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A256579
Integer areas of integer-sided triangles where at least one of the three altitudes is of prime length.
0
6, 12, 30, 60, 84, 168, 330, 546, 660, 1092, 1224, 1710, 2448, 3036, 3420, 6072, 6090, 7440, 12180, 12654, 14880, 17220, 19866, 25308, 25944, 34440, 37206, 39732, 51330, 51888, 56730, 74412, 75174, 89460, 97236, 102660, 113460, 123240, 142926, 150348, 176220
OFFSET
1,1
COMMENTS
Subset of A226453.
The corresponding primes are:
3, 3, 5, 5, 7, 7, 11, 13, 11, 13, 17, 19, 17, 23, 19, 23, 29, 31, 29, 37, 31, 41, 43, 37, 47, 41, 53, 43, 59, 47, 61, 53, 67, 71, 73, 59, 61, 79, 83, 67, 89, ...
The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), 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 (see the formula below);
- The prime altitude of a triangle is the greatest prime divisor of a(n) (the proof is easy if we observe the formula);
- There exists two 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 a1(n) = {6, 30, 84, 330, 546, 1224, ...}. The lengths of the prime altitudes are Ha or Hb = a = p. The sides are of the form (p, q, q+1) with p = sqrt(2q+1) => the sides are equal to (p, (p^2-1)/2, p^2+1)/2) and a(n) = (p^3-p)/4.
Case (ii): A subset with isosceles triangles formed by two right triangles of the sequence. So, the areas are a2(n) = {12, 60, 168, 660, 1092, 2448, ...} = 2*a1(n). The sides are of the form (a, a, 2*(a-1)) = (p^2+1)/2, p^2+1)/2, p^2-1) and Ha = sqrt(2a-1) = p, a2(n) = 2*a1(n) = (p^3-p)/2.
We did not find a class of non-isosceles and non-right triangles (a, b, c) whose three altitudes include one of prime length.
The following table gives the first values (A, a, b, c, Ha, Hb, Hc) where A is the integer area, a, b, c are the sides and Ha <= Hb <= Hc are the altitudes.
+------+-----+-----+-----+----------+----------+---------+
| A | a | b | c | Ha | Hb | Hc |
+------+-----+-----+-----+----------+----------+---------+
| 6 | 3 | 4 | 5 | 12/5 | 3 | 4 |
| 12 | 5 | 5 | 8 | 3 | 24/5 | 24/5 |
| 30 | 5 | 12 | 13 | 5 | 60/13 | 12 |
| 60 | 13 | 13 | 24 | 5 | 120/13 | 120/13 |
| 84 | 7 | 24 | 25 | 168/25 | 7 | 24 |
| 168 | 25 | 25 | 48 | 7 | 336/25 | 336/25 |
| 330 | 11 | 60 | 61 | 660/61 | 11 | 60 |
| 546 | 13 | 84 | 85 | 1092/85 | 13 | 84 |
| 660 | 61 | 61 | 120 | 11 | 1320/61 | 1320/61 |
| 1092 | 85 | 85 | 168 | 13 | 2184/85 | 2184/85 |
| 1224 | 17 | 144 | 145 | 2448/145 | 17 | 144 |
| 1710 | 19 | 180 | 181 | 3420/181 | 19 | 180 |
| 2448 | 145 | 145 | 288 | 4896/145 | 4896/145 | 17 |
+------+-----+-----+-----+----------+----------+---------+
LINKS
Eric Weisstein, Altitude
Eric Weisstein, Isosceles Triangle
Eric Weisstein, Right Triangle
FORMULA
a(n) = (prime(n)^3 - prime(n))/4 for the right triangles;
a(n) = (prime(n)^3 - prime(n))/2 for the isosceles triangles.
MAPLE
# program using the formula
lst:={}:for n from 2 to 50 do:p:=ithprime(n):p1:=(p^3-p)/4:p2:=(p^3-p)/2:lst:=lst union {p1} union {p2}:od:print(lst):
MATHEMATICA
nn = 300; lst = {}; Do[s = (a + b + c)/2; area2 = s (s - a) (s - b) (s - c); If[area2>0 && IntegerQ[Sqrt[area2]]&&(PrimeQ[(2*Sqrt[area2])/a]|| PrimeQ[(2*Sqrt[area2])/b]||PrimeQ[(2*Sqrt[area2])/c]), AppendTo[lst, Sqrt[area2]]], {a, nn}, {b, a}, {c, b}]; Union[lst]
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Apr 02 2015
STATUS
approved