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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 (list; graph; refs; listen; history; text; internal format)
 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 Cf. A188158, A210643, A210645, A226453. Sequence in context: A011987 A036690 A229746 * A322374 A014131 A236539 Adjacent sequences:  A256576 A256577 A256578 * A256580 A256581 A256582 KEYWORD nonn AUTHOR Michel Lagneau, Apr 02 2015 STATUS approved

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Last modified April 11 23:19 EDT 2021. Contains 342895 sequences. (Running on oeis4.)