A330369
Triangle read by rows: T(n,k) (1 <= k <= n) is the total number of right angles of size k in all partitions of n.
Original entry on oeis.org
1, 0, 2, 0, 0, 3, 1, 0, 1, 4, 2, 0, 0, 2, 5, 3, 2, 0, 2, 3, 6, 4, 4, 0, 0, 4, 4, 7, 5, 6, 3, 0, 3, 6, 5, 8, 7, 8, 7, 0, 1, 6, 8, 6, 9, 9, 10, 11, 4, 0, 6, 9, 10, 7, 10, 13, 12, 15, 10, 0, 2, 11, 12, 12, 8, 11
Offset: 1
Triangle begins:
1;
0, 2;
0, 0, 3;
1, 0, 1, 4;
2, 0, 0, 2, 5;
3, 2, 0, 2, 3, 6;
4, 4, 0, 0, 4, 4, 7;
5, 6, 3, 0, 3, 6, 5, 8;
7, 8, 7, 0, 1, 6, 8, 6, 9;
9, 10, 11, 4, 0, 6, 9, 10, 7, 10;
13, 12, 15, 10, 0, 2, 11, 12, 12, 8, 11;
Figure 1 below 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|_|_|
.
There are 5 right angles of size 1, so T(8,1) = 5.
There are 6 right angles of size 2, so T(8,2) = 6.
There are 3 right angles of size 3, so T(8,3) = 3.
There are no right angle of size 4, so T(8,4) = 0.
There are 3 right angles of size 5, so T(8,5) = 3.
There are 6 right angles of size 6, so T(8,6) = 6.
There are 5 right angles of size 7, so T(8,7) = 5.
There are 8 right angles of size 8, so T(8,8) = 8.
Hence the 8th row of triangle is [5, 6, 3, 0, 3, 6, 5, 8].
Note that the sum of the terms after the last zero is 3 + 6 + 5 + 8 = 22, equaling A000041(8) = 22, the number of partitions of 8.
- G. E. Andrews, Theory of Partitions, Cambridge University Press, 1984, page 143 [Defines the right angles in the Ferrers graph of a partition. - N. J. A. Sloane, Nov 20 2020]
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
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.
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.
Original entry on oeis.org
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
Offset: 1
Triangle begins:
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;
...
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 | |
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| | 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.
- G. E. Andrews, Theory of Partitions, Cambridge University Press, 1984, page 143.
Row sums of the terms that are after last zero give
A179862.
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