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%I #25 Nov 20 2020 21:22:32
%S 1,4,9,19,1,33,2,59,7,93,12,150,26,226,43,1,342,76,2
%N 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.
%C Column k starts in row k^2.
%C It appears that column 1 gives A179862.
%e Triangle begins:
%e 1;
%e 4;
%e 9;
%e 19, 1;
%e 33, 2;
%e 59, 7;
%e 93, 12;
%e 150, 26;
%e 226, 43, 1;
%e 342, 76, 2;
%e ...
%e 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.
%e .
%e . Right-angles Right
%e Part Ferrers diagram Part diagram angle
%e _ _ _ _ _ _ _
%e 7 * * * * * * * 7 | _ _ _ _ _ _| 14
%e 6 * * * * * * 6 | | _ _ _ _| 8
%e 3 * * * 3 | | | | 2
%e 3 * * * 3 | | |_|
%e 2 * * 2 | |_|
%e 1 * 1 | |
%e 1 * 1 | |
%e 1 * 1 |_|
%e .
%e Figure 1. Figure 2.
%e .
%e For n = 8 the partitions of 8 and their respective right-angles diagrams are as follows:
%e .
%e _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
%e 1| |8 2| _|8 3| _ _|8 4| _ _ _|8 5| _ _ _ _|8
%e 1| | 1| | 1| | 1| | 1| |
%e 1| | 1| | 1| | 1| | 1| |
%e 1| | 1| | 1| | 1| | 1|_|
%e 1| | 1| | 1| | 1|_|
%e 1| | 1| | 1|_|
%e 1| | 1|_|
%e 1|_|
%e _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
%e 6| _ _ _ _ _|8 7| _ _ _ _ _ _|8 8|_ _ _ _ _ _ _ _|8
%e 1| | 1|_|
%e 1|_|
%e .
%e _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
%e 2| _|7 3| _ _|7 4| _ _ _|7 5| _ _ _ _|7 6| _ _ _ _ _|7
%e 2| |_|1 2| |_| 1 2| |_| 1 2| |_| 1 2|_|_| 1
%e 1| | 1| | 1| | 1|_|
%e 1| | 1| | 1|_|
%e 1| | 1|_|
%e 1|_|
%e .
%e _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
%e 2| _|6 3| _ _|6 3| _ _|6 4| _ _ _|6 4| _ _ _|6 5| _ _ _ _|6
%e 2| | |2 2| | | 2 3| |_ _|2 2| | | 2 3| |_ _| 2 3|_|_ _| 2
%e 2| |_| 2| |_| 1| | 2|_|_| 1|_|
%e 1| | 1|_| 1|_|
%e 1|_|
%e .
%e _ _ _ _ _ _ _ _ _
%e 2| _|5 3| _ _|5 4| _ _ _|5
%e 2| | |3 3| | _|3 4|_|_ _ _|3
%e 2| | | 2|_|_|
%e 2|_|_|
%e .
%e 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.
%e 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.
%Y Row sums give A066186.
%Y Cf. A179862.
%Y Cf. also A000041, A330369, A330378, A330379.
%K nonn,tabf,more
%O 1,2
%A _Omar E. Pol_, Dec 21 2019
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