
COMMENTS

There are triangles with rational square, for instance, with the area 144, we find for (a,b,c)=(6,50,52) the altitudes {Ha,Hb,Hc} = {72/13, 144/25, 48} but with the same area we find also for (a,b,c)=(18,20,34) the altitudes {Ha,Hb,Hc} = {144/17, 72/5, 16}.
The corresponding squares are 4, 4, 9, 16, 9, 9, 16, 9, 16, 36, 9, 16, 16, 36, 16, 9, 36, 16, 16, 36, 16, 36, 36, 36, 16, 16, 25, 36, 36, 36, 36, 36, 16, 36, 36, 36, 144, 144, 36, 36, 36, 36, 100, 64, 144, 36, 36, 36, 144, 36, ...
The subsequence of the primitive triangles are 6, 12, 126, 144, 180, 216, 234, ...
The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s*(sa)*(sb)*(sc)), where s = (a+b+c)/2.
The altitudes of a triangle with sides length a, b, c and area A have length given by Ha= 2A/a, Hb= 2A/b, Hc= 2A/c.
Properties of this sequence:
 The sequence is infinite because from de initial primitive triangle (3,4,5), the area A’ of the triangle (3*3^m, 4*3^m, 5*3^m) is also in the sequence where A’ = 6*3^2m and {Ha, Hb, Hc} = {4*3^m, 3^(m+1), (4*3^(m+1))/5}. The altitude Ha or Hb is square.
 There are three subsets of numbers included into a(n):
Case (i): A subset with right triangles (a,b,c) where a^2+b^2 = c^2 with area a2(n) = {6, 54, 96, 180, 240, 270, ...}
Case (ii): A subset with isosceles triangles of area a1(n)= {12, 108, 192, 360, 480, 540, ...} = 2*a1(n).
Case (iii): A subset with nonisosceles and nonright triangles of area a3(n)= {126, 144, 216, 234, 264, 336, ...}

 A  a  b  c  Ha  Hb  Hc 

 6  3  4  5  12/5  3  4 
 12  5  5  6  4  24/5  24/5 
 54  9  12  15  36/5  9  12 
 96  12  16  20  48/5  12  16 
 108  15  15  24  9  72/5  72/5 
 126  15  28  41  252/41  9  84/5 
 144  18  20  34  144/17  72/5  16 
 180  9  40  41  360/41  9  40 
 192  20  20  24  16  96/5  96/5 
 216  12  39  45  48/5  144/13  36 
 234  15  41  52  9  468/41  156/5 
 240  16  30  34  240/17  16  30 
 264  33  34  65  528/65  264/17  16 
