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A330369
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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.
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5
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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
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OFFSET
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1,3
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COMMENTS
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This triangle has the property that it contains the triangle A049597, since if we replace with zeros the positive terms before the first zero in the row n of this triangle, we get the triangle A049597.
Hence the sum of the terms after the last zero in row n equals A000041(n), the number of partitions of n (see the Example section).
Observation: at least the first 11 terms of column 1 coincide with A188674 (using the same indices).
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REFERENCES
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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]
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LINKS
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EXAMPLE
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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.
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. 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 |_|
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Figure 1. Figure 2.
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For n = 8 the partitions of 8 and their respective right-angles diagrams are as follows:
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_ _ _ _ _ _ _ _ _ _ _ _ _ _ _
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|_|
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_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
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|_|
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_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
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|_|
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_ _ _ _ _ _ _ _ _
2| _|5 3| _ _|5 4| _ _ _|5
2| | |3 3| | _|3 4|_|_ _ _|3
2| | | 2|_|_|
2|_|_|
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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.
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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