%I #10 Dec 30 2020 14:01:54
%S 3,6,0,0,12,6,16,18,16,6,32,6,36,8,28,16,2,26,10,6,10,54,6,18,0,36,0,
%T 132,18,68,12,40,24,12,20,22,20,12,24,48,0,66,30,120,150,24,62,6,4,32,
%U 48,24,8,0,28,16,18,84,90,180,18,144,6,132,52,36,44,54,28,38,14
%N Area of prime triangles.
%H Harvey P. Dale, <a href="/A154924/b154924.txt">Table of n, a(n) for n = 1..1000</a>
%F Take six consecutive primes and group them in ordered pairs (p1,p2) (p3,p4) (p5,p6) and compute the area of the triangle they form in the Cartesian plane.
%e a(1)=3 because the triangle with vertices (2,3)(5,7)(11,13) has an area of 3. a(2)=6 because the triangle with vertices (3,5)(7,11)(13,17) has an area of 6. a(3)=0 because the vertices (5,7)(11,13)(17,19) are collinear and do not form a triangle.
%t artr[{a_,b_,c_,d_,e_,f_}]:=Module[{x=Sqrt[(c-a)^2+(d-b)^2],y=Sqrt[(d-f)^2+(c-e)^2],z=Sqrt[(e-a)^2+(f-b)^2],s},s=(x+y+z)/2;Sqrt[s(s-x)(s-y)(s-z)]]; artr/@Partition[Prime[Range[80]],6,1]//Simplify (* _Harvey P. Dale_, Dec 30 2020 *)
%K easy,nonn
%O 1,1
%A _Gil Broussard_, Jan 17 2009