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%I #16 Apr 09 2017 03:33:30
%S 24,30,36,42,48,54,60,66,84,96,114,120,126,150,156,198,210,270,294,
%T 330,336,390,420,462,504,510,546,570,630,714,726,756,810,840,924,930,
%U 1008,1014,1056,1134,1386,1428,1554,1638,1680,1716,1734,1848,1890,1950,2016
%N Integer areas A of the triangles such that A and the sides are integers, and the length of the inradius is a prime number.
%C Subsequence of A228383.
%C The corresponding inradii r are 2, 2, 2, 2, 3, 3, 2, 3, 3, 3, 3, 3, 3, 5, 3, 3, 3, 5, 7, 5, 7, 5, 7, 7, 7, 5, 7, 5, ...
%C 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 inradius r is given by r = A/s.
%H Mohammad K. Azarian, <a href="http://www.jstor.org/stable/25678790">Solution of problem 125: Circumradius and Inradius</a>, Math Horizons, Vol. 16, No. 2 (Nov. 2008), p. 32.
%e 24 is in the sequence because for (a, b, c) = (6, 8, 10) => s =(6 + 8 + 10)/2 = 12; A = sqrt(12*(12-6)*(12-8)*(12-10)) = sqrt(576)= 24; r = A/s = 2 is prime.
%t nn = 1500; lst = {}; Do[s = (a + b + c)/2; If[IntegerQ[s], area2 = s (s - a) (s - b) (s - c); If[0 < area2 <= nn^2 && IntegerQ[Sqrt[area2]] && PrimeQ[Sqrt[area2]/s], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}]; Union[lst]
%Y Cf. A228383.
%K nonn
%O 1,1
%A _Michel Lagneau_, Oct 11 2013