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EXAMPLE
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Triangle begins:
1;
1,1,1;
3,3,3,2,2,1,1;
15,15,15,12,12,9,9,6,6,4,2,2,1;
108,108,108,93,93,78,78,63,63,51,39,39,30,21,21,15,9,9,5,3,1,1;
1036,1036,1036,928,928,820,820,712,712,619,526,526,448,370,370,307,244,244,193,154,115,115,85,64,43,43,28,19,10,10,5,2,1;
12569,12569,12569,11533,11533,10497,10497,9461,9461,8533,7605,7605,6785,5965,5965,5253,4541,4541,3922,3396,2870,2870,2422,2052,1682,1682,1375,1131,887,887,694,540,425,310,310,225,161,118,75,75,47,28,18,8,8,3,1;
...
Number of terms in rows is given by A136219, which starts:
[1,3,7,13,22,33,47,64,84,106,131,159,190,224,261,301,343,388,...].
To generate row 3, start with row 2:
[3,3,3,2,2,1,1];
insert zeros at positions [0,1,3,5,7,10] to get:
[0,0,3,0,3,0,3,0,2,2,0,1,1],
then take reverse partial sums (from right to left) to obtain row 3:
[15,15,15,12,12,9,9,6,6,4,2,2,1].
For row 4, insert zeros in row 3 at positions [0,1,3,5,7,10,13,16,20]:
[0,0,15,0,15,0,15,0,12,12,0,9,9,0,6,6,0,4,2,2,0,1]
then take reverse partial sums to obtain row 4:
[108,108,108,93,93,78,78,63,63,51,39,39,30,21,21,15,9,9,5,3,1,1].
Continuing in this way will generate all the rows of this triangle.
Amazingly, column 0 of this triangle = column 0 of triangle P=A136220:
1;
1, 1;
3, 2, 1;
15, 10, 3, 1;
108, 75, 21, 4, 1;
1036, 753, 208, 36, 5, 1;
12569, 9534, 2637, 442, 55, 6, 1;
185704, 146353, 40731, 6742, 805, 78, 7, 1; ...
where column k of P^3 = column 0 of P^(3k+3) such that
column 0 of P^3 = column 0 of P shift one place left.
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