A330379 - OEIS

A330379

Triangle read by rows: T(n,k) (1 <= k <= n) is the sum of the sizes of all right angles of size k of all partitions of n.

1, 0, 4, 0, 0, 9, 1, 0, 3, 16, 2, 0, 0, 8, 25, 3, 4, 0, 8, 15, 36, 4, 8, 0, 0, 20, 24, 49, 5, 12, 9, 0, 15, 36, 35, 64, 7, 16, 21, 0, 5, 36, 56, 48, 81, 9, 20, 33, 16, 0, 36, 63, 80, 63, 100, 13, 24, 45, 40, 0, 12, 77, 96, 108, 80, 121

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OFFSET

1,3

COMMENTS

Observation: at least the first 11 terms of column 1 coincide with A188674 (using the same indices).

REFERENCES

G. E. Andrews, Theory of Partitions, Cambridge University Press, 1984, page 143.

LINKS

Table of n, a(n) for n=1..66.

FORMULA

T(n,k) = k*A330369(n,k).

EXAMPLE

Triangle begins:

0, 4;

0, 0, 9;

1, 0, 3, 16;

2, 0, 0, 8, 25;

3, 4, 0, 8, 15, 36;

4, 8, 0, 0, 20, 24, 49;

5, 12, 9, 0, 15, 36, 35, 64;

7, 16, 21, 0, 5, 36, 56, 48, 81;

9, 20, 33, 16, 0, 36, 63, 80, 63, 100;

13, 24, 45, 40, 0, 12, 77, 96, 108, 80, 121;

...

Below the figure 1 shows the Ferrers diagram of the partition of 24: [7, 6, 3, 3, 2, 1, 1, 1]. The 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 |_|

Figure 1. Figure 2.

For n = 8 the partitions of 8 and their respective right-angles diagrams look as shown below:

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _

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|_|

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

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|_|_|

There are 5 right angles of size 1, so T(8,1) = 5*1 = 5.

There are 6 right angles of size 2, so T(8,2) = 6*2 = 12.

There are 3 right angles of size 3, so T(8,3) = 3*3 = 9.

There are no right angle of size 4, so T(8,4) = 0*4 = 0.

There are 3 right angles of size 5, so T(8,5) = 3*5 = 15.

There are 6 right angles of size 6, so T(8,6) = 6*6 = 36.

There are 5 right angles of size 7, so T(8,7) = 5*7 = 35.

There are 8 right angles of size 8, so T(8,8) = 8*8 = 64.

Hence the 8th row of triangle is [5, 12, 9, 0, 15, 36, 35, 64].

The row sum gives A066186(8) = 8*A000041(8) = 8*22 = 176.

CROSSREFS

Row sums give A066186, n >= 1.

Row sums of the terms that are after last zero give A179862.

Cf. A188674.

Cf. A000041, A000290, A330369, A330375, A330378.

Sequence in context: A048728 A002915 A127733 * A364102 A249035 A244121

Adjacent sequences: A330376 A330377 A330378 * A330380 A330381 A330382

KEYWORD

nonn,tabl,more

AUTHOR

Omar E. Pol, Dec 31 2019

STATUS

approved