Search: a330379 -id:a330379
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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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+10
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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A330375
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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.
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+10
3
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1, 4, 9, 19, 1, 33, 2, 59, 7, 93, 12, 150, 26, 226, 43, 1, 342, 76, 2
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OFFSET
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1,2
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COMMENTS
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Column k starts in row k^2.
It appears that column 1 gives A179862.
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LINKS
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EXAMPLE
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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.
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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|_|
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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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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.
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CROSSREFS
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KEYWORD
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nonn,tabf,more
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AUTHOR
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STATUS
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approved
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A330378
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a(n) is the sum over all partitions of n of the number of right angles that are not the largest right angle.
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+10
3
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0, 0, 0, 1, 2, 5, 8, 14, 22, 34, 50, 75, 106, 151, 210, 291, 394, 535, 712, 949, 1246, 1634, 2118, 2745, 3520, 4508, 5728, 7266, 9152, 11512, 14390, 17959, 22298, 27634, 34094, 41993, 51510, 63075, 76966, 93752, 113834, 137992, 166788, 201269, 242248, 291102, 348976, 417727
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OFFSET
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1,5
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COMMENTS
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a(n) is also the sum of the sizes of the Durfee squares of all partitions of n, minus the number of partitions of n.
a(n) is also the sum of positive cranks of all partitions of n, minus the number of partitions of n.
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REFERENCES
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G. E. Andrews, Theory of Partitions, Cambridge University Press, 1984, page 143.
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LINKS
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FORMULA
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EXAMPLE
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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|_|
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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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In total there are 14 right angles that are not the largest right angle of the partitions of 8, so a(8) = 14.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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