-j), {j, 0, n-k}];
    Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* Jean-François Alcover, Jul 23 2018 *)
  • SageMath
    def A112465(n,k): return (-1)^(n+k)*sum((-1)^j*binomial(j+k-1,j) for j in range(n-k+1))
    print(flatten([[A112465(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Apr 18 2025

Formula

Number triangle T(n, k) = Sum_{j=0..n-k} (-1)^(n-k-j)*C(j+k-1, j).
T(2*n, n) = A072547(n) (main diagonal). - Paul Barry, Apr 08 2011
From Reinhard Zumkeller, Jan 03 2014: (Start)
T(n, k) = T(n-1, k-1) + T(n-1, k), 0 < k < n, with T(n, 0) = (-1)^n and T(n, n) = 1.
T(n, k) = A108561(n, n-k). (End)
T(n, k) = T(n-1, k-1) + T(n-2, k) + T(n-2, k-1), T(0, 0) = 1, T(1, 0) = -1, T(1, 1) = 1, T(n, k) = 0 if k < 0 or if k > n. - Philippe Deléham, Jan 11 2014
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(-1 + x + x^2/2! + x^3/3!) = -1 + 2*x^2/2! + 6*x^3/3! + 13*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 21 2014
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