A334763 Ceiling of circumradius of triangle whose sides are consecutive Ulam numbers (A002858).

3, 4, 5, 6, 7, 9, 10, 14, 15, 19, 21, 24, 26, 29, 31, 34, 37, 40, 43, 45, 48, 52, 55, 58, 60, 63, 68, 72, 77, 80, 84, 87, 93, 99, 103, 104, 107, 110, 115, 118, 123, 126, 131, 134, 138, 139, 142, 146, 149, 153, 158, 168, 176, 182, 185, 190, 194, 200, 204, 208

Offset: 2

Author

Frank M Jackson, May 10 2020

Keywords

• nonn

Comments

It has been proved that three consecutive Ulam numbers U(n) for n > 1 satisfy the triangle inequality. See Wikipedia link below. Consequently it is possible to create n-gons using n consecutive Ulam numbers. The sequence starts at offset 2 because using the first Ulam number generates a triangle with sides (1,2,3) that is degenerate with infinite circumradius.

Conjecture: Triangles whose sides are consecutive Ulam numbers are acute apart from (1,2,3), (2,3,4), (3,4,6), (4,6,8), (6,8,11) and (16,18,26).

Examples

			a(2)=3 because a triangle with sides 2,3,4 has area = (1/4)*sqrt((2+3+4)(2+3-4)(2-3+4)(-2+3+4)) = 2.904... and circumradius = 2*3*4/(4A) = 2.065...
		

Links

• Eric Weisstein's World of Mathematics, Circumradius.
• Eric Weisstein's World of Mathematics, Ulam Sequence.
• Wikipedia, Ulam number.

Crossrefs

Cf. A002858, A331676.

Programs

Mathematica

lst1=ReadList["https://oeis.org/A002858/b002858.txt", {Number, Number}]; lst={}; Do[{a, b, c}={lst1[[n]][[2]], lst1[[n+1]][[2]], lst1[[n+2]][[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, 100}]; lst

Formula

Circumradius 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.