Search: a330369 -id:a330369
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A208478
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Triangle read by rows: T(n,k) = number of partitions of n with positive k-th rank.
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+10
12
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0, 1, 1, 1, 1, 1, 2, 1, 2, 1, 3, 1, 3, 2, 1, 5, 2, 4, 4, 2, 1, 6, 3, 5, 6, 4, 2, 1, 10, 5, 7, 9, 7, 4, 2, 1, 13, 7, 9, 11, 11, 7, 4, 2, 1, 19, 11, 12, 15, 16, 12, 7, 4, 2, 1, 25, 16, 15, 19, 22, 18, 12, 7, 4, 2, 1, 35, 24, 20, 26, 29, 27, 19, 12, 7, 4, 2, 1
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OFFSET
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1,7
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COMMENTS
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We define the k-th rank of a partition as the k-th part minus the number of parts >= k. Every partition of n has n ranks. This is a generalization of the Dyson's rank of a partition which is the largest part minus the number of parts. Since the first part of a partition is also the largest part of the same partition so the Dyson's rank of a partition is the case for k = 1.
The sum of the k-th ranks of all partitions of n is equal to zero.
Also T(n,k) = number of partitions of n with negative k-th rank.
1) The k-th part of a partition of n is also the number of parts >= k of its conjugate partition.
2) The k-th rank of a partitions is also the number of parts >= k of its conjugate partition minus the number of parts >= k.
For example: for n = 9 consider the partition [5, 3, 1]. The first part is 5, so the conjugate partition has five parts >= 1. The second part is 3, so the conjugate partition has three parts >= 2. The third part is 1, so the conjugate partition has only one part >= 3. The mentioned conjugate partition is [3, 2, 2, 1, 1]. And conversely, consider the partition [3, 2, 2, 1, 1]. The first part is 3, so the conjugate partition has three parts >= 1. The second part is 2, so the conjugate partition has two parts >= 2. the Third part is 2, so the conjugate partition has two parts >= 3, and so on. In this case the conjugate partition is [5, 3, 1].
3) The difference between the k-th part and the (k+1)-st part of the partition of n is also the number of k's in its conjugate partition. For example: consider the partition [5, 3, 1]. The difference between the first and the second part is 5 - 3 = 2, equals the number of 1's in its conjugate partition. The difference between the second and the third part is 3 - 1 = 2, equals the number of 2's in its conjugate partition. The difference between the third and the fourth (virtual) part is 1 - 0 = 1, equals the number of 3's in its conjugate partition [3, 2, 2, 1, 1]. And conversely, consider the partition [3, 2, 2, 1, 1]. The difference between the first and the second part is 3 - 2 = 1, equals the number of 1's in its conjugate partition. The difference between the second and the third part is 2 - 2 = 0, equals the number of 2's in its conjugate partition. The difference between the third and the fourth part is 2 - 1 = 1, equals the number of 3's in its conjugate partition, and so on.
4) The list of n ranks of a partition of n equals the list of n ranks multiplied by -1 of its conjugate partition. For example the nine ranks of the partition [5, 3, 1] of 9 are [2, 1, -1, -1, -1, -1, 0, 0, 0], and the nine ranks of its conjugate partition [3, 2, 2, 1, 1] are [-2, -1, 1, 1, 1, 1, 0, 0, 0].
For a list of partitions of the positive integers ordered by its k-th ranks see A330370. (End)
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LINKS
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EXAMPLE
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For n = 4 the partitions of 4 and the four types of ranks of the partitions of 4 are
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Partitions First Second Third Fourth
of 4 rank rank rank rank
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4 4-1 = 3 0-1 = -1 0-1 = -1 0-1 = -1
3+1 3-2 = 1 1-1 = 0 0-1 = -1 0-0 = 0
2+2 2-2 = 0 2-2 = 0 0-0 = 0 0-0 = 0
2+1+1 2-3 = -1 1-1 = 0 1-0 = 1 0-0 = 0
1+1+1+1 1-4 = -3 1-0 = 1 1-0 = 1 1-0 = 1
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The number of partitions of 4 with positive k-th ranks are 2, 1, 2, 1 so row 4 lists 2, 1, 2, 1.
Triangle begins:
0;
1, 1;
1, 1, 1;
2, 1, 2, 1;
3, 1, 3, 2, 1;
5, 2, 4, 4, 2, 1;
6, 3, 5, 6, 4, 2, 1;
10, 5, 7, 9, 7, 4, 2, 1;
13, 7, 9, 11, 11, 7, 4, 2, 1;
19, 11, 12, 15, 16, 12, 7, 4, 2, 1;
25, 16, 15, 19, 22, 18, 12, 7, 4, 2, 1;
35, 24, 20, 26, 29, 27, 19, 12, 7, 4, 2, 1;
...
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CROSSREFS
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Cf. A063995, A105805, A181187, A194547, A194549, A195822, A208482, A208483, A209616, A330368, A330369, A330370.
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KEYWORD
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AUTHOR
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EXTENSIONS
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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 |_|
.
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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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|_|
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
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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A330379
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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.
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+10
3
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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
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1,3
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COMMENTS
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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.
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LINKS
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FORMULA
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EXAMPLE
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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.
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For n = 8 the partitions of 8 and their respective right-angles diagrams look as shown below:
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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*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].
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CROSSREFS
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Row sums of the terms that are after last zero give A179862.
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KEYWORD
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AUTHOR
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STATUS
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approved
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A330376
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Irregular triangle read by rows: T(n,k) is the total number of parts in all partitions of n with Durfee square of size k (n>=1; 1<=k<=floor(sqrt(n))).
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+10
1
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1, 3, 6, 10, 2, 15, 5, 21, 14, 28, 26, 36, 50, 45, 80, 3, 55, 130, 7, 66, 190, 19, 78, 280, 41, 91, 385, 80, 105, 532, 143, 120, 700, 248, 136, 924, 399, 4, 153, 1176, 627, 9, 171, 1500, 949, 24, 190, 1860, 1397, 51, 210, 2310, 2003, 107, 231, 2805, 2823, 193
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OFFSET
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1,2
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LINKS
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EXAMPLE
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Triangle begins:
1;
3;
6;
10, 2;
15, 5;
21, 14;
28, 26;
36, 50;
45, 80, 3;
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PROG
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(PARI) \\ by enumeration over partitions.
ds(p)={for(i=2, #p, if(p[#p+1-i]<i, return(i-1))); #p}
row(n)={my(v=vector(sqrtint(n))); forpart(p=n, v[ds(p)] += #p); v}
(PARI) \\ by generating function.
P(n, k, y)={1/prod(j=1, k, 1 - y*x^j + O(x*x^n))}
T(n, k)={my(r=n-k^2); if(r<0, 0, subst(deriv(polcoef(y^k*P(r, k, 1)*P(r, k, y), r)), y, 1))}
{ for(n=1, 10, print(vector(sqrtint(n), k, T(n, k)))) } \\ Andrew Howroyd, Feb 02 2022
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CROSSREFS
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Cf. A114089, A115721, A115722, A115994, A116858, A118198, A208474, A330369, A330640, A330641, A330642, A330643.
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KEYWORD
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nonn,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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