A331676 Ceiling of circumradius of triangle with consecutive prime sides.

5, 8, 7, 9, 10, 12, 15, 17, 20, 22, 24, 26, 28, 32, 34, 37, 39, 41, 44, 46, 49, 53, 56, 58, 60, 62, 64, 68, 72, 77, 79, 82, 85, 89, 91, 94, 97, 101, 103, 107, 109, 112, 114, 117, 123, 128, 131, 133, 135, 138, 141, 145, 149, 152, 155, 158, 160, 162, 166, 171, 176, 180

Offset: 2

Author

Frank M Jackson, Jan 24 2020

Keywords

• nonn

Comments

The sequence starts at offset 2 because using the first three primes yields a triangle with sides (2,3,5) that is degenerate with infinite circumradius.

Also the first two triangles in this sequence with sides (3,5,7) and (5,7,11) are obtuse and do not have their circumcentres within the bounds of the triangle. Thereafter, the triangles are acute and their circumcentres lie within the bounds of the triangle.

Examples

			a(2)=5 because a triangle with sides 3,5,7 has area = (1/4)*sqrt((3+5+7)(3+5-7)(3-5+7)(-3+5+7)) = 6.495... and circumradius = 3*5*7/(4A) = 4.041...
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 2..1000
• Eric Weisstein's World of Mathematics, Circumradius.

Crossrefs

Cf. A096377, A330096, A331227, A331228.

Programs

Mathematica

lst = {}; Do[{a, b, c}={Prime[n], Prime[n+1], Prime[n+2]}; s=(a+b+c)/2; A=Sqrt[s(s-a)(s-b)(s-c)]; R=a*b*c/(4 A); AppendTo[lst, Ceiling@R], {n, 2, 200}]; lst
ccr[{a_,b_,c_}]:=Module[{s=(a+b+c)/2,A},A=Sqrt[s(s-a)(s-b)(s-c)];Ceiling[(a*b*c)/(4A)]]; ccr/@Partition[Prime[Range[2,70]],3,1] (* Harvey P. Dale, Aug 02 2025 *)

Formula

Circumradius R of a triangle with sides a, b, c is given by R = a*b*c/(4A) where the area A is given by Heron's formula A = sqrt(s(s-a)(s-b)(s-c)) and where s = (a+b+c)/2.