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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.

A330375 Irregular triangle read by rows: T(n,k) (n>=1) is the sum of the lengths of all k-th right angles in all partitions of n.

Original entry on oeis.org

1, 4, 9, 19, 1, 33, 2, 59, 7, 93, 12, 150, 26, 226, 43, 1, 342, 76, 2
Offset: 1

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Author

Omar E. Pol, Dec 21 2019

Keywords

Comments

Column k starts in row k^2.
It appears that column 1 gives A179862.

Examples

			Triangle begins:
    1;
    4;
    9;
   19,  1;
   33,  2;
   59,  7;
   93, 12;
  150, 26;
  226, 43, 1;
  342, 76, 2;
...
Figure 1 shows the Ferrers diagram of the partition of 24: [7, 6, 3, 3, 2, 1, 1, 1]. Figure 2 shows the right-angles diagram of the same partition. Note that in this last diagram we can see the size of the three right angles as follows: the first right angle has size 14 because it contains 14 square cells, the second right angle has size 8 and the third right angle has size 2.
.
.                                     Right-angles   Right
Part   Ferrers diagram         Part   diagram        angle
                                      _ _ _ _ _ _ _
  7    * * * * * * *             7   |  _ _ _ _ _ _|  14
  6    * * * * * *               6   | |  _ _ _ _|     8
  3    * * *                     3   | | | |           2
  3    * * *                     3   | | |_|
  2    * *                       2   | |_|
  1    *                         1   | |
  1    *                         1   | |
  1    *                         1   |_|
.
       Figure 1.                      Figure 2.
.
For n = 8 the partitions of 8 and their respective right-angles diagrams are as follows:
.
    _       _ _       _ _ _       _ _ _ _       _ _ _ _ _
  1| |8   2|  _|8   3|  _ _|8   4|  _ _ _|8   5|  _ _ _ _|8
  1| |    1| |      1| |        1| |          1| |
  1| |    1| |      1| |        1| |          1| |
  1| |    1| |      1| |        1| |          1|_|
  1| |    1| |      1| |        1|_|
  1| |    1| |      1|_|
  1| |    1|_|
  1|_|
    _ _ _ _ _ _       _ _ _ _ _ _ _       _ _ _ _ _ _ _ _
  6|  _ _ _ _ _|8   7|  _ _ _ _ _ _|8   8|_ _ _ _ _ _ _ _|8
  1| |              1|_|
  1|_|
.
    _ _       _ _ _       _ _ _ _       _ _ _ _ _       _ _ _ _ _ _
  2|  _|7   3|  _ _|7   4|  _ _ _|7   5|  _ _ _ _|7   6|  _ _ _ _ _|7
  2| |_|1   2| |_|  1   2| |_|    1   2| |_|      1   2|_|_|        1
  1| |      1| |        1| |          1|_|
  1| |      1| |        1|_|
  1| |      1|_|
  1|_|
.
    _ _       _ _ _       _ _ _       _ _ _ _       _ _ _ _       _ _ _ _ _
  2|  _|6   3|  _ _|6   3|  _ _|6   4|  _ _ _|6   4|  _ _ _|6   5|  _ _ _ _|6
  2| | |2   2| | |  2   3| |_ _|2   2| | |    2   3| |_ _|  2   3|_|_ _|    2
  2| |_|    2| |_|      1| |        2|_|_|        1|_|
  1| |      1|_|        1|_|
  1|_|
.
    _ _       _ _ _        _ _ _ _
  2|  _|5   3|  _ _|5    4|  _ _ _|5
  2| | |3   3| |  _|3    4|_|_ _ _|3
  2| | |    2|_|_|
  2|_|_|
.
The sum of the lengths of the first right angles of all partitions of 8 is 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 7 + 7 + 7 + 7 + 7 + 6 + 6 + 6 + 6 + 6 + 6 + 5 + 5 + 5 = 150, so T(8,1) = 150.
The sum of the second right angles of all partitions of 8 is 1 + 1 + 1 + 1 + 1 + 2 + 2 + 2 + 2 + 2 + 2 + 3 + 3 + 3 = 26, so T(8,2) = 26.
		

Crossrefs

Row sums give A066186.
Cf. A179862.