1, 2, 1, 3, 1, 4, 2, 1, 5, 1, 6, 2, 3, 1, 7, 1, 8, 4, 2, 1, 9, 3, 1, 10, 2, 5, 1, 11, 1, 12, 6, 4, 2, 3, 1, 13, 1, 14, 2, 7, 1, 15, 5, 3, 1, 16, 8, 4, 2, 1, 17, 1, 18, 6, 2, 9, 3, 1, 19, 1, 20, 10, 4, 2, 5, 1, 21, 7, 3, 1, 22, 2, 11, 1, 23, 1, 24, 12, 8, 6, 4, 2, 3, 1, 25, 5, 1, 26, 2, 13, 1, 27, 9, 3, 1
Offset: 1
Triangle begins:
1;
2, 1;
3, 1;
4, 2, 1;
5, 1;
6, 2, 3, 1;
7, 1;
8, 4, 2, 1;
9, 3, 1;
10, 2, 5, 1;
11, 1;
12, 6, 4, 2, 3, 1;
13, 1;
14, 2, 7, 1;
15, 5, 3, 1;
16, 8, 4, 2, 1;
17, 1;
18, 6, 2, 9, 3, 1;
19, 1;
20, 10, 4, 2, 5, 1;
21, 7, 3, 1;
22, 2, 11, 1;
23, 1;
24, 12, 8, 6, 4, 2, 3, 1;
25, 5, 1;
26, 2, 13, 1;
27, 9, 3, 1;
28, 14, 4, 2, 7, 1;
...
For n = 12 the divisors of 12 are [1, 2, 3, 4, 6, 12]. The even divisors of 12 in decreasing order are [12, 6, 4, 2], and the odd divisors of 12 in decreasing order are [3, 1], so the 12th row of triangle is [12, 6, 4, 2, 3, 1].
On the other hand, consider the diagram that appears in the Links section (figure 1). Then consider only the semicircumferences that contain the point [12,0]. In the first quadrant, from bottom to top, we can see the curves with diameters [1, 2, 3, 6]. Also we can see these curves in the third quadrant from top to bottom. The associated numbers c = 12/d are [12, 6, 4, 2] respectively. These are the even divisors of n in decreasing order. Then, in the second quadrant, from bottom to top, we can see the curves with diameters [4, 12]. Also we can see these curves in the fourth quadrant from top to bottom. The associated numbers c = 12/d are [3, 1] respectively. These are the odd divisors of 12 in decreasing order. Finally all numbers c obtained are [12, 6, 4, 2, 3, 1] equaling the 12th row of triangle.
From _David A. Corneth_, Jan 17 2019: (Start)
For n = 100, we have n even so first even divisors in decreasing order and then odd divisors in decreasing order. The even divisors of 100 are twice the divisors of 100/2 = 50. Those divisors, in decreasing order, are: 100, 50, 20, 10, 4, 2.
The odd divisors of 100 in decreasing order are the odd divisors of the odd part of 100, which is 25. Those divisors are 25, 5, 1.
Concatenating all these divisors gives: 100, 50, 20, 10, 4, 2, 25, 5, 1. (End)
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