%I #31 Jan 22 2022 18:22:03

%S 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,

%T 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,

%U 13,24,45,40,0,12,77,96,108,80,121

%N 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.

%C Observation: at least the first 11 terms of column 1 coincide with A188674 (using the same indices).

%D G. E. Andrews, Theory of Partitions, Cambridge University Press, 1984, page 143.

%F T(n,k) = k*A330369(n,k).

%e Triangle begins:

%e 1;

%e 0, 4;

%e 0, 0, 9;

%e 1, 0, 3, 16;

%e 2, 0, 0, 8, 25;

%e 3, 4, 0, 8, 15, 36;

%e 4, 8, 0, 0, 20, 24, 49;

%e 5, 12, 9, 0, 15, 36, 35, 64;

%e 7, 16, 21, 0, 5, 36, 56, 48, 81;

%e 9, 20, 33, 16, 0, 36, 63, 80, 63, 100;

%e 13, 24, 45, 40, 0, 12, 77, 96, 108, 80, 121;

%e ...

%e 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.

%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 look as shown below:

%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 There are 5 right angles of size 1, so T(8,1) = 5*1 = 5.

%e There are 6 right angles of size 2, so T(8,2) = 6*2 = 12.

%e There are 3 right angles of size 3, so T(8,3) = 3*3 = 9.

%e There are no right angle of size 4, so T(8,4) = 0*4 = 0.

%e There are 3 right angles of size 5, so T(8,5) = 3*5 = 15.

%e There are 6 right angles of size 6, so T(8,6) = 6*6 = 36.

%e There are 5 right angles of size 7, so T(8,7) = 5*7 = 35.

%e There are 8 right angles of size 8, so T(8,8) = 8*8 = 64.

%e Hence the 8th row of triangle is [5, 12, 9, 0, 15, 36, 35, 64].

%e The row sum gives A066186(8) = 8*A000041(8) = 8*22 = 176.

%Y Row sums give A066186, n >= 1.

%Y Row sums of the terms that are after last zero give A179862.

%Y Cf. A188674.

%Y Cf. A000041, A000290, A330369, A330375, A330378.

%K nonn,tabl,more

%O 1,3

%A _Omar E. Pol_, Dec 31 2019