OFFSET
2,1
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.
LINKS
Eric Weisstein's World of Mathematics, Circumradius.
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.
EXAMPLE
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...
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
CROSSREFS
KEYWORD
nonn
AUTHOR
Frank M Jackson, Jan 24 2020
STATUS
approved