A319845 Irregular triangle read by rows in which row n lists the even divisors of n in increasing order together with the odd divisors of n in decreasing order.
1, 2, 1, 3, 1, 2, 4, 1, 5, 1, 2, 6, 3, 1, 7, 1, 2, 4, 8, 1, 9, 3, 1, 2, 10, 5, 1, 11, 1, 2, 4, 6, 12, 3, 1, 13, 1, 2, 14, 7, 1, 15, 5, 3, 1, 2, 4, 8, 16, 1, 17, 1, 2, 6, 18, 9, 3, 1, 19, 1, 2, 4, 10, 20, 5, 1, 21, 7, 3, 1, 2, 22, 11, 1, 23, 1, 2, 4, 6, 8, 12, 24, 3, 1, 25, 5, 1, 2, 26, 13, 1, 27, 9, 3, 1
Offset: 1
Examples
Triangle begins: 1; 2, 1; 3, 1; 2, 4, 1; 5, 1; 2, 6, 3, 1; 7, 1; 2, 4, 8, 1; 9, 3, 1; 2, 10, 5, 1; 11, 1; 2, 4, 6, 12, 3, 1; 13, 1; 2, 14, 7, 1; 15, 5, 3, 1; 2, 4, 8, 16, 1; 17, 1; 2, 6, 18, 9, 3, 1; 19, 1; 2, 4, 10, 20, 5, 1; 21, 7, 3, 1; 2, 22, 11, 1; 23, 1; 2, 4, 6, 8, 12, 24, 3, 1; 25, 5, 1; 2, 26, 13, 1; 27, 9, 3, 1; 2, 4, 14, 28, 7, 1; ... For n = 12 the divisors of 12 are [1, 2, 3, 4, 6, 12]. The even divisors of 12 in increasing order are [2, 4, 6, 12], and the odd divisors of 12 in decreasing order are [3, 1], so the 12th row of triangle is [2, 4, 6, 12, 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 top to bottom, we can see the curves with diameters [6, 3, 2, 1]. Also we can see these curves in the third quadrant from bottom to top. The associated numbers c = 12/d are [2, 4, 6, 12] respectively. These are the even divisors of n in increasing 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 [2, 4, 6, 12, 3, 1] equaling the 12th row of triangle.
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Crossrefs
Programs
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PARI
row(n) = my(d=divisors(n)); concat(select(x->!(x%2), d), Vecrev(select(x->(x%2), d))); lista(nn) = {for (n=1, nn, my(r = row(n)); for (k=1, #r, print1(r[k], ", ")););} \\ Michel Marcus, Jan 17 2019
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